Beer johnston Statics 11- Solution Manual.pdf

Estática para ingenieros

1. Instructor's and Solutions Manual
to accompany

Vector Mechanics for
Engineers, Statics

Eleventh Edition



Ferdinand P. Beer
Late of Lehigh University

E. Russell Johnston, Jr.
Late of University of Connecticut

David F. Mazurek
United States Coast Guard Academy



Prepared by
Amy Mazurek













PROPRIETARY AND CONFIDENTIAL

This Manual is the proprietary property of McGraw-Hill Education and protected by copyright and other state and
federal laws. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in
any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.

www.elsolucionario.org

3. TO THE INSTRUCTOR
As indicated in its preface, Vector Mechanics for Engineers: Statics is designed for the
first course in statics offered in the sophomore year of college. New concepts have,
therefore, been presented in simple terms and every step has been explained in detail.
However, because of the large number of optional sections which have been included and
the maturity of approach which has been achieved, this text can also be used to teach a
course which will challenge the more advanced student.

The text has been divided into units, each corresponding to a well -defined topic and
consisting of one or several theory sections, one or several Sample Problems, a section
entitled Solving Problems on Your Own, and a large number of problems to be assigned.
To assist instructors in making up a schedule of assignments that will best fit their
classes, the various topics covered in the text have been listed in Table I and a suggested
number of periods to be spent on each topic has been indicated. Both a minimum and a
maximum number of periods have been suggested, and the topics which form the
standard basic course in statics have been separated from those which are optional. The
total number of periods required to teach the basic material varies from 26 to 39, while
covering the entire text would require from 41 to 65 periods. If allowance is made for the
time spent for review and exams, it is seen that this text is equally suitable for teaching a
basic statics course to students with limited preparation (since this can be done in 39
periods or less) and for teaching a more complete statics course to advanced students
(since 41 periods or more are necessary to cover the entire text). In most instances, of
course, the instructor will want to include some, but not all, of the additional material
presented in the text. In addition, it is noted that the text is suitable for teaching an
abridged course in statics which can be used as an introduction to the study of dynamics
(see Table I).

The problems have been grouped according to the portions of material they illustrate and
have been arranged in order of increasing difficulty, with problems requiring special
attention indicated by asterisks. We note that, in most cases, problems have been
arranged in groups of six or more, all problems of the same group being closely related.
This means that instructors will easily find additional problems to amplify a particular
point which they may have brought up in discussing a problem assigned for homework.
Accessible through Connect are problem sets for each chapter that are designed to be
solved with computational software. Solutions for these problems, including analyses of
the problems and problem solutions and output for the most widely used computational
programs, are also available through Connect.

To assist in the preparation of homework assignments, Table II provides a brief
description of all groups of problems and a classification of the problems in each group
according to the units used. It should also be noted that the answers to all problems are
given at the end of the text, except for those with a number in italic. Because of the large
number of problems available in both systems of units, the instructor has the choice of
assigning problems using SI units and problems using U.S. customary units in whatever
proportion is found to be most desirable for a given class. To illustrate this point, sample
lesson schedules are shown in Tables III, IV, and V, together with various alternative lists www.elsolucionario.org

4. of assigned homework problems. Half of the problems in each of the six lists suggested in
Table III and Table V are stated in SI units and half in U.S. customary units. On the other
hand, 75% of the problems in the four lists suggested in Table IV are stated in SI units
and 25% in U.S. customary units.

Since the approach used in this text differs in a number of respects from the approach
used in other books, instructors will be well advised to read the preface to Vector
Mechanics for Engineers, in which the authors have outlined their general philosophy. In
addition, instructors will find in the following pages a description, chapter by chapter, of
the more significant features of this text. It is hoped that this material will help instructors
in organizing their courses to best fit the needs of their students. A cknowledgement and
thanks are given to Amy Mazurek for her careful preparation of the solutions contained in
this manual.

David Mazurek

DESCRIPTION OF THE MATERIAL CONTAINED IN
VECTOR MECHANICS FOR ENGINEERS: STATICS, Eleventh Edition

Chapter 1
Introduction
The material in this chapter can be used as a first assignment or for later reference. The
six fundamental principles listed in Sec. 1.2 are introduced separately and are discussed
at greater length in the following chapters. Section 1.3 deals with the two systems of units
used in the text. The SI metric units are discussed first. The base units are defined and the
use of multiples and submultiples is explained. The various SI prefixes are presented in
Table 1.1, while the principal SI units used in statics and dynamics are listed in Table 1.2.
In the second part of Sec. 1.3, the base U.S. customary units used in mechanics are
defined, and in Sec. l.4, it is shown how numerical data stated in U.S. customary units
can be converted into SI units, and vice versa. The SI equivalents of the principal U.S.
customary units used in statics and dynamics are listed in Table 1.3.

The instructor’s attention is called to the fact that the various rules relating to the use of
SI units have been observed throughout the text. For instance, multiples and submultiples
(such as kN and mm) are used whenever possible to avoid writing more than four digits
to the left of the decimal point or zeros to the right of the decimal point. When 5- digit or
larger numbers involving SI units are used, spaces rather than commas are utilized to
separate digits into groups of three (for example, 20 000 km). Also, prefixes are never
used in the denominator of derived units; for example, the constant of a spring which
stretches 20 mm under a load of 100 N is expressed as 5 kN/m, not as 5 N/mm.

In order to achieve as much uniformity as possible between results expressed respectively
in SI and U.S. customary units, a center point, rather than a hyphen, has been used to
combine the symbols representing U.S. cust omary units (for example, 10 lb∙ft);
furthermore, the unit of time has been represented by the symbol s, rather than sec,
whether SI or U.S. customary units are involved (for example, 5 s, 50 ft/s, 15 m/s). www.elsolucionario.org

5. However, the traditional use of commas to separate digits into groups of three has been
maintained for 5- digit and larger numbers involving U.S. customary units.

Chapter 2
Statics of Particles
This is the first of two chapters dealing with the fundamental properties of force systems.
A simple, intuitive classification of forces has been used: forces acting on a particle
(Chap. 2) and forces acting on a rigid body (Chap. 3).

Chapter 2 begins with the parallelogram law of addition of forces and with the
introduction of the fundamental properties of vectors. In the text, forces and other vector
quantities are always shown in bold- face type. Thus, a force F (boldface), which is a
vector quantity, is clearly distinguished from the magnitude F (italic) of the force, which
is a scalar quantity. On the blackboard and in handwritten work, where bold -face lettering
is not practical, vector quantities can be indicated by underlining. Both the magnitude and
the direction of a vector quantity must be given to completely define that quantity. Thus,
a force F of magnitude F = 280 lb, directed upward to the right at an angle of 25° with
the horizontal, is indicated as F = 280 lb
25° when printed or as F = 280 lb
25° when handwritten. Unit vectors i a nd j are introduced in Sec. 2.7, where the
rectangular components of forces are considered.

In the early sections of Chap. 2 the following basic topics are presented: the equilibrium
of a particle, Newton’s first law, and the concept of the free- body diagram. These first
sections provide a review of the methods of plane trigonometry and familiarize the
students with the proper use of a calculator. A general procedure for the solution of
problems involving concurrent forces is given: when a problem involves only three
forces, the use of a force triangle and a trigonometric solution is preferred; when a
problem involves more than three forces, the forces should be resolved into rectangular
components and the equations Σ F
x = 0, ΣF y = 0 should be used.

The second part of Chap. 2 deals with forces in space and with the equilibrium of
particles in space. Unit vectors are used and forces are expressed in the form F = F
xi + F yj
+ F
zk = Fλ, where i , j, and k are the unit vectors directed respectively along the x , y, and z
axes, and λ is the unit vector directed along the line of action of F.

Note that since this chapter deals only with particles or bodies which can be considered
as particles, problems involving compression members have been postponed with only a
few exceptions until Chap. 4, where students will learn to handle rigid- body problems in
a uniform fashion and will not be tempted to erroneously assume that forces are
concurrent or that reactions are directed along members.

It should be observed that when SI units are used a body is generally specified by its
mass expressed in kilograms. The weight of the body, however, should be expressed in
newtons. Therefore, in many equilibrium problems involving SI units, an additional
calculation is required before a free- body diagram can be drawn (compare the example in
Sec. 2.3C and Sample Probs. 2.5 and 2.9). This apparent disadvantage of the SI system of www.elsolucionario.org

6. units, when compared to the U.S. customary units, will be offset in dynamics, where the
mass of a body expressed in kilograms can be entered directly into the equation F = m a,
whereas with U.S. customary units the mass of the body must first be determined in
lb∙s
2
/ft (or slugs) from its weight in pounds.

Chapter 3
Rigid Bodies:
Equivalent Systems of Forces
The principle of transmissibility is presented as the basic assumption of the statics of
rigid bodies. However, it is pointed out that this principle can be derived from Newton’s
three laws of motion (see Sec. 16.1D of Dynamics ). The vector product is then introduced
and used to define the moment of a force about a point. The convenience of using the
determinant form (Eqs. 3.19 and 3.21) to express the moment of a force about a point
should be noted. The scalar product and the mixed triple product are introduced and used
to define the moment of a force about an axis. Again, the convenience of using the
determinant form (Eqs. 3.41 and 3.44) should be noted. The amount of time which should
be assigned to this part of the chapter will depend on the extent to which vector algebra
has been considered and used in prerequisite mathematics and physics courses. It is felt
that, even with no previous knowledge of vector algebra, a maximum of four periods is
adequate (see Table I).

In Secs. 3.12 through 3.15 couples are introduced, and it is proved that couples are
equivalent if they have the same moment. While this fundamental property of couples is
often taken for granted, the authors believe that its rigorous and logical proof is necessary
if rigor and logic are to be demanded of the students in the solution of their mechanics
problems.

In Section 3.3, the concept of equivalent systems of forces is carefully presented. This
concept is made more intuitive through the extensive use of free -body-diagram equations
(see Figs. 3.34 through 3.41). Note that the moment of a force is either not shown or is
represented by a green vector (Figs. 3.10 and 3.22). A red vector with the symbol is used
only to represent a couple, that is, an actual system consisting of two forces (Figs. 3.33
through 3.41). Section 3.4D is optional; it introduces the concept of a wrench and shows
how the most general system of forces in space can be reduced to this combination of a
force and a couple with the same line of action.

Since one of the purposes of Chap. 3 is to familiarize students with the fundamental
operations of vector algebra, students should be encouraged to solve all problems in this
chapter (two-dimensional as well as three-dimensional) using the methods of vector
algebra. However, many students may be expected to develop solutions of their own,
particularly in the case of two-dimensional problems, based on the direct computation of
the moment of a force about a given point as the product of the magnitude of the force
and the perpendicular distance to the point considered. Such alternative solutions may
occasionally be indicated by the instructor (as in Sample Prob. 3.9), who may then wish
to compare the solutions of the sample problems of this chapter with the solutions of the
same sample problems given in Chaps. 3 and 4 of the parallel text Mechanics for www.elsolucionario.org

7. Engineers. It should be pointed out that in later chapters the use of vector products will
generally be reserved for the solution of three-dimensional problems.

Chapter 4
Equilibrium of Rigid Bodies
In the first part of this chapter, problems involving the equilibrium of rigid bodies in two
dimensions are considered and solved using ordinary algebra, while problems involving
three dimensions and requiring the full use of vector algebra are discussed in the second
part of the chapter. Particular emphasis is placed on the correct drawing and use of free-
body diagrams and on the types of reactions produced by various supports and
connections (see Figs. 4.1 and 4.10). Note that a distinction is made between hinges used
in pairs and hinges used alone; in the first case the reactions consist only of force
components, while in the second case the reactions may, if necessary, include couples.

For a rigid body in two dimensions, it is shown (Sec. 4.1B ) that no more than three
independent equations can be written for a given free body, so that a problem involving
the equilibrium of a single rigid body can be solved for no more than three unknowns. It
is also shown that it is possible to choose equilibrium equations containing only one
unknown to avoid the necessity of solving simultaneous equations. Section 4.1C
introduces the concepts of statical indeterminacy and partial constraints. Section 4.2 is
devoted to the equilibrium of two- and three-force bodies; it is shown how these concepts
can be used to simplify the solution of certain problems. This topic is presented only after
the general case of equilibrium of a rigid body to lessen the possibility of students
misusing this particular method of solution.

The equilibrium of a rigid body in three dimensions is considered with full emphasis
placed on the free-body diagram. While the tool of vector algebra is freely used to
simplify the computations involved, vector algebra does not, and indeed cannot, replace
the free-body diagram as the focal point of an equilibrium problem. Therefore, the
solution of every sample problem in this section begins with a reference to the drawing of
a free-body diagram. Emphasis is also placed on the fact that the number of unknowns
and the number of equations must be equal if a structure is to be statically determinate
and completely constrained.

Chapter 5
Distributed Forces:
Centroids and Centers of Gravity
Chapter 5 starts by defining the center of gravity of a body as the point of application of
the resultant of the weights of the various particles forming the body. This definition is
then used to establish the concept of the centroid of an area or line. Section 5.1C
introduces the concept of the first moment of an area or line, a concept fundamental to
the analysis of shearing stresses in beams in a later study of mechanics of materials. All
problems assigned for the first period involve only areas and lines made of simple
geometric shapes; thus, they can be solved without using calculus.
www.elsolucionario.org

8. Section 5.2A explains the use of differential elements in the determination of centroids by
integration. The theorems of Pappus -Guldinus are given in Sec. 5.2B . Sections 5.3A and
5.3B are optional; they show how the resultant of a distributed load can be determined by
evaluating an area and by locating its centroid. Section 5.4 deals with centers of gravity
and centroids of volumes. Here again the determination of the centroids of composite
shapes precedes the calculation of centroids by integration.

It should be noted that when SI units are used, a given material is generally characterized
by its density (mass per unit volume, expressed in kg/m
3
), rather than by its specific
weight (weight per unit volume, expressed in N/m
3
). The specific weight of the material
can then be obtained by multiplying its density by g = 9.81 m/s
2
(see footnote, page 234
of the text).

Chapter 6
Analysis of Structures
In this chapter students learn to determine the internal forces exerted on the members of
pin-connected structures. The chapter starts with the statement of Newton’ s third law
(action and reaction) and is divided into two parts: (a) trusses, that is, structures
consisting of two- force members only, (b) frames and machines, that is, structures
involving multiforce members.

After trusses and simple trusses have been defined in Sec. 6.1A , the method of joints and
the method of sections are explained in detail in Sec. 6.1B and Sec. 6.2A , respectively.
Since a discussion of Maxwell’ s diagram is not included in this text, the use of Bow’s
notation has been avoided, and a uniform notation has been used in presenting the
method of joints and the method of sections.

In the method of joints, a free-body diagram should be drawn for each pin. Since all
forces are of known direction, their magnitudes, rather than their components, should be
used as unknowns. Following the general procedure outlined in Chap. 2, joints involving
only three forces are solved using a force triangle, while joints involving more than three
forces are solved by summing x and y components. Sections 6.1C and 6.1D are optional.
It is shown in Sec. 6. 1C how the analysis of certain trusses can be expedited by
recognizing joints under special loading conditions, while in Sec. 6.1D the method of
joints is applied to the solution of three-dimensional trusses.

It is pointed out in Sec. 6.1B that forces in a simple truss can be determined by analyzing
the truss joint by joint and that joints can always be found that involve only two unknown
forces. The method of sections (Sec. 6.2A ) should be used (a) if only the forces in a few
members are desired, or (b) if the truss is not a simple truss and if the solution of
simultaneous equations is to be avoided (for example, Fink truss). Students should be
urged to draw a separate free-body diagram for each section used. The free body obtained
should be emphasized by shading and the intersect ed members should be removed and
replaced by the forces they exerted on the free body. It is shown that, through a judicious
choice of equilibrium equations, the force in any given member can be obtained in most
cases by solving a single equation. Section 6.2B is optional; it deals with the trusses www.elsolucionario.org

9. obtained by combining several simple trusses and discusses the statical determinacy of
such structures as well as the completeness of their constraints.

Structures involving multiforce members are separated into frames and machines. Frames
are designed to support loads, while machines are designed to transmit and modify
forces. It is shown that while some frames remain rigid after they have been detached
from their supports, others will collapse (Sec. 6.3B ). In the latter case, the equations
obtained by considering the entire frame as a free body provide necessary but not
sufficient conditions for the equilibrium of the frame. It is then necessary to dismember
the frame and to consider the equilibrium of its component parts in order to determine the
reactions at the external supports. The same procedure is necessary with most machines
in order to determine the output force Q from the input force P or inversely (Sec. 6.4).

Students should be urged to resolve a force of unknown magnitude and direction into two
components but to represent a force of known direction by a single unknown, namely its
magnitude. While this rule may sometimes result in slightly more complicated arithmetic,
it has the advantage of matching the numbers of equations and unknowns and thus makes
it possible for students to know at any time during the computations what is known and
what is yet to be determined.

Chapter 7
Forces in Beams and Cables
This chapter consists of five groups of sections, all of which are optional. The first three
groups deal with forces in beams and the last two groups with forces in cables. Most
likely the instructor will not have time to cover the entire chapter and will have to choose
between beams and cables.

Section 7.1 defines the internal forces in a member. While these forces are limited to
tension or compression in a straight two- force member, they include a shearing force and
a bending couple in the case of multiforce members or curved two- force members.
Problems in this section do not make use of sign conventions for shear and bending
moment and answers should specify which part of the member is used as the free body.

In Section 7.2 the usual sign conventions are introduced and shear and bending -moment
diagrams are drawn. All problems in these sections should be solved by drawing the free-
body diagrams of the various portions of the beams.

The relations among load, shear, and bending moment are introduced in Sec. 7.3.
Problems in this section should be solved by evaluating areas under load and shear curves
or by formal integration (as in Probs. 7.85 through 7.88). Some instructors may feel that
the special methods used in this section detract from the unity achieved in the rest of the
text through the use of the free-body diagram, and they may wish to omit Sec. 7.3. Others
will feel that the study of shear and bending-moment diagrams is incomplete without this
section, and they will want to include it. The latter view is particularly justified when the
course in statics is immediately followed by a course in mechanics of materials.
www.elsolucionario.org

10. Section 7.4 is devoted to cables, first with concentrated loads and then with distributed
loads. In both cases, the analysis is based on free-body diagrams. The differential
equation approach is considered in the last problems of this group (Probs. 7.124 through
7.126). Section 7.5 is devoted to catenaries and requires the use of hyperbolic functions.

Chapter 8
Friction
This chapter not only introduces the general topic of friction but also provides an
opportunity for students to consolidate their knowledge of the methods of analysis
presented in Chaps. 2, 3, 4, and 6. It is recommended that each course in statics include at
least a portion of this chapter.

Section 8.1 is devote d to the presentation of the laws of dry friction and to their
application to various problems. The different cases which can be encountered are
illustrated by diagrams in Figs. 8.2, 8.3, and 8.4. Particular emphasis is placed on the fact
that no relation exists between the friction force and the normal force except when
motion is impending or when motion is actually taking place. Following the general
procedure outlined in Chap. 2, problems involving only three forces are solved by a force
triangle, while problems involving more than three forces are solved by summing x and y
components. In the first case the reaction of the surface of contact should be represented
by the resultant R of the friction force and normal force, while in the second case it
should be resolved into its components F and N.

Special applications of friction are considered in Secs. 8.2 through 8.4. They are divided
into the following groups: wedges and screws (Sec. 8.2); axle and disk friction, rolling
resistance (Sec. 8.3); belt friction (Sec. 8.4). The sections on axle and disk friction and on
rolling resistance are not essential to the understanding of the rest of the text and thus
may be omitted.

Chapter 9
Distributed Forces
Moments of Inertia
The purpose of Sec. 9.1A is to give motivation to the study of moments of inertia of
areas. Two examples are considered: one deals with the pure bending of a beam and the
other with the hydrostatic forces exerted on a submerged circular gate. It is shown in each
case that the solution of the problem reduces to the computation of the moment of inertia
of an area. The other sections in the first assignment are devoted to the definition and the
computation of rectangular moments of inertia, polar moments of inertia, and the
corresponding radii of gyration. It is shown how the same differential element can be
used to determine the moment of inertia of an area about each of the two coordinate axes.

Section 9.2introduces the parallel-axis theorem and its application to the determination of
moments of inertia of composite areas. Particular emphasis is placed on the proper use of
the parallel-axis theorem (see Sample Prob. 9.5). Sections 9.3 and 9.4 are optional; they
are devoted to products of inertia and to the determination of principal axes of in ertia.
www.elsolucionario.org

11. Sections 9.5 and 9.6 deal with the moments of inertia of masses. Particular emphasis is
placed on the moments of inertia of thin plates (Sec. 9.5C ) and on the use of these plates
as differential elements in the computation of moments of inertia of symmetrical three-
dimensional bodies (Sec. 9.5D). Section 9.6 is optional but should be used whenever the
following dynamics course includes the motion of rigid bodies in three dimensions.
Sections 9.6A and 9.6B introduce the moment of inertia of a body with respect to an
arbitrary axis as well as the concepts of mass products of inertia and principal axes of
inertia. Section 9.6C discusses the determination of the principal axes and principal
moments of inertia of a body of arbitrary shape.

When solving many of the problems of Chap. 5, information on the specific weight of a
material was generally required. This information was readily available in problems
stated in U.S. customary units, while it had to be obtained from the density of the
material in problems stated in SI units (see the last paragraph of our discussion of Chap.
5). In Chap. 9, when SI units are used, the mass and mass moment of inertia of a given
body are respectively obtained in kg and kg ∙m
2
directly from the dimensions of the body
in meters and from its density in kg/m
3
. However, if U.S. customary units are used, the
density of the body must first be calculated from its specific weight or, alternatively, the
weight of the body can be obtained from its dimensions and specific weight and then
converted into the corresponding mass expressed in lb∙s
2
/ft (or slugs). The mass moment
of inertia of the body is then obtained in lb∙ft ∙s
2
(or slug∙ft
2
). Sample Problem 9.12
provides an example of such a computation. Attention is also called to the footnote on
page of the 530 regarding the conversion of mass moments of inertia from U.S.
customary units to SI units.

Chapter 10
Method of Virtual Work
While this chapter is optional, the instructor should give serious consideration to its
inclusion in the basic statics course. Indeed, students who learn the method of virtual
work in their first course in mechanics will remember it as a fundamental and natural
principle. They may, on the other hand, consider it as an artificial device if its
presentation is postponed to a more advanced course.

Section 10.1 is devoted to the derivation of the principle of virtual work and to its direct
application to the solution of equilibrium problems. Section 10.2 intro duces the concept
of potential energy and shows that equilibrium requires that the derivative of the potential
energy be zero. Section 10.1D defines the mechanical efficiency of a machine and Sec.
10.2D discusses the stability of equilibrium.

The first groups of problems in each assignment utilize the principle of virtual work as an
alternative method for the computation of unknown forces. Subsequent problems call for
the determination of positions of equilibrium, while other problems combine the
conventional methods of statics with the method of virtual work to determine
displacements (Probs. 10.55 through 10.58). www.elsolucionario.org

12. xiv

TABLE I: LIST OF THE TOPICS COVERED IN VECTOR MECHANICS FOR ENGINEERS: STATICS


Suggested Number of Periods

Additional Abridged Course to be
Sections Topics Basic Course Topics used as an introduction
to dynamics
+


1. INTRODUCTION
1.1–6 This material may be used for the first assignment
or for later reference

2. STATICS OF PARTICLES
2.1 Addition and Resolution of Forces 0.5–1 0.5–1
2.2 Rectangular Components 0.5–1 0.5–1
2.3 Equilibrium of a Particle 1 1
2.4 Forces in Space 1 1
2.5 Equilibrium in Space 1 1

3. RIGID BODIES: EQUIVALENT SYSTEMS OF FORCES
3.1 Vector Product; Moment of a Force about a Point 1–2 1–2
3.2 Scalar Product; Moment of a Force about an Axis 1–2 1–2
3.3 Couples 1 1
3.4A–4C Equivalent Systems of Forces 1–1.5 1–1.5
* 3.4D Reduction of a Wrench 0.5–1

4. EQUILIBRIUM OF RIGID BODIES
4.1A–1B Equilibrium in Two Dimensions 1.5–2 1.5–2
4.1C Indeterminate Reactions; Partial Constraints 0.5–1
4.2 Two- and Three-Force Bodies 1
4.3 Equilibrium in Three Dimensions 2 2

5. CENTROIDS AND CENTERS OF GRAVITY
5.1 Centroids and First Moments of Areas and Lines 1–2
5.2 Centroids by Integration 1–2
5.3 Beams and Submerged Surfaces 1–1.5
5.4 Centroids of Volumes 1–2

6. ANALYSIS OF STRUCTURES
6.1A–1B Trusses by Method of Joints 1–1.5
* 6.1C Joints under Special Loading Conditions 0.25–0.5
* 6.1D Space Trusses 0.5–1
6.2A Trusses by Method of Sections 1–2
6.2B Combined Trusses 0.25–0.5
6.3 Frames 2–3 1–2
6.4 Machines 1–2 0.5–1.5

7. INTERNAL FORCES AND MOMENTS
7.1 Internal Forces in Members 1
7.2 Shear and Moment Diagrams by FB Diagram 1–2
7.3 Shear and Moment Diagrams by Integration 1–2
* 7.4 Cables with Concentrated Loads; Parabolic Cable 1–2
* 7.5 Catenary 1

8. FRICTION
8.1 Laws of Friction and Applications 1–2 1–2
8.2 Wedges and Screws 1
* 8.3 Axle and Disk Friction; Rolling Resistance 1–2
8.4 Belt Friction 1

9. MOMENTS OF INERTIA
9.1 Moments of Inertial of Areas 1
9.2 Composite Areas 1–2
* 9.3 Products of Inertia; Principal Axes 1–2
* 9.4 Mohr’s Circle 1
9.5 Moments of Inertia of Masses
#
1–2
* 9.6 Mass Products of Inertia; Principal Axes and Principal 1–2
Moments of Inertia

10. METHOD OF VIRTUAL WORK
10.1A–1C Principle of Virtual Work 1–2
10.1D Mechanical Efficiency 0.5–1
10.2 Potential Energy; Stability 1–1.5

Total Number of Periods 26–39 15–26 14–21


+ A sample assignment schedule for a course in dynamics including this minimum amount of introductory material in statics is given Table V. It is
recommended that a more complete statics course, such as the one outlined in Tables III and IV of this manual, be used in curricula which include
the study of mechanics of materials.

# Mass moments of inertia have not been included in the basic statics course since this material is often taught in dynamics. www.elsolucionario.org

13. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xv
CHAPTER 2: STATICS OF PARTICLES

FORCES IN A PLANE

Resultant of concurrent forces
2.1, 3 2.2, 4 graphical method
2.5, 8 2.6, 7 law of sines
2.9, 10 2.11, 12
2.14 2.13 special problems
2.16, 18 2.15, 17 laws of cosines and sines
2.19, 20


Rectangular components of force
2.22, 23 2.21, 24 simple problems
2.25, 27 2.26, 28 more advanced problems
2.29, 30
2.32, 34 2.31, 33 Resultant by Σ F
x = 0, ΣF y = 0
2.35, 36 2.37, 38
2.39, 40 2.41, 42 Select force so that resultant has a given direction


2.F1, F4 2.F2, F3 Free Body Practice Problems

Equilibrium. Free-Body Diagram
2.44, 45 2.43, 48 equilibrium of 3 forces
2.46, 47
2.49, 50 2.51, 52 equilibrium of 4 forces
2.53, 54 2.55, 56
2.57, 58 2.59, 60 find parameter to satisfy specified conditions
2.61, 62 2.63, 64
2.65, 66 2.67, 68 special problems
2.69, 70


FORCES IN SPACE

Rectangular components of a force in space
2.71, 72 2.75, 76 given F,
θ, and φ, find components and direction angles
2.73, 74 2.77, 78
2.80, 82 2.81, 82 relations between components and direction angles
2.83, 84
2.85, 86 2.87, 88 direction of force defined by two points on its line of action
2.89, 90
2.91, 92 2.93, 94 resultant of two or three forces
2.95, 96 2.97, 98

2.F5, F6 2.F7, F8 Free Body Practice Problems

Equilibrium of a particle in space
2.99, 100 2.103, 104 load applied to three cables, introductory problems
2.101, 102
www.elsolucionario.org

14. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xvi
2.107, 108 2.105, 106 intermediate problems
2.109, 110 2.111, 112

2.115, 116 2.113, 114 advanced problems
2.117, 118 2.119, 120

2.121, 122 2.123, 124 problems involving cable through ring
2.125, 126 special problems


2.127, 131 2.128, 129 Review problems
2.132, 134 2.130, 133
2.135, 136 2.137, 138


2.C1, C3 2.C2 Computer problems
2.C4 2.C5


CHAPTER 3: RIGID BODIES; EQUIVALENT SYSTEMS OF FORCES

Moment of a force about a point: Two dimensions
3.1, 2 3.3, 6 introductory problems
3.4, 5 3.7, 8
3.9, 10 3.11, 12 direction of a force defined by two points on its line of action
3.13, 14
3.15 derivation of a formula
3.17, 18 3.16 applications of the vector product
Moment of a force about a point: Three dimensions
3.19 3.20 computing M = r × F, introductory problems
3.21, 23 3.22, 25 computing M = r × F, more involved problems
3.24, 26
3.28, 29 3.27, 30 using M to find the perpendicular distance from a point to a line
3.32, 33 3.31, 34


3.35 3.36 Scalar Product
3.37, 38 3.39, 40 Finding the angle between two lines
3.41, 42 3.43, 44
3.45 3.46 Mixed triple product
3.47, 48 3.49, 50 Moment of a force about the coordinate axes
3.51, 52 3.53, 54
3.57, 58 3.55, 56 Moment of a force about an oblique axis
3.59, 60 3.61, 62
3.63
*3.66, *67 *3.64, *65 Finding the perpendicular distance between two lines
*3.68, *69


3.70, 71 3.72, 73 Couples in two dimensions
3.74
3.78, 79 3.75, 76 Couples in three dimensions
3.77, 80
www.elsolucionario.org

15. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xvii
3.81, 83 3.82, 84 Replacing a force by an equivalent force-couple system: two dimensions
3.85, 86
3.87, 88 3.89, 91 Replacing a force-couple system by an equivalent force or forces
3.90, 92
3.93, 95 3.94, 97 Replacing a force by an equivalent force-couple system: three dimensions
3.96, 98 3.99, 100


3.101, 102 3.104 Equivalent force-couple systems
3.103
3.107 3.105, 106 Finding the resultant of parallel forces: two dimensions
3.110, 111 3.108, 109 Finding the resultant and its line of action: two dimensions
3.116, 117 3.112, 113
3.118 3.114, 115
3.119, 120 3.121, 122 Reducing a three-dimensional system of forces to a single force-couple system
3.124, 125 3.123, 126
3.127, 128 3.129, 130 Finding the resultant of parallel forces: three dimensions
*3.131, *132
Reducing three-dimensional systems of forces or forces and couples to a wrench
*3.133, *135 *3.134, *136 axis of wrench is parallel to a coordinate axis or passes through O
*3.137 force-couple system parallel to the coordinate axes
*3.139, 140 *3.138 general, three-dimensional case
*3.141 *3.142 special cases where the wrench reduces to a single force
*3.143, *144 *3.145, *146 special, more advanced problems


3.147, 148 3.149, 150 Review problems
3.151, 153 3.152, 154
3.156, 157 3.155, 158


3.C1, C4 3.C2, C3 Computer problems
3.C5 3.C6


CHAPTER 4: EQUILIBRIUM OF RIGID BODIES

EQUILIBRIUM IN TWO DIMENSIONS
4.F1, F4 4.F2, F3 Free Body Practice Problems

4.1, 2 4.3, 4 Parallel forces
4.5, 6 4.7, 8
4.9, 10 4.12, 14 Parallel forces, find range of values of loads to satisfy multiple criteria
4.11, 13
4.15, 16 4.17, 18 Rigid bodies with one reaction of unknown direction and one of known direction
4.19, 20
4.22, 25 4.21, 23
4.26, 28 4.24, 27
4.29, 30
4.33, 34
4.31, 32
4.35, 36 4.37, 38 Rigid bodies with three reactions of known direction
4.39, 40 4.41, 42 www.elsolucionario.org

16. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xviii
4.45, 46 4.43, 44 Rigid bodies with a couple included in the reactions
4.47, 50 4.48, 49
4.51, 52 4.53, 54 Find position of rigid body in equilibrium
4.56, 58 4.55, 57
4.59 4.60 Partial constraints, statical indeterminacy


Three-force bodies
4.62, 63 4.61, 64 simple geometry, solution of a right triangle required
4.65, 67 4.66, 68 simple geometry, frame includes a two-force member
4.69, 70 4.71, 72 more involved geometry
4.73, 74
4.75, 76 4.77, 79
4.78, 81 4.80
4.82
4.83, 84 4.86, 87 find position of equilibrium
4.85, 88 4.89, 90


EQUILIBRIUM IN THREE DIMENSIONS

4.F6, F7 4.F5 Free Body Practice Problems

4.91, 92 4.93, 94 Rigid bodies with two hinges along a coordinate axis and
4.95, 96 an additional reaction parallel to another coordinate axis
4.99, 100 4.97, 98 Rigid bodies supported by three vertical wires or by vertical reactions
4.101, 102 4.103, 104
4.106, 107 4.105, 110 Derrick and boom problems involving unknown tension in two cables
4.108, 109 4.111, 112
4.113, 114 4.115, 116 Rigid bodies with two hinges along a coordinate axis and an additional
4.117, 118 reaction not parallel to a coordinate axis
4.119, 122 4.120, 121 Problems involving couples as part of the reaction at a hinge
4.123, 124 4.125, 126 Advanced problems
4.129, 130 4.127, 128
4.131, 132
4.133, 134 4.135, 136 Problems involving taking moments about an oblique line passing
4.138, 139 4.137, 140 through two supports
4.141


4.144, 145 4.142, 143 Review problems
4.146, 148 4.147, 149
4.150, 153 4.151, 152


4.C2, C5 4.C1, C3 Computer problems
4.C6 4.C4


CHAPTER 5: DISTRIBUTED FORCES: CENTROIDS AND CENTERS OF GRAVITY

Centroid of an area formed by combining
5.1, 3 5.2 rectangles and triangles
5.4
5.6, 9 5.5, 7 rectangles, triangles, and portions of circular areas
5.8 www.elsolucionario.org

17. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xix
5.11, 12 5.10, 13 triangles, portions of circular or elliptical areas, and areas of analytical
5.14, 15 functions
5.16, 17 5.18, 19 Derive expression for location of centroid
5.20, 22 5.21, 23 First moment of an area
5.24, 25 5.26, 27 Center of gravity of a wire figure
5.29, 30 5.28, 33 Equilibrium of wire figures
5.31, 32 Find dimension to maximize distance to centroid

Use integration to find centroid of
5.36 5.34, 35 simple areas
5.37, 38 5.39 areas obtained by combining shapes of Fig. 5.8A
5.40 5.41
5.42 parabolic area
5.43 5.44 areas defined by different functions over the interval of interest
5.45, 46 *5.47 homogeneous wires
*5.48, *49 areas defined by exponential or cosine functions
5.50, 51 areas defined by a hyperbola
Find areas or volumes by Pappus - Guldinus
5.52, 54 5.53, 56 rotate simple geometric figures
5.55, 57
5.60, 61 5.58, 59 practical applications
5.64, *65 5.62, 63

Distributed load on beams
5.66, 67 resultant of loading
5.71 5.68, 69 reactions at supports
5.70
5.73 5.72 parabolic loadings
5.74, 75 5.76, 77 special problems
5.78, 79
Forces on submerged surfaces
5.81, 82 5.80, 84 reactions on dams or vertical gates
5.83, 86 5.85
5.87
5.88, 89 5.90, 91 reactions on non-vertical gates
5.92, 93 5.94, 95

Centroids and centers of gravity of three-dimensional bodies
5.98, 99 5.96, 97
composite bodies formed from two common shapes
5.100, 101 5.102, 103 composite bodies formed from three or more elements
5.104, 105
5.106, 107 5.108, 109 composite bodies formed from a material of uniform thickness
5.110, 113 5.111, 112
5.114, 116 5.115, 117 composite bodies formed from a wire or structural shape of uniform cross
section
5.118, 121 5.119, 120 composite bodies made of two different materials
use integration to locate the centroid of
5.122 5.123, 124 standard shapes: single integration
5.125, 126 *5.128, *129 bodies of revolution: single integration
5.127
www.elsolucionario.org

18. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xx
5.132 *5.130, 131 special applications: single integration
5.133, *134 bodies formed by cutting a standard shape with an oblique
plane: single integration
5.135, 136 special applications: double integration


5.138, 141 5.137, 139 Review problems
5.142, 144 5.140, 143
5.145, 148 5.146, 147

5.C2, C3 5.C1
, C4 Computer problems
5.C5, C6 *5.C7


CHAPTER 6: ANALYSIS OF STRUCTURES

TRUSSES
Method of joints
6.2, 4 6.1, 3 simple problems
6.6, 8 6.5, 7
6.9, 10 6.11, 12 problems of average difficulty
6.13, 14
6.17, 18 6.15, 16 more advanced problems
6.21, 24 6.19, 20
6.25, 26 6.22, 23
6.27, 28
6.29, 30
designate simple trusses
6.32, 34 6.31, 33 find zero-force members
*6.36, *37 *6.35, *38 space trusses
*6.39, *40 *6.41, *42

Method of sections
6.43, 44 6.45, 46 two of the members cut are parallel
6.47, 48
6.49, 50 6.51, 52 none of the members cut are parallel
6.53, 54 6.57, 58
6.55, 56 6.59, 60
6.61, 62 6.63, 64 K-type trusses
6.65, 66 6.67, 68 trusses with counters
6.69, 70 6.71, 72
Classify trusses according to constraints
6.73, 74


FRAMES AND MACHINES
6.F1, F4 6.F2, F3 Free Body Practice Problems – Frames

Analysis of Frames
6.75, 77 6.76, 78 easy problems
6.79, 80 6.81, 82 www.elsolucionario.org

19. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxi
6.83, 84 6.87, 88 problems where internal forces are changed by repositioning a couple or by
6.85, 86 6.89 moving a force along its line of action
6.90 replacement of pulleys by equivalent loadings
6.91, 92 6.93, 94 analysis of frames supporting pulleys or pipes
6.95, 96 analysis of highway vehicles
6.97, 98 6.99, 100 analysis of frames consisting of multiforce members
6.101, 102 6.103, 104
6.105, 106
6.107, 108 6.109, 110 problems involving the solution of simultaneous equations
6.112, 113 6.111, 114
6.115 6.116
6.117 6.118 unusual floor systems
6.119, 120 6.121 rigid and non-rigid frames


6.F6, F8 6.F5, F7 Free Body Practice Problems – Machines

Analysis of Machines
6.122, 125 6.123, 124 toggle-type machines
6.126, 127
6.128
6.131, 132 6.129, 130 machines involving cranks
6.133, 134
6.137, 138 6.135, 136 machines involving a crank with a collar
6.139, 140 robotic machines
6.143, 144 6.141, 142 tongs
6.147, 148 6.145, 146 pliers, boltcutters, pruning shears
6.149, 150 6.151, 152 wrench, find force to maintain position of toggle
6.155, 156 6.153, 154 force in hydraulic cylinder
6.157, 158
6.159, 160 *6.161, *162 gears and universal joints
*6.163 special tongs


6.164, 165 6.166, 167 Review problems
6.169, 170 6.168, 171
6.173, 174 6.172, 175


6.C2, C4 6.C1, C3 Computer problems
6.C6 6.C5











www.elsolucionario.org

20. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxii
CHAPTER 7: FORCES IN BEAMS AND CABLES

Internal forces in members
7.3, 4 7.1, 2 simple frames
7.5, 6
7.9, 10 7.7, 8 curved members
7.11, 12 7.13, 14
7.15, 16 7.17, 18 frames with pulleys or pipes
7.19, 20
7.21, 22 effect of supports
7.23, 24 7.25, 26 bending moment in circular rods due to their own weight
7.27, 28


BEAMS

Shear and bending-moment diagrams using portions of beam as free bodies
7.29, 30 7.31, 32 problems involving no numerical values
7.33, 34
7.35, 36 7.37, 38 beams with concentrated loads
7.39, 40 7.41, 42 beams with mixed loads
7.43, 44 7.45, 46 beams resting on the ground
7.47, 48
7.49, 50 7.51, 52 beams subjected to forces and couples
7.53, 54
7.55, 56 7.59, 60 find value of parameter to minimize absolute value of bending moment
7.57, 58 7.61, *62


Shear and bending-moment diagrams using relations among ω, V, and M
7.63, 64 7.65, 66 problems involving no numerical values
7.67, 68
7.69, 70 7.73, 74 problems involving numerical values
7.71, 72 7.75, 76
7.77, 78 7.81, 82 find magnitude and location of maximum bending moment
7.79, 80 7.83, 84
7.85, 88 7.86, 87 Determine V and M by integrating ω twice
7.89, 90 *7.91, *92 Find values of loads given bending moment at two points


CABLES

Cables with concentrated loads
7.93, 94 7.95, 96 vertical loads
7.97, 98 7.99, 100
7.101, 102 7.103, 104 horizontal and vertical loads
7.105, 106
Parabolic cables
7.107, 108 7.109, 110 supports at the same elevation
7.112, 113 7.111, 114
7.115, 116 7.117, 118 supports at different elevations


www.elsolucionario.org

21. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxiii
*7.119 Derive analogy between a beam and a cable
7.120, 121 7.122, 123 use analogy to solve previous problems
*7.124 *7.125, *126 Derive or use d
2
y/dx
2
= w(x) T
0

Catenary
7.129, 130 7.127, 128 given length of cable and sag or T
m, find span of cable
7.131, 132
7.133, 136 7.134, 135 given span and length of cable, find sag and/or weight
7.139, 140 7.137, 138 given span, T
m, and w, find sag
7.141, 142 7.143, 144 given T
0, w, and sag or slope, find span or sag
7.145, 146 *7.147, *148
*7.151, *152 7.149
, *150 special problems
*7.153

7.154, 155 7.156, 158 Review problems
7.157, 159 7.160, 161
7.163, 164 7.162
, 165

7.C2, C4 7.C1, C3 Computer problems
7.C5, C6


CHAPTER 8: FRICTION
8.F1, F4 8.F2, F3
Free Body Practice Problems – Frames
8.1, 2 8.3, 4 For given loading, determine whether block is in equilibrium and find friction force
8.7, 9 8.5, 6 Find minimum force required to start, maintain, or prevent motion
8.10 8.8
8.13, 14 8.11, 12 Analyze motion of several blocks
8.15, 16 8.17, 18 Sliding and/or tipping of a rigid body
8.21, 22
8.19, 20 Problems involving wheels and cylinders
8.25, 26 8.23, 24 Problems involving rods
8.27, 28 8.29, 30 Analysis of mechanisms with friction
8.32, 33 8.31, 34
8.35
8.37, 39 8.36, 38 Analysis of more advanced rod and beam problems
8.40 8.41, 42
8.43, 44 8.46, 47 Analysis of systems with possibility of slippage for various loadings
8.45

8.48, 49 8.52, 53 Wedges, introductory problems
8.50, 51 8.57
8.54, 55
8.56
8.59, 64 8.58, 60 Wedges, more advanced problems
8.65 8.61, 62
*8.66, *67 8.63
www.elsolucionario.org

22. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxiv
8.68 8.69, 70 Square-threaded screws
8.72, 73 8.71, 74
8.75, 76
8.77, 79 8.78, 83 Axle friction
8.80, 81 8.84, 87
8.82, 85 8.88, 89
8.86, 91 8.90
8.92, *94 8.93, 97 Disk friction
*8.95, *96
8.100, 101 8.98, 99 Rolling resistance
8.102


Belt friction
8.105, 106 8.103, 104 belt passing over fixed drum
8.107, 108
8.109, 112 8.110, 111 transmission belts and band brakes
8.113, 114 8.115, 116
8.117
8.118, 119 8.120, 121 advanced problems
8.124, 125 8.122, 123
8.126, 127 8.128, 129
8.132, 133 8.130, 131 derivations, V belts

8.134, 135 8.136, 137
Review problems
8.138, 139 8.140, 141
8.143, 144 8.142, 145

8.C1, C3 8.C2, C5
Computer problems
8.C4, C6 8.C7
8.C8


CHAPTER 9: DISTRIBUTED FORCES: MOMENTS OF INERTIA

MOMENTS OF INERTIA OF AREAS
Find by direct integration
9.1, 3 9.2, 4 moments of inertia of an area
9.6, 8 9.5, 7
9.9, 12 9.10, 11
9.13, 14
9.15, 17 9.16, 18 moments of inertia and radii of gyration of an area
9.19, 20
9.21, 22 9.23, 24 polar moments of inertia and polar radii of gyration of an area
9.25, 26 9.27, 28
*9.29 *9.30 Special problems
Parallel-axis theorem applied to composite areas to find
9.31, 33 9.32, 34 moment of inertia and radius of gyration
9.35, 36
www.elsolucionario.org

23. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxv
9.39, 40 9.37, 38 centroidal moment of inertia, given I or J
9.41, 42 9.43, 44 centroidal moments of inertia
9.47, 48 9.45, 46 centroidal polar moment of inertia
centroidal moments of inertia of composite areas consisting of rolled-steel
shapes:
9.49, 52 9.50, 51 doubly-symmetrical composite areas
9.54, 55 9.53, 56 singly-symmetrical composite areas (first locate centroid of area)
Center of pressure
9.58, *60 9.57, 59 for end panel of a trough
9.61, 62 for a submerged vertical gate or cover
*9.64 *9.63 used to locate centroid of a volume
*9.65 *9.66 special problems


Products of inertia of areas found by
9.67, 68 9.69, 70 direct integration
9.71, 72 9.73, 74 parallel-axis theorem
9.75, 78 9.76, 77
Using the equations defining the moments and products of inertia with respect to
rotated axes to find
9.79, 80 9.81, 83 I
x'
, I
y'
, I
x'y'
for a given angle of rotation
9.82, 84
9.85, 86 9.87, 89 principal axes and principal moments of inertia
9.88, 90


Using Mohr's circle to find
9.91, 92 9.93, 95 I
x'
, I
y'
, I
x'y'
for a given angle of rotation
9.94, 96
9.97, 98 9.99, 100 principal axes and principal moments of inertia
9.101, 102
9.104, 105 9.103, *106
9.107, 108 9.109, 110 Special problems


MOMENTS OF INERTIA OF MASSES

Mass moment of inertia
9.111, 112 9.113, 114 of thin plates: two-dimensions
9.117, 118 9.115, 116
9.119, 120 9.121, 122 of simple geometric shapes by direct single integration
9.123, 125 9.124, *126
9.128 9.127 and radius of gyration of composite bodies
9.129, 130 9.131, 134 special problems using the parallel-axis theorem
9.132, 133
9.135, 136 9.138, 139 of bodies formed of sheet metal or of thin plates: three-dimensions
9.137, *140
9.141, 142 9.143, 144 of machine elements and of bodies formed of homogeneous wire
9.145, 148 9.146, 147
Mass products of inertia
9.149, 150 9.151, 152 of machine elements
9.153, 154 of bodies formed of sheet metal or of thin plates
9.155, 156
9.157, 158 9.159, 160 of bodies formed of homogeneous wire
9.162 9.161 Derivation and special problem
www.elsolucionario.org

24. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxvi
9.165, 166 9.163, 164 Mass moment of inertia of a body with respect to a skew axis
9.169, 170 9.167, 168
9.171, 172
9.175, 176 9.173, 174 Ellipsoid of inertia and special problems
9.177, 178
*9.179, *180 *9.182, *183 Principal mass moments of inertia and principal axes of inertia
*9.181, *184


9.187, 188 9.185, 186 Review problems
9.189, 193 9.190, 191
9.194, 195 9.192, 196


9.C3, C5 9.C1, C2 Computer problems
*9.C7, *C8 9.C4, C6


CHAPTER 10: METHOD OF VIRTUAL WORK
For linkages and simple machines, find
10.1, 3 10.2, 4 force or couple required for equilibrium (linear relations among displacements)
10.5, 6 10.7, 8
10.11, 12 10.9, 10
force required for equilibrium (trigonometric relations among displacements)
10.13, 14
10.17, 18 10.15, 16 couple required for equilibrium (trigonometric relations among displacements)
10.21, 22 10.19, 20 force or couple required for equilibrium
Find position of equilibrium
10.23, 24 10.25, 26 for numerical values of loads
10.27, 28
10.29, 32 10.30, 31 linear springs included in mechanism
10.33, 34 10.36, 38
10.35, 37
10.39, 40 torsional spring included in mechanism
10.43, 44 10.41, 42 Problems requiring the use of the law of cosines
10.45, 46
10.48, 50 10.47, 49 Problems involving the effect of friction
10.51, 52
Use method of virtual work to find:
10.53, 54 reactions of a beam
10.55 10.56 internal forces in a mechanism
10.57, 58 movement of a truss joint

Potential energy method used to
10.62, 63 10.59
, 60 solve problems from Sec. 10.4
10.64, 65 10.61, 66
10.67 10.68 establish that equilibrium is neutral
find position of equilibrium and determine its stability for a system involving:
10.69, 72 10.70, 71 gears and drums
10.73, 74 torsional springs
10.77, 78 10.75, 76 linear springs
10.79, 82 10.80, 81
10.85, 86 10.87, 88 www.elsolucionario.org

25. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


Problem Number*
SI Units U.S. Units Problem Description

* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.


xxvii
10.83, 84 two applied forces
determine stability of a known position of equilibrium for a system with
10.89, 93 10.90, 91 one degree of freedom
10.94, 96 10.92, 95
*10.97, *98 *10.99, *100 two degrees of freedom


10.103, 104 10.101, 102 Review problems
10.105, 107 10.106, 108
10.111, 112 10.109, 110

10.C2, C3 10.C1, C4 Computer problems
10.C5, C6
10.C7
www.elsolucionario.org

26. TABLE III: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN STATICS

This schedule includes all of the material of VECTOR MECHANICS FOR ENGINEERS: STATICS with the exception of Sections 3.4D, 6.1D, 7.4–7.5, 8.3, and 9.6C.

50% OF THE PROBLEMS IN EACH OF THE FO LLOWING LISTS USE SI UNITS AND 50% U.S. CUSTOMARY UNITS

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK NO ANSWERS ARE GIVEN IN THE BOOK
FOR ANY OF THESE PROBLEMS

Period Sections Topics List 1 List 2 List 3 List 4 List 5 List 6
1 1.1–6 Introduction
2 2.1–2 Addition and Resolution of Forces 2.5, 15, 27, 31 2.10, 17, 29, 37 2.11, 16, 26, 36 2.6, 19, 28, 35 2.12, 18, 30, 33 2.7, 20, 25, 38
3 2.3 Equilibrium of a Particle 2.43, 50, 63, 65 2.48, 49, 59, 69 2.46, 55, 61, 67 2.44, 51, 62, 68 2.45, 56, 64, 66 2.47, 52, 60, 70
4 2.4 Forces in Space 2.71, 81, 85, 97 2.72, 82, 89, 94 2.75, 84, 88, 92 2.77, 79, 87, 91 2.76, 80, 86, 93 2.78, 83, 90, 98
5 2.5 Equilibrium in Space 2.103, 107, 119, 125 2.104, 108, 113, 121 2.101, 106, 116, 123 2.99, 112, 115, 124 2.102, 105, 120, 126 2.100, 111, 114, 122
6 3.1 Vector Product, Moment of a Force about a Point 3.4, 12, 23, 27 3.5, 11, 26, 30 3.7, 13, 25, 28 3.6, 9, 22, 29 3.3, 14, 21, 31 3.8, 10, 24, 34
7 3.2 Scalar Product, Moment of a Force about an Axis 3.43, 48, 61, *68 3.39, 47, 55, *67 3.41, 53, 57, *64 3.37, 49, 58, *65 3.42, 50, 56, *69 3.38, 54, 62, *66
8 3.3 Couples 3.71, 84, 87, 99 3.70, 82, 90, 97 3.73, 86, 89, 96 3.80, 83, 91, 95 3.75, 85, 88, 100 3.72, 81, 92, 94
9 3.4 Equivalent Systems of Forces 3.106, 111, 126, 127 3.105, 110, 122, 128 3.102, 108, 120, 129 3.101, 114, 119, 130 3.107, 112, 121, 131 3.103, 109, 123, 132
10 EXAM NUMBER ONE

11 4.1 Equilibrium in Two Dimensions 4.1, 14, 19, 23 4.2, 12, 15, 24 4.4, 13, 17, 25 4.7, 9, 18, 22 4.3, 10, 16, 27 4.8, 11, 20, 21
12 4.1 Equilibrium in Two Dimensions 4.34, 36, 43, 52 4.33, 35, 48, 51 4.31, 42, 46, 57 4.32, 41, 47, 55 4.30, 37, 44, 58 4.29, 38, 49, 56
13 4.2 Two- and Three-Force Bodies 4.63, 72, 75, 86 4.62, 71, 81, 90 4.66, 73, 77, 88 4.61, 69, 79, 83 4.68, 74, 76, 87 4.64, 70, 82, 89
14 4.3 Equilibrium in Three Dimensions 4.93, 100, 110, 117 4.94, 99, 105, 113 4.91, 103, 107, 116 4.95, 97, 106, 115 4.92, 104, 111, 118 4.96, 98, 112, 114
15 4.3 Equilibrium in Three Dimensions 4.122, 127, 133 4.119, 128, 138 4.121, 131, 135 4.120, 129, 140 4.126, 132, 134 4.125, 130, 139
16 5.1 Centroids and First Moments 5.1, 10, 24, 33 5.3, 13, 20, 26 5.5, 14, 23, 30 5.2, 11, 21, 31 5.8, 15, 25, 28 5.7, 12, 22, 27
17 5.2 Centroids by Integration 5.34, 43, 58, 64 5.35, 40, 53, 60 5.37, 44, 54, 63 5.39, 41, 52, 62 5.38, 42, 59, *65 5.36, *47, 56, 61
18 5.3 Beams and Submerged Surfaces 5.66, 76, 81, 94 5.67, 77, 82, 90 5.70, 74, 80, 89 5.69, 78, 84, 88 5.72, 75, 83, 95 5.68, 79, 86, 91
19 5.4 Centroids of Volumes 5.102, 107, 119, 122 5.103, 106, 117, 125 5.100, 111, 114, 124 5.104, 109, 116, 123 5.101, 112, 115, 126 5.105, 108, 120, 127
20 EXAM NUMBER TWO

21 6.1 Trusses: Method of Joints 6.2, 11, 21, 30 6.4, 12, 24, 27 6.1, 13, 23, 29 6.3, 9, 22, 28 6.7, 14, 25, 33 6.5, 10, 26, 31
22 6.2 Trusses: Method of Sections 6.45, 54, 68, 69 6.46, 53, 67, 70 6.43, 59, 62, 71 6.44, 60, 61, 72 6.47, 57, 64, 73 6.48, 58, 63, 74
23 6.3 Analysis of Frames 6.75, 88, 102, 109 6.77, 89, 101, 110 6.76, 83, 100, 107 6.81, 85, 99, 108 6.78, 84, 105, 111 6.82, 86, 106, 114
24 6.4 Analysis of Machines 6.123, 134, 145, 149 6.124, 133, 141, 155 6.127, 129, 144, 153 6.122, 130, 143, 154 6.126, 135, 142, 150 6.128, 136, 146, 156
25 6.1–4 Review of Chapter 6 6.15, 52, 91, 151 6.19, 51, 92, 152 6.17, 50, 94, 140 6.18, 49, 93, 137 6.16, 48, 98, 157 6.20, 47, 97, 158
26 7.1 Internal Forces in Members 7.1, 12, 18, 23 7.2, 11, 17, 24 7.3, 8, 19, 26 7.4, 7, 20, 25 7.5, 14, 22, 27 7.6, 13, 21, 28
27 7.2 Beams 7.36, 46, 50, 59 7.35, 45, 49, 59 7.42, 48, 52, 55 7.41, 47, 51, 58 7.38, 43, 54, 60 7.37, 44, 53, 61
28 7.3 Beams 7.65, 70, 81, *89 7.66, 69, 82, *90 7.64, 73, 78, 86 7.63, 74, 77, 87 7.68, 75, 83, 85 7.67, 76, 84, 88
29 EXAM NUMBER THREE

30 8.1 Laws of Friction 8.1, 18, 32, 36 8.2, 17, 27, 41 8.3, 14, 30, 40 8.4, 13, 29, 39 8.8, 16, 33, 38 8.6, 15, 28, 42
31 8.2 Wedges and Screws 8.52, 54, 63, 73 8.53, 55, 62, 72 8.48, 57, 65, 74 8.49, 60, 64, 71 8.50, 58, 70, 76 8.51, 61, 69, 75
32 8.4 Belt Friction 8.104, 109, 121, 126 8.103, 112, 120, 133 8.105, 110, 119, 128 8.106, 111, 118, 129 8.108, 115, 123, 127 8.107, 116, 122, 132
33 9.1 Moments of Inertia of Areas 9.1, 10, 17, 23 9.3, 11, 15, 28 9.4, 12, 16, 21 9.2, 9, 18, 22 9.5, 14, 19, 24 9.7, 13, 20, 27
34 9.2 Composite Areas 9.34, 42, 51, 58 9.32, 41, 50, *60 9.31, 43, 52, 57 9.33, 44, 49, 59 9.35, 45, 56, 61 9.36, 46, 53, 62
35 9.3 Product of Inertia 9.72, 83, 86 9.71, 81, 85 9.76, 79, 87 9.74, 80, 89 9.77, 84, 88 9.73, 82, 90
36 9.4 Mohr’s Circle 9.93, 97, 103 9.95, 98, *106 9.92, 100, 107 9.91, 99, 108 9.94, 102, 110 9.96, 101, 109
37 9.5 Moments of Inertia of Masses 9.112, 121, 135, 147 9.111, 122, 136, 143 9.113, 119, 138, 142 9.114, 120, 139, 141 9.116, 123, *140, 146 9.115, 125, 137, 144
38 9.6 Principal Axes 9.152, 158, 167, 175 9.151, 155, 168, 176 9.149, 159, 165, 174 9.150, 160, 166, 173 9.154, 163, 171, 178 9.153, 164, 172, 177
39 EXAM NUMBER FOUR

40 10.1 Virtual Work 10.1, 9, 18, 20 10.3, 10, 17, 19 10.2, 12, 16, 24 10.4, 14, 15, 23 10.8, 13, 21, 25 10.7, 11, 22, 26
41 10.1 Virtual Work 10.31, 35, 46, 53 10.30, 37, 45, 54 10.32, 36, 43, 57 10.33, 38, 44, 58 10.34, 42, 47, 55 10.29, 41, 51, 56
42 10.2 Potential Energy 10.69, 80, 86, 91 10.72, 81, 83, 92 10.71, 77, 88, 89 10.70, 78, 87, 93 10.76, 79, 84, 95 10.75, 82, 85, 90
www.elsolucionario.org

27. TABLE IV: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN STATICS

This schedule includes all of the material of VECTOR MECHANICS FOR ENGINEERS: STATI CS with the exception of Sections 3.4D, 6.1D, 7.4–7.5, 8.3, and 9.6C.

75% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 25% U.S. CUSTOMARY UNITS

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK

Period Sections Topics List 1a List 2a List 3a List 4a
1 1.1–6 Introduction
2 2.1–2 Addition and Resolution of Forces 2.5, 15, 27, 34 2.10, 17, 29, 32 2.8, 16, 26, 36 2.9, 19, 28, 35
3 2.3 Equilibrium of a Particle 2.43, 50, 57, 65 2.48, 49, 58, 69 2.46, 54, 61, 67 2.44, 53, 62, 68
4 2.4 Forces in Space 2.71, 81, 85, 95 2.72, 82, 89, 96 2.74, 84, 88, 92 2.73, 79, 87, 91
5 2.5 Equilibrium in Space 2.103, 107, 118, 125 2.104, 108, 117, 121 2.101, 109, 116, 123 2.99, 110, 115, 124
6 3.1 Vector Product, Moment of a Force about a Point 3.4, 12, 23, 32 3.5, 11, 26, 33 3.1, 13, 25, 28 3.2, 9, 22, 29
7 3.2 Scalar Product, Moment of a Force about an Axis 3.43, 48, 59, *68 3.39, 47, 60, *67 3.41, 52, 57, *64 3.37, 51, 58, *65
8 3.3 Couples 3.71, 84, 87, 98 3.70, 82, 90, 93 3.74, 86, 89, 96 3.78, 83, 91, 95
9 3.4 Equivalent Systems of Forces 3.106, 111, 125, 127 3.105, 110, 124, 128 3.102, 116, 120, 129 3.101, 117, 119, 130
10 EXAM NUMBER ONE

11 4.1 Equilibrium in Two Dimensions 4.1, 14, 19, 28 4.2, 12, 15, 26 4.5, 13, 17, 25 4.6, 9, 18, 22
12 4.1 Equilibrium in Two Dimensions 4.34, 36, 45, 52 4.33, 35, 50, 51 4.31, 40, 46, 57 4.32, 39, 47, 55
13 4.2 Two- and Three-Force Bodies 4.63, 72, 75, 84 4.62, 71, 81, 85 4.67, 73, 77, 88 4.65, 69, 79, 83
14 4.3 Equilibrium in Three Dimensions 4.93, 100, 108, 117 4.94, 99, 109, 113 4.91, 102, 107, 116 4.95, 101, 106, 115
15 4.3 Equilibrium in Three Dimensions 4.122, 127, 133 4.119, 128, 138 4.124, 131, 135 4.123, 129, 140
16 5.1 Centroids and First Moments 5.1, 10, 24, 29 5.3, 13, 20, 32 5.6, 14, 23, 30 5.9, 11, 21, 31
17 5.2 Centroids by Integration 5.34, 43, 57, 64 5.35, 40, 55, 60 5.37, 46, 54, 63 5.39, 45, 52, 62
18 5.3 Beams and Submerged Surfaces 5.66, 76, 81, 93 5.67, 77, 82, 92 5.73, 74, 80, 89 5.71, 78, 84, 88
19 5.4 Centroids of Volumes 5.102, 107, 118, 122 5.103, 106, 121, 125 5.100, 113, 114, 124 5.104, 110, 116, 123
20 EXAM NUMBER TWO

21 6.1 Trusses: Method of Joints 6.2, 11, 21, 32 6.4, 12, 24, 34 6.6, 13, 23, 29 6.8, 9, 22, 28
22 6.2 Trusses: Method of Sections 6.45, 54, 66, 69 6.46, 53, 65, 70 6.43, 56, 62, 71 6.44, 55, 61, 72
23 6.3 Analysis of Frames 6.75, 88, 102, 113 6.77, 89, 101, 112 6.79, 83, 100, 107 6.80, 85, 99, 108
24 6.4 Analysis of Machines 6.123, 134, 147, 149 6.124, 133, 148, 155 6.127, 132, 144, 153 6.122, 131, 143, 154
25 6.1–4 Review of Chapter 6 6.15, 52, 91, 125 6.19, 51, 92, 139 6.50, 94, 140, 164 6.49, 93, 137, 165
26 7.1 Internal Forces in Members 7.1, 12, 15, 23 7.2, 11, 16, 24 7.3, 10, 19, 26 7.4, 9, 20, 25
27 7.2 Beams 7.36, 46, 50, 57 7.35, 45, 49, 56 7.40, 48, 52, 55 7.39, 47, 51, 58
28 7.3 Beams 7.65, 70, 80, *89 7.66, 69, 79, *90 7.64, 71, 78, 86 7.63, 72, 77, 87
29 EXAM NUMBER THREE

30 8.1 Laws of Friction 8.1, 18, 32, 37 8.2, 17, 27, 43 8.7, 14, 30, 40 8.9, 13, 29, 39
31 8.2 Wedges and Screws 8.52, 54, *67, 73 8.53, 55, *66, 72 8.48, 56, 65, 74 8.49, 59, 64, 71
32 8.4 Belt Friction 8.104, 109, 125, 126 8.103, 112, 124, 133 8.105, 113, 119, 128 8.106, 114, 118, 129
33 9.1 Moments of Inertia of Areas 9.1, 10, 17, 26 9.3, 11, 15, 25 9.8, 12, 16, 21 9.6, 9, 18, 22
34 9.2 Composite Areas 9.34, 42, 55, 58 9.32, 41, 54, *60 9.31, 47, 52, 57 9.33, 48, 49, 59
35 9.3 Product of Inertia 9.72, 83, 86 9.71, 81, 85 9.75, 79, 87 9.78, 80, 89
36 9.4 Mohr’s Circle 9.93, 97, 103 9.95, 98, *106 9.92, 104, 107 9.91, 105, 108
37 9.5 Moments of Inertia of Masses 9.112, 121, 135, 148 9.111, 122, 136, 145 9.117, 119, 138, 142 9.118, 120, 139, 141
38 9.6 Principal Axes 9.152, 158, 170, 175 9.151, 155, 169, 176 9.149, 157, 165, 174 9.150, 156, 166, 173
39 EXAM NUMBER FOUR

40 10.1 Virtual Work 10.1, 9, 18, 28 10.3, 10, 17, 27 10.5, 12, 16, 24 10.6, 14, 15, 23
41 10.1 Virtual Work 10.31, 35, 52, 53 10.30, 37, 48, 54 10.32, 40, 43, 57 10.33, 39, 44, 58
42 10.2 Potential Energy 10.69, 80, 86, 96 10.72, 81, 83, 94 10.74, 77, 88, 89 10.73, 78, 87, 93
www.elsolucionario.org

28. TABLE V: SAMPLE ASSIGNMENT SCHEDULE FOR A COMBINED COURSE IN STATICS AND DYNAMICS

This schedule is intended for a 4-semester-credit-hour course consisting of (1) the abridged Statics course defined in Table I of this manual and (2) the standard Dynamics course defined in Table I of the Dynamics Instructor’s Manual. Note that the Statics portion
of the course is designed to serve only as an introduction to Dynamics; a more complete coverage of Statics is recommended for curricula including courses in Mechanics of Materials (see Tables III and ff. of the Statics Manual).

50% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 50% U.S. CUSTOM ARY UNITS

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK NO ANSWERS ARE GIVEN IN THE BOOK
FOR ANY OF THESE PROBLEMS

Period Sections Topics List 1 List 2 List 3 List 4 List 5 List 6
1 1.1–6 Introduction
2 2.1–2 Addit io n and Resolution of Forces 2.5, 15, 27, 31 2.10, 17, 29, 37 2.11, 16, 26, 36 2.6, 19, 28, 35 2.12, 18, 30, 33 2.7, 20, 25, 38
3 2.3 Equilibrium of a Particle 2.43, 50, 63, 65 2.48, 49, 59, 69 2.46, 55, 61, 67 2.44, 51, 62, 68 2.45, 56, 64, 66 2.47, 52, 60, 70
4 2.4 Forces in Space 2.71, 81, 85, 97 2.72, 82, 89, 94 2.75, 84, 88, 92 2.77, 79, 87, 91 2.76, 80, 86, 93 2.78, 83, 90, 98
5 2.5 Equilibrium in Space 2.103, 107, 119, 125 2.104, 108, 113, 121 2.101, 106, 116, 123 2.99, 112, 115, 124 2.102, 105, 120, 126 2.100, 111, 114, 122
6 3.1 Vector Product, Moment of a Force about a Point 3.4, 12, 23, 27 3.5, 11, 26, 30 3.7, 13, 25, 28 3.6, 9, 22, 29 3.3, 14, 21, 31 3.8, 10, 24, 34
7 3.2 Scalar Product, Moment of a Force about an Axis 3.43, 48, 61, *68 3.39, 47, 55, *67 3.41, 53, 57, *64 3.37, 49, 58, *65 3.42, 50, 56, *69 3.38, 54, 62, *66
8 3.3 Couples 3.71, 84, 87, 99 3.70, 82, 90, 97 3.73, 86, 89, 96 3.80, 83, 91, 95 3.75, 85, 88, 100 3.72, 81, 92, 94
9 3.4 Equivalent Systems of Forces 3.106, 111, 126, 127 3.105, 110, 122, 128 3.102, 108, 120, 129 3.101, 114, 119, 130 3.107, 112, 121, 131 3.103, 109, 123, 132
10 EXAM NUMBER ONE

11 4.1 Equilibrium in Two Dimensions 4.1, 14, 19, 23 4.2, 12, 15, 24 4.4, 13, 17, 25 4.7, 9, 18, 22 4.3, 10, 16, 27 4.8, 11, 20, 21
12 4.1 Equilibrium in Two Dimensions 4.34, 36, 43, 52 4.33, 35, 48, 51 4.31, 42, 46, 57 4.32, 41, 47, 55 4.30, 37, 44, 58 4.29, 38, 49, 56
13 6.3 Analysis of Frames 6.75, 88, 102, 109 6.77, 89, 101, 110 6.76, 83, 100, 107 6.81, 85, 99, 108 6.78, 84, 105, 111 6.82, 86, 106, 114
14 6.4 Analysis of Machines 6.123, 134, 145, 149 6.124, 133, 141, 155 6.127, 129, 144, 153 6.122, 130, 143, 154 6.126, 135, 142, 150 6.128, 136, 146, 156
15 4.3 Equilibrium in Three Dimensions 4.93, 100, 110, 117 4.94, 99, 105, 113 4.91, 103, 107, 116 4.95, 97, 106, 115 4.92, 104, 111, 118 4.96, 98, 112, 114
16 4.3 Equilibrium in Three Dimensions 4.122, 127, 133 4.119, 128, 138 4.121, 131, 135 4.120, 129, 140 4.126, 132, 134 4.125, 130, 139
17 8.1 Laws of Friction 8.1, 18, 32, 36 8.2, 17, 27, 41 8.3, 14, 30, 40 8.4, 13, 29, 39 8.8, 16, 33, 38 8.6, 15, 28, 42
18 8.2 Wedges and Screws 8.52, 54, 63, 73 8.53, 55, 62, 72 8.48, 57, 65, 74 8.49, 60, 64, 71 8.50, 58, 70, 76 8.51, 61, 69, 75
19 EXAM NUMBER TWO

20 11.1 Rectilinear Motion
21 11.2 Uniformly Accelerated Motion
22 11.4 Curvilinear Motion (Rect. Comps)
23 11.5 Curvilinear Motion (Other Comp.)
24 12.1 Equations of Motion
25 12.2 Angular Momentum
26 Review
27 EXAM NUMBER THREE

28 13.1 Work and Energy, Power
29 13.2 Conservation of Energy
30 13.2D Application to Space Mechanics
31 13.3 Impulse and Mo mentum
32 13.4 Impact
33 14.1 Systems of Particles
34 14.2 Systems of Particles
35 EXAM NUMBER FOUR

36 15.1 Translat ion, Rotation
37 15.2 General Plane Motion
38 15.3 Instantaneous Center
39 15.4 Acceleration in Plane Motion
40 15.5 Coriolis Acceleration in Plane Motion
41 9.5 Moments of Inertia of Masses 9.112, 121, 135, 147 9.111, 122, 136, 143 9.113, 119, 138, 142 9.114, 120, 139, 141 9.116, 123, *140, 146 9.115, 125, 137, 144
42 Review
43 EXAM NUMBER FIVE

44 16.1 Plane Motion of Rigid Bodies
45 16.1 Plane Motion of Rigid Bodies
46 16.2 Constrained Plane Motion
47 16.2 Constrained Plane Motion
48 17.1 Work and Energy
49 17.2 Impulse and Mo mentum
50 17.3 Eccentric Impact
51 Review
52 EXAM NUMBER SIX

53 19.1 Free Vibrations of Particles
54 19.2 Free Vibrations of Rigid Bodies
55 19.3 Energy Methods
56 19.4 Forced Vibrations
www.elsolucionario.org

29. CHAPTER 2
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

30.  
  Copyright ? McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

31.  
 
SOLUT
(a) P
(b) T
W

 
TION
arallelogram l
Triangle rule:
We measure:

PRO
Two
magn
(b) th
law:

OBLEM 2.1
o forces are a
nitude and dir
he triangle rul
139R=
1
applied as sho
rection of thei
le.
1kN, 4 7α=
own to a hoo
ir resultant us
7.8°
ok. Determine
ing (a) the pa


1391
N=R
e graphically
arallelogram la
N 47.8°
the
aw,
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

32.  
SOLUT
(a) P
(b) T
W
 
TION
arallelogram l
Triangle rule:
We measure:

PR
Two
grap
(a)
law:
 
OBLEM 2.2
o forces are
phically the
the parallelo
90R=
2
applied as s
magnitude
ogram law, (

06lb, 2α=
shown to a b
and directio
(b) the triang


26.6°
bracket supp
on of their r
gle rule.
906R=
ort. Determi
resultant usi
6lb 26.6°
ine
ing
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

33.  

PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 10 kN and Q = 15 kN,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
 
SOLUTION
(a) Parallelogram law:

(b) Triangle rule:

We measure:
20.1 kN,R
= 21.2α= ° 20.1 kN=R 21.2° W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

34. PROBLEM 2.4
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 6 kips and Q = 4 kips,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
 
SOLUTION
(a) Parallelogram law:

(b) Triangle rule:

We measure:
8.03 kips, 3.8R
α= =° 8.03 kips=R 3.8° W

  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

35. PROBLEM 2.5
A stake is being pulled out of the ground by means of two ropes as shown.
Knowing that α = 30°, determine by trigonometry (a) the magnitude of the
force
P so that the resultant force exerted on the stake is vertical, (b) the
corresponding magnitude of the resultant.
 
SOLUTION

Using the triangle rule and the law of sines:
(a)
120 N
sin30 sin 25
P
=
°°
101.4 NP= W
(b)
30 25 180
180 25 30
125
β
β
°+ + °= °
=°− °− °
=°


120 N
sin 30 sin125
=
°°
R
196.6 N=R W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

36. PROBLEM 2.6
A telephone cable is clamped at A to the pole AB. Knowing that the
tension in the left-hand portion of the cable is T
1 = 800 lb, determine
by trigonometry (a) the required tension T
2 in the right-hand portion if
the resultant
R of the forces exerted by the cable at A is to be vertical,
(b) the corresponding magnitude of
R.
 
SOLUTION

Using the triangle rule and the law of sines:
(a)
75 40 180
180 75 40
65
α
α
°+ °+ = °
=°− °− °
=°


2800 lb
sin 65 sin 75
T
=
°°

2
853 lbT= W
(b)
800 lb
sin 65 sin 40
R
=
°° 567 lbR= W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

37. PROBLEM 2.7
A telephone cable is clamped at A to the pole AB. Knowing that the
tension in the right-hand portion of the cable is T
2 = 1000 lb, determine
by trigonometry (a) the required tension T
1 in the left-hand portion if
the resultant
R of the forces exerted by the cable at A is to be vertical,
(b) the corresponding magnitude of
R.
 
SOLUTION

Using the triangle rule and the law of sines:
(a)
75 40 180
180 75 40
65
β
β
°+ °+ = °
=°− °− °
=°


11000 lb
sin 75° sin 65
T
=
°

1
938 lbT= W
(b)
1000 lb
sin 75° sin 40
R
=
°
665 lbR= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

38. PROBLEM 2.8
A disabled automobile is pulled by means of two
ropes as shown. The tension in rope AB is 2.2 kN,
and the angle α is 25°. Knowing that the resultant
of the two forces applied at A is directed along the
axis of the automobile, determine by trigonometry
(a) the tension in rope AC, (b) the magnitude of the
resultant of the two forces applied at A.

 
SOLUTION

Using the law of sines:

2.2 kN
sin30° sin125sin 25
AC
T R
==
°
α
2.603 kN
4.264 kN
AC
T
R
=
=


( a)
2.60 kN
AC
T= W
(b)
4.26 kNR= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

39. PROBLEM 2.9
A disabled automobile is pulled by means of two
ropes as shown. Knowing that the tension in rope
AB is 3 kN, determine by trigonometry the tension
in rope AC and the value of α so that the resultant
force exerted at A is a 4.8-kN force directed along
the axis of the automobile.

 
SOLUTION

Using the law of cosines:
22 2
(3 kN) (4.8 kN) 2(3 kN)(4.8 kN)cos 30°
2.6643 kN
AC
AC
T
T
=+ −
=
Using the law of sines:
sin sin 30
3 kN 2.6643 kN
34.3
α
α
°
=
=°


2.66 kN
AC
=T 34.3° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

40.  
SOLUT
Using th
(a)

(b)

 
 
TION
he triangle rule

e and law of s
si
5
s
αβ++
sin117
R
 
PROBLEM
Two forces a
magnitude o
angle α if the
be horizonta
ines:
insin25
50 N 35 N
sin 0.6037
α
α
°
=
=
37.13
8α=
25 180
180
117.862
β
+°= °
=°−
=
35 N
7.862 sin2
R
=
°
M 2.10
are applied as
of P is 35 N, d
e resultant R o
l, (b) the corre
74
°

8°
25 37.138
2
°− °
°
N
25°
shown to a ho
determine by
of the two forc
esponding mag

ook support. K
trigonometry
ces applied to
gnitude of R .
Knowing that
(a) the requir
the support is
37.1α=°
73.2 NR=
the
red
s to
W
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

41.  
SOLUT
Using th
(a)

(b)

TION
he triangle rulee and the law
50β+°
4
s
4
s


P
A
th
m
fo
m
of sines:
60 180
180
70
β
°+°= °
=°
=°
425 lb
sin 70 sin 6
P
=
°
425 lb
sin 70 sin5
R
=
°
PROBLEM
A steel tank is
hat α = 20°, d
magnitude of
forces applied
magnitude of R
50 60−°−°

0°
0°
2.11
to be position
determine by
the force P i
at A is to be v
R.

ned in an excav
trigonometry
if the resultan
vertical, (b) th
vation. Knowi
(a) the requir
nt R of the t
he correspondi
392 lbP=
346 lbR=
ing
red
two
ing
W
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

42.  
SOLUT
Using th
(a)

(b)
 
 
TION
he triangle rulee and the law
(3α+
sin
 

P
A
th
tr
th
c
of sines:
30) 60
sin (90 )
425 lbβ
β
β
α
°+ °+
°−
90
α°−
(42.598 30
R
°+
PROBLEM
A steel tank is
hat the mag
rigonometry (
he two force
corresponding
180
180 (
90
sin 60
500 lb
α
α
=°
=°− +
=°−
°
=
47.402=°
500 lb
0)sin60
=
°°
2.12
to be position
gnitude of P
a) the require
s applied at
magnitude of

30 ) 60°− °

ned in an excav
P is 500 lb,
d angle α if th
A is to be v
fR.
vation. Knowi
, determine
he resultant R
vertical, (b)
42.6α=°
551 lbR=
ing
by
R of
the
W
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

43.  
SOLUT
The sma
(a) P
(b) R

TION
allest force P w
(425 lb)coP=
(425 lb)sinR=
will be perpen
os30°
n30°


P
A
D
d
o
c
ndicular to R.
PROBLEM
A steel tank
Determine by
direction of the
of the two f
corresponding
2.13
is to be p
y trigonometr
e smallest forc
forces applied
magnitude of

positioned in
ry (a) the
ce P for which
d at A is v
fR.
an excavati
magnitude a
h the resultant
vertical, (b)
368 lb=P
21R=
on.
and
t R
the
W
13 lb WCopyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

44.  
SOLUT
The sma
(a) P
(b) R

TION
allest force P w
(50 N)sinP=
(50 N)cosR=

P
F
m
R
c
will be perpen
25°
25°

PROBLEM
For the hook s
magnitude and
R of the two
orresponding
ndicular to R.
2.14
support of Pro
d direction of t
o forces appli
magnitude of
ob. 2.10, deter
the smallest fo
ied to the su
fR.

rmine by trigo
force P for wh
upport is hor
onometry (a)
hich the result
rizontal, (b)
21.1 N=P
45.3 NR=
the
tant
the
W
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

45. PROBLEM 2.15
For the hook support shown, determine by trigonometry the
magnitude and direction of the resultant of the two forces applied
to the support.


SOLUTION
Using the law of cosines:

222
(200 lb) (300 lb)
2(200 lb)(300 lb)cos(45 65 )
413.57 lb
R
R
=+
−+°
=
α




Using the law of sines:
sin sin (45 65 )
300 lb 413.57 lb
42.972α
α
+°
=
=°
α


90 25 42.972β
=+− °
αα
414 lb=R 72.0° W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

46.  

PROBLEM 2.16
Solve Prob. 2.1 by trigonometry.
PROBLEM 2.1
Two forces are applied as shown to a hook. Determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law,
(b) the triangle rule.
 
SOLUTION
Using the law of cosines:

22 2
(900 N) (600 N )
2(900 N )(600 N)cos(135 )
1390.57N
R
R
=+
−°
=




Using the law of sines:
sin( 30 ) sin (135 )
600N 1390.57N
30 17.7642
47.764α
α
α
− °
=
−=°
=
α
α
α


1391N=R 47.8°W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

47.  
SOLUT
Using th
We hav



Then
And

 
 
 
TION
he force triang
e:

P
So
PR
A.
Q
re
(b
gle and the law
180
105
γ=
=
2
(4
64
8.
0
R
R
= =
=
4kip
sin(25
sin(25
25
°
°
°
 
ROBLEM 2
olve Problem 2
ROBLEM 2.4
. Knowing tha
= 4 kips, de
sultant force
b) the triangle r
ws of cosines a
0(50 25
5
°− °+ °
°
2
2
kips) (6 ki
p
.423 kips
0264 kips
+
ps8.026 4
)sin1
)0.4813
28.775
3.775
α
α
α
α
=
+
+=
°+=
=°
2.17
2.4 by trigono
4 Two structu
at both memb
etermine graph
exerted on th
rule.
and sines:
)°
2
ps) 2(4 kip−
4kips
05
37
5
°
°
°

ometry.
ural members
bers are in ten
hically the ma
he bracket usin
s)(6 kips)cos1
B and C are b
nsion and that
agnitude and
ng (a) the par
105°
8.03k=R
bolted to brac
t P = 6 kips a
direction of
rallelogram la
kips 3.8°
ket
and
the
aw,
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

48.  

PROBLEM 2.18
For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N,
determine by trigonometry the magnitude and direction of the force
P so
that the resultant is a vertical force of 160 N.
PROBLEM 2.5 A stake is being pulled out of the ground by means of two
ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the
magnitude of the force
P so that the resultant force exerted on the stake is
vertical, (b) the corresponding magnitude of the resultant.
 
SOLUTION

Using the laws of cosines and sines:

222
(120 N) (160 N) 2(120 N)(160 N)cos 25
72.096 N
P
P
=+− °
=

And
sin sin 25
120 N 72.096 N
sin 0.70343
44.703
α
α
α
°
=
=
=°


72.1 N=P 44.7° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

49. SOLUT
Using th
We have
Then
and
Hence:

  
TION
he force triang
e

gle and the law
R
48
sin
sin
 
PROBLE
Two forces
Knowing th
the magnitu
ws of cosines a
180 (2
150
γ=°−
=°
22
(48 N)
2(48
N
104.366 N
R
R
= +
−
=
N 104.366N
sin150
n0.22996
13.2947
α
α
α
=
°
=
=°
180
180 1
3
86.705
φ α=°− =°−
=°

M 2.19
P and Q are a
hat P = 48 N
ude and directi
and sines:
2010)°+ °
2
(60 N)
N)(60 N)cos15
N
+
N
°

80
3.2947 80
−°
°− °
applied to the
and Q = 60 N
ion of the resu
50°

lid of a storag
N, determine
ultant of the tw
104.=R
ge bin as show
by trigonome
wo forces.
4 N 86.7°
wn.
etry
° W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

50. SOLUT
Using th
We have
Then
and
Hence:

 
 
 
TION
he force triang
e

P
T
K
m
gle and the law
γ=
=
2
R
R
=
=
60 N
sin
sin
α
α
α
=
=
=
180
180
83.2
φ=°
=°
=

 
PROBLEM
Two forces P
Knowing that
magnitude and
ws of cosines a
180 (20
150
=°− °+
=°
2
(60 N) (4
2(60 N)(48
104.366 N
= +
−
=
104.366 N
sin150
0.28745
16.7054
=
°
=
= °
180
16.7054
295
α°−− °
°−°−
°
M 2.20
P and Q are ap
P = 60 N and
d direction of
and sines:
10 )+°
2
48N)
8N)cos150°
80°

pplied to the l
d Q = 48 N, d
the resultant o
lid of a storag
determine by
of the two forc
104.4=R
ge bin as show
trigonometry
ces.
4 N 83.3°
wn.
the
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

51.  
 
SOLUT
Comput

29-lb Fo

50-lb Fo
51-lb F
o

 
TION
te the followin
orce:
orce:
orce:
ng distances:
 

PROB
Determi
2
2
2
(84)
116 in.
(28)
100 in.
(48)
102 in.
OA
OB
OC
=
=
=
=
=
=
(29 lb
x
F=+
(29 lb
y
F=+
(50 lb
x
F=−
(50 lb
y
F=+
(51 lb
x
F=+
(51 lb
y
F=−
BLEM 2.21
ine the x and y
2 2
2 2
2 2
(80)
.
(96)
.
(90)
.
+
+
+

84
b)
116
80
b)
116
28
b)
100
96
b)
100
48
b)
102
90
b)
102
y components of each of the
F
F
F
F
e forces shown
21.0 lb
x
F=+
20.0 lb
y
F=+
14.00 lb
x
F=−
48.0 lb
y
F=+
24.0 lb
x
F=+
45.0 lb
y
F=−
n.
W

W

W

W

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

52.  
 
SOLUT
Comput

800-N F

424-N F
408-N
F

 
TION
te the followin
Force:
Force:
Force:
ng distances:
 

PROBL
Determin
2
(600)
1000 m
(560)
1060 m
(480)
1020 m
OA
OB
OC
=
=
=
=
=
=
(800 N
x
F=+
(800 N
y
F=+
(424 N
x
F=−
(424 N
y
F=−
(408 N
x
F=+
(408 N
y
F=−
LEM 2.22
ne the x and y
22
2 2
22
(800)
mm
(900)
mm
(900)
mm
+
+
+

800
N)
1000
600
N)
1000
560
N)
1060
900
N)
1060
480
N)
1020
900
N)
1020
components oof each of the
F
forces shown.
640 N
x
F=+
480 N
y
F=+
224 N
x
F=−
360 N
y
F=−
192.0 N
x
F=+
360 N
y
F=−
.
N W
W
N W

W N W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

53.  

PROBLEM 2.23
Determine the x and y components of each of the forces shown.
 
SOLUTION
80-N Force: (80 N)cos 40
x
F
=+° 61.3 N
x
F= W

(80 N)sin 40
y
F
=+° 51.4 N
y
F= W
120-N Force:
(120 N)cos70
x
F
=+° 41.0 N
x
F= W

(120 N)sin 70
y
F
=+° 112.8 N
y
F= W
150-N Force:
(150 N)cos35
x
F
=−° 122. 9 N
x
F=− W

(150 N)sin35
y
F
=+° 86.0 N
y
F= W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

54.  

PROBLEM 2.24
Determine the x and y components of each of the forces shown.
 
SOLUTION
40-lb Force: (40 lb)cos60
x
F
=+° 20.0 lb
x
F= W

(40 lb)sin60
y
F
=−° 34.6 lb
y
F=− W
50-lb Force:
(50 lb)sin50
x
F
=−° 38.3 lb
x
F=− W

(50 lb)cos50
y
F
=−° 32.1 lb
y
F=− W
60-lb Force:
(60 lb)cos25
x
F
=+° 54.4 lb
x
F= W

(60 lb)sin 25
y
F
=+° 25.4 lb
y
F= W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

55.  

PROBLEM 2.25
Member BC exerts on member AC a force P directed along line BC . Knowing
that
P must have a 325-N horizontal component, determine (a) the magnitude
of the force
P, (b) its vertical component.
 
SOLUTION

22
(650 mm) (720 mm)
970 mm
BC=+
=
(a)
650
970
x
PP
⎛⎞
=
⎜⎟
⎝⎠
or
970
650
970
325 N
650
485 N
x
PP
⎛⎞
=
⎜⎟
⎝⎠
⎛⎞
=
⎜⎟
⎝⎠
=

485 NP= W
(b)
720
970
720
485 N
970
360 N
y
PP
⎛⎞
=
⎜⎟
⎝⎠
⎛⎞
=
⎜⎟
⎝⎠
=

970 N
y
P= W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

56.  
SOLUT
(a)

(b) V


 
 
TION
Vertical compo

PRO
Memb
Know
magni
onent
 
OBLEM 2.26
ber BD exerts
wing that P mu
itude of the fo
sin35P °=
P=
v
P=
=
6
s on member
ust have a 300-
orce P, (b) its v
300 lb
300 lb
sin 35°
cos35P °
(523 lb)cos3
5
r ABC a forc
-lb horizontal
vertical compo

5°
ce P directed
component, d
onent.
along line B
determine (a) t
523 lbP=
428 lb
v
P=
BD.
the
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

57.  
 
SOLUT

(a)

(b)


 
 
TION
 
PR
Th
dir
com
ma
180 45
180
15
α
°= °
=
=°
cos
cos
60
cos
621
x
P
P
P
P
α=
=
=
=

tan
t
(60
160
y
x
yx
P
P
PP
α=
=
=
=

ROBLEM 2
he hydraulic cy
rected along li
mponent perp
agnitude of the
90 30
45 90
α++°+
°− °− °−
s
00N
s15
1.17 N
x
P
α
°

tan
00N)tan15
0.770 N
α
°

2.27
ylinder BC ex
ine BC. Know
pendicular to m
e force P, (b)
0
30
°
−°
xerts on memb
wing that P mu
member AB, d
its component
ber AB a force
ust have a 600
determine (a)
t along line AB
621 NP=
160.8 N
y
P=
e P
0-N
the
B.
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

58.  
 
SOLUT
(a)

(b)

 
 
TION

PRO
Cable
that P
of the
 
OBLEM 2.2
e AC exerts o
P must have a
e force P, (b)
cos 55
y
P
P=
°
350 lb
cos 55
610.21
=
°
=
sin 5
5
x
PP=
(610.21
499.85
=
=
28
on beam AB a
a 350-lb vertic
its horizontal
°

lb
5°
1 lb)sin 55
lb
°
a force P dire
cal component
component.
cted along lin
t, determine (a
ne AC. Knowi
a) the magnitu
610 lbP=
500 lb
x
P=
ing
ude
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

59.  
 
SOLUT
(a)

(b)

 
 
TION

P
Th
al
pe
P,
 
PROBLEM 2
he hydraulic
long line BD
erpendicular to
, (b) its compo
750 N sinP=
2192P=
cos
ABC
PP=
(219=
2.29
cylinder BD e
D. Knowing
o member AB
onent parallel
n20°
2.9 N
s20°
2.9 N)cos 20°
exerts on mem
that P must
C, determine
to ABC.

°
mber ABC a
t have a 75
(a) the magni
A
P
force P direc
0-N compon
tude of the fo
2190 NP=
2060 N
ABC
=
ted
ent
rce
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

60.  

PROBLEM 2.30
The guy wire BD exerts on the telephone pole AC a force P directed along
BD. Knowing that
P must have a 720-N component perpendicular to the pole
AC, determine (a) the magnitude of the force
P, (b) its component along line
AC.
 
SOLUTION
(a)
37
12
37
(720 N)
12
2220 N
=
=
=
x
PP



2.22 kNP= W
(b)
35
12
35
(720 N)
12
2100 N
yx
PP=
=
=


2.10 kN=
y
P W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

61.  
SOLUT
Compon



 
TION
nents of the fo
F
2
5
5

orces were dete
Force x
29 lb
50 lb
51 lb
R
tan

PROBLE
Determine t
PROBLEM
forces show
ermined in Pro
x Comp. (lb)
+21.0
–14.00
+24.0
31.0
x
R=+
(31.0 lb)
n
23.0
31.0
36.573
23.0 lb
sin (36.57
x y
y
x
RR
R
R
R
α
α
=+
=
=
=
=°
=
ij
i
R
38.601 lb=
EM 2.31
the resultant o
M 2.21 Determ
wn.
oblem 2.21:
y Comp
+20
+4
–4
2
y
R=+
(23.0 lb)
b
73)
+
°
i j
b
of the three for
mine the x and
p. (lb)
0.0
8.0
5.0
3.0
rces of Problem
d y componen
38.6=R
m 2.21.
nts of each of
6lb 36.6°
the
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

62.  
 
SOLUT
Compon

 
TION
nents of the fo
F
1
1


orces were det
orce x
80 N
20 N
50 N
R
 
PROBLE
Determine
PROBLEM
forces show
termined in Pr
x Comp. (N)
+61.3
+41.0
–122.9
20.6
x
R=−
(20.
tan
250.2
tan
20.6
tan 12.14
85.29
250
sin85
x
y
x
R
R
R
R
α
α
α
α
=+
=−
=
=
=
=
=
Ri
EM 2.32
the resultant o
M 2.23 Determ
wn.
roblem 2.23:
y Com
+
+1
+
2
y
R=+
.6 N) (250.2
2 N
6 N
456
93
0.2 N
5.293
y
R+
+
°
°
j
i
of the three for
mine the x and
mp. (N)
+51.4
112.8
+86.0
250.2
2 N)j
rces of Proble
d y componen
25=R
em 2.23.
nts of each of
1N 85.3°
the
° W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

63.  
 
SOLUT

 
TION
F
4
5
6
orce x
40 lb
50 lb
60 lb

x
R
tan
tan
tan
R
α
α
α
α
R
 

PROBL
Determin
PROBLE
the forces
x Comp. (lb)
+20.00
–38.30
+54.38
36.08
x
=+
( 36.08 lb
41.42 lb
36.08 lb
1.14800
48.942
41.42 lb
sin 48.942
x y
y
x
RR
R
R
R
α
α
α
α
=+
=+
=
=
=
=°
=
°
R ij
LEM 2.33
e the resultant
EM 2.24 Det e
s shown.
y Comp.
–34.6
–32.1
+25.3
41.4
y
R=−
) ( 41.42 lb+−
°
i
t of the three f
ermine the x
(lb)
64
14
36
42
b)j
forces of Prob
and y compon
54.9=R
lem 2.24.
nents of each
9lb 48.9°
h of
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

64.  

SOLUT
Compon

 
TION
nents of the fo
F
80
42
40
orces were det
orce x
00 lb
24 lb
08 lb
R
tan
 

PROBL
Determine
PROBLE
forces sho
ermined in Pr
x Comp. (N)
+640
–224
+192
608
x
R=+
(608 lb)
n
240
608
21.541
240 N
sin(21.54
653.65 N
x y
y
x
RR
R
R
R
α
α
=+
=
=
=
=°
=
=
Ri
j
i
LEM 2.34
e the resultant
EM 2.22 Deter
own.
oblem 2.22:
y Comp. (
+480
–360
–360
240
y
R=−
( 240 lb)
N
41°)
N
+−
j
j
t of the three f
rmine the x an
(N)
0
0
0
0
forces of Probl
nd y componen
654=R
lem 2.22.
nts of each of
4N 21.5°
the
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

65.  
 
SOLUT
100-N F
150-N F
200-N F


 
 
 
TION
Force:
Force:
Force:
Force
100 N
150 N
200 N

 

PRO
Know
forces
(100
(100
x
y
F
F
=+
=−
(150
(150
x
y
F
F
=+
=−
(200
(200
x
y
F
F
=−
=−
x Com
+
+
−1
x
R=−
ta

OBLEM 2.35
wing that α =
s shown.
N)cos35
N)sin35
°=+
°=−
N)cos65
N)sin65
°=+
°=−
N)cos35 N)sin35
°=
−
°=−
mp. (N)
81.915
63.393
63.830
18.522
(18.52
an
308.02
18.522
86.559
xy
y
x
RR
R
R
α
α
=+
=−
=
=
= °
Ri

308.02
sin86.5
R=
5
35°, determi
81.915 N
57.358 N
+
−

63.393 N
135.946 N
+
−

163.830 N
114.715 N
−
−

y Comp. (N
−57.35
−135.94
−114.71
308.02
y
R=−
22 N) ( 308
y
+−
j
i
N
59
ine the result
N)
58
46
15
2
8.02 N)j

309=R
tant of the th
9 N 86.6°
hree
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

66.  
 
SOLUT
Determi
Cable fo
500-N F

200-N F
and
Further:

  
TION
ine force comp
orce AC:
Force:
Force:

ponents:
(365N
(365N
=−
=−
x
y
F
F
(500 N
)
(500 N)
x
y
F
F
=
=
(200 N
)
(200N
x
y
F
F
= =−
2
(400N
474.37
=Σ =−
=Σ =−
=+
=
=
xx
yy
x
RF
RF
RRR
255
tan
400
32.5
α
α=
=
°
 

PR
Kno
resu
960
N)2 4
1460
1100
N)2 7
1460
=−
=−
24
) 480 N
25
7
) 140 N
25
=
=
4
)160 N
5
3
N)120 N
5
=
=−
2
2
240 N 480
275 N 140
N
N)(255N
N
− +
− +
+−
y
R
°

ROBLEM 2.
owing that the
ultant of the th
40N
75N

N

2
N160N 4
N120 N
N)
+=
−= −
36
tension in rop
hree forces exe
400 N
255 N−

pe AC is 365 N
erted at point C
474=R
N, determine
C of post BC .
4N 32.5°
the
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

67. SOLUT
60-lb Fo
80-lb Fo
120-lb F
and
Further:


 
 
TION
orce:
orce:
Force:

(60l
(60l
x
y
F
F
=
=
(80
l
(80l
x
y
F
F
= =
(120
(12
x
y
F
F
=
=−
(20
202.
x x
yy
RF
RF
R
=Σ
=Σ
=
=
29.
tan
200
α=
tan
8.46α
−
=
=
°
 

PRO
Knowi
forces
lb)cos20 5
lb)sin 20 20
°=
°=
lb)cos60 4
lb)sin 60 69
°=
°=
0lb)cos301
20lb)sin30
°=
°=
2
200.305 lb
29.803 lb
00.305 lb)(
.510 lb
=
=
+
803
.305

129.803
200.305
°
BLEM 2.37
ing that α =
shown.
6.382 lb
0.521lb

0.000 lb
9.282 lb

103.923 lb
60.000 lb=−

2
(29.803 lb)
7
40°, determiine the result
203=R
ant of the th
3lb 8.46°
hree
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

68.  
 
SOLUT
60-lb Fo
80-lb Fo
120-lb F
Then
and


 
 
TION
orce:
orce:
Force:
tan
tan
 

PRO
Know
force
(60 lb)co
(60 lb)si
x
y
F
F
=
=
(80 lb)c
o
(80 lb)si
x
y
F
F
= =
(120 lb)c
(120 lb)s
x
y
F
F
=
=
1
6
11
xx
yy
RF RF
=Σ =
=Σ =
(168.95
201.976
R=
=
110.676
n
168.953
n0.65507
33.228
α
α
α=
=
=°
OBLEM 2.3
wing that α =
es shown.
os 20 56.38
in 20 20.52
°=
°=
os95 6.97
n 95 79.696
°=−
°=
cos 5 119.54
sin 5 10.459
°=
°=
68.953 lb
10.676 lb

2
53 lb) (110.
lb
+
38
= 75°, determ
82 lb
1 lb
725 lb
6 lb

43 lb
9 lb

2
.676 lb)
mine the result
202=R
tant of the th
2 lb 33.2°
ree
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

69. PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value
of
α if the resultant of the three forces shown is to be vertical,
(b) the corresponding magnitude of the resultant.

SOLUTION

(100 N)cos (150 N)cos( 30 ) (200 N)cos
(100 N)cos (150 N)cos( 30 )
xx
x
RF
R
α αα
αα
=Σ
=+ +°−
=− + + °
(1)

(100 N)sin (150 N)sin ( 30 ) (200 N)sin
(300 N)sin (150 N)sin( 30 )
yy
y
RF
R
α αα
αα
=Σ
=− − + ° −
=− − + ° (2)
(a) For
R to be vertical, we must have 0.
x
R
= We make 0
x
R= in Eq. (1):

100cos 150cos( 30 ) 0
100cos 150(cos cos 30 sin sin 30 ) 0
29.904cos 75sin
α α
αα α
α α
−+ +°=
−+ °− °=
=


29.904
tan
75
0.39872
21.738
α
α=
=
= ° 21.7α=° W
(b) Substituting for
α in Eq. (2):

300sin 21.738 150sin51.738
228.89 N
y
R=− °− °
=−

| | 228.89 N
y
RR== 229 NR= W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

70. PROBLEM 2.40
For the post of Prob. 2.36, determine (a) the required
tension in rope AC if the resultant of the three forces
exerted at point C is to be horizontal, (b) the corresponding
magnitude of the resultant.
 
SOLUTION

960 24 4
(500 N) (200 N)
1460 25 5
48
640 N
73
xx AC
xAC
RF T
RT
=Σ =− + +
=− +
(1)

1100 7 3
(500 N) (200 N)
1460 25 5
55
20 N
73
yy AC
yAC
RF T
RT
=Σ =− + −
=− +
(2)
(a) For R to be horizontal, we must have 0.
y
R
=
Set
0
y
R= in Eq. (2):
55
20 N 0
73
AC
T
− +=

26.545 N
AC
T= 26.5 N
AC
T= W
(b) Substituting for
AC
T into Eq. (1) gives

48
(26.545 N) 640 N
73
622.55 N
623 N
=− +
=
==
x
x
x
R
R
RR
623 NR= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

71.  
SOLUT
Using th


(
a) S
e

(
b) S
u


TION
he x and y axes
et 0
y
R= in E
ubstituting for

PRO
Dete
of th
BC,
s shown:
Eq. (2):
r
AC
Tin Eq. (1

OBLEM 2.4
ermine (a) the
he three forces
(b) the corresp
x x
R F=Σ
93.6
yy
y
R F
R
=Σ
=
93.631 lb−
1):
x
R
R
41
e required tens
s exerted at P
ponding magn
sin10
sin10
AC
AC
T
T
=°
+
= °+
(50 lb)sin 3
31 lb co
AC
T
=
−
cos10
AC
AC
T
T
°=
=
(95.075 lb)
94.968 lb
x
R
=
=
=
sion in cable A
Point C of boo
nitude of the r

(50 lb)cos35
78.458 lb
+
+
5(75 lb)sin
s10
°+
°
0
95.075 lb
sin10 78.4
5°+
AC, knowing t
om BC must b
esultant.
5(75 lb)co°+
n60 co
AC
T°−
58 lb
that the result
be directed alo
os60°

s10°

95.1 lb
AC
T=
95.0 lbR=
tant
ong
(1)
(2)

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

72.  

SOLUT
Select th
Then

and

(
a) S
e

D

(
b) S
u


 
 
TION
he x axis to be
et 0
y
R= in E
Dividing each t
ubstituting for
R
e along a a′.
x
R
y
R
Eq. (2).
(80 lb
term by cosα
r α in Eq. (1)
60 lb (80
x
R=+
 

PRO
For th
requir
to be p
the res
(60 l
x
F=Σ =
(80 l
y
F=Σ =
b)sin (120α−
gives:
(80
) gives:
0lb)cos56.31°
OBLEM 2.42
he block of P
ed value of α
parallel to the
sultant.
lb) (80 lb)co+
lb)sin (120α−
lb)cos 0α=
0 lb) tan 12
12
tan
8
56
α
α
α
=
=
=
(120 lb) sin°+
2
Problems 2.37
αif the resultan
incline, (b) th

os (120 lb)α+
0 lb)cosα
20 lb
20 lb
80 lb
6.310°
n56.31 204.°=
7 and 2.38, de
nt of the three
he correspondi
sinα
.22 lb
etermine (a)
e forces shown
ing magnitude
56.3α=°
204 lb
x
R=
the
n is
e of
(1)
(2)
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

73. PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
 
SOLUTION
Free-Body Diagram Force Triangle



Law of sines:
400 lb
sin 60 sin 40 sin80
AC BC
TT
==
°°°




(a)
400 lb
(sin 60 )
sin80
AC
T
= °
° 352 lb
AC
T= W
(b)
400 lb
(sin 40 )
sin80
BC
T
= °
° 261 lb
BC
T= W
 
 
 
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

74. PROBLEM 2.44
Two cables are tied together at C and are loaded as shown. Knowing
that α = 30°, determine the tension (a) in cable AC , (b) in cable BC .

 
SOLUTION
Free-Body Diagram Force Triangle



Law of sines:

6 kN
sin 60 sin35 sin85
AC BC
TT
==
°°°





(a)
6 kN
(sin 60 )
sin85
AC
T
= °
° 5.22 kN
AC
T= W
(b)
6 kN
(sin 35 )
sin85
BC
T
= °
° 3.45 kN
BC
T= W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

75.  

PROBLEM 2.45
Two cables are tied together at C and loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
 
SOLUTION
Free-Body Diagram

1.4
tan
4.8
16.2602
1.6
tan
3
28.073
α
α
β
β=
=°
=
=°

Force Triangle

Law of sines:

1.98 kN
sin 61.927 sin 73.740 sin 44.333
AC BC
TT
==
°°°



(
a)
1.98 kN
sin 61.927
sin 44.333
AC
T
= °
° 2.50 kN
AC
T= W
(
b)
1.98 kN
sin 73.740
sin 44.333
BC
T
= °
° 2.72 kN
BC
T= W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

76. SOLUT


Law of
s
(a)
(
b)
 
 
 
TION
Free-Bod

sines:
dy Diagram
 

PROB
Two cab
Knowing
(a) in ca
sin35s
AC
T
=
°
5
s
AC
T=
5
si
BC
T=
LEM 2.46
bles are tied
g that P = 50
able AC, (b) in
For
500 N
sin 75 sin 70
BC
T
=
°
500 N
sin35
in 70
°
°
500 N
sin 75
in70
°
°
together at C
00 N and α =
n cable BC.
rce Triangle
N
0°

C and are lo
60°, determin

oaded as show
ne the tension
305 N
AC
T=
514 N
BC
T=
wn.
n in
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

77.  
SOLUT


Law of s
(a)
(
b)

TION
Free-Bod
mg
(20
196
W=
=
=
sines:

PRO
Two
tensi
dy Diagram
00kg)(9.81m/
62N
sin
A
T
T
T

OBLEM 2.4
cables are tie
ion (a) in cabl
2
/s)

15 sin 105
ACBC
T
=
° °
(1962 N
sin
AC
T=
(1962 N
sin
BC
T=
47
ed together at
e
AC, (b) in c
a
F
1962 N
sin 60
=
° °

N) sin 15
n60
°
°
N)sin 105
n60
°
°
C and are loa
able BC.
Force Triang
aded as shown
gle

T
n. Determine
586 N
AC
T=
2190 N
BC
T=
the
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

78.  
SOLUT


Law of s
(a)
(
b)
 
 
TION
Free
sines:
e-Body Diagra
sin
 

P
K
AC
am
n110 sin5
ACBC
TT
=
°
1200
sin6
AC
T=
1200
sin6
BC
T=
PROBLEM 2
Knowing that α
C, (b) in rope
1200 lb
5sin 65
=
°°

0 lb
sin 110
65
°
°
lb
sin 5
65
°
°
2.48
20 ,α=° dete
BC.
Force Triang
rmine the ten
gle
T
T
sion (a) in ca

1244 lb
AC
T=
115.4 lb
BC
T=
ble
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

79.  
SOLUT
x
FΣ
For P=


y
FΣ

Solvin

 
 
 
TION
0
CA
T=−
200N= we ha

0=
0.8
ng equations (

PRO
Two
Know
BC.
sin30
CB
T+
α
ve,

0.5
C
A
T−
co
CA
T
86603 0.8
CA
T+
1) and (2) sim
 
BLEM 2.49
cables are
wing that P =
Free-B
sin30 cosP−
α
0.5 2 1
A CB
T++
os30 co
CB
T°−
86603 21
CB
T−
multaneously g

9
tied togeth
= 300 N, de
Body Diagram
s45 200N°− =
12.13 200−=
s30 sin4P−
α
12.13 0= (2)
ives,
her at C an
etermine the
m
0=
0 (1)
50=
α

)
nd are load
tension in c
T
T
ded as show
cables AC a
134.6 N
CA
T=
110.4 N
CB
T=
wn.
and
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

80.  
SOLUT
x
FΣ
For
CA
T


y
FΣ

Adding
Substit
u

And
range of

 
TION
0
CA
T=−
0= we have ,

0= c
CA
T

equations (1)
uting for
CB
T=
d 546.40
CA
T=
f 179.315 N an


PRO
Two
Deter
taut.
sin30
CB
T+
α
,

0.5
CB
T
+
cos30 c
CB
T°−
0.866
and (2) gives
0= into the eq
0N , 669P=
nd 669.20 N.

 
BLEM 2.50
cables are
rmine the ran
Free-B
sin30 cosP−
α
0.707112P+ −
cos30 sinP−
α
603 0.707
CB
T−
s, 1.36603
C
T
quilibrium equ
0.5
0.86603
CA
T
−
9.20N Thus fo

0
tied togeth
nge of value
Body Diagram
s45 200N°− =
200 0= (1)
45 0=
α
; aga
711 0P= (2)
200
CB
= henc
uations and sol
0.70711
0.70711
CA
P
T
+ −
−
for both cables

her at C an
es of P for
m

0=


ain setting
CA
T
)
ce 146.4
CB
T=
lving simultan
200 0
10P
−=
=

s to remain ta

nd are load
which both
0
A
=yields,
410N and P=
neously gives,
aut, load P mu
179.3N
ded as show
cables rema
179.315N=

ust be within
N < < 669 NP
wn.
ain
the

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

81.  
 
SOLUT

Resolvin

Substitu
In the y-

Thus,

In the
x
-
Thus,

 
TION
ng the forces i
uting compone
-direction (one
-direction:
into x- and y-d
=R
ents: =R
e unknown for
−
(65
 

P
Tw
co
an
m
directions:
A
=++ +PQF
(500 lb)
[(650 lb)si
(c
BA
FF
=−+
−
+− j
i
rce):
500 lb (650− −
500 lb
A
F
+
=
1302.70=
50lb)cos50°+
cos50
(1302.70
BA
FF=
=
419.55 l
b=
PROBLEM 2
wo forces P a
onnection. Kn
nd that P=
magnitudes of t
0
B
+=F
[(650 lb)cos5
in50 ]
cos50 ) (
A
F
°
°+j
i
lb)sin50
F°+
(650 lb)sin50
sin50
+
°
lb
cos5
BA
FF+−
0(650 lb)c
0lb)cos50
°−
°−
b
2.51
and Q are ap
nowing that th
500=lb and
the forces exer
50]
sin 50 ) 0
A
°
°=i
j
sin50 0
A
F °=
0°

500°=
cos50
(650 lb)cos50
°
pplied as show
e connection i
650Q=lb,
rted on the rod
Free

0°

wn to an aircr
is in equilibriu
determine
ds A and B.
e-Body Diagr
1303 lb
A
F=
420 lb
B
F=
raft
um
the
ram
W

W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

82.  
 
SOLUT

Resolvin

Substitu
In the x-


In the
y
-

 
TION
ng the forces i
uting compone
-direction (one
-direction:
into x- and y-d
=RP
ents: =−
−
+
R
e unknown for
cos5Q
P−−
P=
=
=
 

P
T
c
e
o
d
directions:
A
++ +PQF F
cos 50
[(750 lb)cos
[(750 lb)sin
PQ−+
−
+
j
rce):
50[(750 lb)°−
(750 lb
127.710
Q=
=
sin 50Q
− °+
sin 50
(127.710 lb
476.70 lb
Q=−° +
=−
=
PROBLEM
Two forces P
connection.
equilibrium an
on rods A a
determine the
0
B
=F
0 sin 50
50 ]
50 ] (400 lb
Q°− °
°
°+
i
i
j
cos 50 ] 40
0°+
b)cos 50 40
cos 50
0 lb
°−
°
(750 lb)sin 5
(750 lb)sin
5
b)sin 50 (75
+
°+
M 2.52
and Q are ap
Knowing th
nd that the ma
and B are F
magnitudes o
b)
°j
i
0 lb 0=
00 lb

00°=
50
50 lb)sin 50
°
°
pplied as show
hat the con
agnitudes of th
750
A
F=lb an
of P and Q.
Free-Bo
477 lb;P=
wn to an aircr
nnection is
he forces exer
nd 400
B
F=
ody Diagram
127.7 lbQ=
raft
in
rted
lb,
W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

83.  
SOLUT

With


With
F
A

 
 
TION
A and FB as abo
F
FΣ
F
F
F
FΣ
ove: F
 

PR
A w
the
B
F=
forc
Free-Body Di
3
0:
5
x B
F F=
8 kN
16 kN
A
B
F
F
=
=
4
(16 kN)
5
C
F=
0:
yD
F F=−
3
(16 kN)
5
D
F=
ROBLEM 2.
welded connec
four forces
16=kN, dete
ces.
iagram of Co
3
5
CA
FF−−
=
4
)(8 kN)
5
−
33
55
B
A
FF+−
3
)(8 kN)
5
−
53
ction is in equ
shown. Kn
ermine the m
nnection
 
0=
0
A
=
uilibrium und
nowing that
magnitudes of
F
F
der the action
8
A
F= kN a
f the other t
6.40 kN
C
F=
4.80 kN
D
F=
of
and
two
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

84.  
 
SOLUT

or
With



 
TION
F
y
FΣ
B
F
A
F
B
F
x
FΣ
C
F
 
PRO
A wel
four fo
determ
Free-Body Di
0:
D
F=−−
3
5
DA
FF=+
5kN,
D
F=
5 3
6kN
3 5
⎡
=+
⎢
⎣
0:
C
F=−+
4
(
5
4
(15 kN
5
BA
FF=−
=−
BLEM 2.54
lded connectio
forces shown.
mine the magn
iagram of Co
33
55
AB
FF+
=
8kN=
3
(5 kN)
5
⎤
⎥
⎦
44
55
BA
FF−
=
)
5kN)
4
on is in equili
Knowing tha
nitudes of the o
nnection

0=
0=
ibrium under t
at 5
A
F=kN
other two forc
F
F
the action of
and
6
D
F
=k
es.
15.00 kN
B
F=
8.00 kN
C
F=
the
kN,
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

85.  
 
SOLUT




(
a) S
u

(
b) F
r
 
TION
ubstitute (1) in
rom (1):
0:
x
FΣ=
0:
y
FΣ=
nto (2): 0.6
A
T
T
 
P
A
th
on
co
an
bo
de
(b
Free-B
: cos 10
ACB
T
0.137
CD
T=
: sin 10
ACB
T °
0.67365
A
T
67365 0
ACB
T+
269.46 lb
CB
=
0.137158
CD
T=
PROBLEM 2
A sailor is be
hat is suspen
n the supp
onstant spee
nd β = 10°
oatswain’s
etermine the
b) in the trac
Body Diagram
0 cos3
ACB
T°−
7158
ACB
T
sin 30
ACB
T°+
0.5
CB CD
T+ =
0.5(0.137158T
b
8(269.46 lb)
2.55
eing rescued
nded from a
ort cable A
d by cable C
and that the
chair and
e tension (a)
ction cable C
m
30cos 3
CD
T°−
0 sin 30
CD
T°+
200
) 200
ACB
T=
using a boa
pulley that
ACB and is
CD. Knowin
e combined
the sailor
in the suppo
CD.

300°=

0200 0°− =
T
T
atswain’s ch
can roll free
s pulled at
ng that α = 3
weight of t
r is 200
ort cable AC
269 lb
ACB
T=
37.0 lb
CD
T=
hair
ely
a
30°
the
lb,
CB,
(1)
(2)

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

86.  
SOLUT






 
TION

 
P
A
su
ca
K
ca
th
su
Free-B
0:
xA
F TΣ=
A
T
0: (30
y
FΣ=
+
PROBLEM
A sailor is bein
uspended from
able ACB and
Knowing that α
able CD is 20
he boatswain’s
upport cable A
Body Diagram
cos 15
ACB
T°−
304.04 l
ACB
=
04.04 lb)sin 1
(20 lb)sin 25+
2.56
ng rescued usi
m a pulley that
d is pulled at a
α = 25° and β
0 lb, determin
s chair and th
ACB.
m
cos 25
ACB
T °−
lb
5 (304.04 l°+
5 0
215.6
W
W°− =
=
ing a boatswa
t can roll freel
a constant spe
β = 15° and th
ne (a) the com
he sailor, (b) th

(20 lb)cos 25
lb)sin 25°

64 lb

(
a)
(
b)
T
ain’s chair that
ly on the supp
eed by cable C
hat the tension
mbined weight
he tension in
50°=

216 lbW=
304 lb
ACB
T=
t is
port
CD.
n in
t of
the
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

87. PROBLEM 2.57
For the cables of prob. 2.44, find the value of α for which the tension
is as small as possible (a) in cable bc, (b) in both cables
simultaneously. In each case determine the tension in each cable.
 
SOLUTION
Free-Body Diagram Force Triangle


(
a) For a minimum tension in cable BC, set angle between cables to 90 degrees.
By inspection,
35.0α=
α

W

(6 kN)cos35
AC
T=
α
4.91 kN
AC
T= W

(6 kN)sin35
BC
T=
α
3.44 kN
BC
T= W

(
b) For equal tension in both cables, the force triangle will be an isosceles.
Therefore, by inspection,
55.0α=
α

W


6 kN
(1 / 2)
cos35
AC BC
TT==
° 3.66 kN
AC BC
TT== W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

88.  
 
SOLUT


(a)
L

(
b)
L


TION
F
Law of cosines
Law of sines
Free-Body D

2
P
P
sin
600 N
β
β


PROB
For the
allowabl
Determi (
b) the c
iagram
2 2
(600) (7=+
784.02 NP=
sin (254
N 784.02 N
°+
=
46.0
β=°
LEM 2.58
cables of Pro
le tension is 6
ne (a) the ma
corresponding
2
750) 2(600−
45)
N
°


46.0α
∴ = °
oblem 2.46, it
600 N in cabl
aximum force
value of α.
For
)(750)cos(25°
25°+°
t is known tha
e AC and 750
e P that can b
rce Triangle
45 )°+°
at the maximu
0 N in cable B
be applied at

784 NP=
71.0
α=°
um
BC.
C,
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

89.  
 
SOLUT


To be s
m
(a) T
(b)
 
TION
F
mallest,
BC
T m
Thus,
Free-Body D
must be perpen
α=
BC
T=
 

P
F
th
as
iagram
ndicular to the
5.00= °
(1200 lb)sin=
PROBLEM
or the situatio
he value of α f
s possible, (b)
e direction of
n5°
2.59
on described i
for which the
) the correspon
For

.
AC
T
n Figure P2.4
tension in rop
nding value of
rce Triangle
α
T
48, determine
pe BC is as sm
f the tension.
5.00α=°
104.6 lb
BC
T=
(a)
mall
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

90.  
SOLUT
Free-Bo

 
 
TION
ody Diagram


 
PROBLE
Two cables
range of va
either cable





Requireme
From Eq. (

Requireme
From Eq. (

EM 2.60
s tied together
alues of Q for
e.
0:
x
FΣ=
0:
y
FΣ=
ent:
(2): sQ
ent:
(1): 75 lb−
r at C are loa
which the ten
co
BC
TQ−−
75 lb
BC
T=
sin
AC
TQ−
sin
AC
TQ=
60 lb
AC
T=
sin 60 60 lb°=
69.3 Q=
60 lb
BC
T=
cos60 6Q− °=
30Q=
aded as shown
nsion will not
os60 75 lb°+ =
bcos60Q−°
60 0°=
n60°
b:

lb b:
0lb
0.0 lb 30.0 lb
n. Determine
t exceed 60 lb
0=
69.3 lbQ≤≤
the
b in
(1)
(2)

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

91. PROBLEM 2.61
A movable bin and its contents have a combined weight of 2.8 kN. Determine
the shortest chain sling
ACB that can be used to lift the loaded bin if the
tension in the chain is not to exceed 5 kN.
 
SOLUTION
Free-Body Diagram


tan
0.6 mα=
h
(1)






Isosceles Force Triangle

Law of sines:
1
2
1
2
(2.8 kN)
sin
5kN
(2.8 kN)
sin
5kN
16.2602
AC
AC
T
T
α
α
α=
=
=
=°

From Eq. (1):
tan16.2602 0.175000 m
0.6 m
h
h°= ∴ =
Half-length of chain
22
(0.6 m) (0.175 m)
0.625 m
AC== +
=
Total length: 2 0.625 m=×
1.250 m W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

92. PROBLEM 2.62
For W = 800 N, P = 200 N, and d = 600 mm,
determine the value of h consistent with
equilibrium.

 
SOLUTION
Free-Body Diagram

800 N
AC BC
TT==

( )
22
ACBC h d== +

22
0: 2(800 N) 0
y
h
FP
hd
Σ=− =
+
2
800 1
2
P d
h
⎛⎞
=+
⎜⎟
⎝⎠

Data:
200 N, 600 mmPd
= = and solving for h

2
200 N 600 mm
800 N 1
2 h
⎛⎞
=+
⎜⎟ ⎝⎠



75.6 mmh= W

 
 
   www.elsolucionario.org

93.  
 
SOLUT
(a) F




(
b)
F



 
 
TION
Free Body: Co
Free Body: Co
ollar A
ollar A
 


PROBLE
Collar A is c
slide on a
magnitude
equilibrium
15 in.x=
Forc
4.5
P
Forc
15
P
EM 2.63
connected as s
frictionless h
of the force
of the colla
ce Triangle
50 lb
520.5
=
ce Triangle
50 lb
25
=
shown to a 50
horizontal rod
P required
ar when (a)
0-lb load and c
d. Determine
to maintain
4.5 in.,x=
10.98 lbP=
30.0 lbP=
can
the
the
(
b)
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

94.  
SOLUT




Similar




TION
Free Bo
Triangles
ody: Collar A


A
2
(50
14
N
N
=
=
48
20 in. 14
x
=
PROBLEM
Collar A is c
slide on a f
distance x fo
P = 48 lb.
F
22
0)(48) 1
.00 lb
−=
8 lb
4 lb

M 2.64
connected as s
frictionless ho
or which the
orce Triangle
196

shown to a 50
orizontal rod.
collar is in eq
e
0-lb load and c
. Determine
quilibrium wh
68.6 in.x=
can
the
hen
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

95. PROBLEM 2.65
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50°. Determine
the range of values of
α for which the magnitude of the resultant of the forces
acting at
A is less than 600 N.
 
SOLUTION
Combine the two 150-N forces into a resultant force Q:

  
2(150 N)cos25
271.89 NQ=°
=
 
Equivalent loading at A:
 
Using the law of cosines:
22 2
(600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 )
cos(55 ) 0.132685
α
α
=+ + °+
°+ =
 
Two values for :
α 55 82.375
27.4
α
α
°+ =
=°
or
55 82.375
55 360 82.375
222.6
α
α
α
°+ =− ° °+ = °− °
=°

For 600 lb:
R< 27.4 222.6α°< <
α
W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

96.  
SOLUT
Free-Bo


TION
ody Diagram:
PROBL
A 200-kg
Determine
the free en
the same o
Ch. 4.)
: Pulley A

LEM 2.66
g crate is to be
e the magnitu
nd of the rope
on each side o

 
 
 
         
cos
F
α
α
Σ
For α=+
y
FΣ=
For α=−
y
FΣ=
e supported b
ude and direct
to maintain e
of a simple pu
0: 2
0.59655
53.377
x
F P
α
α
⎛
=− ⎜
⎝
=
=±°
53.377 :+ ° 
16
0: 2
2
P
⎛
= ⎜
⎝
53.377 :
− ° 
16
0: 2
2
P
⎛
= ⎜ ⎝
by the rope-an
tion of the for
quilibrium. (H
ulley. This can
5
c
281
P
⎛ ⎞
+⎟
⎝ ⎠
6
sin 53
81
P
⎞
+⎟ ⎠
6
sin(5
81
P
⎞
+−⎟ ⎠
nd-pulley arra
rce P that mu
Hint: The tensi
n be proved by
cos 0α=
 
3.377 1962N°−
724=P
53.377 ) 196°−
17=P
ngement show
ust be exerted
ion in the rope
y the methods
N0= 
4N  53.4°
62N 0=  
773 53.4°
wn.
on
e is
s of
 W 
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

97.  
SOLUT
Free-Bo

(a)


(b)


(c)

(d)

(e)



TION
ody Diagram of Pulley
FΣ
FΣ
FΣ
FΣ
FΣ

0: 2
y
F T= −
0: 2
y
F T= −
0: 3
y
F T= −
0: 3
y
F T= −
0: 4
y
F T= −

PR
A 6
and
for
(Se
(600 lb) 0
1
2
T
− =
=
(600 lb) 0
1
2
T
− =
=
(600 lb) 0
1
(
3
T
− =
=
(600 lb) 0
1
(
3
T
− =
=
(600 lb) 0
1
4
T
− =
=
ROBLEM 2
600-lb crate i
d-pulley arrang
each arrange
ee the hint for
(600 lb)

(600 lb)

(600 lb)

(600 lb)

(600 lb)

.67
is supported b
gements as sh
ement the tens
Problem 2.66
by several rop
hown. Determ
sion in the ro
.)
300 lbT=
300 lbT=
200 lbT=
200 lbT=
150.0 lbT=
pe-
ine
pe.
W

W

W

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

98.  
SOLUT
Free-Bo

(b)


(d)



TION
ody Diagram of Pulley and

d Crate
0:
y
FΣ=
0:
y
F
Σ=

PR
Solv
that
crat
PR
by
sho
tens
2.66
3 (600 lbT
T
−
4(600 lbT−
ROBLEM 2.
ve Parts b and
t the free end
te.
OBLEM 2.67
several rope
wn. Determin
sion in the rop
6.)
b)0
1
(600 lb)
3
T
=
=
b)0
1
(600 lb)
4
T
=
=
.68
d d of Problem
of the rope is
7 A 600-lb cr
e-and-pulley a
ne for each a
pe. (See the h

)

m 2.67, assumi
s attached to
rate is suppor
arrangements
arrangement
hint for Probl
200 lbT=
150.0 lbT=
ing
the
rted
as
the em
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

99.  
SOLUT
Free-Bo


TION
ody Diagram:: Pulley C



PRO
A lo
cable
secon
suppo
(a) th



(a)
Σ
Hen c


(b) FΣ
or
OBLEM 2.6
ad Q is appli
e ACB. The p
nd cable CAD
orts a load
he tension in c
0:
x ACB
F TΣ=
ce:
0: (
(1292.88 N)(
yA CB
F T=
69
ied to the pul
pulley is held
D, which pa
P. Knowing
cable ACB, (b)
(cos25 cos
B
°−
129
ACB
T=
(sin 25 sin 5
sin 25 sin5
°+
°+
Q
=
lley C, which
d in the posit
asses over the
that 75P=
) the magnitud
s55 ) (750 N°−
92.88 N
A
T
5 ) (750 N)
5) (750 N)
°+
°+
2219.8 N
=
h can roll on
ion shown by
e pulley A a
50 N, determ
de of load Q.
N)cos55° 0=
1293 N
ACB
T=
sin 55 0
sin 55 0
Q
Q
°− =
°− =
2220 NQ=
the
y a
and
mine
W
0
0


W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

100.  
SOLUT
Free-Bo



 
 
TION
ody Diagram:: Pulley C

 

PRO
An 1
on th
a sec
suppo
(b) th


0
x
FΣ=
or

y
FΣ=
or
(a) Substit
1.24
Hence:

(b) Using (

OBLEM 2.7
800-N load Q
he cable ACB.
cond cable CA
orts a load P
he magnitude o
0:(cos 2
ACB
T
0: (sin2
ACB
T
tute Equation (
4177 0.
ACB
T+
(1), 0.P=
70
Q is applied to
The pulley is
CAD, which p
. Determine (
of load P.
25cos55)°− ° −
25sin55)°+ ° +
1.24177T
(1) into Equat
81915(0.5801
58010(1048.3
o the pulley C
held in the po
passes over th
(a) the tension
cos550P− °=
0P=
sin55 1P+ °−
0.81915
ACB
T+
tion (2):
10 ) 1800
ACB
T=
1048
ACB
T=

A
T
37 N) 608.16=
C, which can r
osition shown
he pulley A a
n in cable AC
0
0.58010
ACB
T
800 N 0=
51800 NP=
0 N
8.37 N
1048 N
ACB
T=
6 N
608 NP=
roll
by
and
CB,
(1)
(2)

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

101. PROBLEM 2.71
Determine (a) the x, y, and z components of the 600-N force,
(
b) the angles θx, θy, and θz that the force forms with the
coordinate axes.
 
SOLUTION


(
a) (600 N)sin 25 cos30
x
F=°
α


219.60 N
x
F= 220 N
x
F= W

(600 N)cos 25°
543.78 N
y
y
F
F
=
= 544 N
y
F= W

(380.36 N)sin 25 sin 30
126.785 N
z
z
F
F
=°
=
α
126.8 N
z
F= W
(b)
219.60 N
cos
600 N
x
x
F
F
θ== 68.5
x
θ=° W

543.78 N
cos
600 Ny
y
F
F
θ== 25.0
y
θ=° W

126.785 N
cos
600 N
z
z
F
F
θ== 77.8
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

102. PROBLEM 2.72
Determine (a) the x, y, and z components of the 450-N force,
(b) the angles
θx, θy, and θz that the force forms with the
coordinate axes.
 
SOLUTION


(a)
(450 N)cos 35 sin 40
x
F=− °
α


236.94 N
x
F=− 237 N
x
F=− W

(450 N)sin 35°
258.11 N
y
y
F
F
=
= 258 N
y
F= W

(450 N)cos35 cos40
282.38 N
z
z
F
F
=°
=
α
282 N
z
F= W
(b)
-236.94 N
cos
450 N
x
x
F
F
θ== 121.8
x
θ=° W

258.11 N
cos
450 Ny
y
F
F
θ== 55.0
y
θ=° W

282.38 N
cos
450 N
z
z
F
F
θ== 51.1
z
θ=° W
Note: From the given data, we could have computed directly
90 35 55 , which checks with the answer obtained.
y
θ=−=
αα α



 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

103. PROBLEM 2.73
A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle
of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z
components of that force, (b) the values of the angles θ
x, θy, and θ z defining the direction of the recoil
force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)  
SOLUTION
Recoil force 400 NF=

(400 N)cos40
306.42 N
H
F∴= °
=








(a)
sin35
(306.42 N)sin35
xH
FF=− °
=− °

175.755 N=−
175.8 N
x
F=− W

sin 40
(400 N)sin 40
257.12 N
y
FF=− °
=− °
=−
257 N
y
F=− W

cos35
(306.42 N)cos35
251.00 N
zH
FF=+ °
=+ °
=+
251 N
z
F=+ W
(b)
175.755 N
cos
400 N
x
x
F
F
θ
−
== 116.1
x
θ=° W

257.12 N
cos
400 Ny
y
F
F
θ
−
==
130.0
y
θ=° W

251.00 N
cos
400 N
z
z
F
F
θ== 51.1
z
θ=°W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

104. PROBLEM 2.74
Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun
forms an angle of 25° with the horizontal.
PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the
gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine
(a) the x, y, and z components of that force, (b) the values of the angles θ
x, θy, and θ z defining the direction
of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)
 
SOLUTION
Recoil force 400 NF=

(400 N)cos25
362.52 N
H
F∴= °
=








(a)
cos15
(362.52 N)cos15
xH
FF=+ °
=+ °

350.17 N=+
350 N
x
F=+ W

sin 25
(400 N)sin 25
169.047 N
y
FF=− °
=− °
=−
169.0 N
y
F=− W

sin15
(362.52 N)sin15
93.827 N
zH
FF=+ °
=+ °
=+
93.8 N
z
F=+ W
(b)
350.17 N
cos
400 N
x
x
F
F
θ
+
== 28.9
x
θ=° W

169.047 N
cos
400 Ny
y
F
F
θ
−
==
115.0
y
θ=° W

93.827 N
cos
400 N
z
z
F
F
θ
+
== 76.4
z
θ=°W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

105. PROBLEM 2.75
The angle between spring AB and the post DA is 30°. Knowing
that the tension in the spring is 50 lb, determine (a) the x, y,
and z components of the force exerted on the circular plate at
B, (b) the angles
θx, θy, and θz defining the direction of the
force at B.
 
SOLUTION




cos 60
(50 lb)cos 60
25.0 lb
h
h
FF
F
=°
=°
=


cos 35 sin60 sin35
( 25.0 lb)cos35 (50.0 lb)sin60 ( 25.0 lb)sin35
20.479 lb 43.301 lb 14.3394 l
xh y zh
xyz
xyz
FF FF FF
FFF
FFF
=− ° = =−
=− = =−
=− = =−
αα
αα α
b

(a)

 20.5 lb
x
F=− W

43.3 lb
y
F= W

14.33 lb
z
F=− W
(b)
20.479 lb
cos
50 lb
x
x
F
F
θ
−
== 114.2
x
θ=° W

43.301 lb
cos
50 lby
y
F
F
θ== 30.0
y
θ=° W

-14.3394 lb
cos
50 lb
z
z
F
F
θ== 106.7
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

106. PROBLEM 2.76
The angle between spring AC and the post DA is
30°. Knowing that the tension in the spring is 40 lb,
determine (a) the x, y, and z components of the force
exerted on the circular plate at C, (b) the angles
θx,
θy, and θz defining the direction of the force at C.
 
SOLUTION




cos 60
(40 lb)cos 60
20.0 lb
h
h
FF
F
=°
=°
=

(a)
cos 35
xh
FF=° sin 60
y
FF
= ° sin 35
zh
FF=−°
(20.0 lb)cos 35°= (40 lb)sin 60°= (20.0 lb)sin35=−°
16.3830 lb
x
F= 34.641 lb
y
F= 11.4715 lb
z
F=−
 16.38 lb
x
F= W

34.6 lb
y
F= W

11.47 lb
z
F=− W
(b)
16.3830 lb
cos
40 lb
x
x
F
F
θ== 65.8
x
θ=° W

34.641 lb
cos
40 lby
y
F
F
θ== 30.0
y
θ=° W

-11.4715 lb
cos
40 lb
z
z
F
F
θ== 106.7
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

107.  
SOLUT

 
 
TION


P
C
D
c
d

From trian
(a)



(b)
Fr o

 
PROBLEM
Cable AB is
Determine (a)
cable on the a
direction of tha
ngle AOB:
F
F
F
cosθ
om above:
2.77
65 ft long, a
the x, y, and
anchor B, (b)
at force.
cos
y
y
θ
θ
sin
(3900 lb
x y
F Fθ=−
=−
cos
y y
F Fθ=+
(3900 lb
z
F=+
18
39
x
x
F
F
θ==−
30.5
y
θ=
cos
z
z
F
F
θ
=
and the tensio
d z component
the angles θ
56 ft
65 ft
0.86154
30.51
y
y
=
=
=°
cos 20
b)sin30.51 co
°
°
(3900 lb)(
0
y
=
b)sin 30.51° si
861 lb
0.47
900 lb
=−
51°
677 lb
3900 lb
=+ =
on in that ca
ts of the force
,
x
θ ,
y
θ and
os20°

F
0.86154) F
in20°
771
0.1736=+
able is 3900
e exerted by
z
θ defining
1861 lb
x
F=−
3360 lb
y
F=+
677 lb
z
F=+
118.5
x
θ=°
30.5
y
θ=°
80.0
z
θ=°
lb.
the
the
W

W

W

W

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

108.  

PROBLEM 2.78
Cable AC is 70 ft long, and the tension in that cable is 5250 lb.
Determine (a) the x, y, and z components of the force exerted by the
cable on the anchor C, (b) the angles
θx, θy, and θz defining the
direction of that force.
 
SOLUTION


In triangle AOB: 70 ft
56 ft
5250 lb
AC
OA
F
=
= =


56 ft
cos
70 ft
36.870
sin
(5250 lb)sin 36.870
3150.0 lb
y
y
Hy
FF
θ
θ
θ=
=°
=
= °
=

(a)
sin50 (3150.0 lb)sin50 2413.0 lb
xH
FF=− °=− °=−



2410 lb
x
F=− W

cos (5250 lb)cos36.870 4200.0 lb
yy
FF θ=+ =+ °=+



4200 lb
y
F=+ W

cos50 3150cos50 2024.8 lb
zH
FF=− °=− °=− 2025 lb
z
F=− W
(b)
2413.0 lb
cos
5250 lb
x
x
F
F
θ
−
== 117.4
x
θ=° W
From above:

36.870
y
θ
= ° 36.9
y
θ=° W

2024.8 lb
5250 lb
z
z
F
F
θ
−
== 112.7
z
θ=° W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

109.  
PROBLEM 2.79
Determine the magnitude and direction of the force F = (240 N)i – (270 N)j + (680 N)k.
 
SOLUTION

222
222
(240 N) ( 270 N) ( 680 N)
xyz
FFFF
F=++
=+−+−
770 NF= W

240 N
cos
770 N
x
x
F
F
θ== 71.8
x
θ=° W

270 N
cos
770 Ny
y
F
F
θ
−
==
110.5
y
θ=° W

680 N
cos
770 N
z
y
F
F
θ== 28.0
z
θ=° W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

110.  
PROBLEM 2.80
Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.
 
SOLUTION

222
22 2
(320 N) (400 N) ( 250 N)
xyz
FFFF
F=++
=++−
570 NF= W

320 N
cos
570 N
x
x
F
F
θ== 55.8
x
θ=° W

400 N
cos
570 Ny
y
F
F
θ== 45.4
y
θ=° W

250 N
cos
570 N
z
y
F
F
θ
−
== 116.0
z
θ=° W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

111.  
PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3° and θz
= 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle
θy, (b) the other
components and the magnitude of the force.
 
SOLUTION
222
222
cos cos cos 1
cos (69.3 ) cos cos (57.9 ) 1
cos 0.7699
xyz
y
y
θθθ
θ
θ++=
°+ + °=
=±

(a) Since
0,
y
F< we choose cos 0.7699
y
θ=− 140.3
y
θ∴= ° W
(b)
cos
174.0 lb ( 0.7699)
yy
FF
F θ=
−=−
226.0 lbF=

226 lbF=
W

cos (226.0 lb)cos69.3
xx
FF θ== ° 79.9 lb
x
F= W

cos (226.0 lb)cos57.9
zz
FF θ== ° 120.1lb
z
F= W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

112.  
PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and
θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the
other components and the magnitude of the force.
 
SOLUTION
222
22 2
cos cos cos 1
cos 70.9 cos 144.9 cos 1
cos 0.47282
xyz
z
z
θθθ
θ
θ++=
+°+°=
=±
α

(a) Since
0,
z
F< we choose cos 0.47282
z
θ=− 118.2
z
θ∴= ° W
(b)
cos
52.0 ( 0.47282)
zz
FF
lb Fθ=
−=−

110.0 lbF=

110.0 lbF= W

cos (110.0 lb)cos70.9
xx
FF θ== ° 36.0 lb
x
F= W

cos (110.0 lb)cos144.9
yy
FF θ== ° 90.0 lb
y
F=− W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

113.  
PROBLEM 2.83
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that F x = 80 N,
θz = 151.2°, and F y < 0, determine (a) the components F y and F z, (b) the angles θx and θy.
 
SOLUTION
(a) cos (210 N)cos151.2
zz
FF θ== °
184.024 N
=− 184.0 N
z
F=− W
Then:
2222
x yz
FFFF=++
So:
222 2
(210 N) (80 N) ( ) (184.024 N)
y
F=++
Hence:
22 2
(210 N) (80 N) (184.024 N)
y
F=− − −
61.929 N=−
62.0 lb
y
F=− W
(b)
80 N
cos 0.38095
210 N
x
x
F
F
θ== = 67.6
x
θ=° W

61.929 N
cos 0.29490
210 Ny
y
F
F
θ== =−


107.2
y
θ=° W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

114.  
PROBLEM 2.84
A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that
θx = 65°, θy = 40°, and F z > 0, determine (a) the components of the force, (b) the angle θz.  
SOLUTION
222
222
cos cos cos 1
cos 65 cos 40 cos 1
cos 0.48432
xyz
z
z
θθθ
θ
θ++=
+°+°=
=±
α

(b) Since
0,
z
F> we choose cos 0.48432, or 61.032
zz
θθ==
α
61.0
z
θ∴=° W
(a)
1200 NF=

cos (1200 N) cos65
xx
FF θ==
α

507 N
x
F= W

cos (1200 N)cos40
yy
FF θ== ° 919 N
y
F= W

cos (1200 N)cos61.032
zz
FF θ== ° 582 N
z
F= W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

115.  

PROBLEM 2.85
A frame ABC is supported in part by cable DBE that
passes through a frictionless ring at B. Knowing that the
tension in the cable is 385 N, determine the components
of the force exerted by the cable on the support at D.
 
SOLUTION

22 2
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm ) (320 mm)
770 mm
385 N
[(480 mm) (510 mm) (320 mm) ]
770 mm
(240 N) (255 N) (160 N)
DB
DB
DB
F
DB
F
DB
=−+
=++
=
=
=
=−+
=−+ ijk
Fλ
ijk
ijk
JJJG
JJJG


240 N, 255 N, 160.0 N
xyz
FFF=+ =− =+ W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

116.  

PROBLEM 2.86
For the frame and cable of Problem 2.85, determine the
components of the force exerted by the cable on the
support at E.
PROBLEM 2.85 A frame ABC is supported in part by
cable DBE that passes through a frictionless ring at B.
Knowing that the tension in the cable is 385 N, determine
the components of the force exerted by the cable on the
support at D.
 
SOLUTION

222
(270 mm) (400 mm) (600 mm)
(270 mm)(400 mm)(600 mm)
770 mm
385 N
[(270 mm) (400 mm) (600 mm) ]
770 mm
(135 N) (200 N) (300 N)
EB
EB
EB
F
EB
F
EB
=−+
=++
=
=
=
=−+
=− + ijk
Fλ
ijk
Fi jk
γγγG
γγγG


135.0 N, 200 N, 300 N
xyz
FFF=+ =− =+ W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

117.  

PROBLEM 2.87
In order to move a wrecked truck, two cables are attached
at A and pulled by winches B and C as shown. Knowing
that the tension in cable AB is 2 kips, determine the
components of the force exerted at A by the cable.
 
SOLUTION


Cable AB :
( 46.765 ft) (45 ft) (36 ft)
74.216 ft
46.765 45 36
74.216
AB
AB AB AB
AB
AB
T
−++
==
−++
==
ijk
λ
ijk
T λ
γγγG


( ) 1.260 kips
AB x
T=− W

( ) 1.213 kips
AB y
T=+ W
( ) 0.970 kips
AB z
T=+ W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

118.  

PROBLEM 2.88
In order to move a wrecked truck, two cables are attached
at A and pulled by winches B and C as shown. Knowing
that the tension in cable AC is 1.5 kips, determine the
components of the force exerted at A by the cable.
 
SOLUTION


Cable AB :
( 46.765 ft) (55.8 ft) ( 45 ft)
85.590 ft
46.765 55.8 45
(1.5 kips)
85.590
AC
AC AC AC
AC
AC
T
−++−
==
−+−
==
ijk
λ
ijk
T λ
γJJG


() 0.820kips
AC x
T =− W

( ) 0.978 kips
AC y
T =+ W
( ) 0.789 kips=−
AC z
T W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

119.  

PROBLEM 2.89
A rectangular plate is supported by three cables as
shown. Knowing that the tension in cable AB is 408 N,
determine the components of the force exerted on the
plate at B.

 
SOLUTION



We have:

(320 mm) (480 mm) (360 mm) 680 mmBA BA=+ = i+ j- k
γγγG
Thus:

8129
F
17 17 17
BBABABA BA
BA
TTT
BA
⎛⎞
===
⎜⎟
⎝⎠
λ i+ j- k
γγγG



8129
0
17 17 17
BA BA BA
TTT
⎛⎞⎛⎞⎛⎞
+ −=
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠
ijk
Setting
408 N
BA
T= yields,

192.0 N, 288 N, 216 N
xyz
FFF=+ =+ =− W

 Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

120. PROBLEM 2.90
A rectangular plate is supported by three cables as
shown. Knowing that the tension in cable AD is 429 N,
determine the components of the force exerted on the
plate at D.

 
SOLUTION



We have:

(250 mm) (480 mm) (360 mm) 650 mmDA DA=− = i+ j+ k
γγγG
Thus:

54836
F
13 65 65
DDADADA DA
DA
TTT
DA
⎛⎞
===−
⎜⎟
⎝⎠
λ i+ j+ k
γγγG



54836
0
13 65 65
DA DA DA
TTT
⎛⎞⎛⎞⎛⎞
−+ + =
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠
ijk
Setting
429 N
DA
T= yields,

165.0 N, 317 N, 238 N
xyz
FFF=− =+ =+ W

 
  
 Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

121. PROBLEM 2.91
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 300 N and Q = 400 N.
 
SOLUTION

(300 N)[ cos30 sin15 sin30 cos30 cos15 ]
(67.243 N) (150 N) (250.95 N)
(400 N)[cos50 cos20 sin50 cos50 sin 20 ]
(400 N)[0.60402 0.76604 0.21985]
(241.61 N) (306.42 N) (87.939 N)
(174.
=−°°+°+°°
=− + +
=°°+°−°°
=+−
=+−
=+
=
Pijk
ij k
Qijk
ij
ijk
RPQ
22 2
367 N) (456.42 N) (163.011 N)
(174.367 N) (456.42 N) (163.011 N)
515.07 N
R
++
=++
=
ij k
515 NR=
W

174.367 N
cos 0.33853
515.07 N
x
x
R
R
θ== = 70.2
x
θ=° W

456.42 N
cos 0.88613
515.07 Ny
y
R
R
θ== = 27.6
y
θ=° W

163.011 N
cos 0.31648
515.07 N
z
z
R
R
θ== = 71.5
z
θ=° W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

122. PROBLEM 2.92
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 400 N and Q = 300 N.
 
SOLUTION

(400 N)[ cos30 sin15 sin 30 cos30 cos15 ]
(89.678 N) (200 N) (334.61 N)
(300 N)[cos50 cos20 sin 50 cos50 sin 20 ]
(181.21 N) (229.81 N) (65.954 N)
(91.532 N) (429.81 N) (268.66 N)
(91.5R
=−°°+°+°°
=− + +
=°°+°−°°
=+−
=+
=++
=
Pijk
ij k
Qijk
ijk
RPQ
ijk
22 2
32 N) (429.81 N) (268.66 N)
515.07 N
++
=
515 NR= W

91.532 N
cos 0.177708
515.07 N
x
x
R
R
θ== = 79.8
x
θ=° W

429.81 N
cos 0.83447
515.07 Ny
y
R
R
θ== = 33.4
y
θ=° W

268.66 N
cos 0.52160
515.07 N
z
z
R
R
θ== = 58.6
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

123. PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.

SOLUTION

222
222
(40in.) (45in.) (60in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40in.) (45in.) (60in.)
(425 lb)
85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
=++=
=−+
=++=
−+
===
ijk
ijk
ijk
T λ
γγγG
JJJG
γγγG
(200 lb) (225 lb) (300 lb)
(100 in.) (45 in.) (60 in.)
(510 lb)
125 in.
(408 lb) (183.6 lb) (244.8 lb)
(608) (408.6 lb) (544.8 lb)
AB
AC AC AC AC
AC
AB AC
AC
TT
AC
⎡ ⎤
⎢ ⎥
⎣ ⎦
=−+
⎡ ⎤−+
== =
⎢ ⎥
⎣ ⎦
=− +
=+= − +
Tijk
ijk
T λ
Tijk
RT T i j k
JJJG

Then: 912.92 lbR= 913 lbR=
W
and
608 lb
cos 0.66599
912.92 lb
x
θ== 48.2
x
θ=° W

408.6 lb
cos 0.44757
912.92 lb
y
θ==− 116.6
y
θ=° W

544.8 lb
cos 0.59677
912.92 lb
z
θ== 53.4
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

124. PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
 
SOLUTION

222
222
(40in.) (45in.) (60in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40in.) (45in.) (60in.)
(510 lb)
85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
=++=
=−+
=++=
−+
===
ijk
ijk
ijk
T λ
γγγG
JJJG
γγγG
(240 lb) (270 lb) (360 lb)
(100 in.) (45 in.) (60 in.)
(425 lb)
125 in.
(340 lb) (153 lb) (204 lb)
(580 lb) (423 lb) (564 lb)
AB
AC AC AC AC
AC
AB AC
AC
TT
AC
⎡ ⎤
⎢ ⎥
⎣ ⎦
=−+
⎡ ⎤−+
== =
⎢ ⎥
⎣ ⎦
=−+
=+= − +
Tijk
ijk
T λ
Tijk
RT T i j k
JJJG

Then: 912.92 lbR= 913 lbR=
W
and
580 lb
cos 0.63532
912.92 lb
x
θ== 50.6
x
θ=° W

423 lb
cos 0.46335
912.92 lb
y
θ
−
==−
117.6
y
θ=° W

564 lb
cos 0.61780
912.92 lb
z
θ== 51.8
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

125. PROBLEM 2.95
For the frame of Problem 2.85, determine the magnitude and
direction of the resultant of the forces exerted by the cable at B
knowing that the tension in the cable is 385 N.
PROBLEM 2.85 A frame ABC is supported in part by cable
DBE that passes through a frictionless ring at B. Knowing that
the tension in the cable is 385 N, determine the components of
the force exerted by the cable on the support at D.
 
SOLUTION

222
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm) (320 mm) 770 mm
BD
BD
=− + −
=++=
ijk
JJJG


(385 N)
[ (480 mm) (510 mm) (320 mm) ]
(770 mm)
(240 N) (255 N) (160 N)
BD BD BD BD
BD
TT
BD
==
=−+−
=− + −
F λ
ijk
ijk
JJJG


222
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm) 770 mm
BE
BE
=− + −
=++=
ijk
JJJG


(385 N)
[ (270 mm) (400 mm) (600 mm) ]
(770 mm)
(135 N) (200 N) (300 N)
BE BE BE BE
BE
TT
BE
==
=−+−
=− + −
F λ
ijk
ijk
JJJG


(375 N) (455 N) (460 N)
BD BE
=+=− + −RF F i j k

222
(375 N) (455 N) (460 N) 747.83 NR=++= 748 NR= W

375 N
cos
747.83 N
x
θ
−
=
120.1
x
θ=° W

455 N
cos
747.83 N
y
θ= 52.5
y
θ=° W

460 N
cos
747.83 N
z
θ
−
=
128.0
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

126.  

PROBLEM 2.96
For the plate of Prob. 2.89, determine the tensions in cables AB
and AD knowing that the tension in cable AC is 54 N and that
the resultant of the forces exerted by the three cables at A must
be vertical.
 
SOLUTION

We have:

()
(320 mm) (480 mm) (360 mm) 680 mm
(450 mm) (480 mm) (360 mm) 750 mm
(250 mm) (480 mm) 360 mm 650 mm
AB AB
AC AC
AD AD
=− − + =
=−+ =
=−− =
ijk
ijk
ijk
γγγG
JJJG
JJJG

Thus:

()
()
()
320 480 360
680
54
450 480 360
750
250 480 360
650
AB
AB AB AB AB
AC AC AC AC
AD
AD AD AD AD
TAB
TT
AB
AC
TT
AC
TAD
TT
AD
===−−+
== = −+
== = −−
T λ ijk
T λ ijk
T λ ijk
γγγG
JJJG
JJJG

Substituting into the Eq.
=ΣRF and factoring , , :ijk

320 250
32.40
680 650
480 480
34.560
680 650
360 360
25.920
680 650
AB AD
AB AD
AB AD
TT
TT
TT
⎛⎞
=− + +
⎜⎟
⎝⎠
⎛⎞
+− − −
⎜⎟
⎝⎠
⎛⎞
++−
⎜⎟
⎝⎠
Ri
j
k




 
 Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

127. PROBLEM 2.96 (Continued)

Since R is vertical, the coefficients of i and k are zero:
:i
320 250
32.40 0
680 650
AB AD
TT−++ = (1)
:k
360 360
25.920 0
680 650
AB AD
TT
+ −= (2)
Multiply (1) by 3.6 and (2) by 2.5 then add:

252
181.440 0
680
AB
T−+=
489.60 N
AB
T=

490 N
AB
T= W
Substitute into (2) and solve for
:
AD
T

360 360
(489.60 N) 25.920 0
680 650
AD
T
+ −=

514.80 N
AD
T=

515 N
AD
T= W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

128.  

PROBLEM 2.97
The boom OA carries a load P and is supported by
two cables as shown. Knowing that the tension in
cable AB is 183 lb and that the resultant of the load P
and of the forces exerted at A by the two cables must
be directed along OA, determine the tension in cable
AC.

SOLUTION

Cable AB:
183 lb
AB
T=


( 48 in.) (29 in.) (24 in.)
(183 lb)
61in.
(144 lb) (87 lb) (72 lb)
AB AB AB AB
AB
AB
TT
AB
−+ +
===
=− + +
ijk
T
Tijk
γγγG
λ

Cable AC:
( 48 in.) (25 in.) ( 36 in.)
65 in.
48 25 36
65 65 65
AC AC AC AC AC
AC AC AC AC
AC
TTT
AC
TTT
−+ +−
===
=− + −
ij k
T
Tijk
γJJG
λ

Load P: P=Pj
For resultant to be directed along OA, i.e., x -axis

36
0: (72 lb) 0
65
zz A C
RF T ′=Σ= − = 130.0 lb
AC
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

129. PROBLEM 2.98
For the boom and loading of Problem. 2.97, determine
the magnitude of the load P.
PROBLEM 2.97 The boom OA carries a load P and is
supported by two cables as shown. Knowing that the
tension in cable AB is 183 lb and that the resultant of
the load P and of the forces exerted at A by the two
cables must be directed along OA, determine the
tension in cable AC.

SOLUTION
See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write

25
0: (87 lb) 0
65
yy A C
RF TP=Σ= + −=

130.0 lb
AC
T= from Problem 2.97.
Then
25
(87 lb) (130.0 lb) 0
65
P+−=

137.0 lbP=
W

Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

130. PROBLEM 2.99
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AB is 6 kN.

SOLUTION
Free-Body Diagram at A:

The forces applied at A are:
,,,and
AB AC AD
TTT W
where .W=Wj To express the other forces in terms of the unit vectors i, j, k, we write

(450 mm) (600 mm) 750 mm
(600 mm) (320 mm) 680 mm
(500 mm) (600 mm) (360 mm) 860 mm
AB AB
AC AC
AD AD
=− + =
=+ − =
=+ + + =
ij
jk
ijk
γγγG
JJJG
JJJG

and
(450mm) (600mm)
750 mm
AB AB AB AB AB
AB
TT T
AB
−+
===
ij
T λ
γγγG


45 60
75 75
AB
T
⎛⎞
=− +
⎜⎟
⎝⎠
ij

(600 mm) (320 mm)
680 mm
60 32
68 68
(500 mm) (600 mm) (360 mm)
860 mm
50 60 36
86 86 86
−
== =
⎛⎞
=−
⎜⎟
⎝⎠
++
== =
⎛⎞
=++
⎜⎟
⎝⎠
γJJG
JJJG
AC AC AC AC AC
AC
AD AD AD AD AD
AD
AC
TT T
AC
T
AD
TT T
AD
T
ij
T λ
jk
ijk
T λ
ijk

Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

131. PROBLEM 2.99 (Continued)

Equilibrium condition:
0: 0
AB AC AD
FΣ= ∴ + + + =TTTW
Substituting the expressions obtained for
,,and;
AB AC AD
TT T factoring i, j, and k ; and equating each of
the coefficients to zero gives the following equations:
From i:
45 50
0
75 86
AB AD
TT−+ = (1)
From j:
60 60 60
0
75 68 86
AB AC AD
TTTW++−= (2)
From k:
32 36
0
68 86
AC AD
TT−+= (3)
Setting
6kN
AB
T= in (1) and (2), and solving the resulting set of equations gives

6.1920 kN
5.5080 kN
AC
AC
T
T
=
=
13.98 kNW=
W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

132. PROBLEM 2.100
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AD is 4.3 kN.

SOLUTION
See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:

45 50
0
75 86
AB AD
TT−+ = (1)

60 60 60
0
75 68 86
AB AC AD
TTTW++−= (2)

32 36
0
68 86
AC AD
TT−+= (3)
Setting
4.3 kN
AD
T= into the above equations gives

4.1667 kN
3.8250 kN
AB
AC
T
T
=
=
9.71 kNW=
W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

133.  
SOLUT
The forc
where P

and
 
 
TION
ces applied at A
.P=Pj To exp


A are:
press the other
AB
AC
AD
=
=
=
γγγG
JJJG
JJJG
AB
AC
AD
=
=
=
T
T
T
 
PROB
Three c
the vert
tension
FREE-BOD
,
AB
T
r forces in term
(4.20 m)
(2.40 m)(
(5.60 m)
=− −
= −
=− −
i
i
j
AB AB A
B
AC AC A
AD AD A
TT
TT
TT
= =
= =
= =
λ
λ
λ
BLEM 2.101
cables are use
tical force P e
in cable AD i
DY DIAGRA
,,and
AC AD
TT
ms of the unit
(5.60 m)
(5.60 m) (4
(3.30 m)
−
+
−
j
j
k
(0.6
(0.32
(0.8
B
AC
AD
AB
AB
AC
AC
AD
AD
=−
=
=−
γγγG
JJJG
JJJG
1
ed to tether a
exerted by the
s 481 N.
AM AT A

dP
vectors i , j, k
.20 m)
AB
AC
A
D
k
0.8 )
2432 0.7567
86154 0.507
AB
T−
−
−
ij
i
j
balloon as sh
balloon at A k
, we write
7.00 m
7.40 m
6.50 mD
=
=
=

76 0.56757
769 )
AD
T
+j k
k
hown. Determ
knowing that
)
AC
Tk

ine
the Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

134. Equilibr
Substitu

Equating



Setting

 
rium condition
uting the expre
(−
g to zero the c
481 N
AD
T=
P
n:
essions obtaine
0.6 0.3
AB
T− +
coefficients of
0.8
AB
T−−
in (2) and (3),
 
PROBLEM 2
0:
AB
FΣ= T
ed for ,
AB A
TT
2432 ) (
(0.56757
AC
T+
+
i
f i, j, k:
0.6
A
T−
0.75676
AC
T−
0.56757T
, and solving t
43
23
AC
AD
T
T
=
=
2.101 (Con
B AC AD
++TT
,and
ACAD
T an
(0.8 0.7
7 0.50769
AB
AC
T
T
−−
−
0.32432
AB
TT+
0.86154
AD
T
−+
0.50769
AC
TT−
the resulting se
30.26 N
32.57 N
ntinued)
0P+=j
nd factoring i,
5676 0.8
9)0
AC
AD
T
T
−
=k
0
AC
T=
0P+=
0
AD
T=
et of equation
, j, and k:
86154 )
AD
TP+
s gives
)j

926 N=P
(1)
(2)
(3)

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

135. PROBLEM 2.102
Three cables are used to tether a balloon as shown. Knowing
that the balloon exerts an 800-N vertical force at A, determine
the tension in each cable.
 
SOLUTION
See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).

0.6 0.32432 0
AB AC
TT
− += (1)

0.8 0.75676 0.86154 0
AB AC AD
TTTP−− − += (2)

0.56757 0.50769 0
AC AD
TT
− = (3)
From Eq. (1):
0.54053
AB AC
TT=
From Eq. (3):
1.11795
AD AC
TT=
Substituting for
AB
T and
AD
T in terms of
AC
T into Eq. (2) gives

0.8(0.54053 ) 0.75676 0.86154(1.11795 ) 0
AC AC AC
TT TP−−− +=

2.1523 ; 800 N
800 N
2.1523
371.69 N
AC
AC
TPP
T
= =
=
=

Substituting into expressions for
AB
T and
AD
T gives

0.54053(371.69 N)
1.11795(371.69 N)
AB
AD
T
T
=
=


201 N, 372 N, 416 N
AB AC AD
TTT=== W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

136. PROBLEM 2.103
A 36-lb triangular plate is supported by three wires as shown. Determine
the tension in each wire, knowing that a = 6 in.
 
SOLUTION
By Symmetry
DBDC
TT= Free-Body Diagram of Point D:

The forces applied at D are:
,,,and
DB DC DA
TTT P
where (36 lb) .P==Pj j To express the other forces in terms of the unit vectors i, j, k, we write

(16 in.) (24 in.) 28.844 in.
(8in.) (24in.) (6in.) 26.0in.
(8 in.) (24 in.) (6 in.) 26.0 in.
DA DA
DB DB
DC DC
=− =
=− − + =
=− − − =
ij
ijk
ijk
JJJG
JJJG
JJJG

and
(0.55471 0.83206 )
( 0.30769 0.92308 0.23077 )
( 0.30769 0.92308 0.23077 )
DA DA DA DA DA
DBDBDBDB DB
DCDCDCDC DC
DA
TT T
DA
DB
TT T
DB
DC
TT T
DC
=== −
== =− − +
== =−− −
T λ ij
T λ ijk
T λ ijk
γJJG
JJJG
JJJG

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

137. PROBLEM 2.103 (Continued)

Equilibrium condition:
0: (36 lb) 0
DA DB DC
FΣ= + + + =TTT j
Substituting the expressions obtained for
,,and
DADB DC
TT T and factoring i, j, and k :

(0.55471 0.30769 0.30769 ) ( 0.83206 0.92308 0.92308 36 lb)
(0.23077 0.23077 ) 0
DA DB DC DA DB DC
DB DC
TTT TTT
TT
−− +−−−+
+− =
ij
k

Equating to zero the coefficients of i, j, k:

0.55471 0.30769 0.30769 0
DA DB DC
TTT
− −= (1)

0.83206 0.92308 0.92308 36 lb 0
DA DB DC
TTT−−− += (2)

0.23077 0.23077 0
DB DC
TT
− = (3)
Equation (3) confirms that
DBDC
TT= . Solving simultaneously gives,

14.42 lb; 13.00 lb
DA DB DC
TTT===

W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

138. PROBLEM 2.104
Solve Prob. 2.103, assuming that a = 8 in.
PROBLEM 2.103 A 36-lb triangular plate is supported by three wires
as shown. Determine the tension in each wire, knowing that a = 6 in.
 
SOLUTION
By Symmetry
DBDC
TT= Free-Body Diagram of Point D:

The forces applied at D are:
,,,and
DB DC DA
TTT P
where (36 lb) .P==Pj j To express the other forces in terms of the unit vectors i, j, k, we write

(16 in.) (24 in.) 28.844 in.
(8 in.) (24 in.) (8 in.) 26.533 in.
(8 in.) (24 in.) (8 in.) 26.533 in.
DA DA
DB DB
DC DC
=− =
=− − + =
=− − − =
ij
ijk
ijk
JJJG
JJJG
JJJG

and
(0.55471 0.83206 )
( 0.30151 0.90453 0.30151 )
( 0.30151 0.90453 0.30151 )
DA DA DA DA DA
DBDBDBDB DB
DCDCDCDC DC
DA
TT T
DA
DB
TT T
DB
DC
TT T
DC
=== −
== =− − +
== =−− −
T λ ij
T λ ijk
T λ ijk
γγγG
JJJG
JJJG

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

139. PROBLEM 2.104 (Continued)

Equilibrium condition:
0: (36 lb) 0
DA DB DC
FΣ= + + + =TTT j
Substituting the expressions obtained for
,,and
DADB DC
TT T and factoring i, j, and k :

(0.55471 0.30151 0.30151 ) ( 0.83206 0.90453 0.90453 36 lb)
(0.30151 0.30151 ) 0
DA DB DC DA DB DC
DB DC
TTT TT T
TT
−− +−−−+
+− =
ij
k

Equating to zero the coefficients of i, j, k:

0.55471 0.30151 0.30151 0
DA DB DC
TTT
− −= (1)

0.83206 0.90453 0.90453 36 lb 0
DA DB DC
TTT−−− += (2)

0.30151 0.30151 0
DB DC
TT
− = (3)
Equation (3) confirms that
DBDC
TT= . Solving simultaneously gives,

14.42 lb; 13.27 lb
DA DB DC
TTT=== W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

140.  
 
Solution

where P

and
Equilibr



n The forces ap
.P=Pj To exp
rium Condition
pplied at A are
press the other
AB
AB
AC
AC
AD
AD
=
=
=
=
=
=
γγγG
JJJG
JJJG
AB
AC
AD
=
=
=
=
=
=
T
T
T
n with =W
FΣ=
PR
A
the
is 5
e:
,,
AB AC A D
TTT
r forces in term
(36 in.)
75 in.
(60 in.) (3
68 in. (40 in.) (
6
77 in.
=−+
=
= +
=
= +
=
i
j
i
(0.48 0.
8
(0.88235
(0.51948
AB AB A
AC AC
A
AD AD A
TT
TT
TT
= =
=−+
= =
= +
= =
= +
λ
i
λ
j
λ
i
W=−j
0:
ABA
= +TT
ROBLEM 2
crate is suppo
e weight of the
544 lb.
and
D
W
ms of the unit
(60 in.) (27
32in.)
60 in.) (27 in
−
−
j
k
j
80.36 )
0.47059 )
0.77922 0.3
B
AB
AC
AC
AD
AB
AB
T
AC
AC
T
AD
AD
−
−
jk
k
j
γγγG
JJJG
JJJG
ACAD
W+−Tj

2.105
orted by three
e crate knowin
vectors i , j, k
in.)
n.)
k
k

35065 )
B
C
AD
Tk

0=
e cables as sh
ng that the ten
, we write
hown. Determ
nsion in cable A
ine
AC Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

141.  
PROBLEM 2.105 (Continued)

Substituting the expressions obtained for
,,and
AB AC AD
TT T and factoring i, j, and k :

( 0.48 0.51948 ) (0.8 0.88235 0.77922 )
( 0.36 0.47059 0.35065 ) 0
AB AD AB AC AD
AB AC AD
TTTTTW
TTT
−+ ++ + −
+− + − =
ij
k

Equating to zero the coefficients of i, j, k:

0.48 0.51948 0
AB AD
TT
− += (1)

0.8 0.88235 0.77922 0
AB AC AD
TTTW++−= (2)

0.36 0.47059 0.35065 0
AB AC AD
TTT−+ − = (3)
Substituting
544 lb
AC
T= in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives:

374.27 lb
345.82 lb
AB
AD
T
T
=
=
1049 lbW= W
 
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

142.  
SOLUT
The forc

where P

and
Equilibr



TION
ces applied at A
.P=Pj To exp
rium Condition
A are:
press the other
AB
AB
AC
AC
AD
AD
=
=
=
=
=
=
γγγG
JJJG
JJJG
AB
AC
AD
=
=
=
=
=
=
T
T
T
n with =W
FΣ=

P
A
D
,,
AB AC A D
TTT
r forces in term
(36 in.)
75 in.
(60 in.) (3
68 in. (40 in.) (
6
77 in.
=−+
=
= +
=
= +
=
i
j
i
(0.48 0.
8
(0.88235
(0.51948
AB AB A
AC AC
A
AD AD A
TT
TT
TT
= =
=−+
= =
= +
= =
= +
λ
i
λ
j
λ
i
W=−j
0:
ABA
= +TT
PROBLEM 2
A 1600-lb crat
etermine the t
and
D
W
ms of the unit
(60 in.) (27
32in.)
60 in.) (27 in
−
−
j
k
j
80.36 )
0.47059 )
0.77922 0.3
B
AB
AC
AC
AD
AB
AB
T
AC
AC
T
AD
AD
−
−
jk
k
j
γγγG
JJJG
JJJG
ACAD
W+−Tj
2.106
te is supporte
tension in each
vectors i , j, k
in.)
n.)
k
k

35065 )
B
C
AD
Tk

0=
ed by three c
h cable.
, we write
ables as showwn. Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

143. PROBLEM 2.106 (Continued)

Substituting the expressions obtained for
,,and
AB AC AD
TT T and factoring i, j, and k :

( 0.48 0.51948 ) (0.8 0.88235 0.77922 )
( 0.36 0.47059 0.35065 ) 0
AB AD AB AC AD
AB AC AD
TTTTTW
TTT
−+ ++ + −
+− + − =
ij
k

Equating to zero the coefficients of i, j, k:

0.48 0.51948 0
AB AD
TT
− += (1)

0.8 0.88235 0.77922 0
AB AC AD
TTTW++−= (2)

0.36 0.47059 0.35065 0
AB AC AD
TTT−+ − = (3)
Substituting
1600 lbW= in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives,

571 lb
AB
T= W

830 lb
AC
T= W

528 lb
AD
T= W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

144. PROBLEM 2.107
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that
0,Q
= find the value of P for
which the tension in cable AD is 305 N.
 
SOLUTION
0: 0
AA BA CA D
Σ= + + +=FTTTP where
P=Pi

(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
(960 mm) (720 mm) (220 mm) 1220 mm
AB AB
AC AC
AD AD
=− − + =
=− − − =
=− + − =
ijk
ijk
ijk
JJJG
JJJG
JJJG


48 12 19
53 53 53
12 3 4
13 13 13
305 N
[( 960 mm) (720 mm) (220 mm) ]
1220 mm
(240 N) (180 N) (55 N)
AB AB AB AB AB
AC AC AC AC AC
AD AD AD
AB
TTT
AB
AC
TT T
AC
T
⎛⎞
===−−+
⎜⎟
⎝⎠
⎛⎞
== =−−−
⎜⎟
⎝⎠
== − + −
=− + −
T λ ijk
T λ ijk
T λ ijk
ijk
JJJG
JJJG

Substituting into
0,
A
Σ=F factoring ,, ,ijk and setting each coefficient equal to
φ gives:

48 12
: 240 N
53 13
AB AC
PT T=++i (1)
:j
12 3
180 N
53 13
AB AC
TT+= (2)

:k
19 4
55 N
53 13
AB AC
TT−= (3)
Solving the system of linear equations using conventional algorithms gives:

446.71 N
341.71 N
AB
AC
T
T
=
= 960 NP= W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

145. PROBLEM 2.108
Three cables are connected at A, where the forces P and Q
are applied as shown. Knowing that
1200 N,P= determine
the values of Q for which cable AD is taut.
 
SOLUTION
We assume that 0
AD
T= and write 0: (1200 N) 0
AA BA C
Q
Σ=+++ =FTTj i

(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
AB AB
AC AC
=− − + =
=− − − =
ijk
ijk
γγγG
JJJG


48 12 19
53 53 53
12 3 4
13 13 13
ABABABAB AB
ACACACAC AC
AB
TT T
AB
AC
TT T
AC
⎛⎞
===−−+
⎜⎟
⎝⎠
⎛⎞
== =−−−
⎜⎟
⎝⎠
T λ ijk
T λ ijk
γγγG
JJJG
Substituting into
0,
A
Σ=F factoring , , ,ijk and setting each coefficient equal to
φ gives:

48 12
:1200 N0
53 13
AB AC
TT
− −+ =i (1)

12 3
:0
53 13
AB AC
TTQ
− −+=j (2)

19 4
:0
53 13
AB AC
TT
− =k (3)
Solving the resulting system of linear equations using conventional algorithms gives:

605.71 N
705.71 N
300.00 N
AB
AC
T
T
Q
=
=
=
0 300 NQ≤< W
Note: This solution assumes that Q is directed upward as shown (0),Q≥ if negative values of Q
are considered, cable AD remains taut, but AC becomes slack for
460 N.Q
=−
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

146.  
SOLUT
We note
Point A.


We have

Thus:

Substitu



TION
e that the we

e:
(4
(2
AB
AC
AD
=−
=
=
γγγG
JJJG
JJJG
uting into the E
ight of the pl
FΣ
(320 mm)
450 mm) (4
250 mm) (4
− −
− −
i
i i
AB
AC
AD
=
=
=
T
T
T
Eq. 0FΣ= an

PROB
A recta
Knowin
weight
late is equal
0:
AB
F= +T
(
(480 mm)(
480 mm) (3
480 mm)3
+
+ −
j
j j
AB AB A
AC AC A
AD AD A
TT
TT
TT
= =
= =
= =
λ
λ
λ
nd factoring ,i
8
17
12
17
9
17
AB
A
B
AB
T
T
T
⎛
−+
⎜
⎝
⎛
+−
⎜
⎝
⎛
+
+
⎜
⎝
BLEM 2.10
angular plate
ng that the te
of the plate.
in magnitude
AC AD
+ ++TT
)
(360 mm)
360 mm)
360 mm
A
A
A
k
k
k
(
8
17
0.6
5
13
AB
AC
AD
AB
AB
AC
AC
AD
AD
⎛
=−
⎜
⎝
=
⎛
=
⎜
⎝
i
i
γγγG
JJJG
JJJG
, , :jk
5
0.6
13
0.64
7.
0.48
13
AC A
B AC
AC
T
T
T
T
+
−−
+ −
9
is supported
ension in cabl
to the force
0P=j
680 mm
750 mm
650 mm
AB
AC
AD
=
=
=
12 9
17 17
0.64 0.48
9.6 7.2
13 13
−+
−+
−−
ij k
j
i j k
9.6
13
.2
0
3
AD
AD
AD
T
TP
T
⎞
⎟
⎠
⎞
+
⎟
⎠
⎞
=
⎟
⎠
i
j
k
d by three ca
le AC is 60 N
P exerted by
Free
m
m
m

)8
AB
AC
AD
T
T
T
⎞
⎟
⎠
⎞
⎟
⎠
k
k
k

j

Dim
ables as show
N, determine
y the support
e Body A:
mensions in m
wn.
the
on
mmCopyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

147. PROBLEM 2.109 (Continued)

Setting the coefficient of i, j, k equal to zero:

:i
85
0.6 0
17 13
AB AC AD
TTT−+ + = (1)
:j
12 9.6
0.64 0
71 3
AB AC AD
TTTP−− − += (2)
:k
97 .2
0.48 0
17 13
AB AC AD
TTT
+ −= (3)
Making
60 N
AC
T= in (1) and (3):

85
36 N 0
17 13
AB AD
TT−++ = (1 ′)

97 .2
28.8 N 0
17 13
AB AD
TT
+ −= (3 ′)
Multiply (1′) by 9, (3′) by 8, and add:

12.6
554.4 N 0 572.0 N
13
AD AD
TT−==
Substitute into (1′ ) and solve for
:
AB
T

17 5
36 572 544.0 N
813
AB AB
TT
⎛⎞
=+× =
⎜⎟
⎝⎠

Substitute for the tensions in Eq. (2) and solve for P:

12 9.6
(544N) 0.64(60 N) (572 N)
17 13
844.8 N
P=+ +
=
Weight of plate 845 NP== W

Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

148. PROBLEM 2.110
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 520 N, determine the
weight of the plate.
 
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:

85
0.6 0
17 13
AB AC AD
TTT−+ + = (1)

12 9.6
0.64 0
17 13
AB AC AD
TTTP−+ − += (2)

97 .2
0.48 0
17 13
AB AC AD
TTT
+ −= (3)
Making
520 N
AD
T= in Eqs. (1) and (3):

8
0.6 200 N 0
17
AB AC
TT−+ += (1 ′)

9
0.48 288 N 0
17
AB AC
TT
+ −= (3 ′)
Multiply (1′) by 9, (3′) by 8, and add:

9.24 504 N 0 54.5455 N
AC AC
TT−= =
Substitute into (1′ ) and solve for
:
AB
T

17
(0.6 54.5455 200) 494.545 N
8
AB AB
TT=× + =
Substitute for the tensions in Eq. (2) and solve for P:

12 9.6
(494.545 N) 0.64(54.5455 N) (520 N)
17 13
768.00 N
P=+ +
=
Weight of plate 768 NP== W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

149. PROBLEM 2.111
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D . If the
tension in wire AB is 840 lb, determine the vertical force
P exerted by the tower on the pin at A.
 
SOLUTION
0: 0
AB AC AD
PΣ= + + + =FTTTj Free-Body Diagram at A:

= 20 100 25 105 ft
60 100 18 118 ft
20 100 74 126 ft
AB AB
AC AC
AD AD
−− + =
=− + =
=− − − =
ijk
ijk
ijk
γγγG
JJJG
JJJG

We write
4205
21 21 21
AB AB AB AB
AB
AB
TT
AB
T
==
⎛⎞
=− − +
⎜⎟
⎝⎠
T λ
ijk
γγγG


30 50 9
59 59 59
AC AC AC AC
AC
AC
TT
AC
T
==
⎛⎞
=−+
⎜⎟
⎝⎠
T λ
ijk
γJJG


10 50 37
63 63 63
AD AD AD AD
AD
AD
TT
AD
T
==
⎛⎞
=− − −
⎜⎟ ⎝⎠
T λ
ijk
γJJG

Substituting into the Eq.
0Σ=F and factoring , , : ijk

Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

150. PROBLEM 2.111 (Continued)

43010
21 59 63
20 50 50
21 59 63
5937
0
21 59 63
AB AC AD
AB AC AD
AB AC AD
TTT
TTTP
TTT
⎛⎞
−+ −
⎜⎟
⎝⎠
⎛⎞
+− − − +
⎜⎟
⎝⎠
⎛⎞
++− =
⎜⎟
⎝⎠
i
j
k


Setting the coefficients of , , ,ijk equal to zero:

:i
43010
0
21 59 63
AB AC AD
TTT−+ − = (1)
:j
20 50 50
0
21 59 63
AB AC AD
TTTP−−−+= (2)
:k
5937
0
21 59 63
AB AC AD
TTT+−= (3)
Set
840 lb
AB
T= in Eqs. (1) – (3):

30 10
160 lb 0
59 63
AC AD
TT−+ − = (1 ′)

50 50
800 lb 0
59 63
AC AD
TTP−− − += (2 ′)

937
200 lb 0
59 63
AC AD
TT+−= (3 ′)
Solving,
458.12 lb 459.53 lb 1552.94 lb
AC AD
TTP=== 1553 lbP= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

151. PROBLEM 2.112
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D . If the
tension in wire AC is 590 lb, determine the vertical force
P exerted by the tower on the pin at A.
 
SOLUTION
0: 0
AB AC AD
PΣ= + + + =FTTTj Free-Body Diagram at A:

= 20 100 25 105 ft
60 100 18 118 ft
20 100 74 126 ft
AB AB
AC AC
AD AD
−− + =
=− + =
=− − − =
ijk
ijk
ijk
γγγG
JJJG
JJJG

We write
4205
21 21 21
AB AB AB AB
AB
AB
TT
AB
T
==
⎛⎞
=− − +
⎜⎟
⎝⎠
T λ
ijk
γγγG


30 50 9
59 59 59
AC AC AC AC
AC
AC
TT
AC
T
==
⎛⎞
=−+
⎜⎟
⎝⎠
T λ
ijk
γJJG


10 50 37
63 63 63
AD AD AD AD
AD
AD
TT
AD
T
==
⎛⎞
=− − −
⎜⎟ ⎝⎠
T λ
ijk
γJJG

Substituting into the Eq.
0Σ=F and factoring , , :ijk Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

152. PROBLEM 2.112 (Continued)

43010
21 59 63
20 50 50
21 59 63
5937
0
21 59 63
AB AC AD
AB AC AD
AB AC AD
TTT
TTTP
TTT
⎛⎞
−+ −
⎜⎟
⎝⎠
⎛⎞
+− − − +
⎜⎟
⎝⎠
⎛⎞
++− =
⎜⎟
⎝⎠
i
j
k

Setting the coefficients of , , ,ijk equal to zero:

:i
43010
0
21 59 63
AB AC AD
TTT−+ − = (1)
:j
20 50 50
0
21 59 63
AB AC AD
TTTP−−−+= (2)
:k
5937
0
21 59 63
AB AC AD
TTT+−= (3)
Set
590 lb
AC
T= in Eqs. (1) – (3):

41 0
300 lb 0
21 63
AB AD
TT−+− = (1 ′)

20 50
500 lb 0
21 63
AB AD
TTP−−−+= (2 ′)

53 7
90 lb 0
21 63
AB AD
TT+− = (3 ′)
Solving,
1081.82 lb 591.82 lb
AB AD
TT== 2000 lbP= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

153. PROBLEM 2.113
In trying to move across a slippery icy surface, a 175-lb man uses
two ropes AB and AC. Knowing that the force exerted on the man
by the icy surface is perpendicular to that surface, determine the
tension in each rope.
 
SOLUTION
Free-Body Diagram at A

16 30
34 34
N
⎛⎞
=+
⎜⎟
⎝⎠
Nij
and (175 lb)W==−
Wj j

( 30 ft) (20 ft) (12 ft)
38 ft
15 10 6
19 19 19
AC AC AC AC AC
AC
AC
TT T
AC
T
−+ −
== =
⎛⎞
=−+−
⎜⎟
⎝⎠
ijk
T λ
ijk
γJJG


( 30 ft) (24 ft) (32 ft)
50 ft
15 12 16
25 25 25
AB AB AB AB AB
AB
AB
TTT
AB
T
−+ +
===
⎛⎞
=−++
⎜⎟ ⎝⎠
ijk
T λ
ijk
γγγG

Equilibrium condition:
0Σ=F

0
AB AC
+ ++ =TTNW








Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

154. PROBLEM 2.113 (Continued)

Substituting the expressions obtained for
,,,
AB AC
TTN and W; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
From
i:
15 15 16
0
25 19 34
AB AC
TT N−−+= (1)
From
j:
12 10 30
(175 lb) 0
25 19 34
AB AC
TT N++− = (2)
From
k:
16 6
0
25 19
AB AC
TT
− = (3)
Solving the resulting set of equations gives:

30.8 lb; 62.5 lb
AB AC
TT== W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

155. PROBLEM 2.114
Solve Problem 2.113, assuming that a friend is helping the man
at A by pulling on him with a force
P = −(45 lb)k.
PROBLEM 2.113 In trying to move across a slippery icy
surface, a 175-lb man uses two ropes AB and AC. Knowing that
the force exerted on the man by the icy surface is perpendicular
to that surface, determine the tension in each rope.

 
SOLUTION
Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3)
being modified to include the additional force (45lb).
=−Pk

15 15 16
0
25 19 34
AB AC
TT N−−+= (1)

12 10 30
(175 lb) 0
25 19 34
AB AC
TT N++− = (2)


16 6
(45 lb) 0
25 19
AB AC
TT
− −= (3)
Solving the resulting set of equations simultaneously gives:

81.3 lb
AB
T= W
22.2 lb
AC
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

156. PROBLEM 2.115
For the rectangular plate of Problems 2.109 and 2.110,
determine the tension in each of the three cables knowing that
the weight of the plate is 792 N.
 
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below. Setting
792 NP= gives:

85
0.6 0
17 13
AB AC AD
TTT−+ + = (1)

12 9.6
0.64 792 N 0
17 13
AB AC AD
TTT−− − += (2)

97 .2
0.48 0
17 13
AB AC AD
TTT
+ −= (3)
Solving Equations (1), (2), and (3) by conventional algorithms gives

510.00 N
AB
T= 510 N
AB
T= W

56.250 N
AC
T= 56.2 N
AC
T= W

536.25 N
AD
T= 536 N
AD
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

157. PROBLEM 2.116
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
2880 NP
= and 0.Q=
 
SOLUTION
0: 0
AA BA CA D
Σ= + + ++=FTTTPQ
Where
P=Pi and Q=Qj

(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
(960 mm) (720 mm) (220 mm) 1220 mm
AB AB
AC AC
AD AD
=− − + =
=− − − =
=− + − =
ijk
ijk
ijk
γγγG
JJJG
JJJG

48 12 19
53 53 53
12 3 4
13 13 13
48 36 11
61 61 61
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TTT
AB
AC
TT T
AC
AD
TT T
AD
⎛⎞
===−−+
⎜⎟
⎝⎠
⎛⎞
== =−−−
⎜⎟
⎝⎠
⎛⎞
== =−+−
⎜⎟
⎝⎠
T λ ijk
T λ ijk
T λ ijk
γγγG
JJJG
JJJG

Substituting into
0,
A
Σ=F setting (2880 N)P= i and 0,Q
= and setting the coefficients of , , ijkequal to
0, we obtain the following three equilibrium equations:

48 12 48
: 2880 N 0
53 13 61
AB AC AD
TT T
− −−+ =i (1)

12 3 36
:0
53 13 61
AB AC AD
TT T
− −+ =j (2)

19 4 11
:0
53 13 61
AB AC AD
TTT
− −=k (3) Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

158. PROBLEM 2.116 (Continued)

Solving the system of linear equations using conventional algorithms gives:

1340.14 N
1025.12 N
915.03 N
AB
AC
AD
T
T
T
=
=
=
1340 N
AB
T= W

1025 N
AC
T= W
915 N
AD
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

159. PROBLEM 2.117
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
2880 NP
= and 576 N.Q=
 
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

48 12 48
0
53 13 61
AB AC AD
TT TP−−− += (1)

12 3 36
0
53 13 61
AB AC AD
TT TQ−−+ += (2)

19 4 11
0
53 13 61
AB AC AD
TTT
− −= (3)
Setting
2880 NP= and 576 NQ= gives:

48 12 48
2880 N 0
53 13 61
AB AC AD
TT T−−− + = (1 ′)

12 3 36
576 N 0
53 13 61
AB AC AD
TT T−−+ += (2 ′)

19 4 11
0
53 13 61
AB AC AD
TTT
− −= (3 ′)
Solving the resulting set of equations using conventional algorithms gives:

1431.00 N
1560.00 N
183.010 N
AB
AC
AD
T
T
T
=
=
=
1431 N
AB
T= W
1560 N
AC
T= W

183.0 N
AD
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

160. PROBLEM 2.118
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
2880 NP
= and 576Q=−N. (Q is directed downward).
 
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

48 12 48
0
53 13 61
AB AC AD
TT TP−−− += (1)

12 3 36
0
53 13 61
AB AC AD
TT TQ−−+ += (2)

19 4 11
0
53 13 61
AB AC AD
TTT
− −= (3)
Setting
2880 NP= and 576 NQ=− gives:

48 12 48
2880 N 0
53 13 61
AB AC AD
TT T−−− + = (1 ′)

12 3 36
576 N 0
53 13 61
AB AC AD
TT T−−+ −= (2 ′)

19 4 11
0
53 13 61
AB AC AD
TTT
− −= (3 ′)
Solving the resulting set of equations using conventional algorithms gives:

1249.29 N
490.31 N
1646.97 N
AB
AC
AD
T
T
T
=
=
=
1249 N
AB
T= W

490 N
AC
T= W

1647 N
AD
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

161. PROBLEM 2.119
For the transmission tower of Probs. 2.111 and 2.112,
determine the tension in each guy wire knowing that the
tower exerts on the pin at A an upward vertical force of
1800 lb.
PROBLEM 2.111 A transmission tower is held by three
guy wires attached to a pin at A and anchored by bolts at
B, C, and D . If the tension in wire AB is 840 lb, determine
the vertical force P exerted by the tower on the pin at A.
 
SOLUTION
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
:i
43010
0
21 59 63
AB AC AD
TTT−+ − = (1)

:j
20 50 50
0
21 59 63
AB AC AD
TTTP−−−+= (2)
:k
5937
0
21 59 63
AB AC AD
TTT+−= (3)

Substituting for
1800 lbP= in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:

43010
0
21 59 63
AB AC AD
TTT−+ − = (1 ′)

20 50 50
1800 lb 0
21 59 63
AB AC AD
TTT−−−+ = (2 ′)

5937
0
21 59 63
AB AC AD
TTT+−= (3 ′)

973.64 lb
531.00 lb
532.64 lb
AB
AC
AD
T
T
T
=
=
=
974 lb
AB
T= W

531 lb
AC
T= W

533 lb
AD
T= W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

162. PROBLEM 2.120
Three wires are connected at point D, which is
located 18 in. below the T-shaped pipe support
ABC. Determine the tension in each wire when
a 180-lb cylinder is suspended from point D as
shown.

SOLUTION
Free-Body Diagram of Point D:



The forces applied at D are:

,, and
DA DB DC
TTT W
where 180.0 lb .=−Wj To express the other forces in terms of the unit vectors i, j, k, we write

(18 in.) (22 in.)
28.425 in.
(24 in.) (18 in.) (16 in.)
34.0 in.
(24in.) (18in.) (16in.)
34.0 in.
DA
DA
DB
DB
DC
DC
=+
=
=− + −
=
=+−
=
jk
ijk
ijk
γγγG
JJJG
JJJG




 Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

163. PROBLEM 2.120 (Continued)
and
(0.63324 0.77397 )
( 0.70588 0.52941 0.47059 )
(0.70588 0.52941 0.47059 )
DA Da DA Da
DA
DB DB DB DB
DB
DC DC DC DC
DC
DA
TT
DA
T
DB
TT
DB
T
DC
TT
DC
T
==
=+
==
=− + −
==
=+−
T λ
jk
T λ
ijk
T λ
ijk
γγγG
JJJG
JJJG

Equilibrium Condition with W=−
Wj

0: 0
DA DB DC
FWΣ= + + − =TTT j


Substituting the expressions obtained for
,,and
DADB DC
TT T and factoring i, j, and k:

( 0.70588 0.70588 )
(0.63324 0.52941 0.52941 )
(0.77397 0.47059 0.47059 )
DB DC
DA DB DC
DA DB DC
TT
TTTW
TTT
−+
++−
−−
i
j
k

Equating to zero the coefficients of
i, j, k:

0.70588 0.70588 0
DB DC
TT
− += (1)

0.63324 0.52941 0.52941 0
DA DB DC
TTTW
+ +−= (2)

0.77397 0.47059 0.47059 0
DA DB DC
TTT
− −= (3)
Substituting
180 lbW= in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms gives,

119.7 lb
DA
T= W

98.4 lb
DB
T= W

98.4 lb
DC
T= W


  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

164. PROBLEM 2.121
A container of weight W is suspended from ring
A, to which cables AC and AE are attached. A
force
P is applied to the end F of a third cable
that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing
that
1000 N,W
= determine the magnitude of P.
(Hint: The tension is the same in all portions of
cable FBAD .)
 
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector
along the cable. That is, with

222
(0.78 m) (1.6 m) (0 m)
(0.78m) (1.6 m) (0)
1.78 m
[ (0.78 m) (1.6 m) (0 m) ]
1.78 m
( 0.4382 0.8989 0 )
AB AB AB
AB
AB AB
AB
AB
AB
TT
AB
T
T
=− + +
=− + +
=
==
=− ++
=− + + ijk
T λ
ijk
Tij k
γγγG
γγγG

and
222
(0) (1.6 m) (1.2 m)
(0 m) (1.6 m) (1.2 m) 2 m
[(0) (1.6 m) (1.2 m) ]
2 m
(0.8 0.6 )
AC
AC AC AC
AC AC
AC
AC
TAC
TT
AC
T
=+ +
=++=
== = + +
=+
ijk
T λ ijk
Tjk
JJJG
JJJG

and
22 2
(1.3 m) (1.6 m) (0.4 m)
(1.3 m) (1.6 m) (0.4 m) 2.1 m
[(1.3 m) (1.6 m) (0.4 m) ]
2.1 m
(0.6190 0.7619 0.1905 )
AD
AD AD AD
AD AD
AD
AD
TAD
TT
AD
T
=++
=++=
== = + +
=++
ijk
T λ ijk
Tijk
JJJG
JJJG
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

165. PROBLEM 2.121 (Continued)

Finally,
22 2
(0.4 m) (1.6 m) (0.86 m)
( 0.4 m) (1.6 m) ( 0.86 m) 1.86 m
[(0.4 m) (1.6m) (0.86m)]
1.86 m
( 0.2151 0.8602 0.4624 )
AE AE AE
AE
AE AE
AE
AE
AE
TT
AE
T
T
=− + −
=− + +− =
==
=−+−
=− + − ij k
T λ
ij k
Tijk
JJJG
JJJG
With the weight of the container ,W
=−Wj at A we have:

0: 0
AB AC AD
W
Σ=++−=FTTTj
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:

0.4382 0.6190 0.2151 0
AB AD AE
TTT−+ − = (1)

0.8989 0.8 0.7619 0.8602 0
AB AC AD AE
TT T TW++ + −= (2)

0.6 0.1905 0.4624 0
AC AD AE
TTT
+ −= (3)
Knowing that
1000 NW= and that because of the pulley system at B ,
AB AD
TT P
= = where P is the
externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely
for P.

378 NP= W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

166. PROBLEM 2.122
Knowing that the tension in cable AC of the
system described in Problem 2.121 is 150 N,
determine (a) the magnitude of the force P, (b)
the weight W of the container.
PROBLEM 2.121 A container of weight W is
suspended from ring A, to which cables AC and
AE are attached. A force P is applied to the end F
of a third cable that passes over a pulley at B and
through ring A and that is attached to a support at
D. Knowing that
1000 N,W
= determine the
magnitude of P. (Hint: The tension is the same in
all portions of cable FBAD.)
 
SOLUTION
Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB,
with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus,
with the condition

AB AD
TT P
= =
and using the linear algebraic equations of Problem 2.131 with
150 N,
AC
T= we obtain
(a)
454 NP= W
(b)
1202 NW= W

  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

167. PROBLEM 2.123
Cable BAC passes through a frictionless ring A and is
attached to fixed supports at B and C, while cables AD and
AE are both tied to the ring and are attached, respectively, to
supports at D and E. Knowing that a 200-lb vertical load P is
applied to ring A, determine the tension in each of the three
cables.
 
SOLUTION
Free Body Diagram at A:

Since
tension in
BAC
T= cable BAC, it follows that

AB AC BAC
TTT==






( 17.5 in.) (60 in.) 17.5 60
62.5 in. 62.5 62.5
(60 in.) (25 in.) 60 25
65 in. 65 65
(80 in.) (60 in.) 4 3
100 in. 5 5
AB BAC AB BAC BAC
AC BAC AC BAC BAC
AD AD AD AD AD
AE AE AE AE
TT T
TT T
TT T
TT
−+ − ⎛⎞
== = +
⎜⎟
⎝⎠
+ ⎛⎞
== = +
⎜⎟
⎝⎠
+ ⎛⎞
== =+
⎜⎟
⎝⎠
==
ij
T λ ij
ik
T λ jk
ij
T λ ij
T λ
(60 in.) (45 in.) 4 3
75 in. 5 5
AE
T
− ⎛⎞
=−
⎜⎟ ⎝⎠
jk
jk


















Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

168. PROBLEM 2.123 (Continued)

Substituting into
0,
A
Σ=F setting ( 200 lb) ,=−Pj and setting the coefficients of i, j, k equal to ,
φ we
obtain the following three equilibrium equations:
From
17.5 4
:0
62.5 5
BAC AD
TT−+=i (1)
From
60 60 3 4
:200lb0
62.5 65 5 5
BAC AD AE
TTT
⎛⎞
+++−=
⎜⎟
⎝⎠
j (2)
From
25 3
:0
65 5
BAC AE
TT−=k (3)
Solving the system of linear equations using conventional algorithms gives:

76.7 lb; 26.9 lb; 49.2 lb
BAC AD AE
TTT=== W
 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

169. PROBLEM 2.124
Knowing that the tension in cable AE of Prob. 2.123 is 75 lb,
determine (a) the magnitude of the load P, ( b) the tension in
cables BAC and AD.
PROBLEM 2.123 Cable BAC passes through a frictionless
ring A and is attached to fixed supports at B and C, while
cables AD and AE are both tied to the ring and are attached,
respectively, to supports at D and E. Knowing that a 200-lb
vertical load P is applied to ring A, determine the tension in
each of the three cables.
 
SOLUTION
Refer to the solution to Problem 2.123 for the figure and analysis leading to the following set of
equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity:

17.5 4
0
62.5 5
BAC AD
TT−+= (1)

60 60 3 4
0
62.5 65 5 5
BAC AD AE
TTTP
⎛⎞
+++−=
⎜⎟
⎝⎠
(2)

25 3
0
65 5
BAC AE
TT−= (3)
Substituting for
75 lb
AE
T= and solving simultaneously gives:

(a)
305 lbP= W
(b)
117.0 lb; 40.9 lb
BAC AD
TT== W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

170.  
SOLUT
For both

Here
or
Thus, w
Now
Where y
From th
Setting t
With
Now, fr
Setting t
And usi
TION
h Problems 2.
when y given, z
y and z are in u
he F.B. Diagra
the j coefficie
rom the free bo
the k coefficie
ing the above r
125 and 2.126
(0
z is determined
λ
units of meter
am of collar A:
ent to zero give
ody diagram o
ent to zero giv
result for
AB
T

P
C
ca
P
te
of
sy
6:
2 2
()ABx=
2
0.525 m) (0=
22
0.yz+=
d,
1
0.525 m
0.38095
AB
AB
AB
=
=
=
λ
i
γγγG
rs, m.
: 0:Σ=F
es (1.9P−
AB
P
T
of collar B :
ves
, we have
PROBLEM 2
ollars A and B
an slide fre
(341 N)=P j i
ension in the w
f the force Q
ystem.
2 22
yz++
22
0.20 m)y+
2
.23563 m
(0.20
1.90476
yz
y
−+
− +
ij
i j
:
x z
NN+ +ik
90476 )
AB
yT=
341 N
341 N
1.90476
P
y
=
=
0:
NΣ=F
(1.9
AB
QT−
(
AB
QTz= =
2.125
B are connecte
eely on fric
is applied to
wire when y=
required to m
Free
2
z+
)m
1.90476
z
z+
k
k

AB AB
PT
λ++j
0

x y
NNQ++ij k
0476 ) 0z=
341 N
(1.
(1.90476)y
ed by a 525-m
ctionless rod
collar A, de
155 mm,= (b
maintain the eq
e-Body Diagr
0=
0
AB AB
T−=k λ
(34
.90476 )z=
mm-long wire a
ds. If a fo
etermine (a)
b) the magnitu
quilibrium of
rams of Colla
0
41N)( )z
y

and
rce
the
ude
the
rs: Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

171. PROBLEM 2.125 (Continued)

Then from the specifications of the problem, 155 mm 0.155 my= =

222
0.23563 m (0.155 m)
0.46 m
z
z
=−
=

and
(a)
341 N
0.155(1.90476)
1155.00 N
AB
T=
=

or
1155 N=
AB
T W
and
(b)
341 N(0.46 m)(0.866)
(0.155 m)
(1012.00 N)
Q=
=

or 1012 N=Q
W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

172. PROBLEM 2.126
Solve Problem 2.125 assuming that 275 mm.y=
PROBLEM 2.125 Collars A and B are connected by a
525-mm-long wire and can slide freely on frictionless
rods. If a force
(341 N)
=Pj is applied to collar A,
determine (a) the tension in the wire when y
155 mm,=
(b) the magnitude of the force Q required to maintain the
equilibrium of the system.
 
SOLUTION
From the analysis of Problem 2.125, particularly the results:

22 2
0.23563 m
341 N
1.90476
341 N
AB
yz
T
y
Qz
y
+=
=
=

With 275 mm 0.275 m,y== we obtain:

222
0.23563 m (0.275 m)
0.40 m
z
z
=−
=

and
(a)
341 N
651.00
(1.90476)(0.275 m)
AB
T==
or
651 N
AB
T= W
and
(b)
341 N(0.40 m)
(0.275 m)
Q=

or 496 NQ=
W

Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

173.  
SOLUT
Using th
we have
Then
and
Hence:

 
TION
he force triang
e
gle and the law
γ
R
R
10 kN
sin
sin
α
α
α
φ
 

PROB
Two st
shown.
that the
determ
resultan
B.
ws of cosines a
180 (40
120
γ=°−
=°
22
2
(15 kN)
2(15 k
N
475 kN
21.794 kN
R
R
= +
−
= =
N21.794 kN
sin120
10 kN
21.794 k
0.39737
23.414
α
α
α
=
°
⎛
=
⎜
⎝
=
=
50
φα=+°=
BLEM 2.12
tructural mem
. Knowing th
e force is 15
mine by trigon
nt of the force
and sines,
020 )°+ °
2
(10 kN)
N)(10 kN)cos1
N
+
N
N
sin120
kN
⎞
°
⎟
⎠

73.414
27
mbers A and
hat both mem
kN in membe
ometry the m
es applied to t
20°

B are bolted
mbers are in c
er A and 10 k
magnitude and
the bracket by
21.8=R
to a bracket
compression a
kN in member
direction of
y members A a
8 kN 73.4°
as
and
r B,
the
and
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

174. PROBLEM 2.128
Determine the x and y components of each of the forces shown.

SOLUTION
Compute the following distances:

22
22
22
(24 in.) (45 in.)
51.0 in.
(28 in.) (45 in.)
53.0 in.
(40 in.) (30 in.)
50.0 in.
OA
OB
OC
=+
=
=+
=
=+
=

102-lb Force:
24 in.
102 lb
51.0 in.
x
F=− 48.0 lb
x
F=− W

45 in.
102 lb
51.0 in.
y
F=+ 90.0 lb
y
F=+ W
106-lb Force:
28 in.
106 lb
53.0 in.
=+
x
F 56.0 lb
x
F=+ W

45 in.
106 lb
53.0 in.
y
F=+ 90.0 lb
y
F=+ W
200-lb Force:
40 in.
200 lb
50.0 in.
x
F=− 160.0 lb
x
F=− W

30 in.
200 lb
50.0 in.
y
F=− 120.0 lb
y
F=− W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

175. SOLUT




(a) Fo
S

(b) Si

 
 
 
TION
or R to be ver
et
ince R is to be

rtical, we must
e vertical:
 
PROBLEM
A hoist troll
α = 40°, det
resultant of
magnitude o
x
R=Σ
1
x
RP=−
y
R=FΣ
410.
y
R=
t have 0.
x
R=
0
x
R= in E
01 7
177.86
P
P
=−
=
=R
M 2.129
ley is subjecte
ermine (a) th
the three forc
f the resultant
(20
x
FPΣ=+
177.860 lb
(200 lb)c
y
F=
32 lb
.
Eq. (1)
77.860 lb
60lb
410 lb==
y
R
ed to the three
he required ma
ces is to be v
t.
0 lb)sin 40°−
cos40 (400°+
e forces show
agnitude of th
vertical, (b) th
(400 lb)cos4
lb)sin 40°
wn. Knowing t
he force P if
he correspondi
40°
177.9 lbP=
410 lbR=
that
the
ing
(1)
(2)

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

176.  
 
SOLUT


Law of s
(a)
(b)
 
TION
Free
sines:

PRO
Know
along
cable
e-Body Diagra
sin
F
 
OBLEM 2.1
wing that α=
g line AC, dete
BC.
am

n35 sin 50
ACBC
F T
=
°
300 lb
sin 95
AC
F=
300 lb
sin 95
BC
T=
30
55° and that
ermine (a) the
300 lb
0sin 95
=
°°

b
sin 35
5
°
°
b
sin 50
5
°
°
boom AC ex
e magnitude o
Force Triang
erts on pin C
of that force, (
gle

F
a force direc
(b) the tension
172.7 lb
AC
F=
231 lb
BC
T=
ted
n in
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

177.  
 
SOLUT
(a)
(b)

 
TION
0:
x
Σ= −F
0:
1
y
Σ=F
T
 
PR
Two
Kno
AC,
Fr
12 4
(3
13 5
AC
T− +
5
(312 N)
13
T+
480 N
BC
T=−
ROBLEM 2.
o cables are
owing thatP=
(b) in cable B
ee Body: C
360 N) 0=
3
(360 N
5
BC
T+
120 N 216 N−
131
tied together
360 N,= determ
BC.

N)480 N0−=
N
at C and lo
mine the tens
0
T
oaded as show
sion (a) in ca
312 N
AC
T=
144.0 N
BC
T=
wn.
able
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

178.  
 
SOLUT


Force tri

(a)
S
(b)

 
 
TION
iangle is isosc
ince 0,P> th
Free-Body D
celes with
he solution is c
 

PROB
Two cab
the max
determin
applied a
Diagram: C
218
4
β
β
=
=
2P=
correct.
1α=
LEM 2.132
bles tied togeth
ximum allow
ne (a) the ma
at C, (b) the co

8085
7.5
°− °
°

2(800 N)cos 47
80 50 47°− °−
2
her at C are lo
wable tension
agnitude of th
orresponding
7.5° 1081 N=
7.5 82.5°= °
oaded as show
n in each ca
he largest forc
value of α.
Force Trian

wn. Knowing t
able is 800
ce P that can
ngle
1081 NP=
82.5
α=°
that
N,
be
W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

179.  

PROBLEM 2.133
The end of the coaxial cable AE is attached to the pole AB,
which is strengthened by the guy wires AC and AD . Knowing
that the tension in wire AC is 120 lb, determine (a) the
components of the force exerted by this wire on the pole, (b)
the angles
θx, θy, and θz that the force forms with the
coordinate axes.
 
SOLUTION
(a) (120 lb)cos 60 cos 20
x
F
= °°

56.382 lb
x
F= 56.4 lb
x
F=+ W

(120 lb)sin 60
103.923 lb
y
y
F
F
=−°
=−
103.9 lb
y
F=− W

(120 lb)cos 60 sin 20
20.521 lb
z
z
F
F
=−°°
=−
20.5 lb
z
F=− W
(b)
56.382 lb
cos
120 lb
x
x
F
F
θ== 62.0
x
θ=° W

103.923 lb
cos
120 lby
y
F
F
θ
−
==
150.0
y
θ=° W

20.52 lb
cos
120 lb
z
z
F
F
θ
−
== 99.8
z
θ=° W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

180.  

PROBLEM 2.134
Knowing that the tension in cable AC is 2130 N, determine
the components of the force exerted on the plate at C.
 
SOLUTION

222
(900 mm) (600 mm) (920 mm)
(900 mm) (600 mm) (920 mm)
1420 mm
2130 N
[ (900 mm) (600 mm) (920 mm) ]
1420 mm
(1350 N) (900 N) (1380 N)
CA CA CA
CA
CA
CA
CA
T
CA
T
CA
=− + −
=++
=
=
=
=−+ −
=− + −
ijk
T λ
Tijk
ij k
γγγG
γγγG


( ) 1350 N, ( ) 900 N, ( ) 1380 N
CA x CA y CA z
TTT=− = =− W
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

181.  

PROBLEM 2.135
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 600 N and Q = 450 N.
 
SOLUTION

(600 N)[sin 40 sin 25 cos 40 sin 40 cos 25 ]
(162.992 N) (459.63 N) (349.54 N)
(450 N)[cos55 cos30 sin 55 cos55 sin 30 ]
(223.53 N) (368.62 N) (129.055 N)
(386.52 N) (828.25 N) (220.49 N)
(3R
=°°+°+°°
=++
=°°+°−°°
=+−
=+
=++
=
Pijk
ijk
Qijk
ij k
RPQ
ijk
22 2
86.52 N) (828.25 N) (220.49 N)
940.22 N
++
=
940 NR= W

386.52 N
cos
940.22 N
x
x
R
R
θ== 65.7
x
θ=° W

828.25 N
cos
940.22 Ny
y
R
R
θ== 28.2
y
θ=° W

220.49 N
cos
940.22 N
z
z
R
R
θ== 76.4
z
θ=° W
Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

182.  
SOLUT


Setting



TION
coefficients of
(13
13
45
AB AB
T
AB
T
AB
T
T
=
=
−
=
⎛
=−
⎜
⎝
T λ
γγγK
(15
15
49
0:
ACAC
AB
T
AC
T
AC
T
T
F
=
=
−
=
⎛
=−
⎜
⎝
Σ=
T λ
T
JJJK
f i, j, k equal t
13
:
45
T− −i
40
:
45
T++
j
16
:
45
T+ −k
0 mm) (400
450
340 16
545 45
+
++
i
ij
0 mm) (400
49
540 24
949 49
B AC
+
+−
++ +
i
ij
TQ
to zero:
15
0
49
TP− +=
40
0
49
TW−=
24
0
49
TQ− +=
PROBLE
A containe
Cable BAC
fixed supp
QQ=k a
container i
376 N,=
the same in
0 mm) (160
0 mm
+
⎞
⎟
⎠
j
k
0 mm) ( 24
90 mm
0
+−
⎞
⎟
⎠
++=
j
k
PW
0.595
1.705
0 0.13424

EM 2.136
er of weight
C passes throu
ports at B and
are applied t
in the positio
determine P
n both portions
0 mm)k

40 mm)k

01TP=
21TW=
40TQ=
W is suspend
ugh the ring an
d C. Two for
to the ring t
on shown. K
and Q. (Hint
s of cable BA C
Free-Bod
ded from ring
nd is attached
rces P=Pi a
to maintain
Knowing that
t: The tension
C.)
dy A:
A.
d to
and
the
W
n is
(1)
(2)
(3) Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

183. PROBLEM 2.136 (Continued)

Data: 376 N 1.70521 376 N 220.50 NWTT= ==

0.59501(220.50 N)
P= 131.2 NP= W
0.134240(220.50 N)Q= 29.6 NQ= W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

184. SOLUT
A:


Collar A
Substitu

Collar B
Substitu

(a)
F r
(b) F
 
TION

A:
ute for
AB
λ an
B:
ute for
AB
λand
rom Eq. (2):
From Eq. (1):
A
λ
Σ
nd set coefficie
0: (6Σ=F
d set coefficien
9 in.x=
 

PROB
Collars A
freely on
B as sh
9 in.x=
required
Free-Body D
B
ABx
AB
−
==
i
γγγG
0:PNΣ=+Fi
ent of i equal t
60 lb)
x
N′++ki
nt of k equal to
60 lb
2
T
−
2
(9 in.)+
60 lb
25
A
T−
(125
P=
LEM 2.137
A and B are co
n frictionless
hown, determ
.,(b) the co
d to maintain th
Diagrams of C
B
(20 in.)
25 in.
z−+i j
yz
NN T++jk
to zero:
25 in
AB
Tx
P−
yABA B
NT′+−j λ
o zero:
0
5 in.
AB
Tz
=
22
(20 in.)z
z
+
(12 in.)
in.
AB
5.0 lb)(9 in.)
25 in.
7
onnected by a
rods. If a 60-l
mine (a) the
orresponding
he equilibrium
Collars:
B:
zk

0
AB AB
T =λ
0
.
=
0
B
=
2
(25 in.)
12 in.z
=
=
25-in.-long w
lb force Q is
tension in
magnitude o
m of the system

T
wire and can sl
applied to col
the wire wh
of the force
m.
125.0 lb
AB
T=
45.0 lbP=
ide
llar
hen
P
(1)
(2)

W

W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

185. PROBLEM 2.138
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. Determine the distances x and z for
which the equilibrium of the system is maintained when
120 lbP
= and 60 lb.Q=

SOLUTION
See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below:
0
25 in.
AB
Tx
P
= = (1)
60 lb 0
25 in.
AB
Tz− = (2)
For
120 lb,P= Eq. (1) yields (25 in.)(20 lb)
AB
Tx= (1 ′)
From Eq. (2):
(25 in.)(60 lb)
AB
Tz= (2 ′)
Dividing Eq. (1′) by (2′),
2
x
z
= (3)
Now write
22 2 2
(20 in.) (25 in.)xz++ = (4)
Solving (3) and (4) simultaneously,

22
2
4 400 625
45
6.7082 in.
zz
z
z
++ =
=
=

From Eq. (3):
2 2(6.7082 in.)
13.4164 in.
xz
==
=


13.42 in., 6.71 in.xz
= = W

 
  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

186. PROBLEM 2F1
Two cables are tied together at C and loaded as shown.
Draw the free-body diagram needed to determine the
tension in AC and BC .


SOLUTION
Free-Body Diagram of Point C:
2
3
(1600 kg)(9.81 m/s )
15.6960(10 ) N
=15.696 kN
W
W
W
=
=




Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

187. PROBLEM 2.F2
Two forces of magnitude T A = 8 kips and T B = 15 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, draw the
free-body diagram
needed to determine the magnitudes of the forces T
C
and T
D.
SOLUTION
Free-Body Diagram of Point E:






Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

188. PROBLEM 2.F3
The 60-lb collar A can slide on a frictionless vertical rod and is connected
as shown to a 65-lb counterweight C. Draw the free-body diagram needed
to determine the value of h for which the system is in equilibrium.

SOLUTION
Free-Body Diagram of Point A:



  Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

189. SOLUT
Free-Bo





Free-B
o










TION
ody Diagram
ody Diagram
of Point B:
of Point
C:
P
A
sh
N
dr
de

1
1
250 N
8
tan
1
tan
2
E
AB
BC
W
θ
θ
−
−
=
=
=
Use this free
11
tan
CD
θ
−
=
Use this free
Then weight
PROBLEM 2
A chairlift has
hown. Knowin
N and that the s
raw the free
etermine the w
765 N 101
8.25
30.510
14
10
22.620
24
+=
= °
=°
body to determ
1.1
10.3889
6
=°
body to deter
m
of skier W S is
2.F4
been stopped
ng that each c
skier in chair E
e-body diagra
weight of the s
15N
°
mine T AB and

mine T CD and
found by
S
W
=
d in the positi
chair weighs 2
E weighs 765
ams needed
skier in chair F
TBC.
W
F.
250 N
F
W
=−
ion
250
N,
to
F.
W Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

190. PROBLEM 2.F5
Three cables are used to tether a balloon as shown. Knowing that
the tension in cable AC is 444 N, draw the free-body diagram
needed to determine the vertical force P exerted by the balloon at
A.

SOLUTION
Free-Body Diagram of Point A:


Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

191. PROBLEM 2.F6
A container of mass m = 120 kg is supported by three cables
as shown. Draw the free-body diagram needed to determine
the tension in each cable

SOLUTION
Free-Body Diagram of Point A:

2
(120 kg)(9.81 m/s )
1177.2 N
W=
=


Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

192. PROBLEM 2.F7
A 150-lb cylinder is supported by two cables AC and BC that are
attached to the top of vertical posts. A horizontal force P, which
is perpendicular to the plane containing the posts, holds the
cylinder in the position shown. Draw the free-body diagram
needed to determine the magnitude of P and the force in each
cable.

SOLUTION
Free-Body Diagram of Point C:




Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

193. PROBLEM 2.F8
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D.
Knowing that the tension in wire AB is 630 lb, draw the
free-body diagram needed to determine the vertical force
P exerted by the tower on the pin at A.

SOLUTION
Free-Body Diagram of point A:



 Copyright © McGraw-Hill Education. Permission required for reproduction or display. www.elsolucionario.org

194. CHAPTER 3
www.elsolucionario.org

195. www.elsolucionario.org

196. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.1
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at B that creates a moment of equal
magnitude and opposite sense about E.

SOLUTION

(a) By definition,
2
80 kg(9.81 m/s ) 784.8 NWmg 
We have
: (784.8 N)(0.25 m)
EE
MM

196.2 N m
E
M 
(b) For the force at B to be the smallest, resulting in a moment (M
E) about E, the line of action of force
F
B must be perpendicular to the line connecting E to B. The sense of F B must be such that the force
produces a counterclockwise moment about E.
Note:
22
(0.85 m) (0.5 m) 0.98615 md
We have
: 196.2 N m (0.98615 m)
EB
MF

198.954 N
B
F
and
10.85 m
tan 59.534
0.5 m





or
199.0 N
B
F 59.5° 
www.elsolucionario.org

197. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.2
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at A that creates a moment of equal
magnitude and opposite sense about E, (c) the magnitude, sense,
and point of application on the bottom of the crate of the smallest
vertical force that creates a moment of equal magnitude and
opposite sense about E.

SOLUTION
First note. . .

2
mg (80 kg)(9.81m/s ) 784.8 NW 


(a) We have
/
(0.25 m)(784.8 N) 196.2 N m
EHE
MrW   or 196.2 N m
E
M 
(b) For F
A to be minimum, it must be perpendicular to the line
joining Points A and E. Then with F
A directed as shown,
we have
/min
() ().
EAEA
MrF


Where
22
/
(0.35 m) (0.5 m) 0.61033 m
AE
r
then
min
196.2 N m (0.61033 m)( )
A
F
or
min
() 321 N
A
F 
Also
0.35 m
tan
0.5 m
 or 35.0
min
() 321 N
A
F 35.0° 
(c) For F
vertical to be minimum, the perpendicular distance from its line of action to Point E must be
maximum. Thus, apply (F
vertical)min at Point D, and then

/min
min
() ( )
196.2 N m (0.85 m)( )
E D E vertical
vertical
MrF
F


or
min
( ) 231 N
vertical
F

at Point D 
www.elsolucionario.org

198. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.3
It is known that a vertical force of 200 lb is required to remove the
nail at C from the board. As the nail first starts moving, determine
(a) the moment about B of the force exerted on the nail, (b) the
magnitude of the force P that creates the same moment about B if
α  10°, (c) the smallest force P that creates the same moment
about B.

SOLUTION





(a) We have
/
(4 in.)(200 lb)
800 lb in.
BCBN
MrF


or
800 lb in.
B
M 
(b) By definition,
/
sin
10 (180 70 )
120
BAB
MrP 

 

Then
800 lb in. (18 in.) sin120P  
or
51.3 lbP 
(c) For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
min
/
B
AB
MdP
dr

or
min
800 lb in. (18 in.)P
or
min
44.4 lbP
min
44.4 lbP 20° 

www.elsolucionario.org

199. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.4
A 300-N force is applied at A as shown. Determine
(a) the moment of the 300-N force about D, (b) the
smallest force applied at B that creates the same moment
about D.

SOLUTION





(a)
(300 N)cos 25
271.89 N
(300 N)sin 25
126.785 N
(271.89 N) (126.785 N)
x
y
F
F




Fi j

(0.1 m) (0.2 m)
[ (0.1 m) (0.2 m) ] [(271.89 N) (126.785 N) ]
(12.6785 N m) (54.378 N m)
(41.700 N m)
D
D
DA 

   
   

rij
MrF
Mij i j
kk
k



41.7 N m
D
M


(b) The smallest force Q at B must be perpendicular to

DB
at 45°

()
41.700 N m (0.28284 m)
D
QDB
Q


M




147.4 NQ 45.0° 
www.elsolucionario.org

200. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.5
A 300-N force is applied at A as shown. Determine
(a) the moment of the 300-N force about D, (b) the
magnitude and sense of the horizontal force applied at C
that creates the same moment about D, (c) the smallest
force applied at C that creates the same moment about D.

SOLUTION
(a) See Problem 3.3 for the figure and analysis leading to the determination of M D

41.7 N m
D
M



(b) Since C is horizontal
CCi

(0.2 m) (0.125 m)
(0.125 m)
41.7 N m (0.125 m)( )
333.60 N
D
DC
CC
C
C
 
 


rij
Mri k

334 NC 
(c) The smallest force C must be perpendicular to DC; thus, it forms
 with the vertical


22
0.125 m
tan
0.2 m
32.0
( ); (0.2 m) (0.125 m)
0.23585 m
D
CDC DC





M


41.70 N m (0.23585 m)C 176.8 NC 58.0° 
www.elsolucionario.org

201. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.6
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α  25°, determine the moment of the force about Point B by
resolving the force into horizontal and vertical components.

SOLUTION
Free-Body Diagram of Rod AB:


(9 in.)cos65
3.8036 in.
(9 in.)sin 65
8.1568 in.
x
y










(20 lb cos25 ) ( 20 lb sin25 )
(18.1262 lb) (8.4524 lb)
xy
FF


Fij
ij
ij


/
( 3.8036 in.) (8.1568 in.)
AB
BA rij


/
( 3.8036 8.1568 ) (18.1262 8.4524 )
32.150 147.852
115.702 lb-in.
BAB

   


Mr F
ij ij
kk
115.7 lb-in.
B
M


www.elsolucionario.org

202. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.7
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α  25°, determine the moment of the force about Point B
by resolving the force into components along AB and in a direction perpendicular
to AB.

SOLUTION
Free-Body Diagram of Rod AB:



90 (65 25 )
50 









(20 lb)cos50
12.8558 lb
(9 in.)
(12.8558 lb)(9 in.)
115.702 lb-in.
B
Q
MQ





115.7 lb-in.
B
M


www.elsolucionario.org

203. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.8
A 20-lb force is applied to the control rod AB as shown. Knowing that the length
of the rod is 9 in. and that the moment of the force about B is 120 lb
· in. clockwise,
determine the value of α.

SOLUTION
Free-Body Diagram of Rod AB:




25







(20 lb)cosQ

and
()(9 in.)
B
MQ
Therefore,
120 lb-in. (20 lb)(cos )(9 in.)
120 lb-in.
cos
180 lb-in.




or
48.190

Therefore,
48.190 25
 23.2



www.elsolucionario.org

204. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.9
Rod AB is held in place by the cord AC. Knowing that the tension in
the cord is 1350 N and that c = 360 mm, determine the moment about
B of the force exerted by the cord at point A by resolving that force
into horizontal and vertical components applied (a) at point A, (b) at
point C.

SOLUTION
Free-Body Diagram of Rod AB:
(a)
22
1350 N (450) (600) 750 mmFAC

450 600
cos 0.6 sin 0.8
750 750
 

cos sin
(1350 N)0.6 (1350 N)0.8
(810 N) (1080N)
FF


Fi+j
ij
ij



/(0.45 m) (0.24 m)
AB rij

/ ( 0.45 0.24 ) (810 1080 )
BAB   Mr F i j i j
  

486 194.4 kk
(291.6 N m)  k 292 N m
B
M 
(b)
(810 N) (1080 N)Fi j

/
(0.36 m)
CB
rj

/
0.36 (810 1080 )
(291.6 N m)
BCB
  
 
Mr F j i j
k 292 N m
B
M 
www.elsolucionario.org

205. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.10
Rod AB is held in place by the cord AC. Knowing that c = 840 mm
and that the moment about B of the force exerted by the cord at point
A is 756 N·m, determine the tension in the cord.

SOLUTION
Free-Body Diagram of Rod AB:

22
(756 N m ) (450) (1080) 1170 mm
B
AC    Mk

450 1080
cos sin
1170 1170


cos sin
450 1080
1170 1170
FF
FF

Fi+j
ij



/(0.45 m) (0.24 m)
AB rij

/ ( 0.45 0.24 ) (450 1080 )
1170
BAB
F
   Mr F i j i j
  
(486 108)
1170
F
 kk
378
1170
F




k
Substituting for :
B
M

378
756
1170
F
2340 NF 
www.elsolucionario.org

206. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.11
The tailgate of a car is supported by the hydraulic lift BC. If
the lift exerts a 125-lb force directed along its centerline on
the ball and socket at B, determine the moment of the force
about A.

SOLUTION
First note
22
(12.0 in.) (2.33 in.)
12.2241in.
CB
d

Then
12.0 in.
cos
12.2241in.
2.33 in.
sin
12.2241in.



and
cos sin
125 lb
[(12.0 in.) (2.33 in.) ]
12.2241in.
CB CB CB
FF

Fij
ij

Now
/ABACB
Mr F
where
/
(15.3 in.) (12.0 in. 2.33 in.)
(15.3 in.) (14.33 in.)
BA


ri j
ij

Then
125 lb
[(15.3 in.) (14.33 in.) ] (12.0 2.33 )
12.2241in.
(1393.87 lb in.)
A
  

Mij ij
k

(116.156 lb ft) k or 116.2 lb ft
A
M


www.elsolucionario.org

207. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.12
The tailgate of a car is supported by the hydraulic lift BC. If
the lift exerts a 125-lb force directed along its centerline on
the ball and socket at
B, determine the moment of the force
about
A.

SOLUTION
First note
22
(17.2 in.) (7.62 in.)
18.8123 in.
CB
d

Then
17.2 in.
cos
18.8123 in.
7.62 in.
sin
18.8123 in.



and
(cos)(sin)
125 lb
[(17.2 in.) (7.62 in.) ]
18.8123 in.
CB CB CB
FF

Fij
ij

Now
/ABACB
Mr F
where
/
(20.5 in.) (4.38 in.)
BA
rij
Then
125 lb
[(20.5 in.) (4.38 in.) ] (17.2 7.62 )
18.8123 in.
A
 Mij ij

(1538.53 lb in.)
(128.2 lb ft )


k
k
or 128.2 lb ft
A
M



www.elsolucionario.org

208. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.13
It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force
directed down and to the left along the centerline of
AB. Determine the moment of
the force about
C.

SOLUTION
Using (a):

11
() ()
724
(0.056 m) 2500 N (0.042 m) 2500 N
25 25
140.0 N m
CABxABy
MyF xF










( a)

140.0 N m
C
M 

Using (b):

2
()
7
(0.2 m) 2500 N
25
140.0 N m
CABx
MyF









(b)

140.0 N m
C
M 
www.elsolucionario.org

209. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.14
It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force
directed down and to the left along the centerline of AB. Determine the moment of
the force about C.

SOLUTION
Using (a):

11
() ()
724
(0.056 m) 2500 N (0.042 m) 2500 N
25 25
61.6 N m
CABxABy
MyF xF 

   


 



(a)

61.6 N m
C
M 



Using (b):

2
()
7
(0.088 m) 2500 N
25
61.6 N m
CABx
MyF




 



(b)


61.6 N m
C
M 
www.elsolucionario.org

210. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.15
Form the vector products B × C and B × C, where B  B, and use the
results obtained to prove the identity
11
sin cos sin ( ) sin ( ).
22
 

SOLUTION
Note: (cos sin )
(cos sin )
(cos sin )
B
B
C
 
 
 



Bij
Bij
Cij
By definition,
||sin()BC
 BC (1)

||sin()BC
 BC (2)
Now
(cos sin ) (cos sin )BC
    BC i j i j

(cos sin sin cos )BC
 k (3)
and
(cos sin ) (cos sin )BC
   BC i j i j

(cos sin sin cos )BC
 k (4)
Equating the magnitudes of
BCfrom Equations (1) and (3) yields:

sin( ) (cos sin sin cos )BC BC
 (5)
Similarly, equating the magnitudes of BC from Equations (2) and (4) yields:

sin( ) (cos sin sin cos )BC BC
 (6)
Adding Equations (5) and (6) gives:

sin( ) sin( ) 2cos sin
  
or
11
sin cos sin( ) sin( )
22
    


www.elsolucionario.org

211. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.16
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram
when (a) P = –8i + 4j – 4k and Q = 3i + 3j + 6k, (b) P = 7i – 6j – 3k and Q = –3i + 6j – 2k

SOLUTION
(a) We have ||APQ
where
844
 Pijk

336
Qijk
Then 84 4
336
[(24 12) ( 12 48) ( 24 12) ]
(36) (36) ( 36)
 


ijk
PQ
ijk
ij k


22 2
(36) (36) ( 36)A or 62.4A 
(
b) We have ||A
PQ
where
763
Pijk

362
 Qijk
Then 763
36 2
[(12 18) (9 14) (42 18) ]
(30) (23) (24)
  



ijk
PQ
ij k
ijk


222
(30) (23) (24)A or 44.8A 


www.elsolucionario.org

212. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.17
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are
equal to, respectively, (
a) 2i + 3j – 6k and 5i – 8j – 6k,
(
b) 4i – 4j + 3k and –3i + 7j – 5k.

SOLUTION
(a) We have
||



AB
λ
AB

where
236
Aijk

586
Bijk
Then 23 6
586
(18 15) (30 12) (16 15)
(33 18 31)
 

     
  
ijk
AB
ijk
ijk

and
222
| | (33) (18) (31) 2374   AB

(33 18 31)
2374


ijk
λ
or 0.677 0.369 0.636
 λ ijk 
(
b) We have
||



AB
λ
AB

where
443
Aijk

375
 Bijk
Then 443
37 5
(20 21) ( 9 20) (28 12)
(1116)
 

  
 
ijk
AB
ijk
ijk

and
22
| | ( 1) (11) (16) 378  AB

(1116)
378


ijk
λ
or 0.0514 0.566 0.823
λ ijk 

www.elsolucionario.org

213. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.18
A line passes through the points (12 m, 8 m) and (–3 m, –5 m). Determine the perpendicular distance d from
the line to the origin O of the system of coordinates.

SOLUTION

22
[12 m ( 3 m)] [8 m ( 5 m)]
394 m
AB
d


Assume that a force
F, or magnitude F(N), acts at Point A and is
directed from A to B.
Then
AB
FF
where
1
(15 13 )
394
BA
AB
AB
d



rr
ij


By definition,
||
OA
dFMrF
where
(3 m) (5 m)
A
 ri+j
Then
[( 3 m) ( 5 m) ] [(15 m) (13 m) ]
394m
[39 75] Nm
394
36
Nm
394
O
F
F
F
   
 



Mi+j ij
kk
k

Finally,
36
()
394
FdF





36
m
394
d 1.184 md 

www.elsolucionario.org

214. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.19
Determine the moment about the origin O of the force F  4i  3j  5k that acts at a Point A. Assume that
the position vector of
A is (a) r  2i  3j  4k, ( b) r  8i  6j  10k, (c) r  8i  6j  5k.

SOLUTION

O
MrF
(
a)
23 4
435
O


i
jk
M


(15 12) ( 16 10) ( 6 12)  ijk 32618
O
 Mijk 
(
b)
86 10
435
O
 

i
jk
M


(30 30) ( 40 40) (24 24)  ijk 0
O
M 
(
c)
865
435
O


i
jk
M


(30 15) (20 40) (24 24)     ij k 15 20
O
 Mij 
Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the
determinant are proportional.

www.elsolucionario.org

215. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.20
Determine the moment about the origin O of the force F  2i  3j  4k that acts at a Point A. Assume that
the position vector of
A is (a) r  3i  6j  5k, ( b) r  i  4j  2k, ( c) r  4i  6j  8k.

SOLUTION

O
MrF
(
a)
365
23 4
O
 
ijk
M


(24 15) (10 12) (9 12)  ijk 92221
O
 Mijk 
(
b)
142
23 4
O
 
ijk
M


(16 6) ( 4 4) (3 8)ijk 22 11
O
Mik 
(
c)
46 8
23 4
O
 
ijk
M


( 24 24) ( 16 16) (12 12)     ijk 0
O
M 
Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the
determinant are proportional.

www.elsolucionario.org

216. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.21
Before the trunk of a large tree is felled, cables AB and BC are
attached as shown. Knowing that the tensions in cables
AB
and BC are 555 N and 660 N, respectively, determine the
moment about
O of the resultant force exerted on the tree by
the cables at
B.

SOLUTION













222
222
( 0.75 m) ( 7 m) (6 m) 9.25 m
(4.25 m) ( 7 m) (1 m) 8.25 m
BA
BC
d
d
   



Have
AA
A
555 N
( 0.75 m 7 m 6 m )
9.25 m
BB
B
BA
T
d
Tijk




(45 N) (420 N) (360 N)
BA
  Tijk


660 N
(4.25 m 7 m )
8.25 m
BC BC
BC
BC
T
d
 Ti jk



(340 N) (540 N) (80 N)
BC
Tijk

BA BC
RT T

   295 N 980 N 440 N  Ri j k
//
where (7 m)
OBO BO
 Mr R r j
070Nm
295 980 440
O


ijk
M

  3080 N m 2065 N m ik
  3080 N m 2070 N m
O
Mik 
www.elsolucionario.org

217. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.22
The 12-ft boom AB has a fixed end A. A steel cable is stretched from
the free end
B of the boom to a point C located on the vertical wall. If
the tension in the cable is 380 lb, determine the moment about
A of the
force exerted by the cable at
B.

SOLUTION
First note
222
(12) (4.8) (8)
15.2 ft
BC
d  

Then
380 lb
(12 4.8 8)
15.2
BC
Tijk
We have
/
A BA BC
Mr T
where
/
(12 ft )
BA
ri
Then
380 lb
(12 ft) ( 12 4.8 8 )
15.2
A
 Mi ijk
or
(2400 lb ft) (1440 lb ft)
A
Mjk 


www.elsolucionario.org

218. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.23
A 200-N force is applied as shown to the bracket ABC.
Determine the moment of the force about
A.

SOLUTION
We have
/A CA C
Mr F
where
/
(0.06 m) (0.075 m)
(200 N)cos 30 (200 N)sin 30
CA
C

   
rij
Fjk
Then
200 0.06 0.075 0
0 cos30 sin 30
200[(0.075sin 30 ) (0.06sin 30 ) (0.06cos 30 ) ]
A

 

ij k
M
ijk

or
(7.50 N m) (6.00 N m) (10.39 N m)
A
 Mijk 
www.elsolucionario.org

219. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.24
The wire AE is stretched between the corners A and E of a
bent plate. Knowing that the tension in the wire is 435 N,
determine the moment about
O of the force exerted by the
wire (
a) on corner A, (b) on corner E.

SOLUTION

222
(0.21 m) (0.16 m) (0.12 m)
(0.21 m) ( 0.16 m) (0.12 m) 0.29 m
AE
AE


ijk


(
a)
0.21 0.16 0.12
(435 N)
0.29
(315 N) (240 N) (180 N)
AAAE
AE
FF
AE




F
ijk
ijk



/
(0.09 m) (0.16 m)
AO
 rij

0.09 0.16 0
315 240 180
O


i
jk
M


28.8 16.20 (21.6 50.4) ij k (28.8 N m) (16.20 N m) (28.8 N m)
O
 Mijk 
(b)
(315 N) (240 N) (180 N)
EA
   FF i j k

/
(0.12 m) (0.12 m)
EO
rik

0.12 0 0.12
315 240 180
O


i
jk
M


28.8 ( 37.8 21.6) 28.8    ijk (28.8 N m) (16.20 N m) (28.8 N m)
O
     Mijk 


www.elsolucionario.org

220. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.25
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the
resulting force in the line is 6 lb. Determine the moment about
A of the force exerted by the line at B.


SOLUTION
We have (6 lb)cos 8 5.9416 lb
xz
T
Then
sin 30 2.9708 lb
sin 8 0.83504 lb
cos 30 5.1456 lb
xxz
yBC
zxz
TT
TT
TT



Now
/ABABC
Mr T
where
/
(6sin 45 ) (6cos 45 )
6 ft
()
2
BA


rjk
jk
Then
6
01 1
2
2.9708 0.83504 5.1456
666
( 5.1456 0.83504) (2.9708) (2.9708)
222
A


   
ijk
M
ijk

or
(25.4 lb ft) (12.60 lb ft) (12.60 lb ft)
A
     Mijk 


www.elsolucionario.org

221. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.26
A precast concrete wall section is temporarily held by two
cables as shown. Knowing that the tension in cable BD is 900
N, determine the moment about Point O of the force exerted by
the cable at B.

SOLUTION
where 900 N
BD
FF
BD
F



222
(1 m) (2 m) (2 m)
( 1 m) ( 2 m) (2 m)
3 m
BD
BD
  
  

ijk



/
22
(900 N)
3
(300 N) (600 N) (600 N)
(2.5 m) (2 m)
BO
 

  

ijk
F
ijk
rij


/OBO
Mr F

2.5 2 0
300 600 600
1200 1500 ( 1500 600)



ijk
ij k

(1200 N m) (1500 N m) (900 N m)
O
Mijk 

www.elsolucionario.org

222. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.27
In Prob. 3.22, determine the perpendicular distance from point A to
cable
BC.
PROBLEM 3.22 The 12-ft boom
AB has a fixed end A. A steel cable
is stretched from the free end
B of the boom to a point C located on
the vertical wall. If the tension in the cable is 380 lb, determine the
moment about
A of the force exerted by the cable at B.

SOLUTION
From the solution to problem 3.22
22
(2400) (1440)
2798.9 lb ft
A
M


But
AB
M Fd

A
B
M
d
F

2798.9 lb ft
380 lb
d


or
7.37 ftd 


www.elsolucionario.org

223. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.28
In Prob. 3.24, determine the perpendicular distance from point
O to wire AE.
PROBLEM 3.24 The wire
AE is stretched between the
corners
A and E of a bent plate. Knowing that the tension in
the wire is 435 N, determine the moment about
O of the force
exerted by the wire (
a) on corner A, (b) on corner E.

SOLUTION
From the solution to Prob. 3.24

222
(28.8 N m) (16.20 N m) (28.8 N m)
(28.8) (16.20) (28.8)
43.8329 N m
O
O
M
 


Mijk
But
or
O
OA
A
M
MFd d
F

43.8329 N m
435 N
0.100765 m
d





100.8 mmd 


www.elsolucionario.org

224. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.29
In Prob. 3.24, determine the perpendicular distance from point
B to wire AE.
PROBLEM 3.24 The wire
AE is stretched between the
corners
A and E of a bent plate. Knowing that the tension in
the wire is 435 N, determine the moment about
O of the force
exerted by the wire (
a) on corner A, (b) on corner E.

SOLUTION
From the solution to Prob. 3.24

/
/
(315 N) (240 N) (180 N)
(0.210 m)
0.21 (315 240 180 )
A
AB
BABA


   
Fijk
ri
Mr F i i j k


50.4 37.8kj

22
(50.4) (37.8)
63.0 N m
B
M



or
B
BA
A
M
MFd d
F

63.0 N m
435 N
0.144829 m
d





144.8 mmd 
www.elsolucionario.org

225. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.30
In Prob. 3.25, determine the perpendicular distance from point A to a line drawn through points B and C.
PROBLEM 3.25 A 6-ft-long fishing rod
AB is securely anchored in the sand of a beach. After a fish
takes the bait, the resulting force in the line is 6 lb. Determine the moment about
A of the force exerted by
the line at
B.


SOLUTION
From the solution to Prob. 3.25:

222
(25.4 lb ft) (12.60 lb ft) (12.60 lb ft)
( 25.4) ( 12.60) ( 12.60)
31.027 lb ft
A
A
M
     
   

Mijk


or
A
ABC
BC
M
MTd d
T


31.027 lb ft
6 lb
5.1712 ft





5.17 ftd 
www.elsolucionario.org

226. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.31
In Prob. 3.25, determine the perpendicular distance from point D to a line drawn through points B and C.
PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish
takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by
the line at B.


SOLUTION


6 ft
6 lb
BC
AB
T







We have
||
DBC
TdM
where d  perpendicular distance from D to line BC.

//
(6sin 45 ft) (4.2426 ft)
DBDBC BD
  Mr T r j

: ( ) (6 lb)cos8 sin 30 2.9708 lb
BC BC x
T T

( ) (6 lb)sin8 0.83504 lb
( ) (6 lb)cos8 cos30 5.1456 lb
BC y
BC z
T
T
 
  


(2.9708 lb) (0.83504 lb) (5.1456 lb)
04.2426 0
2.9708 0.83504 5.1456
(21.831 lb ft) (12.6039 lb ft)
BC
D
 


   
Tijk
ijk
M
i


22
| | ( 21.831) ( 12.6039) 25.208 lb ft
D
M   

25.208 lb ft (6 lb)d
 4.20 ftd 
www.elsolucionario.org

227. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.32
In Prob. 3.26, determine the perpendicular distance from point
O to cable BD.
PROBLEM 3.26 A precast concrete wall section is
temporarily held by two cables as shown. Knowing that the
tension in cable BD is 900 N, determine the moment about
Point O of the force exerted by the cable at B.

SOLUTION
From the solution to Prob. 3.26 we have

222
(1200 N m) (1500 N m) (900 N m)
(1200) ( 1500) ( 900) 2121.3 N m
O
O
O
O
M
M
MFd d
F



Mijk


2121.3 N m
900 N



2.36 md 
www.elsolucionario.org

228. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.33
In Prob. 3.26, determine the perpendicular distance from point C
to cable
BD.
PROBLEM 3.26 A precast concrete wall section is
temporarily held by two cables as shown. Knowing that the
tension in cable
BD is 900 N, determine the moment about
Point
O of the force exerted by the cable at B.

SOLUTION
From the solution to Prob. 3.26 we have

/
/
(300 N) (600 N) (600 N)
(2 m)
(2 m) ( 300 N 600 N 600 N )
BC
CBC
  

   
Fijk
rj
Mr F j i j k


(600 N m) (1200 N m)
  ki

22
(600) (1200) 1341.64 N m
C
M 

C
C
M
MFd d
F

1341.64 N m
900 N



1.491 md 
www.elsolucionario.org

229. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.34
Determine the value of a that minimizes the
perpendicular distance from Point
C to a
section of pipeline that passes through Points
A and B.

SOLUTION
Assuming a force F acts along AB,

/
||| | ()
CAC
FdMr F
where
dperpendicular distance from C to line AB

222
/
(24ft) (24ft) (28ft)
(24) (24) (28) ft
(6) (6) (7)
11
(3 ft) (10 ft) ( 10 ft)
31010
11
66 7
[(10 6 ) (81 6 ) 78 ]
11
AB
AC
C
F
F
F
a
F
a
F
aa





 


  
Fλ
ijk
ijk
rij k
ij k
M
ijk

Since
222
//
||||or||()
CAC AC
dF MrF rF

22221
(10 6 ) (81 6 ) (78)
121
aa d 

Setting
2
()0
d
da
d to find a to minimize d:

1
[2(6)(10 6 ) 2( 6)(81 6 )] 0
121
aa  

Solving
5.92 fta or 5.92 fta 
www.elsolucionario.org

230. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.35
Given the vectors P = 2i + 3j – k, Q = 5i – 4j + 3k, and S = –3i + 2j – 5k, compute the scalar products
P · Q, P · S, and Q · S.

SOLUTION

(2 3 ) (5 4 3 )
(2)(5) (3)( 4) ( 1)(3)
10 12 3
   


PQ i j k i j k


5PQ 

(2 3 ) ( 3 2 5 )
(2)( 3) (3)(2) ( 1)( 5)
665
    
 
  
PS i j k i j k


5PS 

(5 4 3 ) ( 3 2 5 )
(5)( 3) ( 4)(2) (3)( 5)
15 8 15
    
 
  
QS i j k i j k


38QS 


www.elsolucionario.org

231. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.36
Form the scalar product B · C and use the result obtained to prove the
identity
cos (
α  β)  cos α cos β  sin α sin β .

SOLUTION
cos sinBB Bij (1)

cos sinCC
 Cij (2)
By definition:

cos( )BC
 BC (3)
From (1) and (2):

( cos sin ) ( cos sin )
(cos cos sin sin )
BBCC
BC
  
 
   

BC i j i j
(4)
Equating the right-hand members of (3) and (4),

cos( ) cos cos sin sin
 


www.elsolucionario.org

232. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.37
Three cables are used to support a container as shown.
Determine the angle formed by cables
AB and AD.

SOLUTION
First note:
22
222
(450 mm) (600 mm)
750 mm
( 500 mm) (600 mm) (360 mm)
860 mm
AB
AD


  


and
(450 mm) (600 mm)
( 500 mm) (600 mm) (360 mm)
AB
AD

  
ij
ijk


By definition,
()()cosAB AD AB AD



(450 600 ) ( 500 600 360 ) (750)(860)cos
 ij ijk

(450)( 500) (600)(600) (0)(360) (750)(860)cos
  
or
cos 0.20930
 77.9 
www.elsolucionario.org

233. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.38
Three cables are used to support a container as shown.
Determine the angle formed by cables AC and AD.

SOLUTION
First note:
22
222
(600 mm) ( 320 mm)
680 mm
( 500 mm) (600 mm) (360 mm)
860 mm
AC
AD


  


and
(600 mm) ( 320 mm)
( 500 mm) (600 mm) (360 mm)
AC
AD

  
jk
ijk



By definition,
()()cosAC AD AC AD

 

(600 320 ) ( 500 600 360 ) (680)(860)cos
jk i jk

0( 500) (600)(600) ( 320)(360) (680)(860)cos
  

cos 0.41860
 65.3 
www.elsolucionario.org

234. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.39
Knowing that the tension in cable AC is 280 lb, determine
(a) the angle between cable AC and the boom AB, (b) the
projection on AB of the force exerted by cable AC at point
A.

SOLUTION
(a) First note
222
222
(6) (2) (3)
7.00 ft
(6) (4.5) (0)
7.50 ft
AC
AB
  

  

and (6 ft) (2 ft) (3 ft)
(6 ft) (4.5 ft)
AC
AB
  
 
ijk
ij


By definition
()()cosAC AB AC AB

 
or
( 6 2 3 ) ( 6 4.5 ) (7.00)(7.50) cos
    ijk i j
or
( 6)( 6) (2)( 4.5) (3)(0) 52.50cos
   
or
cos 0.51429
 or 59.0 
(b) We have
()
cos
(280 lb)(0.51429)
AC AB AC AB
AC
T
T




T
or
( ) 144.0 lb
AC AB
T  

www.elsolucionario.org

235. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.40
Knowing that the tension in cable AD is 180 lb, determine
(a) the angle between cable AD and the boom AB, (b) the
projection on AB of the force exerted by cable AD at point
A.

SOLUTION
(a) First note
22 2
222
(6) (3) (6)
9.00 ft
(6) (4.5) (0)
7.50 ft
AD
AB
  

  

and (6 ft) (3 ft) (6 ft)
(6 ft) (4.5 ft)
AD
AB
  
 
ijk
ij


By definition,
()()cosAD AB AD AB

 


( 6 3 6 ) ( 6 4.5 ) (9.00)(7.50)cos
( 6)( 6) (3)( 4.5) ( 6)(0) 67.50cos


   
   ijk i j

1
cos
3

 70.5 
(b)
()
cos
AD AB AD AB
AD
T
T

 T

1
(180 lb)
3




( ) 60.0 lb
AD AB
T  

www.elsolucionario.org

236. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.41
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope AB is 540 N,
determine (a) the angle between rope AB and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.

SOLUTION
First note:
22 2
222
( 3) (3) ( 1.5) 4.5 m
( 0.08) (0.38) (0.16) 0.42 m
BA
BD
   
   

Then
(3 3 1.5)
4.5
(2 2 )
3
1
( 0.08 0.38 0.16 )
0.42
1
(4 19 8)
21
BA
BA
BA
BD
T
T
BD
BD


   

Tijk
ijk
ijk
ijk



(a) We have
cos
BA BD BA
T
T
or
1
(2 2 ) (4 19 8) cos
321
BA
BA
T
T
     ijk i jk
or
1
cos [( 2)( 4) (2)(19) ( 1)(8)]
63
0.60317
 


or
52.9
 
(b) We have
()
cos
(540 N)(0.60317)
BA BD BA BD
BA
T
T




T
or
() 326 N
BA BD
T  
www.elsolucionario.org

237. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.42
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope BC is 490 N,
determine (a) the angle between rope BC and
the stake, (b) the projection on the stake of the
force exerted by rope BC at Point B.

SOLUTION
First note:
22 2
222
(1) (3) ( 1.5) 3.5 m
( 0.08) (0.38) (0.16) 0.42 m
BC
BD

   


(31.5)
3.5
(2 6 3 )
7
1
( 0.08 0.38 0.16 )
0.42
1
(4 19 8)
21
BC
BC
BC
BD
T
T
BD
BD



   
Tijk
ijk
ijk
ijk


(a)
cos
BC BD BC
T
 T

1
(263) (4198) cos
721
BC
BC
T
T
    ijk i jk

1
cos [(2)( 4) (6)(19) ( 3)(8)]
147
0.55782




56.1
 
(b)
()
cos
(490 N)(0.55782)
BC BD BC BD
BC
T
T 



T

( ) 273 N
BC BD
T  
www.elsolucionario.org

238. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.43
The 20-in. tube AB can slide along a horizontal rod. The ends A
and B of the tube are connected by elastic cords to the fixed point
C. For the position corresponding to x  11 in., determine the
angle formed by the two cords (a) using Eq. (3.32), (b) applying
the law of cosines to triangle ABC.

SOLUTION
(a) Using Eq. (3.32):

222
222
11 12 24
(11) ( 12) (24) 29 in.
31 12 24
(31) ( 12) (24) 41 in.
CA
CA
CB
CB




ijk
ijk




cos
()()
(1112 24)(3112 24)
(29)(41)
(11)(31) ( 12)( 12) (24)(24)
(29)(41)
0.89235
CA CB
CA CB


  

  


ijk ijk
 

26.8
 
(b) Law of cosines:

222
222
( ) ( ) ( ) 2( )( )cos
(20) (29) (41) 2(29)(41)cos
cos 0.89235AB CA CB CA CB








26.8
 
www.elsolucionario.org

239. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.44
Solve Prob. 3.43 for the position corresponding to x  4 in.
PROBLEM 3.43 The 20-in. tube AB can slide along a horizontal
rod. The ends A and B of the tube are connected by elastic cords
to the fixed point C. For the position corresponding to x  11 in.,
determine the angle formed by the two cords (a) using Eq. (3.32),
(b) applying the law of cosines to triangle ABC.

SOLUTION
(a) Using Eq. (3.32):

222
222
412 24
(4) ( 12) (24) 27.129 in.
24 12 24
(24) ( 12) (24) 36 in.
CA
CA
CB
CB
 



ijk
ijk




cos
()()
(4 12 24 ) (24 12 24 )
(27.129)(36)
0.83551
CA CB
CA CB


  


i
jki jk
 
33.3 
(b) Law of cosines:

222
222
()()()2()()cos
(20) (27.129) (36) 2(27.129)(36)cos
cos 0.83551AB CA CB CA CB 







33.3
 
www.elsolucionario.org

240. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.45
Determine the volume of the parallelepiped of Fig. 3.25 when
(
a) P  4i  3j  2k, Q  2i  5j  k, and S  7i  j  k,
(
b) P  5i  j  6k, Q  2i  3j  k, and S  3i  2j  4k.

SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
(
a)
3
Vol. ( )
432
251in.
711
(202147064)
67
 

 

   

PQ S
or Volume
67.0 
(
b)
3
Vol. ( )
516
231in.
324
(60 3 24 54 8 10)
111
 





PQS
or Volume
111.0 
www.elsolucionario.org

241. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.46
Given the vectors P = 3i – j + k, Q = 4i + Q
yj – 2k, and S = 2i – 2j + 2k, determine the
value of Q
y for which the three vectors are coplanar.

SOLUTION
If P, Q, and S are coplanar, then P must be perpendicular to ().QS

()0PQ S
(or, the volume of a parallelepiped defined by P, Q, and S is zero).
Then
311
4Q 2 0
222
y



or
64828120
yy
QQ   2
y
Q
 
www.elsolucionario.org

242. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.47
A crane is oriented so that the end of the 25-m boom AO lies in
the yz plane. At the instant shown, the tension in cable AB is
4 kN. Determine the moment about each of the coordinate axes
of the force exerted on A by cable AB.

SOLUTION


22
22
()( )
(25 m) (15 m)
20 m
OC OA AC





(15 m) (20 m)
A
rjk
22
(2.5 m) (15 m) 15.2069 m
2.5 15
(4 kN)
15.2069
(0.65760 kN) (3.9456 kN)
AB
AB
P
AB





P
ij
ij


Using Eq. (3.11):

01520
0.65760 3.9456 0
(78.912 kN m) (13.1520 kN m) (9.8640 kN m)
OA
O

  
ijk
MrP
Mi+j-k

But
Ox y z
MMMMi
jk
Therefore,
78.9 kN m, 13.15 kN m, 9.86 kN m
xy z
MM M
   

www.elsolucionario.org

243. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.48
The 25-m crane boom AO lies in the yz plane. Determine the
maximum permissible tension in cable AB if the absolute value
of moments about the coordinate axes of the force exerted on A
by cable AB must be
|Mx| ≤ 60 kN·m, |M y| ≤ 12 kN·m, |M z| ≤ 8 kN·m


22
22
()()
(25 m) (15 m)
20 m
OC OA AC




(15 m) (20 m)
A
rjk
22
(2.5 m) (15 m) 15.2069 m
2.5 15
( )
15.2069
(0.164399 kN) (0.98639 kN)AB
AB
P
AB
P
PP




P
ij
ij


Using Eq. (3.11):

01520
0.163499 0.98639 0
(19.7278 ) (3.2700 ) (2.4525 )
OA
O
PP
PPP



ijk
MrP
Mi+j-k

But
60 kN m: 19.7278 60 3.04 kN
12 kN m: 3.270 12 3.67 kN
8 kN m: 2.4525 8 3.15 kN
x
y
z
MPP
MPP
MPP

  
  
<

Therefore, the maximum permissible tension is
3.04 kNP 

www.elsolucionario.org

244. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.49
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that
25 ,

M
x61lb ft,
 and 43 lb ft,
z
M determine  and d.

SOLUTION
We have
/
:
OAO O
Mr FM
where
/
(4 in.) (11in.) ( )
(cos cos sin cos sin )
AO
d
F
 
   
rijk
Fijk
For
70 lb, 25F


(70 lb)[(0.90631cos ) 0.42262 (0.90631sin ) ]
(70 lb) 4 11 in.
0.90631cos 0.42262 0.90631sin
(70 lb)[(9.9694sin 0.42262 ) ( 0.90631 cos 3.6252sin )
(1.69048 9.9694cos ) ] in.
O
d
dd
 

 


 



Fijk
ijk
M
ij
k

and
(70 lb)(9.9694sin 0.42262 ) in. (61 lb ft)(12 in./ft)
x
Md  (1)

(70 lb)( 0.90631 cos 3.6252sin ) in.
y
Md   (2)

(70 lb)(1.69048 9.9694cos ) in. 43 lb ft(12 in./ft)
z
M   (3)
From Equation (3):
1634.33
cos 24.636
697.86





or 24.6
 
From Equation (1):
1022.90
34.577 in.
29.583
d




or 34.6 in.d 
www.elsolucionario.org

245. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.50
When a force F is applied to the handle of the valve
shown, its moments about the
x and z axes are,
respectively,
77 lb ft
x
M
 and 81 lb ft.
z
M  For
27din., determine the moment My of F about the y axis.

SOLUTION
We have
/
:
OAO O
Mr FM
Where
/
(4 in.) (11in.) (27 in.)
(cos cos sin cos sin )
AO
F
 
   
rijk
Fijk

411 27lbin.
cos cos sin cos sin
[(11cos sin 27sin )
(27cos cos 4cos sin )
(4sin 11cos cos ) ](lb in.)
O
F
F
  
 
 

 


 
 
ijk
M
i
j
k

and
(11cos sin 27sin )(lb in.)
x
MF  
  (1)

( 27cos cos 4cos sin ) (lb in.)
y
MF  
  (2)

(4sin 11cos cos ) (lb in.)
z
MF 
  (3)
Now, Equation (1)
1
cos sin 27sin
11
x
M
F
 




(4)
and Equation (3)
1
cos cos 4sin
11
z
MF
 




(5)
Substituting Equations (4) and (5) into Equation (2),

11
27 4sin 4 27sin
11 11
xz
y
MM
MF
FF

    
      
    

or
1
(27 4 )
11
yzx
M MM www.elsolucionario.org

246. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.50 (Continued)

Noting that the ratios
27
11
and
4
11
are the ratios of lengths, have

27 4
(81lbft) (77lbft)
11 11
226.82 lb ft
y
M  
 or
227 lb ft
y
M  

www.elsolucionario.org

247. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.51
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook
B. Knowing that the moments about
the
y and the z axes of the force exerted at B by portion AB of the
rope are, respectively, 120 N  m and 460 N  m, determine the
distance
a.

SOLUTION
First note (2.2 m) (3.2 m) ( m)BA a ijk

Now
/
D AD BA
Mr T
where
/
(2.2 m) (1.6 m)
(2.2 3.2 ) (N)
AD
BA
BA
BA
T
a
d


rij
Tijk
Then
2.2 1.6 0
2.2 3.2
{ 1.6 2.2 [(2.2)( 3.2) (1.6)(2.2)] }
BA
D
BA
BA
BA
T
d
a
T
aa
d


  
ijk
M
ij k

Thus
2.2 (N m)
10.56 (N m)
BA
y
BA
BA
z
BA
T
Ma
d
T
M
d

 

Then forming the ratio
y
z
M
M


2.2 (N m)
120 N m
460 N m 10.56 (N m)
BA
BA
BA
BA
T
d
T
d




 
or 1.252 ma 
www.elsolucionario.org

248. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.52
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook
B. Knowing that the man applies a
195-N force to end
A of the rope and that the moment of that force
about the
y axis is 132 N  m, determine the distance a.

SOLUTION
First note
222
2
(2.2) ( 3.2) ( )
15.08 m
BA
da
a


and
195 N
(2.2 3.2 )
BA
BA
a
d
Tijk
Now
/
()
yADBA
M jr T
where
/0
(2.2 m) (1.6 m)
A
rij
Then
010
195
2.2 1.6 0
2.2 3.2
195
(2.2 ) (N m)
y
BA
BA
M
d
a
a
d
 


Substituting for
My and dBA

2
195
132 N m (2.2 )
15.08
a
a



or
2
0.30769 15.08aa
Squaring both sides of the equation

22
0.094675(15.08 )aa

or
1.256 ma 
www.elsolucionario.org

249. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.53
A farmer uses cables and winch pullers B and E to plumb
one side of a small barn. If it is known that the sum of the
moments about the
x-axis of the forces exerted by the
cables on the barn at Points
A and D is equal to 4728 lb 
ft, determine the magnitude of T
DE when TAB  255 lb.

SOLUTION
The moment about the x-axis due to the two cable forces can be found using the z components of each
force acting at their intersection with the
xy plane (A and D). The x components of the forces are parallel
to the
x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y
components produce a moment about the
x-axis.
We have
: ( )( ) ( )( )
xABzA DEzD x
MT y T yM
where
()
()
12 12
255 lb
17
180 lb
AB z AB
AB AB
T
T



  



kT
k
ijk
k

()
()
1.5 14 12
18.5
0.64865
12 ft
14 ft
4728 lb ft
DE z DE
DE DE
DE
DE
A
D
x
T
T
T
T
y
y
M



 







kT
k
ijk
k


(180 lb)(12 ft) (0.64865 )(14 ft) 4728 lb ft
DE
T
and
282.79 lb
DE
T or 283 lb
DE
T 
www.elsolucionario.org

250. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.54
Solve Problem 3.53 when the tension in cable AB is 306
lb.
PROBLEM 3.53 A farmer uses cables and winch pullers
B and E to plumb one side of a small barn. If it is known
that the sum of the moments about the x-axis of the forces
exerted by the cables on the barn at Points A and D is equal
to 4728 lb  ft, determine the magnitude of T
DE when
T
AB  255 lb.

SOLUTION
The moment about the x-axis due to the two cable forces can be found using the z components of each
force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to
the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y
components produce a moment about the x-axis.
We have
: ( )( ) ( )( )
xABzA DEzD x
MT y T yM
Where
()
()
12 12
306 lb
17
216 lb
AB z AB
AB AB
T
T 

  



kT
k λ
ijk
k

()
()
1.5 14 12
18.5
0.64865
12 ft
14 ft
4728 lb ft
DE z DE
DE DE
DE
DE
A
D
x
T
T
T
T
y
y
M


 







kT
k λ
ijk
k


(216 lb)(12 ft) (0.64865 )(14 ft) 4728 lb ft
DE
T
and
235.21 lb
DE
T or 235 lb
DE
T 

www.elsolucionario.org

251. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.55
The 23-in. vertical rod CD is welded to the midpoint C of
the 50-in. rod
AB. Determine the moment about AB of the
235-lb force
P.

SOLUTION
(32 in.) (30 in.) (24 in.)ABijk



222
(32) ( 30) ( 24) 50 in.AB
0.64 0.60 0.48
AB
AB
AB
  
ik


We shall apply the force
P at Point G:

/
(5 in.) (30 in.)
GB
rik

(21 in.) (38 in.) (18 in.)DGijk



222
(21) ( 38) (18) 47 in.DG

21 38 18
(235 lb)
47
DG
P
DG

ijk
P



(105 lb) (190 lb) (90 lb)
 Pijk
The moment of
P about AB is given by Eq. (3.46):

/
0.64 0.60 0.48
()5 in.030 in.
105 lb 190 lb 90 lb
AB AB G B
P

 

Mr

0.64[0 (30 in.)( 190 lb)]
0.60[(30 in.)(105 lb) (5 in.)(90 lb)]
0.48[(5 in.)( 190 lb) 0]
2484 lb in.
AB
 


 
M

207 lb ft
AB
 M 
www.elsolucionario.org

252. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.56
The 23-in. vertical rod CD is welded to the midpoint C of
the 50-in. rod
AB. Determine the moment about AB of the
174-lb force Q.

SOLUTION
(32 in.) (30 in.) (24 in.)ABijk



222
(32) ( 30) ( 24) 50 in.AB
0.64 0.60 0.48
AB
AB
AB
  
ijk


We shall apply the force Q at Point
H:

/
(32 in.) (17 in.)
HB
 rij

(16 in.) (21 in.) (12 in.)DH  ijk



22 2
(16) (21) (12) 29 in.DH

16 21 12
(174 lb)
29
DH
DH
 
ijk
Q



(96 lb) (126 lb) (72 lb)Q
 ijk
The moment of Q about
AB is given by Eq. (3.46):

/
0.64 0.60 0.48
()32 in.17 in.0
96 lb 126 lb 72 lb
AB AB H B

 
 
MrQ
0.64[(17 in.)( 72 lb) 0]
0.60[(0 ( 32 in.)( 72 lb)]
0.48[( 32 in.)( 126 lb) (17 in.)( 96 lb)]
2119.7 lb in.
AB


   
 
M

176.6 lb ft
AB
 M 
www.elsolucionario.org

253. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.57
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at
B and is attached to
hooks at
G and H. Knowing that the tension in the cable is
450 N, determine the moment about the diagonal
AD of
the force exerted on the frame by portion
BH of the cable.

SOLUTION

/
()
ADADBABH
M rT
Where
/
1
(4 3 )
5
(0.5 m)
AD
BA


ik
ri

and
22 2
(0.375) (0.75) ( 0.75)
1.125 m
BH
d

Then
450 N
(0.375 0.75 0.75 )
1.125
(150 N) (300 N) (300 N)
BH


Tijk
ijk
Finally,
40 3
1
0.5 0 0
5
150 300 300
1
[( 3)(0.5)(300)]
5
AD
M




or
90.0 N m
AD
M  
www.elsolucionario.org

254. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.58
In Problem 3.57, determine the moment about the diagonal
AD of the force exerted on the frame by portion BG of the
cable.
PROBLEM 3.57 The frame
ACD is hinged at A and D
and is supported by a cable that passes through a ring at
B
and is attached to hooks at
G and H. Knowing that the
tension in the cable is 450 N, determine the moment about
the diagonal
AD of the force exerted on the frame by
portion
BH of the cable.

SOLUTION

/
()
ADADBABG
M rT
Where
/
1
(4 3 )
5
(0.5 m)
AD
BA


ik
rj

and
222
( 0.5) (0.925) ( 0.4)
1.125 m
BG  

Then
450 N
( 0.5 0.925 0.4 )
1.125
(200 N) (370 N) (160 N)
BG

  Tijk
ijk
Finally,
40 3
1
0.5 0 0
5
200 370 160
AD
M




1
[( 3)(0.5)(370)]
5

111.0 N m
AD
M  
www.elsolucionario.org

255. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.59
The triangular plate ABC is supported by ball-and-socket
joints at
B and D and is held in the position shown by
cables
AE and CF. If the force exerted by cable AE at A is
55 N, determine the moment of that force about the line
joining Points
D and B.

SOLUTION
First note:
AE AE
AE
T
AE

T



222
(0.9) ( 0.6) (0.2) 1.1 mAE
Then
55 N
(0.9 0.6 0.2 )
1.1
5[(9 N) (6 N) (2 N) ]
AE

Tijk
ijk
Also,
222
(1.2) ( 0.35) (0)
1.25 m
DB

Then
1
(1.2 0.35 )
1.25
1
(24 7 )
25
DB
DB
DB



ij
ij


Now
/
()
DBDBADAE
M  rT
where
/
(0.1 m) (0.2 m)
AD
 rjk
Then
24 7 0
1
(5) 0 0.1 0.2
25
962
1
( 4.8 12.6 28.8)
5
DB
M



 
or
2.28 N m
DB
M 
www.elsolucionario.org

256. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.60
The triangular plate ABC is supported by ball-and-socket
joints at B and D and is held in the position shown by
cables AE and CF. If the force exerted by cable CF at C is
33 N, determine the moment of that force about the line
joining Points D and B.

SOLUTION
First note:
CF CF
CF
T
CF

T



222
(0.6) ( 0.9) ( 0.2) 1.1 mCF
Then
33 N
(0.6 0.9 0.2 )
1.1
3[(6 N) (9 N) (2 N) ]
CF

Tijk
ijk
Also,
222
(1.2) ( 0.35) (0)
1.25 m
DB

Then
1
(1.2 0.35 )
1.25
1
(24 7 )
25
DB
DB
DB



ij
ij


Now
/
()
DBDBCDCF
M  rT
where
/
(0.2 m) (0.4 m)
CD
rjk
Then
24 7 0
1
(3) 0 0.2 0.4
25
692
3
( 9.6 16.8 86.4)
25
DB
M



 

or
9.50 N m
DB
M  
www.elsolucionario.org

257. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.61
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
about edge OA.

SOLUTION
We have
/
()
OA OA C O
M rP
From triangle OBC: ()
2
1
() ()tan30
2323
x
zx
a
OA
aa
OA OA


 

Since
222 2
() ()() ( )
xyz
OA OA OA OA
or
22
22
22
2
()
2 23
2
()
412 3
y
y
aa
aOA
aa
OA a a

  
 


Then
/
2
23 23
AO
aa
a
 rijk
and
121
23 23
OA
 ij k
/
(sin30) (cos30)
() ( 3)
2
BC
CO
aa P
PP
a
a
 
 

ik
Pik
ri



121
23 23
()
10 0 2
10 3
2
(1)( 3 )
23 2
OA
P
Ma
aP aP






 



2
OA
aP
M

www.elsolucionario.org

258. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.62
A regular tetrahedron has six edges of length a. (a) Show that
two opposite edges, such as OA and BC, are perpendicular to
each other. (b) Use this property and the result obtained in
Problem 3.61 to determine the perpendicular distance between
edges OA and BC.

SOLUTION
(a) For edge OA to be perpendicular to edge BC,

0OA BC
 
From triangle OBC:
()
2
1
() ()tan30
2323
()
2 23
x
zx
y
a
OA
aa
OA OA
aa
OA OA


 


  
 
ij k


and
(sin30) (cos30)
3
(3)
22 2
BC a a
aa a

  
ik
ikik

Then
() ( 3) 0
22 23
y
aa a
OA
 
  

ij kik

or
22
()(0) 0
44
0
y
aa
OA
OA BC


 

so that
OA
is perpendicular to .BC
 
(b) We have ,
OA
MPd with P acting along BC and d the perpendicular distance from OA
to .BC

From the results of Problem 3.57,

2
2
OA
Pa
M
Pa
Pd

 or
2
a
d

www.elsolucionario.org

259. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.63
Two forces
1
F and
2
F in space have the same magnitude F. Prove that the moment of
1
F about the line
of action of
2
F is equal to the moment of
2
F about the line of action of
1
F.

SOLUTION

First note that
111 2 22
andFFFF
Let
12
moment of M F about the line of action of
1
Fand
2
momentM of
1
F about the line of
action of
2
F.
Now, by definition,
11 / 2
1/ 22
22 / 1
2/ 11
()
()
()
()
BA
BA
AB
AB
M
F
M
F
 
 
 
 
rF
r
rF
r



Since
12 / /
11 / 2
22 / 1
and
()
()
AB BA
BA
BA
FF F
MF
MF
 
 
 
rr
r
r



Using Equation (3.39):
1/ 2 2 / 1
()( )
BA BA
rr 
so that
21/ 2
()
BA
MF r 
12 21
MM 


www.elsolucionario.org

260. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.64
In Prob. 3.55, determine the perpendicular distance
between rod AB and the line of action of P.
PROBLEM 3.55 The 23-in. vertical rod CD is welded to
the midpoint C of the 50-in. rod AB. Determine the
moment about AB of the 235-lb force P.

SOLUTION
(32 in.) (30 in.) (24 in.)ABijk



222
(32) ( 30) ( 24) 50 in.AB
0.64 0.60 0.48
AB
AB
AB
  
ijk



105 190 90
235
P
P
 
Pijk

Angle
 between AB and P:

cos
105 190 90
(0.64 0.60 0.48 )
235
0.58723
AB P


 

ijk
ijk



54.039
 
The moment of P about
AB may be obtained by multiplying the projection of P on a plane perpendicular
to
AB by the perpendicular distance d from AB to P:

(sin)
AB
P dM
From the solution to Prob. 3.55:
207 lb ft 2484 lb in.
AB
 M
We have
2484 lb in. (235 lb)(sin 54.039)d

13.06 in.d 
www.elsolucionario.org

261. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.65
In Prob. 3.56, determine the perpendicular distance
between rod
AB and the line of action of Q.
PROBLEM 3.56 The 23-in. vertical rod
CD is welded to
the midpoint
C of the 50-in. rod AB. Determine the
moment about
AB of the 174-lb force Q.

SOLUTION
(32 in.) (30 in.) (24 in.)ABijk



222
(32) ( 30) ( 24) 50 in.AB
0.64 0.60 0.48
AB
AB
AB
  
ijk



96 126 72
174
Q
Q

Qijk

Angle
 between AB and Q:

cos
( 96 126 72 )
(0.64 0.60 0.48 )
174
0.28000
AB Q

 
 

ijk
ijk


73.740
 
The moment of Q about
AB may be obtained by multiplying the projection of Q on a plane perpendicular
to
AB by the perpendicular distance d from AB to Q:

(sin)
AB
Qd
M
From the solution to Prob. 3.56:
176.6 lb ft 2119.2 lb in.
AB
 M

2119.2 lb in. (174 lb)(sin 73.740 )d 

12.69 in.d 
www.elsolucionario.org

262. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.66
In Problem 3.57, determine the perpendicular distance
between portion
BH of the cable and the diagonal AD.
PROBLEM 3.57 The frame
ACD is hinged at A and D
and is supported by a cable that passes through a ring at
B
and is attached to hooks at
G and H. Knowing that the
tension in the cable is 450 N, determine the moment about
the diagonal
AD of the force exerted on the frame by
portion
BH of the cable.

SOLUTION
From the solution to Problem 3.57: 450 N
(150 N) (300 N) (300 N)
BH
BH
T
Tijk

| | 90.0 N m
AD
M
 

1
(4 3 )
5
AD
ik
Based on the discussion of Section 3.11, it follows that only the perpendicular component of T
BH will
contribute to the moment of T
BH about line .
AD


Now
parallel
()
1
(150 300 300 ) (4 3 )
5
1
[(150)(4) ( 300)( 3)]
5
300 N
BH BH AD
T 
 


T
ijkik

Also,
parallel perpendicular
() ()
BH BH BH
TT T
so that
22
perpendicular
( ) (450) (300) 335.41 N
BH
T 
Since
AD
 and
perpendicular
()
BH
T are perpendicular, it follows that

perpendicular
()
AD BH
MdT
or
90.0 N m (335.41 N)d

0.26833 md
 0.268 md 
www.elsolucionario.org

263. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.67
In Problem 3.58, determine the perpendicular distance
between portion
BG of the cable and the diagonal AD.
PROBLEM 3.58 In Problem 3.57, determine the moment
about the diagonal
AD of the force exerted on the frame by
portion
BG of the cable.
PROBLEM 3.57 The frame
ACD is hinged at A and D
and is supported by a cable that passes through a ring at
B
and is attached to hooks at
G and H. Knowing that the
tension in the cable is 450 N, determine the moment about
the diagonal
AD of the force exerted on the frame by
portion
BH of the cable.

SOLUTION
From the solution to Problem 3.58: 450 N
(200 N) (370 N) (160 N)
BG
BG

  Tijk


||111 Nm
AD
M
 

1
(4 3 )
5
AD
ik
Based on the discussion of Section 3.11, it follows that only the perpendicular component of T
BG will
contribute to the moment of T
BG about line .
AD


Now
parallel
()
1
( 200 370 160 ) (4 3 )
5
1
[( 200)(4) ( 160)( 3)]
5
64 N
BG BG AD
T 
    
  

T
ijkik

Also,
parallel perpendicular
() ()
BG BG BG
TT T
so that
22
perpendicular
( ) (450) ( 64) 445.43 N
BG
T
Since
AD
 and
perpendicular
()
BG
T are perpendicular, it follows that

perpendicular
()
AD BG
MdT
or
111 N m (445.43 N)d

0.24920 md
 0.249 md 
www.elsolucionario.org

264. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.68
In Problem 3.59, determine the perpendicular distance
between cable
AE and the line joining Points D and B.
PROBLEM 3.59 The triangular plate
ABC is supported by
ball-and-socket joints at
B and D and is held in the position
shown by cables
AE and CF. If the force exerted by cable
AE at A is 55 N, determine the moment of that force about
the line joining Points
D and B.

SOLUTION
From the solution to Problem 3.59: 55 N
5[(9 N) (6 N) (2 N) ]
AE
AE

Tijk


||2.28 Nm
DB
M
 

1
(24 7 )
25
DB
 ij
Based on the discussion of Section 3.11, it follows that only the perpendicular component of T
AE will
contribute to the moment of T
AE about line .DB


Now
parallel
()
1
5(962) (247)
25
1
[(9)(24) ( 6)( 7)]
5
51.6 N
AE AE DB
T 
 


T
ijk ij

Also,
parallel perpendicular
() ()
AE AE AE
TT T
so that
22
perpendicular
( ) (55) (51.6) 19.0379 N
AE
 T
Since
DB
 and
perpendicular
()
AE
T are perpendicular, it follows that

perpendicular
()
DB AE
MdT
or
2.28 N m (19.0379 N)d


0.119761d
 0.1198 md 
www.elsolucionario.org

265. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.69
In Problem 3.60, determine the perpendicular distance
between cable
CF and the line joining Points D and B.
PROBLEM 3.60 The triangular plate
ABC is supported by
ball-and-socket joints at
B and D and is held in the position
shown by cables
AE and CF. If the force exerted by cable
CF at C is 33 N, determine the moment of that force about
the line joining Points
D and B.

SOLUTION
From the solution to Problem 3.60: 33 N
3[(6 N) (9 N) (2 N) ]
CF
CF

Tijk


||9.50 Nm
DB
M
 

1
(24 7 )
25
DB
 ij
Based on the discussion of Section 3.11, it follows that only the perpendicular component of T
CF will
contribute to the moment of T
CF about line .DB


Now
parallel
()
1
3(6 9 2 ) (24 7 )
25
3
[(6)(24) ( 9)( 7)]
25
24.84 N
CF CF DB

 


TT
ijk ij

Also,
parallel perpendicular
() ()
CF CF CF
TT T
so that
22
perpendicular
() (33)(24.84)
21.725 N
CF


T
Since
DB
 and
perpendicular
()
CF
T are perpendicular, it follows that

perpendicular
||()
DB CF
MdT
or
9.50 N m 21.725 Nd
or
0.437 md 
www.elsolucionario.org

266. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.70
Two 80-N forces are applied as shown to the corners B and D of
a rectangular plate. (a) Determine the moment of the couple
formed by the two forces by resolving each force into horizontal
and vertical components and adding the moments of the two
resulting couples. (b) Use the result obtained to determine the
perpendicular distance between lines BE and DF.

SOLUTION
(a) Resolving forces into components:

(80 N)sin50 61.284 NP



(80 N)cos50 51.423 NQ



(51.423 N)(0.5 m) (61.284 N)(0.3 m)
7.3263 N mM





7.33 N mM 
(b) Distance between lines BE and DF

FdM
or
7.3263 N m=(80 N)
0.091579 m
d
d


91.6 mmd 


www.elsolucionario.org

267. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.71
Two parallel 40-N forces are applied to a lever as shown.
Determine the moment of the couple formed by the two forces
(
a) by resolving each force into horizontal and vertical
components and adding the moments of the two resulting
couples, (
b) by using the perpendicular distance between the
two forces, (
c) by summing the moments of the two forces
about Point
A.

SOLUTION
(a) We have
12
:
Bxy
dC dCMM
where
1
2
(0.270 m)sin 55°
0.22117 m
(0.270 m)cos55
0.154866 m
d
d





(40 N)cos20
37.588 N
(40 N)sin 20
13.6808 N
x
y
C
C






(0.22117 m)(37.588 N) (0.154866 m)(13.6808 N)
(6.1946 N m)
 
 Mk k
k
or 6.19 N mM 
(
b) We have ()
40 N[(0.270 m)sin(55 20 )]( )
Fd

Mk
k

(6.1946 N m)
 k or 6.19 N mM 
(
c) We have
//
:( )
AA BABCAC
MrFrFrFM

(0.390 m)(40 N) cos55 sin 55 0
cos 20 sin 20 0
(0.660 m)(40 N) cos55 sin 55 0
cos 20 sin 20 0
(8.9478 N m 15.1424 N m)
(6.1946 N m)
M




 
ijk
ijk
k
k
or 6.19 N mM 
www.elsolucionario.org

268. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.72
Four 1
1
2
-in.-diameter pegs are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. (a) Determine the resultant couple acting on
the board. (b) If only one string is used, around which pegs
should it pass and in what directions should it be pulled to
create the same couple with the minimum tension in the string?
(c) What is the value of that minimum tension?

SOLUTION






(a) (60 lb)(10.5 in.) (40 lb)(13.5 in.)
630 lb in. 540 lb in.
M



1170 lb in.M


(b) With only one string, pegs A and D, or B and C should be used. We
have

9
tan 36.9 90 53.1
12
 
Direction of forces:
With pegs A and D:
53.1 
With pegs B and C:
53.1 
(c) The distance between the centers of the two pegs is

22
12 9 15 in.
Therefore, the perpendicular distance d between the forces is

3
15 in. 2 in.
4
16.5 in.
d






We must have
1170 lb in. (16.5 in.)MFd F



70.9 lbF 
www.elsolucionario.org

269. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.73
Four pegs of the same diameter are attached to a board as
shown. Two strings are passed around the pegs and pulled with
the forces indicated. Determine the diameter of the pegs
knowing that the resultant couple applied to the board is
1132.5 lb·in. counterclockwise.

SOLUTION
1132.5 lb in. [(9 ) in.](60 lb) [(12 ) in.](40 lb)
AD AD BC BC
MdF dF
dd

   
1.125 in.d 
www.elsolucionario.org

270. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.74
A piece of plywood in which several holes are being drilled
successively has been secured to a workbench by means of two
nails. Knowing that the drill exerts a 12-N·m couple on the
piece of plywood, determine the magnitude of the resulting
forces applied to the nails if they are located (
a) at A and B,
(
b) at B and C, (c) at A and C.

SOLUTION








(a)
12 N m (0.45 m)
MFd
F




26.7 NF 



(
b)
12 N m (0.24 m)
MFd
F




50.0 NF 

(
c)
22
(0.45 m) (0.24 m)
0.510 m
MFd d 


12 N m (0.510 m)F


23.5 NF 
www.elsolucionario.org

271. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.75
The two shafts of a speed-reducer unit are subjected to
couples of magnitude
M1  15 lb·ft and M2  3 lb·ft,
respectively. Replace the two couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.

SOLUTION

1
(15 lb ft )M k

2
(3 lb ft)M i

22
12
22
(15) (3)
15.30 lb ft
MMM



15
tan 5
3
x


78.7
x


90
y


90 78.7
11.30
z
  


15.30 lb ft; 78.7 , 90.0 , 11.30
xyz
M 
  
www.elsolucionario.org

272. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.76
If 0,P
 replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.

SOLUTION

121 2
1
/2 /
22 2
2
2
;16 lb, 40 lb
(30 in.) [ (16 lb) ] (480 lb in.)
; (15 in.) (5 in.)
(0) (5) (10) 5 5 in.
40 lb
(5 10 )
55
85[(1 lb) (2lb)]
8515 5 0
012
8 5[(10 lb in.)
C
EB EB
DE
FF
d
F  
   
  






1
2
MM M
MrF i j k
Mr Fr i j
jk
jk
ijk
M
i
(30 lb in.) (15 lb in.) ]
(480 lb in.) 8 5[(10 lb in.) (30 lb in.) (15 lb in.) ]
(178.885 lb in.) (536.66 lb in.) (211.67 lb in.)

       

jk
Mk ijk
ijk

22 2
(178.885) (536.66) ( 211.67)
603.99 lb in
M

604 lb in.M 

axis
z
0.29617 0.88852 0.35045
cos 0.29617
cos 0.88852
cos 0.35045
x
y
M



  



M
ijk



z
72.8 27.3 110.5
xy

   
www.elsolucionario.org

273. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.77
If 20 lb,P
 replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.

SOLUTION
From the solution to Problem. 3.78:
16-lb force:
1
(480 lb in.)M  k
40-lb force:
2
8 5[(10 lb in.) (30 lb in.) (15 lb in.) ]M i j k
P20 lb
3
(30 in.) (20 lb)
(600 lb in.)
C
MP


r
ik
j

123
22 2
(480) 8 5 (10 30 15 ) 600
(178.885 lb in.) (1136.66 lb in.) (211.67 lb in.)
(178.885) (113.66) (211.67)
1169.96 lb in.
M

    



MM M M
kijkj
ijk

1170 lb in.M 

axis
0.152898 0.97154 0.180921
cos 0.152898
cos 0.97154
cos 0.180921
x
y
z
M



  



M
ij k




81.2 13.70 100.4
xy z
 
    
www.elsolucionario.org

274. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.78
Replace the two couples shown with a single equivalent
couple, specifying its magnitude and the direction of its axis.

SOLUTION

Replace the couple in the ABCD plane with two couples P and Q shown:

160 mm
(50 N) (50 N) 40 N
200 mm
CD
P
CG

 




120 mm
(50 N) (50 N) 30 N
200 mm
CF
Q
CG

 



Couple vector M
1 perpendicular to plane ABCD:

1
(40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM
 
Couple vector M
2 in the xy plane:

2
(12.5 N)(0.192 m) 2.40 N mM


144 mm
tan 36.870
192 mm

 

1
(4.80 cos36.870 ) (4.80 sin 36.870 )
3.84 2.88
  
M
j k
jk


2
2.40Mj

12
1.44 2.88  MM M j k

3.22 N m; 90.0 , 53.1 , 36.9
xyz

 M 
www.elsolucionario.org

275. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.79
Solve Prob. 3.78, assuming that two 10-N vertical forces
have been added, one acting upward at C and the other
downward at B.
PROBLEM 3.78 Replace the two couples shown with a
single equivalent couple, specifying its magnitude and the
direction of its axis.

SOLUTION

Replace the couple in the ABCD plane with two couples P and Q shown.

160 mm
(50 N) (50 N) 40 N
200 mm
CD
P
CG

 




120 mm
(50 N) (50 N) 30 N
200 mm
CF
Q
CG

 



Couple vector M
1 perpendicular to plane ABCD.

1
(40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM
 

144 mm
tan 36.870
192 mm

 

1
(4.80cos36.870 ) (4.80sin 36.870 )
3.84 2.88
  
M
j k
jk


2
(12.5 N)(0.192 m) 2.40 N m
2.40
M


j


3/ 3/
; (0.16 m) (0.144 m) (0.192 m)
(0.16 m) (0.144 m) (0.192 m) ( 10 N)
1.92 1.6
BC BC
M   
  
 Mr r i j k
ijkj
ik
www.elsolucionario.org

276. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.79 (Continued)


123
(3.84 2.88 ) 2.40 ( 1.92 1.6 )
(1.92 N m) (1.44 N m) (1.28 N m)
MM M M    
     jk j ik
i
j k


222
( 1.92) (1.44) (1.28) 2.72 N mM     2.72 N mM 

cos 1.92/2.72
cos 1.44/2.72
cos 1.28/2.72
x
y
z






134.9 58.0 61.9
xyz

  
www.elsolucionario.org

277. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.80
Shafts A and B connect the gear box to the wheel
assemblies of a tractor, and shaft C connects it to the
engine. Shafts A and B lie in the vertical yz plane,
while shaft C is directed along the x axis. Replace
the couples applied to the shafts by a single
equivalent couple, specifying its magnitude and the
direction of its axis.

SOLUTION

1200sin 20 1200cos20 410.42 1127.63
900sin 20 900cos20 307.82 845.72
840
The single equivalent couple is the sum of the three moments
(840 lb ft) (102.60 lb ft) (1973.35 lb ft
A
B
C
   
 

     
Mjkjk
Mjkjk
Mi
Mi j


)k

222
( 840) ( 102.60) (1973.35)
2147.3 lb ftM  

2150 lb ftM 

axis
z
0.39119 0.047921 0.91899
cos 0.39119
cos 0.047921
cos 0.91899
x
y
M



  



M
ijk


z
113.0 92.7 23.2
xy

  


www.elsolucionario.org

278. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.81
A 500-N force is applied to a bent plate as shown. Determine
(a) an equivalent force-couple system at B, (b) an equivalent
system formed by a vertical force at A and a force at B.

SOLUTION

(a) Force-couple system at B

(500 N)sin30 (500 N)cos30
(250.0 N) (433.01 N)


Fik
Fik



(0.3 0.175 ) (250 433.01 )
86.153 N m
B
  
 
Mi ji j
k

The equivalent force-couple system at
B is

500 N
B
F 60.0 86.2 N m
B
M 
(
b) Require

Equivalence requires

86.153 N m (0.125 m); (689.22 N)
or ; (250 N) (433.01 N) (689.22 N)
BB
MM
 

AA j
FA+B BF-A B i j j


(250 N) (1122.23 N) 1149.74 N Bi j 77.4 1150 NB 77.4 
689 N A 
www.elsolucionario.org

279. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.82
The tension in the cable attached to the end
C of an adjustable boom ABC is 560 lb.
Replace the force exerted by the cable at
C
with an equivalent force-couple system
(
a) at A, (b) at B.

SOLUTION



(a) Based on : 560 lb
A
FFT
or
560 lb
A
F 20.0° 

:(sin50)()
AA A
MM T d

(560 lb)sin 50 (18 ft)
7721.7 lb ft



or
7720 lb ft
A
M 
(
b) Based on : 560 lb
B
FFT
or
560 lb
B
F 20.0° 

:(sin50)()
BB B
MM T d

(560 lb)sin50°(10 ft)
4289.8 lb ft



or
4290 lb ft
B
M 
www.elsolucionario.org

280. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.83
A dirigible is tethered by a cable attached to its cabin at B.
If the tension in the cable is 1040 N, replace the force
exerted by the cable at
B with an equivalent system formed
by two parallel forces applied at
A and C.

SOLUTION


Require the equivalent forces acting at A and C be parallel and at an
angle of
 with the vertical.
Then for equivalence,

: (1040 N)sin 30 sin sin
xAB
FFF  (1)

: (1040 N)cos30 cos cos
yAB
FFF    (2)
Dividing Equation (1) by Equation (2),

()sin(1040 N)sin30
(1040 N)cos30 ( )cos
AB
AB
FF
FF 




Simplifying yields
30 .
Based on

: [(1040 N)cos30 ](4 m) ( cos30 )(10.7 m)
CA
MF

388.79 N
A
F
or
389 N
A
F 60.0° 
Based on

: [(1040 N)cos30 ](6.7 m) ( cos30 )(10.7 m)
AC
MF   

651.21 N
C
F
or
651 N
C
F 60.0° 
www.elsolucionario.org

281. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.84
A 30-lb vertical force P is applied at A to the bracket shown, which is held
by screws at
B and C. (a) Replace P with an equivalent force-couple
system at
B. (b) Find the two horizontal forces at B and C that are
equivalent to the couple obtained in part
a.

SOLUTION
(a) (30 lb)(5 in.)
150.0 lb in.
B
M



30.0 lbF

, 150.0 lb in.
B
M 
(
b)
150 lb in.
50.0 lb
3.0 in.
BC

 


50.0 lbB ; 50.0 lbC 


www.elsolucionario.org

282. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.85
A worker tries to move a rock by applying a 360-N
force to a steel bar as shown. (
a) Replace that force
with an equivalent force-couple system at
D. (b) Two
workers attempt to move the same rock by applying a
vertical force at
A and another force at D. Determine
these two forces if they are to be equivalent to the
single force of Part
a.

SOLUTION

(
a) We have : 360 N( sin 40° cos40 ) (231.40 N) (275.78 N)  FijijF

or 360 NF 50° 
We have
/
:
DBD
Mr RM
where
/
[(0.65 m)cos30 ] [(0.65 m)sin 30 ]
(0.56292 m) (0.32500 m)
0.56292 0.32500 0 N m
231.40 275.78 0
[155.240 75.206)N m]
BD
   
 
 

 
rij
ij
ijk
M
k

(230.45 N m) k or 230 N mM


(
b) We have
/
:
DADA
MMr F
where
/
[(1.05 m)cos30 ] [(1.05 m)sin 30 ]
(0.90933 m) (0.52500 m)
0.90933 0.52500 0 N m
010
[230.45 N m]
BD
A
F
   
 
 


rij
ij
ijk
k

(a) (b) www.elsolucionario.org

283. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.85 (Continued)

or
(0.90933 ) 230.45
A
Fkk

253.42 N
A
F or 253 N
A
F 
We have
:
A D
FFF F

(231.40 N) (275.78 N) (253.42 N) ( cos sin )
D
F
 ij jij
From
:231.40N cos
D
F
i (1)

:j 22.36 N sin
D
F
 (2)
Equation (2) divided by Equation (1)

tan 0.096629
5.5193 or 5.52


  
Substitution into Equation (1)

231.40
232.48 N
cos5.5193
D
F
 or 232 N
D
F 5.52°  www.elsolucionario.org

284. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.86
A worker tries to move a rock by applying a 360-N
force to a steel bar as shown. If two workers attempt
to move the same rock by applying a force at
A and a
parallel force at
C, determine these two forces so that
they will be equivalent to the single 360-N force
shown in the figure.

SOLUTION
: -360sin40 = sin sin (1)
XAC
FFF 



: -360cos 40 cos F cos (2)
YAC
FF 


Dividing (1) by (2) yields
40

and then 360 N (3)
AC
FF


: 0 (0.4 m) cos10 (0.35 m) cos10
7
(4)
8
Substituting into (3) ....
7
360
8
BA C
AC
CC
FF
FF
FF
 


M


Finally
168.0 N
A
F 50.0°; 192.0 N
C
F 50.0°; 
www.elsolucionario.org

285. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.87
The shearing forces exerted on the cross section of a steel channel can be
represented by a 900-N vertical force and two 250-N horizontal forces as
shown. Replace this force and couple with a single force F applied at
Point
C, and determine the distance x from C to line BD. (Point C is
defined as the
shear center of the section.)

SOLUTION
Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about
H is equal to the moment of the couple


(0.18)(250 N)
45 N m
H
M


Then
(900 N)
H
Mx
or
45 N m (900 N)
0.05 m
x
x



900 NF 50.0 mmx 
www.elsolucionario.org

286. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.88
A force and a couple are applied as shown to the end of a
cantilever beam. (
a) Replace this system with a single force
F applied at Point
C, and determine the distance d from C to
a line drawn through Points
D and E. (b) Solve part a if the
directions of the two 360-N forces are reversed.

SOLUTION


(a) We have : (360 N) (360 N) (600 N)  FF j j k
or
(600 N)Fk 
and
: (360 N)(0.15 m) (600 N)( )
D
Md

0.09 md
or
90.0 mm belowdED 
(
b) We have from part a: (600 N)Fk 
and
: (360 N)(0.15 m) (600 N)( )
D
Md 

0.09 md
or
90.0 mm abovedED 
www.elsolucionario.org

287. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.89
Three control rods attached to a lever ABC exert on it the
forces shown. (
a) Replace the three forces with an
equivalent force-couple system at
B. (b) Determine the
single force that is equivalent to the force-couple system
obtained in Part
a, and specify its point of application on
the lever.

SOLUTION
(a) First note that the two 20-lb forces form A couple. Then
48 lbF 
where
180 (60 55 ) 65
and
(30 in.)(48 lb)cos55 (70 in.)(20 lb)cos20
489.62 lb in
B
MM

 

The equivalent force-couple system at
B is

48.0 lbF 65.0°; 490 lb in.M 
(
b) The single equivalent force Fis equal to F. Further, since the sense of M is clockwise, F must be
applied between
A and B. For equivalence.

: cos 55
B
MMaF  
where
a is the distance from B to the point of application of F. Then

489.62 lb in. (48.0 lb)cos 55a 
or
17.78 in.a 48.0 lbF 65.0° 
and is applied to the lever 17.78 in. to the left of pin
B 

www.elsolucionario.org

288. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.90
A rectangular plate is acted upon by the force and couple
shown. This system is to be replaced with a single equivalent
force. (
a) For   40, specify the magnitude and the line of
action of the equivalent force. (
b) Specify the value of  if the line
of action of the equivalent force is to intersect line
CD 300 mm
to the right of
D.

SOLUTION
(a) The given force-couple system (F, M) at B is

48 NF
and
(0.4 m)(15 N)cos 40 (0.24 m)(15 N)sin 40
B
MM
 
or
6.9103 N mM
The single equivalent force
F is equal to F. Further for equivalence

:
B
MMdF 
or
6.9103 N m 48 Nd
or
0.14396 md 48.0 NF 
and the line of action of
F intersects line AB 144.0 mm to the right of A. 
(
b) Following the solution to Part a but with 0.1 md and  unknown, have

: (0.4 m)(15 N)cos (0.24 m)(15 N)sin
B
M 

(0.1 m)(48 N)
or
5cos 3sin 4
Rearranging and squaring
22
25 cos (4 3 sin )
Using
22
cos 1 sin and expanding

22
25(1 sin ) 16 24 sin 9 sin 
or
2
34 sin 24 sin 9 0
Then
2
24 ( 24) 4(34)( 9)
sin
2(34)
sin 0.97686 or sin 0.27098


  




77.7 or 15.72   
www.elsolucionario.org

289. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.91
While tapping a hole, a machinist applies the horizontal
forces shown to the handle of the tap wrench. Show that
these forces are equivalent to a single force, and specify, if
possible, the point of application of the single force on the
handle.

SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one
of the forces equal in magnitude to the force at
A except with an opposite sense, resulting in a force-
couple.
We have
2.9 lb 2.65 lb 0.25 lb,
B
F  where the 2.65-lb force is part of the couple. Combining the two
parallel forces,

couple
(2.65 lb)[(3.2 in. 2.8 in.)cos 25 ]
14.4103 lb in.M 


and
couple
14.4103 lb in.M

A single equivalent force will be located in the negative
z direction.
Based on
: 14.4103 lb in. [(0.25 lb)cos 25 ]( )
B
Ma  

63.600 in.a
F
(0.25 lb)(cos 25 sin 25 ) ik
F
(0.227 lb) (0.1057 lb) ik and is applied on an extension of handle BD at a
distance of 63.6 in. to the right of
B. 
www.elsolucionario.org

290. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.92
A hexagonal plate is acted upon by the force P and the couple shown.
Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at
E.

SOLUTION
From the statement of the problem, it follows that 0
E
M for the given force-couple system. Further,
for P
min, we must require that P be perpendicular to
/
.
BE
r Then

min
: (0.2 sin 30 0.2)m 300 N
(0.2 m)sin 30 300 N
(0.4 m) 0
E
M
P




or
min
300 NP

min
300 NP 30.0° 

www.elsolucionario.org

291. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.93
An eccentric, compressive 250-kN force P is applied to the end
of a column. Replace P with an equivalent force-couple system
at
G.

SOLUTION


Have
: 250 kN FjF
or
250 kNFj 
Also have
:
GP
MrPM

0.030 0 0.060 kN m =
02500


ijk
M

 15 kN m 7.5 kN m   Mi k
or
15.00 kN m 7.50 kN mMik 


www.elsolucionario.org

292. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.94
A 2.6-kip force is applied at Point D of the cast iron post
shown. Replace that force with an equivalent force-couple
system at the center A of the base section.

SOLUTION
(12 in.) (5 in.) ; 13.00 in.DE DE  jk


(2.6 kips)
DE
DE
F



12 5
(2.6 kips)
13


jk
F


(2.40 kips) (1.000 kip)
Fjk 

/ADA
Mr F
where
/
(6 in.) (12 in.)
DA
rij

6 in. 12 in. 0
0 2.4 kips 1.0 kips
A


ij k
M

(12.00 kip in.) (6.00 kip in.) (14.40 kip in.)
A
     Mijk 
www.elsolucionario.org

293. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.95
Replace the 150-N force with an equivalent force-couple
system at A.

SOLUTION
Equivalence requires : (150 N)( cos 35 sin 35 )
(122.873 N) (86.036 N)
 
 FF
j k
j k


/
:
ADA
MMr F
where
/
(0.18 m) (0.12 m) (0.1 m)
DA
rijk
Then
0.18 0.12 0.1 N m
0 122.873 86.036
[( 0.12)( 86.036) (0.1)( 122.873)]
[ (0.18)( 86.036)]
[(0.18)( 122.873)]
(22.6 N m) (15.49 N m) (22.1 N m)
 

   
 

 
ij k
M
i
j
k
ijk
The equivalent force-couple system at A is

(122.9 N) (86.0 N)
Fjk 

(22.6 N m) (15.49 N m) (22.1 N m)
   Mi jk 
www.elsolucionario.org

294. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.96
To keep a door closed, a wooden stick is wedged between the
floor and the doorknob. The stick exerts at
B a 175-N force
directed along line
AB. Replace that force with an equivalent
force-couple system at
C.

SOLUTION

We have

:
AB C
FP F
where

(33 mm) (990 mm) (594 mm)
(175 N)
1155.00 mm
AB AB AB
P


P λ
ijk

or
(5.00 N) (150.0 N) (90.0 N)
C
 Fijk 
We have
/
:
CBCAB C
Mr P M

5 0.683 0.860 0 N m
13018
(5){( 0.860)( 18) (0.683)( 18)
[(0.683)(30) (0.860)(1)] }
C
 

  

ijk
M
ij
k

or
(77.4 N m) (61.5 N m) (106.8 N m)
C
 Mijk 
www.elsolucionario.org

295. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.97
A 46-lb force F and a 2120-lbin. couple
M are applied to corner A of the block
shown. Replace the given force-couple
system with an equivalent force-couple
system at corner H.

SOLUTION
We have
222
(18) ( 14) ( 3) 23 in.
AJ
d
Then
46 lb
(18 14 3 )
23
(36 lb) (28 lb) (6 lb)


Fijk
ijk

Also
22 2
( 45) (0) ( 28) 53 in.
AC
d   
Then
2120 lb in.
(45 28)
53
(1800 lb in.) (1120 lb in.)


   Mik
ik
Now
/AH
 MMr F
where
/
(45 in.) (14 in.)
AH
rij
Then
( 1800 1120 ) 45 14 0
36 28 6
  

ijk
Mik


( 1800 1120 ) {[(14)( 6)] [ (45)( 6)] [(45)( 28) (14)(36)] }
( 1800 84) (270) ( 1120 1764)
(1884 lb in.) (270 lb in.) (2884 lb in.)
(157 lb ft) (22.5 lb ft) (240 lb ft)
        
    
     
     ik i j k
ij k
ij k
ijk

The equivalent force-couple system at H is
(36.0 lb) (28.0 lb) (6.00 lb)
 Fijk 

(157.0 lb ft) (22.5 lb ft) (240 lb ft)
Mijk


www.elsolucionario.org

296. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.98
A 110-N force acting in a vertical plane parallel to the yz-
plane is applied to the 220-mm-long horizontal handle
AB
of a socket wrench. Replace the force with an equivalent
force-couple system at the origin
O of the coordinate
system.

SOLUTION


We have :
B
FP F
where
110 N[ (sin15 ) (cos15 ) ]
(28.470 N) (106.252 N)
B

 Pjk
jk

or (28.5 N) (106.3 N) Fjk 
We have
/
:
OBOB O
MrPM
where
/
[(0.22cos35 ) (0.15) (0.22sin 35 ) ] m
(0.180213 m) (0.15 m) (0.126187 m)
BO


rijk
ij k


0.180213 0.15 0.126187 N m
0 28.5 106.3
O


ijk
M

[(12.3487) (19.1566) (5.1361) ] N m
O
 Mijk

or (12.35 N m) (19.16 N m) (5.13 N m)
O
Mijk 

www.elsolucionario.org

297. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.99
An antenna is guyed by three cables as shown.
Knowing that the tension in cable AB is 288 lb, replace
the force exerted at A by cable AB with an equivalent
force-couple system at the center O of the base of the
antenna.

SOLUTION
We have
222
( 64) ( 128) (16) 144 ft
AB
d   
Then
288 lb
(64 128 16)
144
(32 lb)( 4 8 )
AB

Tijk
ijk
Now
/
128 32( 4 8 )
(4096 lb ft) (16,384 lb ft)
OAO AB


 
MM r T
jijk
ik
The equivalent force-couple system at O is

(128.0 lb) (256 lb) (32.0 lb)
Fijk 

(4.10 kip ft) (16.38 kip ft)
  Mi k 
www.elsolucionario.org

298. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.100
An antenna is guyed by three cables as shown.
Knowing that the tension in cable AD is 270 lb, replace
the force exerted at A by cable AD with an equivalent
force-couple system at the center O of the base of the
antenna.

SOLUTION
We have
222
(64) (128) (128)
192 ft
AD
d  

Then
270 lb
( 64 128 128 )
192
(90 lb)( 2 2 )
AD

Tijk
ijk
Now
/
128 90( 2 2 )
(23,040 lb ft) (11,520 lb ft)
OAOAD


   
MM r T
jijk
ik
The equivalent force-couple system at O is

(90.0 lb) (180.0 lb) (180.0 lb)
 Fijk 

(23.0 kip ft) (11.52 kip ft)
 Mik 
www.elsolucionario.org

299. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.101
A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-
couple system at end
A of the beam. (b) Which of the loadings are equivalent?


SOLUTION


(
a) (a) We have : 300 N 200 N
Ya
FR  
or
500 N
a
R 
and
:400 Nm(200 N)(3 m)
Aa
MM  
or
1000 N m
a
M 
(
b) We have : 200 N 300 N
Yb
FR
or
500 N
b
R 
and
: 400 N m (300 N)(3 m)
Ab
MM  
or
500 N m
b
M 
(
c) We have : 200 N 300 N
Yc
FR  
or
500 N
c
R 
and
: 400 N m (300 N)(3 m)
Ac
MM 
or
500 N m
c
M 




www.elsolucionario.org

300. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.101 (Continued)

(
d) We have : 500 N
Yd
FR 
or
500 N
d
R 
and
: 400 N m (500 N)(3 m)
Ad
MM 
or
1100 N m
d
M 
( e) We have :300 N800 N
Ye
FR
or
500 N
e
R 
and
: 400 N m 1000 N m (800 N)(3 m)
Ae
MM 
or
1000 N m
e
M 
( f ) We have :300 N200 N
Yf
FR  
or
500 N
f
R 
and
: 400 N m (200 N)(3 m)
Af
MM 
or
200 N m
f
M 
( g) We have :800 N300 N
Yg
FR  
or
500 N
g
R 
and
: 1000 Nm 400 Nm (300 N)(3 m)
Ag
MM 
or
2300 N m
g
M 
(
h) We have : 250 N 250 N
Yh
FR  
or
500 N
h
R 
and
: 1000 Nm 400 Nm (250 N)(3 m)
Ah
MM 
or
650 N m
h
M 
(
b) Therefore, loadings (a) and (e) are equivalent.

www.elsolucionario.org

301. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.102
A 3-m-long beam is loaded as shown. Determine the
loading of Prob. 3.101 that is equivalent to this loading.

SOLUTION


We have
: 200 N 300 N
Y
FR  
or
500 NR
and
: 500 N m 200 N m (300 N)(3 m)
A
MM 
or
200 N mM
Problem 3.101 equivalent force-couples at
A:

Case
R

M


(a) 500 N 1000 Nm
(b) 500 N 500 Nm
(c) 500 N 500 Nm
(d) 500 N 1100 Nm
(e) 500 N 1000 Nm
(f ) 500 N 200 Nm
(g) 500 N 2300 Nm
(h) 500 N 650 Nm

Equivalent to case (
f ) of Problem 3.101 
www.elsolucionario.org

302. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.103
Determine the single equivalent force and the distance from Point A to its line of action for the beam and
loading of (
a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102.

SOLUTION









For equivalent single force at distance d from A:
(
a) We have : 300 N 200 N
Y
FR  
or
500 NR 
and
: 400 N m (300 N)( )
(200 N)(3 ) 0
C
Md
d
 

or
2.00 md 



(
b) We have : 200 N 300 N
Y
FR
or
500 NR 
and
: 400 N m (200 N)( )
(300 N)(3 ) 0
C
Md
d
 

or
1.000 md 



(
c) We have : 200 N 300 N
Y
FR  
or
500 NR 
and
: 500 N m 200 N m
(200 N)( ) (300 N)(3 ) 0
C
M
dd


or
0.400 md 
www.elsolucionario.org

303. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into
the shape shown. Determine which of these systems is equivalent to a force F  (10 lb)i and a couple of
moment M  (15 lb  ft)j  (15 lb  ft)k located at the origin.


SOLUTION
First note that the force-couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces
remain unchanged.

: (5 lb ft) (15 lb ft) (2 ft) (10 lb)
AO
A       MM j k k i

(25 lb ft) (15 lb ft)
  jk

: (5 lb ft) (25 lb ft)
[(4.5 ft) (1 ft) (2 ft) ] 10 lb)
(15 lb ft ) (15 lb ft )
DO
D     


MM j k
ijk i
ik


: (15 lb ft) (15 lb ft)
: (15 lb ft) (5 lb ft)
[(4.5 ft) (1 ft) ] (10 lb)
GO
II
G
I
    
    

MM i j
MM j k
ij j


(15 lb ft) (15 lb ft)
  jk
The equivalent force-couple system is the system at corner D.




www.elsolucionario.org

304. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.105
The weights of two children sitting at ends A and B of a
seesaw are 84 lb and 64 lb, respectively. Where should a
third child sit so that the resultant of the weights of the
three children will pass through
C if she weighs (a) 60 lb,
(
b) 52 lb.

SOLUTION

(
a) For the resultant weight to act at C, 060 lb
CC
MW 
Then
(84 lb)(6 ft) 60 lb( ) 64 lb(6 ft) 0d 

2.00 ft to the right of dC 
(
b) For the resultant weight to act at C, 052 lb
CC
MW 
Then
(84 lb)(6 ft) 52 lb( ) 64 lb(6 ft) 0d 

2.31 ft to the right of dC 


www.elsolucionario.org

305. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.106
Three stage lights are mounted on a pipe as
shown. The lights at
A and B each weigh 4.1 lb,
while the one at
C weighs 3.5 lb. (a) If d  25 in.,
determine the distance from
D to the line of
action of the resultant of the weights of the three
lights. (
b) Determine the value of d so that the
resultant of the weights passes through the
midpoint of the pipe.

SOLUTION

For equivalence

: 4.1 4.1 3.5 or 11.7 lb
y
FR  R

: (10 in.)(4.1 lb) (44 in.)(4.1 lb)
[(4.4 )in.](3.5 lb) ( in.)(11.7 lb)
D
F
dL 
 

or
375.4 3.5 11.7 ( , in in.)dLdL
(
a) 25 in.d
We have
375.4 3.5(25) 11.7 or 39.6 in.LL 
The resultant passes through a Point 39.6 in. to the right of
D. 
(
b) 42 in.L
We have
375.4 3.5 11.7(42)d or 33.1 in.d 
www.elsolucionario.org

306. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.107
A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of
position. If
b  1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the
value of
a so that the distance from support A to the line of action of the equivalent force is maximum,
(
b) the magnitude of the equivalent force and its point of application on the beam.


SOLUTION

For equivalence,
: 1300 400 400 600
y
a
FR
b
 
or
2300 400 N
a
R
b



 (1)

: 400 (400) ( )(600)
2
A
aa
MaabLR
b





or
2
1000 600 200
2300 400
a
ab
b
L
a
b




Then with
24
10 9
3
1.5 m
8
23
3
aa
bL
a



(2)
where a, L are in m.
(a) Find value of a to maximize L.

2
288 48
10 23 10 9
33 33
8
23
3
aaaa
dL
da
a
 

 
 





www.elsolucionario.org

307. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.107 (Continued)

or
22184 80 64 80 32
230 24 0
339 3 9
aaa a a   
or
2
16 276 1143 0aa
 
Then
2
276 ( 276) 4(16)(1143)
2(16)
a
 


or
10.3435 m and 6.9065 maa
Since
9 m,AB a must be less than 9 m 6.91 ma 
(b) Using Eq. (1),
6.9065
2300 400
1.5
R or 458 NR 
and using Eq. (2),
24
10(6.9065) 9 (6.9065)
3
3.16 m
8
23 (6.9065)
3
L




R is applied 3.16 m to the right of A.

www.elsolucionario.org

308. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.108
A 6 × 12-in. plate is subjected to four loads as shown. Find the resultant
of the four loads and the two points at which the line of action of the
resultant intersects the edge of the plate.

SOLUTION
We have

For equivalence,
(40 lb) (20 lb) Rij

44.721 lbR 44.7 lbR 26.6° 

/
= (6 in.) (50 lb)(cos45 sin 45 )
= (212.13 lb in.)
CBC
M
 

rB
iij
k


Intersection with edge AC

()(20)
212.13 (20 )
10.61 in.
C
Mx
x
x
 

 ij
kk
10.61 in. to the left of C. 
Intersection with edge CD

()(40)
213.13 (40 )
5.30 in.
C
My
y
y
  


ji
kk


5.30 in. below .C 
www.elsolucionario.org

309. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.109
A 32-lb motor is mounted on the floor. Find the resultant of
the weight and the forces exerted on the belt, and determine
where the line of action of the resultant intersects the floor.

SOLUTION



We have

: (60 lb) (32 lb) (140 lb)(cos30 sin30 ) Fij ijR

(181.244 lb) (38.0 lb)Rij
or
185.2 lbR 11.84° 
We have
:
OOy
MMxR

[(140 lb)cos30 ][(4 2cos30 )in.] [(140 lb)sin 30 ][(2 in.)sin 30 ]
    


(60lb)(2in.) (38.0lb)x

1
( 694.97 70.0 120) in.
38.0
x 

and
23.289 in.x
Or resultant intersects the base (x-axis) 23.3 in. to the left of
the vertical centerline (y-axis) of the motor.


www.elsolucionario.org

310. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.110
To test the strength of a 625  500-mm suitcase, forces are
applied as shown. If P = 88 N, (a) determine the resultant of
the applied forces, (b) locate the two points where the line
of action of the resultant intersects the edge of the suitcase.


SOLUTION
We have

First replace the applied forces and couples with an equivalent force-couple system at B.
(a )Thus,
45
: 100 (212) 180
53
xx
FR  ; 100 N
x
R

28
: (212) 88
53
yy
FR  ; 200 N
y
R

(100 N) (200 N)Rij

224 NR 63.4° 
(b)

28
: (0.1 m)(100 N) (0.53 m) (212 N)
53
(0.08 m)(88 N) (0.28 m)(180 N)
B
B
M
M




 


26.0 N m
B
M (1)
The single equivalent force
R must then act as indicated. Then with R at E:

:26.0 Nm (200 N)
B
Mx

130 mmx

2
or 260 mm
1
y
y
x


Therefore, the line of action of
R intersects top AB 130.0 mm to the left of B and
intersects side BC 260 mm below B.

www.elsolucionario.org

311. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.111
Solve Prob. 3.110, assuming that P = 138 N.
Problem 3.110
To test the strength of a 625  500-mm
suitcase, forces are applied as shown. If P = 88 N, (a)
determine the resultant of the applied forces, (b) locate the
two points where the line of action of the resultant
intersects the edge of the suitcase.

SOLUTION
We have

First replace the applied forces and couples with an equivalent force-couple system at B.
(a)
Thus,
45
: 100 (212) 180
53
xx
FR  ; 100 N
x
R

28
: (212) 138
53
yy
FR  ; 250 N
y
R

269 NR 68.2° 
(b)

28
: (0.1 m)(100 N) (0.53 m) (212 N)
53
(0.08 m)(138 N) (0.28 m)(180 N)
B
B
M
M







30.0 N m
B
M
The single equivalent force R must then act as indicated. Then with R at E:

: 30 N m (250 N)
B
Mx

120 mmx

5
or 300 mm
2
y
y
x


Therefore, the line of action of
R intersects top AB 120 mm to the left of B and
intersects side BC 300 mm below B.

www.elsolucionario.org

312. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.112
Pulleys A and B are mounted on bracket CDEF. The
tension on each side of the two belts is as shown. Replace
the four forces with a single equivalent force, and
determine where its line of action intersects the bottom
edge of the bracket.

SOLUTION
Equivalent force-couple at A due to belts on pulley A
We have
: 120 lb 160 lb
A
R  F

280 lb
A
R
We have
:
A
M 40 lb(2 in.)
A
M

80 lb in.
A
M
Equivalent force-couple at B due to belts on pulley B
We have
: (210 lb 150 lb)F 25
B
R

360 lb
B
R 25°
We have
: 60 lb(1.5 in.)
BB
M M

90 lb in.
B
M
Equivalent force-couple at F
We have
: ( 280 lb) (360 lb)(cos 25 sin 25 )
F
  FR j i j

22
22
1
1
(326.27 lb) (127.857 lb)
(326.27) (127.857)
350.43 lb
tan
127.857
tan
326.27
21.399
F
Fx Fy
Fy
Fx
RR
RR
R
R









 





 
ij
or 350 lb
F
RR 21.4° 

PROBLEM 3.112 (Continued) www.elsolucionario.org

313. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


We have
:(280lb)(6in.)80lbin.
FF
M M

[(360 lb)cos 25 ](1.0 in.)
[(360 lb)sin 25 ](12 in.) 90 lb in.
 
 


(350.56 lb in.)
F
 Mk
To determine where a single resultant force will intersect line FE,

350.56 lb in.
127 857 lb
2.7418 in.
Fy
F
y
MdR
M
d
R


 


 or
2.74 in.d 


www.elsolucionario.org

314. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.113
A truss supports the loading shown. Determine the
equivalent force acting on the truss and the point of
intersection of its line of action with a line drawn through
Points A and G.

SOLUTION
We have
(240 lb)(cos70 sin 70 ) (160 lb)
(300 lb)( cos40 sin 40 ) (180 lb)



RF
Rijj
ij j

22
22
1
1
(147.728 lb) (758.36 lb)
(147.728) (758.36)
772.62 lb
tan
758.36
tan
147.728
78.977
xy
y
x
RRR
R
R



 




 






Rij
or 773 lbR 79.0 
We have
Ay
MdR
where
[240 lbcos70 ](6 ft) [240 lbsin 70 ](4 ft)
(160 lb)(12 ft) [300 lbcos40 ](6 ft)
[300 lbsin 40 ](20 ft) (180 lb)(8 ft)
7232.5 lb ft
A
M   


 

7232.5 lb ft
758.36 lb
9.5370 ft
d




or 9.54 ft to the right of dA 


www.elsolucionario.org

315. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.114
A couple of magnitude M = 80 lb·in. and the three forces shown are
applied to an angle bracket. (a) Find the resultant of this system of
forces. (b) Locate the points where the line of action of the resultant
intersects line AB and line BC.

SOLUTION
(a) We have :(10)(25cos60)
(25 sin 60 ) ( 40 )
(27.50 lb) (11.6506 lb)
 

 FR j i
ji
ij
or
29.9 lbR 23.0° 
(b) First reduce the given forces and couple to an equivalent force-couple system
(, )
B
RM at B.
We have : (80 lb in) (12 in.)(10 lb) (8 in.)(40 lb)
120.0 lb in.
BB
MM 
 

Then with
R at D, : 120.0 lb in (11.6506 lb)
B
Ma 
or
10.30 in.a
and with
R at E, : 120.0 lb in (27.5 lb)
B
Mc 
or
4.36 in.c
The line of action of
R intersects line AB 10.30 in. to the left of B and intersects line BC 4.36 in.
below
B. 
www.elsolucionario.org

316. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.115
A couple M and the three forces shown are applied to an angle
bracket. Find the moment of the couple if the line of action of the
resultant of the force system is to pass through (a) point A, (b) point
B, (c) point C.

SOLUTION
In each case, we must have
1
0
R
M
(a)
(12 in.)[(25 lb)sin 60 ] (8 in.)(40 lb) 0
B
AA
MMM     

60.192 lb in.M  60.2 lb in.M 
(b)
(12 in.)(10 lb) (8 in.)(40 lb) 0
R
BB
MMM    

200 lb in.M  200 lb in.M


(c)
(12 in.)(10 lb) (8 in.)[(25 lb) cos 60 ] 0
R
CC
MMM     

20.0 lb in.M  20.0 lb in.M 
www.elsolucionario.org

317. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.116
A machine component is subjected to the forces and
couples shown. The component is to be held in place by a
single rivet that can resist a force but not a couple. For
P  0, determine the location of the rivet hole if it is to be
located (
a) on line FG, (b) on line GH.

SOLUTION
We have

First replace the applied forces and couples with an equivalent force-couple system at
G.
Thus,
:200cos15 120cos70
xx
FPR
or
(152.142 ) N
x
RP

: 200sin 15 120sin 70 80
yy
FR    
or
244.53 N
y
R

: (0.47 m)(200 N)cos15 (0.05 m)(200 N)sin15
(0.47 m)(120 N)cos70 (0.19 m)(120 N)sin 70
(0.13 m)( N) (0.59 m)(80 N) 42 N m
40 N m
G
G
M
P
M
  

 


or
(55.544 0.13 ) N m
G
MP   (1)
Setting
0P in Eq. (1):
Now with
R at I, : 55.544 N m (244.53 N)
G
Ma 
or
0.227 ma
and with
R at J, : 55.544 N m (152.142 N)
G
Mb 
or
0.365 mb
(
a) The rivet hole is 0.365 m above G. 
(
b) The rivet hole is 0.227 m to the right of G. 
www.elsolucionario.org

318. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.117
Solve Problem 3.116, assuming that P  60 N.
PROBLEM 3.116 A machine component is subjected to
the forces and couples shown. The component is to be held
in place by a single rivet that can resist a force but not a
couple. For P  0, determine the location of the rivet hole
if it is to be located (a) on line FG, (b) on line GH.

SOLUTION
See the solution to Problem 3.116 leading to the development of Equation (1):
and
(55.544 0.13 ) N m
(152.142 ) N
G
x
MP
RP
 


For
60 NP

we have
(152.142 60)
212.14 N
[55.544 0.13(60)]
63.344 N m
x
G
R
M


 
 
Then with R at I,
: 63.344 N m (244.53 N)
G
Ma 
or
0.259 ma

and with R at J,
: 63.344 N m (212.14 N)
G
Mb 
or
0.299 mb

(a) The rivet hole is 0.299 m above G.

(b) The rivet hole is 0.259 m to the right of G.



www.elsolucionario.org

319. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.118
As follower AB rolls along the surface of member C, it exerts
a constant force
F perpendicular to the surface. (a) Replace
F with an equivalent force-couple system at Point D
obtained by drawing the perpendicular from the point of
contact to the
x-axis. (b) For a  1 m and b  2 m, determine
the value of
x for which the moment of the equivalent force-
couple system at
D is maximum.

SOLUTION
(a) The slope of any tangent to the surface of member C is

2
22
2
1
dy d x b
bx
dx dx aa
 




Since the force
F is perpendicular to the surface,

1 2
1
tan
2dy a
dx b x


 
 
 
 

For equivalence,

:FFR

:(cos)( )
DAD
MF yM 
where

22 2
2
2
3
2
2
422
2
cos
() (2)
1
2
4
A
D
bx
abx
x
yb
a
x
Fb x
a
M
abx












Therefore, the equivalent force-couple system at D is

FR
2
1
tan
2
a
bx







3
2
2
422
2
4
x
Fb x
a
abx





M
 www.elsolucionario.org

320. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.118 (Continued)

(b) To maximize M, the value of x must satisfy
0
dM
dx

where for
1m, 2mab
 

3
2
22 3 21/2
2
8( )
116
1
116 (13 ) ( ) (32)(116 )
2
80
(1 16 )
Fx x
M
x
xxxx x x
dM
F
dx x





 






22 3
(1 16 )(1 3 ) 16 ( ) 0xxxxx
  
or
42
32 3 1 0xx


222
3 9 4(32)( 1)
0.136011 m and 0.22976 m
2(32)
x
  


Using the positive value of x
2
: 0.36880 mx
 or 0.369 mx 


www.elsolucionario.org

321. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.119
A machine component is subjected to the forces shown, each
of which is parallel to one of the coordinate axes. Replace
these forces with an equivalent force-couple system at A.

SOLUTION











For equivalence
:
BC D A
FF F F R

240 N 125 N 300 N 150 N
A
   Rjkik


 300 N 240 N 25.0 N
A
  Rijk 


Also for equivalence

// /
:
BA B CA C DA D A
rFrFrFM
or
0 0.12 m 0 0.06 m 0.03 m 0.075 m 0.06 m 0.08 m 0.75 m
0 240 N 125 N 300 N 0 0 0 0 150 N
A
M
 
ij k i j k i j k


 15 N m 22.5 N m 9 N m 12 N m 9 N m           
  ijkij
or
 3.00 N m 13.50 N m 9.00 N m
A
     Mijk 


www.elsolucionario.org

322. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.120
Two 150-mm-diameter pulleys are mounted
on line shaft
AD. The belts at B and C lie in
vertical planes parallel to the
yz-plane.
Replace the belt forces shown with an
equivalent force-couple system at
A.

SOLUTION
Equivalent force-couple at each pulley:
Pulley
B: (145 N)( cos 20 sin 20 ) 215 N
(351.26 N) (49.593 N)
(215 N 145 N)(0.075 m)
(5.25 N m)
B
B

 
 
 
Rjkj
jk
Mi
i

Pulley
C: (155 N 240 N)( sin10 cos10 )
(68.591 N) (389.00 N)
(240 N 155 N)(0.075 m)
(6.3750 N m)
C
C
 
 


Rjk
jk
Mi
i

Then
(419.85 N) (339.41)
BC
 RR R j k or (420 N) (339 N)Rjk 

//
(5.25 N m) (6.3750 N m) 0.225 0 0 N m
0 351.26 49.593
+0.45 0 0 N m
0 68.591 389.00
(1.12500 N m) (163.892 N m) (109.899 N m)
ABCBABCAC

   




MMMr Rr R
ijk
ii
ij k
ijk

or
(1.125 N m) (163.9 N m) (109.9 N m)
A
Mijk 


www.elsolucionario.org

323. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.121
As an adjustable brace BC is used to bring a wall into plumb,
the force-couple system shown is exerted on the wall. Replace
this force-couple system with an equivalent force-couple system
at
A if 21.2 lbR and 13.25 lb · ft.M

SOLUTION

We have
:
ABC
FRR R
where
(42in.) (96in.) (16in.)
106 in.
BC


ijk
λ

21.2 lb
(42 96 16 )
106
A
Rijk
or
(8.40 lb) (19.20 lb) (3.20 lb)
A
 Rijk 
We have
/
:
ACA A
MrRMM
where
/
1
(42 in.) (48 in.) (42 48 )ft
12
(3.5 ft) (4.0 ft)
CA
 

rikik
ik


(8.40 lb) (19.50 lb) (3.20 lb)
42 96 16
(13.25 lb ft)
106
(5.25 lb ft) (12 lb ft) (2 lb ft)
BC
M
 

 

     Ri jk
M
ijk
ijk
Then
3.5 0 4.0 lb ft ( 5.25 12 2 ) lb ft
8.40 19.20 3.20
A
   

ijk
ijk M

(71.55 lb ft) (56.80 lb ft) (65.20 lb ft)
A
Mijk
or
(71.6 lb ft) (56.8 lb ft) (65.2 lb ft)
A
Mijk 
www.elsolucionario.org

324. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.122
In order to unscrew the tapped faucet A, a
plumber uses two pipe wrenches as shown.
By exerting a 40-lb force on each wrench, at
a distance of 10 in. from the axis of the pipe
and in a direction perpendicular to the pipe
and to the wrench, he prevents the pipe from
rotating, and thus avoids loosening or further
tightening the joint between the pipe and the
tapped elbow C. Determine (a) the angle θ
that the wrench at A should form with the
vertical if elbow C is not to rotate about the
vertical, (b) the force-couple system at C
equivalent to the two 40-lb forces when this
condition is satisfied.

SOLUTION
We first reduce the given forces to force-couple systems at A and B, noting that

||||(40 lb)(10 in.)
400 lb in.
AB

MM

We now determine the equivalent force-couple system at C.

(40 lb)(1 cos ) (40 lb)sin
  Rij (1)

(15 in.) [ (40 lb)cos (40 lb)sin ]
(7.5 in.) (40 lb)
400 400 600cos 600sin 300
(600 lb in.)sin (300 lb in.)(1 2cos )
R
CAB
 

  

    
 
MMM k i j
ki
jij
ij (2)
(a) For no rotation about vertical, y component of
R
C
M must be zero.

12cos 0
cos 1/2






60.0
 
(b) For
60.0
 in Eqs. (1) and (2),

(20.0 lb) (34.641 lb) ; (519.62 lb in.)
R
C
  Ri
jMi

(20.0 lb) (34.6 lb) ; (520 lb in.)
R
C
 Ri
jMi 

www.elsolucionario.org

325. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.123
Assuming θ  60° in Prob. 3.122, replace the
two 40-lb forces with an equivalent force-
couple system at D and determine whether
the plumber’s action tends to tighten or
loosen the joint between (a) pipe CD and
elbow D, (b) elbow D and pipe DE.
Assume all threads to be right-handed.
PROBLEM 3.122 In order to unscrew the
tapped faucet A, a plumber uses two pipe
wrenches as shown. By exerting a 40-lb
force on each wrench, at a distance of 10 in.
from the axis of the pipe and in a direction
perpendicular to the pipe and to the wrench,
he prevents the pipe from rotating, and thus
avoids loosening or further tightening the
joint between the pipe and the tapped elbow
C. Determine (a) the angle θ that the
wrench at A should form with the vertical if
elbow C is not to rotate about the vertical,
(b) the force-couple system at C equivalent
to the two 40-lb forces when this condition
is satisfied.

SOLUTION
The equivalent force-couple system at C for 60 was obtained in the solution to Prob. 3.122:

(20.0 lb) (34.641 lb)
(519.62 lb in.)
R
C
 

Ri j
Mi

The equivalent force-couple system at D is made of R and
R
D
M where

/
(519.62 lb in.) (25.0 in.) [(20.0 lb) (34.641 lb) ]
(519.62 lb in.) (500 lb in.)
RR
DCCD


MMr R
iji j
ik

Equivalent force-couple at D:

(20.0 lb) (34.6 lb) ; (520 lb in.) (500 lb in.)
R
C
 Ri
jMik 
(a) Since
R
D
M has no component along the y-axis, the plumber’s action will neither loosen
nor tighten the joint between pipe CD and elbow.

(b) Since the x component of
R
D
M is , the plumber’s action will tend to tighten
the joint between elbow and pipe DE.

www.elsolucionario.org

326. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.124
Four forces are applied to the machine component
ABDE as shown. Replace these forces with an
equivalent force-couple system at
A.

SOLUTION

(50 N) (300 N) (120 N) (250 N)
(420 N) (50 N) (250 N)
(0.2 m)
(0.2 m) (0.16 m)
(0.2 m) (0.1 m) (0.16 m)
B
D
E
   
  



Rjiik
Rijk
ri
rk i
rijk


[ (300 N) (50 N) ]
( 250 N) ( 120 N)
0.2 m 0 0 0.2 m 0 0.16 m
300 N 50 N 0 0 0 250 N
0.2 m 0.1 m 0.16 m
120 N 0 0
(10 N m) (50 N m) (19.2 N m) (12 N m)
R
AB
D
 
 

 


     
Mr i j
rkri
ijkijk
ijk
kj jk

Force-couple system at
A is

(420 N) (50 N) (250 N) (30.8 N m) (22.0 N m)
R
A
      RijkM j k 
www.elsolucionario.org

327. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.125
A blade held in a brace is used to tighten a screw at
A. (a) Determine the forces exerted at B and C,
knowing that these forces are equivalent to a force-
couple system at A consisting of R = (25 N)i + R
yj
+ R
zk and M
R
A
= – (13.5 N·m)i. (b) Find the
corresponding values of R
y and R z. (c) What is the
orientation of the slot in the head of the screw for
which the blade is least likely to slip when the brace
is in the position shown?

SOLUTION
(a) Equivalence requires :FRBC
or
(25 N) ( )
yz xyz
RR B CCCi
jkk i jk
Equating the i coefficients:
: 25 N or 25 N
xx
CC i
Also,
//
:
R
AABA CA
MMr Br C
or
(13.5 N m) [(0.2 m) (0.15 m) ] ( )
(0.4 m) [ (25 N) ]
yz
B
CC
   
iijk
iijk
Equating coefficients:
: 13.5 N m (0.15 m) or 75 N
:0 (0.4 m) or 0
: 0 (0.2 m)(75 N) (0.4 m) or 37.5 N
yy
zz
BB
CC
CC
  

 i
k
j

(75.0 N) (25.0 N) (37.5 N)
BkCik 
(b) Now we have for the equivalence of forces

(25 N) (75 N) [( 25 N) (37.5 N) ]
yz
RR i
jkk ik
Equating coefficients:
:0
y
R
j 0
y
R 
:7537.5
z
R k or 37.5 N
z
R 
(c) First note that
(25 N) (37.5 N) . Rik Thus, the screw is best able to resist the lateral force
z
R
when the slot in the head of the screw is vertical.

www.elsolucionario.org

328. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.126
A mechanic uses a crowfoot wrench to loosen a bolt at C.
The mechanic holds the socket wrench handle at Points A
and B and applies forces at these points. Knowing that
these forces are equivalent to a force-couple system at C
consisting of the force
(8 lb) + (4 lb)
Ci k and the couple
(360 lb·
C
M in.)i, determine the forces applied at A and
at B when
2 lb.
z
A

SOLUTION
We have :F ABC
or
:8lb
xxx
FAB 

(8lb)
xx
BA  (1)


:0
yyy
FAB
or
y y
A B (2)


: 2 lb 4 lb
zz
FB
or
2 lb
z
B (3)
We have
//
:
CBC AC C
Mr Br AM

8 0 2 8 0 8 lb in. (360 lb in.)
22
xy xy
BB AA

ijk ijk
i

or
(2 8 ) (2 16 8 16)
yy x x
BA B Ai
j

(8 8 ) (360 lb in.)
yy
BA   ki
From i-coefficient:
2 8 360 lb in.
yy
BA  (4)
j-coefficient:
2832lbin.
xx
BA   (5)
k-coefficient:
880
yy
BA (6) www.elsolucionario.org

329. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.126 (Continued)

From Equations (2) and (4):
28()360
yy
BB 

36 lb 36 lb
yy
BA
From Equations (1) and (5):
2( 8) 8 32
xx
AA
 

1.6 lb
x
A
From Equation (1):
(1.6 8) 9.6 lb
x
B  

(1.600 lb) (36.0 lb) (2.00 lb)
 Aijk 

(9.60 lb) (36.0 lb) (2.00 lb)
 Bijk 
www.elsolucionario.org

330. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.127
Three children are standing on a 55-m raft. If the
weights of the children at Points
A, B, and C are 375 N,
260 N, and 400 N, respectively, determine the magnitude
and the point of application of the resultant of the three
weights.

SOLUTION

We have
:
ABC
FF F F R

(375 N) (260 N) (400 N)
(1035 N)
  

jjjR
jR


or 1035 NR 
We have
:()()()()
xAABBCC D
MFzFzFz Rz

(375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (1035 N)( )
D
z 

3.0483 m
D
z or 3.05 m
D
z 
We have
:()()()()
zAABBCC D
MFxFxFx Rx

375 N(1 m) (260 N)(1.5 m) (400 N)(4.75 m) (1035 N)( )
D
x 

2.5749 m
D
x or 2.57 m
D
x 
www.elsolucionario.org

331. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.128
Three children are standing on a 55-m raft. The
weights of the children at Points
A, B, and C are 375 N,
260 N, and 400 N, respectively. If a fourth child of weight
425 N climbs onto the raft, determine where she should
stand if the other children remain in the positions shown
and the line of action of the resultant of the four weights
is to pass through the center of the raft.

SOLUTION

We have
:
ABC
FF F F R

(375 N) (260 N) (400 N) (425 N)   jjjjR

(1460 N)Rj
We have
:()()() ()()
xAA BB CC DD H
MFzFzFzFz Rz

(375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m)
(425 N)( ) (1460 N)(2.5 m)
D
z




1.16471 m
D
z or 1.165 m
D
z 
We have
:()()() ()()
zAABBCC DD H
MFxFxFxFx Rx

(375 N)(1 m) (260 N)(1.5 m) (400 N)(4.75 m)
(425 N)( ) (1460 N)(2.5 m)
D
x




2.3235 m
D
x or 2.32 m
D
x 
www.elsolucionario.org

332. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.129
Four signs are mounted on a frame spanning a
highway, and the magnitudes of the horizontal
wind forces acting on the signs are as shown.
Determine the magnitude and the point of
application of the resultant of the four wind forces
when
1 fta and 12 ft.b

SOLUTION
We have

Assume that the resultant R is applied at Point
P whose coordinates are (x, y, 0).
Equivalence then requires

: 105 90 160 50
z
FR
or 405 lbR 

: (5 ft)(105 lb) (1 ft)(90 lb) (3 ft)(160 lb)
(5.5 ft)(50 lb) (405 lb)
x
M
y


or
2.94 fty

: (5.5 ft)(105 lb) (12 ft)(90 lb) (14.5 ft)(160 lb)
(22.5 ft)(50 lb) (405 lb)
y
M
x



or
12.60 ftx
R acts 12.60 ft to the right of member
AB and 2.94 ft below member BC. 
www.elsolucionario.org

333. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.130
Four signs are mounted on a frame spanning a
highway, and the magnitudes of the horizontal
wind forces acting on the signs are as shown.
Determine a and b so that the point of application
of the resultant of the four forces is at G.

SOLUTION
Since R acts at G, equivalence then requires that
G
M of the applied system of forces also be zero. Then
at

: : ( 3) ft (90 lb) (2 ft)(105 lb)
(2.5 ft)(50 lb) 0
x
GM a 

or
0.722 fta 

: (9 ft)(105 ft) (14.5 ) ft (90 lb)
(8 ft)(50 lb) 0
y
Mb 


or 20.6 ftb 

www.elsolucionario.org

334. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.131
A concrete foundation mat of 5-m radius supports four equally spaced
columns, each of which is located 4 m from the center of the mat.
Determine the magnitude and the point of application of the resultant of
the four loads.

SOLUTION
Have: :
AC D E
FFFFFR
100 kN 125 kN 25kN 75 kN   Rjjjj

325 kN j
(325 kN)RP j or 325 kNR 

Have:
  :
xCCEE G
MFzFzRz

125 kN (4 m) (75 kN)(4 m)

325 kN
G
z

0.61539 m
G
z
or
0.615 m
G
z 

  :
zAADD G
MFxFxRx

 100kN (4 m) (25kN) 4m 325 kN
Gx
or
0.923 m
G
x 
www.elsolucionario.org

335. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.132
Determine the magnitude and the point of application of the smallest
additional load that must be applied to the foundation mat of Prob. 3.131
if the resultant of the five loads is to pass through the center of the mat.

SOLUTION
The smallest load P will be placed on the edge of the mat at a radius of 5 m. Therefore
22 2
(5 m) (1)
PP
xz
Have:
   :
xCCEE P G
MFzFzPzRz

125kN 4 m 75kN (4 m) ( ) (0)
P
Pz R

()200kNm
G
Pz (2)
Also
  :()
zAADD P G
MFxFxPxRx

100 kN (4 m) (25 kN)(4 m) ( ) (0)
P
Px R

( ) 300 kN m
P
Px
(3)
Substituting for
x and z from equations (2) and (3) into (1),
22
2
300 200
(5)
PP





22
2
2
(300) (200)
5200
(5)
P


or 72.111 kNP 72.1 kNP 
Finally, from (2) and (3)
4.16 m ; z 2.77 m
PP
x 
www.elsolucionario.org

336. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.133*
Three forces of the same magnitude P act on a cube of side a as
shown. Replace the three forces by an equivalent wrench and
determine (
a) the magnitude and direction of the resultant force
R, (
b) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION
Force-couple system at O:

()PPP P  Rijk ijk

R
O
aPa PaP
Pa Pa Pa
   
  
Mjikjik
kij


()
R
O
Pa  Mijk
Since R and
R
O
M have the same direction, they form a wrench with
1
.
R
O
MM Thus, the axis of the
wrench is the diagonal
OA. We note that

1
cos cos cos
33
xyz
a
a


1
3 54.7
3
xyz
R
O
RP
MM Pa



1 3
Pitch
3M Pa
pa
R P
  

(
a)
354.7
xyz
RP  
(
b) – a
(
c) Axis of the wrench is diagonal OA.


www.elsolucionario.org

337. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.134*
A piece of sheet metal is bent into the shape shown and is
acted upon by three forces. If the forces have the same
magnitude
P, replace them with an equivalent wrench and
determine (
a) the magnitude and the direction of the
resultant force R, (
b) the pitch of the wrench, (c) the axis
of the wrench.

SOLUTION
First reduce the given forces to an equivalent force-couple system ,
R
O
RM at the origin.
We have

:PPP RF jjk
or
PRk

5
:() ()
2
R
OO
aP aP aP M
 
 
 

Mjik
or
5
2
R
O
aP




Mijk
(
a) Then for the wrench,

RP 
and
axis
R

R
λ k

cos 0 cos 0 cos 1
xyz

or
90 90 0
xyz
   
(
b) Now

1axis
5
2
5
2
R
O
M
aP
aP

 



M
kijk


Then
5
12
aP
P
RP

M
or
5
2
Pa  www.elsolucionario.org

338. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.134* (Continued)

(
c) The components of the wrench are
1
(, ),RM where
11axis
,MM  and the axis of the wrench is
assumed to intersect the
xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require

zQ R
MrR
where
1zO
MMM
Then

55
()
22
aP aP x y P

     


ij k k i j k

Equating coefficients:

:or
:or
aP yP y a
aP xP x a
 
 
i
j

The axis of the wrench is parallel to the
z-axis and intersects the xy-plane at ,.xay a 
www.elsolucionario.org

339. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.135*
The forces and couples shown are applied to two screws
as a piece of sheet metal is fastened to a block of wood.
Reduce the forces and the couples to an equivalent
wrench and determine (
a) the resultant force R, ( b) the
pitch of the wrench, (
c) the point where the axis of the
wrench intersects the
xz-plane.

SOLUTION
First, reduce the given force system to a force-couple system.
We have
: (20N) (15N) 25N R   FijR
We have
:( )
R
OO CO
MrFMM

20 N(0.1 m) (4 N m) (1 N m)
(4 N m) (3 N m)
R
O
    
   
Mjij
ij

(
a) (20.0 N) (15.00 N) Rij 
(
b) We have
1
(0.8 0.6)[(4Nm) (3Nm)]
5 N m
R
RO
M
R
 
     

R
M
ij i j
Pitch:
15N m
0.200 m
25 NM
p
R 
 
or
0.200 mp 
(
c) From above, note that

1
R
O
MM
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in
the
xy-plane with a slope of
3
4
yx 
www.elsolucionario.org

340. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.136*
The forces and couples shown are applied to two screws as a
piece of sheet metal is fastened to a block of wood. Reduce the
forces and the couples to an equivalent wrench and determine
(
a) the resultant force R, ( b) the pitch of the wrench, (c) the point
where the axis of the wrench intersects the
xz-plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin.
We have
: (10 lb) (11 lb)  FjjR

(21lb)Rj
We have
:( )
R
OO CO
MrFMM

0 0 20 lb in. 0 0 15 lb in. (12 lb in)
0100 0110
(35 lb in.) (12 lb in.)
R
O



ijk ij k
Mj
ij

(
a) (21lb)Rj or (21.0 lb)Rj 
(
b) We have
1
1
( ) [(35 lb in.) (12 lb in.) ]
12 lb in. and (12 lb in.)
R
ROR
M
R 
    
  
R
λM λ
jij
Mj

and pitch
112 lb in.
0.57143 in.
21 lbM
p
R 
  or
0.571in.p 
www.elsolucionario.org

341. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.136* (Continued)

(c) We have
12
21
(35 lb in.)
R
O
R
O

 
MMM
MMM i
We require
2/
(35 lb in.) ( ) [ (21 lb) ]
35 (21 ) (21 )
QO
xz
xz


 Mr R
iik j
iki
From i:
35 21
1.66667 in.
z
z



From k:
021
0
x
z



The axis of the wrench is parallel to the y-axis and intersects the xz-plane at
0, 1.667 in.xz 
www.elsolucionario.org

342. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.137*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (
a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the
xz-plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin.
We have
:(84N)(80N) 116N R   FjkR
and
:( )
R
OO CO
MrFMM

0.6 0 0.1 0.4 0.3 0 ( 30 32 ) N m
0840 0 080
R
O

ijk ijk
jk M


(15.6 N m) (2 N m) (82.4 N m)
R
O
     Mijk
(
a) (84.0 N) (80.0 N) Rjk 
(
b) We have
1
84 80
[ (15.6 N m) (2 N m) (82.4 N m) ]
116
55.379 N m
R
ROR
M
R 

       

R
λM λ
jk
ij k

and
11
(40.102 N m) (38.192 N m)
R
M  Mjk
Then pitch
155.379 N m
0.47741 m
116 NM
p
R 
  or
0.477 mp  www.elsolucionario.org

343. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.137* (Continued)

(c) We have
12
21
[( 15.6 2 82.4 ) (40.102 38.192 )] N m
(15.6 N m) (42.102 N m) (44.208 N m)
R
O
R
O

    
     
MMM
MMM ij k j k
ijk
We require
2/
( 15.6 42.102 44.208 ) ( ) (84 80 )
(84 ) (80 ) (84 )
QO
xz
zxx

  
Mr R
ijkikjk
ijk
From i:
15.6 84
0.185714 m
z
z
  
or
0.1857 mz

From k:
44.208 84
0.52629 m
x
x
  
or
0.526 mx

The axis of the wrench intersects the xz-plane at

0.526 m 0 0.1857 mxyz
  
www.elsolucionario.org

344. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.138*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a single
equivalent wrench and determine (
a) the resultant R,
(
b) the pitch of the single equivalent wrench, (c) the point
where the axis of the wrench intersects the
xz-plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin at
B.
(
a) We have
815
: (26.4 lb) (17 lb)
17 17

   


FkijR


(8.00 lb) (15.00 lb) (26.4 lb)  Rijk 
and
31.4 lbR
We have
/
:
R
BABA A B B
Mr FMMM

815
0 10 0 220 238 264 220 14(8 15 )
17 17
00 26.4
(152 lb in.) (210 lb in.) (220 lb in.)
R
B
R
B 
       




ij k
Mkijikij
Mijk

(
b) We have
1
8.00 15.00 26.4
[(152 lb in.) (210 lb in.) (220 lb in.) ]
31.4
246.56 lb in.
R
ROR
M
R 
 


R
λM λ
ijk
ijk
www.elsolucionario.org

345. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.138* (Continued)

and
11
(62.818 lb in.) (117.783 lb in.) (207.30 lb in.)
R
M   Mijk
Then pitch
1246.56 lb in.
7.8522 in.
31.4 lbM
p
R

  or 7.85 in.p 
(c) We have
12
21
(152 210 220 ) ( 62.818 117.783 207.30 )
(214.82 lb in.) (92.217 lb in.) (12.7000 lb in.)
R
B
R
B

     
 
MMM
MMM i j k i j k
ij k
We require
2/QB
Mr R

214.82 92.217 12.7000 0
81526.4
(15 ) (8 ) (26.4 ) (15 )
xz
zz x x
 
 
 
ij k
ij k
ij j k

From i:
214.82 15 14.3213 in.zz
 
From k:
12.7000 15 0.84667 in.xx
  
The axis of the wrench intersects the xz-plane at
0.847 in. 0 14.32 in.xyz
  
www.elsolucionario.org

346. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.139*
Two ropes attached at A and B are used to move the trunk
of a fallen tree. Replace the forces exerted by the ropes
with an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the
axis of the wrench intersects the yz-plane.

SOLUTION
(a) First replace the given forces with an equivalent force-couple system  ,
R
O
RM at the origin.
We have

222
222
(6) (2) (9) 11 m
(14) (2) (5) 15 m
AC
BD
d
d


Then

1650 N
(6 2 9 )
11
(900 N) (300 N) (1350 N)
AC
T
 ijk
ij k

and

1500 N
(14 2 5 )
15
(1400 N) (200 N) (500 N)
BD
T
 ijk
ijk

Equivalence then requires

:
(900 300 1350 )
(1400 200 500 )
(2300 N) (500 N) (1850 N)
AC BD



FRT T
ij k
ijk
ij k


:
R
OOAACBBD
MMrT rT

(12 m) [(900 N) (300 N) (1350 N) ]
(9 m) [(1400 N) (200 N) (500 N) ]
(3600) (10,800 4500) (1800)
(3600 N m) (6300 N m) (1800 N m)
 
  
   
     kij k
iijk
ijk
ijk

The components of the wrench are
1
(, ),RM where

(2300 N) (500 N) (1850 N)
 Rijk  www.elsolucionario.org

347. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.139* (Continued)

(b) We have

22 2
100 (23) (5) (18.5) 2993.7 NR
Let

axis
1
(23 5 18.5 )
29.937
R
 
R
ij k
Then
1axis
1
(23 5 18.5 ) ( 3600 6300 1800 )
29.937
1
[(23)( 36) (5)(63) (18.5)(18)]
0.29937
601.26 N m
R
O
M


 
M
ij k i j k

Finally,
1601.26 N m
2993.7 NM
P
R


or
0.201 mP 
(c) We have
11axis

1
( 601.26 N m) (23 5 18.5 )
29.937
MM
    
λ
ij k

or
1
(461.93 N m) (100.421 N m) (371.56 N m)     Mijk
Now
21
( 3600 6300 1800 )
( 461.93 100.421 371.56 )
(3138.1 N m) (6400.4 N m) (2171.6 N m)
R
O

  
  
     
MMM
ijk
ijk
ijk
For equivalence:




www.elsolucionario.org

348. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.139* (Continued)

Thus, we require
2
()
P
yz  MrR r jk
Substituting:

3138.1 6400.4 2171.6 0
2300 500 1850 yz  
ijk
ijk

Equating coefficients:

: 6400.4 2300 or 2.78 m
: 2171.6 2300 or 0.944 m
zz
yy
 
 
j
k

The axis of the wrench intersects the yz-plane at
0.944 m 2.78 myz
 

www.elsolucionario.org

349. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.140*
A flagpole is guyed by three cables. If the tensions in the
cables have the same magnitude P, replace the forces
exerted on the pole with an equivalent wrench and
determine (a) the resultant force R, (b) the pitch of the
wrench, (c) the point where the axis of the wrench
intersects the xz-plane.

SOLUTION

(a) First reduce the given force system to a force-couple at the origin.
We have
:
BA DC DE
PP PF λλ λ R

43 34 9412
55 55 25525
P
  

 
 
Rjkijijk


3
(2 20 )
25P
Rijk 

2223275
(2)(20)(1)
25 25
P
RP
We have
:( )
R
OO
PMr M

43 34 9412
(24 ) (20 ) (20 )
55 55 25525
R
OPP PP PP P
aaa
 
 
 
 
jjk jij jijkM


24
()
5
R
O Pa
Mik

www.elsolucionario.org

350. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.140* (Continued)

(b) We have
1
R
RO
MM
where
3251
(2 20 ) (2 20 )
25 27 5 9 5
R
P
R P
     
R
ijk ijk

Then
1
1248
(2 20 ) ( )
595 155
PaPa
M

ijk ik
and pitch
1825 8
8115 5 27 5
M Paa
p
R P
 
  

or 0.0988pa 
(c)
11
81 8
(2 20 ) ( 2 20 )
67515 5 9 5
R
Pa Pa
M

  

Mijkijk

Then
21
24 8 8
( ) ( 2 20 ) ( 430 20 406 )
5 675 675
R
O Pa Pa Pa
   MMM ik i jk i j k
We require
2/QO
 Mr R

83
( 403 20 406 ) ( ) (2 20 )
675 25
3
[20 ( 2 ) 20 ]
25
Pa P
xz
P
zxz x
 
  
 
 




ij k ik ijk
ijk

From i:
3
8( 403) 20 1.99012
675 25
Pa P
zz a

 



From k:
3
8( 406) 20 2.0049
675 25
Pa P
x xa

 


The axis of the wrench intersects the xz-plane at

2.00 , 1.990
x az a  
www.elsolucionario.org

351. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.141*
Determine whether the force-and-couple system shown can
be reduced to a single equivalent force R. If it can,
determine R and the point where the line of action of R
intersects the yz-plane. If it cannot be so reduced, replace
the given system with an equivalent wrench and determine
its resultant, its pitch, and the point where its axis intersects
the yz-plane.

SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
:
AG
FF F R

(40 mm) (60 mm) (120 mm)
(50 N) 70 N
140 mm
(20N) (30N) (10N)
 

 
 

ij k
Rk
ijk

and
37.417 NR
We have
:( )
R
OO CO
MrFMM

0
[(0.12 m) (50 N) ] {(0.16 m) [(20 N) (30 N) (60 N) ]}
(160 mm) (120 mm)
(10 N m)
200 mm
(40 mm) (120 mm) (60 mm)
(14 N m)
140 mm
(18 N m) (8.4 N m) (10.8 N m)
R
O
R

 



 



  
Mjkiijk
ij
ijk
Mijk

To be able to reduce the original forces and couples to a single equivalent force, R and M must be
perpendicular. Thus,
0.RM
Substituting

?
(20 30 10 ) (18 8.4 10.8 ) 0    ijk i j k
or
?
(20)(18) (30)( 8.4) ( 10)(10.8) 0
  
or 00


R and M are perpendicular so that the given system can be reduced to the single equivalent force.

(20.0 N) (30.0 N) (10.00 N)
 Rij k  www.elsolucionario.org

352. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.141* (Continued)

Then for equivalence,

Thus, we require
R
Op p
yz MrRr jk
Substituting:

18 8.4 10.8 0
20 30 10
yz 

ijk
ij k
Equating coefficients:

: 8.4 20 or 0.42 m
:10.8 20 or 0.54 m
zz
yy
 
 
j
k

The line of action of R intersects the yz-plane at
0 0.540 m 0.420 mxy z  
www.elsolucionario.org

353. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.142*
Determine whether the force-and-couple system shown
can be reduced to a single equivalent force R. If it can,
determine R and the point where the line of action of R
intersects the
yz-plane. If it cannot be so reduced, replace
the given system with an equivalent wrench and
determine its resultant, its pitch, and the point where its
axis intersects the
yz-plane.

SOLUTION
First determine the resultant of the forces at D. We have

222
22 2
( 12) (9) (8) 17 in.
(6) (0) (8) 10 in.
DA
ED
d
d   
   

Then

34 lb
(12 9 8)
17
(24 lb) (18 lb) (16 lb)
DA

  Fijk
ijk

and

30 lb
(6 8)
10
(18 lb) (24 lb)
ED

 Fik
ik

Then

:
DAED
FRF F

(24 18 16 (18 24)
(42 lb) (18 lb) (8 lb)
  
  
ijkik
ijk

For the applied couple

222
(6) (6) (18) 611 in.
AK
d   
Then

160 lb in.
(6 6 18)
611
160
[ (1 lb in.) (1 lb in.) (3 lb in.) ]
11



Mijk
ijk

To be able to reduce the original forces and couple to a single equivalent force, R and M
must be perpendicular. Thus

?
0
RM www.elsolucionario.org

354. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.142* (Continued)

Substituting

?160
(42 18 8) ( 3) 0
11
    ijk ijk

or
?160
[( 42)( 1) (18)( 1) ( 8)(3)] 0
11
  

or
00


R and M are perpendicular so that the given system can be reduced to the single equivalent force.

(42.0 lb) (18.00 lb) (8.00 lb)  Rijk 
Then for equivalence,

Thus, we require
/PD
Mr R
where
/
(12 in.) [( 3)in.] ( in.)
PD
yz   rijk
Substituting:

160
(3)12(3)
11
42 18 8
[( 3)( 8) ( )(18)]
[( )( 42) ( 12)( 8)]
[( 12)(18) ( 3)( 42)]
yz
yz
z
y
    



   
ijk
ij k
i
j
k

Equating coefficients:

160
:4296or1.137in.
11
480
: 216 42( 3) or 11.59 in.
11
zz
yy
 
   
j
k

The line of action of R intersects the yz-plane at
0 11.59 in. 1.137 in.xy z  


www.elsolucionario.org

355. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.143*
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y-axis and applied respectively at A and B.

SOLUTION

Express the forces at A and B as

xz
xz
AA
BB

Aik
Bik

Then, for equivalence to the given force system,

:0
xxx
FAB (1)

:
zzz
FABR (2)

:()()0
xz z
MAaBab (3)

:()()
zx x
MAaBabM (4)
From Equation (1),
xx
BA
Substitute into Equation (4):

() ( )
xx
Aa Aa b M

and
xx
MM
AB
bb

From Equation (2),
zz
BRA
and Equation (3),
()()0
zz
Aa R A a b 

1
z
a
AR
b





and
1
z
a
BRR
b

 




z
a
BR
b

Then
1
Ma
R
bb
  

  
  
Ai k 

Ma
R
bb
 
 
 
 
Bik

www.elsolucionario.org

356. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.144*
Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force
passes through a given point while the other force lies in a given plane.

SOLUTION

First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as

and
xyz x y z
RRR MM M M   Rijk i j k
while the unknown forces A and B can be expressed as

and
xyz xz
AAA BB Aijk Bik
Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the
position vector r
P are also known.
Then, for equivalence of the two systems,

:
xxxx
FRAB (1)

:
yyy
FRA (2)

:
zzzz
FRAB (3)

:
xxzy
MMyAzA (4)

:
yyxzz
MMzAxAbB (5)

:
zzyx
M M xA yA (6)
Based on the above six independent equations for the six unknowns
(, , , , ,),
xyzxz
AAABBb there
exists a unique solution for A and B.
From Equation (2),
yy
AR  www.elsolucionario.org

357. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.144* (Continued)

Equation (6):
1
()
x yz
A xR M
y


 
Equation (1):
1
()
x xyz
BR xRM
y

 
 
Equation (4):
1
()
zxy
A MzR
y


 
Equation (3):
1
()
zz x y
BR MzR
y

 
 
Equation (5):
()
()
x yz
xzy
xMyMzM
b
MyRzR



www.elsolucionario.org

358. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.145*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given
point.

SOLUTION

First, observe that it is always possible to construct a line perpendicular to a given line so that the
constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes
of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through
the given point.
See Figures a and b.
We have
andRMRj M j and are known.
The unknown forces A and B can be expressed as

and
xyz xyz
AAA BBB A i jk B i jk
The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for
equivalence,

:
x
F 0
xx
AB (1)

:
y
F
yy
RA B (2)

:
z
F 0
zz
AB (3)

:
x
M 0
y
zB (4)

:
yzzx
MMaAxBzB (5)

:
z
M 0
yy
aA xB (6)
Since A and B are made perpendicular,

0or 0
xx yy zz
AB AB AB   AB (7)
There are eight unknowns:
,,,,,,,
xyzxyz
AAABBBxz
But only seven independent equations. Therefore, there exists an infinite number of solutions.
Next, consider Equation (4):
0
y
zB
If
0,
y
B Equation (7) becomes 0
xx zz
AB AB
Using Equations (1) and (3), this equation becomes
22
0
xz
AA www.elsolucionario.org

359. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.145* (Continued)

Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that
0,
y
B so that from Equation (4), 0.z
To obtain one possible solution, arbitrarily let
0.
x
A

(
Note: Setting ,,or
yzz
AAB equal to zero results in unacceptable solutions.)
The defining equations then become

0
x
B (1) 

y y
RA B (2)

0
zz
AB (3)

zz
M aA xB  (5) 

0
y y
aA xB (6)

0
yy zz
AB AB
  (7) 
Then Equation (2) can be written
y y
ARB
Equation (3) can be written
zz
BA

Equation (6) can be written
y
y
aA
x
B

Substituting into Equation (5),

()
y
zz
y
RB
M aA a A
B

   



or
zy
M
A B
aR
 (8)
Substituting into Equation (7),
() 0
yy y y
MM
RBB B B
aR aR

 



or
23
22 2
y
aR
B
aR M



Then from Equations (2), (8), and (3),

22 2
22 2 22 2
23 2
22 2 22 2
2
22 2
y
z
z
aR RM
AR
aR M aR M
MaR aRM
A
aRaR M aR M
aR M
B
aR M
 


 






PROBLEM 3.145* (Continued) www.elsolucionario.org

360. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


In summary,

22 2
()
RM
MaR
aR M


Ajk 

2
22 2
()
aR
aR M
aR M


Bjk 
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is
applied at a given point.
Lastly, if
0R and 0,M it follows from the equations found for A and B that 0
y
A and 0.
y
B
From Equation (6),
0x (assuming 0).a Then, as a consequence of letting 0,
x
A force A lies in a
plane parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yz-
plane but to the left to the origin, as shown in the figure below.

www.elsolucionario.org

361. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.146*
Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.

SOLUTION

First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action (AA). Note that it has been assumed that the line of
action of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA by

Ax y z
ijk
it follows that force A can be expressed as

()
Axyz
AA Aijk
Force B can be expressed as

xyz
BBBBijk
Next, observe that since the axis of the wrench and the prescribed line of action AA are known, it follows
that the distance a can be determined. In the following solution, it is assumed that a is known.
Then for equivalence,

:0
xxx
FAB  (1)

:
yyy
FRA B (2)

:0
zzz
FAB  (3)

:0
xy
MzB (4)

:
yzxz
MMaA zBxB (5)

:0
xyy
MaAxB  (6)
Since there are six unknowns (A, B
x, By, Bz, x, z) and six independent equations, it will be possible to
obtain a solution.
www.elsolucionario.org

362. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.146* (Continued)

Case 1
: Let 0z to satisfy Equation (4).
Now Equation (2):
y y
A RB
Equation (3):
zz
BA

Equation (6): ()
y
y
yy
aA a
x RB
BB
 
  



Substitution into Equation (5):

()()
1
zyz
y
y
z
a
MaA RBA
B
M
AB
aR
 

 
     

 
 




Substitution into Equation (2):

2
1
yy y
z
z
y
zy
M
R BB
aR
aR
B
aR M










Then
zy
yz
x
xx
zy
z
zz
zy
MR R
A
aRaR M
M
MR
BA
aR M
MR
BA
aR M

 






 


 

 

In summary,
A
yz
P
aR
M



A 

()
xz z
zy
R
MaRM
aR M


Bijk 
and
2
1
1
y
zy
z
R
xa
B
aR M
aR
aR






 
 

  

or
y
z
M
x
R


 
Note that for this case, the lines of action of both
A and B intersect the x-axis. www.elsolucionario.org

363. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.146* (Continued)

Case 2: Let
0
y
B to satisfy Equation (4).
Now Equation (2):
y
R
A


Equation (1):
x
x
y
BR






Equation (3):
z
z
y
BR






Equation (6):
0
y
aA which requires 0a
Substitution into Equation (5):

or
x z
zx y
yy M
MzR xR x z
R 
 

 

    

 

 

This last expression is the equation for the line of action of force B.
In summary,

A
y
R






A 

()
xx
y
R






Bik 
Assuming that
,, 0,
xyz
. the equivalent force system is as shown below.

Note that the component of A in the xz-plane is parallel to B.

www.elsolucionario.org

364. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.147
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving it
into horizontal and vertical components. (b) Using the result of
part (a), determine the perpendicular distance from O to the
line of action of P.

SOLUTION


(0.2 m)cos 40
0.153209 m
(0.2 m)sin 40
0.128558 m
x
y






/
(0.153209 m) (0.128558 m)
AO
 rij




(a)
(300 N)sin 30
150 N
(300 N)cos30
259.81 N
x
y
F
F




(150 N) (259.81 N)Fi j

/
(0.153209 0.128558 ) m (150 259.81 ) N
(39.805 19.2837 ) N m
(20.521 N m)
OAO

 
 

Mr F
ijij
kk
k
20.5 N m
O
M


(b)
O
MFd

20.521 N m (300 N)( )
0.068403 m
d
d


68.4 mmd 
www.elsolucionario.org

365. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.148
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.


SOLUTION
(a) Slope of line:
0.875 m 5
1.90 m 0.2 m 12
EC

Then
12
()
13
ABx AB
TT

12
(1040 N)
13
960 N

 ( a)
and
5
(1040 N)
13
400 N
ABy
T


Then
(0.875 m) (0.2 m)
(960 N)(0.875 m) (400 N)(0.2 m)
D ABx ABy
MT T


760 N m or 760 N m
D
M 




(b) We have
() ()
D ABx ABx
M TyTx

(960 N)(0) (400 N)(1.90 m)
760 N m


( b)

or
760 N m
D
M 
www.elsolucionario.org

366. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.149
A small boat hangs from two davits, one of which is shown in
the figure. The tension in line ABAD is 82 lb. Determine the
moment about C of the resultant force
RA exerted on the davit
at A.

SOLUTION
We have 2
A AB AD
RFF
where
(82 lb)
AB
Fj
and
67.753
(82 lb)
10.25
(48 lb) (62 lb) (24 lb)
AD AD
AD
AD
AD


 ijk
FF
Fijk


Thus
2 (48 lb) (226 lb) (24 lb)
AABAD
  RFF i j k
Also
/
(7.75 ft) (3 ft)
AC
rjk
Using Eq. (3.21):
07.75 3
48 226 24
(492 lb ft) (144.0 lb ft) (372 lb ft)
C


 
ijk
M
ijk


(492 lb ft) (144.0 lb ft) (372 lb ft)
C
 Mijk 

www.elsolucionario.org

367. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.150
Consider the volleyball net
shown. Determine the angle
formed by guy wires AB and
AC.

SOLUTION
First note:
222
222
( 6.5) ( 8) (2) 10.5 ft
(0) ( 8) (6) 10 ft
AB
AC
   


and
(6.5 ft) (8 ft) (2 ft)
(8 ft) (6 ft)AB
AC
  
  ijk
jk



By definition,
()()cosAB AC AB AC


or
( 6.5 8 2 ) ( 8 6 ) (10.5)(10)cos
( 6.5)(0) ( 8)( 8) (2)(6) 105cos


 

ijk jk
or
cos 0.72381
 or 43.6 

www.elsolucionario.org

368. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.151
A single force P acts at C in a direction perpendicular to
the handle BC of the crank shown. Determine the moment
M
x of P about the x-axis when θ  65°, knowing that
M
y  15 N · m and M z  36 N · m.

SOLUTION
See the solution to Prob. 3.53 for the derivation of the following equations:

(0.2) sin( )
x
MP
 (1)
tan
z
y
M
M
 (4)

22
4
yz
P MM (5)
Substituting for known data gives:

22
36 N m
tan
15 N m
67.380
4 ( 15) ( 36)
156.0 N
0.2 m(156.0 N)sin(65 67.380 )
23.047 N m
x
P
P
M






 





23.0 N m
x
M 
www.elsolucionario.org

369. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.152
A small boat hangs from two davits, one of which is shown in
the figure. It is known that the moment about the z-axis of the
resultant force
A
R exerted on the davit at A must not exceed
279 lbft in absolute value. Determine the largest allowable
tension in line ABAD when
6xft.

SOLUTION
First note: 2
AABAD
RTT
Also note that only
TAD will contribute to the moment about the z-axis.
Now
222
(6)(7.75)(3)
10.25 ft
AD


Then
(6 7.75 3 )
10.25
AD
AD
T
AD
T


T
ijk


Now
/
()
zACAD
M kr T
where
/
(7.75 ft) (3 ft)
AC
rjk
Then for
max
,T
max
max
00 1
279 0 7.75 3
10.25
67.753
|(1)(7.75)(6)|
10.25
T
T




or
max
61.5 lbT 
www.elsolucionario.org

370. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.153
In a manufacturing operation, three holes are drilled
simultaneously in a workpiece. If the holes are
perpendicular to the surfaces of the workpiece, replace the
couples applied to the drills with a single equivalent
couple, specifying its magnitude and the direction of
its axis.

SOLUTION

123
222
(1.5 N m)( cos20 sin 20 ) (1.5 N m)
(1.75 N m)( cos 25 sin 25 )
(4.4956 N m) (0.22655 N m)
(0) ( 4.4956) (0.22655)
4.5013 N m
M



   
 

MM M M
jk j
jk
jk
4.50 N mM 

axis
(0.99873 0.050330 )
cos 0
cos 0.99873
cos 0.050330
x
y
z
M



 



M
jk
90.0 , 177.1 , 87.1
xy z
 
  
www.elsolucionario.org

371. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.154
A 260-lb force is applied at A to the rolled-steel section shown. Replace that
force with an equivalent force-couple system at the center C of the section.

SOLUTION

22
(2.5 in.) (6.0 in.) 6.50 in.AB

2.5 in. 5
sin
6.5 in. 13
6.0 in. 12
cos 22.6
6.5 in. 13


 

sin cos
512
(260 lb) (260 lb)
13 13
(100.0 lb) (240 lb)
FF 
 
 
Fij
ij
ij


/
(2.5 4.0 ) ( 100.0 240 )
400 600
(200 lb in.)
CAC

 

 Mr F
ij i j
kk
k


260 lbF 67.4°; 200 lb in.
C
M 
www.elsolucionario.org

372. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 3.155
The force and couple shown are to be replaced by an equivalent single force.
Knowing that
P  2Q, determine the required value of α if the line of action
of the single equivalent force is to pass through (
a) Point A, (b) Point C.

SOLUTION











(
a) We must have 0
A
M


(sin) () 0PaQa
 

1
sin
22
QQ
PQ



30.0
 

(
b) We must have 0
C
M


(sin) (cos) () 0PaP aQa
  

1
sin cos
22
QQ
PQ

  

1
sin cos
2

  (1)

22 1
sin cos cos
4

 

22 1
1cos cos cos
4

 

2
2cos cos 0.75 0
  (2)
Solving the quadratic in
cos :


17
cos 65.7 or 155.7
4


  
Only the first value of
 satisfies Eq. (1),
therefore
65.7
 


www.elsolucionario.org

373. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.156
A 77-N force F1 and a 31-N  m
couple
M1 are applied to corner E of
the bent plate shown. If
F1 and M1
are to be replaced with an equivalent
force-couple system (
F2, M2) at
corner
B and if (M2)z  0, determine
(
a) the distance d, (b) F2 and M2.

SOLUTION
(a) We have
2
:0
Bz z
MM 

/1 1
() 0
HB z
Mkr F (1)
where
/
(0.31 m) (0.0233)
HB
ri
j

11
11
11
2
(0.06 m) (0.06 m) (0.07 m)
(77 N)
0.11 m
(42 N) (42 N) (49 N)
(0.03 m) (0.07 m)
(31 N m)
0.0058 m
EH
z
EJ
F
M
M
d
d






 


Fλ
ijk
ijk
kM
Mλ
ijk

Then from Equation (1),

2
001
(0.07m)(31Nm)
0.31 0.0233 0 0
0.0058
42 42 49
d

 


Solving for d, Equation (1) reduces to

2
2.17 N m
(13.0200 0.9786) 0
0.0058d

 

from which
0.1350 md
 or 135.0 mmd  www.elsolucionario.org

374. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.156 (Continued)

(b)
21
(42 42 49 ) N  FF i j k or
2
(42.0 N) (42.0 N) (49.0 N)Fijk 

2/11
2
(0.1350) 0.03 0.07
0.31 0.0233 0 (31 N m)
0.155000
42 42 49
(1.14170 15.1900 13.9986 ) N m
( 27.000 6.0000 14.0000 ) N m
(25.858 N m) (21.190 N m)
HB


  

 
   
   
Mr FM
ijk
ijk
ijk
ij k
Mij

or
2
(25.9 N m) (21.2 N m)   Mij 


www.elsolucionario.org

375. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.157
Three horizontal forces are applied as shown to a vertical cast iron
arm. Determine the resultant of the forces and the distance from the
ground to its line of action when (a) P  200 N, (b) P  2400 N,
(c) P  1000 N.

SOLUTION
(a)






200 N 600 N 400 N 800 N
D
R   

(200 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m)
150.0 N m
D
M  
 

150 N m
0.1875 m
800 N
D
M
y
R

 


800 NR
; 187.5 mmy 

(b)






2400 N 600 N 400 N 1400 N
D
R   

(2400 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m)
840 N m
D
M  
 

840 N m
0.600 m
1400 N
D
M
y
R

 


1400 NR ; 600 mmy  www.elsolucionario.org

376. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.157 (Continued)

(c)


1000 600 400 0
D
R   

(1000 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m)
210 N m
D
M  
 



y System reduces to a couple.

210 N m
D
M 


www.elsolucionario.org

377. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.158
While using a pencil sharpener, a
student applies the forces and couple
shown. (a) Determine the forces
exerted at B and C knowing that these
forces and the couple are equivalent
to a force-couple system at A
consisting of the force
(2.6 lb) +
Ri (0.7
y
Rj lb)k and
the couple

R
Ax
MMi
(1.0 lb · ft)
j(0.72 lb · ft) .k

(b) Find
the corresponding values of
y
R and
.
x
M

SOLUTION
(a) From the statement of the problem, equivalence requires

:FBCR
or
:2.6 lb
xxx
FBC (1)

:
y yy
F CR (2)

: 0.7 lb or 0.7 lb
zz z
FC C 
and
//
:( )
R
A BA B CA A
MMrBMrC
or
1.75
: (1 lb ft) ft ( )
12
x yx
CM

 


M (3)

3.75 1.75 3.5
: ft( ) ft( ) ft(0.7 lb) 1 lbft
12 12 12
yx x
MBC




or
3.75 1.75 9.55
xx
BC
Using Eq. (1):
3.75 1.75(2.6 ) 9.55
xx
BB
or
2.5 lb
x
B
and
0.1 lb
x
C

3.5
:ft()0.72 lbft
12
zy
MC

  


or
2.4686 lb
y
C

(2.50 lb) (0.1000 lb) (2.47 lb) (0.700 lb)
 BiC ij k 
(b) Eq. (2) 
2.47 lb
y
R 
Using Eq. (3):
1.75
1 (2.4686)
12
x
M




or 1.360 lb ft
x
M  www.elsolucionario.org

378. CHAPTER 4
www.elsolucionario.org

379. www.elsolucionario.org

380. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.1
A gardener uses a 60-N wheelbarrow to transport a 250-N bag
of fertilizer. What force must she exert on each handle?

SOLUTION
Free-Body Diagram:








0: (2 )(1 m) (60 N)(0.15 m) (250 N)(0.3 m) 0
A
MF   


42.0 NF 
www.elsolucionario.org

381. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.2
The gardener of Problem 4.1 wishes to transport a second
250-N bag of fertilizer at the same time as the first one.
Determine the maximum allowable horizontal distance from
the axle A of the wheelbarrow to the center of gravity of the
second bag if she can hold only 75 N with each arm.
PROBLEM 4.1 A gardener uses a 60-N wheelbarrow to
transport a 250-N bag of fertilizer. What force must she exert on
each handle?

SOLUTION
Free-Body Diagram:









0: 2(75 N)(1 m) (60 N)(0.15 m)
(250 N)(0.3 m) (250 N) 0
A
M
x
 



0.264 mx 


www.elsolucionario.org

382. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.3
A 2100-lb tractor is used to lift 900 lb of gravel.
Determine the reaction at each of the two (a) rear wheels
A, (b) front wheels B.

SOLUTION

(a) Rear wheels
0: (2100 lb)(40 in.) (900 lb)(50 in.) 2 (60 in.) 0
B
MA    
325 lb A 325 lb A 
(b) Front wheels
: (2100 lb)(20 in.) (900 lb)(110 in.) 2 (60 in.) 0
A
MB    1175 lb B 1175 lb B 
Check:
0: 2 2 2100 lb 900 lb 0
y
FAB     
2(325 lb) 2(1175 lb) 2100 lb 900 0
00(Checks)  
www.elsolucionario.org

383. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.4
For the beam and loading shown, determine (a) the reaction at A,
(b) the tension in cable BC
.

SOLUTION
Free-Body Diagram:

(a) Reaction at A:
0: 0
xx
FA 
0: (15 lb)(28 in.) (20 lb)(22 in.) (35 lb)(14 in.)
(20 lb)(6 in.) (6 in.) 0
B
y
M
A
  


245 lb
y
A 245 lb A 
(b) Tension in BC:
0: (15lb)(22in.) (20lb)(16in.) (35lb)(8in.)
(15 lb)(6 in.) (6 in.) 0
A
BC
M
F
  

140.0 lb
BC
F 140.0 lb
BC
F 
Check: 0: 15 lb 20 lb 35 lb 20 lb 0
105 lb 245 lb 140.0 0
yBC
FAF      
 

00(Checks) 


www.elsolucionario.org

384. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.5
A load of lumber weighing 25 kNW is being raised by a
mobile crane. The weight of the boom ABC and the
combined weight of the truck and driver are as shown.
Determine the reaction at each of the two (a) front wheels
H, (b) rear wheels K.

SOLUTION
Free-Body Diagram:








(a)

   0: 25 kN 5.4 m + 3 kN 3.4 m 2 2.5 m + 50 kN 0.5 m 0
KH
MF  

2 68.080 kN
H
F or 34.0 kN
H
F

(b)

 0: 25 kN 2.9 m + 3 kN 0.9 m 50 kN 2.0 m + 2 2.5 m 0
H K
MF  

2 9.9200 kN
K
F or 4.96 kN
K
F


www.elsolucionario.org

385. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.6
A load of lumber weighing 25 kNW is being raised as
shown by a mobile crane. Knowing that the tension is 25
kN in all portions of cable AEF and that the weight of boom
ABC is 3 kN, determine (a) the tension in rod CD, (b) the
reaction at pin B.

SOLUTION
Free-Body Diagram: (boom)







(a)

   0: 25 kN 2.6 m + 3 kN 0.6 m 25 kN 0.4 m 0.7 m 0
B CD
MT   

81.143 kN
CD
T or 81.1 kN
CD
T 
(b)
0: 0
xx
FB  so that
y
BB


0: 25 3 25 81.143 kN 0
y
FB   
134.143 kN B or 134.1 kN B



www.elsolucionario.org

386. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.7
A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if
10 in.,a (b) if 7 in.a

SOLUTION
Free-Body Diagram: 0: 0
xx
FB 

0: (40lb)(6in.) (30lb) (10lb)( 8in.) (12in.) 0
B
MaaA     

(40 160)
12
a
A


(1)

0: (40lb)(6in.) (50lb)(12in.) (30lb)( 12in.)
(10 lb)( 20 in.) (12 in.) 0
A
y
Ma
aB
    
 

(1400 40 )
12
y
a
B



Since
(1400 40 )
0,
12
x
a
BB


(2)
(a)
For 10 in.,a
Eq. (1):
(40 10 160)
20.0 lb
12
A


20.0 lb A 
Eq. (2):
(1400 40 10)
150.0 lb
12
B


150.0 lb B 
(b)
For 7 in.,a
Eq. (1):
(40 7 160)
10.00 lb
12
A


10.00 lb A 
Eq. (2):
(1400 40 7)
140.0 lb
12
B


140.0 lb B 
www.elsolucionario.org

387. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.8
For the bracket and loading of Problem 4.7, determine the
smallest distance a if the bracket is not to move.
PROBLEM 4.7 A T-shaped bracket supports the four loads
shown. Determine the reactions at A and B (a) if
10 in.,a (b) if
7 in.a

SOLUTION
Free-Body Diagram:

For no motion, reaction at A must be downward or zero; smallest distance a for no motion
corresponds to
0.A

0: (40lb)(6in.) (30lb) (10lb)( 8in.) (12in.) 0
B
MaaA     

(40 160)
12
a
A



0: (40 160) 0 Aa 4.00 in.a 
www.elsolucionario.org

388. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.9
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Neglecting the weight of the beam,
determine the range of values of Q for which neither cable becomes
slack when
0.P

SOLUTION
Free-Body Diagram:







For
min
, 0
D
QT 

 
min
0: 7.5 kN 0.5 m 3 m 0
B
MQ  

min
1.250 kNQ
For
max
, 0
B
QT 

 
max
0: 7.5 kN 2.75 m 0.75 m 0
D
MQ  

max
27.5 kNQ
Therefore:
1.250 kN 27.5 kNQ 

www.elsolucionario.org

389. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.10
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Knowing that the maximum allowable
tension in each cable is 12 kN and neglecting the weight of the
beam, determine the range of values of Q for which the loading is
safe when 0.P

SOLUTION
Free-Body Diagram:








0: 7.5 kN 2.75 m 2.25 m 0.75 m 0
DB
MTQ   

27.5 3
B
QT (1)

0: 7.5 kN 0.5 m + 2.25 m 3 m 0
BD
MTQ  

1.25 0.75
D
QT (2)
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
Thus, making 0
 TB  12 kN in (1), we have
-8.50 kN
 Q  27.5 kN (3)
And making 0
 TD  12 kN in (2), we have
1.25
 Q  10.25 kN (4)
(3) and (4) now give: 1.250 kN
 Q  10.25 kN 
www.elsolucionario.org

390. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.11
For the beam of Prob. 4.10, determine the range of values of Q for
which the loading is safe when P = 5 kN.

SOLUTION
Free-Body Diagram:








0: 7.5kN 2.75m 2.25m + 5kN 1.5m 0.75 m 0
DB
MTQ   

37.5 3 kN
B
QT (1)

0: 7.5 kN 0.5 m 5 kN 0.75 m + 2.25 m 3 m 0
BD
MTQ   

0.75 kN
D
QT (2)
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
Thus, making 0
 TB  12 kN in (1), we have
1.500 kN
 Q  37.5 kN (3)
And making 0
 TD  12 kN in (2), we have
0
 Q  9.00 kN (4)
(3) and (4) now give: 1.500 kN
 Q  9.00 kN 
www.elsolucionario.org

391. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.12
For the beam of Sample Prob. 4.2, determine the range of values
of P for which the beam will be safe, knowing that the maximum
allowable value of each of the reactions is 25 kips and that the
reaction at A must be directed upward.

SOLUTION
Free-Body Diagram:




  0: (3 ft) 9 ft (6 kips) 11ft 6 kips 13 ft 0
Ay
MPB     


3B 48 kips
y
Q (1)


  0: 9 ft 6 ft (6 kips) 2 ft (6 kips) 4 ft 0
By
MAP     

1.5 +6 kips
y
PA (2)
For the loading to meet the design criteria, the reactions must not exceed 25 kips.
Thus, making
25 kips
y
B in (1), we have
P
 27.0 kips (3)
And making 0
 Ay  25 kips in (2), we have
6.00
 P  43.5 kips (4)
(3) and (4) now give: 6.00 kips
 P  27.0 kips
www.elsolucionario.org

392. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.13
The maximum allowable value of each of the reactions is 180 N.
Neglecting the weight of the beam, determine the range of the
distance d for which the beam is safe.

SOLUTION


0: 0
xx
FB 

y
BB

0: (50 N) (100 N)(0.45 m ) (150 N)(0.9 m ) (0.9 m ) 0
A
Md d dBd    

50 45 100 135 150 0.9 0dd dBBd     

180 N m (0.9 m)
300
B
d
AB



(1)

0: (50 N)(0.9 m) (0.9 m ) (100 N)(0.45 m) 0
B
MAd   

45 0.9 45 0AAd

(0.9 m) 90 N mA
d
A


(2)
Since 180 N,B Eq. (1) yields

180 (0.9)180 18
0.15 m
300 180 120
d



150.0 mm d 
Since 180 N,A Eq. (2) yields

(0.9)180 90 72
0.40 m
180 180
d


400 mm d 
Range: 150.0 mm 400 mm d


www.elsolucionario.org

393. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.14
For the beam and loading shown, determine the range of
the distance a for which the reaction at B does not exceed
100 lb downward or 200 lb upward.

SOLUTION
Assume B is positive when directed .


Sketch showing distance from D to forces.


0: (300 lb)(8 in. ) (300 lb)( 2 in.) (50 lb)(4 in.) 16 0
D
Maa B     
600 2800 16 0 aB 

(2800 16 )
600
B
a


(1)
For 100 lbB 100 lb, Eq. (1) yields:

[2800 16( 100)] 1200
2in.
600 600
a


2.00 in. a 
For
200B 200 lb, Eq. (1) yields:

[2800 16(200)] 6000
10 in.
600 600
a


10.00 in. a 
Required range: 2.00 in. 10.00 in. a 


www.elsolucionario.org

394. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.15
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction
at C.

SOLUTION
Free-Body Diagram:


0: (100 mm) (120 mm) 0
CAB DE
MF F  

5
6
DE AB
FF (1)

(a) For
720 N
AB
F

5
(720 N)
6
DE
F 600 N
DE
F 
(b)
3
0: (720 N) 0
5
xx
FC   

432 N
x
C

4
0: (720 N) 600 N 0
5
1176 N
yy
y
FC
C
    



1252.84 N
69.829
C




1253 N C 69.8° 
www.elsolucionario.org

395. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.16
Two links AB and DE are connected by a bell crank as
shown. Determine the maximum force that may be
safely exerted by link AB on the bell crank if the
maximum allowable value for the reaction at C is
1600 N.

SOLUTION
See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1).

5
6
DE AB
FF (1)

33
0: 0
55
xABxxAB
FFCCF    

4
0: 0
5
yAByDE
FFCF    

45
0
56
49
30
AB y AB
yAB
FC F
CF




22
22
1
(49) (18)
30
1.74005
xy
AB
AB
CCC
F
CF




For
1600 N, 1600 N 1.74005
AB
CF 920 N
AB
F 


www.elsolucionario.org

396. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.17
The required tension in cable AB is 200 lb. Determine (a) the
vertical force P that must be applied to the pedal, (b) the
corresponding reaction at C.

SOLUTION
Free-Body Diagram:

7in. BC
(a)
0: (15 in.) (200 lb)(6.062 in.) 0  
C
MP
80.83 lb P 80.8 lb P 
(b)
0: 200 lb 0
yx
FC   200 lb
x
C

0: 0 80.83 lb 0
yy y
FCPC    80.83 lb
y
C

22.0
215.7 lbC



216 lbC 22.0 
www.elsolucionario.org

397. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.18
Determine the maximum tension that can be developed in cable
AB if the maximum allowable value of the reaction at C is 250 lb.

SOLUTION
Free-Body Diagram:

7in. BC

0: (15 in.) (6.062 in.) 0 0.40415
C
MP T P T   

0: 0 0.40415 0
yy y
FPC PC     

0.40415
y
CT

0: 0
xx
FTC  
x
CT

22 2 2
(0.40415 )
1.0786
xy
CCC T T
CT



For 250 lb C
250 lb 1.0786 T 232 lb T 

www.elsolucionario.org

398. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.19
The bracket BCD is hinged at C and attached to a control
cable at B. For the loading shown, determine (a) the
tension in the cable, (b) the reaction at C.

SOLUTION

At B:
0.18 m
0.24 my
x
T
T


3
4
yx
TT (1)
(a)
0: (0.18 m) (240 N)(0.4 m) (240 N)(0.8 m) 0
Cx
MT   

1600 N
x
T
From Eq. (1):
3
(1600 N) 1200 N
4
y
T

22 2 2
1600 1200 2000 N
xy
TTT   2.00 kNT 
(b)
0: 0
xxx
FCT 

1600 N 0 1600 N
xx
CC 1600 N
x
C

0: 240N 240N 0
yyy
FCT   

1200 N 480 N 0
y
C

1680 N
y
C 1680 N
y
C

46.4
2320 NC

 2.32 kNC 46.4° 

www.elsolucionario.org

399. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.20
Solve Problem 4.19, assuming that 0.32 m.a
PROBLEM 4.19 The bracket BCD is hinged at C and
attached to a control cable at B. For the loading shown,
determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B:
0.32 m
0.24 m
4
3y
x
yx
T
T
TT




0: (0.32m)(240N)(0.4m)(240N)(0.8m)0
Cx
MT   

900 N
x
T
From Eq. (1):
4
(900 N) 1200 N
3
y
T

22 2 2
900 1200 1500 N
xy
TTT   1.500 kNT 

0: 0
xxx
FCT 

900 N 0 900 N
xx
CC 900 N
x
C

0: 240 N 240 N 0
yyy
FCT   

1200 N 480 N 0
y
C

1680 N
y
C 1680 N
y
C

61.8
1906 NC

 1.906 kN 
C 61.8° 

www.elsolucionario.org

400. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.21
The 40-ft boom AB weighs 2 kips; the distance from the
axle A to the center of gravity G of the boom is 20 ft. For
the position shown, determine (a) the tension T in the cable,
(b) the reaction at A.

SOLUTION


( )sin10 (40 ft)sin10
6.9459 ft
AD AB
AD





0: ( ) 2(20cos30 ) 5(40cos30 ) 0
A
MTAD   


29.924 kips T
29.9 kipsT


0: (29.924)cos 20 0
xx
FA  



28.119 kips
x
A

0: (29.924)sin 20 2 5 0
yy
FA  



17.2346 kips
y
A

22
28.119 17.2346 32.980 kipsA

117.2346
tan
28.119
31.5





 33.0 kips
A 31.5° 


www.elsolucionario.org

401. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.22
A lever AB is hinged at C and attached to a control cable at A. If
the lever is subjected to a 500-N horizontal force at B,
determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

Triangle ACD is isosceles with 90 30 120C 

1
(180 120 ) 30 .
2
AD 

Thus, DA forms angle of 60° with the horizontal axis.
(a) We resolve
AD
F into components along AB and perpendicular to AB.


0: ( sin 30 )(250 mm) (500 N)(100 mm) 0
CAD
MF    400 N
AD
F 
(b)
0: (400 N)cos60 500 N 0
xx
FC     300 N
x
C

0: (400 N)sin 60° 0
yy
FC    346.4 N
y
C
458 N 
C 49.1° 
www.elsolucionario.org

402. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.23
For each of the plates and loadings shown, determine the reactions at A and B.


SOLUTION
(a) Free-Body Diagram:


0: (20 in.) (50 lb)(4 in.) (40 lb)(10 in.) 0
A
MB   
30 lb B 30.0 lb
B



0: 40 lb 0
xx
FA  

40 lb
x
A 40.0 lb
x
A

0: 50 lb 0
yy
FAB  

30 lb 50 lb 0
y
A

20 lb
y
A 20.0 lb
y
A


26.56
44.72 lbA

 44.7 lb 
A 26.6° 




www.elsolucionario.org

403. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.23 (Continued)

(b)
Free-Body Diagram:


0: ( cos30 )(20 in.) (40 lb)(10 in.) (50 lb)(4 in.) 0
A
MB    
34.64 lb B 34.6 lb 
B 60.0° 

0: sin30 40 lb
xx
FAB  

(34.64 lb)sin 30 40 lb 0
x
A

22.68 lb
x
A 22.68 lb
x
A

0: cos30 50 lb 0
yy
FAB   

(34.64 lb)cos30 50 lb 0
y
A

20 lb
y
A 20.0 lb
y
A

41.4
30.24 lbA

 30.2 lb 
A 41.4° 

www.elsolucionario.org

404. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.24
For each of the plates and loadings shown, determine the reactions at A and B.


SOLUTION
(a) Free-Body Diagram:


0: (20 in.) (50 lb)(16 in.) (40 lb)(10 in.) 0
B
MA   
20 lb A 20.0 lb 
A 

0: 40 lb 0
xx
FB  

40 lb
x
B 40 lb
x
B

0: 50 lb 0
yy
FAB   

20 lb 50 lb 0
y
B 

30 lb
y
B 30 lb
y
B

36.87 50 lbB 50.0 lb  B 36.9° 




www.elsolucionario.org

405. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.24 (Continued)

(b)


0: ( cos30 )(20 in.) (40 lb)(10 in.) (50 lb)(16 in.) 0
A
MA     
23.09 lb A 23.1lb 
A 60.0° 

0: sin30 40 lb 0
xx
FA B   

(23.09 lb)sin 30 40 lb 8 0
x
  

51.55 lb
x
B 51.55 lb
x
B

0: cos30 50 lb 0
yy
FA B   

(23.09 lb)cos30 50 lb 0
y
B  

30 lb
y
B 30 lb
y
B

30.2
59.64 lbB

 59.6 lb
B 30.2° 
www.elsolucionario.org

406. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.25
A rod AB hinged at A and attached at B to cable BD
supports the loads shown. Knowing that
200 mm,d
determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION
Free-Body Diagram:

(a) Move
T along BD until it acts at Point D.

0: ( sin 45 )(0.2 m) (90 N)(0.1 m) (90 N)(0.2 m) 0
A
MT    
190.919 N T 190.9 N T


(b)
0: (190.919 N)cos45 0
xx
FA  

135.0 N
x
A 135.0 N
x
A

0: 90 N 90 N (190.919 N)sin 45° 0
yy
FA    

45.0 N
y
A 45.0 N
y
A
142.3 N 
A 18.43°

www.elsolucionario.org

407. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.26
A rod AB, hinged at A and attached at B to cable BD,
supports the loads shown. Knowing that
150 mm,d
determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION
Free-Body Diagram:


10
tan ; 33.690°
15

(a) Move
T along BD until it acts at Point D.


0: ( sin 33.690 )(0.15 m) (90 N)(0.1 m) (90 N)(0.2 m) 0
A
MT    
324.50 N T 324 N T

(b)
0: (324.50 N)cos33.690 0
xx
FA  

270 N
x
A 270 N
x
A

0: 90 N 90 N (324.50 N)sin 33.690 0
yy
FA    

0
y
A 270 N  A 

www.elsolucionario.org

408. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.27
Determine the reactions at A and B when (a)   0, (b)   90,
(c)
  30.

SOLUTION
(a) 0

0: (20 in.) 75 lb(10 in.) 0
37.5 lb
A
MB
B
  



0: 0
xx
FA 

 0: 75 lb 37.5 lb 0
37.5 lb
yy
y
FA
A
   


37.5 lbAB 
(b)
90

0: (12 in.) 75 lb(10 in.) 0
62.5 lb
A
MB
B
  



0: 0
62.5 lb
xx
x
FAB
A
 


0: 75 lb 0
75 lb
yy
y
FA
A
  



22
22
(62.5 lb) (75 lb)
97.6 lb
xy
AAA




75
tan
62.5
50.2





97.6 lbA 50.2 ; 62.51 lbB 

www.elsolucionario.org

409. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.27 (Continued)
(c) 30

0: ( cos 30°)(20 in.) ( sin 30 )(12 in.)
(75 lb)(10 in.) 0
A
MB B  



32.161 lbB

0: (32.161)sin 30 0
16.0805 lb
xx
x
FA
A
  


0: (32.161)cos30 75 0
47.148 lb
yy
y
FA
A
   



22
22
(16.0805) (47.148)
49.8 lb
xy
AAA




47.148
tan
16.0805
71.2





49.8 lbA 71.2 ; 32.2 lbB 60.0 
www.elsolucionario.org

410. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.28
Determine the reactions at A and C when (a) 0,
(b)
30 .

SOLUTION
(a) 0
From F.B.D. of member ABC:

0: (300 N)(0.2 m) (300 N)(0.4 m) (0.8 m) 0
C
MA   

225 NA or 225 N  A


0: 225 N 0
yy
FC  

225 N or 225 N
yy
C  C
0: 300 N 300 N 0
xx
FC   

600 N or 600 N
xx
C  C
Then
22 2 2
(600) (225) 640.80 N
xy
CCC  
and
11 225
tan tan 20.556
600y
x
C
C


 
  


or 641 N 
C 20.6 
(b)
30
From F.B.D. of member ABC:
0: (300 N)(0.2 m) (300 N)(0.4 m) ( cos30 )(0.8 m)
( sin 30 )(20 in.) 0
C
MA
A
   


365.24 NA or 365 N  A 60.0 
0: 300 N 300 N (365.24 N)sin 30 0
xx
FC    

782.62
x
C


www.elsolucionario.org

411. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.28 (Continued)


0: (365.24 N)cos30 0
yy
FC  

316.31 N or 316 N
yy
C  C
Then
22 2 2
(782.62) (316.31) 884.12 N
xy
CCC  
and
11 316.31
tan tan 22.007
782.62y
x
C
C


 
  


or 884 N C
22.0 




www.elsolucionario.org

412. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.29
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the
  30, determine the reaction (a) at B, (b) at C.

SOLUTION
Free-Body Diagram: 0: () () 0
Dx
MCRPR  


x
CP


0: sin 0
xx
FCB   


sin 0
/sin
PB
BP




sin
P
B 

0: cos 0
yy
FCBP   

(/sin)cos 0
1
1
tan
y
y
CP P
CP 






For
30 ,
(a) /sin 30 2 BP P
2PB 60.0° 
(b)
xx
CPCP 

(1 1/tan 30 ) 0.732/
y
CP P  0.7321
y
PC

1.239PC 36.2° 

www.elsolucionario.org

413. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.30
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that
  60, determine the reaction (a) at B, (b) at C.

SOLUTION
See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following
expressions:

1
1
tan
sin
x
y
CP
CP
P
B









For
60 ,
(a) /sin 60 1.1547 BP P
1.155PB 30.0° 
(b)
xx
CPCP 

(1 1/tan 60 ) 0.4226
y
CP P  0.4226
y
PC

1.086PC 22.9° 


www.elsolucionario.org

414. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the
reaction at C when
  60.

SOLUTION
0: (2 cos ) 0
C
MTaaTaPa   


1cos
P
T


(1)

0: sin 0
xx
FCT   


sin
sin
1cos
x
P
CT






0: cos 0
yy
FCTT P   


1cos
(1 cos )
1cos
0
y
y
CPT PP
C



  



Since
0,
yx
CCC
sin
1cos
P



C
(2)
For

60 :

Eq. (1):
1
2
1 cos60 1
PP
T


2
3
TP

Eq. (2):
1
2
sin 60 0.866
1 cos60 1
CP P



0.577PC 

www.elsolucionario.org

415. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.32
Neglecting friction, determine the tension in cable ABD and the
reaction at C when
  45.

SOLUTION
Free-Body Diagram:

Equilibrium for bracket:

0: () () ( sin45)(2sin45)
( cos 45 )( 2 cos 45 ) 0
C
MTaPaTa
Taa
     



0.58579T or 0.586TP



0: (0.58579 )sin 45 0
xx
FC P  


0.41422
x
CP

0: 0.58579 (0.58579 )cos 45 0
yy
FC PP P   

0
y
C or 0.414PC 

www.elsolucionario.org

416. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.33
A force P of magnitude 90 lb is applied to member ACDE, which is
supported by a frictionless pin at D and by the cable ABE. Since the
cable passes over a small pulley at B, the tension may be assumed to
be the same in portions AB and BE of the cable. For the case when
a  3 in., determine (a) the tension in the cable, (b) the reaction at D.

SOLUTION
Free-Body Diagram:

(a)
512
0: (90 lb)(9 in.) (9 in.) (7 in.) (3 in.) 0
13 13
D
MTTT    

117 lbT 117.0 lbT 
(b)
5
0: 117 lb (117 lb) 90 0
13
xx
FD    


72 lb
x
D

12
0: lb) 0
13
yy
FD  

108 lb
y
D 129.8 lb D 56.3° 


www.elsolucionario.org

417. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.34
Solve Problem 4.33 for a  6 in.
PROBLEM 4.33 A force P of magnitude 90 lb is applied to member
ACDE, which is supported by a frictionless pin at D and by the cable
ABE. Since the cable passes over a small pulley at B, the tension may
be assumed to be the same in portions AB and BE of the cable. For
the case when a  3 in., determine (a) the tension in the cable, (b) the
reaction at D.

SOLUTION
Free-Body Diagram:

(a)
512
0: (90 lb)(6 in.) (6 in.) (7 in.) (6 in.) 0
13 13
D
MTTT    

195 lbT 195.0 lbT 
(b)
5
0: 195 lb (195 lb) 90 0
13
xx
FD    


180 lb
x
D

12
0: lb) 0
13
yy
FD    

180 lb
y
D 255 lbD 45.0° 

www.elsolucionario.org

418. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.35
Bar AC supports two 400-N loads as shown. Rollers at A
and C rest against frictionless surfaces and a cable BD is
attached at B. Determine (a) the tension in cable BD, (b) the
reaction at A, (c) the reaction at C.

SOLUTION

Similar triangles: ABE and ACD

0.15 m
; ; 0.075 m
0.5 m 0.25 m
AE BE BE
BE
AD CD


(a)
0.075
0: (0.25 m) (0.5 m) (400 N)(0.1 m) (400 N)(0.4 m) 0
0.35
Ax x
MT T

    



1400 N
x
T

0.075
(1400 N)
0.35
300 N
y
T

1432 N T  www.elsolucionario.org

419. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.35 (Continued)

(b)
0: 300 N 400 N 400 N 0
y
FA    
1100 N A 1100 N 
A 
(c)
0: 1400 N 0
x
FC  


1400 NC

1400 N
C



www.elsolucionario.org

420. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.36
A light bar AD is suspended from a cable BE and supports a
20-kg block at C. The ends A and D of the bar are in contact
with frictionless vertical walls. Determine the tension in cable
BE and the reactions at A and D.

SOLUTION
Free-Body Diagram:
2
(20 kg)(9.81 m/s ) 196.20 NW
0:
x
FAD 

0:
yBE
FTW  196.2 N
BE
T 
We note that the forces shown form two couples.

0: (200 mm) (196.20 N)(75 mm) 0
B
MA  
73.575 N A
73.6 N 
A , 73.6 ND 


www.elsolucionario.org

421. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.37
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine the reactions at C, D, and E
when
30 .

SOLUTION
Free-Body Diagram:
0: cos30 20 40 0
y
FE   

60 lb
69.282 lb
cos 30°
E
69.3 lbE 60.0 

0: (20 lb)(4 in.) (40 lb)(4 in.)
(3 in.) sin 30 (3 in.) 0
 

D
M
CE
80 3 69.282(0.5)(3) 0 C  

7.9743 lbC 7.97 lbC 

0: sin30 0
x
FE CD 
(69.282 lb)(0.5) 7.9743 lb 0 D

42.615 lbD 42.6 lbD 

www.elsolucionario.org

422. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.38
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine (a) the smallest value of
 for
which the equilibrium of the bracket is maintained, (b) the corresponding reactions
at C, D, and E.


SOLUTION
Free-Body Diagram:
0: cos 20 40 0
y
FE    

60
cos
E

 (1)

0: (20 lb)(4 in.) (40 lb)(4 in.) (3 in.)
60
+sin3in.0
cos
D
MC


  




1
(180 tan 80)
3
C

(a) For
0,C 180tan 80

4
tan 23.962
9
  24.0 
From Eq. (1):
60
65.659
cos23.962
E



0: sin 0
x
FDCE   
(65.659) sin 23.962 26.666 lbD
(b) 0, 26.7 lb, 
CD 65.7 lbE 66.0 
www.elsolucionario.org

423. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.39
A movable bracket is held at rest by a cable attached at C and by
frictionless rollers at A and B. For the loading shown, determine
(a) the tension in the cable, (b) the reactions at A and B.

SOLUTION
Free-Body Diagram:


(a)
0: 600 N 0
y
FT  
600 N T

(b)
0: 0
x
FBA BA   

Note that the forces shown form two couples.

0: (600 N)(600 mm) (90 mm) 0MA  


4000 NA
4000 N B

4.00 kNA ; 4.00 kNB 

www.elsolucionario.org

424. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.40
A light bar AB supports a 15-kg block at its midpoint C. Rollers at
A and B rest against frictionless surfaces, and a horizontal cable
AD is attached at A. Determine (a) the tension in cable AD, (b) the
reactions at A and B.

SOLUTION
Free-Body Diagram:


2
(15 kg)(9.81 m/s )
147.150 N
W


(a)
0: 105.107 N 0
xAD
FT  

105.1 N
AD
T 
(b)
0: 0
y
FAW 


147.150 N 0A

147.2 N 
A 

0: (350 mm) (147.150 N)(250 mm) 0
105.107 N
A
MB
B
  

105.1 N 
B 
www.elsolucionario.org

425. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.41
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless
pins A and B. Knowing that P  15 lb, determine (a) the
force each pin exerts on the plate, (b) the reaction at F.

SOLUTION
Free-Body Diagram:

(a)
0: 15 lb sin30 0
x
FB  

30.0 lb
B 60.0°

(b)
0: (30 lb)(4 in.) sin 30 (3 in.) cos30 (11in.) (13 in.) 0
A
MBBF       
120 lb in. (30 lb)sin 30 (3 in.) (30 lb)cos30 (11in.) (13 in.) 0F    
16.2145 lb F 16.21 lb 
F 
(a)
0: 30 lb cos30 0
y
FA B F   


30 lb (30 lb)cos30 16.2145 lb 0A  
20.23 lb A 20.2 lb 
A 
www.elsolucionario.org

426. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.42
For the plate of Problem 4.41 the reaction at F must be
directed downward, and its maximum allowable value
is 20 lb. Neglecting friction at the pins, determine the
required range of values of P.
PROBLEM 4.41 Two slots have been cut in plate
DEF, and the plate has been placed so that the slots fit
two fixed, frictionless pins A and B. Knowing that
P  15 lb, determine (a) the force each pin exerts on the
plate, (b) the reaction at F.

SOLUTION
Free-Body Diagram:


0: sin30 0
x
FPB  

2PB 60°

0: (30 lb)(4 in.) sin 30 (3 in.) cos30 (11in.) (13 in.) 0
A
MBBF       


120 lb in.+2 sin30 (3 in.) 2 cos30 (11in.) (13 in.) 0
120 3 19.0525 13 0
PPF
PPF
    
  


13 120
22.0525
F
P


(1)

For 0:F
13(0) 120
5.442 lb
22.0525
P



For 20 lb:P
13(20) 120
17.232 lb
22.0525
P



For 0 20 lb: F 5.44 lb 17.23 lb P


www.elsolucionario.org

427. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.43
The rig shown consists of a 1200-lb horizontal member ABC
and a vertical member DBE welded together at B. The rig is
being used to raise a 3600-lb crate at a distance x  12 ft
from the vertical member DBE. If the tension in the cable is
4 kips, determine the reaction at E, assuming that the cable is
(a) anchored at F as shown in the figure, (b) attached to the
vertical member at a point located 1 ft above E.

SOLUTION
Free-Body Diagram:


0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0
EE
MM x T   


3.75 3600 7800
E
MTx
(1)
(a) For 12 ft and 4000 lbs,xT

3.75(4000) 3600(12) 7800
36,000 lb ft
E
M



00
xx
FE  

0: 3600 lb 1200 lb 4000 0
yy
FE    

8800 lb
y
E
8.80 kips 
E ; 36.0 kip ft
E
M 



www.elsolucionario.org

428. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.43 (Continued)


(b)
0: (3600 lb)(12 ft) (1200 lb)(6.5 ft) 0
EE
MM   

 51,000 lb ft
E
M 

 00
xx
FE  

0: 3600 lb 1200 lb 0
yy
FE   

4800 lb
y
E
4.80 kips 
E ; 51.0 kip ft
E
M 

www.elsolucionario.org

429. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.44
For the rig and crate of Prob. 4.43, and assuming that cable
is anchored at F as shown, determine (a) the required tension
in cable ADCF if the maximum value of the couple at E as x
varies from 1.5 to 17.5 ft is to be as small as possible, (b) the
corresponding maximum value of the couple.

SOLUTION
Free-Body Diagram:


0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0
EE
MM x T   


3.75 3600 7800
E
MTx
(1)

For 1.5 ft, Eq. (1) becomesx


1
( ) 3.75 3600(1.5) 7800
E
MT 
(2)
For 17.5 ft, Eq. (1) becomesx


2
( ) 3.75 3600(17.5) 7800
E
MT 
(a) For smallest max value of
||,
E
M we set

12
()()
EE
MM



3.75 13,200 3.75 70,800TT

11.20 kipsT 
(b) From Equation (2), then

3.75(11.20) 13.20
E
M

| | 28.8 kip ft
E
M 
www.elsolucionario.org

430. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.45
A 175-kg utility pole is used to support at C the end of an electric
wire. The tension in the wire is 600 N, and the wire forms an angle
of 15° with the horizontal at C. Determine the largest and smallest
allowable tensions in the guy cable BD if the magnitude of the
couple at A may not exceed 500 N·m.

SOLUTION
Free-Body Diagram:


Geometry:
Distance BD =
22
(1.5) (3.6)  3.90 m
Note also that: W  mg  (175 kg)(9.81 m/s
2
)  1716.75 N
With
MA  500 N m clockwise: (i.e. corresponding to T max )

0:
A
M 500 N m – [(600 N) cos 15
o
](4.5 m) +
max
1.5
3.90
T



(3.6 m)  0
T
max  2244.7 N or T max  2240 N
With
MA  500 N m counter-clockwise: (i.e. corresponding to T min )

0:
A
M 500 N m – [(600 N) cos 15
o
](4.5 m) +
min
1.5
3.90
T



(3.6 m)  0
T
min  1522.44 N or T min  1522 N
www.elsolucionario.org

431. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.46
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.

SOLUTION

1300 N
5
13
500 N
12
13
1200 N
x
y
T
TT
TT







0: 450N 500N 0 50N
xx x
MC C     50 N
x
C

0: 750 N 1200 N 0 1950 N
yy y
FC C    
1950 N
y
C


1951 N
C 88.5° 

0: (750 N)(0.5 m) (4.50 N)(0.4 m)
(1200 N)(0.4 m) 0
CC
MM  


75.0 N m
C
M  75.0 N m
C
M 
www.elsolucionario.org

432. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.47
Determine the range of allowable values of the tension in wire
BD if the magnitude of the couple at the fixed support C is not
to exceed
100 N · m.

SOLUTION


5
0: (750 N)(0.5 m) (450 N)(0.4 m) (0.6 m)
13
12
(0.15 m) 0
13
C
C
MT
TM

  







4.8
375 N m 180 N m m 0
13
C
TM

   



13
(555 )
4.8
C
TM
For
13
100 N m: (555 100) 1232 N
4.8
C
MT    
For
13
100 N m: (555 100) 1774 N
4.8
C
MT    
For
||100Nm:
C
M 1.232 kN 1.774 kN T 

www.elsolucionario.org

433. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.48
Beam AD carries the two 40-lb loads shown. The beam is held by a
fixed support at D and by the cable BE that is attached to the
counterweight W. Determine the reaction at D when (a) 100W
lb, (b)
90 lb.W

SOLUTION
(a) 100 lb W
From F.B.D. of beam AD:

0: 0
xx
FD 

0: 40 lb 40 lb 100 lb 0
yy
FD  

20.0 lb
y
D or 20.0 lb  D 

0: (100 lb)(5 ft) (40 lb)(8 ft)
(40 lb)(4 ft) 0
DD
MM  


20.0 lb ft
D
M or 20.0 lb ft
D
M



(b) 90 lb W
From F.B.D. of beam AD:

0: 0
xx
FD 

0: 90 lb 40 lb 40 lb 0
yy
FD    

10.00 lb
y
D or 10.00 lb D 

0: (90 lb)(5 ft) (40 lb)(8 ft)
(40 lb)(4 ft) 0
DD
MM  


30.0 lb ft
D
M  or 30.0 lb ft
D
M 


www.elsolucionario.org

434. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.49
For the beam and loading shown, determine the range of values of
W for which the magnitude of the couple at D does not exceed 40
lb  ft.

SOLUTION
For
min
,W 40 lb ft
D
M 
From F.B.D. of beam AD:
min
0: (40 lb)(8 ft) (5 ft)
(40 lb)(4 ft) 40 lb ft 0
D
MW 


min
88.0 lbW
For
max
,W 40 lb ft
D
M
From F.B.D. of beam AD:
max
0: (40 lb)(8 ft) (5 ft)
(40 lb)(4 ft) 40 lb ft 0
D
MW 


max
104.0 lbW or 88.0 lb 104.0 lb W 

www.elsolucionario.org

435. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.50
An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a
100-mm radius, determine the reaction at A in each case.


SOLUTION
















2
(8 kg)(9.81 m/s ) 78.480 NWmg 
(a)
0: 0
xx
FA 

0: 0
yy
FAW 

78.480 N
y
A

0: (1.6 m) 0
AA
MMW  


(78.480 N)(1.6 m)
A
M

125.568 N m
A
M
78.5 N 
A , 125.6 N m
A
M 
(b)
0: 0
xx
FAW 

78.480
x
A



0: 0
yy
FAW 

78.480
y
A

(78.480 N) 2 110.987 NA 45°

0: (1.6 m) 0
AA
MMW  

(78.480 N)(1.6 m) 125.568 N m
AA
M   M
111.0 N 
A 45°, 125.6 N m
A
M 
(c)
0: 0
xx
FA 

0: 2 0
yy
FAW  

2 2(78.480 N) 156.960 N
y
AW 

0: 2 (1.6 m) 0
AA
MMW  

2(78.480 N)(1.6 m)
A
M 251.14 N m
A
M
157.0 N 
A , 251 N m
A
M 
www.elsolucionario.org

436. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.51
A uniform rod AB with a length of l and weight of W is suspended
from two cords AC and BC of equal length. Determine the angle θ
corresponding to the equilibrium position when a couple
M is applied
to the rod.

SOLUTION
Free-Body Diagram:








Using h 
tan
2
l


0:
C
M (W)( sin ) 0Mh 
or
sin
M
Wh

2
cot
M
Wl






or
1 cot
sin 2M
Wl







www.elsolucionario.org

437. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.52
Rod AD is acted upon by a vertical force P at end A and by two equal
and opposite horizontal forces of magnitude Q at points B and C.
Neglecting the weight of the rod, express the angle
 corresponding to
the equilibrium position in terms of P and Q.

SOLUTION
Free-Body Diagram:




0:
D
M (3 sin ) ( cos ) (2 cos ) 0Pa Qa Qa   
3 sin cosPQ


tan
3
Q
P
or
1
tan
3
Q
P






www.elsolucionario.org

438. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.53
A slender rod AB, of weight W, is attached to blocks A and B,
which move freely in the guides shown. The blocks are
connected by an elastic cord that passes over a pulley at C.
(a) Express the tension in the cord in terms of W and
.
(b) Determine the value of
 for which the tension in the cord is
equal to 3W.

SOLUTION
(a) From F.B.D. of rod AB:

1
0: ( sin ) cos ( cos ) 0
2
C
MTlW Tl 

   




cos
2(cos sin )
W
T




Dividing both numerator and denominator by cos
,

1
21tan
W
T





or

2
(1 tan )
W
T


 
(b) For
3,TW

2
3
(1 tan )
1
1tan
6
W
W





or
15
tan 39.806
6





or 39.8 
www.elsolucionario.org

439. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.54
A vertical load P is applied at end B of rod BC. (a) Neglecting the
weight of the rod, express the angle
 corresponding to the
equilibrium position in terms of P, l, and the counterweight W. (b)
Determine the value of
 corresponding to equilibrium if
P  2W.

SOLUTION
Free-Body Diagram:

(a) Triangle ABC is isosceles. We have
()cos cos
22
CD BC l
  

 
 


0: ( cos ) cos 0
2
C
MPlWl


  



Setting
2
cos 2cos 1:
2



2
2cos 1 cos 0
22
Pl Wl

 




2
2
2 1
cos cos 0
22 22
1
cos 8
24
W
P
WW
P P










2
1
2
1
2cos 8
4
WW
P P


 
 
 
 

www.elsolucionario.org

440. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.54 (Continued)

(b) For 2,
PW 
11 1 1
cos 8 1 33
242 4 8





cos 0.84307 and cos 0.59307
22
32.534 126.375
22



 


65.1
 252.75 (discard) 

65.1
 

www.elsolucionario.org

441. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.55
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight
of the rod, express the angle
 corresponding to the equilibrium position in
terms of P, l, and the counterweight W. (b) Determine the value of

corresponding to equilibrium if P  2W.


SOLUTION
(a) Triangle ABC is isosceles. We have
()cos cos
22
CD BC l


0: cos ( sin ) 0
2
C
MWl Pl


  



Setting
sin 2sin cos : cos 2 sin cos 0
22 2 22
Wl Pl
  



2sin 0
2
WP


1
2sin
2
W
P

 



(b) For
2,PW
sin 0.25
22 4
WW
PW

 

14.5
2

 29.0 
or
165.5 331 (discard)
2



www.elsolucionario.org

442. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.56
A collar B of weight W can move freely along the vertical rod shown. The
constant of the spring is k, and the spring is unstretched when
0.
 (a)
Derive an equation in
, W, k, and l that must be satisfied when the collar is
in equilibrium. (b) Knowing that
300 N,W l500 mm, and
800 N/m,k determine the value of  corresponding to equilibrium.

SOLUTION
First note: Tks
where spring constant
elongation of spring
cos
(1 cos )
cos
(1 cos )
cos
k
s
l
l
l
kl
T










(a) From F.B.D. of collar B:

0: sin 0
y
FTW  
or
(1 cos ) sin 0
cos
kl
W
 or
tan sin
W
kl 
(b) For
3lb
6in.
8lb/ft
6in.
0.5 ft
12 in./ft
3lb
tan sin 0.75
(8 lb/ft)(0.5 ft)
W
l
k
l





 
Solving numerically,
57.957 or 58.0 
www.elsolucionario.org

443. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.57
Solve Sample Problem 4.5, assuming that the spring is
unstretched when
90 .

SOLUTION
First note: tension in springTks
where
deformation of springs
r
Fkr





From F.B.D. of assembly:
0
0: ( cos ) ( ) 0MWl Fr  
or
2
2
cos 0
cos
Wl kr
kr
Wl



For
2
250 lb/in.
3in.
8in.
400 lb
(250 lb/in.)(3 in.)
cos
(400 lb)(8 in.)
k
r
l
W






or cos 0.703125

Solving numerically, 0.89245 rad

or 51.134

Then
90 51.134 141.134   or 141.1  
www.elsolucionario.org

444. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.58
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position terms of P, k, and l. (b) Determine the value of θ
corresponding to equilibrium if P =
4
1kl.

SOLUTION
Free-Body Diagram:

(a) Triangle ABC is isosceles. We have
2( ) 2 sin ; cos
22
AB AD l CD l
  
 
 
 

Elongation of spring:
()() 60
2 sin 2 sin 30
2
xAB AB
ll









1
2sin
22
Tkx kl
 




0: cos ( sin ) 0
2
C
MTl Pl


  





www.elsolucionario.org

445. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.58 (Continued)


1
2
1
2sin cos 2sincos 0
22 2 2 2
cos 0 or 2( )sin 0
22
180 (trivial) sin
2
kl l Pl
kl P kl
kl
kl P 


  
 
  
  

 



11
2sin /( )
2
kl kl P






(b) For
1
,
4
Pkl
1
2
3
4
2
sin
23
kl
kl
 
41.8
2

 83.6 
www.elsolucionario.org

446. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.59
Eight identical 500 750-mm rectangular plates, each of mass 40 kg,m are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate
or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever
possible, compute the reactions.


SOLUTION
1. Three non-concurrent, non-parallel reactions:
(a) Plate: completely constrained
(b) Reactions: determinate
(c) Equilibrium maintained
196.2 N 
AC
2. Three non-concurrent, non-parallel reactions:
(a) Plate: completely constrained
(b) Reactions: determinate
(c) Equilibrium maintained
0, 196.2 N 
BCD
3. Four non-concurrent, non-parallel reactions:
(a) Plate: completely constrained
(b) Reactions: indeterminate
(c) Equilibrium maintained

294 N
x
A , 294 N
x
D

(392N
yy
AD )
4. Three concurrent reactions (through D):
(a) Plate: improperly constrained
(b) Reactions: indeterminate
(c) No equilibrium
(0)
D
M www.elsolucionario.org

447. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.59 (Continued)

5. Two reactions:
(a) Plate: partial constraint
(b) Reactions: determinate
(c) Equilibrium maintained
196.2 N 
CD
6. Three non-concurrent, non-parallel reactions:
(a) Plate: completely constrained
(b) Reactions: determinate
(c) Equilibrium maintained
294 N 
B , 491 ND 53.1°
7. Two reactions:
(a) Plate: improperly constrained
(b) Reactions determined by dynamics
(c) No equilibrium
(0)
y
F
8. Four non-concurrent, non-parallel reactions:
(a) Plate: completely constrained
(b) Reactions: indeterminate
(c) Equilibrium maintained

196.2 N
y
BD

(0)
x
CD
www.elsolucionario.org

448. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.60
The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth
pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever
possible, compute the reactions, assuming that the magnitude of the force
P is 100 lb.


SOLUTION
1. Three non-concurrent, non-parallel reactions:
(a) Bracket: complete constraint
(b) Reactions: determinate
(c) Equilibrium maintained
120.2 lb 
A 56.3°, 66.7 lb B
2. Four concurrent, reactions (through A):
(a) Bracket: improper constraint
(b) Reactions: indeterminate
(c) No equilibrium
(0)
A
M
3. Two reactions:
(a) Bracket: partial constraint
(b) Reactions: indeterminate
(c) Equilibrium maintained
50 lb 
A , 50 lbC
4. Three non-concurrent, non-parallel reactions:
(a) Bracket: complete constraint
(b) Reactions: determinate
(c) Equilibrium maintained
50 lb 
A , 83.3 lbB 36.9°, 66.7 lb C
www.elsolucionario.org

449. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.60 (Continued)

5. Four non-concurrent, non-parallel reactions:
(a) Bracket: complete constraint
(b) Reactions: indeterminate
(c) Equilibrium maintained
(0) 50lb
Cy
M  A
6. Four non-concurrent, non-parallel reactions:
(a) Bracket: complete constraint
(b) Reactions: indeterminate
(c) Equilibrium maintained

66.7 lb
x
A 66.7 lb
x
B

( 100 lb
yy
AB )
7. Three non-concurrent, non-parallel reactions:
(a) Bracket: complete constraint
(b) Reactions: determinate
(c) Equilibrium maintained

50 lbAC
8. Three concurrent, reactions (through A)
(a) Bracket: improper constraint
(b) Reactions: indeterminate
(c) No equilibrium
(0)
A
M



www.elsolucionario.org

450. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.61
A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft
obstruction. A cable is wrapped around the tank and pulled
horizontally as shown. Knowing that the corner of the obstruction at A
is rough, find the required tension in the cable and the reaction at A.

SOLUTION
Free-Body Diagram:

Force triangle

2 ft
cos 0.5 60
4 ft
GD
AG
 

1
30 ( 60 )
2
(500 lb) tan 30°T
  


T=289 lb


500 lb
cos30
A

577 lbA 60.0


www.elsolucionario.org

451. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.62
Determine the reactions at A and B when 180 mm.a

SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.

:BCD
240
tan
180
53.13




Force triangle

(300 N) tan 53.13
400 N
A

400 NA 

300 N
cos 53.13°
500 N
B

 500 NB 53.1 

Free-Body Diagram:
(Three-force member) www.elsolucionario.org

452. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.63
For the bracket and loading shown, determine the range of values
of the distance a for which the magnitude of the reaction at B does
not exceed 600 N.

SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.



240 mm
tan
a

 (1)
Force Triangle
(with 600 N) B

300 N
cos 0.5
600 N
60.0




Eq. (1)
240 mm
tan 60.0°
138.56 mm
a

For
600 NB 138.6 mma 

Free-Body Diagram:
(Three-force member) www.elsolucionario.org

453. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.64
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60-lb force
P is exerted on the spanner at D,
find the reactions at A and B.

SOLUTION
Free-Body Diagram:
(Three-force body)

The line of action of
A must pass through D, where B and P intersect.

3sin50
tan
3cos50 15
0.135756
7.7310









60 lb
sin 7.7310°
446.02 lb
60 lb
tan 7.7310°
441.97 lb
A
B




Force triangle
446 lbA 7.73 
 442 lbB 
www.elsolucionario.org

454. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.65
Determine the reactions at B and C when a = 30 mm.

SOLUTION
Since CD is a two-force member, the force it exerts on member ABD is directed along DC.
Free-Body Diagram of ABD: (Three-Force member)

The reaction at B must pass through E, where
D and the 250-N load intersect.
Triangle CFD:
60
tan 0.6
100
30.964



Triangle EAD:
200tan 120 mm
120 30 90 mm
AE
GE AE AG

Triangle EGB:
60
tan
90
33.690
GB
GE



Force triangle
250 N
sin120.964 sin 33.690 sin 25.346°
BD




500.7 N
323.9 N
B
D


501 NB 56.3 
 324 NCD 31.0 
www.elsolucionario.org

455. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.66
A 12-ft wooden beam weighing 80 lb is supported by a pin and
bracket at A and by cable BC. Find the reaction at A and the tension
in the cable.


SOLUTION
Since CB is a two-force member, the force it exerts on member AB is directed along CB.
Free-Body Diagram of AB: (Three-Force member)

The reaction at B must pass through E, where
T and the 80-lb load intersect.
Triangle CFD:


2
6 ft 1.50 ft
8
BG
DG AC
AB
DG



Triangle EAD:
1.50
tan
6
14.0362
DG
AE





Triangle EGB: Force triangle
80 lb
sin53.13 sin 75.964 sin 50.906°
AT




82.5 lbA 14.04 
 100.0 lbT  www.elsolucionario.org

456. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.67
Determine the reactions at B and D when 60 mm.b

SOLUTION
Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.
Free-Body Diagram:
(Three-force body)

Reaction at B must pass through E, where the reaction at D and the 80-N force intersect.

220 mm
tan
250 mm
41.348




Force triangle

Law of sines:

80 N
sin 3.652° sin 45 sin131.348
888.0 N
942.8 N
BD
B
D




 888 NB 41.3 943 ND 45.0 
www.elsolucionario.org

457. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.68
For the frame and loading shown, determine the reactions at C and D.

SOLUTION
Since BD is a two-force member, the reaction at D must pass through Points B and D.
Free-Body Diagram:
(Three-force body)

Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect.
Triangle CEF:
4.5 ft
tan 56.310
3 ft

 
Triangle ABE:
1
tan 26.565
2

 




www.elsolucionario.org

458. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.68 (Continued)
Force Triangle

Law of sines:
150 lb
sin 29.745 sin116.565 sin 33.690
CD




270.42 lb,
167.704 lb
C
D




270 lbC 56.3 ; 167.7 lbD 26.6 
www.elsolucionario.org

459. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.69
A 50-kg crate is attached to the trolley-beam system shown.
Knowing that
1.5 m,a determine (a) the tension in cable CD,
(b) the reaction at B.

SOLUTION
Three-force body: W and
CD
Tintersect at E.

0.7497 m
tan
1.5 m
26.556




Three forces intersect at E.

2
(50 kg) 9.81 m/s
490.50 N
W
 Force triangle
Law of sines:
490.50 N
sin 61.556° sin 63.444 sin 55
498.99 N
456.96 N
CD
CD
T B
T
B






(a)
499 N
CD
T 
(b)
457 NB 26.6 



www.elsolucionario.org

460. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.70
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 150-N load at its midpoint C, find the
reaction at A and the tension in the cord.

SOLUTION
Free-Body Diagram: (Three-force body) Dimensions in mm

The line of action of reaction at A must pass through E, where
T and the 150-N load intersect.
Force triangle

460
tan
240
62.447
100
tan
240
22.620
EF
AF
EH
DH









150 N
sin 67.380 sin 27.553 sin 85.067°
AT




139.0 NA 62.4 

69.6 NT 

www.elsolucionario.org

461. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.71
For the boom and loading shown, determine (a) the tension in cord
BD, (b) the reaction at C.

SOLUTION
Three-force body: 3-kip load and T intersect at E.
Geometry:

32
;
12 48
8 in.
BF CF BF
AG CG
BF




32 32 8 24 in.
48
; ; 36 in.
24 32
36 32 4 in.
BH BF
JE DJ JE
JE
BH DH
EG JE JG
 

  


4 in.
tan
48 in.
4.7636
24 in.
tan
32 in.
36.870








Force Triangle

Law of sines:
3 kips
sin 94.764 sin 32.106sin53.13
BD
T C




(a)
5.63 kips
BD
T 
(b) 4.52 kips 
C 4.76 

Free-Body Diagram: www.elsolucionario.org

462. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.72
A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness
of each tile is 0.3 in., determine the force
P required to move the roller
onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.

SOLUTION


Force Triangle 


Force Triangle




Geometry: For each case as roller comes into contact with
tile,


13.7 in.
cos
4 in.
22.332







(a) Roller pushed to left (three-force body):
Forces must pass through O.

Law of sines:
40 lb
;24.87 lb
sin37.668 sin 22.332
P
P




24.9 lbP 30.0° 

(b) Roller pulled to right (three-force body):
Forces must pass through O.
Law of sines:
40 lb
; 15.3361 lb
sin 97.668 sin 22.332
P
P




15.34 lbP 30.0° 

www.elsolucionario.org

463. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.73
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α  45°.


SOLUTION
Free-Body Diagram:
(Three-force body)

The line of action of
C must pass through E, where A and the 300-N force intersect.
Triangle ABE is isosceles:
400 mmEA AB

In triangle CEF:

150 mm
tan
700 mm
CF CF
EF EA AF
 
 12.0948




www.elsolucionario.org

464. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.73 (Continued)
Force Triangle







Law of sines:
300 N
sin 32.905 sin135 sin12.0948
AC

 
778 NA ;1012 NC 77.9  www.elsolucionario.org

465. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.74
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α  60°.


SOLUTION
Free-Body Diagram:


(400 mm) tan 30°
230.94 mm
EA




In triangle CEF: tan
CF CF
EFEAAF



150
tan
230.94 300
15.7759








www.elsolucionario.org

466. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.74 (Continued)
Force Triangle









Law of sines:
300 N
sin 44.224 sin120 sin15.7759
770 N
956 N
AC
A
C

 




770 NA ; 956 NC 74.2 
www.elsolucionario.org

467. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.75
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.

SOLUTION

The reaction at A must pass through D where
C and the 170-N force intersect.

160 mm
tan
300 mm
28.07




We note that triangle ABD is isosceles (since AC  BC) and, therefore,
28.07 CAD
 
Also, since
,CD CB reaction C forms angle 28.07 with the horizontal axis.

Force triangle
We note that A forms angle 2 with the vertical axis. Thus, A and C form angle
180 (90 ) 2 90
     
Force triangle is isosceles, and we have

170 N
2(170 N)sin
160.0 N
A
C





170.0 N 
A 33.9°; 160.0 N C 28.1° 



Free-Body Diagram:
(Three-force body) www.elsolucionario.org

468. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.76
Solve Problem 4.75, assuming that the 170-N force applied at B is horizontal
and directed to the left.
PROBLEM 4.75 Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a
170-N vertical force is applied at B.

SOLUTION
Free-Body Diagram: (Three-Force body)

The reaction at A must pass through D, where
C and the 170-N force intersect.

160 mm
tan
300 mm
28.07




We note that triangle ADB is isosceles (since AC  BC). Therefore 90 .AB

Also 2 ADB

Force triangle
The angle between A and C must be 2

Thus, force triangle is isosceles and
28.07


170.0 N
2(170 N)cos 300 N
A
C





170.0 N
A
56.1° 300 N C 28.1° 

www.elsolucionario.org

469. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.77
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a
frictionless pulley at D. The tension may be assumed to be the
same in portions AD and CD of the cord. For the loading shown
and neglecting the size of the pulley, determine the tension in the
cord and the reaction at B.

SOLUTION
Reaction at B must pass through D.

7 in.
tan
12 in.
30.256
7 in.
tan
24 in.
16.26








Force Triangle

Law of sines:
72 lb
sin 59.744 sin13.996 sin106.26
(sin13.996 ) ( 72 lb)(sin 59.744°)
(0.24185) ( 72)(0.86378)
TT B
TT
TT



 



100.00 lbT 100.0 lbT 

sin 106.26°
(100 lb)
sin 59.744
111.14 lb
B


111.1 lbB
30.3 



Free-Body Diagram: www.elsolucionario.org

470. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.78
Using the method of Section 4.7, solve Problem 4.22.
PROBLEM 4.22 A lever AB is hinged at C and attached to a
control cable at A. If the lever is subjected to a 500-N horizontal
force at B, determine (a) the tension in the cable, (b) the
reaction at C.

SOLUTION


Reaction at C must pass through E, where
FAD and 500-N force intersect.
Since 250 mm,AC CD triangle
ACD is isosceles.
We have 90 30 120 C 

and
1
(180 120 ) 30
2
AD 
 Dimensions in mm
On the other hand, from triangle BCF:

( )sin 30 200 sin 30 100 mm
250 100 150 mm
CF BC
FD CD CF



From triangle EFD, and since 30 :D

( ) tan 30 150 tan 30 86.60 mmEF FD
From triangle EFC:
100 mm
tan
86.60 mm
49.11
CF
EF



Force triangle
Law of sines
500 N
sin 49.11 sin 60 sin 70.89°
400 N, 458 N
AD
AD
F C
FC




(a) 400 N
AD
F 
(b) 458 N 
C 49.1° 

Free-Body Diagram:
(Three-Force body) www.elsolucionario.org

471. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.79
Knowing that   30, determine the reaction (a) at B, (b) at C.

SOLUTION

Reaction at C must pass through D where force
P and reaction at B intersect.
In  CDE:
(3 1)
tan
31
36.2
R
R







Force Triangle

Law of sines:
sin 23.8 sin126.2 sin 30
2.00
1.239
PBC
BP
CP





(a)
2PB 60.0° 
(b)
1.239PC 36.2° 


Free-Body Diagram:
(Three-force body) www.elsolucionario.org

472. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.80
Knowing that   60, determine the reaction (a) at B, (b) at C.

SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In CDE:
3
tan
1
1
3
22.9
R
R
R







Force Triangle

Law of sines:
sin 52.9 sin 67.1 sin 60
1.155
1.086
PBC
BP
CP





(a)
1.155PB 30.0° 
(b)
1.086PC 22.9° 


Free-Body Diagram:
(Three-force body) www.elsolucionario.org

473. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.81
Determine the reactions at A and B when 50°. 

SOLUTION
Free-Body Diagram: (Three-force body)
Reaction
A must pass through Point D where the 100-N force
and
B intersect.
In right  BCD:

90 75 15
250 tan 75 933.01 mmBD


In right  ABD:

150 mm
tan
933.01 mm
9.1333
AB
BD




Force Triangle
Law of sines:

100 N
sin 9.1333° sin15 sin155.867
163.1 N; 257.6 N
AB
AB




163.1 N 
A 74.1° 258 N B 65.0° 

Dimensions in mm www.elsolucionario.org

474. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.82
Determine the reactions at A and B when   80.

SOLUTION
Free-Body Diagram:
(Three-force body)

Reaction
A must pass through D where the 100-N force and B intersect.
In right triangle BCD:
90 75 15
  

tan 75 250 tan75
933.01 mm
BD BC
BD
  


In right triangle ABD:
150 mm
tan
933.01 mm
AB
BD
 9.1333
 








www.elsolucionario.org

475. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.82 (Continued)
Force Triangle

Law of sines:

100 N
sin 9.1333° sin15 sin155.867
AB



 163.1 NA 55.9 

258 NB 65.0 


www.elsolucionario.org

476. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.83
Rod AB is bent into the shape of an arc of circle and is lodged between two
pegs D and E. It supports a load
P at end B. Neglecting friction and the weight
of the rod, determine the distance c corresponding to equilibrium when a  20
mm and R  100 mm.

SOLUTION
Free-Body Diagram:
Since ,
ED ED
yxa
slope of ED is
45 ;
slope of HC is
45 .
Also
2DE a
and
1
2 2
a
DH HE DE

 



For triangles DHC and EHC,
2
sin
2
a
a
R R

Now sin(45 ) cR

For
20 mm and 100 mm
20 mm
sin
2(100 mm)
0.141421
8.1301
aR







and (100 mm)sin(45 8.1301 )c
60.00 mm  or 60.0 mm c


www.elsolucionario.org

477. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.84
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle
 in terms of the angle .

SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry:

Free-Body Diagram:

tan
GB
AB
x
y


where

cos
AB
yL 

and

1
sin
2
GB
xL 

1
2
sin
tan
cos
1
tan
2
L
L





or tan 2 tan  

www.elsolucionario.org

478. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.85
An 8-kg slender rod of length L is attached to collars that can slide freely
along the guides shown. Knowing that the rod is in equilibrium and that
  30, determine (a) the angle  that the rod forms with the vertical,
(b) the reactions at A and B.

SOLUTION
(a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the
geometry of the forces:

Free-Body Diagram:

tan
CB
BC
x
y

where

1
sin
2
CB
xL 
and

cos
1
tan tan
2
BC
yL 



or tan 2 tan

For 30


tan 2tan30
1.15470
49.107



or 49.1 
(b)
2
(8 kg)(9.81 m/s ) 78.480 NWmg 

From force triangle:
tan
(78.480 N) tan 30
AW

45.310 N  or 45.3 N 
A 
and
78.480 N
90.621 N
cos cos30
W
B

 

or 90.6 N B 60.0 

www.elsolucionario.org

479. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.86
A uniform rod AB of length 2R rests inside a hemispherical bowl of
radius R as shown. Neglecting friction, determine the angle

corresponding to equilibrium.

SOLUTION
Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the
three forces. Since force A passes through O, the center of the circle, and since force
C is perpendicular to
the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.
Note that the angle
 of triangle DOA is the central angle corresponding to the inscribed angle
of
triangle DCA.

2
The horizontal projections of
,( ),
AE
AE x and ,( ),
AG
AG x are equal.

AE AG A
xxx
or ( )cos 2 ( )cos AE AG

and (2 )cos2 cos RR

Now
2
cos 2 2cos 1
then
2
4cos 2 cos
or
2
4cos cos 2 0
Applying the quadratic equation,
cos 0.84307 and cos 0.59307

32.534 and 126.375 (Discard)
   or 32.5 

www.elsolucionario.org

480. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.87
A slender rod BC of length L and weight W is held by two cables as
shown. Knowing that cable AB is horizontal and that the rod forms
an angle of 40 with the horizontal, determine (a) the angle
 that
cable CD forms with the horizontal, (b) the tension in each cable.

SOLUTION
Free-Body Diagram:
(Three-force body)

(a) The line of action of
TCD must pass through E, where TAB and W intersect.

1
2
tan
sin 40
cos 40
2tan40
59.210
CF
EF
L
L



 
 


59.2
 
(b) Force Triangle
tan 30.790
0.59588
AB
TW
W
 



0.596
AB
TW 

cos30.790
1.16408
CD
W
T
W





1.164
CD
TW 
www.elsolucionario.org

481. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.88
A thin ring of mass 2 kg and radius r  140 mm is held against a frictionless
wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in
the string, (c) the reaction at C.

SOLUTION
Free-Body Diagram:
(Three-force body)



The force T exerted at B must pass through the center G of the ring, since
C and W intersect at that point.
Thus, points A, B, and G are in a straight line.
(a) From triangle ACG:
22
22
()()
(265 mm) (140 mm)
225.00 mm
dAGCG


225 mmd

Force Triangle

2
(2 kg)(9.81 m/s ) 19.6200 NW



Law of sines:
19.6200 N
265 mm 140 mm 225.00 mm
TC


(b)
23.1 NT 
(c) 12.21 N C 

www.elsolucionario.org

482. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.89
A slender rod of length L and weight W is attached to a
collar at A and is fitted with a small wheel at B. Knowing
that the wheel rolls freely along a cylindrical surface of
radius R, and neglecting friction, derive an equation in
, L,
and R that must be satisfied when the rod is in equilibrium.

SOLUTION
Free-Body Diagram (Three-force body)
Reaction
B must pass through D where B and W intersect.
Note that ABC and BGD are similar.

cosAC AE L
In 
ABC:
22 2
222
2
22
2
22
2
2
()() ()
(2 cos ) ( sin )
4cos sin
4cos 1 cos
3cos 1
CE BE BC
LLR
R
L
R
L
R
L
















2
2
1
cos 1
3
R
L



 



www.elsolucionario.org

483. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.90
Knowing that for the rod of Problem 4.89, L  15 in., R  20
in., and W  10 lb, determine (a) the angle
 corresponding
to equilibrium, (b) the reactions at A and B.

SOLUTION
See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following
equation:

2
2
1
cos 1
3
R
L



 


For 15 in., 20 in., and 10 lb,LR W 
(a)
2
2
120in.
cos 1 ; 59.39
315in.







59.4



In 
ABC:
sin 1
tan tan
2cos 2
1
tan tan 59.39 0.8452
2
40.2
BE L
CE L




 



Force Triangle


tan (10 lb) tan 40.2 8.45 lb
(10 lb)
13.09 lb
cos cos 40.2
AW
W
B
 
 


(b) 8.45 lb 
A 
13.09 lb 
B 49.8° 
www.elsolucionario.org

484. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.91
Two transmission belts pass over a double-sheaved pulley
that is attached to an axle supported by bearings at A and
D. The radius of the inner sheave is 125 mm and the radius
of the outer sheave is 250 mm. Knowing that when the
system is at rest, the tension is 90 N in both portions of
belt B and 150 N in both portions of belt C, determine the
reactions at A and D. Assume that the bearing at D does
not exert any axial thrust.

SOLUTION
We replace
B
T and
B
T by their resultant ( 180 N) j and
C
T and
C
T by their resultant ( 300 N) . k

Dimensions in mm
We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but
equilibrium is maintained
(0).
x
M

0: (150 ) ( 180 ) (250 ) ( 300 ) (450 ) ( ) 0
Ayz
DD       Mijikijk

33
27 10 75 10 450 450 0
yz
DD     k
j k j
Equating coefficients of
j and k to zero,
:
j
3
75 10 450 0
z
D
 166.7 N
z
D

3
:2710450 0
y
D  k 60.0 N
y
D

0:
x
F 0
x
A


0: 180 N 0
yyy
FAD    180 60 120.0 N
y
A

0: 300 N 0
zzz
FAD    300 166.7 133.3 N
z
A 
(120.0 N) (133.3 N) ; (60.0 N) (166.7 N)
 AjkDjk 

www.elsolucionario.org

485. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.92
Solve Problem 4.91, assuming that the pulley rotates at a
constant rate and that T
B  104 N, T′ B  84 N, T C  175 N.
PROBLEM 4.91 Two transmission belts pass over a
double-sheaved pulley that is attached to an axle supported
by bearings at A and D. The radius of the inner sheave is
125 mm and the radius of the outer sheave is 250 mm.
Knowing that when the system is at rest, the tension is 90 N
in both portions of belt B and 150 N in both portions of
belt C, determine the reactions at A and D. Assume that the
bearing at D does not exert any axial thrust.

SOLUTION

Dimensions in mm

We have six unknowns and six equations of equilibrium. —OK

0: (150 250 ) ( 104 ) (150 250 ) ( 84 )
(250 125 ) ( 175 ) (250 125 ) ( )
450 ( ) 0
A
C
yz
T
DD
     

 
Mikjikj
jkiji
ijk


150(104 84) 250(104 84) 250(175 ) 125(175 )
450 450 0
CC
yz
TT
DD
   

ki j
kj

Equating the coefficients of the unit vectors to zero,

: 250(104 84) 125(175 ) 0 175 40 135;
CCC
TTT
    i

: 250(175 135) 450 0
z
D j 172.2 N
z
D

: 150(104 84) 450 0
y
Dk 62.7 N
y
D




www.elsolucionario.org

486. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.92 (Continued)


0:
x
F 0
x
A


0:
y
F 104 84 62.7 0
y
A
  125.3 N
y
A

0:
z
F 175 135 172.2 0
z
A
  137.8 N
z
A

(125.3 N) (137.8 N) ; (62.7 N) (172.2 N) AjkDjk 
www.elsolucionario.org

487. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.93
A small winch is used to raise a 120-lb load.
Find (a) the magnitude of the vertical force
P
that should be applied at C to maintain
equilibrium in the position shown, (b) the
reactions at A and B, assuming that the bearing at
B does not exert any axial thrust.

SOLUTION

Dimensions in in.

We have six unknowns and six equations of equilibrium.

(32 in.) (10 in.)cos30 (10 in.)sin 30
32 8.6603 5
C
 
 ri j k
ijk



0: (10 4 ) ( 120 ) (20 ) ( ) (32 8.6603 5 ) ( ) 0
1200 480 20 20 32 5 0
Ayz
yz
BB P
BBPP
       
 
Mikjijk jkj i
ki k jki


Equating the coefficients of the unit vectors to zero,
: 480 5 ) 0 96.0 lb; PP 
i ( ) 96.0 lbaP 

: 20 0
z
Bj 0
z
B

: 1200 20 32(96.0) 0
y
B  k 213.6 lb
y
B

0:
x
F 0
x
A

0:
y
F 120 213.6 96.0 0
y
A   2.40 lb
y
A

0:
z
F 0
zz
AB 0
zz
AB 

(2.40lb); (214lb) 
AjBj 
www.elsolucionario.org

488. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.94
A 48-ft sheet of plywood weighing 34 lb has been
temporarily placed among three pipe supports. The lower
edge of the sheet rests on small collars at A and B and its
upper edge leans against pipe C. Neglecting friction at all
surfaces, determine the reactions at A, B, and C.

SOLUTION


/
3.75 1.3919
22
GB
 rijk
We have 5 unknowns and 6 Eqs. of equilibrium.
Plywood sheet is free to move in z direction, but equilibrium is maintained
(0).
z
F

///
0: ( ) ( ) ( ) 0
BABxyCB GB
MrAArCrw   ij i j

0 0 4 3.75 1.3919 2 1.875 0.696 1 0
0000340
xy
AA C


ijk i j k i jk


4 4 2 1.3919 34 63.75 0
yx
AAC C   ijj ki k
Equating coefficients of unit vectors to zero:
:
i 4340
y
A 8.5 lb
y
A
:
j 24 0
x
CA 
11
(45.80) 22.9 lb
22
x
AC 
: 1.3919 63.75 0 C
k 45.80 lbC 45.8 lbC

0:
x
F 0:
xx
ABC 45.8 22.9 22.9 lb
x
B

0:
y
F 0:
yy
ABW 34.0 8.50 25.5 lb
y
B
(22.9 lb) (8.50 lb) (22.9 lb) (25.5 lb) (45.8 lb)  
AijBijC i 
www.elsolucionario.org

489. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.95
A 250 × 400-mm plate of mass 12 kg and a 300-mm-diameter
pulley are welded to axle AC that is supported by bearings at A and
B. For
 = 30°, determine (a) the tension in the cable, (b) the
reactions at A and B. Assume that the bearing at B does not exert
any axial thrust.

SOLUTION
Free-Body Diagram:

2
(12 kg)(9.81 m/s ) 117.720 NWmg   j
//A
/
570 ; 150 770 ;
(200sin ) (200cos ) 285
BA D
GA

  
 
rkrjk
rijk


/D//
0: ( ) ( ) 0
ABAxyA GA
MBBTW    rijrir j
0 0 570 0 150 770 200sin 200cos 285 0
0000 0
xy
BB T W
   

ij k i j k i j k

570 570 770 150 285 (200sin ) 0
yx
BBTTW W   ijjki k
Equating the coefficients of the unit vectors to zero,
:
i 570 285(117.72) 0
y
B 58.860 N
y
B
:
j 570 770 0
x
BT
77
; 106.017 N
57
xx
BTB
:
k 150 (200sin30 )117.720 0T

78.480 NT ( ) 78.5 NaT 

0:
y
F 0
yy
ABW 117.720 58.860 58.860 N
y
A

0:
z
F 0
z
A

0:
x
F 0; 27.537 N
xx x
ABT A 
(b) (27.5 N) (58.9 N) ; (106.0 N) (58.9 N)   
AijBij  www.elsolucionario.org

490. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.96
Solve Prob. 4.95 for  = 60°.
PROBLEM 4.95 A 250 × 400-mm plate of mass 12 kg and a
300-mm-diameter pulley are welded to axle AC that is supported
by bearings at A and B. For
 = 30°, determine (a) the tension in
the cable, (b) the reactions at A and B. Assume that the bearing at
B does not exert any axial thrust.

SOLUTION
Free-Body Diagram:

2
(12 kg)(9.81 m/s ) 117.720 NWmg   j
//A
/
570 ; 150 770 ;
(200sin ) (200cos ) 285
BA D
GA

  
 
rkrjk
rijk


/D//
0: ( ) ( ) 0
ABAxyA GA
MBBTW    rijrir j
0 0 570 0 150 770 200sin 200cos 285 0
0000 0
xy
BB T W
   

ij k i j k i j k

570 570 770 150 285 (200sin ) 0
yx
BBTTW W   ijjki k
:
i 570 285(117.72) 0
y
B 58.860 N
y
B
:
j 570 770 0
x
BT
77
; 183.625 N
57
xx
BTB
:
k 150 (200sin 60 )117.720 0T

135.931 NT
( ) 135.9 NaT


0:
y
F 0
yy
ABW 117.720 58.860 58.860 N
y
A

0:
z
F 0
z
A

0:
x
F 0; 47.694 N
xx x
ABT A 
(b) (47.7 N) (58.9 N) ; (183.6 N) (58.9 N)   
AijBij 
www.elsolucionario.org

491. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.97
The 20 × 20-in. square plate shown weighs 56 lb and is
supported by three vertical wires. Determine the tension in
each wire.

SOLUTION

0
b
Wj and (56.0 lb)W j

///
0: ( ) 0
ABABCACGA
MTTW   rjrjr j

(16 10 ) (16 10 ) T 10 ( ) 0
BC
TWik j ik ji j

16 10 16 10 10 0
BBCC
TTTTW  kikik
Equate coefficients of unit vectors to zero:

:10 10 0;
BC BC
TT TT i

:16 16 10 0
16 16 10(56.0 lb)=0
BC
BB
TTW
TT


k


17.50 lb
BC
TT


0: 0
y ABC
F TTTW   

17.50 17.50 56.0 0
21.0 lb
A
A
T
T





21.0 lb
A
T 

17.50 lb
BC
TT 


www.elsolucionario.org

492. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.98
The 20 × 20-in. square plate shown weighs 56 lb and is
supported by three vertical wires. Determine the weight and
location of the lightest block that should be placed on the
plate if the tensions in the three wires are to be equal.

SOLUTION

Let be the weight of block; and its coordinates.
b
Wxzj
Since the tensions in the wires are equal, let
ABC
TTTT

0: ( ) ( ) ( ) 0
OABCG b
MTTTWxzW       rjrjrjr j ik j

(10 ) (16 ) (16 20 ) (10 10 ) ( ) ( ) ( ) 0
b
TT T WxzW         kj ij i kj ik j ik j

10 16 16 20 10 10 0
bb
TT T TW WWxWz       ikk i k i k i
Equate coefficients of unit vectors to zero:

:3010 0;(1)
b
TWWz i

: 32 10 0 (2)
b
TWWxk
Also,

0: 3 0 (3)
yb
FTWW  

(1)+10(3): ( 10) 0 since and must be 20 in. 10 in.
b
zW xz z  

3(2) 32(3) : 2 ( 3 32) 0
22
; since 20 in.
3 32 3(20) 32
1
Finally, with 56 lb; (56 lb) 4.00 lb
14
b
bb
b
WxW
WW
x
Wx W
WW








4.00 lb at 20.0 in., 10.00 in.
b
Wxz 
www.elsolucionario.org

493. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.99
An opening in a floor is covered by a 1 1.2-m sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a
small block C. Determine the vertical component of the
reaction (a) at A, (b) at B, (c) at C.

SOLUTION

/
/
/
0.6
0.8 1.05
0.3 0.6
BA
CA
GA


ri
rik
rik


(18 kg)9.81
176.58 N
Wmg
W




///
0: ( ) 0
ABACAGA
MBCW rjrjr j
(0.6 ) (0.8 1.05 ) (0.3 0.6 ) ( ) 0BC W     
ij i kj i k j

0.6 0.8 1.05 0.3 0.6 0BC CWWkk iki
Equate coefficients of unit vectors to zero:

0.6
: 1.05 0.6 0 176.58 N 100.90 N
1.05
CW C

 


i
:0.6 0.8 0.3 0 BCW 
k
0.6 0.8(100.90 N) 0.3(176.58 N) 0 46.24 NBB
0: 0
y
FABCW 
46.24 N 100.90 N 176.58 N 0 121.92 NAA  
( ) 121.9 N ( ) 46.2 N ( ) 100.9 NaA bB cC

www.elsolucionario.org

494. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.100
Solve Problem 4.99, assuming that the small block C is
moved and placed under edge DE at a point 0.15 m from
corner E.
PROBLEM 4.99 An opening in a floor is covered by a
11.2-m sheet of plywood of mass 18 kg. The sheet is
hinged at A and B and is maintained in a position slightly
above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.

SOLUTION

/
/
/
0.6
0.65 1.2
0.3 0.6
BA
CA
GA


ri
rik
rik


2
(18 kg) 9.81 m/s
176.58 N
Wmg
W




///
0: ( ) 0
ABACAGA
MBCW   rjrjr j
0.6 (0.65 1.2 ) (0.3 0.6 ) ( ) 0BC W     
ij i k j i k j

0.6 0.65 1.2 0.3 0.6 0BCCWWkkiki
Equate coefficients of unit vectors to zero:

0.6
: 1.2 0.6 0 176.58 N 88.29 N
1.2
CW C

   


i
: 0.6 0.65 0.3 0 BCW
k
0.6 0.65(88.29 N) 0.3(176.58 N) 0 7.36 NBB
0: 0
y
FABCW 
7.36 N 88.29 N 176.58 N 0 95.648 NAA   
( ) 95.6 N ( ) 7.36 N ( ) 88.3 N aA bB cC


www.elsolucionario.org

495. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.101
Two steel pipes AB and BC, each having a mass per unit
length of 8 kg/m, are welded together at B and supported
by three wires. Knowing that
0.4 m,a determine the
tension in each wire.

SOLUTION



1
2
0.6
1.2
Wmg
Wmg







//1/2/
0: ( ) ( ) 0
DADAED FD CDC
MTWWT  rjr jr jrj

12
( 0.4 0.6 ) ( 0.4 0.3 ) ( ) 0.2 ( ) 0.8 0
AC
TWWT     ikj ik j i j ij

112
0.4 0.6 0.4 0.3 0.2 0.8 0
AA C
TTWWWT   kikikk
Equate coefficients of unit vectors to zero:

11
11
: 0.6 0.3 0; 0.6 0.3
22
AA
T W T W mg mg   i

12
:0.4 0.4 0.2 0.8 0
AC
TWWT k

0.4(0.3 ) 0.4(0.6 ) 0.2(1.2 ) 0.8 0
C
mg mg mg T  

(0.12 0.24 0.24)
0.15
0.8
C
mg
Tmg




12
0: 0
yACD
FTTTWW     

0.3 0.15 0.6 1.2 0
1.35
D
D
mg mg T mg mg
Tmg
    



2
(8 kg/m)(9.81m/s ) 78.480 N/mmg

0.3 0.3 78.480
A
Tmg  23.5 N
A
T 

0.15 0.15 78.480
C
Tmg  11.77 N
C
T 

1.35 1.35 78.480
D
Tmg  105.9 N
D
T 
www.elsolucionario.org

496. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.102
For the pipe assembly of Problem 4.101, determine (a) the
largest permissible value of a if the assembly is not to tip,
(b) the corresponding tension in each wire.

SOLUTION




1
2
0.6
1.2
Wmg
Wmg







//1/2/
0: ( ) ( ) 0
DADAED FD CDC
MTWWT  rjr jr jrj

12
( 0.6 ) ( 0.3 ) ( ) (0.6 ) ( ) (1.2 ) 0
A C
aTa W aW aT          ikjik j i j ij

112
0.6 0.3 (0.6 ) (1.2 ) 0
AA C
Ta T Wa W W a T a     kiki k k
Equate coefficients of unit vectors to zero:

11
11
: 0.6 0.3 0; 0.6 0.3
22
AA
T W T W mg mg   i

12
:(0.6)(1.2)0
AC
Ta Wa W a T a  k

0.3 0.6 1.2 (0.6 ) (1.2 ) 0
C
mga mga mg a T a  

0.3 0.6 1.2(0.6 )
1.2
C
aa a
T
a
 


For maximum a and no tipping, 0.
C
T
(a)
0.3 1.2(0.6 ) 0
0.3 0.72 1.2 0
aa
aa
 
  

1.5 0.72a 0.480 m a 


www.elsolucionario.org

497. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.102 (Continued)

(b) Reactions:
2
(8 kg/m) 9.81 m/s 78.48 N/mmg

0.3 0.3 78.48 23.544 N
A
Tmg
 23.5 N
A
T 

12
0: 0
yACD
FTTTWW     

00.61.20
AD
TTmgmg
   

1.8 1.8 78.48 23.544 117.72
DA
TmgT
 117.7 N
D
T 

www.elsolucionario.org

498. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.103
The 24-lb square plate shown is supported by three
vertical wires. Determine (a) the tension in each wire
when 10a
in., (b) the value of a for which the tension in
each wire is 8 lb.

SOLUTION

/
/
/
30
30
15 15
BA
CA
GA
a
a


rik
rik
rik




By symmetry,
.BC

//
0: ( ) 0
ABAC GA
MBCW rjrjr j
(30) (30 ) (1515)()0 aBaB W
ikj ikj ik j
30 30 15 15 0 Ba B B Ba W W   
kikiki
Equate coefficient of unit vector
i to zero:
:30 15 0 BBa W 
i

15 15
30 30
WW
BCB
aa


(1)
0: 0
y
FABCW 

15
20;
30 30
WaW
AWA
aa




(2)
(a) For 10 in. a
From Eq. (1):
15(24 lb)
9.00 lb
30 10
CB 


From Eq. (2):
10(24 lb)
6.00 lb
30 10
A

6.00 lb; 9.00 lbABC 

www.elsolucionario.org

499. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.103 (Continued)

(b) For tension in each wire  8 lb,
From Eq. (1):
15(24 lb)
8lb
30a



30 in. 45 a
 15.00 in. a 
www.elsolucionario.org

500. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.104
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported
by three legs equally spaced around the edge. A vertical load
P of
magnitude 100 lb is applied to the top of the table at D. Determine the
maximum value of a if the table is not to tip over. Show, on a sketch,
the area of the table over which
P can act without tipping the table.

SOLUTION
2ft sin30 1ft rbr
We shall sum moments about AB.
()() 0 brC abPbW
(1 2) ( 1)100 (1)30 0 Ca  

1
[30 ( 1)100]
3
Ca

If table is not to tip,
0.C

[30 ( 1)100] 0
30 ( 1)100
a
a
 


1 0.3 1.3 ft 1.300 ftaa a  

Onlydistance from P to AB matters. Same condition must be satisfied for each leg. P must be located
in shaded area for no tipping.
www.elsolucionario.org

501. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.105
A 10-ft boom is acted upon by the 840-lb force shown. Determine
the tension in each cable and the reaction at the ball-and-socket
joint at A.

SOLUTION
We have five unknowns and six equations of equilibrium, but equilibrium is maintained (0).
x
M
Free-Body Diagram:


( 6 ft) (7 ft) (6 ft) 11 ft
( 6 ft) (7 ft) (6 ft) 11 ft
(6 7 6)
11
(6 7 6)
11
BD
BD BD
BE
BE BE
BD BD
BE BE
TBD
TT
BD
TBE
TT
BE
   
   

 ijk
ijk
ijk
ijk






0: ( 840 ) 0
ABBDBBEC
MTT  rrrj

6 (676)6 (676)10(840)0
11 11
BD BE
TT
iijkiijkij


42 36 42 36
8400 0
11 11 11 11
BD BD BE BE
TTTT kjkjk
Equate coefficients of unit vectors to zero:

36 36
:0
11 11
BD BE BE BD
TT TT i

42 42
:84000
11 11
BD BE
TTk

42
28400
11
BD
T



 1100 lb
BD
T 

1100 lb
BE
T 
www.elsolucionario.org

502. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.105 (Continued)


66
0: (1100 lb) (1100 lb) 0
11 11
xx
FA   

1200 lb
x
A

77
0: (1100 lb) (1100 lb) 840 lb 0
11 11
yy
FA    
560 lb
y
A

66
0: (1100 lb) (1100 lb) 0
11 11
zz
FA   

0
z
A
 (1200 lb) (560 lb) Aij 
www.elsolucionario.org

503. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.106
The 6-m pole ABC is acted upon by a 455-N force as shown.
The pole is held by a ball-and-socket joint at A and by two
cables BD and BE. For
3 m,a determine the tension in
each cable and the reaction at A.

SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium, but equilibrium is maintained

(0)
AC
M

3
6
B
C


rj
rj


362 7m
1.5 3 3 4.5 m
1.5 3 3 4.5 m
CF CF
BD BD
BE BE
   
 
 
ijk
ijk
ijk





(3 6 2)
7
(1.5 3 3 ) ( 2 2 )
4.5 3
(22)
3
BD BD
BD BD
BD
BE BE
CF P
P
CE
TTBD
T
BD
TBE
T
BE

 
 
Pijk
Tijkijk
Tijk





0: 0
03 0 030 0 60 0
337
122 122 362
ABBDBBEC
BD BE
M
TT P
   

  
rT rT rP
ijk ijk i jk

Coefficient of i:
12
22 0
7
BD BE
TT P  (1)
Coefficient of k:
18
0
7
BD BF
TT P   (2)
www.elsolucionario.org

504. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.106 (Continued)

Eq. (1)  2 Eq. (2):
48 12
40
77
BD BD
TPTP 
Eq. (2):
12 18 6
0
77 7
BE BE
PTPTP 
Since
12
445 N (455)
7
BD
PT 780 N
BD
T 

6
(455)
7
BE
T 390 N
BE
T 

0: 0
BD BE
FTTPA
Coefficient of i:
780 390 455
(3) 0
337
x
A
 

260 130 195 0 195.0 N
xx
AA 
Coefficient of j:
780 390 455
(2) (2) (6) 0
337
y
A
520 260 390 0 1170 N
yy
AA   
Coefficient of k:
780 390 455
(2) (2) (2) 0
337
z
A

520 260 130 0 130.0 N
zz
AA  
(195.0 N) (1170 N) (130.0 N)
Aijk 
www.elsolucionario.org

505. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.107
Solve Problem 4.106 for 1.5 m.a
PROBLEM 4.106 The 6-m pole ABC is acted upon by a 455-
N force as shown. The pole is held by a ball-and-socket joint
at A and by two cables BD and BE. For
3 m,a determine
the tension in each cable and the reaction at A.

SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is maintained

(0)
AC
M

3
6
B
C


rj
rj


1.5 6 2 6.5 m
1.5 3 3 4.5 m
1.5 3 3 4.5 m
CF CF
BD BD
BE BE
   
 
 
ijk
ijk
ijk





(1.562) (3124)
6.5 13
(1.5 3 3 ) ( 2 2 )
4.5 3
(22)
3
BD BD
BD BD
BD
BE BE
CF P P
P
CE
TTBD
T
BD
TBE
T
BE

 
 
Pijkijk
Tijkijk
Tijk





0: 0
03 0 030 0 6 0 0
33 13
122 122 312 4
ABBDBBEC
BD BE
M
TT P
   
 
   
rT rT rP
ijk ijk i j k

Coefficient of i:
24
22 0
13
BD BE
TT P   (1)
Coefficient of k:
18
0
13
BD BE
TT P  (2)
www.elsolucionario.org

506. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.107 (Continued)

Eq. (1) 2 Eq. (2):
60 15
40
13 13
BD BD
TPTP  
Eq (2):
15 18 3
0
13 13 13
BE BE
PTPTP 
Since
15
445 N (455)
13
BD
PT 525 N
BD
T 

3
(455)
13
BE
T 105.0 N
BE
T 
0: 0
BD BE
    FTTPA
Coefficient of i:
525 105 455
(3) 0
3313
x
A
 

175 35 105 0 105.0 N
xx
AA   
Coefficient of j:
525 105 455
(2) (2) (12) 0
3313
y
A 
350 70 420 0 840 N
yy
AA  
Coefficient of k:
525 105 455
(2) (2) (4) 0
3313
z
A

350 70 140 0 140.0 N
zz
AA 
(105.0 N) (840 N) (140.0 N)
Aijk 

www.elsolucionario.org

507. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.108
A 2.4-m boom is held by a ball-and-socket joint at C and by
two cables AD and AE. Determine the tension in each cable
and the reaction at C.

SOLUTION
Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained
(0).
AC
M

1.2
2.4
B
A

rk
rk


0.8 0.6 2.4 2.6 m
0.8 1.2 2.4 2.8 m
AD AD
AE AE
   
 
ijk
ijk




( 0.8 0.6 2.4 )
2.6
(0.8 1.2 2.4 )
2.8
AD
AD
AE
AE
TAD
T
AD
TAE
T
AE
 
 
ijk
ijk




0: ( 3 kN) 0
CAADAAEB
M    rT rT r j

0 0 2.4 0 0 2.4 1.2 ( 3.6 kN) 0
2.6 2.8
0.8 0.6 2.4 0.8 1.2 2.4
AD AE
TT

 
ijk ijk
kj

Equate coefficients of unit vectors to zero:

: 0.55385 1.02857 4.32 0
: 0.73846 0.68671 0
AD AE
AD AE
TT
TT


i
j (1)

0.92857
AD AE
TT (2)
From Eq. (1):
0.55385(0.92857) 1.02857 4.32 0
AE AE
TT

1.54286 4.32
2.800 kN
AE
AE
T
T


2.80 kN
AE
T 
www.elsolucionario.org

508. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.108 (Continued)

From Eq. (2):
0.92857(2.80) 2.600 kN
AD
T 2.60 kN
AD
T 

0.8 0.8
0: (2.6 kN) (2.8 kN) 0 0
2.6 2.8
0.6 1.2
0: (2.6 kN) (2.8 kN) (3.6 kN) 0 1.800 kN
2.6 2.8
2.4 2.4
0: (2.6 kN) (2.8 kN) 0 4.80 kN
2.6 2.8
xx x
yy y
zz z
FC C
FC C
FC C
    
     
    

(1.800 kN) (4.80 kN)
 Cjk 
www.elsolucionario.org

509. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.109
Solve Problem 4.108, assuming that the 3.6-kN load is
applied at Point A.
PROBLEM 4.108 A 2.4-m boom is held by a ball-and-
socket joint at C and by two cables AD and AE. Determine
the tension in each cable and the reaction at C.

SOLUTION
Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained
(0).
AC
M

0.8 0.6 2.4 2.6 m
0.8 1.2 2.4 2.8 m
AD AD
AE AE
   
 
ijk
ijk




( 0.8 0.6 2.4 )
2.6
(0.8 1.2 2.4 )
2.8
AD
AD
AE
AE
TAD
T
AD
TAE
T
AE
 
 
ijk
ijk




0: ( 3.6 kN)
CAADAAEA
M   rT rT r j

Factor
:
A
r ((3.6kN))
AADAE
rT T j
or
(3 kN) 0
AD AE
 TT j (Forces concurrent at A)
Coefficient of
i:
(0.8) (0.8) 0
2.6 2.8
AD AE
TT



2.6
2.8
AD AE
TT (1)
Coefficient of
j:
(0.6) (1.2) 3.6 kN 0
2.6 2.8
2.6 0.6 1.2
3.6 kN 0
2.8 2.6 2.8
0.6 1.2
3.6 kN
2.8
AD AE
AE AE
AE
TT
TT
T










5.600 kN
AE
T 5.60 kN
AE
T www.elsolucionario.org

510. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.109 (Continued)

From Eq. (1):
2.6
(5.6) 5.200 kN
2.8
AD
T 5.20 kN
AD
T 

0.8 0.8
0: (5.2 kN) (5.6 kN) 0 0
2.6 2.8
0.6 1.2
0: (5.2 kN) (5.6 kN) 3.6 kN 0 0
2.6 2.8
2.4 2.4
0: (5.2 kN) (5.6 kN) 0 9.60 kN
2.6 2.8
xx x
yy y
zz z
FC C
FC C
FC C
    
     
    

(9.60 kN) Ck

Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
www.elsolucionario.org

511. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.110
The 10-ft flagpole AC forms an angle of 30 with the z axis. It
is held by a ball-and-socket joint at C and by two thin braces
BD and BE. Knowing that the distance BC is 3 ft, determine
the tension in each brace and the reaction at C.

SOLUTION
Free-Body Diagram:







Express all forces in terms of rectangular components:


3ft 3ft
E
rij


    3 ft sin30 3 ft cos30 1.5 ft 2.598 ft
B
 rjkjk


3ft 3ft
D
 rij


   10 ft sin30 10 ft cos30 5 ft 8.66 ft
A
 rjkjk


  3 ft 3 ft 1.5 ft 2.598 ft
EB
BE   rr i j j k


or

  3 ft 1.5 ft 2.598 ft ,BE ij k

and 4.243 ftBE


  3 ft 3 ft 1.5 ft 2.598 ft
DB
BD   rr i j j k


or

  3ft 1.5ft 2.598ft ,BD  ij k

and 4.243 ftBD
Then
0.707 0.3535 0.6123BD
BD BD
BD
TT T
BD
 ijk




0.707 0.3535 0.6123BE
BE BE
BE
TT T
BE
  ijk



www.elsolucionario.org

512. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.110 (Continued)

   0: 5ft 8.66ft 75lb
CBBDBBE
     

MrTrT j k j
or
0 1.5 2.598 0 1.5 2.598 649.5 0
0.707 0.3535 0.6123 0.707 0.3535 0.6123
BD
T
 
ijk ijk
i


Equating the coefficients of the unit vectors to zero:

: 1.837 1.837 0
BD BE
TT j

: 1.837 1.837 649.5 lb 0
BD BE
TT i


176.8 lb
BD
T 

176.8 lb
BE
T 
Force equations:


   176.8 0.707 176.8 0.707 0,
x
C  or 0
x
C


   176.8 0.3535 176.8 0.3535 75 lb 0,
y
C or 50 lb
y
C


   176.8 0.6123 176.8 0.6123 0,
z
C or 216.5 lb
z
C
  50 lb 216.5 lb Cj k 
www.elsolucionario.org

513. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.111
A 48-in. boom is held by a ball-and-socket joint at C and by
two cables BF and DAE; cable DAE passes around a
frictionless pulley at A. For the loading shown, determine
the tension in each cable and the reaction at C.

SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
(0).
AC
M
TTension in both parts of cable DAE.

30
48
B
A

rk
rk


20 48 52 in.
20 48 52 in.
16 30 34 in.
AD AD
AE AE
BF BF
  
 
 
ik
jk
ik





(20 48) (5 12)
52 13
(20 48 ) (5 12 )
52 13
(16 30 ) (8 15 )
34 17
AD
AE
BF BF
BF BF
AD T T
T
AD
AE T T
T
AE
TTBF
T
BF

 
 
Tikik
Tjkjk
Tikik





0: ( 320 lb) 0
CAADAAEBBFB
   M rTrTrTr j

0 0 48 0 0 48 0 0 30 (30 ) ( 320 ) 0
13 13 17
5 0 12 0 5 12 8 0 15
BF
TTT
 

ijk ijk ijk
kj

Coefficient of
i:
240
9600 0 520 lb
13
TT 



www.elsolucionario.org

514. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.111 (Continued)

Coefficient of j:
240 240
0
13 17
BD
TT 

17 17
(520) 680 lb
13 13
BD BD
TT T 

0: 320 0
AD AE BF
C     FTTT j
Coefficient of i:
20 8
(520) (680) 0
52 17
x
C

200 320 0 120 lb
xx
CC  
Coefficient of j:
20
(520) 320 0
52
y
C
200 320 0 120 lb
yy
CC 
Coefficient of k:
48 48 30
(520) (520) (680) 0
52 52 34
z
C

480 480 600 0
z
C


1560 lb
z
C
Answers:
DAE
TT 520 lb
DAE
T 
680 lb
BD
T 
(120.0 lb) (120.0 lb) (1560 lb)
 Cijk 
www.elsolucionario.org

515. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.112
Solve Problem 4.111, assuming that the 320-lb load is
applied at A.
PROBLEM 4.111 A 48-in. boom is held by a ball-and-
socket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
loading shown, determine the tension in each cable and the
reaction at C.

SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
(0).
AC
M
Ttension in both parts of cable DAE.

30
48
B
A


rk
rk


20 48 52 in.
20 48 52 in.
16 30 34 in.
AD AD
AE AE
BF BF
  
 
 
ik
jk
ik





(20 48) (5 12)
52 13
(20 48 ) (5 12 )
52 13
(16 30 ) (8 15 )
34 17
AD
AE
BF BF
BF BF
AD T T
T
AD
AE T T
T
AE
TTBF
T
BF

 
 
Tikik
Tjkjk
Tikik





0: ( 320 lb) 0
C A AD A AE B BF A
M     rT rT rT r j

0 0 48 0 0 48 0 0 30 48 ( 320 ) 0
13 13 17
5 0 12 0 5 12 8 0 15
BF
TTT
 
  
ijk ijk ijk
kj

Coefficient of
i:
240
15,360 0 832 lb
13
TT  



www.elsolucionario.org

516. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.112 (Continued)

Coefficient of j:
240 240
0
13 17
BD
TT 

17 17
(832) 1088 lb
13 13
BD BD
TT T 

0: 320 0
AD AE BF
     FTTT jC
Coefficient of i:
20 8
(832) (1088) 0
52 17
x
C
 

320 512 0 192 lb
xx
CC  
Coefficient of j:
20
(832) 320 0
52
y
C

320 320 0 0
yy
CC 
Coefficient of k:
48 48 30
(832) (852) (1088) 0
52 52 34
z
C 

768 768 960 0 2496 lb
zz
CC   
Answers:
DAE
TT
 832 lb
DAE
T 
1088 lb
BD
T 
(192.0 lb) (2500 lb) Cik 
www.elsolucionario.org

517. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.113
A 10-kg storm window measuring 900 × 1500 mm is held by hinges at
A and B. In the position shown, it is held away from the side of the
house by a 600-mm stick CD. Assuming that the hinge at A does not
exert any axial thrust, determine the magnitude of the force exerted by
the stick and the components of the reactions at A and B.

SOLUTION
Free-Body Diagram: Since CD is a two-force member,
CD
F
is directed along CD and triangle ACD is isosceles. We have
0.3 m
sin 0.2; 11.5370 and 2 23.074
1.5 m
   


2
(10 kg)(9.81 m/s ) (98.10 N) Wj
(0.75 m)sin 23.074 (0.75 m)cos 23.074 (0.45 m)
(0.29394 m) (0.690 m) (0.45 m)
(cos11.5370 sin11.5370 )
(0.97980 0.20 )
G
G
CD CD
CD CD
F
F
 



rijk
rijk
Fij
Fij



0: 0
BAG CCD
M  rArWrF

0.9 ( ) (0.29394 0.690 0.45 ) ( 98.10 )
(1.5 0.9 ) (0.97980 0.20 ) 0
xy
CD
AA
F
    
   kij i j k j
jk i j

0.90 0.90 28.836 44.145 1.46970 0.88182 0.180 0
x y CD CD CD
AA F F F    jik i k j i
Equating the coefficients of the unit vectors to zero,

28.836 1.46970 0; 19.6203 N
CD CD
FF  k: www.elsolucionario.org

518. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.113 (Continued)


0.90 0.88182(19.6203 N) 0; 19.2240 N
xx
AA  j:
0.90 44.145 0.180(19.6203 N) 0; 45.123 N
yy
AA i:

19.62 N
CD
F 
(19.22N) (45.1 N)
Aij 
0: cos11.5370 0
19.2240 19.6203cos11.5370 0
0
xxxCD
x
x
FABF
B
B
   
  



0: sin11.5370 0
45.122 19.6230sin11.5370 98.1 0
49.053 N
yyyCD
y
y
FABF W
B
B
   






0: 0
zz
FB 
(49.1 N) Bj

www.elsolucionario.org

519. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.114
The bent rod ABEF is supported by bearings at C and D and
by wire AH. Knowing that portion AB of the rod is 250 mm
long, determine (a) the tension in wire AH, (b) the reactions
at C and D. Assume that the bearing at D does not exert any
axial thrust.


SOLUTION
Free-Body Diagram: ABH is equilateral.
Dimensions in mm


/
/
/
50 250
300
350 250
(sin 30 ) (cos30 ) (0.5 0.866 )
HC
DC
FC
TT T
 



rij
ri
rik
Tjkjk


//
0: ( 400 ) 0
CHCDFC
  MrTrDr j

50 250 0 300 0 0 350 0 250 0
0 0.5 0.866 0 0 400 0
yz
T
DD


ij k ijki jk
Coefficient
i:
3
216.5 100 10 0T
461.9 N T 462 N T

Coefficient of
j: 43.3 300 0
z
TD 

43.3(461.9) 300 0 66.67 N
zz
DD
www.elsolucionario.org

520. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.114 (Continued)

Coefficient of k:
3
25 300 140 10 0
y
TD


3
25(461.9) 300 140 10 0 505.1 N
yy
DD   
(505 N) (66.7 N)
 Djk 
0: 400 0     FCDTj
Coefficient i:
0
x
C 0
x
C

Coefficient j: (461.9)0.5 505.1 400 0 336 N
yy
CC
Coefficient k:
(461.9)0.866 66.67 0
z
C 467 N
z
C (336 N) (467 N)
Cjk 

www.elsolucionario.org

521. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.115
The horizontal platform ABCD weighs 60 lb and supports a 240-lb
load at its center. The platform is normally held in position by
hinges at A and B and by braces CE and DE. If brace DE is
removed, determine the reactions at the hinges and the force
exerted by the remaining brace CE. The hinge at A does not exert
any axial thrust.


SOLUTION
Free-Body Diagram:








Express forces, weight in terms of rectangular components:

3ft 4ft 2ftECijk



  
222
342
342
CE CE CE
EC
FF
EC



ijk
F



0.55709 0.74278 0.37139
CE CE CE
FFF i
j k
() (300lb)mg Wj j


    

0: 4 ft 1.5 ft 2 ft 300 lb
3ft 4ft 0
B
CE
   

 

MkAikj
ikF

or

     4 ft 4 ft 1.5 ft 300 lb 2 ft 300 lb
yz
AA  i j ki
304 ft0
0.55709 0.74278 0.37139
CE
F
ijk
www.elsolucionario.org

522. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.115 (Continued)

Setting the coefficients of the unit vectors equal to zero:

 
  : 0.74278 3 ft 300 lb 1.5 ft 0
CE
F k

201.94 lb
CE
F
or
202 lb
CE
F 


    : 4 ft 0.55709 201.94 lb 4 ft 0.37139 201.94 lb 3 ft 0
x
A 

j

56.250 lb
x
A


   : 4 ft 0.74278 201.94 lb 4 ft 300 lb 2 ft 0
y
A   

i
0
y
A

 0: 56.250 lb 0.55709 201.94 lb 0
xx
FB    

56.249 lb
x
B

 0: 0 300 lb 0.74278 201.94 lb 0
yy
FB    
150.003 lb
y
B


0: 0.371391 201.94 lb 0
zz
FB  

74.999 lb
z
B
Therefore:


56.3 lbAi 
   56.2 lb 150.0 lb 75.0 lb  Bijk 



www.elsolucionario.org

523. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.116
The lid of a roof scuttle weighs 75 lb. It is hinged at corners
A and B and maintained in the desired position by a rod CD
pivoted at
C; a pin at end D of the rod fits into one of several
holes drilled in the edge of the lid. For the position shown,
determine (
a) the magnitude of the force exerted by rod CD,
(
b) the reactions at the hinges. Assume that the hinge at B
does not exert any axial thrust.

SOLUTION
Free-Body Diagram:







Geometry:
Using triangle
ACD and the law of sines

sin sin 50
or 20.946
7 in. 15 in.






20.946 70.946
     
Expressing
CD
Fin terms of its rectangular coordinates:

sin cos
CD CD CD
FF
F j k

sin 70.946 cos70.946
CD CD
FF
j k

0.94521 0.32646
CD CD CD
FFF
j k
        0: 26 in. 13 in. 16 in. sin50 16 in. cos50 75 lb
B
     

MiAi j k j


26 in. 7 in. 0
CD
   

ikF
or

       26 in. 26 in. 13 in. 75 lb 16 in. 75 lb cos50
yz
AA  kj k i


    26 in. 0.94521 26 in. 0.32646 7 in. 0.94521 0
CD CD CD
FFF kji
www.elsolucionario.org

524. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.116 (Continued)

(
a) Setting the coefficients of the unit vectors to zero:


   : 75lb 16in. cos50 0.94521 7in. 0
CD
F  

i

116.6 lb
CD
F 
(
b) 0: 0
xx
FA 


     : 0.94521 116.580 lb 26 in. 75 lb 13 in. 26 in. 0
y
A

k
72.693 lb
y
A


   : 0.32646 116.580 lb 26 in. 26 in. 0
z
A  

j

38.059 lb
z
A

 0: 72.693 lb 0.94521 116.580 lb 75 lb 0
y y
FB     
37.500 lb
y
B

 0: 38.059 lb 0.32646 116.580 lb 0
zz
FB    

0
z
B
Therefore:

  72.7 lb 38.1 lb Ajk 

37.5 lbBj 

www.elsolucionario.org

525. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.117
A 100-kg uniform rectangular plate is supported in the
position shown by hinges A and B and by cable DCE that
passes over a frictionless hook at C. Assuming that the
tension is the same in both parts of the cable, determine (a)
the tension in the cable, (b) the reactions at A and B.
Assume that the hinge at B does not exert any axial thrust.

SOLUTION

/
/
/
(960 180) 780
960 450
90
22
390 225
600 450
BA
GA
CA







rii
rik
ik
rik Dimensions in mm

TTension in cable DCE

690 675 450 1065 mm
270 675 450 855 mm
CD CD
CE CE
   
 
ijk
ijk




2
( 690 675 450 )
1065
(270 675 450 )
855
(100 kg)(9.81 m/s ) (981 N)
CD
CE
T
T
mg


  
Tijk
Tijk
Wi j j


/// /
0: ( ) 0
ACACDCACEGA BA
W  MrTrTr jrB

600 0 450 600 0 450
1065 855
690 675 450 270 675 450
390 0 225 780 0 0 0
0 981 0 0
yz
TT
BB

 


ijk ijk
ijkijk

www.elsolucionario.org

526. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.117 (Continued)

Coefficient of i:
3
(450)(675) (450)(675) 220.73 10 0
1065 855
TT

(a) 344.64 NT 345 NT 
Coefficient of j:
344.64 344.64
( 690 450 600 450) (270 450 600 450) 780 0
1065 855
z
B    

185.516 N
z
B
Coefficient of k:
3344.64 344.64
(600)(675) (600)(675) 382.59 10 780 0 113.178 N
1065 855
yy
BB
(
b) (113.2 N) (185.5 N)
 Bjk 

0: 0
CD CE
     FABTTW
Coefficient of i:
690 270
(344.64) (344.64) 0
1065 855
x
A 114.454 N
x
A
Coefficient of j:
675 675
113.178 (344.64) (344.64) 981 0 377.30 N
1065 855
y y
AA  
Coefficient of k:
450 450
185.516 (344.64) (344.64) 0
1065 855
z
A   141.496 N
z
A
(114.4 N) (377 N) (141.5 N)
 Aijk 
www.elsolucionario.org

527. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.118
Solve Problem 4.117, assuming that cable DCE is replaced
by a cable attached to Point
E and hook C.
PROBLEM 4.117 A 100-kg uniform rectangular plate is
supported in the position shown by hinges
A and B and by
cable
DCE that passes over a frictionless hook at C.
Assuming that the tension is the same in both parts of the
cable, determine (
a) the tension in the cable, (b) the
reactions at
A and B. Assume that the hinge at B does not
exert any axial thrust.

SOLUTION
See solution to Problem 4.113 for free-body diagram and analysis leading to the following:

1065 mm
855 mm
CD
CE



Now,
2
( 690 675 450 )
1065
(270 675 450 )
855
(100 kg)(9.81 m/s ) (981 N)
CD
CE
T
T
mg


  
Tijk
Tijk
Wi j j


// /
0: ( ) 0
ACACEGA BA
TWB
Mr r jr
600 0 450 390 0 225 780 0 0 0
855
270 675 450 0 981 0 0
yz
T
BB
 

ijk i jk ijk
Coefficient of i:
3
(450)(675) 220.73 10 0
855
T


621.31 NT
 621 NT 
Coefficient of j:
621.31
(270 450 600 450) 780 0 364.74 N
855
zz
BB   
Coefficient of k:
3621.31
(600)(675) 382.59 10 780 0 113.186 N
855
yy
BB
(113.2 N) (365 N)
 Bjk 
www.elsolucionario.org

528. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.118 (Continued)


0: 0
CE
FABTW
Coefficient of i:
270
(621.31) 0
855
x
A 196.2 N
x
A
Coefficient of j:
675
113.186 (621.31) 981 0
855
y
A  377.3 N
y
A
Coefficient of k:
450
364.74 (621.31) 0
855
z
A  37.7 N
z
A 

(196.2 N) (377 N) (37.7 N)
Aijk 

www.elsolucionario.org

529. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.119
Solve Prob. 4.113, assuming that the hinge at A has been removed and
that the hinge at B can exert couples about axes parallel to the x and y
axes.
PROBLEM 4.113A 10-kg storm window measuring 900 × 1500 mm is
held by hinges at A and B. In the position shown, it is held away from
the side of the house by a 600-mm stick CD. Assuming that the hinge at
A does not exert any axial thrust, determine the magnitude of the force
exerted by the stick and the components of the reactions at A and B.

SOLUTION
Free-Body Diagram: Since CD is a two-force member,
CD
F
is directed along CD and triangle ACD is isosceles. We have

2
(10 kg)(9.81 m/s ) (98.10 N) Wj
(0.75 m)sin 23.074 (0.75 m)cos 23.074 (0.45 m)
(0.29394 m) (0.690 m) (0.45 m)
(cos11.5370 sin11.5370 )
(0.97980 0.20 )
G
G
CD CD
CD CD
F
F
 



rijk
rijk
Fij
Fij



With hinge at A removed, hinge at B will develop two couples to prevent rotation about the x and z axes.
0:() () 0
BBxByGCCD
MMM   ijrWrF

( ) ( ) (0.29394 0.690 0.45 ) ( 98.10)
(1.5 0.9 ) (0.97980 0.20 ) 0
Bx By
CD
MM
F
 
   ij ijk j
jk i j

( ) ( ) 28.836 44.145 1.46970 0.88182 0.180 0
B x B y CD CD CD
MM F F F    ijki k j i

www.elsolucionario.org

530. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.119 (Continued)

Equating the coefficients of the unit vectors to zero,

28.836 1.46970 0; 19.6203 N
CD CD
FF  k:
( ) 0.88182(19.6203) 0; ( ) 17.3016 N m
By By
MM
 j:

( ) 44.145 0.180(19.6203) 0; ( ) 40.613 N m
Bx Bx
MM
 i:

19.62 N
CD
F 

(40.6 N m) (17.30 N m)
B
   Mi
j 

0: cos11.5370 0
19.6203cos11.5370 0
19.22 N
xxCD
x
x
FBF
B
B
  
 



0: sin11.5370 0
19.6230sin11.5370 98.1 0
94.2 N
yyCD
y
y
FBF W
B
B
  
 





0: 0
zz
FB 
(19.22 N) (94.2 N)
Bij 
www.elsolucionario.org

531. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.120
Solve Prob. 4.115, assuming that the hinge at B has been removed
and that the hinge at A can exert an axial thrust, as well as couples
about axes parallel to the x and y axes.
PROBLEM 4.115 The horizontal platform ABCD weighs 60 lb
and supports a 240-lb load at its center. The platform is normally
held in position by hinges at A and B and by braces CE and DE. If
brace DE is removed, determine the reactions at the hinges and
the force exerted by the remaining brace CE. The hinge at A does
not exert any axial thrust.

SOLUTION
Free-Body Diagram:

0
xyz
BBB






Express forces, weight in terms of rectangular components:

3ft 4ft 2ftECijk



  
222
342
342
CE CE CE
EC
FF
EC



ijk
F



0.55709 0.74278 0.37139
CE CE CE
FFFi
j k
() (300lb)mg Wj j

//
0: ( ) ( ) ( ) 0
xy
AGA CACEA A
WFMM    Mr
jri j

0: (1.5 2 ) ( 300 ) 3 (0.55709 0.74278 0.37139 )
( ) ( ) 0
xy
ACE
AA
F
MM
    

Mikji ijk
ij

or
450 600 2.2283 1.11417 0
xy
CE CE A A
FFMM   ki k j i j
www.elsolucionario.org

532. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.120 (Continued)

Setting the coefficients of the unit vectors equal to zero:


: 450 2.2283 0
CE
F k

201.95 lb
CE
F
or
202 lb
CE
F 

: 1.11417(201.95) ( ) 0
y
A
Mj

225.00 lb ft
y
A
M
:600 0
x
A
M i
600 lb ft
x
A
M


0: 0.55709 201.94 lb 0
xx
FA  

112.499 lb
x
A

 0: 300 lb 0.74278 201.94 lb 0
yy
FA   
150.003 lb
y
A

0: 0.371391 201.94 lb 0
zz
FA  

74.999 lb
z
A
Therefore:
 600 lb ft (225 lb ft)
A
Mij 
   112.5 lb 150.0 lb 75.0 lb  Aijk 



www.elsolucionario.org

533. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.121
The assembly shown is used to control the tension T in a tape that
passes around a frictionless spool at E. Collar C is welded to rods
ABC and CDE. It can rotate about shaft FG but its motion along the
shaft is prevented by a washer S. For the loading shown, determine
(a) the tension T in the tape, (b) the reaction at C.


SOLUTION
Free-Body Diagram:

/
/
4.2 2
1.6 2.4
AC
EC


rjk
rij


//
0: ( 6 ) ( ) ( ) ( ) 0    
CAC EC CyCz
MTMM rjrik jk
(4.2 2 ) ( 6 ) (1.6 2.4 ) ( ) ( ) ( ) 0
Cy Cz
TMM    jk j i j ik j k


Coefficient of
i: 12 2.4 0T 5.00 lbT 
Coefficient of
j: 1.6(5 lb) ( ) 0 ( ) 8 lb in.
Cy Cy
MM 
Coefficient of
k: 2.4(5 lb) ( ) 0 ( ) 12 lb in.
Cz Cz
MM 

(8.00 lb in.) (12.00 lb in.)
C
 Mjk 
0: (6 lb) (5 lb) (5 lb) 0  
xyz
FCCC ijk j i k
Equate coefficients of unit vectors to zero.
5 lb 6 lb 5 lb
xyz
CCC   

(5.00 lb) (6.00 lb) (5.00 lb)  Cijk 
www.elsolucionario.org

534. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.122
The assembly shown is welded to collar A that fits on the
vertical pin shown. The pin can exert couples about the x
and z axes but does not prevent motion about or along the y-
axis. For the loading shown, determine the tension in each
cable and the reaction at A.


SOLUTION
Free-Body Diagram:
First note:
22
22
(0.08 m) (0.06 m)
(0.08) (0.06) m
(0.8 0.6)
(0.12 m) (0.09 m)
(0.12) (0.09) m
(0.8 0.6 )








CF CF CF CF
CF
DE DE DE DE
DE
TT
T
TT
T
ij
T λ
ij
jk
T λ
jk
(a) From F.B.D. of assembly:
0: 0.6 0.8 480 N 0   
yCFDE
FTT
or
0.6 0.8 480 N
CF DE
TT (1)
0: (0.8 )(0.135 m) (0.6 )(0.08 m) 0   
yCF DE
MT T
or
2.25
DE CF
TT (2)
Substituting Equation (2) into Equation (1),

0.6 0.8[(2.25) ] 480 N
CF CF
TT

200.00 N
CF
T
or
200 N
CF
T 

and from Equation (2): 2.25(200.00 N) 450.00
DE
T
or
450 N
DE
T 

www.elsolucionario.org

535. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.122 (Continued)

(b) From F.B.D. of assembly:

0: (0.6)(450.00 N) 0 270.00 N   
zz z
FA A

0: (0.8)(200.00 N) 0 160.000 N   
xx x
FA A
or (160.0 N) (270 N)
 Aik 

0: (480 N)(0.135 m) [(200.00 N)(0.6)](0.135 m)
[(450 N)(0.8)](0.09 m) 0
  

x
xA
MM
16.2000 N m
x
A
M 

0: (480 N)(0.08 m) [(200.00 N)(0.6)](0.08 m)
[(450 N)(0.8)](0.08 m) 0
z
zA
MM  


0
z
A
M
or
(16.20 N m)
A
 Mi 
www.elsolucionario.org

536. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.123
The rigid L-shaped member ABC is
supported by a ball-and-socket joint at A and
by three cables. If a 1.8-kN load is applied at
F, determine the tension in each cable.

SOLUTION
Free-Body Diagram: Dimensions in mm

In this problem:
210 mma

We have (240 mm) (320 mm) 400 mm
(420 mm) (240 mm) (320 mm) 580 mm
(420 mm) (320 mm) 528.02 mm
CD CD
BD BD
BE BE
 
   
 
jk
ijk
ik




Thus,
(0.6 0.8 )
( 0.72414 0.41379 0.55172 )
(0.79542 0.60604 )
CD CD CD
BD BD BD
BE BE BE
CD
TT T
CD
BD
TT T
BD
BE
TT T
BE
 
 
 
jk
ijk
ik





0: ( ) ( ) ( ) ( ) 0    
ACCDBBDBBEW
M rTrTrTrW www.elsolucionario.org

537. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.123 (Continued)

Noting that (420 mm) (320 mm)
(320 mm)
(320 mm)
C
B
W
a
 

 
rik
rk
ri k

and using determinants, we write

420 0 320 0 0 320
0 0.6 0.8 0.72414 0.41379 0.55172
0 0 320 0 320 0
0.79542 0 0.60604 0 1.8 0
CD BD
BE
TT
Ta

 


ijk i j k
ijk ijk

Equating to zero the coefficients of the unit vectors,
i:
192 132.413 576 0
CD BD
TT  (1)
: j
336 231.72 254.53 0
CD BD BE
TTT   (2)
k:
252 1.8 0
CD
Ta
  (3)
Recalling that
210 mm,a Eq. (3) yields

1.8(210)
1.500 kN
252
CD
T 1.500 kN
CD
T 
From Eq. (1):
192(1.5) 132.413 576 0
BD
T 

2.1751 kN
BD
T 2.18 kN
BD
T 
From Eq. (2):
336(1.5) 231.72(2.1751) 254.53 0
BE
T  
 3.9603 kN
BE
T  3.96 kN
BE
T 
www.elsolucionario.org

538. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.124
Solve Problem 4.123, assuming that the 1.8-
kN load is applied at C.
PROBLEM 4.123 The rigid L-shaped
member ABC is supported by a ball-and-
socket joint at A and by three cables. If a 1.8-
kN load is applied at F, determine the
tension in each cable.

SOLUTION
See solution of Problem 4.123 for free-body diagram and derivation of Eqs. (1), (2), and (3):

192 132.413 576 0
CD BD
TT  (1)

336 231.72 254.53 0
CD BD BE
TTT   (2)

252 1.8 0
CD
Ta
  (3)

In this problem, the 1.8-kN load is applied at C and we have
420 mm.a
 Carrying into Eq. (3) and
solving for
,
CD
T

3.00
CD
T 3.00 kN
CD
T 
From Eq. (1):
(192)(3) 132.413 576 0
BD
T  0
BD
T
 
From Eq. (2):
336(3) 0 254.53 0
BE
T  3.96 kN
BE
T 
www.elsolucionario.org

539. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.125
The rigid L-shaped member ABF is supported by
a ball-and-socket joint at A and by three cables.
For the loading shown, determine the tension in
each cable and the reaction at A.

SOLUTION
Free-Body Diagram:

/
/
/
/
/
12
12 8
12 16
12 24
12 32




ri
rjk
rik
rik
rik
BA
FA
DA
EA
FA


12 9
15 in.
0.8 0.6
BG
BG
BG
 

 
ik
λ ik



12 16 ; 20 in.; 0.6 0.8
12 16 ; 20 in.; 0.6 0.8
DH
FJ
DH DH
FJ FJ 
    
    ij i j
ij i j




//
//
0:
( 24) ( 24) 0
12 0 0 12 0 16 12 0 32
0.8 0 0.6 0.6 0.8 0 0.6 0.8 0
12 0 8 12 0 24 0
0240 0240
A BABGBGDH DHDHFAFJFJ
FA EA
BG DH FJ
TTT
   
   

 


MrTrT rT
rjrj
ijk i jk i jk
ijk ijk

Coefficient of i:
12.8 25.6 192 576 0
DH FJ
TT (1)
Coefficient of k:
9.6 9.6 288 288 0
DH FJ
TT (2)
3
4
Eq. (1)  Eq. (2): 9.6 0
FJ
T 0
FJ
T  www.elsolucionario.org

540. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.125 (Continued)

From Eq. (1):
12.8 268 0
DH
T
 60 lb
DH
T 
Coefficient of j:
7.2 (16 0.6)(60.0 lb) 0
  
BG
T 80.0 lb
BG
T 

0: 24 24 0
BG BG DH DH FJ
TT T      FA
jj
Coefficient of i:
(80)( 0.8) (60.0)( 0.6) 0 100.0 lb  
xx
AA
Coefficient of j: (60.0)(0.8) 24 24 0
y
A
  0
y
A
Coefficient of k:
(80.0)( 0.6) 0
z
A
  48.0 lb
z
A
(100.0 lb) (48.0 lb)  Aik 
Note: The value 0
y
A can be confirmed by considering 0.
BF
M
 
www.elsolucionario.org

541. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.126
Solve Problem 4.125, assuming that the load
at C has been removed.
PROBLEM 4.125 The rigid L-shaped
member ABF is supported by a ball-and-
socket joint at A and by three cables. For the
loading shown, determine the tension in each
cable and the reaction at A.

SOLUTION
Free-Body Diagram:

/
/
/
/
12
12 16
12 24
12 32
BA
BA
EA
FA



ri
rik
rik
rik







12 9 ; 15 in.; 0.8 0.6
12 16 ; 20 in.; 0.6 0.8
12 16 ; 20 in.; 0.6 0.8
BG
DH
FJ
BG BG
DH DH
FJ FJ
    
    
    
ik i k
ij i j
ij i j







//
0:
A BA BG BG DA DH
TT  Mr r 
//
(24) 0
DH F A FJ FJ E A
T rrj
12 0 0 12 0 16 12 0 32 12 0 24 0
0.8 0 0.6 0.6 0.8 0 0.6 0.8 0 0 24 0
BG DH FJ
TT T
  
ijk i jk i jk ijk

: 12.8 25.6 576 0
DH FJ
TTi (1)
:
k 9.6 9.6 288 0
DH FJ
TT
  (2)
www.elsolucionario.org

542. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.126 (Continued)

Multiply Eq. (1) by
3
4
and subtract Eq. (2):

9.6 144 0
FJ
T
 15.00 lb
FJ
T 
From Eq. (1):
12.8 25.6(15.00) 576 0
DH
T
  15.00 lb
DH
T 
:
j 7.2 (16)(0.6)(15) (32)(0.6)(15) 0
BG
T  

7.2 432 0
BG
T
  60.0 lb
BG
T 

0: 24 0
BG BG DA DH FJ FJ
FTTT     Aj 

: (60)( 0.8) (15)( 0.6) (15)( 0.6) 0
x
Ai 66.0 lb
x
A
:
j (15)(0.8) (15)(0.8) 24 0
y
A
  0
y
A
:
k (60)(0.6) 0
z
A
  36.0 lb
z
A
(66.0 lb) (36.0 lb)  Aik 
www.elsolucionario.org

543. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.127
Three rods are welded together to form a “corner” that is
supported by three eyebolts. Neglecting friction, determine the
reactions at A, B, and C when
240Plb, 12 in., 8 in.,ab and
10 in.c

SOLUTION
From F.B.D. of weldment:

///
0: 0
OAOBOCO
 MrArBrC

120 0 080 0 0100
000
yz x z xy
AA B B CC
 
ijk ijk ijk

(12 12 ) (8 8 ) (10 10 ) 0
zy zx yx
AA BB CC   jk ik i j
From i-coefficient: 810 0
zy
BC
or 1.25
zy
BC (1)
j-coefficient:
12 10 0
zx
AC 
or
1.2
xz
CA (2)
k-coefficient: 12 8 0
yx
AB
or 1.5
xy
BA (3)
 0: 0    FABCP 
or ()( 240lb)()0
xx yy zz
BC AC AB  ijk 
From i-coefficient: 0
xx
BC
or
xx
CB (4)
j-coefficient: 240 lb 0
yy
AC 
or 240 lb
yy
AC (5)
www.elsolucionario.org

544. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.127 (Continued)


k-coefficient: 0
zz
AB

or
zz
A B (6)
Substituting
x
C from Equation (4) into Equation (2),

1.2
zz
BA (7)
Using Equations (1), (6), and (7),

1
1.25 1.25 1.25 1.2 1.5
x xzz
y
BBBA
C
 
 


(8)
From Equations (3) and (8):

1.5
or
1.5
y
yyy
A
CCA
and substituting into Equation (5),
2 240 lb
y
A
120 lb
yy
AC (9)
Using Equation (1) and Equation (9),

1.25(120 lb) 150.0 lb
z
B
Using Equation (3) and Equation (9),

1.5(120 lb) 180.0 lb
x
B
From Equation (4):
180.0 lb
x
C
From Equation (6):
150.0 lb
z
A
Therefore, (120.0 lb) (150.0 lb)
 Ajk 

(180.0 lb) (150.0 lb)
 Bik 

(180.0 lb) (120.0 lb)
Cij 
www.elsolucionario.org

545. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.128
Solve Problem 4.127, assuming that the force P is removed and is
replaced by a couple (600 lb in.) Mj acting at B.
PROBLEM 4.127 Three rods are welded together to form a
“corner” that is supported by three eyebolts. Neglecting friction,
determine the reactions at A, B, and C when
240Plb,
12 in., 8 in.,ab and c10 in.

SOLUTION
From F.B.D. of weldment:

///
0: 0
OAOBOCO
 MrArBrCM

12 0 0 0 8 0 0 0 10 (600 lb in.) 0
000
yz x z x y
AA B B CC
 
ijk ijk ijk
j

(12 12 ) (8 8 ) (10 10 ) (600lbin.) 0
zy zx yx
AA BB CC    jk jk i j j
From i-coefficient: 810 0
zy
BC
or 0.8
yz
CB (1)
j-coefficient:
12 10 600 0
zx
AC 
or
1.2 60
xz
CA (2)
k-coefficient: 12 8 0
yx
AB
or 1.5
xy
BA (3)
 0: 0   FABC 

()()()0
xx yy zz
BC AC AB ijk 
From i-coefficient:
xx
CB (4)
j-coefficient:
yy
CA (5)
k-coefficient:
zz
AB (6)


www.elsolucionario.org

546. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.128 (Continued)

Substituting
x
C from Equation (4) into Equation (2),
50
1.2
x
z
B
A




(7)
Using Equations (1), (6), and (7),

2
0.8 0.8 40
3
yz z x
CB A B

   

 (8)
From Equations (3) and (8):
40
yy
CA

Substituting into Equation (5), 2 40
y
A

20.0 lb
y
A
From Equation (5): 20.0 lb
y
C
Equation (1):
25.0 lb
z
B
Equation (3):
30.0 lb
x
B
Equation (4):
30.0 lb
x
C
Equation (6):
25.0 lb
z
A
Therefore, (20.0 lb) (25.0 lb)
 Ajk 

(30.0 lb) (25.0 lb)
 Bik 

(30.0 lb) (20.0 lb)
Cij 
www.elsolucionario.org

547. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.129
Frame ABCD is supported by a ball-and-
socket joint at A and by three cables. For
150 mm,a determine the tension in
each cable and the reaction at A.

SOLUTION
First note:
22
(0.48 m) (0.14 m)
(0.48) (0.14) m
DG DG DG DG
TT



ij
T


0.48 0.14
0.50
(24 7 )
25
DG
DG
T
T



ij
ij


22
(0.48 m) (0.2 m)
(0.48) (0.2) m
BE BE BE BE
TT



ik
T


0.48 0.2
0.52
(12 5)
13
BE
BE
T
T



ik
jk

From F.B.D. of frame ABCD:

7
0: (0.3 m) (350 N)(0.15 m) 0
25
xDG
MT

  


or
625 N
DG
T 

24 5
0: 625 N (0.3 m) (0.48 m) 0
25 13
yBE
MT

    
 
or
975 N
BE
T 

7
0: (0.14 m) 625 N (0.48 m) (350 N)(0.48 m) 0
25

    


zCF
MT
or
600 N
CF
T 
www.elsolucionario.org

548. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.129 (Continued)


0: ( ) ( ) 0
xxCFBExDGx
FATTT    

12 24
600 N 975 N 625 N 0
13 25
x
A

  



2100 N
x
A
0: ( ) 350 N 0
yyDGy
FAT   

7
625 N 350 N 0
25
y
A

  


175.0 N
y
A

0: ( ) 0
zzBEz
FAT  

5
975 N 0
13
z
A

 



375 N
z
A
Therefore, (2100 N) (175.0 N) (375 N)
 Ai jk 
www.elsolucionario.org

549. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.130
Frame ABCD is supported by a ball-and-
socket joint at A and by three cables.
Knowing that the 350-N load is applied at
D (300 mm),a determine the tension in
each cable and the reaction at A.

SOLUTION
First note
22
(0.48 m) (0.14 m)
(0.48) (0.14) m
DG DG DG DG
TT



ij
T


0.48 0.14
0.50
(24 7 )
25
DG
DG
T
T



ij
ij


22
(0.48 m) (0.2 m)
(0.48) (0.2) m
BE BE BE BE
TT



ik
T


0.48 0.2
0.52
(12 5)
13
BE
BE
T
T



ik
ik

From f.b.d. of frame ABCD

7
0: (0.3 m) (350 N)(0.3 m) 0
25
xDG
MT

  


or
1250 N
DG
T 

24 5
0: 1250 N (0.3 m) (0.48 m) 0
25 13
yBE
MT

    
 
or
1950 N
BE
T 

7
0: (0.14 m) 1250 N (0.48 m) (350 N)(0.48 m) 0
25

    


zCF
MT
or
0
CF
T 
www.elsolucionario.org

550. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.130 (Continued)


0: ( ) ( ) 0
xxCFBExDGx
FATTT    

12 24
0 1950 N 1250 N 0
13 25
x
A

  

 

3000 N
x
A 
0: ( ) 350 N 0
yyDGy
FAT   

7
1250 N 350 N 0
25
y
A

 


0
y
A


0: ( ) 0
zzBEz
FAT  

5
1950 N 0
13
z
A





750 N
z
A
Therefore (3000 N) (750 N)
 Aik 
www.elsolucionario.org

551. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.131
The assembly shown consists of an 80-
mm rod AF that is welded to a cross
consisting of four 200-mm arms. The
assembly is supported by a ball-and-
socket joint at F and by three short links,
each of which forms an angle of 45 with
the vertical. For the loading shown,
determine (a) the tension in each link,
(b) the reaction at F.

SOLUTION



/
/
/
/
200 80
()/2 80200
()/2 20080
()/2 80200
 
 
  
 
rij
Tij r jk
Tjkr ij
Tijrjk
EF
BB BF
CC CF
DD DE
T
T
T


// //
0: ( ) 0
FBFBCFCDFDEF
MTP    rTrTr r j

0 80 200 200 80 0 0 80 200 200 80 0 0
22 2
11 0 0 11 110 0 0
CBD
TTT
P
   

ij k i jk i j k i jk

Equate coefficients of unit vectors to zero and multiply each equation by
2.
: i
200 80 200 0
BC D
TT T  (1)
: j
200 200 200 0
BC D
TTT   (2)
: k
80 200 80 200 2 0
BCD
TTT P    (3)

80
(2):
200
80 80 80 0
BC D
TTT   (4)
Eqs. (3) (4):
160 280 200 2 0
BC
TT P   (5)
Eqs. (1) (2):
400 120 0
BC
TT 

120
0.3
400
BCC
TTT  (6)
www.elsolucionario.org

552. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.131 (Continued)

Eqs. (6) (5):
160( 0.3 ) 280 200 2 0
CC
TT P   
232 200 2 0
C
TP 

1.2191
C
TP 1.219
C
TP 
From Eq. (6):
0.3(1.2191 ) 0.36574
B
TPP   0.366
B
TP 
From Eq. (2):
200( 0.3657 ) 200(1.2191 ) 200 0
D
PPT

   

0.8534
D
TP 0.853
D
TP 
0: F
0
BC D
PFT T T j

( 0.36574 ) ( 0.8534 )
:0
22
x
PP
F

i

0.3448 0.345
xx
FPFP 

( 0.36574 ) (1.2191 ) ( 0.8534 )
: 200 0
222
y
PP P
F

j

yy
FP FP

(1.2191 )
:0
2
z
P
Fk

0.8620 0.862
zz
FPFP  0.345 0.862 PP P  Fijk 
www.elsolucionario.org

553. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.132
The uniform 10-kg rod AB is supported by a ball-and-socket
joint at A and by the cord CG that is attached to the midpoint G
of the rod. Knowing that the rod leans against a frictionless
vertical wall at B, determine (a) the tension in the cord, (b) the
reactions at A and B.

SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
(0).
AB
M
(a)

2
(10 kg)9.81m/s
98.1 N



Wmg
W


/
/
300 200 225 425 mm
( 300 200 225 )
425
600 400 150 mm
300 200 75 mm
BA
GA
GC GC
GC T
T
GC
   

  
  
ijk
Tijk
rij
rij




///
0: ( ) 0
ABAGAGA
MW    rBrTr j

600 400 150 300 200 75 300 200 75
425
0 0 300 200 225 0 98.1 0
T
B
 
 
ijk ijk i jk
Coefficient of : ( 105.88 35.29) 7357.5 0T  i
52.12 N T 52.1 N T



www.elsolucionario.org

554. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.132 (Continued)
(b)
Coefficient of
52.12
: 150 (300 75 300 225) 0
425
B 
j
73.58 N B
 (73.6 N)  Bi 
0: 0 W    
FABTj
Coefficient of :
i
300
73.58 52.15 0
425
x
A  36.8 N
x
A 
Coefficient of :
j
200
52.15 98.1 0
425
y
A 73.6 N
y
A 
Coefficient of :
k
225
52.15 0
425
z
A 27.6 N
z
A 


www.elsolucionario.org

555. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.133
The frame ACD is supported by ball-and-socket joints at A
and D and by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the frame
supports at Point C a load of magnitude P  268 N,
determine the tension in the cable.

SOLUTION
Free-Body Diagram:



(1 m) (0.75 m)
1.25 m
0.8 0.6
0.5 0.925 0.4
1.125
0.375 0.75 0.75
1.125
AD
AD
BG BG
BG
BH BH
BH
AD
AD
BG
T
BG
T
BH
TT
BH
T




 




ik
ik
T
ijk
ijk






www.elsolucionario.org

556. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.133 (Continued)


//
(0.5 m) ; (1 m) ; (268 N)
BA CA
ririPj
To eliminate the reactions at A and D, we shall write

0:
AD
M
// /
()()()0
AD B A BG AD B A BH AD C A
  rT rT rP (1)
Substituting for terms in Eq. (1) and using determinants,

0.8 0 0.6 0.8 0 0.6 0.8 0 0.6
0.5 0 0 0.5 0 0 1 0 0 0
1.125 1.125
0.5 0.925 0.4 0.375 0.75 0.75 0 268 0
BG BH
T T

 
  
Multiplying all terms by (–1.125),

0.27750 0.22500 180.900
BG BH
TT (2)
For this problem,
BG BH
TTT
 
(0.27750 0.22500) 180.900 T  360 N T 
www.elsolucionario.org

557. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.134
Solve Prob. 4.133, assuming that cable GBH is replaced by
a cable GB attached at G and B.
PROBLEM 4.133 The frame ACD is supported by ball-
and-socket joints at A and D and by a cable that passes
through a ring at B and is attached to hooks at G and H.
Knowing that the frame supports at Point C a load of
magnitude P  268 N, determine the tension in the cable.

SOLUTION
Free-Body Diagram:


(1 m) (0.75 m)
1.25 m
0.8 0.6
0.5 0.925 0.4
1.125
0.375 0.75 0.75
1.125
AD
AD
BG BG
BG
BH BH
BH
AD
AD
BG
TT
BG
T
BH
TT
BH
T




 



 ik
ik
ijk
ijk







www.elsolucionario.org

558. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.134 (Continued)


//
(0.5 m) ; (1 m) ; (268 N)
BA CA
ririPj
To eliminate the reactions at A and D, we shall write

// /
0: ( ) ( ) ( ) 0
AD AD B A BG AD B A BH AD C A
 MrTrTrP (1)
Substituting for terms in Eq. (1) and using determinants,

0.8 0 0.6 0.8 0 0.6 0.8 0 0.6
0.5 0 0 0.5 0 0 1 0 0 0
1.125 1.125
0.5 0.925 0.4 0.375 0.75 0.75 0 268 0
BG BH
T T

 
  
Multiplying all terms by (–1.125),

0.27750 0.22500 180.900
BG BH
TT (2)
For this problem,
0.
BH
T
Thus, Eq. (2) reduces to

0.27750 180.900
BG
T 652 N
BG
T 

www.elsolucionario.org

559. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.135
The bent rod ABDE is supported by ball-and-socket joints
at A and E and by the cable DF. If a 60-lb load is applied
at C as shown, determine the tension in the cable.

SOLUTION
Free-Body Diagram:

/
/
16 11 8 21in.
(16 11 8)
21
16
16 14
724
25
DE
CE
EA
DF DF
DE T
T
DF
EA
EA
   





ijk
Tijk
ri
rik
ik
λ





//
0: ( ) ( ( 60 )) 0
EA EA B E EA C E
M  λrT λrj

7024 70 24
1
16 0 0 16 0 14 0
21 25 25
16 11 8 0 60 0
24 16 11 7 14 60 24 16 60
0
21 25 25
201.14 17,160 0
T
T
T



 
   
 



85.314 lb T 85.3 lb T

www.elsolucionario.org

560. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.136
Solve Problem 4.135, assuming that cable DF is replaced
by a cable connecting B and F.
PROBLEM 4.135 The bent rod ABDE is supported by
ball-and-socket joints at A and E and by the cable DF. If a
60-lb load is applied at C as shown, determine the tension
in the cable.

SOLUTION
Free-Body Diagram:

/
/
9
910
BA
CA


ri
rik


16 11 16 25.16 in.
(16 11 16)
25.16
724
AE
BF BF
BF T
T
BF
AE
AE
   
 

 ijk
Tijk
ik
λ
25





//
0: ( ) ( ( 60 )) 0
AE AF B A AE C A
M  λrT λrj

7024 70 24
1
900 9010 0
25 25.16 25
16 11 16 0 60 0
T






24 9 11 24 9 60 7 10 60
0
25 25.16 25
T
   
 


94.436 17,160 0 T 181.7 lb T

www.elsolucionario.org

561. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.137
Two rectangular plates are welded together to form the
assembly shown. The assembly is supported by ball-and-
socket joints at B and D and by a ball on a horizontal
surface at C. For the loading shown, determine the reaction
at C.

SOLUTION
First note:
22 2
/
/
(6 in.) (9 in.) (12 in.)
(6) (9) (12) in.
1
(6 9 12)
16.1555
(6 in.)
(80 lb)
(8 in.)
()
 






ij k
ij k
ri
Pk
ri
Cj
BD
AB
CD
C

From the F.B.D. of the plates:

//
0: ( P C 0
BD BD A B BD C D
M   rr

6912 6912
6(80) (8)
10 0 1 0 0 0
16.1555 16.1555
001 010
( 9)(6)(80) (12)(8) 0
C
C
 
 
 
 
 


45.0 lb C or (45.0 lb)
Cj 
www.elsolucionario.org

562. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.138
The pipe ACDE is supported by ball-and-socket joints at A and E
and by the wire DF. Determine the tension in the wire when a
640-N load is applied at B as shown.

SOLUTION
Free-Body Diagram:
Dimensions in mm

/
/
480 160 240
560 mm
480 160 240
560
623
7
200
480 160
AE
AE
BA
DA
AE
AE
AE
AE








ijk
ijk
λ
ijk
λ
ri
rij




480 330 240 ; 630 mm
480 330 240 16 11 8
630 21
DF DF DF DF
DF DF
DF
TT T
DF
   
  
  ijk
ijk ijk
T




//
()((600))0
AE AE D A DF AE B A
M  λrT λrj

623 6 2 3
1
480 160 0 200 0 0 0
21 7 7
16 11 8 0 640 0
DF
T



 


6 160 8 2 480 8 3 480 11 3 160 16 3 200 640
0
21 7 7
DF
T
 



3
1120 384 10 0
DF
T

342.86 N
DF
T 343 N
DF
T 
www.elsolucionario.org

563. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.139
Solve Problem 4.138, assuming that wire DF is replaced by a
wire connecting C and F.
PROBLEM 4.138 The pipe ACDE is supported by ball-and-
socket joints at A and E and by the wire DF. Determine the
tension in the wire when a 640-N load is applied at B as shown.

SOLUTION
Free-Body Diagram:
Dimensions in mm

/
/
480 160 240
560 mm
480 160 240
560
623
7
200
480
480 490 240 ; 726.70 mm
480 490 240
726.70
AE
AE
BA
CA
CF CF
AE
AE
AE
AE
CF CF
CE
T
CF








   
 

ijk
ijk
λ
ijk
λ
ri
ri
ijk
ijk
T






//
0: ( ) ( ( 600 )) 0
AE AE C A CF AE B A
M   λrT λrj

62 3 623
1
480 0 0 200 0 0 0
726.7 7 7
480 490 240 0 640 0
CF
T



 


2 480 240 3 480 490 3 200 640
0
726.7 7 7
CF
T
 



3
653.91 384 10 0
CF
T 587 N
CF
T 

www.elsolucionario.org

564. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.140
Two 24-ft plywood panels, each of weight 12 lb, are nailed
together as shown. The panels are supported by ball-and-
socket joints at A and F and by the wire BH. Determine
(a) the location of H in the xy plane if the tension in the wire
is to be minimum, (b) the corresponding minimum tension.

SOLUTION
Free-Body Diagram:


1
2
/
/
/
424 6ft
1
(2 2 )
3
2
42
4
 



ijk
λ ij k
rij
rijk
ri

AF
GA
GA
BA
AF AF



12
///
0: ( ( 12 ) ( ( 12 )) ( ) 0
AF AF G A AF G A AF B A
MT   λrj λrj λr

/
212 212
11
210 412 ( )0
33
0120 0120
 


λ rT
AF B A


/
11
(2212) (2212 2412) ( ) 0
33
AF B A
       λ rT

///
( ) 32 or ( ) 32     λ rT T λ r
AF BA AF BA
(1)


www.elsolucionario.org

565. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.140 (Continued)

Projection of
T on
/
()
AF B A
λr is constant. Thus,
min
T is parallel to

/
11
(2 2 ) 4 ( 8 4 )
33
AF B A
 λrijkijk
Corresponding unit vector is
1
5
(2 ).jk

min
1
(2 )
5
TT
jk (2)
From Eq. (1):
1
(2 ) (2 2) 4 32
35
1
(2 ) (8 4) 32
35
T
T

    


   
jk ij k i
jk j k


35(32)
(16 4) 32 4.8 5
2035
T
T  


10.7331 lbT
From Eq. (2):
min
min
1
(2 )
5
1
4.8 5( 2 )
5
(9.6 lb) (4.8 lb )
TT

 
jk
jk
Tjk

Since
min
Thas no i component, wire BH is parallel to the yz plane, and 4 ft.x





( a) 4.00 ft; 8.00 ftxy

( b)
min
10.73 lbT 
www.elsolucionario.org

566. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.141
Solve Problem 4.140, subject to the restriction that H must lie
on the y-axis.
PROBLEM 4.140 Two
24-ft plywood panels, each of weight
12 lb, are nailed together as shown. The panels are supported
by ball-and-socket joints at A and F and by the wire BH.
Determine (a) the location of H in the xy plane if the tension
in the wire is to be minimum, (b) the corresponding minimum
tension.

SOLUTION
Free-Body Diagram:

1
2
/
/
/
424
1
(2 2 )
3
2
42
4
AF
GA
GA
BA
AF



 ijk
λ ij k
rij
rijk
ri


2
///
0:((12)((12))()0
AF AF G A AF G A AF B A
MT   λrj λrj λr

/
212 212
11
210 412 ( )0
33
0120 0120



λ rT
AF B A


/
11
(2212) (2212 2412) ( ) 0
33
AF B A
       λ rT

/
()32
AF B A
λ rT (1)

21/2
44 (32)BH y BH y    ijk



21/2
44
(32 )
BH y
TT
BH y
 


ijk
T



www.elsolucionario.org

567. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.141 (Continued)

From Eq. (1):

/ 21/2
212
()400 32
3(32 )
44
AF B A
T
T
y
y

 


λr

21/2
21/2
(32 )
( 16 8 ) 3 32(32 ) 96
816

   

y
yT y T
y
(2)

2 1/2 2 1/21
2
2
(8 +16) (32 ) (2 ) (32 ) (8)
0: 96
(8 16)
yyyydT
dy y





Numerator  0:
2
(8 16) (32 )8yy y

22
8163288yy y 0 ft; 16.00 ft xy 
From Eq. (2):
21/2
(32 16 )
96 11.3137 lb
81616
T




min
11.31 lbT 

www.elsolucionario.org

568. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.142
A 3200-lb forklift truck is used to lift a 1700-lb crate.
Determine the reaction at each of the two (a) front wheels A,
(b) rear wheels B.

SOLUTION
Free-Body Diagram:


(a) Front wheels:
0: (1700 lb)(52 in.) (3200 lb)(12 in.) 2 (36 in.) 0
B
MA   
1761.11lb A 1761lb  A 
(b) Rear wheels: 0: 1700 lb 3200 lb 2(1761.11 lb) 2 0
y
FB      688.89 lb B 689 lb  B 

www.elsolucionario.org

569. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.143
The lever BCD is hinged at C and attached to a control rod at B. If
P  100 lb, determine (a) the tension in rod AB, (b) the reaction at C.

SOLUTION
Free-Body Diagram:

(a)
0: (5 in.) (100 lb)(7.5 in.) 0
C
MT  

150.0 lbT 
(b)
3
0: 100 lb (150.0 lb) 0
5
xx
FC   

190 lb
x
C 190 lb
x
C

4
0: lb) 0
5
yy
FC    
120 lb
y
C 120 lb
y
C

32.3 225 lbC 225 lb C 32.3° 
www.elsolucionario.org

570. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.144
Determine the reactions at A and B when (a) 0,h
(b)
200 mm.h

SOLUTION
Free-Body Diagram:


0: ( cos60 )(0.5 m) ( sin 60 ) (150 N)(0.25 m) 0
A
MB Bh    

37.5
0.25 0.866
B
h


(1)
(a) When
0,h
From Eq. (1):
37.5
150 N
0.25
B
150.0 N B 30.0° 
0: sin60 0
yx
FAB  

(150)sin 60 129.9 N
x
A 129.9 N
x
A
0: 150 cos60 0
yy
FA B   
150 (150)cos60 75 N
y
A  75 N
y
A

30
150.0 NA

 150.0 N A 30.0°

(b) When h  200 mm  0.2 m,
From Eq. (1):
37.5
488.3 N
0.25 0.866(0.2)
B

488 N B 30.0°


0: sin60 0
xx
FAB  

(488.3)sin 60 422.88 N
x
A 422.88 N
x
A
0: 150 cos60 0
yy
FA B   
150 (488.3)cos60 94.15 N
y
A  94.15 N
y
A

12.55
433.2 NA

 433 N A 12.55°

www.elsolucionario.org

571. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.145
Neglecting friction and the radius of the pulley,
determine (a) the tension in cable ADB, (b) the reaction
at C.

SOLUTION
Free-Body Diagram:
Dimensions in mm

Geometry:
Distance:
22
(0.36) (0.150) 0.39 mAD
Distance:
22
(0.2) (0.15) 0.25 mBD


Equilibrium for beam:
(a)
0.15 0.15
0: (120 N)(0.28 m) (0.36 m) (0.2 m) 0
0.39 0.25
C
MTT
 
   
 
 
130.000 N T
or 130.0 N T

(b)
0.36 0.2
0: (130.000 N) (130.000 N) 0
0.39 0.25
xx
FC
 
   
 
 

224.00 N
x
C

0.15 0.15
0: (130.00 N) (130.00 N) 120 N 0
0.39 0.25
yy
FC
 
    
 
 
8.0000 N
y
C
Thus,
22 2 2
( 224) ( 8) 224.14 N
xy
CCC 
and
11 8
tan tan 2.0454
224y
x
C
C


  224 NC 2.05 

www.elsolucionario.org

572. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.146
Bar AD is attached at A and C to collars that can move
freely on the rods shown. If the cord BE is vertical (
  0),
determine the tension in the cord and the reactions at A and
C.

SOLUTION
Free-Body Diagram:

0: cos30 (80 N)cos30 0
y
FT   
80 N T 80.0 N T


0: ( sin 30 )(0.4 m) (80 N)(0.2 m) (80 N)(0.2 m) 0
C
MA    
160 N A 160.0 NA 30.0°


0: (80 N)(0.2 m) (80 N)(0.6 m) ( sin 30 )(0.4 m) 0
A
MC    


160 NC 160.0 NC 30.0°



www.elsolucionario.org

573. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.147
A slender rod AB, of weight W, is attached to blocks A and B
that move freely in the guides shown. The constant of the spring
is k, and the spring is unstretched when
0. (a) Neglecting the
weight of the blocks, derive an equation in W, k, l, and
 that
must be satisfied when the rod is in equilibrium. (b) Determine
the value of
 when 75 lb,W 30 in.,l and 3 lb/in.k

SOLUTION
Free-Body Diagram:

Spring force:
(cos)(1cos)
s
Fkskll kl    
(a)
0: ( sin ) cos 0
2
D s
l
MFlW


  




(1 cos ) sin cos 0
2
W
kl l l 
(1 cos ) tan 0
2
W
kl or (1 cos ) tan
2
W
kl 
(b) For given values of
75 lb
30 in.
3 lb/in.
(1 cos ) tan tan sin
75 lb
2(3 lb/in.)(30 in.)
0.41667
W
l
k
  







Solving numerically,
49.710 or 49.7 
www.elsolucionario.org

574. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.148
Determine the reactions at A and B when a  150 mm.

SOLUTION
Free-Body Diagram:


Force triangle
80 mm 80 mm
tan
150 mm
28.072
a





320 N
sin 28.072
A

680 NA 28.1 


320 N
tan 28.072
B

600 NB 

www.elsolucionario.org

575. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.149
For the frame and loading shown, determine the reactions at A and C.

SOLUTION

Since member AB is acted upon by two forces,
A and B, they must be colinear, have the same magnitude,
and be opposite in direction for AB to be in equilibrium. The force
B acting at B of member BCD will be
equal in magnitude but opposite in direction to force
B acting on member AB. Member BCD is a three-
force body with member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the
above conditions. The force triangle for member BCD is also shown. The angle
 is found from the
member dimensions:

16in.
tan 30.964
10 in.






Applying the law of sines to the force triangle for member BCD,

30 lb
sin(45 ) sin sin135
BC


 

or
30 lb
sin14.036 sin 30.964 sin135
BC




(30 lb)sin 30.964
63.641 lb
sin14.036
AB

 


or
63.6 lbA 45.0 
and
(30 lb)sin135
87.466 lb
sin14.036
C




or
87.5 lbC 59.0 
www.elsolucionario.org

576. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.150
A 200-mm lever and a 240-mm-diameter pulley
are welded to the axle BE that is supported by
bearings at C and D. If a 720-N vertical load is
applied at A when the lever is horizontal,
determine (a) the tension in the cord, (b) the
reactions at C and D. Assume that the bearing at
D does not exert any axial thrust.

SOLUTION
Dimensions in mm

We have six unknowns and six equations of equilibrium. —OK
0: ( 120 ) ( ) (120 160 ) (80 200 ) ( 720 ) 0
Cxy
DD T         Mkijjkikij

33
120 120 120 160 57.6 10 144 10 0
xy
DDTTjikj i k
Equating to zero the coefficients of the unit vectors:
:
k
3
120 144 10 0T ( a) 1200 NT 
:
i
3
120 57.6 10 0 480 N
yy
DD 

: 120 160(1200 N) 0
x
D j 1600 N
x
D

0:
x
F 0
xx
CDT 1600 1200 400 N
x
C
0:
y
F 720 0
yy
CD  480 720 1200 N
y
C

0:
z
F 0
z
C
(b) (400 N) (1200 N) ; (1600 N) (480 N)  
Ci jD ij 
www.elsolucionario.org

577. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.151
The 45-lb square plate shown is supported by three vertical
wires. Determine the tension in each wire.

SOLUTION
Free-Body Diagram:


// /
0: ( 45 lb) 0
BCBCABAGB
MTT    rjrjr j

[ (20 in.) (15 in.) ] (20 in.)
CA
TT  ikjkj


[ (10 in.) (10 in.) ] [ (45 lb) ] 0  ik j

20 15 20 450 450 0
CCA
TTT
 kiiki
Equating to zero the coefficients of the unit vectors,
:
k 20 450 0
C
T
  ; 22.5 lb
C
T
:
i 15(22.5) 20 450 0
A
T ; 5.625 lb
A
T
0:
y
F 45 lb 0
ABC
TTT
 

5.625 lb 22.5 lb 45 lb 0
B
T
 ; 16.875 lb
B
T 
5.63 lb; 16.88 lb; 22.5 lb
AB C
TT T  

www.elsolucionario.org

578. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.152
The rectangular plate shown weighs 75 lb and is held in the
position shown by hinges at A and B and by cable EF.
Assuming that the hinge at B does not exert any axial
thrust, determine (a) the tension in the cable, (b) the
reactions at A and B.

SOLUTION

/
/
/
// /
(38 8) 30
(30 4) 20
26 20
38
10
2
19 10
82520
33 in.
(8 25 20 )
33
0: ( 75 ) 0





 


 


BA
EA
GA
AEAGA BA
EF
EF
AE T
T
AE
TB
rii
rik
ik
rik
ik
ijk
Tijk
Mr r jr


26 0 20 19 0 10 30 0 0 0
33
82520 0 750 0
yz
T
BB


ijk i jk ijk

Coefficient of i:
(25)(20) 750 0:
33
T
 49.5 lbT 
Coefficient of j:
49.5
(160 520) 30 0: 34 lb
33
zz
BB
Coefficient of k:
49.5
(26)(25) 1425 30 0: 15 lb
33
yy
BB   (15.00 lb) (34.0 lb)Bjk 


www.elsolucionario.org

579. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.152 (Continued)

 0: (75 lb) 0    FABT j 
Coefficient of i:
8
(49.5) 0 12.00 lb
33
xx
AA
Coefficient of
j:
25
15 (49.5) 75 0 22.5 lb
33
yy
AA  
Coefficient of
k:
20
34 (49.5) 0 4.00 lb
33
zz
AA   

(12.00 lb) (22.5 lb) (4.00 lb)
Aijk 


www.elsolucionario.org

580. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.153
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each
case, if possible, determine the reactions at the supports.


SOLUTION
(a) 0: ( sin 45 )2 (cos45 ) 0
Aa
MPCa a   
3
2
C
P
2
3
CP
0.471PC 45° 

21
0:
3 2
xx
FA P

  



3
x
P
A


21 2
0:
33 2
yy y
P
FAPPA

   




0.745PA 63.4° 
(b)
0: ( cos 30 )2 ( sin 30 ) 0   
C
MPaAaAa
(1.732 0.5) 0.812 APAP 

0.812PA 60.0° 

0: (0.812 )sin 30 0
xx
FPC  

0.406
x
CP
0: (0.812 )cos30 0
yy
FPPC  

0.297
y
CP

0.503PC 36.2°  www.elsolucionario.org

581. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 4.153 (Continued)

(c)
0: ( cos30 )2 ( sin 30 ) 0
C
MPaAaAa   
(1.732 0.5) 0.448 APAP 

0.448PA 60.0° 

0: (0.448 )sin 30 0 0.224
xxx
FPCCP    
0: (0.448 )cos30 0 0.612
yyy
FPPCCP   

0.652PC 69.9° 

(d) Force T exerted by wire and reactions A and C all intersect at Point D.


0: 0
Da
MP 
Equilibrium is not maintained.
Rod is improperly constrained.

www.elsolucionario.org

582. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.F1
Two crates, each of mass 350 kg, are placed as shown in
the bed of a 1400-kg pick-up truck. Draw the free-body
diagram needed to determine the reactions at each of the
two rear wheels A and front wheels B.

SOLUTION
Free-Body Diagram of Truck:

Weights:
2
(350 kg)(9.81 m/s ) 3.4335 NW
2
(1400 kg)(9.81 m/s ) 13.734 N
T
W






www.elsolucionario.org

583. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 4.F2
A lever AB is hinged at C and attached to a control cable at A.
If the lever is subjected to a 75-lb vertical force at B, draw the
free-body diagram needed to determine the tension in the cable
and the reaction at C.

SOLUTION
Free-Body Diagram of Lever AB:


Geometry:

11
(10 in.)cos20 9.3969 in.
(10 in.)sin20 3.4202 in.
12 in. 3.4202 in. 8.5798 in.
8.5798 in.
=tan tan 42.398
9.3969 in.
90 20 42.398 27.602
AC
AC
AD
AD
AC
x
y
y
y
x





 
 
 

 



  


www.elsolucionario.org

584. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.F3
A light rod AD is supported by frictionless pegs at B and C and
rests against a frictionless wall at A. A vertical 120-lb force is
applied at D. Draw the free-body diagram needed to determine
the reactions at A, B, and C.

SOLUTION
Free-Body Diagram of Rod AD:


www.elsolucionario.org

585. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.F4
A tension of 20 N is maintained in a tape as it passes
through the support system shown. Knowing that the
radius of each pulley is 10 mm, draw the free-body
diagram needed to determine the reaction at C.

SOLUTION
Free-Body Diagram of Support System:

www.elsolucionario.org

586. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.F5
Two tape spools are attached to an axle supported by
bearings at A and D. The radius of spool B is 1.5 in. and
the radius of spool C is 2 in. Knowing that T
B = 20 lb and
that the system rotates at a constant rate, draw the free-
body diagram needed to determine the reactions at A and
D. Assume that the bearing at A does not exert any axial
thrust and neglect the weights of the spools and axle.

SOLUTION
Free-Body Diagram of Axle-Spool System:

www.elsolucionario.org

587. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.F6
A 12-m pole supports a horizontal cable CD and is held by a
ball and socket at A and two cables BE and BF. Knowing that
the tension in cable CD is 14 kN and assuming that CD is
parallel to the x axis (
 = 0), draw the free-body diagram
needed to determine the tension in cables BE and BF and the
reaction at A.

SOLUTION
Free-Body Diagram of Pole:

www.elsolucionario.org

588. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 4.F7
A 20-kg cover for a roof opening is hinged at corners A
and B. The roof forms an angle of 30 with the
horizontal, and the cover is maintained in a horizontal
position by the brace CE. Draw the free-body diagram
needed to determine the magnitude of the force exerted
by the brace and the reactions at the hinges. Assume that
the hinge at A does not exert any axial thrust.

SOLUTION
Geometry of Brace CE:

Free-Body Diagram of Brace CE:

Cover Weight:
2
(20 kg)(9.81 m/s ) 196.2 NW



www.elsolucionario.org

589. CHAPTER 5
www.elsolucionario.org

590. www.elsolucionario.org

591. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.1
Locate the centroid of the plane area shown.

SOLUTION

Area 1: Rectangle 72 mm by 45 mm.
Area 2: Triangle b = 27 mm, h = 45 mm.


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 3240 36 22.5 116,640 72,900
2 -607.5 9 215 -5467.5 -9112.5
 2632.5 11,172.5 63,787.5

XAxA

23
(2632.5 mm ) 11,172.5 mmX  42.2 mmX 
YA yA

23
(2632.5 mm ) 63787.5 mmY  24.2 mmY 

www.elsolucionario.org

592. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.2
Locate the centroid of the plane area shown.

SOLUTION




2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1 8 0.5 4 4 32
2 3 2.5 2.5 7.5 7.5
 11 11.5 39.5

XAxA

23
(11in ) 11.5 inX  1.045 in.X 
YA yA

23
(11 in ) 39.5 inY  3.59 in.Y 

www.elsolucionario.org

593. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.3
Locate the centroid of the plane area shown.

SOLUTION
Area 1: Rectangle 42 mm by 44 mm.
Area 2: Triangle b = 18 mm, h = 12 mm.
Area 3: Triangle b = 24 mm, h = 12 mm.

2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 42 44 1848 3 22 5544 40,656
2
1
18 12 108
2
  
-12 4 1296 -432
3
1
24 12 144
2
  
16 4 -2304 -576
 1596 4536 39,648

Then XAxA

23
(1596 mm ) 4536 mmX  or 2.84 mmX 
and YA yA

23
(1596 mm ) 39,648 mmY  or 24.8 mmY 

www.elsolucionario.org

594. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.4
Locate the centroid of the plane area shown.

SOLUTION
Area 1: Triangle b = 60 mm, h = 75 mm.
Area 2: Four triangles b = 60 mm, h = 75 mm forming the diamond.

2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1
1
60 75 2250
2

20 25 45,000 56,250
2
1
460759000
2
  
60 75 540,000 675,000
 11,250 585,000 731,250

Then XAxA

23
(11,250 mm ) 585,000 mmX  or 52.0 mmX 
and YA yA

23
(11,250 m ) 731,250 mmY  or 65.0 mmY 

www.elsolucionario.org

595. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.5
Locate the centroid of the plane area shown.

SOLUTION

Area 1: Rectangle 6 in. by 4 in.
Area 2: Triangle b = 6 in., h = 3 in.


2
,inA
,inx ,iny
3
,inxA
3
,inyA
1 24 3 2 72 48
2 9 4 5 36 45
 33 108 93

XAxA

23
(33 in ) 108 inX  3.27 in.X 
YA yA

23
(33 in ) 93 inY  2.82 in.Y 

www.elsolucionario.org

596. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.6
Locate the centroid of the plane area shown.

SOLUTION


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 (60)(120) 7200 –30 60
3
216 10
3
432 10
2
2
(60) 2827.4
4

 25.465 95.435
3
72.000 10
3
269.83 10
3
2
(60) 2827.4
4

 –25.465 25.465
3
72.000 10
3
72.000 10
 7200
3
72.000 10
3
629.83 10

Then
233
(7200 mm ) 72.000 10 mmXA x A X   10.00 mmX 

233
(7200 mm ) 629.83 10 mmYA y A Y   87.5 mmY 
www.elsolucionario.org

597. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.7
Locate the centroid of the plane area shown.

SOLUTION
Area 1: Rectangle 16 in. by 8 in.
Area 2: Semicircle radius of 5 in.



2
,inA
,inx ,iny
3
,inxA
3
,inyA
1 128 0 4 0 512
2 -39.27 0 5.878 0 -230.83
 88.73 0 281.17

XAxA
0in.X 
YA yA

23
(88.73 in ) 281.17 inY  3.17 in.Y 



www.elsolucionario.org

598. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.8
Locate the centroid of the plane area shown.

SOLUTION


2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1
2
(38) 2268.2
2

 0 16.1277 0 36,581
2 20 16 320  10 8 3200 2560
 1948.23 3200 34,021

Then
3
2
3200 in
1948.23 in
xA
X
A




1.643 in.X 

3
2
34,021 in
1948.23 in
yA
Y
A




17.46 in.Y 

www.elsolucionario.org

599. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.9
Locate the centroid of the plane area shown.

SOLUTION
Area 1: Square 75 mm by 75 mm.
Area 2: Quarter circle radius of 75 mm.

2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 75 75 5625 37.5 37.5 210,938 210,938
2
2
75 4417.9
4

 
43.169

43.169 -190,715

 1207.10 20,223 20,223

Then by symmetry
xA
XY
A





23
(1207.14 mm ) 20,223 mmX  16.75 mmXY 


www.elsolucionario.org

600. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.10
Locate the centroid of the plane area shown.

SOLUTION
Area 1: Parabolic Spandrel, refer to fig 5.8A for centroid location.
Area 2: Rectangle 16 in. by 3 in.

2
,in.A
,in.x ,in.y
3
,inxA
3
,inyA
1
1
(10)(16) 53.333
3


12 6 640 320
2 (3)(16) 48.0 8 1.5 384 72
 101.333 1024 392

Then XAxA

23
(101.333 in ) 1024 inX  10.11in.X 

23
(101.333 in ) 393 in
YA yA
Y

 3.88 in.Y 
www.elsolucionario.org

601. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.11
Locate the centroid of the plane area shown.

SOLUTION


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1
2
(75)(120) 6000
3

28.125 48 168,750 288,000
2
1
(75)(60) 2250
2

25 20 –56,250 –45,000
 3750 112,500 243,000

Then XA xA

23
(3750 mm ) 112,500 mmX  or 30.0 mmX 
and YA yA

23
(3750 mm ) 243,000 mmY  or 64.8 mmY 

www.elsolucionario.org

602. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.12
Locate the centroid of the plane area shown.

SOLUTION
Area 1: Outer semicircle diameter 120 mm.
Area 2: Inner semicircle diameter 72 mm.

2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1
2
(120) 22,620
2

 -50.93 0
3
1152.04 10 0
2
2
(72) 8143.0
2

 -30.558 0
3
248.80 10 0
 14,477
3
903.20 10 0

Then XAxA

233
(14,477 mm ) 903.20 10 mmX   62.4 mmX 
YA yA 0Y 
www.elsolucionario.org

603. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.13
Locate the centroid of the plane area shown.

SOLUTION

Dimensions in in.

2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1
2
(4)(8) 21.333
3

4.8 1.5 102.398 32.000
2
1
(4)(8) 16.0000
2

5.3333 1.33333 85.333 21.333
 5.3333 17.0650 10.6670

Then XA xA

23
(5.3333 in ) 17.0650 inX  or 3.20 in.X 
and YA yA

23
(5.3333 in ) 10.6670 inY  or 2.00 in.Y 

www.elsolucionario.org

604. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.14
Locate the centroid of the plane area shown.

SOLUTION

First note that symmetry implies 0X 




2
,mA
,my
3
,myA
1
4
4.5 3 18
3

1.2 21.6
2
2
(1.8) 5.0894
2

 0.76394 3.8880
 12.9106 17.7120

Then
3
2
17.7120 m
12.9106 m
yA
Y
A



or
1.372 mY 
www.elsolucionario.org

605. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.15
Locate the centroid of the plane area shown.

SOLUTION


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1
1
(120)(90) 5400
2

0 30 0 162,000
2
2
(60 2) 5654.9
4

 0 9.07 0 51,290
3
1
(120)(60) 3600
2

0 20 0 -72,000
 7454.9 0 141,290

Then XA xA 0mmX 

23
(7454.9 mm ) 141,290 mmYA yA Y  18.95 mmY 
www.elsolucionario.org

606. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.16
Determine the y coordinate of the centroid of the shaded area
in terms of r
1, r2, and .

SOLUTION

First, determine the location of the centroid.
From Figure 5.8A:



2 2
22 2 2
2
2
2
sin2
32
2cos
3
yr A r
r



 







 



Similarly,

2
11 1 1
22cos
32
yr A r



 


 

Then
 

22
2211
22
33
212 cos 2 cos
3232
2
cos
3
yA r r r r
rr

 


  
   
  
  


and

22
21
22
21
22
2
Ar r
rr




   



 



Now
YA yA
 
22 33
21 21 2
cos
23
Yrrrr





33
21
22
21
22cos
32
rr
Y
rr 

 


www.elsolucionario.org

607. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.17
Show that as r 1 approaches r 2, the location of the centroid
approaches that for an arc of circle of radius
12
()/2.rr

SOLUTION

First, determine the location of the centroid.
From Figure 5.8A:



2 2
22 2 2
2
2
2
sin2
32
2cos
3
yr A r
r



 







 



Similarly,

2
11 1 1
22cos
32
yr A r



 


 

Then
 

22
2211
22
33
212cos 2cos
3232
2
cos
3
yA r r r r
rr

 


  
   
  
  


and

22
21
22
21
22
2
Ar r
rr




   



 



Now
YA yA

 
22 33
21 21
33
21
22
21 2
cos
23
22cos
32
Yrrrr
rr
Y
rr







 



www.elsolucionario.org

608. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.17 (Continued)

Using Figure 5.8B,
Yof an arc of radius
12
1
()
2
rris

2
12
2
12
2
sin( )1
()
2()
1cos
()
2()
Yrr
rr











(1)
Now

22
33
212 121
21
22
212121
22
2121
21
()
()()
rrr rrr
rr
rrrrrr
rrrr
rr








Let
2
1
rr
rr


Then
12
1
()
2
rrr
and
33 22
21
22
21
22
()()()()
()()
3
2
rr rrrr
rrrr
r
r
 

  



In the limit as
 0 (i.e.,
12
),rr then

33
21
22
21
12
3
2
31
()
22
rr
r
rr
rr



 

So that
12
2
23 cos
()
34
Yrr



 
 or
12
cos
()
2
Yrr


 
which agrees with Equation (1).

www.elsolucionario.org

609. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.18
Determine the x coordinate of the centroid of the trapezoid shown in
terms of h
1, h2, and a.

SOLUTION

Area 1:
A x xA
1
1
1
2
ha

1
3
a

2
11
6
ha

2
2
1
2
ha

2
3
a

2
22
6
ha

 12
1
()
2
ah h

2
121
(2)
6
ah h



21
126
1
122
(2)
()
ah hxA
X
Aahh




12
12
2
3
hha
X
hh






www.elsolucionario.org

610. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.19
For the semiannular area of Prob. 5.12, determine the ratio r 1 to r2 so that the centroid
of the area is located at x =
2
1
2
r and y = 0.

SOLUTION

First, determine the location of the centroid.
2
2
4
3
x r


Similarly, 1
1
4
3
x r



22
21
2
A rr

 
Then

22 21
21
33
2144
23 23
2
3
rr
xA r r
rr
 
 
 
 
 
 
Now
XAxA

  
 
22 33
21 21 21
22
21 2 211
22
2211
21 2
; Dividing by
23
2
(1)
23
4
3
Xrr rr rr
Xrr rrrr
rrrr
X
rr


  



  



 


By symmetry,
0Y





www.elsolucionario.org

611. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.19 (Continued)

Now substituting for
2
1
2
Xr in equation (1) gives.


22
221 212112
(r)
22 3
rrr rrr



(1)
Dividing through by
2
and 2/3,r

2
111
222
3
11
24
rrr
rrr  
  
  

Then
2
11
22
0.1781 0.1781 0
rr
rr
 
  
 

Solving the quadratic yields,
1
2
0.520
r
r


 


www.elsolucionario.org

612. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.20
A composite beam is constructed by bolting
four plates to four 60  60  12-mm angles as
shown. The bolts are equally spaced along the
beam, and the beam supports a vertical load.
As proved in mechanics of materials, the
shearing forces exerted on the bolts at A and B
are proportional to the first moments with
respect to the centroidal x-axis of the red-
shaded areas shown, respectively, in parts a
and b of the figure. Knowing that the force
exerted on the bolt at A is 280 N, determine
the force exerted on the bolt at B.

SOLUTION

From the problem statement, F is proportional to
.
x
Q
Therefore,
()
,or
() () ()
xBAB
BA
xA xB xA
QFF
FF
QQ Q


For the first moments,
3
3
12
( ) 225 (300 12)
2
831,600 mm
12
( ) ( ) 2 225 (48 12) 2(225 30)(12 60)
2
1, 364, 688 mm
xA
xB x
Q
QQA

 




 




Then
1,364,688
(280 N)
831,600
B
F or 459 N
B
F 
www.elsolucionario.org

613. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.21
The horizontal x-axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A
1 and A 2. Determine the
first moment of each component area with respect to the x-axis, and
explain the results obtained.

SOLUTION







Length of BD:
0.84in.
0.48 in. (1.44 in. 0.48 in.) 0.48 0.56 1.04 in.
0.84 in. 0.60 in.
BD  


Area above x-axis (consider two triangular areas):

1
33
11
(0.28 in.) (0.84 in.)(1.04 in.) (0.56 in.) (0.84 in.)(0.48 in.)
22
0.122304 in 0.112896 in
A
Qy

  





3
1
0.235 inQ 
Area below x-axis:

2
33
11
(0.40 in.) (0.60 in.)(1.44 in.) (0.20 in.) (0.60 in.)
22
0.1728 in 0.0624 in
QyA

  


 


3
2
0.235 inQ 

2
|| | |,QQ since C is centroid and thus,
0QyA 

www.elsolucionario.org

614. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.22
The horizontal x-axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A
1 and A 2. Determine the
first moment of each component area with respect to the x-axis, and
explain the results obtained.

SOLUTION


Area above x-axis (Area A
1):
1
33
(25)(20 80) (7.5)(15 20)
40 10 2.25 10
QyA    
  


33
1
42.3 10 mmQ 
Area below x-axis (Area A
2):

2
( 32.5)(65 20)QyA   
33
2
42.3 10 mmQ  
12
||||,QQ since C is centroid and thus,
0QyA 

Dimensions in mm www.elsolucionario.org

615. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.23
The first moment of the shaded area with respect to the x-axis is denoted by
Q
x. (a) Express Q x in terms of b, c, and the distance y from the base of the
shaded area to the x-axis. (b) For what value of y is
x
Q maximum, and what is
that maximum value?

SOLUTION
Shaded area:
()
1
()[()]
2
x
Abcy
QyA
cybcy


 
(a)
221
()
2
x
Qbcy 
(b) For
max
,Q
1
0or (2)0
2
dQ
by
dy

0y 
For
0,y
21
()
2
x
Qbc
21
()
2
x
Qbc 
www.elsolucionario.org

616. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.24
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.

SOLUTION




,mmL
,mmx ,mmy
2
,mmxL
2
,mmyL
1 45 49.5 0 2227.5 0
2 45 72 22.5 3240 1012.5
3 72 36 45 2592 3240
4 52.479 13.5 22.5 708.47 1180.78
 2632.5 8768.0 5433.3

XLxL

2
(214.479 mm) 8768.0 mmX  40.9 mmX 
YL yL

2
(214.479 mm) 5433.3 mmY  25.3 mmY 

www.elsolucionario.org

617. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.25
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION

L, mm ,mmx ,mmy
2
,mmxL
2
,mmyL
1
22
24 12 26.833 12  322 
2 32 24 28 768 896
3 42 3 44 126 1848
4 32 18 28 576 896
5
22
18 12 21.633 9 6 194.697 129.798
 154.466   3930.8
Then XL xL
(154.466) 445.30X  or 2.88 mmX 


YL yL


(154.466) 3930.8Y  or 25.4 mmY 
www.elsolucionario.org

618. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.26
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.

SOLUTION


L, in. ,in.x ,in.y
2
,inxL
2
,inyL
1 4 0 2 0 8
2
22
3 6 6.7082 3 5.5 20.125 36.894
3 7 6 3.5 42 24.5
4 6 3 0 18 0
 23.708 80.125 69.394
Then (23.708) 80.125XL xL X  3.38 in.X 
(23.708) 69.394YL yL Y  2.93 in.Y 
www.elsolucionario.org

619. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.27
A thin, homogeneous wire is bent to form the perimeter of the
figure indicated. Locate the center of gravity of the wire figure
thus formed.

SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of
the corresponding line.


6
2
(38 in.)Y






,in.L
,in.x , in.y
2
,in.xL
2
,in.yL
1 18 –29 0 –522 0
2 16 –20 8 –320 128
3 20 –10 16 –200 320
4 16 0 8 0 128
5 38 19 0 722 0
6 (38) 119.381 0 24.192 0 2888.1
 227.38 –320 3464.1

Then
320
227.38
xL
X
L




1.407 in.X 

3464.1
227.38
yL
Y
L




15.23 in.Y 
www.elsolucionario.org

620. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.28
The homogeneous wire ABC is bent into a semicircular arc and a straight
section as shown and is attached to a hinge at A. Determine the value of
 for
which the wire is in equilibrium for the indicated position.

SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A.
Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line. Thus,
0X
so that 0xL
Then
12
cos ( ) cos ( ) 0
2
r
rr r r


 

 
 

or
4
cos
12
0.54921





or
56.7 


www.elsolucionario.org

621. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.29
The frame for a sign is fabricated from thin, flat steel bar stock of mass
per unit length 4.73 kg/m. The frame is supported by a pin at
C and by a
cable
AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION
First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide
with the centroid of the corresponding line.


L, m
,mx
2
,mxL
1 1.35 0.675 0.91125
2 0.6 0.3 0.18
3 0.75 0 0
4 0.75 0.2 0.15
5 (0.75) 1.17810
2

 1.07746 1.26936
 4.62810 2.5106

Then
(4.62810) 2.5106
XL xL
X



or 0.54247 mX
The free-body diagram of the frame is then
where
2
()
4.73 kg/m 4.62810 m 9.81 m/s
214.75 N
WmLg
 

www.elsolucionario.org

622. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.29 (Continued)

Equilibrium then requires
(
a)
3
0: (1.55 m) (0.54247 m)(214.75 N) 0
5
CBA
MT

  


or
125.264 N
BA
T or 125.3 N
BA
T 
(
b)
3
0: (125.264 N) 0
5
xx
FC  
or
75.158 N
x
C

4
0: (125.264 N) (214.75 N) 0
5
yy
FC   
or
114.539 N
y
C
Then
137.0 NC 56.7° 

www.elsolucionario.org

623. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.30
The homogeneous wire ABCD is bent as shown and is attached to
a hinge at
C. Determine the length L for which portion BCD of the
wire is horizontal.

SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C.
Further, because the wire is homogeneous, the center of gravity of the wire will coincide with the
centroid of the corresponding line. Thus,
0X so that 0XL



Then ( 40 mm)(80 mm) ( 40 mm)(100 mm) 0
2
  
L

22
14,400 mmL 120.0 mmL 
www.elsolucionario.org

624. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.31
The homogeneous wire ABCD is bent as shown and is attached to
a hinge at
C. Determine the length L for which portion AB of the
wire is horizontal.

SOLUTION


III III
80 100WwW wWLw 

0: (80 )(32) (100 )(14) ( )(0.4 ) 0
C
MwwLwL   


2
9900L 99.5 mmL 
www.elsolucionario.org

625. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.32
Determine the distance h for which the centroid of the shaded
area is as far above line
BB′ as possible when (a) k  0.10,
(
b) k  0.80.

SOLUTION


A
y yA
1
1
2
ba

1
3
a

21
6
ab

2
1
()
2
kb h

1
3
h

21
6
kbh

 ()
2
b
akh
22
()
6
b
akh

Then
22
()( )
26
YA yA
bb
Yakh akh


 



or
22
3( )
akh
Y
akh



(1)
and
22
2
12 ( )( )( )
0
3 ()dY kh a kh a kh k
dh akh



or
22
2( ) 0ha kh a kh  (2)
Simplifying Eq. (2) yields
22
20kh ah a
www.elsolucionario.org

626. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.32 (Continued)
Then
22
2(2)4()()
2
11
aaka
h
k
a
k
k
 




Note that only the negative root is acceptable since
.ha
 Then
(a)
0.10k

1 1 0.10
0.10
a
h 
 or 0.513ha 
(b)
0.80k

1 1 0.80
0.80
a
h 
 or 0.691ha 

www.elsolucionario.org

627. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.33
Knowing that the distance h has been selected to maximize
the distance
y from line BB′ to the centroid of the shaded
area, show that 2/3.yh

SOLUTION
See solution to Problem 5.32 for analysis leading to the following equations:

22
3( )
akh
Y
akh



(1)

22
2( ) 0ha kh a kh  (2)
Rearranging Eq. (2) (which defines the value of h which maximizes
)Yyields

22
2( )akh hakh 
Then substituting into Eq. (1) (which defines
),Y

1
2( )
3( )
Yhakh
akh


or
2
3
Yh


www.elsolucionario.org

628. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.34
Determine by direct integration the centroid of the area shown. Express
your answer in terms of
a and h.

SOLUTION
We have
h
yx
a


and
()
1
dA h y dx
x
hdx
a






1
()
2
1
2
EL
EL
xx
yhy
hx
a







Then
2
0
0
1
1
22
a
a
xx
AdA h dxhx ah
aa

    
 


and
23
2
0
0
1
1
23 6
a
a
EL
xxx
xdA xh dx h ah
aa

 
 



0
2223
2
22
0
0
11
2
1
1
223 3
a
EL
a
ahx x
ydA h dx
aa
hxhx
dx x ah
aa 
 


  
 

  




211
:
26
EL
xA x dA x ah a h






2
3
xa


211
:
23
EL
yA y dA y ah ah






2
3
yh


www.elsolucionario.org

629. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.35
Determine by direct integration the centroid of the area shown.
Express your answer in terms of
a and h.

SOLUTION
At (, ),ah
2
1
:yhka
or
2
h
k
a



2
:yhma
or
h
m
a


Now
12
1
()
2
EL
EL
xx
yyy


and
2
21 2
2
2
()
()
hh
dA y y dx x x dx
aa
h
ax x dx
a

  



Then
223
22
0
0 11
()
23 6
a
a
hha
A dA ax x dx x x ah
aa

    




and

2342
22
0
0
22
12 21 21 11
()
34 12
11
()[()]
22
a
a
EL
EL
hha
xdA x ax xdx x x ah
aa
ydA y y y ydx y ydx




  

 





www.elsolucionario.org

630. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.35 (Continued)


22
24
24
0
22
35
4
0
2
1
2
11
235
1
15
a
ahh
x xdx
aa
ha
xx
a
ah












211
:
612
EL
xAxdAxah ah






1
2
xa 

211
:
615
EL
yA y dA y ah ah






2
5
yh



www.elsolucionario.org

631. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.36
Determine by direct integration the centroid of the area shown.
Express your answer in terms of
a and h.


SOLUTION

For the element (el) shown at

2, xayh or 
2
2hka

2
=
4
h
k
a



el
xx
1

2
el
yy
dA ydx 2
3
22
220
0
Then
344
7
12
a
a
hhx
AdA xdx
aa
ah

  





2
4
2 2
22
2
and
444
15
16
a
a
el a
a
hhx
xdA x x dx
aa
ha

  
 



2 2
2
22
24
24
2
25
2
41
2
1
2432
31
516032a
el a
a
a
a
a
ydA y dx
hx h
dx x dx
aa
hx
ah
a






 



Then
2715
:
12 16
el
xA x dA x ah a h



 or
1.607xa 

2731
:
12 160
EL
yA y dA y ah ah



 or
0.332yh 
www.elsolucionario.org

632. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.37
Determine by direct integration the centroid of the area shown.

SOLUTION
First note that symmetry implies 0y 


EL
dA adx
xx



1
()
2
2
cos
3
EL
dA a ad
xa 



Then
2
0
2
2
01
2
[] [] (1 )
2







 
 
 
a
a
AdA adx ad
a
ax a

and
2
0
2
3
0
321
() cos
32
1
[sin ]
23
12
sin
23




















 
a
EL
a
x dA x adx a a d
x
aa
a


23 12
:[(1)] sin
23
EL
xA x dA x a a 





or
34sin
6(1 )
xa






www.elsolucionario.org

633. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.38
Determine by direct integration the centroid of the area shown.

SOLUTION
First note that symmetry implies 0x 
For the element (EL) shown

2
(Figure 5.8B)
EL
r
y
dA rdr





Then

2
2
1
1
2
22
21
22
r
r
r
r
r
AdA rdr rr

    




and

2
2
1
1
333
21212
()2
33
r
r
EL
r
r
r
ydA rdr r r r








So
 
22 33
21 21 2
:
23
EL
yA y dA y r r r r





or
33
21
22
21
4
3
rr
y
rr







www.elsolucionario.org

634. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.39
Determine by direct integration the centroid of the area shown.

SOLUTION
For the element (EL) shown,
22b
yax
a


and
 
 
22
22
()
1
()
2
2
EL
EL
dA b y dx
b
aaxdx
a
xx
yyb
b
aax
a





Then
 
22
0
ab
AdA aaxdx
a
 


To integrate, let
22
sin : cos , cosxa a x a dxa d
Then
/2
0
/2
22
0
(cos)(cos)
2
sin sin
24
1
4
b
Aaaad
a
b
aa
a
ab






 

 







and
 
22
0
/2
2223/2
0
3
1
()
23
1
6
a
ELb
xdA x a a x dx
a
ba
xax
a
ab













www.elsolucionario.org

635. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.39 (Continued)


  
22 22
0
223
2
22
0
0
2
2
()
322
1
6
a
EL
a
abb
ydA a a x a a x dx
aa
bbx
xdx
aa
ab

 



 







21
:1
46
EL
xAxdAxab ab






2
or
3(4 )
a
x





21
:1
46
EL
yA y dA y ab ab






2
or
3(4 )
b
y


 

www.elsolucionario.org

636. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.40
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.

SOLUTION
At
2
2
0,
(0 ) or
xyb
b
bk a k
a

 
Then
2
2
()
b
yxa
a


Now
2
2
()
22
EL
EL
xx
yb
yxa
a

 
and
2
2
()
b
dA ydx x a dx
a
 

Then

32
22
0 0 1
()
33
aa
bb
AdA xadx xa ab
aa

    



and

2322
22
00
42
32 2
2
2
22 5
22 4
0
0
2
() 2
21
43 2 12
1
() () ()
522
1
10
aa
EL
a
a
ELbb
xdA x xadx x ax axdx
aa
bx a
ax x a b
a
bb b
ydA xa xadx xa
aa a
ab









 



 


Hence
211
:
312
EL
xA x dA x ab a b






1
4
xa


211
:
310
EL
yA y dA y ab ab






3
10
yb


www.elsolucionario.org

637. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.41
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.

SOLUTION

222
11 1 1 2
333
22 2 2 3
3
2
21 2
12
3
2
2
but
but
()
1
()
2
2
EL
EL
b
ykx bka y x
a
b
ykx bkay x
a
bx
dA y y dx x dx
aa
xx
yyy
bx
x
aa



   










3
2
2
0
34
2
0
3
2
2
0
4
3
2
0
45
2
0
2
34
1
12
45
1
20
a
a
a
EL
a
abx
AdA x dx
aa
bx x
aa
ba
bx
xdA x x dx
aa
bx
xdx
aa
bx x
aa
ab

  







 



 














































www.elsolucionario.org

638. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.41 (Continued)


33
22
22
0
26
4
42
0
25 7
2
42
0
2
2
1
53527
a
EL
a
abxbx
ydA x x dx
aaaa
bx
xdx
aa
bx x
ab
aa

 



 









211
:
12 20
EL
xAxdAx ba ab






3
5
xa 

211
:
12 35
EL
yA y dA y ba ab






12
35
yb


www.elsolucionario.org

639. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.42
Determine by direct integration the centroid of the area shown.

SOLUTION
We have
2
2
2
2
1
1
22
1
EL
EL
xx
axx
yy
LL
xx
dA y dx a dx
LL


 



 


Then
2
223
2
22
0
0
1
23
8
3
L
L
xx x x
AdA a dxax
LLLL
aL
 
   

 



and
2
2234
2
22
0
0
2
1
23 4
10
3
L
L
EL
xx x x x
xdA xa dx a
LLLL
aL
 
 

 




22
2
22
0
2234
234
0
2
22345
234
0
2
11
2
12 3 2
2
2 25
11
5
L
EL
EL
Laxx xx
ydA a dx
LLLL
axxxx
dx
LLLL
axxxx
x
LLLL
aL

  

 











Hence,
2810
:
33
EL
xA x dA x aL aL






5
4
xL


2111
:
85
EL
yA y dA y a a






33
40
ya



www.elsolucionario.org

640. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.43
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.

SOLUTION
For y 2 at,xa
2
2
,,or
a
yb akb k
b
 

Then
1/ 2
2b
yx
a


Now
EL
xx
and for 0,
2
a
x

1/ 2
2
1/ 2
2
22
EL
ybx
y
a
x
dA y dx b dx
a



For
,
2
a
xa
1/ 2
12
11
()
222
EL
bx x
yyy
a a

 




1/ 2
21
1
()
2
xx
dA y y dx b dx
aa

   



Then
1/ 2 1/ 2
/2
0/2
1
2
aa
axxx
AdA b dx b dx
aaa

   


 

/2 3/2 2
3/2
0
/2
3/2 3/2
3/2
2
2
221
3322
2
()
32 2
11
() ()
2222
13
24
a
a
a
bxx
xb x
aaa
ba a
a
a
aa
ba a
a
ab

 
 

 

 
 

   
     
  



www.elsolucionario.org

641. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.43 (Continued)

and
1/ 2 1/ 2
/2
0/2
1
2
aa
EL
a xxx
xdA x b dx x b dx
aaa
   
    
  
     
 


/2 5/2 3 4
5/2
0
/2
5/2 5/2
5/2
32
32
2
22
5534
2
()
52 2
11
() ()
3242
71
240
a
a
a
bxxx
xb
aaa
ba a
a
a
aa
ba a
a
ab

 
 

 

 
 


  
     
 



1/ 2 1/ 2
/2
0
1/ 2 1/ 2
/2
/2 3222
2
0
/2
22 2
2
2
11
22 2
111
22 2 2 3 2
1
()
42 2 622
a
EL
a
a
a
a
abx x
ydA b dx
aa
bx x x x
bdx
aa aa
bbxx
x
aaaa
ba a ba
a
aa

 

 
   

   

  



  


  

 
 



3
2
11
48
ab





Hence,
213 71
:
24 240
EL
xAxdAx ab ab






17
130
x a 

213 11
:
24 48
EL
yA y dA y ab ab






11
26
yb



www.elsolucionario.org

642. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.44
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and b.

SOLUTION
For y 1 at ,xa
2
2 2
2, 2 , orb
yb bka k
a
 

Then
2
1 22b
yx
a


By observation,
2
(2) 2
bx
yxbb
aa

   


Now
EL
xx
and for
0,xa
22
11 2212
and
2
EL
bb
y y x dA y dx x dx
aa
  

For
2,ax a
22
1
2and 2
22
EL
bx x
yy dAydxb dx
aa
 
  
 
 

Then
2
2
2
0
2
23
2
0 02
2
27
2
32 6
aa
a
aabx
AdA xdx b dx
aa
bx a x
bab
aa

  


 
  
 
 

and
2
2
2
0
2
43
2
2
00
22223
22
2
2
43
11
(2 ) ( ) (2 ) ( )
23
7
6
aa
EL
a
aabx
xdA x xdx xb dx
aa
bx x
bx
aa
ab b a a a a
a
ab
 

 
  
  
  
  

  



 



www.elsolucionario.org

643. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.44 (Continued)


2
22
22
00
2
325 2
4
0
22
22
2
2
2
523
17
30
aa
EL
aa
abb bx x
ydA x xdx b dx
aaaa
bx b a x
aa
ab
 

 
  
 
  
 

 

Hence,
277
:
66
EL
xAxdAxab ab






xa 

2717
:
630
EL
yA y dA y ab ab






17
35
yb


www.elsolucionario.org

644. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.45
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.

SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line.
Now
322
cos and
EL
x a dL dx dy
where
32
32
cos : 3 cos sin
sin : 3 sin cos
xa dx a d
ya dy a d


Then
222 21/2
221/2
/2
/2
2
0
0
[( 3 cos sin ) (3 sin cos ) ]
3cos sin (cos sin )
3cos sin
1
3 cos sin 3 sin
2
3
2
dLadad
ad
ad
LdL a d a
a


 
   

  



 





and
/2
3
0
/2
25 2
0
cos (3 cos sin )
13
3cos
55
EL
xdL a a d
aa





 




Hence,
233
:
25
EL
xL x dL x a a






2
5
xa

Alternative Solution:
2/3
32
2/3
32
cos cos
sin sin
x
xa
a
y
ya
a










www.elsolucionario.org

645. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.45 (Continued)

2/3 2/3
2/3 2/3 3/2
1or ( )
xy
ya x
aa
 
 
 
 
Then
2/3 2/3 1/2 1/3
()()
dy
ax x
dx

 
Now
EL
xx
and

2
1/ 2
2
2/3 2/3 1/2 1/3
1
1( )( )
dy
dL
dx
dx a x x dx





  


Then
1/3
1/3 2/3
1/3
0
0
33
22
a
a
a
LdL dxa x a
x

  




and
1/3
1/3 5/3 2
1/3
0
0
33
55
a
a
EL
a
xdL x dx a x a
x
 
 



Hence
233
:
25
EL
xLxdLxa a






2
5
xa 



www.elsolucionario.org

646. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.46
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.

SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line.
Now cos and
EL
xr dLrd
Then
7/4
7/4
/4
/4 3
[]
2
LdL rdr r




   


and
7/4
/4
7/42
/4
2
2
cos ( )
[sin ]
11
22
2
EL
xdL r rd
r
r
r














Thus
23
:2
2
xL x dL x r r







22
3
xr

 



www.elsolucionario.org

647. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.47*
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid. Express your answer in terms of a.

SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
We have at
,xa
3/2 1
,,orya aka k
a
 

Then
3/21
yx
a


and
1/ 23
2
dy
x
dx a


Now
EL
xx
and
2
1/ 2
2
1/ 2
1
3
1
2
1
49
2
dy
dL dx
dx
xdx
a
axdx
a










Then
0
3/2
0
3/2
1
49
2
121
(4 9 )
392
[(13) 8]
27
1.43971
a
a
LdL axdx
a
ax
a
a
a
 








and
0
1
49
2
a
EL
xdL x a xdx
a







www.elsolucionario.org

648. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.47* (Continued)

Use integration by parts with

3/2
49
2
(4 9 )
27
u x dv a x dx
du dx v a x



Then
3/2 3/2
0
0
3/2
25/2
0
2
3/2 5/2
212 2
(4 9 ) (4 9 )
27 272
(13) 1 2
(4 9 )
27 4527
2
(13) [(13) 32]
27 45
0.78566
a
a
EL
a
xdL x a x a x dx
a
aax
a
a
a


 


 









2
: (1.43971 ) 0.78566
EL
xLxdLx a a

or 0.546x a 



www.elsolucionario.org

649. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.48*
Determine by direct integration the centroid of the area shown.

SOLUTION
We have
22
cos cos
33
22
sin sin
33
EL
EL
xr ae
yr ae





and
2211
()( )
22
dA r rd a e d


Then
22 2 2
0
0
22
2111
222
1
(1)
4
133.623
AdA aed a e
ae
a






 






and
22
0
33
021
cos
32
1
cos
3
EL
xdA ae aed
ae d














To proceed, use integration by parts, with

3
ue

 and
3
3du e d



cosdv d and sinv
Then
33 3
cos sin sin (3 )ede ed
 
   


Now let
3
ue

 then
3
3du e d



sin ,dv d then cosv
Then
33 3 3
cos sin 3 cos ( cos )(3 )ede e ed
  
    






www.elsolucionario.org

650. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.48* (Continued)

so that
3
3
3
3
0
3
33
cos (sin 3cos )
10
1
(sin 3cos )
310
( 3 3) 1239.26
30
EL
e
ed
e
xdA a
a
ea





  




 



Also,
22
0
33
021
sin
32
1
sin
3
EL
ydA ae aed
ae d




 









Use integration by parts, as above, with

3
ue

 and
3
3du e d



sindv d


and cosv 
Then
33 3
sin cos ( cos )(3 )ede ed
 
  


so that
3
3
3
3
0
3
33
sin ( cos 3sin )
10
1
(cos 3sin)
310
( 1) 413.09
30
EL
e
ed
e
ydA a
a
ea





  
 







Hence,
23
: (133.623 ) 1239.26
EL
xAxdAx a a

or 9.27x a 

23
: (133.623 ) 413.09
EL
yAydAy a a

or 3.09ya 


www.elsolucionario.org

651. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.49*
Determine by direct integration the centroid of the area shown.

SOLUTION



/2 /2
00
/2
0
sin
1
,,
2
sin
cos
EL EL
LL
L
x
ya
L
x x y y dA y dx
x
A y dx a dx
L
LxaL
Aa
L




 


 





/2 /2
00
sin
LL
EL x
x dA xydx xa dx
L

 

Setting
,
x
u
L

 we have
,
L
xu

 ,
L
dx du



2
/2 /2
00
sin sin
EL
LLL
xdA ua u du a u xdu


  

  
  
 

Integrating by parts,
2 2
/2
/2
0 2
0
2
/2 /2 /2
222 2
00 0
/222
/2
2
2
0
0
[ cos ] cos
11
sin sin
22 2
111
(1 cos 2 ) sin 2
24282
EL
LL
EL
LaL
xdAa u u udu
xaL
y dA y dx a dx u du
L
aL aL
udu u u aL





 




  

 





  



2
2
:
 





EL
aL aL
xA x dA x



L
x 

21
:
8






EL
aL
yA y dA y a L

8

ya 

www.elsolucionario.org

652. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.50
Determine the centroid of the area shown in terms of a.


SOLUTION
We have
11 1
1
22
EL
EL
xx
yy
x


 


and
1
1dA ydx dx
x

 


Then
 
1
1
1 1
0
0
ln (1 2ln )
a
a a
a
a
a
dx
AdA adx ax x a
x
    
 

and

1
1
22
1
1
0
0
1121
22 2
aa
a
a
EL
a
a
dx ax a
xdA xdx x x a
xaaa
 
  





1
22
1
22
1 2 0
00
1
11 11 121
22 2 2 2222 2
a
aa
a a
EL
a
a
dx a a a a
ydA ydx adx x
xaax




 

Because of symmetry, computation of only one coordinate was necessary.


2
21
:12ln
2
EL
a
xA x dA x a
a





2
21
212lna
xy
aa





www.elsolucionario.org

653. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.51
Determine the centroid of the area shown when a = 4 in.

SOLUTION
We have
11 1
1
22
EL
EL
xx
yy
x


 


and
1
1dA ydx dx
x

 


Then
 
1
1
1 1
0
0
ln (1 2ln )
a
a a
a
a
a
dx
AdA adx ax x a
x
    
 

and

1
1
22
1
1
0
0
1121
22 2
aa
a
a
EL
a
a
dx ax a
xdA xdx x x a
xaaa
 
  





1
22
1
22
1 2 0
00
1
11 11 121
22 2 2 2222 2
a
aa
a a
EL
a
a
dx a a a a
ydA ydx adx x
xaax




 

Because of symmetry, computation of only one coordinate was necessary.



2
2
21
:12ln
2
21
212ln
EL
a
xA x dA x a
a
a
xy
aa







Find
xand y when 4in.a
We have
 
2
2(4) 1
24 1 2ln4
xy



or
1.027 in.xy 
www.elsolucionario.org

654. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.52
Determine the volume and the surface area of the solid obtained by rotating
the area of Prob. 5.1 about (a) the x axis, (b) the line x = 72 mm.

SOLUTION
From the solution of Problem 5.1, we have

2
3
3
2632.5 mm
111,172.5 mm
63,787.5 mm
A
xA
yA




Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x-axis:

area
3
Volume 2 2
2(63,787.5mm)
yA yA

 or
33
Volume 401 10 mm 


line line
22 33 44
22
Area 2 2 ( )
2( )
2 [(22.5) 45 (45)(72) (22.5) 27 45 ]
yL y L
yL yL yL




or
32
Area 34.1 10 mm 
(b) Rotation about the line x=72 mm:

  
area
23
Volume 2 (72 ) 2 (72 )
2 72 mm 2632.5 mm 111,172.5 mm
xA AxA
  


or
33
Volume 492 10 mm 

 
line line 1 3 4
11 3 3 4 4
22
Area 2 2 ( ) where , and are measured wrt 72 mm
2( )
45 72
2 [(22.5)(45) (36)(72) 27 45 ]
2
xL x L xx x x
xL xL xL
 

 


 



or
32
Area 41.9 10 mm 

www.elsolucionario.org

655. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.53
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.2 about (a) the x-axis, (b) the y-axis.

SOLUTION
From the solution of Problem 5.2, we have

2
3
3
11in
11.5 in
39.5 in
A
xA
yA




Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x-axis:

area
3
Volume 2 2
2(39.5in)
yA yA


or
3
Volume 248 in 

line line
22 33 44 55 66 77 88
Area 2 2 ( )
2( )
2 [(1)(2) (2)(3) (2.5)(1) (3)(3) (5.5)(5) (8)(1) (4)(8)]




yL y L
yL yL yL yL yL yL yL
or
2
Area 547 in 
(b) Rotation about the y-axis:

area
3
Volume 2 2
2 (11.5 in )
xAxA


or
3
Volume 72.3 in 

line line
11 22 33 44 55 66 77
Area 2 2 ( )
2( )
2 [(0.5)(1) (1)(2) (2.5)(3) (4)(1) (2.5)(3) (1)(5) (0.5)(1)]
xL x L
xL xL xL xL xL xL xL





or
2
Area 169.6 in 

www.elsolucionario.org

656. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.54
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.6 about (a) the line x  60 mm, (b) the line y  120 mm.

SOLUTION
From the solution of Problem 5.6, we have

2
33
33
7200 mm
72 10 mm
629.83 10 mm
A
xA
yA


 

Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about line 60 mm:x

3
Volume 2 ( 60) 2 ( 60 )
2 [ 72 10 60(7200)]
x AxAA 



63
Volume 2.26 10 mm 

line line
11 2 2 3 3
Area 2 2 ( )
2( )
2(60) (60) 2(60) (60)
2 60 60 (60)(120)
22
xL x L
xL xL xL
 





  
  
  
  

where
123
,,xxx are measured with respect to line 60 mm.x
32
Area 116.3 10 mm 
(b) Rotation about line
120 mm:y

3
Volume 2(120 ) 2(120 )
2 [120(7200) 629.83 10 ]yA A yA
 
 


63
Volume 1.471 10 mm 

line line
11 22 44
Area 2 2 ( )
2( )
yLyL
yLyLyL




where
124
,,yyy are measured with respect to line 120 mm.y

2(60) (60) 2(60) (60)
Area 2 120 (60)(120)
22

 
  
 
 

32
Area 116.3 10 mm 

www.elsolucionario.org

657. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.55
Determine the volume and the surface area of the chain link
shown, which is made from a 6-mm-diameter bar, if R  10 mm
and L  30 mm.

SOLUTION
The area A and circumference C of the cross section of the bar are

2
and .
4
A dCd


Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V,

side end
2( ) 2( )
2( ) 2( )
2( )
VV V
ALRA
LRA





or
2
2[30mm (10mm)] (6mm)
4
V







3
3470 mm
3
or 3470 mmV 
For the area A,
side end
2( ) 2( )
2( ) 2( )
2( )
AA A
CL RC
LRC





or
2[30 mm (10 mm)][ (6 mm)]A
 

2
2320 mm
2
or 2320 mmA 

www.elsolucionario.org

658. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.56
Determine the volume of the solid generated by rotating the
parabolic area shown about (a) the x-axis, (b) the axis AA.

SOLUTION
First, from Figure 5.8a, we have
4
3
2
5
Aah
yh



Applying the second theorem of Pappus-Guldinus, we have
(a) Rotation about the x-axis:

Volume 2
24
2
53
yA
hah



 

 
 
or
216
Volume
15
ah
 
(b) Rotation about the line
:
AA

Volume 2 (2 )
4
2(2)
3
aA
aah







or
216
Volume
3
ah
 



www.elsolucionario.org

659. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.57
Verify that the expressions for the volumes of the first four shapes in Fig. 5.21 are correct.

SOLUTION




Following the second theorem of Pappus-Guldinus, in each case, a specific
generating area A will be rotated about the x-axis to produce the given
shape. Values of
y are from Figure 5.8a.
(1) Hemisphere: the generating area is a quarter circle.
We have
24
22
34
a
VyA a

 

 
 
or
32
3
Va
 

(2) Semiellipsoid of revolution: the generating area is a quarter ellipse.
We have
4
22
34
a
VyA ha

 

 
 
or
22
3
Vah 

(3) Paraboloid of revolution: the generating area is a quarter parabola.
We have
32
22
83
VyA aah

 

 
 
or
21
2
Vah
 

(4) Cone: the generating area is a triangle.
We have
1
22
32
a
VyA ha

 

 
 
or
21
3
Vah
 

www.elsolucionario.org

660. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.58
Knowing that two equal caps have been removed from a 10-in.-diameter
wooden sphere, determine the total surface area of the remaining portion.

SOLUTION
The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of
Pappus-Guldinus, we have

11 2 2
22
2(2 )
AXL xL
xL xL



Now
4
tan
3

or
53.130
Then
2
180
5 in. sin 53.130
53.130
4.3136 in.
x







and
2
2 53.130 (5 in.)
180
9.2729 in.
3
2 2 in. (3 in.) (4.3136 in.)(9.2729 in.)
2
L
A











or
2
308 inA 

www.elsolucionario.org

661. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.59
Three different drive belt profiles are to be
studied. If at any given time each belt
makes contact with one-half of the
circumference of its pulley, determine the
contact area between the belt and the
pulley for each design.

SOLUTION










Applying the first theorem of Pappus-Guldinus, the contact area
C
A of a
belt is given by

C
AyL yL
where the individual lengths are the lengths of the belt cross section that
are in contact with the pulley.
(a)

11 2 2
[2( ) ]
0.125 0.125 in.
2 3 in. [(3 0.125) in.](0.625 in.)
2cos20
C
AyLyL


  


or
2
8.10 in
C
A 
(b)
11
[2( )]
0.375 0.375 in.
2 3 0.08 in.
2cos20
C
AyL





or
2
6.85 in
C
A 
(c)
11
[2( )]
2(0.25)
3 in. [ (0.25 in.)]
C
AyL






or
2
7.01in
C
A 

www.elsolucionario.org

662. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.60
Determine the capacity, in liters, of the punch bowl
shown if R  250 mm.

SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying
the second theorem of Pappus-Guldinus and using Figure 5.8a, we have

11 2 2
2
6
33
3
3
3
22
2( )
11 11 3 2sin30
2
32 22 2 3 6
2
16 3 2 3
33
8
33
(0.25 m)
8
0.031883 m
VxA xA
xA xA
R
RRR R
RR
R









   
    
   








Since
33
10 l 1 m

3
3
3
10 l
0.031883 m
1m
V
31.9 litersV 


www.elsolucionario.org

663. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.61
Determine the volume and total surface area of the bushing shown.

SOLUTION

Volume:
The volume can be obtained by rotating the triangular area shown through  radians about the y axis.
The area of the triangle is:

21
52 60 1560 mm
2
A
Applying the theorems of Pappus-Guldinus, we have

 
2
52 mm 1560 mmVxA or
33
255 10 mmV 
The surface area can be obtained by rotating the triangle shown through an angle of  radians about the y
axis.

Considering each line BD, DE, and BE separately:
Line
22
1
: 22 60 63.906 mmBD L
1
22
20 31 mm
2
x 

Line
2
:52mmDE L
2
20 22 26 68 mmx
Line
22
3
: 74 60 95.268 mmBE L
1
74
20 57 mm
2
x 

www.elsolucionario.org

664. Copyright © McGraw-Hill Education. Permission required for reproduction or display.





PROBLEM 5.61 (Continued)
Then applying the theorems of Pappus-Guldinus for the part of the surface area generated by the lines:

    
32
31 63.906 68 52 57 95.268 10947.6 34.392 10 mm
L
AxA       


The area of the “end triangles”:

321
25260 3.1210mm
2
E
A





Total surface area is therefore:



32
34.392 3.12 10 mm
LE
AA A  
32
or 37.5 10 mmA  www.elsolucionario.org

665. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.62
Determine the volume and weight of the solid brass knob shown, knowing
that the specific weight of brass is 0.306 lb/in
3
.

SOLUTION

Area, in
2
,in.y
3
,inyA
1
2
(0.75) 0.4418
4

 0.8183 0.3615
2 (0.5)(0.75) 0.375 0.25 0.0938
3 (1.25)(0.75) 0.9375 0.625 0.5859
4
2
(0.75) 0.4418
4


0.9317 0.4116
 0.6296


33
2 2 (0.6296 in ) 3.9559 inVyA  
3
3.96 inV 

33
(0.306 lb/in )(3.9559 in )WV 1.211lbW 


Volume of knob is obtained by rotating area
at left about the x-axis.
Consider area as made
of components shown below. www.elsolucionario.org

666. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.63
Determine the total surface area of the solid brass knob shown.

SOLUTION

Area is obtained by rotating lines shown about the x-axis.
L, in.

,in.y
2
,inyL
1 0.5 0.25 0.1250
2 (0.75) 1.1781
2

 0.9775 1.1516
3 (0.75) 1.1781
2

 0.7725 0.9101
4 0.5 0.25 0.1250
 2.3117


2
22(2.3117in)AyL 
2
14.52 inA 

www.elsolucionario.org

667. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.64
The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing
that the density of aluminum is 2800 kg/m
3
, determine the mass of the shade.


SOLUTION
The mass of the lamp shade is given by

mV At

where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have

11 22 33 44
22
2( )
AyL yL
yL yL yL yL



or
22
22
22
213 mm 13 16
2 (13 mm) mm (32 mm) (3 mm)
22
16 28
mm (8 mm) (12 mm)
2
28 33
mm (28 mm) (5 mm)
2
2 (84.5 466.03 317.29 867.51)
10,903.4 mm
A


 

 








 



Then
332
(2800 kg/m )(10.9034 10 m )(0.001 m)
mAt


 or

0.0305 kgm 
www.elsolucionario.org

668. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.65*
The shade for a wall-mounted light is formed from a thin
sheet of translucent plastic. Determine the surface area of the
outside of the shade, knowing that it has the parabolic cross
section shown.

SOLUTION
First note that the required surface area A can be generated by rotating the parabolic cross section through 
radians about the y-axis. Applying the first theorem of Pappus-Guldinus, we have
AxL
Now at
21
100 mm, 250 mm
250 (100) or 0.025 mm
xy
kk



and
2
1
EL
xx
dy
dL dx
dx





where
2
dy
kx
dx

Then
22
14dL k x dx
We have  
100
22
0
14
EL
xL x dL x k x dx 




100
223/2
2
0
223/23/2
2
2
11
(1 4 )
34
11
[1 4(0.025) (100) ] (1)
12(0.025)
17,543.3 mm
xL k x
k




 


Finally,
2
(17,543.3 mm )A or
32
55.1 10 mmA 


www.elsolucionario.org

669. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.66
For the beam and loading shown, determine (a) the magnitude
and location of the resultant of the distributed load, (b) the
reactions at the beam supports.

SOLUTION

I
II
1
( ) (400 N/m)(6 m)
2
1200 N
(1600 N/m)(6m)
4800 N
aR
R





III
F0:
1200 N 4800 N
6000 N
y
RRR
R
R
    



0 : 6000 2 1200 4 4800
3.60 m
A
MX
X
   



6000 N , 3.60 mXR 
(b) Reactions 0: 0
0 : 6000 0 6000 N
xx
yy y
FA
FA A
 
   

6000 NA 

0: (6000 N)(3.60 m) 0
AA
MM  

21.6 kN m
A
M

www.elsolucionario.org

670. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.67
For the beam and loading shown, determine (a) the
magnitude and location of the resultant of the
distributed load, (b) the reactions at the beam supports.

SOLUTION

(a)
I
II
III
1
(1100 N/m)(6 m) 2200 N
3
(900 N/m)(6m) 5400 N
2200 5400 7600 N
R
R
RR R


   


: (7600) (2200)(1.5) (5400)(3)XR xR X  

2.5658 mX 7.60 kNR
,

2.57 mX 
(b)
0: (6 m) (7600 N)(2.5658 m) 0
3250.0 N
A
MB
B
  


3.25 kNB 

0: 3250.0 N 7600 N 0
4350.0 N
   

y
FA
A

4.35 kNA 
www.elsolucionario.org

671. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.68
Determine the reactions at the beam supports for the given
loading.

SOLUTION

I
II
(200 lb/ft)(4 ft) 800 lb
1
(150 lb/ft)(3 ft) 225 lb
2
R
R






0: 800 lb 225 lb 0
y
FA   

575 lbA



0: (800 lb)(2 ft) (225 lb)(5 ft) 0
AA
MM   

475 lb ft
A
M



www.elsolucionario.org

672. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.69
Determine the reactions at the beam supports for the given
loading.

SOLUTION

I
II
1
(50 lb/in.)(12 in.)
2
300 lb
(50 lb/in.)(20in.)
1000 lb
R
R






0: 300 lb 1000 lb 400 lb 0
yy
FA    

900 lbA 

0: (300 lb)(8 in.) (1000 lb)(22 in.) 400 lb 38 in. 0
AA
MM    

9200 lb in.
A
M 


www.elsolucionario.org

673. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.70
Determine the reactions at the beam supports for the given
loading.

SOLUTION
We have
I
II
III
1
(3ft)(480 lb/ft) 720 lb
2
1
(6 ft)(600 lb/ft) 1800 lb
2
(2ft)(600 lb/ft) 1200 lb
R
R
R




Then
0: 0
xx
FB 

0: (2 ft)(720 lb) (4 ft)(1800 lb) (6 ft) (7 ft)(1200 lb) 0
By
MC    

or
2360 lb
y
C

2360 lbC



0: 720 lb 1800 lb 2360 lb 1200 lb 0
yy
FB      

or 1360 lb
y
B

1360 lbB





www.elsolucionario.org

674. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.71
Determine the reactions at the beam supports for the given loading.

SOLUTION
First replace the given loading by the loadings shown below. Both loadings are equivalent since they are
both defined by a linear relation between load and distance and have the same values at the end points.


1
2
1
(900 N/m)(1.5 m) 675 N
2
1
(400 N/m)(1.5 m) 300 N
2
R
R




0: (675 N)(1.4 m) (300 N)(0.9 m) (2.5 m)
A
MBC   

270 NB 270 NB 

0: 675 N 300 N 270 N 0
y
FA    

105.0 NA 105.0 NA 
www.elsolucionario.org

675. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.72
Determine the reactions at the beam supports for the given
loading.

SOLUTION

We have
I
II
1
(12 ft)(200 lb/ft) 800 lb
3
1
(6 ft)(100 lb/ft) 200 lb
3
R
R



Then

0: 0
xx
FA 

0: 800 lb 200 lb 0
yy
FA   
or
1000 lb
y
A 1000 lbA



0: (3 ft)(800 lb) (16.5 ft)(200 lb) 0
AA
MM   

or

5700 lb ft
A
M 5700 lb ft
A
M 


www.elsolucionario.org

676. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.73
Determine the reactions at the beam supports for the given
loading.

SOLUTION
First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the
same.

We have
I
II
(6 m)(300 N/m) 1800 N
2
(6 m)(1200 N/m) 4800 N
3
R
R


Then

0: 0
xx
FA 

0: 1800 N 4800 N 0
yy
FA   
or
3000 N
y
A 3.00 kNA



15
0: (3 m)(1800 N) m (4800 N) 0
4
AA
MM

   


or
12.6 kN m
A
M 12.60 kN m
A
M




www.elsolucionario.org

677. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.74
Determine (a) the distance a so that the vertical reactions at
supports A and B are equal, (b) the corresponding reactions at
the supports.

SOLUTION
(a)

We have
I
II
1
( m)(1800 N/m) 900 N
2
1
[(4 ) m](600 N/m) 300(4 ) N
2
Ra a
Ra a

  

Then
0: 900 300(4 ) 0
yy y
FAa aB    
or
1200 600
yy
AB a 
Now
600 300 (N)
yy yy
AB AB a  (1)
Also,
0: (4 m) 4 m [(900 ) N]
3
By
a
MA a
  




1
(4 ) m [300(4 ) N] 0
3
aa

  



or
2
400 700 50
y
Aaa  (2)
Equating Eqs. (1) and (2),
2
600 300 400 700 50aaa
or
2
840aa
Then
2
8(8)4(1)(4)
2
a
 


or
0.53590 ma 7.4641 ma Now 4ma 0.536 ma 
www.elsolucionario.org

678. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.74 (Continued)

(b) We have
0: 0
xx
FA 
From Eq. (1):
600 300(0.53590)
yy
AB


761 N 761 NAB 


www.elsolucionario.org

679. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.75
Determine (a) the distance a so that the reaction at support B is
minimum, (b) the corresponding reactions at the supports.

SOLUTION
(a)

We have
I
II
1
( m)(1800 N/m) 900 N
2
1
[(4 )m](600 N/m) 300(4 ) N
2
Ra a
Ra a

  

Then
8
0: m (900 N) m [300(4 )N] (4 m) 0
33
A y
aa
Ma aB
  
     
  
  

or
2
50 100 800
y
Ba a (1)
Then
100 100 0
y
dB
a
da

or 1.000 ma 
(b) From Eq. (1):
2
50(1) 100(1) 800 750 N
y
B 750 NB 
and
0: 0
xx
FA 

0: 900(1) N 300(4 1) N 750 N 0
yy
FA     
or
1050 N
y
A 1050 NA 

www.elsolucionario.org

680. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.76
Determine the reactions at the beam supports for the given
loading when w
O  400 lb/ft.

SOLUTION



I
II
11
(12 ft) (400 lb/ft)(12 ft) 2400 lb
22
1
(300 lb/ft)(12 ft) 1800 lb
2
O
Rw
R
 


0: (2400 lb)(1 ft) (1800 lb)(3 ft) (7 ft) 0   
B
MC

428.57 lbC
429 lbC 

0: 428.57 lb 2400 lb 1800 lb 0    
y
FB


3771lbB 3770 lbB 

www.elsolucionario.org

681. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.77
Determine (a) the distributed load w O at the end A of the beam
ABC for which the reaction at C is zero, (b) the corresponding
reaction at B.

SOLUTION


For
,
O
w
I
II
1
(12 ft ) 6
2
1
(300 lb/ft)(12 ft) 1800 lb
2
OO
Rw w
R


(a) For
0,C 0: (6 )(1 ft) (1800 lb)(3 ft) 0
BO
Mw   900 lb/ft
O
w 
(b) Corresponding value of
I
:R

I
6(900) 5400 lbR

0: 5400 lb 1800 lb 0   
y
FB 7200 lbB 


www.elsolucionario.org

682. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.78
The beam AB supports two concentrated loads and rests on soil that
exerts a linearly distributed upward load as shown. Determine the
values of
A and B corresponding to equilibrium.

SOLUTION


I
II
1
(1.8 m) 0.9
2
1
(1.8 m) 0.9
2
AA
BB
R
R




0: (24 kN)(1.2 ) (30 kN)(0.3 m) (0.9 )(0.6 m) 0
DA
Ma     (1)
For
0.6 m,a 24(1.2 0.6) (30)(0.3) 0.54 0
a
  

14.4 9 0.54 0
A
  10.00 kN/m
A
 

0: 24 kN 30 kN 0.9(10 kN/m) 0.9 0
yB
F       50.0 kN/m
B
 


www.elsolucionario.org

683. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.79
For the beam and loading of Problem 5.78, determine (a) the
distance a for which
A  20 kN/m, (b) the corresponding value of
B.
PROBLEM 5.78 The beam AB supports two concentrated loads
and rests on soil that exerts a linearly distributed upward load as
shown. Determine the values of
A and B corresponding to
equilibrium.

SOLUTION

We have
I
II
1
(1.8 m)(20 kN/m) 18 kN
2
1
(1.8 m)( kN/m) 0.9 kN
2
BB
R
R



(a)
0: (1.2 )m 24 kN 0.6 m 18 kN 0.3 m 30 kN 0
C
Ma       
or
0.375 ma 
(b)
0: 24 kN 18 kN (0.9 ) kN 30 kN 0
yB
F      
or
40.0 kN/m
B
 


www.elsolucionario.org

684. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.80
The cross section of a concrete dam is as shown. For a 1-ft-wide
dam section determine (a) the resultant of the reaction forces exerted
by the ground on the base AB of the dam, (b) the point of application
of the resultant of part a, (c) the resultant of the pressure forces
exerted by the water on the face BC of the dam.

SOLUTION
The free body shown consists of a 1-ft thick section of the dam and the triangular section of water above
the dam.

Note:


1
2
3
4
6 ft
(9 3)ft 12 ft
15 2 ft 17 ft
15 4 ft 19 ft
x
x
x
x

 
 
 

(a) Now
WV
So that


 

3
1
3
2
3
3
3
41
(150 lb/ft ) (9 ft) 15 ft (1 ft) 10,125 lb
2
(150 lb/ft )[(6 ft)(18 ft)(1 ft)] 16,200 lb
1
(150 lb/ft ) 6 ft 18 ft (1 ft) 8100 lb
2
1
(62.4 lb/ft ) (6 ft) 18 ft (1 ft) 3369.6 lb
2
W
W
W
W














Also
311
[(18 ft)(1 ft)][(62.4 lb/ft )(18 ft)] 10,108.8 lb
22
PAp 

Then
0: 10,108.8 lb 0
x
FH   or 10.11 kipsH  www.elsolucionario.org

685. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.80 (Continued)


0: 10,125 lb 16,200 lb 8100 lb 3369.6 lb 0
y
FV     
or
37,795 lbV 37.8 kipsV 
(b) We have
0: (37,794.6 lb) (6 ft)(10,125 lb) (12 ft)(16,200 lb)
(17 ft)(8100 lb) (19 ft)(3369.6 lb)
+(6 ft)(10,108.8 lb) 0
A
Mx  


or
37,794.6 60,750 194,400 137,700 64,022.4 60,652.8 0x    
or
10.48 ftx 
(c) Consider water section BCD as the free body
We have
0F ,
4
3369.6 lb, 10,108.8 lbWP

Then
10.66 kipsR 18.43 or 10.66 kipsR 18.43 
www.elsolucionario.org

686. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.81
The cross section of a concrete dam is as shown. For a 1-m-wide
dam section, determine (a) the resultant of the reaction forces
exerted by the ground on the base AB of the dam, (b) the point of
application of the resultant of part a, (c) the resultant of the pressure
forces exerted by the water on the face BC of the dam.

SOLUTION
(a) Consider free body made of dam and section BDE of water. (Thickness  1 m)




32
(3 m)(10 kg/m )(9.81 m/s )p


33 2
1
33 2
2
33 2
3
33 2
(1.5 m)(4 m)(1 m)(2.4 10 kg/m )(9.81 m/s ) 144.26 kN
1
(2 m)(3 m)(1 m)(2.4 10 kg/m )(9.81 m/s ) 47.09 kN
3
2
(2 m)(3 m)(1 m)(10 kg/m )(9.81 m/s ) 39.24 kN
3
11
(3 m)(1 m)(3 m)(10 kg/m )(9.81 m/s ) 44.145 kN
22
 


 
W
W
W
PAp


0: 44.145 kN 0  
x
FH


44.145 kNH

44.1 kNH


0: 141.26 47.09 39.24 0    
y
FV


227.6 kNV
228 kNV

www.elsolucionario.org

687. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.81 (Continued)


1
2
3
1
(1.5 m) 0.75 m
2
1
1.5 m (2 m) 2 m
4
5
1.5 m (2 m) 2.75 m
8
x
x
x

 
 


0: (1m) 0  
A
MxVxWP

(227.6 kN) (141.26 kN)(0.75 m) (47.09 kN)(2 m)
(39.24 kN)(2.75 m) (44.145 kN)(1 m) 0
(227.6 kN) 105.9 94.2 107.9 44.145 0
(227.6) 263.9 0
x
x
x


 



1.159 mx

(to right of A)


(b) Resultant of face BC:
Consider free body of section BDE of water.




59.1 kNR 41.6

59.1 kNR 41.6 

www.elsolucionario.org

688. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.82
The dam for a lake is designed to withstand the additional force caused by
silt that has settled on the lake bottom. Assuming that silt is equivalent to
a liquid of density
33
1.76 10 kg/m
s
 and considering a 1-m-wide
section of dam, determine the percentage increase in the force acting on
the dam face for a silt accumulation of depth 2 m.

SOLUTION
First, determine the force on the dam face without the silt.
We have
33 2
11
()
22
1
[(6.6 m)(1 m)][(10 kg/m )(9.81 m/s )(6.6 m)]
2
213.66 kN
ww
PAp Agh 


Next, determine the force on the dam face with silt
We have
33 2
33 2
I
33 2
II1
[(4.6 m)(1 m)][(10 kg/m )(9.81 m/s )(4.6 m)]
2
103.790 kN
( ) [(2.0 m)(1 m)][(10 kg/m )(9.81 m/s )(4.6 m)]
90.252 kN
1
( ) [(2.0 m)(1 m)][(1.76 10 kg/m )(9.81 m/s )(2.0 m)]
2
34.531 kN
w
s
s
P
P
P







Then
III
( ) ( ) 228.57 kN
ws s
PP P P  
The percentage increase, % inc., is then given by

%inc. 100%
(228.57 213.66)
100%
213.66
6.9874%





w
w
PP
P

% inc. 6.98% 
www.elsolucionario.org

689. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.83
The base of a dam for a lake is designed to resist up to 120 percent of the
horizontal force of the water. After construction, it is found that silt (that is
equivalent to a liquid of density
33
1.76 10 kg/m )
s
 is settling on the
lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of
dam, determine the number of years until the dam becomes unsafe.

SOLUTION
First determine force on dam without the silt,

33 2
allow
11
()
22
1
[(6.6 m)(1 m)][(10 kg/m )(9.81 m/s )(6.6 m)]
2
213.66 kN
1.2 (1.5)(213.66 kN) 256.39 kN
w
wp
w
PA Agh
PP 


 

Next determine the force
Pon the dam face after a depth d of silt has settled.
We have
33 2
2
33 2
I
2
33 2
II
21
[(6.6 ) m (1 m)][(10 kg/m )(9.81 m/s )(6.6 ) m]
2
4.905(6.6 ) kN
( ) [ (1 m)][(10 kg/m )(9.81 m/s )(6.6 ) m]
9.81(6.6 ) kN
1
( ) [ (1 m)][(1.76 10 kg/m )(9.81 m/s )( ) m]
2
8.6328 kN
w
s
s
Pd d
d
Pd d
dd
Pd d
d
 






2
III
22
2
( ) ( ) [4.905(43.560 13.2000 )
9.81(6.6 ) 8.6328 ] kN
[3.7278 213.66] kN
ws s
PP P P dd
dd d
d
    


Now it’s required that
allow
PP to determine the maximum value of d.

2
(3.7278 213.66) kN 256.39 kNd
or
3.3856 md
Finally,
3m
3.3856 m 12 10 N
year

  or 282 yearsN 
www.elsolucionario.org

690. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.84
An automatic valve consists of a 9  9-in. square plate that is pivoted about
a horizontal axis through A located at a distance h  3.6 in. above the lower
edge. Determine the depth of water d for which the valve will open.

SOLUTION
Since valve is 9 in. wide, 99,wp h where all dimensions are in inches.

12
I1
II 2
9( 9), 9
11
(9 in.) (9)(9 )( 9)
22
11
(9 in.) (9)(9 )
22
wd wd
Pw d
Pw d

 
 


Valve opens when
0.B

III
0: (6 in. 3.6 in.) (3.6 in. 3 in.) 0    
A
MP P


11
(9)(9 )( 9) (2.4) (9)(9 ) (0.6) 0
22
dd


 




( 9)(2.4) (0.6) 0
1.8 21.6 0


dd
d
12.00 in.d 

www.elsolucionario.org

691. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.85
An automatic valve consists of a 9  9-in. square plate that is pivoted about
a horizontal axis through A. If the valve is to open when the depth of water
is d  18 in., determine the distance h from the bottom of the valve to the
pivot A.

SOLUTION
Since valve is 9 in. wide,99,wp h where all dimensions are in inches.

1
2
9( 9)
9
wd
wd


For
18 in.,d
1
2
I
II
9 (18 9) 81
9(18) 162
11
(9)(9 )(18 9) (729 )
22
1
(9)(9 )(18) 729
2
w
w
P
P








Valve opens when
0.B

1II
0: (6 ) ( 3) 0  
A
MPhPh

1
729 (6 ) 729( 3) 0
2
1
330
2
61.5 0
hh
hh
h



  4.00 in.h 

www.elsolucionario.org

692. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.86
The 3  4-m side AB of a tank is hinged at its bottom A and is held in place by a
thin rod BC. The maximum tensile force the rod can withstand without breaking
is 200 kN, and the design specifications require the force in the rod not to exceed
20 percent of this value. If the tank is slowly filled with water, determine the
maximum allowable depth of water d in the tank.

SOLUTION
Consider the free-body diagram of the side.
We have
11
()
22
PApAgd

Now
0: 0
3
A
d
MhTP
  
where
3 mh
Then for
,
max
d
3332 1
(3 m)(0.2 200 10 N) (4 m ) (10 kg/m 9.81 m/s ) 0
32
max
max max
d
dd

      



or

32
120 N m 6.54 N/m 0
max
d 

or
2.64 m
max
d 


www.elsolucionario.org

693. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.87
The 3  4-m side of an open tank is hinged at its bottom A and is held in place by
a thin rod BC. The tank is to be filled with glycerine, whose density is 1263
kg/m
3
. Determine the force T in the rod and the reactions at the hinge after the
tank is filled to a depth of 2.9 m.

SOLUTION
Consider the free-body diagram of the side.
We have
32
11
()
22
1
[(2.9 m)(4 m)] [(1263 kg/m )(9.81 m/s )(2.9 m)]
2
= 208.40 kN
PApAgd


Then
0: 0
yy
FA 

2.9
0: (3 m) m (208.4 kN) 0
3
A
MT

  


or
67.151 kNT 67.2 kNT 

0: 208.40 kN 67.151 kN 0
xx
FA   
or
141.249 kN
x
A 141.2 kNA 

www.elsolucionario.org

694. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.88
A 0.5  0.8-m gate AB is located at the bottom of a tank filled with
water. The gate is hinged along its top edge A and rests on a
frictionless stop at B. Determine the reactions at A and B when cable
BCD is slack.

SOLUTION
First consider the force of the water on the gate.
We have
11
()
22
PApAgh
so that
33 2
I1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.45 m)]
2
882.9 N
P



33 2
II1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.93 m)]
2
1824.66 N
P


Reactions at A and B when T  0:
We have


12
0: (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) (0.8 m) 0
33
A
MB  
or
1510.74 NB

or
1511 NB 53.1 

0: 1510.74 N 882.9 N 1824.66 N 0FA    
or
1197 NA 53.1 

www.elsolucionario.org

695. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.89
A 0.5  0.8-m gate AB is located at the bottom of a tank filled with
water. The gate is hinged along its top edge A and rests on a
frictionless stop at B. Determine the minimum tension required in
cable BCD to open the gate.

SOLUTION
First consider the force of the water on the gate.
We have
11
()
22
PApAgh
so that
33 2
I1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.45 m)]
2
882.9 N
P



33 2
II1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.93 m)]
2
1824.66 N
P


T to open gate:
First note that when the gate begins to open, the reaction at B 0.
Then
12
0: (0.8 m)(882.9 N)+ (0.8 m)(1824.66 N)
33
8
(0.45 0.27)m 0
17
A
M
T


  


or
235.44 973.152 0.33882 0T 
or
3570 NT 

www.elsolucionario.org

696. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.90
A 42-ft gate is hinged at A and is held in position by rod CD. End
D rests against a spring whose constant is 828 lb/ft. The spring is
undeformed when the gate is vertical. Assuming that the force
exerted by rod CD on the gate remains horizontal, determine the
minimum depth of water d for which the bottom B of the gate will
move to the end of the cylindrical portion of the floor.

SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor.
We have
2
sin 30
4

Then
(3 ft) tan 30
SP
x
and
828 lb/ft 3 ft tan30°
1434.14 lb
SP SP
Fkx


Assume
4ftd
We have
11
()
22
PApAh
Then
3
I1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4) ft]
2
249.6( 4) lb
Pd
d
 



3
II1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4 4cos30 )]
2
249.6( 0.53590 ) lb
Pd
d
 


For
min
d so that the gate opens, 0W
Using the above free-body diagrams of the gate, we have

48
0: ft [249.6( 4) lb] ft [249.6( 0.53590) lb]
33
A
Md d
 
   
 
 

(3 ft)(1434.14 lb) 0
or
(332.8 1331.2) (665.6 356.70) 4302.4 0dd 
or
6.00 ftd
4ftd assumption correct 6.00 ftd 
www.elsolucionario.org

697. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.91
Solve Problem 5.90 if the gate weighs 1000 lb.
PROBLEM 5.90 A 42-ft gate is hinged at A and is held in
position by rod CD. End D rests against a spring whose constant is
828 lb/ft. The spring is undeformed when the gate is vertical.
Assuming that the force exerted by rod CD on the gate remains
horizontal, determine the minimum depth of water d for which the
bottom B of the gate will move to the end of the cylindrical portion
of the floor.

SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor.
We have
2
sin 30
4

Then
(3 ft) tan 30
SP
x
and
828 lb/ft 3 ft tan30°
1434.14 lb
SP SP
Fkx 

Assume
4ftd
We have
11
()
22
PApAh
Then
3
I1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4) ft]
2
249.6( 4) lb
Pd
d
 



3
II1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4 4cos30 )]
2
249.6( 0.53590 ) lb
Pd
d
 


For
min
d so that the gate opens,1000 lbW
Using the above free-body diagrams of the gate, we have

48
0: ft [249.6( 4) lb] ft [249.6( 0.53590) lb]
33
(3 ft)(1434.14 lb) (1 ft)(1000 lb) 0
A
Md d
 
   
 
 


or
(332.8 1331.2) (665.6 356.70) 4302.4 1000 0 dd
or
7.00 ftd
4ftd assumption correct 7.00 ftd 
www.elsolucionario.org

698. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.92
A prismatically shaped gate placed at the end of a freshwater channel is
supported by a pin and bracket at A and rests on a frictionless support at
B. The pin is located at a distance
0.10 mh below the center of gravity
C of the gate. Determine the depth of water d for which the gate will
open.

SOLUTION
First note that when the gate is about to open (clockwise rotation is impending),
y
B 0 and the line of
action of the resultant
P of the pressure forces passes through the pin at A. In addition, if it is assumed that
the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area.
Then


(0.25 )
3
d
ah 
and
28
(0.4)
3153
d
b





Now
8
15a
b


so that

3
82
3153
(0.25 )8
15(0.4)
d
d
h




Simplifying yields

289 70.6
15
45 12
dh
(1)


Alternative solution:
Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water
above the gate.
Now
2
2
11
(1m)( )
22
1
(N)
2
18
1m
215
4
(N)
15
PAp d gd
gd
WgVg dd
gd






 



www.elsolucionario.org

699. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.92 (Continued)

Then with
0
y
B (as explained above), we have

222184 1
0: (0.4) (0.25 ) 0
3 3 15 15 3 2
A
d
Mdgdhgd

    
    
     
   

Simplifying yields
289 70.6
15
45 12
dh

as above.
Find d: 0.10 mh

Substituting into Eq. (1),
289 70.6
15(0.10)
45 12
d
or 0.683 md 

www.elsolucionario.org

700. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.93
A prismatically shaped gate placed at the end of a freshwater channel is
supported by a pin and bracket at A and rests on a frictionless support at
B. Determine the distance h if the gate is to open when
0.75 m.d

SOLUTION
First note that when the gate is about to open (clockwise rotation is impending),
y
B 0 and the line of
action of the resultant
P of the pressure forces passes through the pin at A. In addition, if it is assumed that
the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area.
Then


(0.25 )
3
d
ah 
and
28
(0.4)
3153
d
b





Now
8
15
a
b



so that 
3
82
3153
(0.25 )8
15(0.4)
d
d
h




Simplifying yields

289 70.6
15
45 12
dh (1)


Alternative solution:
Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water
above the gate.
Now
2
2
11
(1m)( )
22
1
(N)
2
18
1m
215
4
(N)
15
PAp d gd
gd
WgVg dd
gd






 





www.elsolucionario.org

701. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.93 (Continued)

Then with
0
y
B (as explained above), we have

222184 1
0: (0.4) (0.25 ) 0
3 3 15 15 3 2
A
d
Mdgdhgd

    
    
     
   

Simplifying yields
289 70.6
15
45 12
dh

as above.
Find h: 0.75 md

Substituting into Eq. (1),
289 70.6
(0.75) 15
45 12
h
or 0.0711 mh 


www.elsolucionario.org

702. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.94
A long trough is supported by a continuous hinge along
its lower edge and by a series of horizontal cables
attached to its upper edge. Determine the tension in
each of the cables, at a time when the trough is
completely full of water
.

SOLUTION
Consider free body consisting of 20-in. length of the trough and water.

20-in.l length of free body

2
2
4
11 1
()
22 2
A
A
Wv rl
Pr
PPrl rrl rl








 


1
0: 0
3
A
MTrWrPr

  



2241 1
0
432 3
r
Tr r l r l r







22 2111
362
Trlrlrl
Data:
3 24 20
62.4 lb/ft ft 2 ft ft
12 12
rl

Then
32120
(62.4 lb/ft )(2 ft) ft
212
T






208.00 lb 208 lbT 

www.elsolucionario.org

703. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.95
The square gate AB is held in the position shown by hinges along its top edge
A and by a shear pin at B. For a depth of water
3.5dft, determine the force
exerted on the gate by the shear pin.

SOLUTION
First consider the force of the water on the gate. We have

1
2
1
()
2
PAp
Ah




Then
23
I
23
II1
(1.8 ft) (62.4 lb/ft )(1.7 ft)
2
171.850 lb
1
(1.8 ft) (62.4 lb/ft ) (1.7 1.8cos30 ) ft
2
329.43 lb
P
P





Now
III
12
0: 0
33
AABABABB
MLPLPLF

   


or
12
(171.850 lb) (329.43 lb) 0
33
B
F
or
276.90 lb
B
F 277 lb
B
F 30.0 

www.elsolucionario.org

704. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.96
Consider the composite body shown. Determine (a) the
value of
x when /2,hL (b) the ratio h/L for which
.xL

SOLUTION

V x xV
Rectangular
prism
Lab
1
2
L
21
2
Lab
Pyramid
1
32
b
ah




1
4
Lh
11
64
abh L h






Then
2
1
6
11
3
64
VabL h
xVabLhLh

 


 
  
 


Now
XVxV
so that
2211 1
3
66 4
XabL h ab L hL h
  
 
  
  

or
2
2
11 1
13
66 4
hhh
XL
LL L

   
 
(1)
(a)
?Xwhen
1
.
2
hL
Substituting
1
into Eq. (1),
2
h
L



2
11 1 1 11
13
62 6 2 42
XL
  
   
   

or
57
104
XL

0.548XL 




www.elsolucionario.org

705. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.96 (Continued)

(b)
?
h
L
 when .XL
Substituting into Eq. (1),
2
2
11 1
13
66 4
hhh
LL
LL L

   
 

or
2
2
111 1
1
62624
hhh
LL L


or
2
2
12
h
L

23
h
L
 

www.elsolucionario.org

706. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.97
Determine the location of the centroid of the composite body shown
when (a)
2,hb (b) 2.5 .hb

SOLUTION


V
x xV
Cylinder I
2
ab
1
2
b

221
2
ab

Cone II
21
3
ah

1
4
bh

211
34
ah b h







2
22 21
3
11 1
2312
Vabh
xV a b hb h







  



(a) For
2,hb
2215
(2 )
33
Vab b ab


 




22 2
22 2211 1
(2 ) (2 )
23 12
121 3
233 2
xV a b b b b
ab ab



  








22253 9
:
32 10
XV xV X a b a b X b


  



Centroid is
1
10
b to the left of base of cone. 


www.elsolucionario.org

707. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.97 (Continued)

(b) For
2.5 ,hb
221
(2.5 ) 1.8333
3
Vab b ab


 




22 2
22
2211 1
(2.5 ) (2.5 )
23 12
[0.5 0.8333 0.52083]
1.85416
xVab bb b
ab
ab


 
  
 
 




222
: (1.8333 ) 1.85416 1.01136XVxVX ab abX b  
Centroid is 0.01136 b to the right of base of cone.

Note: Centroid is at base of cone for
6 2.449 .hb b 

www.elsolucionario.org

708. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.98
The composite body shown is formed by removing a semiellipsoid
of revolution of semimajor axis h and semiminor axis a/2 from a
hemisphere of radius a. Determine (a) the y coordinate of the
centroid when h  a/2, (b) the ratio h/a for which
y  0.4a.

SOLUTION


V
y yV
Hemisphere
32
3
a
3
8
a
41
4
a

Semiellipsoid
2
2
21
32 6
a
hah






3
8
h

221
16
ah


Then
2
222
(4 )
6
(4 )
16
Vaah
yV a a h


 
 

Now YV yV
So that
2222
(4 ) (4 )
616
Yaah aah

 



or
2
3
44
8
hh
Ya
aa
 
 
   
 
   
 
(1)
(a) ?when
2
a
Yh 
Substituting
1
2h
a
 into Eq. (1)

2
13 1
44
28 2
Ya
 
 
   
 
   
 

or
45
112
Ya

0.402Ya 
www.elsolucionario.org

709. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.98 (Continued)

(b)
?when 0.4
h
Ya
a

Substituting into Eq. (1)

2
3
(0.4)4 4
8
hh
aa
aa
 
 
  
 
   
 

or
2
33.20.80
hh
aa
 

 
 
Then
2
3.2 ( 3.2) 4(3)(0.8)
2(3)
h
a
 


3.2 0.8
6

 or
22
and
53
hh
aa




www.elsolucionario.org

710. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.99
Locate the centroid of the frustum of a right circular cone when r 1 = 40 mm,
r
2 = 50 mm, and h = 60 mm.

SOLUTION


By similar triangles:

11
12
60
240 mm, therefore 240 60 300 mm
40 50
hh
hh

 

   
12 2 1
22 33
12
1300 1 240
75 mm 60 60 120 mm
44 4 4
50 300 250 10 40 240 128 10
33
yh y h
VV


 




3
mmV
mmy
4
mmyV
Triangle 1
3
250 10

75
6
18.75 10

Triangle 2
3
128 10

120
6
15.36 10



3
122 10


6
3.39 10



36
122 10 3.39 10
YV yV
Y  




27.787 mmY Centroid is 27.8 mm above base of cone. 
www.elsolucionario.org

711. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.100
For the machine element shown, locate the x coordinate of
the center of gravity.

SOLUTION


3
,mmV
,mmx ,mmz
4
,mmxV
4
,mmzV
I Rectangular plate
3
(10)(90)(38) 34.2 10

19 45 649.8  10
3
1539  10
3
II Half cylinder
23
(20) (10) 6.2832 10
2



46.5 20 292.17  10
3
125.664  10
3

III –(Cylinder)
23
(12) (10) 4.5239 10

38 20 171.908  10
3
90.478  10
3
IV Rectangular prism
3
(30)(10)(24) 7.2 10

5 78 36  10
3
561.6  10
3

V Triangular prism
31
(30)(9)(24) 3.24 10
2


13 78 42.12  10
3
252.72  10
3
 46.399  10
3
848.18  10
3
2388.5  10
3



Dimensions in mm www.elsolucionario.org

712. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.100 (Continued)


34
33
848.18 10 mm
46.399 10 mm



 
XV xV
xV
X
V

18.28 mmX 
www.elsolucionario.org

713. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.101
For the machine element shown, locate the z coordinate of
the center of gravity.

SOLUTION


3
,mmV
,mmx ,mmz
4
,mmxV
4
,mmzV
I Rectangular plate
3
(10)(90)(38) 34.2 10

19 45 649.8  10
3
1539  10
3
II Half cylinder
23
(20) (10) 6.2832 10
2



46.5 20 292.17  10
3
125.664  10
3

III –(Cylinder)
23
(12) (10) 4.5239 10

38 20 171.908  10
3
90.478  10
3
IV Rectangular prism
3
(30)(10)(24) 7.2 10 5 78 36  10
3
561.6  10
3

V Triangular prism
31
(30)(9)(24) 3.24 10
2


13 78 42.12  10
3
252.72  10
3
 46.399  10
3
848.18  10
3
2388.5  10
3



Dimensions in mm www.elsolucionario.org

714. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.101 (Continued)


34
33
2388.5 10 mm
46.399 10 mm



 
ZV zV
zV
Z
V

51.5 mmZ 


www.elsolucionario.org

715. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.102
For the machine element shown, locate the y coordinate of
the center of gravity.

SOLUTION
For half-cylindrical hole,


III
0.95 in.
4(0.95)
1.5
3
1.097 in.
r
y

 



For half-cylindrical plate,
IV
1.5 in.
4(1.5)
5.25 5.887 in.
3
r
z


 

3
,inV ,in.y ,in.z
4
,inyV
4
,inzV
I Rectangular plate (5.25)(3)(0.5) 7.875 0.25 2.625 1.9688

20.672
II Rectangular plate (3)(1.5)(0.75) 3.375 0.75 1.5 2.5313 5.0625
III –(Half cylinder)
2
(0.95) (0.75) 1.063
2

 1.097 1.5 1.1664 1.595
IV Half cylinder
2
(1.5) (0.5) 1.767
2

 0.25 5.887 0.4418 10.403
V –(Cylinder)
2
(0.95) (0.5) 1.418 0.25 5.25 0.3544 7.443
 10.536 0.6910 27.10




34
(10.536 in ) 0.6910 in
YV yV
Y

 0.0656 in.Y 
www.elsolucionario.org

716. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.103
For the machine element shown, locate the z coordinate of
the center of gravity.

SOLUTION
For half-cylindrical hole,


III
0.95 in.
4(0.95)
1.5
3
1.097 in.
r
y

 



For half-cylindrical plate,
IV
1.5 in.
4(1.5)
5.25 5.887 in.
3
r
z


 



3
,inV
,in.y ,in.z
4
,inyV
4
,inzV
I Rectangular plate (5.25)(3)(0.5) 7.875 0.25 2.625 1.9688

20.672
II Rectangular plate (3)(1.5)(0.75) 3.375 0.75 1.5 2.5313 5.0625
III –(Half cylinder)
2
(0.95) (0.75) 1.063
2

 1.097 1.5 1.1664 1.595
IV Half cylinder
2
(1.5) (0.5) 1.767
2

 0.25 5.887 0.4418 10.403
V –(Cylinder)
2
(0.95) (0.5) 1.418 0.25 5.25 0.3544 7.443
 10.536 0.6910 27.10

Now ZVzV

34
(10.536 in ) 27.10 inZ  2.57 in.Z 
www.elsolucionario.org

717. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.104
For the machine element shown, locate the y coordinate of the
center of gravity.

SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.



3
,mmV
,mmx ,mmy
4
,mmxV
4
,mmyV
I
3
(120)(100)(10) 120 10 5 -50
6
0.60 10
6
6.00 10
II
3
(120)(50)(10) 60 10 35 -5
6
2.10 10
6
0.30 10
III
23
(60) (10) 56.549 10
2



85.5 -5
6
4.8349 10
6
0.28274 10
IV
23
(40) (10) 50.266 10  60 5
6
3.0160 10
6
0.25133 10
V 
2 3
30 10 28.274 10 5 -50
6
0.141370 10

6
1.41370 10

3
258.54 10
6
10.4095 10
6
4.9177 10

We have YV yV

33 64
(258.54 10 mm ) 4.9177 10 mmY  or 19.02 mmY www.elsolucionario.org

718. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.105
For the machine element shown, locate the x coordinate of the
center of gravity.

SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.



3
,mmV
,mmx ,mmy
4
,mmxV
4
,mmyV
I
3
(120)(100)(10) 120 10 5 -50
6
0.60 10
6
6.00 10
II
3
(120)(50)(10) 60 10 35 -5
6
2.10 10
6
0.30 10
III
23
(60) (10) 56.549 10
2



85.5 -5
6
4.8349 10
6
0.28274 10
IV
23
(40) (10) 50.266 10  60 5
6
3.0160 10
6
0.25133 10
V 
2 3
30 10 28.274 10 5 -50
6
0.141370 10
6
1.41370 10

3
258.54 10
6
10.4095 10
6
4.9177 10

We have XV xV

33 64
(258.54 10 mm ) 10.4095 10 mmY  or 40.3 mmX 

www.elsolucionario.org

719. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.106
Locate the center of gravity of the sheet-metal form shown.

SOLUTION
First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide
with the centroid of the corresponding area. Now note that symmetry implies
125.0 mmX 


II
II
III
280
150
200.93 mm
280
50.930 mm
4125
230
3
283.05 mm
y
z
y














2
,mmA
,mmy ,mmz
3
,mmyA
3
,mmzA
I (250)(170) 42500 75 40 3187500 1700000
II (80)(250) 31416
2

 200.93 50930 6312400 1600000
III
2
(125) 24544
2

 283.05 0 6947200 0
 98460 16447100 3300000

We have
23
: (98460 mm ) 16447100 mmYA yA Y  or 167.0 mmY 

263
: (98460 mm ) 3.300 10 mmZA zA Z   or 33.5 mmZ 
www.elsolucionario.org

720. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.107
Locate the center of gravity of the sheet-metal form shown.

SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide
with the centroid of the corresponding area.


I
I
III
1
(1.2) 0.4 m
3
1
(3.6) 1.2 m
3
4(1.8) 2.4
m
3
y
z
x

 

 



2
,mA
,mx ,my ,mz
3
,mxA
3
,myA
3
,mzA
I
1
(3.6)(1.2) 2.16
2

1.5 0.4 1.2 3.24 0.864 2.592
II (3.6)(1.7) 6.12 0.75 0.4 1.8 4.59 2.448 11.016
III
2
(1.8) 5.0894
2


2.4

 0.8 1.8 3.888 4.0715 9.1609
 13.3694 3.942 5.6555 22.769

We have
23
: (13.3694 m ) 3.942 mXV xV X  or 0.295 mX 

23
: (13.3694 m ) 5.6555 mYV yV Y  or 0.423 mY 

23
: (13.3694 m ) 22.769 mZV zV Z  or 1.703 mZ 

www.elsolucionario.org

721. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.108
A corner reflector for tracking by radar has two sides in the
shape of a quarter circle with a radius of 15 in. and one side in
the shape of a triangle. Locate the center of gravity of the
reflector, knowing that it is made of sheet metal of uniform
thickness.

SOLUTION
By symmetry, XZ

For I and II (Quarter-circle ),
22 2
44(15)
6.3662 in.
33
(15) 1.76715 in
44
r
xy
Ar


  
 


2
,inA
,in.x ,in.x
3
,inxA
3
,inyA
I 176.715 6.3662 6.3662 1125.0 1125.0
II 176.715 0 6.3662 0 1125.0
III

21
15 112.50
2


5.0 0 562.50 0
 465.93 1687.50 2250.0

:XAxA
23
(465.93 in ) 1687.50 inX 
3.62 in.XZ 
:YA yA
23
(465.93 in ) 2250.0 inY 
4.83 in.Y 
www.elsolucionario.org

722. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.109
A wastebasket, designed to fit in the corner of a room, is 16 in.
high and has a base in the shape of a quarter circle of radius 10
in. Locate the center of gravity of the wastebasket, knowing
that it is made of sheet metal of uniform thickness.

SOLUTION
By symmetry, XZ
For III (Cylindrical surface),
2
22(10)
6.3662 in.
(10)(16) 251.33 in
22
r
x
Arh


 
 
For IV (Quarter-circle bottom),
22 2
44(10)
4.2441in.
33
(10) 78.540 in
44
r
x
Ar


 
 

2
,inA
,in.x ,in.x
3
,inxA
3
,inyA
I (10)(16) 160 5 8 800 1280
II (10)(16) 160 0 8 0 1280
III 251.33 6.3662 8 1600.0 2010.6
IV 78.540 4.2441 0 333.33 0
 649.87 2733.3 4570.6

:XA xA
23
(649.87 in ) 2733.3 inX 
4.2059 in.X 4.21in.XZ 
:YA yA
23
(649.87 in ) 4570.6 inY 
7.0331in.Y 7.03 in.Y 
www.elsolucionario.org

723. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.110
An elbow for the duct of a ventilating system is made of sheet
metal of uniform thickness. Locate the center of gravity of the
elbow.

SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with
the centroid of the corresponding area. Also, note that the shape of the duct implies

38.0 mmY 

II
II
II
IV IV
V
V
2
Note that 400 (400) 145.352 mm
2
400 (200) 272.68 mm
2
300 (200) 172.676 mm
4
400 (400) 230.23 mm
3
4
400 (200) 315.12 mm
3
4
300 (200) 215.12 mm
3
xz
x
z
xz
x
z






  
 
 
  
 
 

Also note that the corresponding top and bottom areas will contribute equally when determining
and .xz

Thus,

2
,mmA ,mmx ,mmz
3
,mmxA
3
,mmzA
I (400)(76) 47,752
2

 145.352 145.352 6,940,850 6,940,850
II (200)(76) 23,876
2

 272.68 172.676 6,510,510 4,122,810
III 100(76) 7600 200 350 1,520,000 2,660,000
IV
2
2 (400) 251,327
4




230.23 230.23 57,863,020 57,863,020
V
2
2 (200) 62,832
4




315.12 215.12 –19,799,620 –13,516,420
VI 2(100)(200) 40,000 300 350 –12,000,000 –14,000,000
 227,723 41,034,760 44,070,260
www.elsolucionario.org

724. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.110 (Continued)

We have
23
: (227,723 mm ) 41,034,760 mmXA xA X  or 180.2 mmX 

23
: (227,723 mm ) 44,070,260 mmZA zA Z  or 193.5 mmZ 

www.elsolucionario.org

725. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.111
A window awning is fabricated from sheet metal of uniform
thickness. Locate the center of gravity of the awning.

SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides
with the centroid of the corresponding area.

II VI
II VI
IV
IV
22
II VI
2
IV
(4)(25)
4 14.6103 in.
3
(4)(25) 100
in.
33
(2)(25)
4 19.9155 in.
(2)(25) 50
in.
(25) 490.87 in
4
(25)(34) 1335.18 in
2
yy
zz
y
z
AA
A






 
 
 

 




2
, inA
,in.y ,in.z
3
, inyA
3
,inzA
I (4)(25) 100 2 12.5 200 1250
II 490.87 14.6103
100
3

7171.8 5208.3
III (4)(34) 136 2 25 272 3400
IV 1335.18 19.9155
50

26,591 21,250
V (4)(25) 100 2 12.5 200 1250
VI 490.87 14.6103
100
3

7171.8 5208.3
 2652.9 41,607 37,567
www.elsolucionario.org

726. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.111 (Continued)

Now, symmetry implies
17.00 in.X 
and
23
: (2652.9 in ) 41,607 inYA yA Y  or 15.68 in.Y 

23
: (2652.9 in ) 37,567 inZA zA Z  or 14.16 in.Z 

www.elsolucionario.org

727. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.112
A mounting bracket for electronic components is formed
from sheet metal of uniform thickness. Locate the center of
gravity of the bracket.

SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram)


V
2
V
2
4(0.625)
2.25
3
1.98474 in.
(0.625)
2
0.61359 in
z
A









2
,inA
,in.x ,in.y ,in.z
3
,inxA
3
,inyA
3
,inzA
I (2.5)(6) 15 1.25 0 3 18.75 0 45
II (1.25)(6) 7.5 2.5 –0.625 3 18.75 –4.6875 22.5
III (0.75)(6) 4.5 2.875 –1.25 3 12.9375 –5.625 13.5
IV
5
(3) 3.75
4



1.0 0 3.75 3.75 0 –14.0625
V 0.61359 1.0 0 1.9847 40.61359 0 –1.21782
 22.6364 46.0739 10.3125 65.7197

We have XA xA

23
(22.6364 in ) 46.0739 inX  or 2.04 in.X 

23
(22.6364 in ) 10.3125 in
YA yA
Y

 or 0.456 in.Y 

23
(22.6364 in ) 65.7197 in
ZA zA
Z

 or 2.90 in.Z 
www.elsolucionario.org

728. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.113
A thin sheet of plastic of uniform thickness is bent to form
a desk organizer. Locate the center of gravity of the
organizer.

SOLUTION
First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide
with the centroid of the corresponding area. Now note that symmetry implies
30.0 mmZ 


2
4
8
10
24810
6
2
24810
2
6
26
6 2.1803 mm
26
36 39.820 mm
26
58 54.180 mm
26
133 136.820 mm
26
6 2.1803 mm
25
75 78.183 mm
6 60 565.49 mm
2
5 60 942.48 mm
x
x
x
x
yyyy
y
AAAA
A









 

 

 

 

  

 
 
 



www.elsolucionario.org

729. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.113 (Continued)


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 (74)(60) 4440 0 43 0 190,920
2 565.49 2.1803 2.1803 1233 1233
3 (30)(60) 1800 21 0 37,800 0
4 565.49 39.820 2.1803 22,518 1233
5 (69)(60) 4140 42 40.5 173,880 167,670
6 942.48 47 78.183 44,297 73,686
7 (69)(60) 4140 52 40.5 215,280 167,670
8 565.49 54.180 2.1803 30,638 1233
9 (75)(60) 4500 95.5 0 429,750 0
10 565.49 136.820 2.1803 77,370 1233
 22,224.44 1,032,766 604,878
We have
23
: (22,224.44 mm ) 1,032,766 mmXA xA X  or 46.5 mmX 

23
: (22,224.44 mm ) 604,878 mmYA yA Y  or 27.2 mmY 



www.elsolucionario.org

730. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.114
A thin steel wire of uniform cross section is bent into the shape
shown. Locate its center of gravity.

SOLUTION
First assume that the wire is homogeneous so that its center of gravity
will coincide with the centroid of the corresponding line.


22
22.4 4.8
mxz


 


,mL
,mx ,my ,mz
2
,mxL
2
,myL
2
,mzL
1 2.6 1.2 0.5 0 3.12 1.3 0
2 2.4 1.2
2
 
4.8

0
4.8

5.76 0 5.76
3 2.4 0 0 1.2 0 0 2.88
4 1.0 0 0.5 0 0 0.5 0
 9.7699 8.88 1.8 8.64
We have
2
: (9.7699 m) 8.88 m XL xL X or 0.909 mX 

2
: (9.7699 m) 1.8 m YL yL Y or 0.1842 mY 

2
: (9.7699 m) 8.64 m ZL zL Z or 0.884 mZ 

www.elsolucionario.org

731. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.115
The frame of a greenhouse is constructed from uniform aluminum
channels. Locate the center of gravity of the portion of the frame shown.

SOLUTION
First assume that the channels are homogeneous so that the center of gravity
of the frame will coincide with the centroid of the corresponding line.

89
89
23 6
ft
23
56.9099ft
xx
yy



 

 


,ftL
,ftx ,fty ,ftz
2
,ftxL
2
,ftyL
2
,ftzL
1 2 3 0 1 6 0 2
2 3 1.5 0 2 4.5 0 6
3 5 3 2.5 0 15 12.5 0
4 5 3 2.5 2 15 12.5 10
5 8 0 4 2 0 32 16
6 2 3 5 1 6 10 2
7 3 1.5 5 2 4.5 15 6
8 3 4.7124
2


6

6.9099 0 9 32.562 0
9 3 4.7124
2


6

6.9099 2 9 32.562 9.4248
10 2 0 8 1 0 16 2
 39.4248 69 163.124 53.4248
We have
2
: (39.4248 ft) 69 ftXL xL X  or 1.750 ftX 

2
: (39.4248 ft) 163.124 ftYL yL Y  or 4.14 ftY 

2
: (39.4248 ft) 53.4248 ftZL zL Z  or 1.355 ftZ 
www.elsolucionario.org

732. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.116
Locate the center of gravity of the figure shown, knowing that it is
made of thin brass rods of uniform diameter.

SOLUTION
Uniform rod:


22 2 2
(1 m) (0.6 m) (1.5 m)AB 

1.9 mAB


,mL
,mx ,my ,mz
2
,mxL
2
,myL ,mL
AB 1.9 0.5 0.75 0.3 0.95 1.425 0.57
BD 0.6 1.0 0 0.3 0.60 0 0.18
DO 1.0 0.5 0 0 0.50 0 0
OA 1.5 0 0.75 0 0 1.125 0
 5.0 2.05 2.550 0.75


2
:(5.0m)2.05mXL xL X  0.410 mX 

2
:(5.0m)2.55mYL yL Y  0.510 mY 

2
:(5.0m)0.75mZL zL Z  0.1500 mZ 

www.elsolucionario.org

733. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.117
Locate the center of gravity of the figure shown, knowing that it is
made of thin brass rods of uniform diameter.

SOLUTION
By symmetry, 0X 

L, in. ,in.y ,in.z
2
,inyL
2
,inzL
AB
22
30 16 34 15 0 510 0
AD
22
30 16 34 15 8 510 272
AE
22
30 16 34 15 0 510 0
BDE (16) 50.265 0
2(16)
10.186

 0 512
 152.265 1530 784


2
: (152.265 in.) 1530 inYL yL Y 
10.048 in.Y 10.05 in.Y 

2
: (152.265 in.) 784 inZL zL Z  
 5.149 in.Z  5.15 in.Z 

www.elsolucionario.org

734. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.118
A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is
1030 kg/m
3
and of steel is
3
7860 kg/m ,locate the center of gravity of the awl.


SOLUTION
First, note that symmetry implies 0YZ 


I
33
I
3
II
32
II
3
III
32
III
5
(12.5 mm) 7.8125 mm
8
2
(1030 kg/m ) (0.0125 m)
3
4.2133 10 kg
52.5 mm
(1030 kg/m ) (0.025 m) (0.08 m)
4
40.448 10 kg
92.5 mm 25 mm 67.5 mm
(1030 kg/m ) (0.0035 m) (0.
4
x
W
x
W
x
W






















3
05 m)
0.49549 10 kg

 


IV
3223
IV
V
32 3
V
182.5 mm 70 mm 112.5 mm
(7860 kg/m ) (0.0035 m) (0.14 m) 10.5871 10 kg
4
1
182.5 mm (10 mm) 185 mm
4
(7860 kg/m ) (0.00175 m) (0.01 m) 0.25207 10 kg
3
x
W
x
W









 








www.elsolucionario.org

735. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.118 (Continued)

, kgW
, mmx , kg mmxW 
I
3
4.123 10

 7.8125
3
32.916 10


II
3
40.948 10

 52.5
3
2123.5 10


III
3
0.49549 10

 67.5
3
33.447 10


IV
3
10.5871 10

 112.5
3
1191.05 10


V
3
0.25207 10

 185
3
46.633 10



3
55.005 10


3
3360.7 10



We have
33
: (55.005 10 kg) 3360.7 10 kg mmXW xW X

    
or 61.1 mmX 
(from the end of the handle)


www.elsolucionario.org

736. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.119
A bronze bushing is mounted inside a steel sleeve. Knowing that the
specific weight of bronze is 0.318 lb/in
3
and of steel is 0.284 lb/in
3
,
determine the location of the center of gravity of the assembly.

SOLUTION
First, note that symmetry implies 0XZ 

Now
()WgV

3222
II
3222
II II
32
III III
0.20 in. (0.284 lb/in ) [(1.8 0.75 ) in ](0.4 in.) 0.23889 lb
4
0.90 in. (0.284 lb/in ) [(1.125 0.75 ) in ](1in.) 0.156834 lb
4
0.70 in. (0.318 lb/in ) [(0.75 0
4
yW
yW
yW



   


   


 


22
.5 ) in ](1.4 in.) 0.109269 lb




We have
(0.20 in.)(0.23889 lb) (0.90 in.)(0.156834 lb) (0.70 in.)(0.109269 lb)
0.23889 lb 0.156834 lb 0.109269 lb
YW yW
Y



or
0.526 in.Y 
(above base)

www.elsolucionario.org

737. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.120
A brass collar, of length 2.5 in., is mounted on an aluminum rod
of length 4 in. Locate the center of gravity of the composite body.
(Specific weights: brass  0.306 lb/in
3
, aluminum  0.101 lb/in
3
)

SOLUTION

Aluminum rod:
32
(0.101lb/in ) (1.6 in.) (4 in.)
4
0.81229 lb
WV








Brass collar:
322
(0.306 lb/in. ) [(3 in.) (1.6 in.) ](2.5 in.)
4
3.8693 lb
WV




Component W(lb) (in.)y (lb in.)yW
Rod 0.81229 2 1.62458
Collar 3.8693 1.25 4.8366
 4.6816 6.4612

: (4.6816 lb) 6.4612 lb in.YW yW Y  
1.38013 in.Y 1.380 in.Y 
www.elsolucionario.org

738. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.121
The three legs of a small glass-topped table are equally spaced and are
made of steel tubing, which has an outside diameter of 24 mm and a
cross-sectional area of
2
150 mm . The diameter and the thickness of the
table top are 600 mm and 10 mm, respectively. Knowing that the
density of steel is
3
7860 kg/m and of glass is
3
2190 kg/m , locate the
center of gravity of the table.

SOLUTION
First note that symmetry implies 0XZ 
Also, to account for the three legs, the masses of components I and II will each bex
multiplied by three

I
2180
12 180
77.408 mm
y


 


362
II
7860 kg/m (150 10 m ) (0.180 m)
2
0.33335 kg
ST
mV



  


II
2280
12 180
370.25 mm
y


 


362
II II
7860 kg/m (150 10 m ) (0.280 m)
2
0.51855 kg
ST
mV



  


III
24 180 280 5
489 mm
y  

32
III III
2190 kg/m (0.6 m) (0.010 m)
4
6.1921 kg
GL
mV


  


, kgm
, mmy , kg mmym
I 3(0.33335) 77.408 77.412
II 3(0.51855) 370.25 515.98
III 6.1921 489 3027.9
 8.7478 3681.3
We have : (8.7478 kg) 3681.3 kg mmYm ym Y  
or 420.8 mmY
The center of gravity is 421 mm

(above the floor)


www.elsolucionario.org

739. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.122
Determine by direct integration the values of
x for the two volumes obtained by passing a vertical
cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given
shape and divides the shape into two volumes of equal height.
A hemisphere

SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then

2
,
EL
dV r dx x x
The equation of the generating curve is
222
xya so that
222
rax and then

22
()dV a x dx
Component 1:
/2
3
/2
22 2
1
0
0
3
()
3
11
24
a
a
x
Vaxdxax
a



 




and
/2
22
10
/2
24
2
0
4
()
24
7
64
a
EL
a
xdV x a xdx
xx
a
a 









Now

34
11 1
111 7
:
24 64
EL
xV x dV x a a 






1
21
or
88
xa

Component 2:


3
22 2
2
/2
/2
3
3
222
3
()
3
()
323
5
24
a
a
a
a
a
x
Vaxdxax
aa
aa a
a




 

 
 

 






www.elsolucionario.org

740. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.122 (Continued)

and 
24
22 2
2/2
/2
24
24
2222
4
()
24
() ()
24 24
9
64
a
a
EL
a
a
aa
xx
xdV x a xdx a
aa
aa
a



 
  

 
  
 
 
 
 
 
 



Now

34
22 2
259
:
24 64
EL
xVxdVx a a  






2
27
or
40
x a 

www.elsolucionario.org

741. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.123
Determine by direct integration the values of
x for the two volumes obtained by passing a vertical
cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given
shape and divides the shape into two volumes of equal height.
A semiellipsoid of revolution

SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then

2
,
EL
dV r dx x x
The equation of the generating curve is
22
22
1
xy
ha
 so that

2
222
2
()
a
rhx
h

and then

2
22
2
()
a
dV h x dx
h
Component 1:
/2
223
/2
22 2
1 22
0
0
2
()
3
11
24
h
h
aax
Vhxdxhx
hh
ah



 




and
2
/2
22
2
10
/2
224
2
2
0
22
()
24
7
64
h
EL
h a
xdV x h xdx
h
axx
h
h
ah




 







Now
222
11 1
111 7
:
24 64
EL
xV x dV x ah ah






1
21
or
88
xh

Component 2:


223
22 2
2 22
/2
/2
3
23
222
2
2
()
3
()
323
5
24
h
h
h
h
h
aax
Vhxdxhx
hh
ahh
hh h
h
ah




 

 
 

 





www.elsolucionario.org

742. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.123 (Continued)

and
2
22
2
2/2
224
2
2
/2
()
24
h
EL
h
h
h a
xdV x h x dx
h
axx
h
h 

 
 
 







24
224
2222
2
22
() ()
24 24
9
64
hh
ahh
hh
h
ah


  
 
 
 
 
 
 


Now

222
22 2
259
:
24 64
EL
xVxdVx ah ah 






2
27
or
40
x h 

www.elsolucionario.org

743. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.124
Determine by direct integration the values of
x for the two volumes obtained by passing a vertical
cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given
shape and divides the shape into two volumes of equal height.
A paraboloid of revolution

SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then

2
,
EL
dV r dx x x
The equation of the generating curve is
2
2h
xh y
a

so that
2
2
().
a
rhx
h

and then

2
()
a
dV h x dx
h
Component 1:
2
/2
1
0
/2
22
0
2
()
2
3
8
h
ha
Vhxdx
h
ax
hx
h
ah









and
2
/2
10
/2
223
22
0
()
1
23 12
h
EL
h a
xdV x h xdx
h
axx
hah
h



 






Now
222
11 1
131
:
812
EL
xV x dV x ah ah






1
2
or
9
xh


Component 2:


222
2
/2
/2
2
22
2
2
()
2
()
()
222
1
8
h
h
h
h
h
aax
Vhxdxhx
hh
ahh
hh h
h
ah




 

 
 

 







www.elsolucionario.org

744. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.124 (Continued)

and 
2223
2/2
/2
23
223
22
22
()
23
() ()
23 23
1
12
h
h
EL
h
h
hh
aaxx
xdV x h xdx h
hh
ahh
hh
h
ah



  
  
  
 
 

 

 




Now

222
22 2
211
:
812
EL
xVxdVxah ah 






2
2
or
3
x h 

www.elsolucionario.org

745. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.125
Locate the centroid of the volume obtained by rotating the shaded area
about the x-axis.

SOLUTION
First, note that symmetry implies 0y 
0z 
Choose as the element of volume a disk of radius r and thickness dx.
Then
2
,
EL
dV r dx x x
Now
1
1r
x

so that
2
2
1
1
21
1
dV dx
x
dx
xx











Then
3
3
2
1
1
3
21 1
12ln
11
32ln3 12ln1
31
(0.46944 ) m
Vdxxx
xxx



 

 
 







and
3
2
3
2
1
1
23
21
12ln
2
31
2(3) ln 3 2(1) ln1
22
(1.09861 ) m
EL
x
xdV x dx x x
xx




 
 






Now
34
: (0.46944 m ) 1.09861 m
EL
xV x dV x 

or 2.34 mx 

www.elsolucionario.org

746. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.126
Locate the centroid of the volume obtained by rotating the shaded area
about the x-axis.

SOLUTION
First note that symmetry implies 0y 
and 0z 
We have
2
()ykXh
At
0, ,xya
2
()akh
or
2
a
k
h


Choose as the element of volume a disk of radius r and thickness dx. Then

2
,
EL
dV r dx X x
Now
2
2
()
a
rxh
h

so that

2
4
4
()
a
dV x h dx
h
Then
22
45
044
0
2
() [()]
5
1
5
h
haa
Vxhdxxh
hh
ah 






and
2
4
4
0
2
5 4 23 32 4
4
0
2
6 5 24 33 42
4
0
22
()
(4 6 4 )
14 3 4 1
65 2 3 2
1
30
h
EL
h
h a
xdV x xhdx
h
a
xhxhxhxhxdx
h
a
xhxhxhxhx
h
ah
















Now
222
:
530
EL
xV x dV x ah ah






1
or
6
xh

www.elsolucionario.org

747. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.127
Locate the centroid of the volume obtained by rotating the shaded
area about the line
.xh

SOLUTION
First, note that symmetry implies xh 
0z 
Choose as the element of volume a disk of radius r and thickness dx. Then

2
,
EL
dV r dy y y
Now
2
222
2
()
h
xay
a
 so that
22
.
h
rh a y
a
 
Then  
2 2
22
2
h
dV a a y dy
a

and

 
2 2
22
2
0
a
h
Vaaydy
a



Let sin cosya dya d
Then
 
2 2/2
222
2
0
2
/2
2222
2
0
/2
222
0
sin cos
2(cos ) ( sin ) cos
(2cos 2cos sin cos )
h
Vaaaad
a
h
aaa aa a d
a
ah d














/2
23
0
2 2
2
sin 2 1
2sin 2 sin
24 3
1
22
23
0.095870
ah
ah
ah



 

 

 











www.elsolucionario.org

748. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.127 (Continued)

and

 
 
2 2
22
2
0
2
2223
2
0
22
a
EL
a h
ydV y a a y dy
a
h
ay ay a y y dy
a










2
22 2 23/2 4
2
0
2
22 4 23/2
2
22
21
()
34
12
() ( )
43
1
12
a
h
ay aa y y
a
h
aa a aa
a
ah



 

 
 
   
   
  


Now
:
EL
yVydV


2221
(0.095870 )
12
yahah
 or
0.869ya 


www.elsolucionario.org

749. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.128*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the x-axis.

SOLUTION
First, note that symmetry implies 0y 
0z 
Choose as the element of volume a disk of radius r and thickness dx.
Then
2
,
EL
dV r dx x x
Now sin
2
x
rb
a


so that
22
sin
2
x
dV b dx
a



Then

2
22
2
2 2
22
22
2
sin
2
sin
22
1
2
a
a
a
x
a
a
a
aa x
Vb dx
a
x
b
b
ab
















and
2
22
sin
2
a
EL
a x
xdV x b dx
a






Use integration by parts with

2
2
sin
2
sin
2
x
a
a
x
ux dV
a
x
du dx V









www.elsolucionario.org

750. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.128* (Continued)

Then
2
2
2
22
2
2
22
2
sin sin
22
21
2cos
224 2
a
xx
a
aa
EL
a
aa
a
a
a
xx
xdV b x dx
aa a x
ba a x
a












 
  
  
   




22
22 2 2
22
22
2
22
31 1
(2 ) ( )
24 4 22
31
4
0.64868
aa
ba a a
ab
ab





   
  
  






Now
2221
: 0.64868
2
EL
xVxdVx ab ab





or
1.297x a 

www.elsolucionario.org

751. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.129*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the y-axis. (Hint: Use a thin
cylindrical shell of radius r and thickness dr as the element of
volume.)

SOLUTION
First note that symmetry implies 0x 
0z
Choose as the element of volume a cylindrical shell of radius r and thickness dr.
Then
1
(2 )( )( ),
2
EL
dV r y dr y y
Now
sin
2
r
yb
a


so that
2sin
2
r
dV br dr
a



Then
2
2sin
2
a
a r
Vbrdr
a




Use integration by parts with

sin
2
2
cos
2
r
urd dv dr
a
ar
du dr v
a






Then

2
2
2
2
2
22
2() cos cos
22
24
2()()sin21
2
a
a
a
a
a
a
ar ar
Vbr dr
aa
aar
b a
a




 


 



  





22
2
2
2
44
2
1
81
5.4535
aa
Vb
ab
ab














www.elsolucionario.org

752. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.129* (Continued)

Also
2
1
2
2
22
sin 2 sin
22
sin
2
a
EL
a
a
a rr
ydV b br dr
aa
r
br dr
a


 

 
 




Use integration by parts with

2
2
sin
2
sin
2
r
a
a
r
ur dv dr
a
r
du dr v






Then
2
2
2
22
2
22
2
2
sin sin
()
22
2
(2 ) ( ) cos
224 2
a
rr
a
aa
EL
a
aa
a
a
a
rr
ydV b r dr
aarar
ba a
a












 
  
  
   




22 22
22
22
22
2
22
3(2) ()
24 4 22
31
4
2.0379
aaaa
ba
ab
ab




   
  
  






Now
222
: (5.4535 ) 2.0379
EL
yVydVy ab ab

or 0.374yb 

www.elsolucionario.org

753. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.130*
Show that for a regular pyramid of height h and n sides (3,4,)n the centroid of the volume of the
pyramid is located at a distance h/4 above the base.

SOLUTION
Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the
base of the pyramid is given by

2
base
Akb
where
();kkN see note below. Using similar triangles, we have

shy
bh



or
()
b
shy
h

Then
2
22
slice 2
()
b
dV A dy ks dy k h y dy
h

and
22
23
22
0
0
2
1
() ()
3
1
3
h
h
bb
Vkhydyk hy
hh
kb h







Also,
EL
yy
so that
22
2223
22
00
2
22 3 4 22
2
0
() ( 2 )
121 1
23412
hh
EL
h bb
ydV yk h ydy k hy hy ydy
hh
b
khyhyy kbh
h







 

Now

22211
:
312
EL
yV y dV y kb h kb h





or
1
Q.E.D.
4
yh

Note:
2
base
tan
2
21
2
4tan
()
N
b
N
ANb
N
b
kNb









www.elsolucionario.org

754. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.131
Determine by direct integration the location of the centroid of one-half of
a thin, uniform hemispherical shell of radius R.

SOLUTION
First note that symmetry implies 0x
The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the
xy plane. Now

()( )
2
EL
dA r Rd
r
y



where
sinrR
so that
2
sin
2
sin
EL
dA R d
R
y



Then
/2
22/2
0
0
2
sin [ cos ]ARdR
R


 




and
/2
2
0
/2
3
0
32
sin ( sin )
sin 2
2
24
2
EL
R
ydA R d
R
R


 








 





Now
23
:( )
2
EL
yA y dA y R R




or
1
2
yR 
Symmetry implies
zy
1
2
zR


www.elsolucionario.org

755. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.132
The sides and the base of a punch bowl are of
uniform thickness t. If
tR
 and R  250 mm,
determine the location of the center of gravity of
(a) the bowl, (b) the punch.

SOLUTION
(a) Bowl:
First note that symmetry implies 0x 
0z 
for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the
center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl,
an element of area is obtained by rotating the arc ds about the y-axis. Then

wall
(2 sin )( )dA R R d
and
wall
() cos
EL
yR 
Then
/2
2
wall
/6
/22
/6
2
2sin
2[cos]
3
ARd
R
R











and
wall wall wall
/2
2
/6
/232
/6
3
()
(cos)(2 sin )
[cos ]
3
4
EL
yA y dA
RRd
R
R




 









By observation,
2
base base 3
,
42
ARy R

Now
yA yA
or
22 32 33
3
4442
yRR RR R
 
  
 

or
0.48763 250 mmyRR 121.9 mmy 




www.elsolucionario.org

756. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.132 (Continued)

(b) Punch:
First note that symmetry implies 0x 
0z 
and that because the punch is homogeneous, its center of gravity will coincide with the centroid of
the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy.
Then

2
,
EL
dV x dy y y
Now
22 2
xyR
so that
22
()dV R y dy
Then
0
22
3/2
0
23
3/2
3
23
()
1
3
313 3
3
232 8
R
R
VRydy
Ry y
RR R R











    






and

0
22
3/2
0
22 4
3/2
24
24
()
11
24
1313 15
22 42 64
EL
R
R
ydV y R y dy
Ry y
RR R R 












    






Now
34315
:3
864
EL
yV y dV y R R






or
5
250 mm
83
yRR 

90.2 mmy 

www.elsolucionario.org

757. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.133
Locate the centroid of the section shown, which was cut from a thin circular
pipe by two oblique planes.

SOLUTION
First note that symmetry implies 0x 
Assume that the pipe has a uniform wall thickness t and choose as the element of volume a vertical strip
of width ad
 and height
21
().yy Then


21 12
1
(), ()
2
EL EL
dV y y ta d y y y z z   
Now
3
1
26

h
h
yz
a

2
3
2
2
23
 
h
yzh
a

()
6

h
za
a
(2)
3

h
za
a
and
cosza
Then
21
()(cos2)(cos)
36
(1 cos )
2
hh
yy a a a a
aa
h 
    

and
12
()(cos)(cos2)
63
(5 cos )
6
(1 cos) (5 cos), cos
212
EL EL
hh
yy a a a a
aa
h
aht h
dV d y z a


    

  
www.elsolucionario.org

758. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.133 (Continued)

Then
0
0
2(1cos) [sin]
2
aht
V d aht
aht







and
0
2
2
0
2
0
2
2 (5 cos ) (1 cos )
12 2
(5 6cos cos )
12
sin 2
56sin
12 2 4
11
24
EL
haht
ydV d
ah t
d
ah t
ah t



 




 
 
 
 










0
2
0
2
2cos (1cos)
2
sin 2
sin
24
1
2
EL
aht
zdV a d
aht
aht


 



 

 
 







Now
211
:( )
24
EL
yV y dV y aht ah t

or
11
24
yh

and
21
:( )
2


EL
zV z dV z aht a ht or
1
2
za



www.elsolucionario.org

759. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.134*
Locate the centroid of the section shown, which was cut from an
elliptical cylinder by an oblique plane.

SOLUTION
First note that symmetry implies 0x 

Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then

1
2, ,
24
EL EL
dV xy dz y z z
Now
22a
xbz
b


and
/2
()
22
   
hhh
yz bz
bb

Then
22
2()
2



 


b
bah
Vbzbzdz
bb

Let
sin coszb dzb d
Then
/2
2
/2
/2
22
/2
/2
3
/2
( cos )[ (1 sin )] cos
(cos sin cos )
sin 2 1
cos
243
1
2
ah
Vbbbd
b
abh d
abh
Vabh























www.elsolucionario.org

760. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.134* (Continued)

and
22
2
22 2
31
()2 ()
22 2
1
()
4


 
     
 


b
EL
b
b
b hah
ydV bz b z bzdz
bb b
ah
bz b zdz
b

Let sin coszb dzb d
 
Then
2
/2
2
3
/2
/2
22 222
/2
1
[(1 sin )]( cos ) ( cos )
4
1
(cos 2sin cos sin cos )
4





  


 



EL
ah
ydV b b b d
b
abh d

Now
2211
sin (1 cos 2 ) cos (1 cos 2 )
22
   
so that
22 2 1
sin cos (1 cos 2 )
4

Then
/2
22 2 2
/2
/2
23
/2
211
cos 2sin cos (1 cos 2 )
44
1sin2111sin4
cos
4243 4428
5
32
EL
y dV abh d
abh
abh




 
 








 

 
 



Also,
22
22
2
2()
2
()


 
 



b
EL
b
b
b ah
zdV z a z bzdz
bb
ah
zb z b z dz
b

Let sin coszb dzb d
 
Then
/2
2
/2
/2
2222
/2
( sin )[ (1 sin )]( cos ) ( cos )
(sin cos sin cos )
EL
ah
zdV b b b b d
b
ab h d




  
 







Using
22 2 1
sin cos (1 cos 2 )
4
 from above,

/2
222
/2 1
sin cos (1 cos 2 )
4
EL
zdV abh d




 

 
 



/2
23 2
/2
111sin4 1
cos
34428 8
ab h ab h



 

 
  
 


www.elsolucionario.org

761. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.134* (Continued)

Now
215
:
232
EL
yV y dV y abh abh





or
5
16
yh

and
211
:
28
EL
z V z dV z abh ab h





or
1
4
zb



www.elsolucionario.org

762. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.135
Determine by direct integration the location of the centroid
of the volume between the xz plane and the portion shown
of the surface y  16h(ax  x
2
)(bz  z
2
)/a
2
b
2
.

SOLUTION

First note that symmetry implies
2
a
x


2
b
z

Choose as the element of volume a filament of base
dx dz and height y. Then

1
,
2
EL
dV y dx dz y y
or
22
2216
()()
h
dV ax x bz z dx dz
ab


Then
22
22
0016
()()
ba h
Vaxxbzzdxdz
ab




223
22
0
0
2323
22
0
23
216 1
()
3
16 1 1
() ()
2323
81
() ()
233
4
9
a
b
b
ha
Vbzzxxdz
zab
ha b
aazz
ab
ah b
bb
b
abh

















www.elsolucionario.org

763. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.135 (Continued)

and
22 22
22 22
00
2
22 3 4 22 3 4
44
00
22
22 3 4 3 4 5
24
0
0116 16
()() ()()
2
128
(2 )(2 )
128 1
(2 )
325
ba
EL
ba
a
b hh
y dV ax x bz z ax x bz z dx dz
ab ab
h
a x ax x b z bz z dx dz
ab
haa
bz bz z x x x dz
ab
  

  
  


 






22 2
345345
44
0
23
345 2
4
128 1 1
() () ()
3253 5
64 1 32
() () ()
3 2 5 22515
b
ha a b b
aaa zzz
zab
ah b b
b b b abh
b
  
  
  




Now

2432
:
9225
EL
yV y dV y abh abh




 or
8
25
yh


www.elsolucionario.org

764. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.136
After grading a lot, a builder places four stakes to
designate the corners of the slab for a house. To provide a
firm, level base for the slab, the builder places a minimum
of 3 in. of gravel beneath the slab. Determine the volume
of gravel needed and the x coordinate of the centroid of
the volume of the gravel. (Hint: The bottom surface of
the gravel is an oblique plane, which can be represented
by the equation y  a  bx  cz.)

SOLUTION
The centroid can be found by integration. The equation for the bottom of the gravel is,yabxcz 
where the constants a, b, and c can be determined as follows:
For
0x and 0,z 3 in., and therefore,y

31
ft , or ft
12 4
aa 

For
30 ft and 0, 5 in.,xzy and therefore,

51 1
ft ft (30 ft), or
12 4 180
bb 

For
0and 50ft, 6in.,xz y  and therefore,

61 1
ft ft (50 ft), or
12 4 200
cc 

Therefore,
11 1
ft
4 180 200
yxz  

Now
EL
xdV
x
V



A volume element can be chosen as

||dV y dx dz
or
111
1
44550
dV x z dx dz

 



and
EL
xx



www.elsolucionario.org

765. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.136 (Continued)

Then
50 30
00 11
1
44550
EL
x
xdV x z dx dz







30
2
50
32
0
0
50
0
50
2
0
4
11
4 2 135 100
1
(650 9 )
4
19
650
42
10937.5 ft
xz
x xdz
zdz
zz












The volume is
50 30
00111
1
44550
VdV x zdxdz







30
50
2
0
0
50
0
50
2
0
3
11
49050
13
40
45
13
40
410
687.50 ft
z
x xxdz
zdz
zz
















Then
4
3
10937.5ft
15.9091 ft
687.5 ftEL
xdV
x
V
 


Therefore,
3
688 ftV 

15.91ftx 


www.elsolucionario.org

766. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.137
Locate the centroid of the plane area shown.

SOLUTION


2
,inA
,inx ,iny
3
,inxA
3
,inyA
1
1
(12)(6) 36
2

4 4 144 144
2 (6)(3) 18 9 7.5 162 135
 54 306 279

Then XAxA
(54) 306X  5.67 in.X 
(54) 279
YA yA
Y

 5.17 in.Y 

www.elsolucionario.org

767. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.138
Locate the centroid of the plane area shown.

SOLUTION


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1
1
(120)(75) 4500
2

80 25
3
360 10
3
112.5 10
2 (75)(75) 5625 157.5 37.5
3
885.94 10
3
210.94 10
3
2
(75) 4417.9
4

 163.169 43.169
3
720.86 10
3
190.716 10
 5707.1
3
525.08 10
3
132.724 10

Then
3
(5707.1) 525.08 10XA xA X   92.0 mmX 

3
(5707.1) 132.724 10YA yA Y   23.3 mmY 

www.elsolucionario.org

768. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.139
A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C
and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION
For quarter circle,
2r
r


(a)
2
0: 0
C
r
MWTr


 




22
(8 lb)TW

 

 
 
5.09 lbT 
(b)
0: 0 5.09 lb 0
xx x
FTC C   5.09 lb
x
C

0: 0 8 lb 0
yy y
FCW C    8 lb
y
C


9.48 lbC 57.5° 

www.elsolucionario.org

769. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.140
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.

SOLUTION

3
(1 )yh kx
For
,0.xay

3
0(1 )hka

3
1
k
a



3
3
1
x
yh
a





1
,
2
EL EL
xxy ydAydx 

34
33
00
0
3
1
44
a
aa
xx
AdA ydx h dxhx ah
aa
 
      

 
 


425
2
33
00
0
3
2105
a
aa
EL
xxx
xdA xydx hx dx h ah
aa
 
  

 



32 36
2
336
00 0
11 2
11
22 2
aa a
EL xh xx
y dA y ydx h dx dx
aaa
  
      
   
  


247
2
36
0
9
22827
a
hxx
xah
aa

 



233
:
410
EL
xA x dA x ah a h






2
5
xa


239
:
428
EL
yA y dA y ah ah






3
7
yh


www.elsolucionario.org

770. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.141
Determine by direct integration the centroid of the area shown.

SOLUTION


1
2
EL EL
xxy ydAydx 

223
22
0
0
223
22
00
234
2
2
0
25
12
23 6
12 2
12 1
23 4 3
L
L
LL
EL
L
xx x x
A dA h dx h x hL
LLLL
xx x x
xdA xh dx h x dx
LLLL
xxx
hhL
L L
 
    

 
 
  
 
 

 


 


251
:
63





EL
xA x dA x hL hL
2
5
xL


2
2
51
12
62

 


EL
xx
AhLy yyh
LL


2
22
2
2
0
224 23
24 23
0
235234
2
24 23
0
1
12
22
14244
2
44 4
21035 3
L
EL
L
Lhxx
ydA ydx dx
LL
hxxxxx
dx
LLL LL
hxxxxx
xhL
LLL LL

 













254
:
610
EL
yA y dA y hL h L






12
25
yh

www.elsolucionario.org

771. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.142
The escutcheon (a decorative plate placed on a pipe where
the pipe exits from a wall) shown is cast from brass.
Knowing that the density of brass is 8470 kg/m
3
,
determine the mass of the escutcheon.

SOLUTION
The mass of the escutcheon is given by (density) ,mV where V is the volume. V can be generated by
rotating the area A about the x-axis.
From the figure:
22
1
2
75 12.5 73.9510 m
37.5
76.8864 mm
tan 26
L
L




21
1
2.9324 mm
12.5
sin 9.5941
75
26 9.5941
8.2030 0.143168 rad
2
aL L





 


Area A can be obtained by combining the following four areas:

Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have

22VyA yA
Seg.
2
, mmA
, mmy
3
,mmyA
1
1
(76.886)(37.5) 1441.61
2


1
(37.5) 12.5
3

18,020.1
2
2
(75) 805.32
2(75)sin
sin ( ) 15.2303
3 


 12,265.3
3
1
(73.951)(12.5) 462.19
2


1
(12.5) 4.1667
3

1925.81
4 (2.9354)(12.5) 36.693
1
(12.5) 6.25
2


229.33
 3599.7

www.elsolucionario.org

772. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 5.142 (Continued)

Then
3
3
363
2
2 (3599.7 mm )
22,618 mm
(density)
(8470 kg/m )(22.618 10 m )
VyA
mV








0.191574 kg or 0.1916 kgm


www.elsolucionario.org

773. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.143
Determine the reactions at the beam supports for the given loading.

SOLUTION

I
I
II
II
(200 lb/ft)(15 ft)
3000 lb
1
(200 lb/ft)(6 ft)
2
600 lb
R
R
R
R






0: (3000 lb)(1.5 ft) (600 lb)(9 ft 2ft) (15 ft) 0
A
MB     


740 lbB

740 lbB



0: 740 lb 3000 lb 600 lb 0
y
FA    


2860 lbA

2860 lbA


www.elsolucionario.org

774. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.144
A beam is subjected to a linearly distributed downward load
and rests on two wide supports BC and DE, which exert
uniformly distributed upward loads as shown. Determine
the values of w
BC and w DE corresponding to equilibrium
when
600
A
w N/m.

SOLUTION

We have
I
II
1
(6 m)(600 N/m) 1800 N
2
1
(6 m)(1200 N/m) 3600 N
2
(0.8 m)( N/m) (0.8 ) N
(1.0 m) ( N/m) ( ) N
BC BC BC
DE DE DE
R
R
Rw w
Rww





Then
0: (1 m)(1800 N) (3 m)(3600 N) (4 m)( N) 0
GDE
Mw    

or
3150 N/m
DE
w 
and
0: (0.8 ) N 1800 N 3600 N 3150 N 0
yBC
Fw    
or
2812.5 N/m
BC
w 2810 N/m
BC
w 


www.elsolucionario.org

775. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.145
A tank is divided into two sections by a 1  1-m square gate that is
hinged at A. A couple of magnitude 490 N · m is required for the
gate to rotate. If one side of the tank is filled with water at the rate of
0.1 m
3
/min and the other side is filled simultaneously with methyl
alcohol (density
ma  789 kg/m
3
) at the rate of 0.2 m
3
/min,
determine at what time and in which direction the gate will rotate.

SOLUTION
Consider the free-body diagram of the gate.
First note
base
VAd and .Vrt
Then
3
3
0.1 m / min (min)
0.25 (m)
(0.4 m)(1m)
0.2 m / min (min)
(m)
(0.2 m)(1m)
W
MA
t
dt
t
dt





Now
11
()
22
PApAgh
 so that

33 2
2
32
21
[(0.25 ) m (1 m)][(10 kg/m )(9.81 m/s )(0.25 ) m]
2
306.56 N
1
[( ) m (1 m)][(789 kg/m )(9.81 m/s )( ) m]
2
3870 N
W
MA
Pt t
t
Pt t
t





Now assume that the gate will rotate clockwise and when
0.6 m.
MA
d When rotation of the gate is
impending, we require

11
:0.6m 0.6m
33

 


AR MAMA WW
MM dP dP
Substituting

2211
490 N m 0.6 m (3870 ) N 0.6 0.25 m (306.56 ) N
33
tt t t
  
     
  
  







www.elsolucionario.org

776. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.145 (Continued)

Simplifying
32
1264.45 2138.1 490 0
 tt
Solving (positive roots only)
0.59451 mint and 1.52411 mint 
Now check assumption using the smaller root. We have

( ) m 0.59451 m 0.6 m
MA
dt  0.59451min 35.7 st
 
and the gate rotates clockwise.


www.elsolucionario.org

777. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.146
Determine the y coordinate of the centroid of the body shown.

SOLUTION
First note that the values of Ywill be the same for the given body and the body shown below. Then


V
y yV
Cone
21
3
ah

1
4
h

221
12
ah

Cylinder
2
2
1
24
a
bab






1
2
b

221
8
ab


2
(4 3 )
12
ahb



22 2
(2 3 )
24
ah b



We have YV yV
Then
2222
(4 3 ) (2 3 )
12 24
Y ahb ah b

 


or
22
23
2(4 3 )
hb
Y
hb





www.elsolucionario.org

778. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 5.147
An 8-in.-diameter cylindrical duct and a 4  8-in.
rectangular duct are to be joined as indicated.
Knowing that the ducts were fabricated from the
same sheet metal, which is of uniform thickness,
locate the center of gravity of the assembly.

SOLUTION
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.
By symmetry,

0.Z



2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1 (8)(12) 96  0 6 0 576
2 (8)(4) 16
2


2(4) 8

 10 128 160
3
2
(4) 8
2


4(4) 16
33

 12 42.667 96
4 (8)(12) 96 6 12 576 1152
5 (8)(12) 96 6 8 576 768
6
2
(4) 8
2


4(4) 16
33

 8 42.667 64
7 (4)(12) 48 6 10 288 480
8 (4)(12) 48 6 10 288 480
 539.33 1514.6 4287.4

Then
1514.67
in.
539.33
xA
X
A



or
2.81in.X 

4287.4
in.
539.33
yA
Y
A



or
7.95 in.Y 
www.elsolucionario.org

779. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 5.148
Three brass plates are brazed to a steel pipe to form the flagpole
base shown. Knowing that the pipe has a wall thickness of 8 mm
and that each plate is 6 mm thick, determine the location of the
center of gravity of the base. (Densities: brass  8470 kg/m
3
;
steel  7860 kg/m
3
.)

SOLUTION
Since brass plates are equally spaced, we note that
the center of gravity lies on the y-axis.
Thus, 0XZ 


Steel pipe:
22
63
363
[(0.064 m) (0.048 m) ](0.192 m)
4
270.22 10 m
(7860 kg/m )(270.22 10 m )
2.1239 kg
V
mV






 

Each brass plate:
63
3631
(0.096 m)(0.192 m)(0.006 m) 55.296 10 m
2
(8470 kg/m )(55.296 10 m ) 0.46836 kg




  
V
mV

Flagpole base:

2.1239 kg 3(0.46836 kg) 3.5290 kg
(0.096 m)(2.1239 kg) 3[(0.064 m)(0.46836 kg)] 0.29382 kg m
: (3.5290 kg) 0.29382 kg m
m
ym
Ym ym Y
  
   
  



0.083259 mY 83.3 mmY above the base 

www.elsolucionario.org

780. CHAPTER 6
www.elsolucionario.org

781. www.elsolucionario.org

782. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.1
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION
Free body: Entire truss:
0: 0 0
yy y
FB   B

0: (16 in.) (240 lb)(36 in.) 0
Cx
MB   


540 kN 540 lb
xx
B  B

0: 540 lb 240 lb 0
x
FC   


300 lbC

300 lbC
Free body: Joint B:

540 lb
54 3
BCAB
FF




900 lb
AB
FT 


720 lb
BC
FT 
www.elsolucionario.org

783. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.1 (Continued)

Free body: Joint C:



300 lb
13 12 5
AC BC
FF


780 lb
AC
FC 
720 lb (checks)
BC
F
www.elsolucionario.org

784. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.2
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.

SOLUTION
Free body: Entire truss:
0: 2.8 kN 0 2.8 kN
yy y
FA A    

2.8 kN
y
A

0: (1.4 m) (2.8 kN)(0.75 m) 0
A
MC  


1.500 kN 1.500 kNC C

0: 1.500 kN 0
xx
FA  


1.500 kN
x
A

1.500 kN
x
A
Free body: Joint C:

1.500 kN
1.25 1 0.75
BC AC
FF




2.00 kN
AC
FT 


2.50 kN
BC
FT  www.elsolucionario.org

785. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.2 (Continued)

Free body: Joint A:


7.5
0: 1.500 kN 0
8.5
xAB
FF

  




1.700 kN
AB
FT 
www.elsolucionario.org

786. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.3
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.

SOLUTION
Reactions:
0: 1260 lb
A
M  C

0: 0
xx
F  A


0: 960 lb
yy
F  A

Joint B:

300 lb
12 13 5
BCAB
FF




720 lb
AB
FT 


780 lb
BC
FC 
Joint A:

4
0: 960 lb 0
5
yAC
FF   

1200 lb
AC
F 1200 lb
AC
FC 

3
0: 720 lb (1200 lb) 0 (checks)
5
x
F  
www.elsolucionario.org

787. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.4
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is in
tension or compression.

SOLUTION
Reactions:


0: (3 m) (24 kN 8 kN)(1.5 m) (7 kN)(3 m) 0
Dy
MF    


23.0 kN
y
F


0: 0
xx
F  F


0: (7 24 8 7) 23 0
y
FD   

23.0 kND
Joint A:
0: 0
xAB
FF  0
AB
F 

0: 7 0
yAD
FF  

7kN
AD
F 7.00 kN
AD
FC 
Joint D:
8
0: 7 23.0 0
17
yBD
FF   
34.0 kN
BD
F 34.0 kN
BD
FC www.elsolucionario.org

788. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.4 (Continued)


15
0: ( 34.0) 0
17
xDE
FF   


30.0 kN
DE
F

30.0 kN
DE
FT


Joint E:



0: 8 0
yBE
FF 


8.00 kN
BE
F

8.00 kN
BE
FT


Truss and loading symmetrical about
c.L

www.elsolucionario.org

789. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.5
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is in
tension or compression.

SOLUTION
Reactions:


0: (5 ft) (10 kips)(10 ft) (10 kips)(20 ft) 0
Bx
MA   


60.0 kips
x
A


0: 0
xx
FAB 

60 kipsB


0: 10 kips 10 kips 0
yy
FA   

20.0 kips
y
A
Joint D:


10 kips
14 17
DC DA
F F



41.2 kips
DA
FT 


 40.0 kips
DC
FC


www.elsolucionario.org

790. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.5 (Continued)

Joint C:
1
0: 10 kips 0
5
yCA
FF  
22.4 kips
CA
F 22.4 kips
CA
FT 

2
0: (22.4 kips) 40 kips 0
5
xCB
FF   


60.0 kips
CB
F

60.0 kips
CB
FC



Joint B:



0: 0
yBA
FF 


0
BA
F


www.elsolucionario.org

791. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.6
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.

SOLUTION
Reactions:  
22
6 m 3.2 m 6.80 mDE 


4
0: (4.5 m) (24 kN)(12 m) 0
5
BAC
MF

  




80.0 kN
AC
TF



4
0: 80 kN 0
5
xx
FB

  




64 kN
x
B



3
0: 80 kN 24 kN 0
5
yy
FB

   



24 kN
y
B
Joint E:

24 kN
66.83.2
CE DE
F F
 45.0 kN
CE
FT 

51.0 kN
DE
FC 
www.elsolucionario.org

792. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.6 (Continued)
Joint D:

66
0: 51.0 kN 0
6.8 6.8
xBD
FF
 
  
 
 


51.0 kN
BD
FC 


48.0 kN
CD
FT 
Joint C:



 
4
0: 80 kN 45.0 kN 0
5
xBC
FF

    




19.00 kN
BC
FC





3
0 : 80 kN 48.0 kN 0 (checks)
5
y
F

  

 


www.elsolucionario.org

793. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.7
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is in
tension or compression.

SOLUTION


22
22
512 13ft
12 16 20 ft
AD
BCD



Reactions:
0: 0
xx
FD 
0: (21 ft) (693 lb)(5 ft) 0
Ey
MD  

165 lb
y
D


0: 165 lb 693 lb 0
y
FE  

528 lbE


Joint D:
54
0: 0
13 5
xADDC
FFF  
(1)

12 3
0: 165 lb 0
13 5
yADDC
FFF   
(2)

Solving Eqs. (1) and (2) simultaneously,

260 lb
AD
F 260 lb
AD
FC 

125 lb
DC
F 125.0 lb
DC
FT 


www.elsolucionario.org

794. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.7 (Continued)

Joint E:

54
0: 0
13 5
xBECE
FFF   (3)

12 3
0: 528 lb 0
13 5
yBECE
FFF   
(4)

Solving Eqs. (3) and (4) simultaneously,

832 lb
BE
F 832 lb
BE
FC 

400 lb
CE
F 400 lb
CE
FT 
Joint C:

Force polygon is a parallelogram (see Fig. 6.11, p. 290).

400 lb
AC
FT 

125.0 lb
BC
FT 
Joint A:
54
0: (260 lb) (400 lb) 0
13 5
xAB
FF   
420 lb
AB
F 420 lb
AB
FC 

12 3
0: (260 lb) (400 lb) 0
13 5
00(Checks)
y
F  

www.elsolucionario.org

795. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.8
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.

SOLUTION
Free body: Entire truss:

0: 2(5 kN) 0
xx
FC  

10 kN 10 kNx
xC  C



0: (2 m) (5 kN)(8 m) (5 kN)(4 m) 0
C
MD   

30 kN 30 kND  D



0: 30 kN 0 30 kN 30 kN y
yy yFC C     C



Free body: Joint A:

5 kN
4117
AB AD
FF





20.0 kN
AB
FT 

20.6 kN
AD
FC 

Free body: Joint B:

1
0: 5 kN 0
5
xBD
FF  


55 kN
BD
F

11.18 kN
BD
FC




2
0: 20 kN ( 5 5 kN) 0
5
yBC
FF    


30 kN
BC
F

30.0 kN
BC
FT

www.elsolucionario.org

796. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.8 (Continued)

Free body: Joint C:


0: 10 kN 0
xCD
FF  


10 kN
CD
F

10.00 kN
CD
FT




0: 30 kN 30 kN 0 (checks)
y
F  


www.elsolucionario.org

797. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.9
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.

SOLUTION
Reactions for Entire truss:



By inspection: 4 kN 4 kNAE


Free body: Joint A:

4 kN
sin 30
AB
F



(4 kN)(cot 30 )
AF
F





8.00 kN
AB
FC 

6.93 kN
AF
FT 
Free body: Joint F:
0:
xFGAF
FFF 


6.93 kN
FG
FT




0: 4.00 kN
yFB
FF T 


4.00 kN
FB
FT


www.elsolucionario.org

798. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.9 (Continued)

Free body: Joint B:

0: 8cos30 cos30 cos30 0 (1)
xBCBG
FFF   



0 : 8sin 30 4 sin 30 sin 30 0 (2)
yBCBG
FFF   


Solving (1) and (2) simultaneously yields:

4.00 kN
BC
FC






4.00 kN
BG
FC




Free body: Joint C:
By symmetry, 4.00 kN
CD CB
FF C 


0: 4sin30 4sin30 0
yCG
FF   



4.00 kN
CG
FT 
By symmetry about the truss centerline, we note:

4.00 kN , 6.93 kN , 4.00 kN , 8.00 kN , 6.93 kN
GD GH DH DE EH
FCFTFTFCFT 


www.elsolucionario.org

799. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.10
Determine the force in each member of the truss shown. State whether each
member is in tension or compression.

SOLUTION
Free body: Entire truss:

0: 0
xx
F  F

0: 5 kN
F
M  H

0: 0
yy
F  F


Free body: Joint C:
5.00 kN, F 5 2 7.07 kN
CB CE
F



5.00 kN
CB
FT 

7.07 kN
CE
FC 

By symmetry about the horizontal centerline: 5.00 kN , 7.07 kN
HG HE
FTF C


Joint E: Members lie in two intersecting straight lines, thus we have:
7.07 kN ; 7.07 kN
EG EC EB EH
FF CFF C  



www.elsolucionario.org

800. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.10 (Continued)
Free body: Joint B:

11
0: 7.07 0
22
yBD
FF  


7.07 kN
BD
FT





11
0: 5 (7.07) 7.07 0
22
xBA
FF    


5.00 kN
BA
FC


By symmetry about the horizontal centerline: 7.07 kN , 5.00 kN
DG GF
FTFC 
Joint D: 7.07 kN
DF DB DA DG
FFFF T 


Free body: Joint A:



5.00 kN
AF
FC



www.elsolucionario.org

801. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.11
Determine the force in each member of the Gambrel roof
truss shown. State whether each member is in tension or
compression.

SOLUTION
Free body: Truss:
0: 0
xx
F  H
Because of the symmetry of the truss and loading,

1
total load
2
y
AH

1200 lb
y
AH
Free body: Joint A:

900 lb
54 3
ACAB
FF

1500 lb
AB
CF 

1200 lb
AC
TF 

Free body: Joint C:
BC is a zero-force member.

0
BC
F 1200 lb
CE
FT 
Free body: Joint B:

24 4 4
0: (1500 lb) 0
25 5 5
xBDBE
FFF   

or
24 20 30,000 lb
BD BE
FF (1)

733
0: (1500) 600 0
25 5 5
yBDBE
FFF    
or
715 7,500lb
BD BE
FF (2)

www.elsolucionario.org

802. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.11 (Continued)

Multiply Eq. (1) by 3, Eq. (2) by 4, and add:

100 120,000 lb
BD
F 1200 lb
BD
FC 
Multiply Eq. (1) by 7, Eq. (2) by –24, and add:

500 30,000 lb
BE
F 60.0 lb
BE
FC 
Free body: Joint D:

24 24
0: (1200 lb) 0
25 25
xDF
FF  

1200 lb
DF
F 1200 lb
DF
FC 

77
0: (1200 lb) ( 1200 lb) 600 lb 0
25 25
yDE
FF     

72.0 lb
DE
F 72.0 lb
DE
FT 
Because of the symmetry of the truss and loading, we deduce that


EF BE
FF 60.0 lb
EF
FC 

EG CE
FF 1200 lb
EG
FT 

FG BC
FF 0
FG
F 

FH AB
FF 1500 lb
FH
FC 

GH AC
FF 1200 lb
GH
FT 
www.elsolucionario.org

803. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.12
Determine the force in each member of the Howe roof
truss shown. State whether each member is in tension or
compression.

SOLUTION
Free body: Truss:


0: 0
xx
F  H
Because of the symmetry of the truss and loading,

1
total load
2
y
AH

1200 lb
y
AH
Free body: Joint A:

900 lb
54 3
ACAB
FF

1500 lb
AB
CF 

1200 lb
AC
TF 

Free body: Joint C:
BC is a zero-force member.

0
BC
F 1200 lb
CE
TF 
www.elsolucionario.org

804. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.12 (Continued)

Free body: Joint B:

444
0: (1500 lb) 0
555
xBDBC
FFF   

or
1500 lb
BD BE
FF (1)

333
0: (1500 lb) 600 lb 0
555
yBDBE
FFF    
or
500 lb
BD BE
FF (2)
Add Eqs. (1) and (2):
2 2000 lb
BD
F 1000 lb
BD
FC 
Subtract Eq. (2) from Eq. (1):
2 1000 lb
BE
F 500 lb
BE
FC 
Free Body: Joint D:

44
0: (1000 lb) 0
55
xDF
FF  

1000 lb
DF
F 1000 lb
DF
FC 

33
0: (1000 lb) ( 1000 lb) 600 lb 0
55
yDE
FF    

600 lb
DE
F 600 lb
DE
FT 
Because of the symmetry of the truss and loading, we deduce that


EF BE
FF 500 lb
EF
FC 

EG CE
FF 1200 lb
EG
FT 

FG BC
FF 0
FG
F 

FH AB
FF 1500 lb
FH
FC 

GH AC
FF 1200 lb
GH
FT 
www.elsolucionario.org

805. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.13
Using the method of joints, determine the force in each
member of the roof truss shown. State whether each
member is in tension or compression.

SOLUTION
Free body: Truss:
0: 0
xx
F  A
From symmetry of loading:

1
total load
2
y
AE

3.6 kN
y
AE
We note that DF is a zero-force member and that EF is aligned with the load. Thus,

0
DF
F 

1.2 kN
EF
FC 
Free body: Joint A:

2.4 kN
13 12 5
ACAB
FF

6.24 kN
AB
FC 

2.76 kN
AC
FT 
Free body: Joint B:

31212
0: (6.24 kN) 0
3.905 13 13
xBCBD
FFF    (1)

2.5 5 5
0: (6.24 kN) 2.4 kN 0
3.905 13 13
yBCBD
FFF      (2)

www.elsolucionario.org

806. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.13 (Continued)

Multiply Eq. (1) by 2.5, Eq. (2) by 3, and add:

45 45
(6.24 kN) 7.2 kN 0, 4.16 kN,
13 13
BD BD
FF 4.16 kN
BD
FC 
Multiply Eq. (1) by 5, Eq. (2) by –12, and add:

45
28.8 kN 0, 2.50 kN,
3.905
BC BC
FF 2.50 kN
BC
FC 
Free body: Joint C:

52.5
0: (2.50 kN) 0
5.831 3.905
yCD
FF  

1.867 kN
CD
FT 

33
0: 5.76 kN (2.50 kN) (1.867 kN) 0
3.905 5.831
xCE
FF    

2.88 kN
CE
FT
Free body: Joint E:

5
0: 3.6 kN 1.2 kN 0
7.81
yDE
FF   

3.75 kN
DE
F 3.75 kN
DE
FC 

6
0: ( 3.75 kN) 0
7.81
xCE
FF    

2.88 kN 2.88 kN
CE CE
FFT  (Checks)
www.elsolucionario.org

807. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.14
Determine the force in each member of the Fink roof truss
shown. State whether each member is in tension or
compression.

SOLUTION
Free body: Truss:
0: 0
xx
F  A
Because of the symmetry of the truss and loading,

1
total load
2
y
AG

6.00 kN
y
AG
Free body: Joint A:

4.50 kN
2.462 2.25 1
ACAB
FF


11.08 kN
AB
FC 

10.125 kN
AC
F 10.13 kN
AC
FT 

Free body: Joint B:

3 2.25 2.25
0: (11.08 kN) 0
5 2.462 2.462
xBC BD
FF F   
(1)


4 11.08 kN
0: 3 kN 0
5 2.462 2.462
BD
yBC
F
FF     
(2)


Multiply Eq. (2) by –2.25 and add to Eq. (1):

12
6.75 kN 0 2.8125
5
BC BC
FF  2.81 kN
BC
FC 
Multiply Eq. (1) by 4, Eq. (2) by 3, and add:

12 12
(11.08 kN) 9 kN 0
2.462 2.462
BD
F

9.2335 kN
BD
F 9.23 kN
BD
FC  www.elsolucionario.org

808. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.14 (Continued)

Free body: Joint C:
44
0: (2.8125 kN) 0
55
yCD
FF  

2.8125 kN,
CD
F

2.81 kN
CD
FT



33
0: 10.125 kN (2.8125 kN) (2.8125 kN) 0
55
xCE
FF    


6.7500 kN
CE
F

6.75 kN
CE
FT


Because of the symmetry of the truss and loading, we deduce that

DE CD
FF 2.81 kN
CD
FT 

DF BD
FF 9.23 kN
DF
FC 

EF BC
FF 2.81 kN
EF
FC 

EG AC
FF 10.13 kN
EG
FT 

FG AB
FF 11.08 kN
FG
FC 
www.elsolucionario.org

809. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.15
Determine the force in each member of the Warren bridge
truss shown. State whether each member is in tension or
compression.

SOLUTION
Free body: Truss: 0: 0
xx
FA 
Due to symmetry of truss and loading,

1
total load 6 kips
2
y
AG 
Free body: Joint A:
6kips
53 4
ACAB
FF

7.50 kips
AB
FC 
4.50 kips
AC
FT 



Free body: Joint B:
7.5 kips
56 5
BC BD
F F

7.50 kips
BC
FT 
9.00 kips
BD
FC 


Free body: Joint C:

44
0: (7.5) 6 0
55
yCD
FF  

0
CD
F 

3
0: 4.5 (7.5) 0
5
xCE
FF   

9kips
CE
F 9.00 kips
CE
FT 
Truss and loading is symmetrical aboutc.L
www.elsolucionario.org

810. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.16
Solve Problem 6.15 assuming that the load applied at E
has been removed.
PROBLEM 6.15 Determine the force in each member of
the Warren bridge truss shown. State whether each
member is in tension or compression.

SOLUTION
Free body: Truss: 0: 0
xx
FA 
0: 6(36) (54) 0 4 kips
Gyy
MA    A

0: 4 6 0 2 kips
y
FG   G

Free body: Joint A:
4kips
53 4
ACAB
FF

5.00 kips
AB
FC 
3.00 kips
AC
FT 


Free body Joint B:
5kips
56 5
BC BD
F F

5.00 kips
BC
FT 
6.00 kips
BD
FC 


Free body Joint C:

44
0: (5) 6 0
55
yCD
MF   2.50 kips
CD
FT 

33
0: (2.5) (5) 3 0
55
xCE
FF   
4.50 kips
CE
FT 




www.elsolucionario.org

811. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.16 (Continued)

Free body: Joint D:

44
0: (2.5) 0
55
yDE
FF   

2.5 kips
DE
F 2.50 kips
DE
FC 

33
0: 6 (2.5) (2.5) 0
55
xDF
FF   

3kips
DF
F 3.00 kips
DF
FC 
Free body: Joint F:
3kips
55 6
FGEF
FF

2.50 kips
EF
FT 
2.50 kips
FG
FC 



Free body: Joint G:
2kips
34
EG
F

1.500 kips
EG
FT 

Also,
2kips
54
FG
F


2.50 kips (Checks)
FG
FC


www.elsolucionario.org

812. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.17
Determine the force in each member of the Pratt
roof truss shown. State whether each member is
in tension or compression.

SOLUTION
Free body: Truss:


0: 0
xx
FA 
Due to symmetry of truss and load,

1
total load 21 kN
2
y
AH 
Free body: Joint A:


15.3 kN
37 35 12
ACAB
FF


47.175 kN 44.625 kN
AB AC
FF 47.2 kN
AB
FC 

44.6 kN
AC
FT 
Free body: Joint B:

From force polygon:
47.175 kN, 10.5 kN
BD BC
FF 10.50 kN
BC
FC 

47.2 kN
BD
FC  www.elsolucionario.org

813. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.17 (Continued)

Free body: Joint C:
3
0: 10.5 0
5
yCD
FF  
17.50 kN
CD
FT



4
0: (17.50) 44.625 0
5
xCE
FF   


30.625 kN
CE
F

30.6 kN
CE
FT



Free body: Joint E:
DE is a zero-force member. 0
DE
F 

Truss and loading symmetrical about .cL
www.elsolucionario.org

814. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.18
The truss shown is one of several supporting an advertising panel.
Determine the force in each member of the truss for a wind load
equivalent to the two forces shown. State whether each member is in
tension or compression.

SOLUTION
Free body: Entire truss:

0: (800 N)(7.5 m) (800 N)(3.75 m) (2 m) 0
F
MA   

2250 NA 2250 NA

0: 2250 N 0
yy
FF  

2250 N 2250 N
yy
F  F


0: 800 N 800 N 0
xx
FF    


1600 N 1600 N
xx
F  F
Joint D:

800 N
81517
DE BD
FF


1700 N
BD
FC 

1500 N
DE
FT 
Joint A:

2250 N
15 17 8
ACAB
FF



2250 N
AB
FC 

1200 N
AC
FT 
Joint F:

0: 1600 N 0
xCF
FF  

1600 N
CF
F 1600 N
CF
FT 

0: 2250 N 0
yEF
FF  

2250 N
EF
F 2250 N
EF
FT  www.elsolucionario.org

815. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.18 (Continued)

Joint C:
8
0: 1200 N 1600 N 0
17
xCE
FF   

850 N
CE
F

850 N
CE
FC



15
0: 0
17
yBCCE
FFF  


15 15
(850 N)
17 17
BC CE
FF  


750 N
BC
F

750 N
BC
FT


Joint E:
8
0: 800 N (850 N) 0
17
xBE
FF    


400 N
BE
F

400 N
BE
FC



15
0: 1500 N 2250 N (850 N) 0
17
00(checks)
y
F   

www.elsolucionario.org

816. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.19
Determine the force in each member of the Pratt bridge
truss shown. State whether each member is in tension or
compression.

SOLUTION
Free body: Truss:

0: 0
zx
F  A

0: (36 ft) (4 kips)(9 ft)
(4 kips)(18 ft) (4 kips)(27 ft) 0
A
MH 
 

6kipsH


0: 6 kips 12 kips 0 6 kips
yy y
FA     A


Free body: Joint A:

6kips
53 4
ACAB
FF


7.50 kips
AB
FC 
4.50 kips
AC
FT 

Free body: Joint C:
0:
x
F 4.50 kips
CE
FT 
0:
y
F 4.00 kips
BC
FT 

Free body: Joint B:

44
0: (7.50 kips) 4.00 kips 0
55
yBE
FF     
2.50 kips
BE
FT 

83
0: (7.50 kips) (2.50 kips) 0
55
xBD
FF    
6.00 kips
BD
F 6.00 kips
BD
FC www.elsolucionario.org

817. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.19 (Continued)

Free body: Joint D: We note that DE is a zero-force member: 0
DE
F 
Also,
6.00 kips
DF
F C 
From symmetry:

FE BE
F F 2.50 kips
EF
F T 

EG CE
F F 4.50 kips
EG
F T 

FG BC
F F 4.00 kips
FG
F T 

FH AB
F F 7.50 kips
FH
F C 

GH AC
F F 4.50 kips
GH
F T 
www.elsolucionario.org

818. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.20
Solve Problem 6.19 assuming that the load applied at G has
been removed.
PROBLEM 6.19 Determine the force in each member of
the Pratt bridge truss shown. State whether each member is
in tension or compression.

SOLUTION
Free body: Truss:

0: 0
xx
F  A

0: (36 ft) (4 kips)(9 ft) (4 kips)(18 ft) 0
A
H   M

3.00 kipsH


0: 5.00 kips
yy
F  A

We note that DE and FG are zero-force members.
Therefore,

0,
DE
F 0
FG
F 
Also,
BD DF
FF (1)
and
EG GH
FF (2)

Free body: Joint A:

5 kips
53 4
ACAB
FF


6.25 kips
AB
FC 
3.75 kips
AC
FT 

Free body: Joint C:

0:
x
F 3.75 kips
CE
FT 
0:
y
F 4.00 kips
BC
FT 

Free body: Joint B:

44
0: (6.25 kips) 4.00 kips 0
55
xBE
FF   

1.250 kips
BE
FT 

33
0: (6.25 kips) (1.250 kips) 0
55
xBD
FF    
4.50 kips
BD
F 4.50 kips
BD
FC www.elsolucionario.org

819. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.20 (Continued)

Free body: Joint F: We recall that 0,
FG
F and from Eq. (1) that

DFBD
F F 4.50 kips
DF
F C 

4.50 kips
55 6
EF FH
FF



3.75 kips
EF
F T 

3.75 kips
FH
F C 

Free body: Joint H:


3.00 kips
34
GH
F



2.25 kips
GH
F T 


Also,

3.00 kips
54
FH
F



3.75 kips (checks)
FH
FC
From Eq. (2):

EG GH
F F 2.25 kips
EG
F T 

www.elsolucionario.org

820. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.21
Determine the force in each of the members located to the
left of FG for the scissors roof truss shown. State whether
each member is in tension or compression.

SOLUTION
Free Body: Truss:


0: 0
xx
F  A

0: (1 kN)(12 m) (2 kN)(10 m) (2 kN)(8 m) (1 kN)(6 m) (12 m) 0
L y
MA     

4.50 kN
y
A
We note that BC is a zero-force member:
0
BC
F 
Also,
CE AC
FF (1)
Free body: Joint A:
12
0: 0
25
xABAC
FFF   (2)

11
0: 3.50 kN 0
25
yABAC
FFF    (3)

Multiply Eq. (3) by –2 and add Eq. (2):

1
7kN 0
2
AB
F 9.90 kN
AB
FC  www.elsolucionario.org

821. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.21 (Continued)

Subtract Eq. (3) from Eq. (2):

1
3.50 kN 0 7.826 kN
5
AC AC
FF 7.83 kN
AC
FT 
From Eq. (1):
7.826 kN
CE AC
FF 7.83 kN
CE
FT 
Free body: Joint B:

11
0: (9.90kN) 2kN 0
22
yBD
FF   

7.071 kN
BD
F 7.07 kN
BD
FC 

1
0: (9.90 7.071) kN 0
2
xBE
FF   

2.000 kN
BE
F 2.00 kN
BE
FC 
Free body: Joint E:
2
0: ( 7.826 kN) 2.00 kN 0
5
xEG
FF   
5.590 kN
EG
F 5.59 kN
EG
FT 

1
0: (7.826 5.590) kN 0
5
yDE
FF   

1.000 kN
DE
F 1.000 kN
DE
FT 
Free body: Joint D:

21
0: ( ) (7.071kN)
52
xDFDG
FFF  
or
5.590 kN
DF DG
FF (4)

11
0: ( ) (7.071 kN) 2 kN 1 kN 0
52
yDFDG
FFF     
or
4.472
DE DG
FF (5)
Add Eqs. (4) and (5):
2 10.062 kN
DF
F 5.03 kN
DF
FC 
Subtract Eq. (5) from Eq. (4):
2 1.1180 kN
DG
F 0.559 kN
DG
FC 
www.elsolucionario.org

822. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.22
Determine the force in member DE and in each of the
members located to the left of DE for the inverted Howe
roof truss shown. State whether each member is in tension
or compression.

SOLUTION
Free body: Truss:


0: 0
xx
F  A

0: (400 lb)(4 ) (800 lb)(3 ) (800 lb)(2 ) (800 lb) (4 ) 0
H y
MddddAd     


1600 lb
y
A
Angles:
6.72
tan 32.52
10.54
6.72
tan 16.26
23.04



Free body: Joint A:


1200 lb
sin 57.48 sin106.26 sin16.26
ACAB
FF




3613.8 lb
4114.3 lb
AB
AC
FC
FT


3610 lb , 4110 lb
AB AC
FCFT  www.elsolucionario.org

823. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.22 (Continued)

Free body: Joint B: 0: (800 lb)cos16.26 0
yBC
FF   

768.0 lb 768 lb
BC BC
FFC  

0: 3613.8 lb (800 lb) sin16.26° 0
xBD
FF   

3837.8 lb
BD
F 3840 lb
BD
FC 
Free body: Joint C:


0: sin 32.52 (4114.3 lb)sin 32.52 (768 lb)cos16.26 0
yCE
FF    

2742.9 lb
CE
F 2740 lb
CE
FT 

0: (4114.3 lb) cos 32.52° (2742.9 lb)cos32.52° (768 lb)sin16.26 0
xCD
FF    

1371.4 lb
CD
F 1371 lb
CD
FT 
Free body: Joint E:


2742.9 lb
sin 32.52 sin73.74°
DE
F



1536 lb
DE
FC  www.elsolucionario.org

824. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.23
Determine the force in each of the members located to the
right of DE for the inverted Howe roof truss shown. State
whether each member is in tension or compression.

SOLUTION
Free body: Truss


0: (4 ) (800 lb) (800 lb)(2 ) (800 lb)(3 ) (400 lb)(4 ) 0
A
Hd d d d d     M


1600 lbH
Angles:
6.72
tan 32.52
10.54
6.72
tan 16.26
23.04



Free body: Joint H:


(1200 lb)cot16.26
GH
F

4114.3 lb
GH
FT 4110 lb
GH
FT 

1200 lb
4285.8 lb
sin16.26
FH
F

4290 lb
FH
FC 

www.elsolucionario.org

825. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.23 (Continued)

Free body Joint F: 0: (800 lb)cos16.26 0
yFG
FF   

768.0 lb
FG
F 768 lb
FG
FC 

0: 4285.8 lb (800 lb) sin 16.26° 0
xDF
FF    

4061.8 lb
DF
F 4060 lb
DF
FC 
Free body: Joint G:
0: sin 32.52 (768.0 lb)cos16.26 0
1371.4 lb
yDG
DG
FF
F
  


1371 lb
DG
FT 

0: 4114.3 lb (768.0 lb)sin16.26 (1371.4 lb)cos32.52 0
xEG
FF     

2742.9 lb
EG
F 2740 lb
EG
FT 
www.elsolucionario.org

826. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.24
The portion of truss shown represents the upper
part of a power transmission line tower. For the
given loading, determine the force in each of the
members located above HJ. State whether each
member is in tension or compression.

SOLUTION










Free body: Joint A:

1.2 kN
2.29 2.29 1.2
ACAB
FF

2.29 kN
AB
FT 

2.29 kN
AC
FC 
Free body: Joint F:

1.2 kN
2.29 2.29 2.1

DF EF
FF

2.29 kN
DF
FT 

2.29 kN
EF
FC 
Free body: Joint D:

2.29 kN
2.21 0.6 2.29
BD DE
FF

2.21 kN
BD
FT 

0.600 kN
DE
FC 
Free body: Joint B:

42.21
0: 2.21 kN (2.29 kN) 0
52.29
xBE
FF   


0
BE
F 

30.6
0: (0) (2.29 kN) 0
52.29
yBC
FF    


0.600 kN
BC
F

0.600 kN
BC
FC


www.elsolucionario.org

827. Copyright © McGraw-Hill Education. Permission required for reproduction or display.










PROBLEM 6.24 (Continued)

Free body: Joint C:

2.21
0: (2.29 kN) 0
2.29
xCE
FF  

2.21 kN
CE
F 2.21 kN
CE
FC 

0.6
0: 0.600 kN (2.29 kN) 0
2.29
yCH
FF    

1.200 kN
CH
F 1.200 kN
CH
FC 
Free body: Joint E:

2.21 4
0: 2.21 kN (2.29 kN) 0
2.29 5
   
xEH
FF

0
EH
F 

0.6
0: 0.600 kN (2.29 kN) 0 0
2.29
yEJ
FF    

1.200 kN
EJ
F 1.200 kN
EJ
FC 
www.elsolucionario.org

828. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.25
For the tower and loading of Problem 6.24 and
knowing that F
CH  FEJ  1.2 kN C and F EH  0,
determine the force in member HJ and in each of
the members located between HJ and NO. State
whether each member is in tension or
compression.
PROBLEM 6.24 The portion of truss shown
represents the upper part of a power transmission
line tower. For the given loading, determine the
force in each of the members located above HJ.
State whether each member is in tension or
compression.

SOLUTION








Free body: Joint G:

1.2 kN
3.03 3.03 1.2
GH GI
FF

3.03 kN
GH
FT 

3.03 kN
GI
FC 
Free body: Joint L:

1.2 kN
3.03 3.03 1.2

JL KL
F F

3.03 kN
JL
FT 

3.03 kN
KL
FC 
Free body: Joint J:

2.97
0: (3.03 kN) 0
3.03
   
xHJ
FF

2.97 kN
HJ
FT 

0.6
0: 1.2 kN (3.03 kN) 0
3.03
yJK
FF
  


1.800 kN
JK
F 1.800 kN
JK
FC 
Free body: Joint H:

42.97
0: 2.97 kN (3.03 kN) 0
53.03
xHK
FF   

0
HK
F 

0.6 3
0: 1.2 kN (3.03) kN (0) 0
3.03 5
yHI
FF     


1.800 kN
HI
F 1.800 kN
HI
FC  www.elsolucionario.org

829. Copyright © McGraw-Hill Education. Permission required for reproduction or display.










PROBLEM 6.25 (Continued)

Free body: Joint I:

2.97
0: (3.03 kN) 0
3.03
xIK
FF  

2.97 kN
IK
F 2.97 kN
IK
FC 

0.6
0: 1.800 kN (3.03 kN) 0
3.03
yIN
FF    

2.40 kN
IN
F 2.40 kN
IN
FC 
Free body: Joint K:

42.97
0: 2.97 kN (3.03 kN) 0
53.03
xKN
FF    

0
KN
F 

0.6 3
0: (3.03 kN) 1.800 kN (0) 0
3.03 5
yKO
FF     

2.40 kN
KO
F 2.40 kN
KO
FC 
www.elsolucionario.org

830. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.26
Solve Problem 6.24 assuming that the cables
hanging from the right side of the tower have
fallen to the ground.
PROBLEM 6.24 The portion of truss shown
represents the upper part of a power transmission
line tower. For the given loading, determine the
force in each of the members located above HJ.
State whether each member is in tension or
compression.

SOLUTION












Zero-Force Members:
Considering joint F, we note that DF and EF are zero-force
members:

0
DF EF
FF 
Considering next joint D, we note that BD and DE are zero-force
members:

0
BD DE
FF 
Free body: Joint A:

1.2 kN
2.29 2.29 1.2
ACAB
FF

2.29 kN
AB
FT 

2.29 kN
AC
FC 
Free body: Joint B:

42.21
0: (2.29 kN) 0
52.29
  
xBE
FF

2.7625 kN
BE
F 2.76 kN
BE
FT 

0.6 3
0: (2.29 kN) (2.7625 kN) 0
2.29 5
yBC
FF    

2.2575 kN
BC
F 2.26 kN
BC
FC 
www.elsolucionario.org

831. Copyright © McGraw-Hill Education. Permission required for reproduction or display.










PROBLEM 6.26 (Continued)

Free body: Joint C:

2.21
0: (2.29 kN) 0
2.29
  
xCE
FF

2.21 kN
CE
FC 

0.6
0: 2.2575 kN (2.29 kN) 0
2.29
yCH
FF  

2.8575 kN
CH
F 2.86 kN
CH
FC 
Free body: Joint E:

44
0: (2.7625 kN) 2.21 kN 0
55
xEH
FF    

0
EH
F 

33
0: (2.7625 kN) (0) 0
55
yEJ
FF    

1.6575 kN
EJ
F 1.658 kN
EJ
FT 
www.elsolucionario.org

832. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.27
Determine the force in each member of the truss shown.
State whether each member is in tension or compression.

SOLUTION
Reactions: 0:
x
F 0
x
E
0:
F
M 45 kips
y
E


0:
y
F 60 kipsF

Joint D:
15 kips
12 13 5
CD DH
F F


36.0 kips
CD
FT 
39.0 kips
DH
FC 
Joint H:
0:F 0
CH
F 

0:F 39.0 kips
GH
FC 
Joint C:
0:F 0
CG
F 
0:F 36.0 kips
BC
FT 
Joint G:
0:F 0
BG
F 

0:F 39.0 kips
FG
FC 
Joint B:
0:F 0
BF
F 
0:F 36.0 kips
AB
FT 
www.elsolucionario.org

833. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.27 (Continued)

Joint A:

22
12 15 19.21 ftAE

36 kips
tan38.7
AF
F

45.0 kips
AF
FC 

36 kips
sin 38.7
AE
F

57.6 kips
AE
FT 

Joint E:

0: (57.6 kips)sin 38.7 0
xEF
FF   

36.0 kips
EF
F 36.0 kips
EF
FC 

0: (57.6 kips)cos38.7 45 kips 0
y
F  
(Checks)
www.elsolucionario.org

834. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.28
Determine the force in each member of the truss shown.
State whether each member is in tension or compression.

SOLUTION
Free body: Truss
0: 0
xx
F  H

0: 48(16) (4) 0 192 kN
H
MG    G


0: 192 48 0
yy
FH  

144 kN 144 kN
yy
H  H
Zero-Force Members:
Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC,
BE, DE, DG, FG
Thus,
0
BC BE DE DG FG
FFFFF 
Also note:
AB BD DF FH
FFFF (1)

AC CE EG
FFF (2)
Free body: Joint A:
48 kN
8373
ACAB
FF




128 kN
AB
F 128.0 kN
AB
FT 

136.704 kN
AC
F 136.7 kN
AC
FC 
From Eq. (1):
128.0 kN
BD DF FH
FFF T 
From Eq. (2):  136.7kN
CE EG
FF C 


www.elsolucionario.org

835. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.28 (Continued)

Free body: Joint H
144 kN
9145
GH
F

192.7 kN
GH
FC 

Also
144 kN
89
FH
F


128.0 kN (Checks)
FH
FT 
www.elsolucionario.org

836. Copyright © McGraw-Hill Education. Permission required for reproduction or display.





PROBLEM 6.29
Determine whether the trusses of Problems 6.31a,
6.32a, and 6.33a are simple trusses.

SOLUTION







Truss of Problem 6.31a:
Starting with triangle HDI and adding two members at a time, we
obtain successively joints A, E, J, and B, but cannot go further. Thus,
this truss
is not a simple truss.


Truss of Problem 6.32a:
Starting with triangle ABC and adding two members at a time, we obtain
joints D, E, G, F, and H, but cannot go further. Thus, this truss
is not a simple truss.


Truss of Problem 6.33a:
Starting with triangle ABD and adding two members at a time, we
obtain successively joints H, G, F, E, I, C, and J, thus completing the
truss.
Therefore, this is a simple truss.

www.elsolucionario.org

837. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.30
Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses.








SOLUTION






Truss of Problem 6.31b:
Starting with triangle CGM and adding two members at a time, we obtain
successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus
completing the truss.
Therefore, this truss is a simple truss.


Truss of Problem 6.32b:
Starting with triangle ABC and adding two members at a time, we obtain
successively joints E, D, F, G, and H, but cannot go further. Thus, this truss
is not a simple truss.






Truss of Problem 6.33b:
Starting with triangle GFH and adding two members at a time, we obtain
successively joints D, E, C, A, and B, thus completing the truss.
Therefore, this is a simple truss.

www.elsolucionario.org

838. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.31
For the given loading, determine the zero-force members in
each of the two trusses shown.

SOLUTION
Truss (a): : Joint : 0
BC
FB B F
: Joint : 0
CD
FB C F

:Joint : 0
IJ
FB J F

:Joint : 0
IL
FB I F

:Joint : 0
MN
FB N F
:Joint : 0
LM
FB M F
The zero-force members, therefore, are
,,,, ,BC CD IJ IL LM MN 
Truss (b):
: Joint : 0
BC
FB C F
: Joint : 0
BE
FB B F

:Joint : 0
FG
FB G F

:Joint : 0
EF
FB F F

:Joint : 0
DE
FB E F

:Joint : 0
IJ
FB I F
:Joint : 0
MN
FB M F

: Joint : 0
KN
FB N F
The zero-force members, therefore, are
,,,,,, ,BC BE DE EF FG IJ KN MN 

(a)
(b) www.elsolucionario.org

839. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.32
For the given loading, determine the zero-force members
in each of the two trusses shown.

SOLUTION
Truss (a): : Joint : 0
BJ
FB B F
: Joint : 0
DI
FB D F

: Joint : 0
EI
FB E F

:Joint : 0
AI
FB I F

:Joint : 0
FK
FB F F

:Joint : 0
GK
FB G F

: Joint : 0
CK
FB K F
The zero-force members, therefore, are
,, ,,, ,AI BJ CK DI EI FK GK 
Truss (b):
: Joint : 0
FK
FB K F
: Joint :FB O 0
IO
F

The zero-force members, therefore, are
andFK IO 
All other members are either in tension or compression.

(b)
(a) www.elsolucionario.org

840. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.33
For the given loading, determine the
zero-force members in each of the
two trusses shown.

SOLUTION
Truss (a):




Note: Reaction at F is vertical
(0).
x
F

Joint :G 0,F 0
DG
F 

Joint :D 0,F 0
DB
F 

Joint :F 0,F 0
FG
F 

Joint :G 0,F 0
GH
F 

Joint :J 0,F 0
IJ
F 

Joint :I 0,F 0
HI
F 
Truss (b):




Joint :A 0,F 0
AC
F 

Joint :C 0,F 0
CE
F 

Joint :E 0,F 0
EF
F 

Joint :F

0,F 0
FG
F 

Joint :G 0,F 0
GH
F 
www.elsolucionario.org

841. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.21, (b) Problem 6.27.


SOLUTION
(a) Truss of Problem 6.21:
:Joint : 0
BC
FB C F 

:Joint : 0
IJ
FB J F


:Joint : 0
JK
FB K F


:Joint : 0
HI
FB I F


The zero-force members, therefore, are
,,,
BCHI IJ JK 

(b) Truss of Problem 6.27:
: Joint : 0
BF
FB B F 

: Joint : 0
BG
FB B F


: Joint : 0
CG
FB C F


: Joint : 0
CH
FB C F

The zero-force members, therefore, are ,,,BFBGCGCH 
www.elsolucionario.org

842. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.35*
The truss shown consists of six members and is supported
by a short link at A, two short links at B, and a ball and
socket at D. Determine the force in each of the members
for the given loading.

SOLUTION
Free body: Truss:
From symmetry:

and
xx yy
DB DB

0: (10 ft) (400 lb)(24 ft) 0
z
MA   

960 lbA

0: 0
xxx
FBDA  

2 960 lb 0, 480 lb
xx
BB 

0: 400 lb 0
yyy
FBD   

2400lb
200 lb
y
y
B
B



Thus,
(480 lb) (200 lb)Bij 
Free body: C:

(24 10)
26
(24 7)
25
(24 7)
25
AC
CA AC
BC
CB BC
CD
CD CD
FCA
FF
CA
FCB
FF
CB
FCD
FF
CD



ij
ik
ik





0: (400 lb) 0
CA CB CD
    FFFF j


PROBLEM 6.35* (Continued) www.elsolucionario.org

843. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



Substituting for
,,,
CA CB CD
FFF and equating to zero the coefficients of ,, :ijk
i:
24 24
()0
26 25
 
AC BC CD
FFF (1)
j:
10
400 lb 0
26
AC
F 1040 lb
AC
FT 
k:
7
()0
25
BC CD CD BC
FF FF 
Substitute for
AC
F and
CD
F in Eq. (1):

24 24
(10.40 lb) (2 ) 0 500lb
26 25
BC BC
FF 500 lb
BC CD
FF C 
Free body: B:

(500 lb) (480 lb) (140 lb)
(10 7 )
12.21
BC
AB
BA AB
BD BD
CB
F
CB
FBA
FF
BA
FF

 

ik
jk
k




0: (480 lb) (200 lb) 0
BA BD BC
   FFFF i j
Substituting for
,,
BA BD BC
FFF and equating to zero the coefficients of j and k:
j:
10
200 lb 0 244.2 lb
12.21
 
AB AB
FF 244 lb
AB
FC 
k:
7
140 lb 0
12.21
AB BD
FF

7
( 244.2 lb) 140 lb 280 lb
12.21
BD
F    280 lb
BD
FT 
From symmetry:
AD AB
FF 244 lb
AD
FC 
www.elsolucionario.org

844. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.36*
The truss shown consists of six members and is supported by
a ball and socket at B, a short link at C, and two short links at
D. Determine the force in each of the members for
P  (2184 N)j and Q  0.

SOLUTION
Free body: Truss:
From symmetry:

and
xx yy
DB DB

0: 2 0
xx
FB 

0
xx
BD

0: 0
zz
FB 

0: 2 (2.8 m) (2184 N)(2 m) 0
cz y
MB  

780 N
y
B
Thus,
(780 N)Bj 
Free body: A:

( 0.8 4.8 2.1 )
5.30
(2 4.8 )
5.20
( 0.8 4.8 2.1 )
5.30
AB
AB AB
AC
AC AC
AD
AD AD
FAB
FF
AB
FAC
FF
AC
FAD
FF
AD



ijk
ij
ijk





0: (2184 N) 0
AB AC AD
    FFFF j


www.elsolucionario.org

845. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.36* (Continued)

Substituting for
,,,
AB AC AD
FFF and equating to zero the coefficients of ,, :ijk
i:
0.8 2
() 0
5.30 5.20
AB AD AC
FF F (1)
j:
4.8 4.8
( ) 2184 N 0
5.30 5.20
AB AD AC
FF F (2)
k:
2.1
()0
5.30
AB AD
FF
AD AB
FF
Multiply Eq. (1) by –6 and add Eq. (2):

16.8
2184 N 0, 676 N
5.20
AC AC
FF


 676 N
AC
FC 
Substitute for
AC
Fand
AD
Fin Eq. (1):

0.8 2
2(676N)0,861.25N
5.30 5.20
 
 
 
AB AB
FF 861 N
AB AD
FF C 
Free body: B:

(861.25 N) (130 N) (780 N) (341.25 N)
2.8 2.1
(0.8 0.6 )
3.5
AB
BC BC BC
BD BD
AB
AB
FF
F






Fijk
ik
Fik
Fk



0: (780 N) 0
AB BC BD
  FFFF j
Substituting for
,,
AB BC BD
FFF and equating to zero the coefficients of i and ,k
i:
130 N 0.8 0 162.5 N
BC BC
FF   162.5 N
BC
FT 
k:
341.25 N 0.6 0
BC BD
FF

341.25 0.6(162.5) 243.75 N 
BD
F 244 N
BD
FT 
From symmetry:
CD BC
FF 162.5 N
CD
FT 
www.elsolucionario.org

846. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.37*
The truss shown consists of six members and is supported by
a ball and socket at B, a short link at C, and two short links at
D. Determine the force in each of the members for P  0 and
Q  (2968 N)i.

SOLUTION
Free body: Truss:
From symmetry:

and
xx yy
DB DB

0: 2 2968 N 0
xx
FB  

1484 N
xx
BD

0: 2 (2.8 m) (2968 N)(4.8 m) 0
cz y
MB
  

2544 N
y
B
Thus,
(1484 N) (2544 N) Bij 
Free body: A:

( 0.8 4.8 2.1 )
5.30
(2 4.8 )
5.20
( 0.8 4.8 2.1 )
5.30
AB AB
AB
AC
AC AC
AD AD
AD
AB
FF
AB
F
FAC
FF
AC
AD
FF
AD
F





ijk
ij
ijk





0: (2968 N) 0
AB AC AD
    FFFF i www.elsolucionario.org

847. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.37* (Continued)

Substituting for
,,,
AB AC AD
FFF and equating to zero the coefficients of ,, ,ijk
i:
0.8 2
( ) 2968 N 0
5.30 5.20
AB AD AC
FF F (1)
j:
4.8 4.8
() 0
5.30 5.20
AB AD AC
FF F (2)
k:
2.1
()0
5.30
AB AD
FF
AD AB
FF
Multiply Eq. (1) by –6 and add Eq. (2):

16.8
6(2968 N) 0, 5512 N
5.20
AC AC
FF


 5510 N
AC
FC 
Substitute for
AC
F and
AD
F in Eq. (2):

4.8 4.8
2 ( 5512 N) 0, 2809 N
5.30 5.20
AB AB
FF
 
 
 

2810 N
AB AD
FF T 
Free body: B:

(2809 N) (424 N) (2544 N) (1113 N)
2.8 2.1
(0.8 0.6 )
3.5
AB
BC BC BC
BD BD
BA
BA
FF
F






Fijk
ik
Fik
Fk



0: (1484 N) (2544 N) 0
AB BC BD
     FFFF i j
Substituting for
,,
AB BC BD
FFF and equating to zero the coefficients of i and ,k
i:
24 N 0.8 1484 N 0, 1325 N
BC BC
FF    1325 N
BC
FT 
k:
1113 N 0.6 0
BC BD
FF

1113 N 0.6(1325 N) 1908 N,
BD
F   1908 N
BD
FC 
From symmetry:
CD BC
FF 1325 N
CD
FT 
www.elsolucionario.org

848. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.38*
The truss shown consists of nine members and is
supported by a ball and socket at A, two short
links at B, and a short link at C. Determine the
force in each of the members for the given
loading.

SOLUTION
Free body: Truss:
From symmetry:

0
zz
AB

0: 0
xx
FA 

0: (6 ft) (1600 lb)(7.5 ft) 0
BC y
MA  

2000 lb
y
A (2000 lb)Aj 
From symmetry:
y
BC

0: 2 2000 lb 1600 lb 0
yy
FB   

1800 lb
y
B (1800 lb)Bj 
Free body: A:
0: (2000 lb) 0
AB AC AD
    FFFF j
(0.6 0.8 ) (2000 lb) 0
22

 
ik ik
ij j
AB AC AD
FFF
Factoring i, j, k and equating their coefficient to zero,

11
0.6 0
22
AB AC AD
FFF (1)

0.8 2000 lb 0
AD
F 2500 lb
AD
FT 

11
0
22
AB AC
FF
AC AB
FF

www.elsolucionario.org

849. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.38* (Continued)

Substitute for
AD
F and
AC
F into Eq. (1):

2
0.6(2500 lb) 0, 1060.7 lb,
2
AB AB
F F 1061 lb
AB AC
FF C 
Free body: B:
(1060.7 lb) (750 lb)( )
2
(0.8 0.6 )
(7.5 8 6 )
12.5
BA AB
BC BC
BD BD
BE
BE BE
BA
F
BA
F
F
FBE
F
BE

  


 
ik
Fik
Fk
Fjk
Fijk



0: (1800 lb) 0
BA BC BD BE
     FFFFF j
Substituting for
,,,and
BA BC BD BE
FFF F and equating to zero the coefficients of ,, ,ijk
i:
7.5
750lb 0, 1250lb
12.5
BE BE
FF


 1250 lb
BE
FC 
j:
8
0.8 ( 1250 lb) 1800 lb 0
12.5
BD
F


 1250 lb
BD
FC 
k:
6
750lb 0.6( 1250lb) ( 1250lb) 0
12.5
BC
F   

2100 lb
BC
FT 
From symmetry:
1250 lb
BD CD
FF C 
Free body: D:
0: 0
DA DB DC DE
    FFFFFi


We now substitute for
,,
DADBDC
FFF and equate to zero the coefficient of i. Only
DA
F contains i and its
coefficient is

0.6 0.6(2500 lb) 1500 lb
AD
F 
i :
1500 lb 0
DE
F 1500 lb
DE
FT 
www.elsolucionario.org

850. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.39*
The truss shown consists of nine members and is supported by a
ball and socket at B, a short link at C, and two short links at D.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) Determine the force in each member for
P  (1200 N)j and Q  0.

SOLUTION
Free body: Truss:
0: 1.8 (1.8 3 ) ( )
By
CDD   Mijikjk

(0.6 0.75 ) ( 1200 ) 0  ik j

1.8 1.8 1.8
yz
CD D kkj

37209000
y
D iki
Equate to zero the coefficients of
,, :ijk
i:
39000, 300N
yy
DD 
j:
0,
z
D (300 N)Dj 
k:
1.8 1.8(300) 720 0C (100 N)Cj 

0: 300 100 1200 0    FBjj j (800 N)Bj 
Free body: B:
0: (800 N) 0,
BA BC BE
    FFFF j with
(0.6 3 0.75 )
3.15
AB
BA AB
FBA
BA
 FF ij k



BC BC
FFi
BE BE
FFk
Substitute and equate to zero the coefficient of
,, :jik
j:
3
800 N 0, 840 N,
3.315
AB AB
FF

 
 840 N
AB
FC 
i:
0.6
(840N) 0
3.15
BC
F


 160.0 N
BC
FT 
k:
0.75
( 840 N) 0
3.15
BE
F


 200 N
BE
FT  www.elsolucionario.org

851. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.39* (Continued)

Free body: C: 0: (100 N) 0,
CA CB CD CE
     FFFFF j with
( 1.2 3 0.75 )
3.317
AC
CA AC
FCA
F
CA
Fijk



(160 N)
CB
F i

(1.8 3)
3,499
CE
CD CD CE CE
FCE
F
CE
    FkFF ik


Substitute and equate to zero the coefficient of
,, :jik
j:
3
100 N 0, 110.57 N
3.317
AC AC
FF

 
 110.6 N
AC
FC 
i:
1.2 1.8
( 110.57) 160 0, 233.3
3.317 3.499
CE CE
FF    233 N
CE
FC 
k:
0.75 3
(110.57) (233.3) 0
3.317 3.499
CD
F  225 N
CD
FT 
Free body: D:
0: (300 N) 0,
DA DC DE
    FFFF j with
( 1.2 3 2.25 )
3.937
AD
DA AD
FDA
F
DA
Fijk



(225 N)
DC CD
FFk k
DE DE
FFi
Substitute and equate to zero the coefficient of
,, :jik
j:
3
300 N 0,
3.937
AD
F


 393.7 N,
AD
F 394 N
AD
FC 
i:
1.2
( 393.7 N) 0
3.937
DE
F


 120.0 N
DE
FT 
k:
2.25
( 393.7 N) 225 N 0
3.937



(Checks)
Free body: E:
Member AE is the only member at E which does not lie in the xz plane.
Therefore, it is a zero-force member.

0
AE
F 
www.elsolucionario.org

852. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.40*
Solve Problem 6.39 for P  0 and Q  (900 N)k.
PROBLEM 6.39* The truss shown consists of nine
members and is supported by a ball and socket at B, a short
link at C, and two short links at D. (a) Check that this truss
is a simple truss, that it is completely constrained, and that
the reactions at its supports are statically determinate. (b)
Determine the force in each member for P  (1200 N)j
and Q  0.

SOLUTION
Free body: Truss:
0: 1.8 (1.8 3 ) ( )
Byz
CDD   Mijikjk

(0.6 3 0.75 ) ( 900N) 0  ij k k

1.8 1.8 1.8
yz
CD Dkkj

3 540 2700 0
y
D ij i
Equate to zero the coefficient of
,, :ijk

327000 900N
yy
DD 

1.8 540 0 300 N
zz
DD 

1.8 1.8 0 900 N
yy
CD CD 
Thus,
(900 N) (900 N) (300 N)  CjDjk 

0: 900 900 300 900 0     FBjjkk (600 N)Bk 
Free body: B:

Since
Bis aligned with member BE,

0,
AB BC
FF 600 N
BE
FT 
Free body: C:
 0: (900 N) 0,
CA CD CE
    FFFF j with


www.elsolucionario.org

853. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.40* (Continued)


( 1.2 3 0.75 )
3.317
AC
CA AC
FCA
F
CA
Fijk



(1.8 3)
3.499
CE
CD CD CE CE
FCE
FF
CE
    FkF ik


Substitute and equate to zero the coefficient of
,, :jik
j:
3
900 N 0,
3.317
AC
F


 995.1 N
AC
F 995 N
AC
FT 
i:
1.2 1.8
(995.1) 0, 699.8 N
3.317 3.499
CE CE
FF 700 N
CE
FC 
k :
0.75 3
(995.1) ( 699.8) 0
3.317 3.499
CD
F 375 N
CD
FT 
Free body: D:
0: (375 N) +(900 N) (300 N) 0
DA DE
    FFF k j k
with ( 1.2 3 2.25 )
3.937
AD
DA AD
FDA
DA
FF ij k


and
DE DE
FF i
Substitute and equate to zero the coefficient
,, :jik
j:
3
900 N 0, 1181.1 N
3.937
AD AD
FF

 
 1181 N
AD
FC 
i:
1.2
( 1181.1 N) 0
3.937
DE
F

 
 360 N
DE
FT 
k:
2.25
( 1181.1 N 375 N 300 N 0)
3.937



(Checks)
Free body: E:
Member AE is the only member at E which
does not lie in the xz plane. Therefore, it is
a zero-force member.

0
AE
F 
www.elsolucionario.org

854. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.41*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at E.

SOLUTION
(a) Check simple truss.
(1) Start with tetrahedron BEFG.
(2) Add members BD, ED, GD joining at D.
(3) Add members BA, DA, EA joining at A.
(4) Add members DH, EH, GH joining at H.
(5) Add members BC, DC, GC joining at C.
Truss has been completed: It is a simple truss.
Free body: Truss:
Check constraints and reactions.
Six unknown reactions—ok; however, supports at A
and B constrain truss to rotate about AB and support
at G prevents such a rotation. Thus,
Truss is completely constrained and reactions are statically determinate.
Determination of reactions:

0: 11 ( ) (11 9.6 )
(10.08 9.6 ) (275 240 ) 0
Ayz
BB G    
  
Mijkikj
jk i k


11 11 11 9.6 (10.08)(275)
(10.08)(240) (9.6)(275) 0
yy
BBGG

kjki k
ij

Equate to zero the coefficient of i, j, k:

: 9.6 (10.08)(240) 0 252 lbi GG ( 252 lb)Gj 
: 11 (9.6)(275) 0 240 lb  j
zz
BB

: 11 11( 252) (10.08)(275) 0, 504 lb
yy
BB   k (504 lb) (240 lb)Bjk 
www.elsolucionario.org

855. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.41* (Continued)


0: (504 lb) (240 lb) (252 lb)
(275 lb) (240 lb) 0
   
 
FA j k j
ik


(275 lb) (252 lb) Aij 
Zero-force members.
The determination of these members will facilitate our solution.
FB: C: Writing
0, 0, 0
xyz
FFF   yields 0
BC CD CG
FFF 
FB : F: Writing 0, 0, 0
xyz
FFF   yields 0
BF EF FG
FFF 
FB : A: Since 0,
z
A writing 0
z
F yields 0
AD
F 
FB : H: Writing 0
y
F yields 0
DH
F 
FB : D: Since 0,
AD CD DH
FFF  we
need consider only members DB, DE, and
DG.
Since
DE
F is the only force not contained in plane BDG, it must
be zero. Simple reasonings show that the other two forces are also
zero.

0
BD DE DG
FFF 
The results obtained for the reactions at the supports and for the zero-force members are shown on
the figure below. Zero-force members are indicated by a zero (“0”).

(b) Force in each of the members joined at E. We already found that 0
DE EF
FF 
Free body: A:
0
y
F yields 252 lb
AE
FT  Free body: H: 0
z
F yields 240 lb
EH
FC 
www.elsolucionario.org

856. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.41* (Continued)

Free body: E:
0: (240 lb) (252 lb) 0
EB EG
    FFF k j
(11 10.08 ) (11 9.6 ) 240 252 0
14.92 14.6
EGBE
FF
 ij ikkj
Equate to zero the coefficient of y and k:

10.08
: 252 0
14.92
BE
F



j 373 lb
BE
FC 

:k
9.6
240 0
14.6
EG
F




365 lb
EG
FT 
www.elsolucionario.org

857. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.42*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at
G. (a) Check that this truss is a simple truss, that it is
completely constrained, and that the reactions at its supports
are statically determinate. (b) For the given loading,
determine the force in each of the six members joined at G.

SOLUTION
See solution to Problem 6.41 for part (a) and for reactions and zero-force members.
(b) Force in each of the members joined at G.
We already know that

0
CG DG FG
FFF 
Free body: H: 0
x
F yields 275 lb
GH
FC  Free body: G: 0: (275 lb) (252 lb) 0
GB GE
    FFF i j
( 10.08 9.6 ) ( 11 9.6 ) 275 252 0
13.92 14.6

BG EG
FF
jk ik i j
Equate to zero the coefficient of i, j, k:

11
:2750
14.6
EG
F



i 365 lb
EG
FT 

10.08
:2520
13.92
BG
F



j 348 lb
BG
FC 

9.6 9.6
: ( 348) (365) 0
13.92 14.6
 
  
 
k
(Checks)
www.elsolucionario.org

858. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.43
A Mansard roof truss is loaded as shown. Determine the force in
members DF, DG, and EG.

SOLUTION
Reactions:
Because of the symmetry of the truss and loadings, 
1
0, (1.2 kN) 5 3 kN
2
xy
AAL 
We pass a section through DF, DG, and EG and use the free body shown:

Member DF:

 
 0: (1.2 kN) 8 m 1.2 kN 4 m
3 kN 10.25 m (3 m) 0
G
DF
M
F
 



5.45 kN
DF
F 5.45 kN
DF
FC 
Member DG:


3
0: 3 kN 1.2 kN 1.2 kN 0
5
yDG
FF

  




1.000 kN
DG
F

1.000 kN
DG
FT 
Member EG:
0: 1.2 kN 4 m (3 m) (3 kN)(6.25 m) 0
DEG
MF   

4.65 kN
EG
FT 
www.elsolucionario.org

859. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.44
A Mansard roof truss is loaded as shown. Determine the force in
members GI, HI, and HJ.

SOLUTION
Reactions:
Because of the symmetry of the truss and loadings, 
1
0, (1.2 kN) 5 3 kN
2
xy
AAL 
We pass a section through GI, HI, and HJ and use the free body shown:

Member DF:
  0: 3 kN 6.25 m 1.2 kN 4 m (3 m) 0
HGI
MF   

4.65 kN
GI
F 4.65 kN
GI
FT 
Member HI:


0: 3 kN 1.2 kN 0
yHI
FF   

1.800 kN
HI
F

1.800 kN
HI
FC 
Member HJ:
 0: 3 kN 6.25 m (3 m) (1.2 kN)(4 m) 0
IHJ
MF   
4.65 kN
HJ
F



4.65 kN
HJ
FC 

Check 0 : 4.65 4.65 0 o.k.
x
  F
www.elsolucionario.org

860. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.45
Determine the force in members BD and CD of the truss
shown.

SOLUTION








Reactions from Free body of entire truss:
27 kips
y
AA



45 kipsH



We pass a section through members BD, CD, and CE and use the
free body shown.
0: (7.5 ft) (27 kips)(10 ft) 0
CBD
MF   

36.0 kips
BD
F 36.0 kips
BD
FC 

3
0: 27 kips+ 0
5
yCD
FF

 




45.0 kips
CD
F 45.0 kips
CD
FC 



www.elsolucionario.org

861. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.46
Determine the force in members DF and DG of the truss
shown.

SOLUTION








Reactions from Free body of entire truss:
27 kips
y
AA



45 kipsH



We pass a section through members DF, DG, and EG and use the
free body shown.

0: (7.5 ft) (45 kips)(10 ft) 0
GFD
MF  

60.0 kips
FD
F 60.0 kips
FD
FC 

3
0: 45 kips+ 36 kips 0
5
yGD
FF

  




15.00 kips
GD
F 15.00 kips
GD
FC 




www.elsolucionario.org

862. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.47
Determine the force in members CD and DF of the truss shown.

SOLUTION
Reactions:







0: (12 kN)(4.8 m) (12 kN)(2.4 m) (9.6 m) 0
J
MB   

9.00 kNB

0: 9.00 kN 12.00 kN 12.00 kN 0
y
FJ   

15.00 kNJ


Member CD:


0: 9.00 kN 0
yCD
FF
 9.00 kN
CD
F C 






Member DF:
0: (1.8 m) (9.00 kN)(2.4 m) 0
CDF
MF   12.00 kN
DF
F T 
www.elsolucionario.org

863. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.48
Determine the force in members FG and FH of the truss shown.

SOLUTION
Reactions:







0: (12 kN)(4.8 m) (12 kN)(2.4 m) (9.6 m) 0
J
MB   

9.00 kNB

0: 9.00 kN 12.00 kN 12.00 kN 0
y
FJ 

15.00 kNJ

Member FG:



3
0: 12.00 kN 15.00 kN 0
5
yFG
FF
   

5.00 kN
FG
F T 




Member FH:
0: (15.00 kN)(2.4 m) (1.8 m) 0
GFH
MF

20.0 kN
FH
F T 
www.elsolucionario.org

864. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.49
Determine the force in members CD and DF of the truss shown.

SOLUTION



5
tan 22.62
12

 
512
sin cos
13 13

Member CD:
0: (9 m) (10 kN)(9 m) (10 kN)(6 m) (10 kN)(3 m) 0
ICD
MF    

20 kN
CD
F 20.0 kN
CD
F C 
Member DF:

0: ( cos )(3.75 m) (10 kN)(3 m) (10 kN)(6 m) (10 kN)(9 m) 0
CDF
MF     

cos 48 kN
DF
F 

12
48 kN 52.0 kN
13
DF DF
FF

 

 52.0 kN
DF
F C 
www.elsolucionario.org

865. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.50
Determine the force in members CE and EF of the truss shown.

SOLUTION


Member CE:
0: (2.5 m) (10 kN)(3 m) (10 kN)(6 m) 0
FCE
MF   

36 kN
CE
F 36.0 kN
CE
F T 
Member EF:

0: (6 m) (10 kN)(6 m) (10 kN)(3 m) 0
IEF
MF   

15 kN
EF
F 15.00 kN
EF
F C 
www.elsolucionario.org

866. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.51
Determine the force in members DE and DF of the truss
shown when P  20 kips.

SOLUTION
Reactions:

2.5PCK


7.5
tan
18
22.62





Member DE:
0: (2.5 )(6 ft) (12 ft) 0
ADE
MPF  

1.25
DE
FP
For
20 kips,P 1.25(20) 25 kips
DE
F  25.0 kips
DE
FT 
Member DF:

0: (12 ft) (2.5 )(6 ft) cos (5 ft) 0
EDF
MP PF    

12 15 cos 22.62 (5 ft) 0
DF
PPF  

0.65
DF
FP
For
20 kips,P 0.65(20) 13 kips
DF
F 

13.00 kips
DF
FC 
www.elsolucionario.org

867. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.52
Determine the force in members EG and EF of the truss
shown when P
 20 kips.

SOLUTION

Reactions:

2.5
7.5
tan
6
51.34
P





CK



Member EG:
0 : (18 ft ) 2.5 (12 ft ) (6 ft ) (7.5 ft ) 0
FEG
MP PPF    

0.8 ;
EG
F P
For
20 kips,P 0.8(20) 16 kips
EG
F 16.00 kips
EG
F T 
Member EF:

0: 2.5 (6 ft) (12 ft) sin 51.34 (12 ft) 0
AEF
MPPF    
0.320 ;
EF
F P
For 20 kips,P
 0.320(20) 6.4 kips
EF
F 
6.40 kips
EF
F C 

www.elsolucionario.org

868. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.53
Determine the force in members DF and DE of the truss shown.

SOLUTION

11
tan 7.1
8




 

Member CE:
0: (1.75 m) (30 kN)(4 m) (20 kN)(2 m) 0
FDF
MF    

91.4 kN
DF
F 91.4 kN
DF
FT 
Member EF:
0: (14 m) (30 kN)(10 m) 20 kN)(12 m) 0
ODE
MF    

38.6 kN
DE
F 38.6 kN
DE
FC 
www.elsolucionario.org

869. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.54
Determine the force in members CD and CE of the truss shown.

SOLUTION

1
11
tan 7.1
8
1.5
tan 36.9
2










 


Member CE:
0: sin 36.9 (14 m) (30 kN)(10 m) (20 kN)(12 m) 0
OCD
MF   



64.2 kN
CD
F 64.2 kN
CD
FT 
Member EF: 0: cos 7.1 (1.75 m) (30 kN)(4 m) (20 kN)(2 m) 0
OCE
MF   



92.1 kN
CE
F 92.1 kN
CE
FC 
www.elsolucionario.org

870. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.55
A monosloped roof truss is loaded as shown. Determine the
force in members CE, DE, and DF.

SOLUTION
Reactions at supports: Because of the symmetry of the loading,

0
11
(Total load) (8 kN)
22
x
y
A
AI

 
4kNAI 
We pass a section through members CD, DE, and DF, and use the free body shown.
(We moved
DE
Fto E and
DF
Fto F)
Slope
2.16 m 9
9.6 m 40
BJ

Slope
9
40
1m 5
2.4 m 12
0.46 m 0.46 m
2.0444 m
Slope
DE
a
BJ





0 : (1 m) (1 kN)(2.4 m) (4 kN)(2.4 m) 0   
DCE
MF


7.20 kN
CE
F

7.20 kN
CE
FT



0: (4 kN)(2.0444 m) (1 kN)(2.0444 m)
5
(2 kN)(4.4444 m) (6.8444 m) 0
13
K
DE
M
F
 

 


1.047 kN
DE
F 1.047 kN
DE
FC 

0: (1 kN)(4.8 m) (2kN)(2.4 m) (4 kN)(4.8 m)
40
(1.54 m) 0
41
E
DF
M
F
  





6.39 kN
DF
F 6.39 kN
DF
FC 
www.elsolucionario.org

871. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.56
A monosloped roof truss is loaded as shown. Determine the
force in members EG, GH, and HJ.

SOLUTION
Reactions at supports: Because of the symmetry of the loading,

0
11
(Total load) (8 kN)
22
x
y
A
AI

 
4kNAI


We pass a section through members EG, GH, and HJ, and use the free body shown.

0: (4 kN)(2.4 m) (1 kN)(2.4 m) (2.08 m) 0
HEG
MF   

3.4615 kN
EG
F 3.46 kN
EG
FT 

0: (2.4 m) (2.62 m) 0
JGH EG
MF F   

2.62
(3.4615 kN)
2.4
GH
F

3.7788 kN
GH
F 3.78 kN
GH
FC 

2.4
0: 0
2.46
xEGHJ
FF F   


2.46 2.46
(3.4615 kN)
2.4 2.4
HJ EG
FF 


3.548 kN
HJ
F

3.55 kN
HJ
FC




www.elsolucionario.org

872. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.57
A Howe scissors roof truss is loaded as shown.
Determine the force in members DF, DG, and EG.

SOLUTION
Reactions at supports.
Because of symmetry of loading,

11
0, (total load) (9.60 kips) 4.80 kips
22
xy
AAL  

4.80 kipsAL


We pass a section through members DF, DG, and EG, and use the free body shown.
We slide
DF
F to apply it at F:

22
0: (0.8 kip)(24 ft) (1.6 kips)(16 ft) (1.6 kips)(8 ft)
8
(4.8 kips)(24 ft) (6 ft) 0
83.5
G
DF
M
F
  
 



10.48 kips, 10.48 kips
DF DF
FFC  

22 22
0: (1.6 kips)(8 ft) (1.6 kips)(16 ft)
2.5 8
(16 ft) (7 ft) 0
8 2.5 8 2.5
A
DG DG
M
FF
  




3.35 kips, 3.35 kips
DG DG
FFC  

22
8
0: (0.8 kips)(16 ft) (1.6 kips)(8 ft) (4.8 kips)(16 ft) (4 ft) 0
81.5
EG
D
F
M
    

 13.02 kips, 13.02 kips
EG EG
FFT  
www.elsolucionario.org

873. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.58
A Howe scissors roof truss is loaded as shown.
Determine the force in members GI, HI, and HJ.

SOLUTION
Reactions at supports. Because of symmetry of loading,
0,
x
A
1
(Total load)
2
1
(9.60 kips)
2
y
AL


4.80 kips 4.80 kipsAL 
We pass a section through members GI, HI, and HJ, and use the free body shown.


22
16
0: (4 ft) (4.8 kips)(16 ft) (0.8 kip)(16 ft) (1.6 kips)(8 ft) 0
16 3
GI
H
F
M    



13.02 kips
GI
F 13.02 kips
GI
FT 

0: (1.6 kips)(8 ft) (16 ft) 0
0.800 kips
LHI
HI
MF
F
  


0.800 kips
HI
FT 
We slide
HG
F to apply it at H.

22
8
0: (4 ft) (4.8 kips)(16 ft) (1.6 kips)(8 ft) (0.8 kip)(16 ft) 0
83.5
HJ
I
F
M
    


13.97 kips
HJ
F 13.97 kips
HJ
FC 
www.elsolucionario.org

874. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.59
Determine the force in members AD, CD, and CE of
the truss shown.

SOLUTION
Reactions:


0: 9 kips(8 ft) (45 ft) 5 kips(30 ft) 5 kips(15 ft) 0
k
MB    

6.6 kipsB

0: 9 kips 0 9 kips
xxx
FK     K


0: 6.6 kips 5 kips 5 kips 0 3.4 kips
yyy
FK      K


0: 9 kips(4 ft) 6.6 kips(7.5 ft) (4 ft) 0
CAD
MF   

3.375 kips
AD
F 3.38 kN
AD
FC 

8
0: (15 ft) 0
17
ACD
MF

 


0
CD
F 

15
0: (8 ft) 6.6 kips(15 ft) 0
17
DCE
MF

  




14.025 kips
CE
F 14.03 kips
CE
FT 
www.elsolucionario.org

875. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.60
Determine the force in members DG, FG, and FH of the
truss shown.

SOLUTION
Reactions:


0: 9 kips(8 ft) (45 ft) 5 kips(30 ft) 5 kips(15 ft) 0
k
MB    

6.6 kipsB

0: 9 kips 0 9 kips
xxx
FK     K


0: 6.6 kips 5 kips 5 kips 0 3.4 kips
yyy
FK      K


0: 9 kips(4 ft) 6.6 kips(22.5 ft) 5 kips(7.5 ft) (4 ft) 0
F DG
MF    

18.75 kips
DG
F 18.75 kips
DG
FC 

8
0: 6.6 kips(15 ft) (15 ft) 0
17
DFG
MF

  




14.025 kips
FG
F 14.03 kips
FG
FT 

15
0: 5 kips(15 ft) 6.6 kips(30 ft) (8 ft) 0
17
GFH
MF

   




17.425 kips
FH
F

17.43 kips
FH
FT 
www.elsolucionario.org

876. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.61
Determine the force in members DG and FI of the truss shown. (Hint:
Use section aa.)

SOLUTION



0: (4 m) (5 kN)(3 m) 0
FDG
MF



3.75 kN
DG
F 3.75 kN
DG
F T 

0: 3.75 kN 0
yFI
FF  


3.75 kN
FI
F

3.75 kN
FI
F C



www.elsolucionario.org

877. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.62
Determine the force in members GJ and IK of the truss shown. (Hint:
Use section bb.)

SOLUTION






0: (4 m) (5 kN)(6 m) (5 kN)(3 m) 0
IGJ
MF



11.25 kN
GJ
F 11.25 kN
GJ
F T 

0: 11.25 kN 0
yIK
FF  


11.25 kN
IK
F

11.25 kN
IK
F C



www.elsolucionario.org

878. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.63
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)

SOLUTION
Reactions:


0: 0
xx
FA 

0: 12 kips(45 ft) 12 kips(30 ft) 12 kips(15 ft) (90 ft) 0
P y
MA    

12 kips
y
A

0: 12 kips 12 kips 12 kips 12 kips 0
y
FP  24 kipsP

0 : (12 kips)(30 ft ) (16 ft ) 0
GEH
MF   


22.5 kips
EH
F

22.5 kips
EH
FC




0: 22.5 kips 0
xGI
FF  

22.5 kips
GI
FT



www.elsolucionario.org

879. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.64
Determine the force in members HJ and IL of the
truss shown. (Hint: Use section bb.)

SOLUTION
See the solution to Problem 6.63 for free body diagram and analysis to determine the reactions at supports
A and P.

0; 12.00 kips
xy
AA ; 24.0 kipsP


0 : (16 ft ) (12 kips)(15 ft ) (24 kips)(30 ft ) 0
LHJ
MF   


33.75 kips
HJ
F 33.8 kips
HJ
FC 

0: 33.75 kips 0
xIL
FF  


33.75 kips
IL
F

33.8 kips
IL
FT



www.elsolucionario.org

880. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.65
The diagonal members in the center
panels of the power transmission line
tower shown are very slender and can act
only in tension; such members are known
as counters. For the given loading,
determine (a) which of the two counters
listed below is acting, (b) the force in
that counter.
Counters CJ and HE

SOLUTION
Free body: Portion ABDFEC of tower.
We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and
EJ.


4
0: 2(1.2 kN)sin 20 0 1.026 kN
5
xCJ CJ
FF F   
Since CJ is found to be in tension, our assumption was correct. Thus, the answers are
( a) CJ

( b)
1.026 kNT 
www.elsolucionario.org

881. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.66
The diagonal members in the center
panels of the power transmission line
tower shown are very slender and can act
only in tension; such members are known
as counters. For the given loading,
determine (a) which of the two counters
listed below is acting, (b) the force in that
counter.
Counters IO and KN

SOLUTION
Free body: Portion of tower shown.

We assume that counter IO is acting and show the forces exerted by that counter and by members IN and
KO.

4
0 : 4(1.2 kN)sin 20 0 2.05 kN
5
xIO IO
FF F   

Since IO is found to be in tension, our assumption was correct. Thus, the answers are
( a) IO

( b)
2.05 kNT 

www.elsolucionario.org

882. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.67
The diagonal members in the center panels of the truss shown are
very slender and can act only in tension; such members are known
as counters. Determine the force in member DE and in the counters
that are acting under the given loading.

SOLUTION
Reactions from free body of entire truss:

0: 12 kips
H
M  A

0 : 15 kips
y
F  H
Portion to the left of vertical cut through panel BCDE:



Required vertical component of bar forces

and
BE CD
FF is downward. We must have
tension and 0.
BE CD
FF

3
0: 12 kips 6 kips 0
5
yBE
FF

   

 10.00 kips
BE
FT 


Required vertical component of bar forces
EF
and
DG
FF is downward. We must have
tension and 0.
EF DG
FF
www.elsolucionario.org

883. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.67 (Continued)


3
0: 12 kips+15 kips 0
5
yEF
FF

   




5.00 kips
EF
F

5.00 kips
EF
FT


Member DE: Free body of joint D 0: 0
yDE
FF 

Joint E: (check)

 
33
0 : 10 kips 5 kips 9 kips+ 0
55
yDE
FF
 
   
 
 


0 (checks)
DE
F

www.elsolucionario.org

884. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.68
Solve Prob. 6.67 assuming that the 9-kip load has been removed.
Problem 6.67 The diagonal members in the center panels of the
truss shown are very slender and can act only in tension; such
members are known as counters. Determine the force in member
DE and in the counters that are acting under the given loading.

SOLUTION
Reactions from free body of entire truss:

0: 7.5 kips
H
M  A

0 : 10.5 kips
y
F  H
Portion to the left of vertical cut through panel BCDE:



Required vertical component of bar forces

and
BE CD
FF is downward. We must have
tension and 0.
BE CD
FF

3
0: 7.5 kips 6 kips 0
5
yBE
FF

   

 2.50 kips
BE
FT 
Portion to the right of vertical cut through panel DFEG:



Required vertical component of bar forces
EF
and
DG
FF is downward. We must have
tension and 0.
DG EF
FF
www.elsolucionario.org

885. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.68 (Continued)


3
0: 12 kips+10.5 kips+ 0
5
yDG
FF

  




2.5 kips
DG
F

2.50 kips
DG
FT


Member DE: Free body of joint D

3
0 : 2.5 kips 0
5
yDE
FF  



1.500 kips
DE
F

1.500 kips
DE
FC



www.elsolucionario.org

886. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.69
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)


SOLUTION
Structure (a)
Number of members:

16m
Number of joints:

10n
Reaction components:

4
20, 2 20
r
mr n

 

Thus,
2mr n 
To determine whether the structure is actually completely constrained and determinate, we must try to
find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body
diagram of each truss.

This is a properly supported simple truss – O.K.

This is an improperly supported simple
truss. (Reaction at C passes through B. Thus,
Eq.
0
B
M cannot be satisfied.)
Structure is impr operly constrained.


Structure (b)

16
10
4
20, 2 20
m
n
r
mr n



 

Thus,
2mr n 
(a)
(b) www.elsolucionario.org

887. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.69 (Continued)

We must again try to find the reactions at the supports dividing the structure as shown.

Both portions are simply supported simple trusses.
Structure is completely constrained and determinate.

Structure (c)

17
10
4
21, 2 20
m
n
r
mr n



 

Thus,
2mr n 
This is a simple truss with an extra support which causes reactions (and forces in members) to be
indeterminate.
Structure is completely co nstrained and indeterminate.


(c) www.elsolucionario.org

888. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.70
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)


SOLUTION
Structure (a):
Nonsimple truss with 4,r 16,m 10,n
so
20 2 ,mr n  but we must examine further.

FBD Sections:


FBD I:
0
A
M 
1
T
II:
0
x
F 
2
T
I:
0
x
F 
x
A
I:
0
y
F 
y
A
II:
0
E
M 
y
C
II:
0
y
F 
y
E
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate.

Structure (b):
Nonsimple truss with 3,r 16,m 10,n
so
19 2 20mr n   Structure is partially constrained. www.elsolucionario.org

889. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.70 (Continued)

Structure (c):

Simple truss with
3,r 17,m 10,n 20 2 ,mr n  but the horizontal reaction forces and
xx
AE
are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained
and indeterminate.

www.elsolucionario.org

890. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.71
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)


SOLUTION
Structure (a):

Nonsimple truss with 4,r 12,m 8n so 16 2 .rm n 
Check for determinacy:
One can solve joint F for forces in EF, FG and then solve joint
E for
y
E and force in DE.
This leaves a simple truss ABCDGH with

3, 9, 6 so 12 2rmn rm n  Structure is completely constrained and determinate. 
Structure (b):

Simple truss (start with ABC and add joints alphabetically to complete truss) with
4,r 13,m 8n
so
17 2 16rm n   Constrained but indeterminate 
www.elsolucionario.org

891. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.71 (Continued)

Structure (c):

Nonsimple truss with
3,r 13,m 8n so 16 2 . rm n To further examine, follow procedure in
part (a) above to get truss at left.
Since
1
0F (from solution of joint F),
1A
MaF

0 and there is no equilibrium.
Structure is improp erly constrained.


www.elsolucionario.org

892. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.72
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)


SOLUTION
Structure (a)
Number of members:

12m
Number of joints:

8n
Reaction components:

3
15, 2 16
r
mr n

 

Thus,
2mr n 
Structure is partially constrained.


Structure (b)

13, 8
3
16, 2 16


 
mn
r
mr n

Thus,
2mr n 
To verify that the structure is actually completely constrained and determinate, we observe that it is a
simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a
roller. Thus,
structure is completely constrained and determinate.



www.elsolucionario.org

893. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.72 (Continued)

Structure (c)
13, 8
4
17, 2 16
mn
r
mr n


 
Thus,
2mr n 
Structure is completely co nstrained and indeterminate.

This result can be verified by observing that the structure is a simple truss (follow lettering to check this),
therefore it is rigid, and that its supports involve four unknowns.

www.elsolucionario.org

894. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.73
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)


SOLUTION
Structure (a): Rigid truss with 3,r 14,m 8,n
so 17 2 16rm n  
so completely constrained but indeterminate

Structure (b): Simple truss (start with ABC and add joints alphabetically), with 3, 13, 8, so 16 2rm n rm n 
so completely constrained and determinate


Structure (c):

Simple truss with 3,r 13,m 8,n so 16 2 ,rm n  but horizontal reactions ( and )
xx
AD are
collinear, so cannot be resolved by any equilibrium equation.
Structure is improp erly constrained.

www.elsolucionario.org

895. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.74
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)


SOLUTION
Structure (a):
No. of members 12m
No. of joints
8162nmr n
No. of reaction components
4 unknows equationsr
FBD of EH:


0
H
M ;
DE
F 0
x
F ;
GH
F 0
y
F
y
H
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate.

Structure (b):
No. of members 12m
No. of joints
815216nmr n
No. of reaction components
3 unknows equationsr
partially constrained

Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents
this action. www.elsolucionario.org

896. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.74 (Continued)

Structure (c): No. of members 13m
No. of joints
817216nmr n
No. of reaction components
4 unknows equationsr
completely constrained but indeterminate

www.elsolucionario.org

897. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.75
Determine the force in member BD and the components of the reaction
at C.

SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:

We have

22
1.92 0.56 2 mBD

0:
C
M

0.56 1.92
(310 N)(2.1m) (1.4 m) 1.4 m 0
22
BD BD
FF





375 N
BD
F 375 N
BD
FC 

0.56
0: 310 N ( 375 N) 0
2
xx
FC    

205 N
x
C 205 N
x
C 


1.92
0: ( 375 N) 0
2
yy
FC   

360 N
y
C 360 N
y
C 
www.elsolucionario.org

898. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.76
Determine the force in member BD and the components
of the reaction at C.

SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:



22
(24) (10) 26 in.BD

10
0 : (160 lb)(30 in.) (16 in.) 0
26
CBD
MF

  




780 lb
BD
F 780 lb
BD
FT 

24
0: (780 lb) 0
26
xx
MC  


720 lb
x
C

720 lb
x
C 


10
0: 160 lb (780 lb) 0
26
yy
FC   


140.0 lb
y
C

140.0 lb
y
C



www.elsolucionario.org

899. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.77
For the frame and loading shown, determine the force
acting on member ABC (a) at B, (b) at C.

SOLUTION
FBD ABC:

: is two-force memberNote BD
(a)
3
0: (0.09 m)(200 N) (2.4 m) 0
5
CBD
MF

  



125.0 N
BD
F 36.9 
(b)
4
0: 200 N (125 N) 0 100 N
5
xxx
FC     C


33
0: 0 (125 N) 75 N
55
yBDy y
FFC     C


125.0 NC 36.9 

www.elsolucionario.org

900. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.78
Determine the components of all forces acting on member ABCD of
the assembly shown.

SOLUTION
Free body: Entire assembly:
0: (6 in.) (120 lb)(4 in.) 0
B
MD  

80.0 lbD 

0: 120 lb 0
xx
FB  

120.0 lb
x
B 

0: 80 lb 0
yy
FB  

80.0 lb
y
B 
Free body: Member ABCD:
0: (80 lb)(10 in.) (8 in.) (120 lb)(2 in.)
(80 lb)(4 in.) 0
A
MC  


30.0 lbC 

0: 120 lb 0
xx
FA  

120.0 lb
x
A 

0: 80 lb 30 lb 80 lb 0
yy
FA    

30.0 lb
y
A 
www.elsolucionario.org

901. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.79
For the frame and loading shown, determine the components of all forces
acting on member ABC.

SOLUTION
Free body: Entire frame:
0: 18 kN 0
xx
FA  

18 kN
x
A 18.00 kN
x
A 

0: (18 kN)(4 m) (3.6 m) 0
Ey
MA   

20 kN
y
A 20.0 kN
y
A 

0: 20 kN 0
y
FF  

20 kNF

20 kNF
Free body: Member ABC
Note: BE is a two-force member, thus B is directed along line BE.

0: (4 m) (18 kN)(6 m) (20 kN)(3.6 m) 0
C
MB   

9kNB 9.00 kNB 

0: 18 kN 9 kN 0
xx
FC   

9kN
x
C 9.00 kN
x
C 

0: 20 kN 0
yy
FC  

20 kN
y
C 20.0 kNC
y

www.elsolucionario.org

902. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.80
Solve Problem 6.79 assuming that the 18-kN load is replaced by a clockwise
couple of magnitude 72 kN · m applied to member CDEF at Point D.
PROBLEM 6.79 For the frame and loading shown, determine the
components of all forces acting on member ABC.

SOLUTION
Free body: Entire frame
0: 0
xx
FA 

0: 72 kN m (3.6 m) 0
Fy
MA   

20 kN 20 kN
yy
A  A 20.0 kNA 

Free body: Member ABC
Note: BE is a two-force member. Thus B is directed along line BE.

0: (4 m) (20 kN)(3.6 m) 0  
C
MB

18 kNB 18.00 kNB 

0: 18 kN 0
xx
FC 

18 kN
x
C 18.00 kN
x
C 


0: 20 kN 0
yy
FC  

20 kN
y
C 20.0 kN
y
C 

www.elsolucionario.org

903. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.81
Determine the components of all forces acting on member
ABCD when
0.

SOLUTION
Free body: Entire assembly


0: (8 in.) (60 lb)(20 in.) 0  
B
MA

150 lbA 150.0 lbA 

0: 150.0 lb 0 150 lb
xx x
FB B    150.0 lb
x
B 

0: 60.0 lb 0 60.0 lb
yy y
FB B    60.0 lb
y
B 
Free body: Member ABCD We note that
D is directed along DE, since DE is a two-force member.
0: (12) (60 lb)(4) (150 lb)(8) 0
C
MD   

80 lbD 80.0 lbD 

0: 150.0 150.0 0 0
xx x
FC C    

0: 60.0 80.0 0 20.0 lb
yy y
FC C    

20.0 lbC 


www.elsolucionario.org

904. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.82
Determine the components of all forces acting on member
ABCD when
90 .

SOLUTION
Free body: Entire assembly


0: (8 in.) (60 lb)(8 in.) 0  
B
MA


60.0 lbA 60.0 lbA 

0: 60 lb 60 lb 0 0    
xx x
FB B

0: 0
yy
FB  0B 
Free body: Member ABCD We note that
D is directed along DE, since DE is a two-force member.
0: (12 in.) (60 lb)(8 in.) 0
C
MD  

40.0 lbD 40.0 lbD 

0: 60 lb 0  
xx
FC

60 lb
x
C 60.0 lb
x
C 


0: 40 lb 0
yy
FC  

40 lb
y
C 40.0 lb
y
C 

www.elsolucionario.org

905. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.83
Determine the components of the reactions at A and
E, (a) if the 800-N load is applied as shown, (b) if
the 800-N load is moved along its line of action and
is applied at point D.

SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the 800-N force
on its line of action is immaterial.


0: (1200) (800 N)(900) 0
Ay
ME  

600 N
y
E 

0: 0 (1)
xxx
FAE  


0: 600 800 0
yy
FA    200 N
y
A 

(a) Free body: Member BE:
We note that E is directed along EB, since BE is a two-force member.

900
or 600 N
900 200 200yx
x
EE
E 





2700 N
x
E 

From equation (1): 2700 N 0
x
A

2700 N
x
A

2700 N
x
A 
www.elsolucionario.org

906. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.83 (Continued)

(b) Free body: Member ABC: We note that A is directed along AB, since ABC is a two-force member.

300
or 200 N
300 200 200yx
x
AA
A 





300 N
x
A 
From equation (1): 300 N 0
x
E

300 N
x
E

300 N
x
E 

www.elsolucionario.org

907. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.84
Determine the components of the reactions at D and E if the
frame is loaded by a clockwise couple of magnitude 150 N·m
applied (a) at A, (b) at B.

SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the couple is
immaterial.


0: (0.6 m) (150 N m) 0
Dy
ME  

250 N
y
E 

0: 0 (1)
xxx
FDE  


0: 250 N 0
yy
FD   250 N
y
D 

(a) Free body: Member BCE:
We note that E is directed along EC, since BCE is a two-force member.

1.2
or 250 N
1.2 0.40 0.40yx
x
EE
E 





750 N
x
E 

From equation (1): D 750 N 0
x


750 N
x
D

750 N
x
D  www.elsolucionario.org

908. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.84 (Continued)

(b) Free body: Member ACD:
We note that D is directed along DC, since ACD is a two-force member.

0.6
or D 250 N
0.6 0.4 0.4yx
x
DD 





375 N
x
D 


From equation (1): 375 N 0
x
E

375 N
x
E

375 N
x
E 
www.elsolucionario.org

909. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.85
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.


SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since position of load on its line of action is
immaterial.

0: (750 N)(80 mm) (200 mm) 0   
Ex
MA

300 N 300 N
xx
A  A

0: 300 N 0 300 N 300 N
xx x x
FE E     E

0: 750 N 0
yyy
FAE   
(1)


(a) Load applied at B.
Free body: Member CE:
CE is a two-force member. Thus, the reaction at E must be directed along CE.

75 mm
90 N
300 N 250 mmy
y
E
E

From Eq. (1):
90 N 750 N 0
y
A  660 N
y
A
Thus, reactions are

300 N
x
A , 660 N
y
A 

300 N
x
E , 90.0 NE
y 
(b) Load applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.

125 mm
150 N
300 N 250 mmy
y
A
A

www.elsolucionario.org

910. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.85 (Continued)

From Eq. (1):
750 N 0
150 N 750 N 0
yy
y
AE
E
 
 

600 N 600 N
yy
E E
Thus, reactions are
300 N
x
A , 150.0 N
y
A 

300 N
x
E , 600 N
y
E 
www.elsolucionario.org

911. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.86
Determine the components of the reactions at A and E if the frame is
loaded by a clockwise couple of magnitude 36 N · m applied (a) at B,
(b) at D.

SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since the point of application of the couple is
immaterial.

0: 36 N m (0.2 m) 0   
Ex
MA

180 N 180 N
xx
A  A

0: 180 N+ 0  
xx
FE

180 N 180 N E
xx
E

0: 0
yyy
FAE  
(1)

(a) Couple applied at B.
Free body: Member CE:
AC is a two-force member. Thus, the reaction at E must be directed along EC.

0.075 m
54 N
180 N 0.25 m

E
y
y
E

From Eq. (1):
54 N 0
y
A

54 N 54.0 N
yy
A  A
Thus, reactions are

180.0 N
x
A , 54.0 N
y
A 

180.0 N
x
E , 54.0 NE
y 
(b) Couple applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be
directed along EC.

www.elsolucionario.org

912. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.86 (Continued)




0.125 m
90 N
180 N 0.25 my
y
A
A




From Eq. (1):
0
90 N 0
yy
y
AE
E



90 N 90 N
yy
E  E
Thus, reactions are

180.0 N
x
A , 90.0 N
y
A 

180.0 N
x
E , 90.0 N
y
E 
www.elsolucionario.org

913. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.87
Determine the components of the reactions at A and B, (a) if the 500-N load is
applied as shown, (b) if the 500-N load is moved along its line of action and is
applied at Point F.

SOLUTION
Free body: Entire frame:

Analysis is valid for either parts (a) or (b), since position of 100-lb load on its line of action is immaterial.

0: (10) (100 lb)(6) 0 60 lb
Ay y
MB B   

0: 60 100 0 40 lb
yy y
FA A   

0: 0
xxx
FAB   (1)
(a) Load applied at E.
Free body: Member AC:

Since AC is a two-force member, the reaction at A must be directed along CA. We have

40 lb
10 in. 5 in.
x
A

80.0 lb
x
A , 40.0 lb
y
A 
From Eq. (1):
80 0 80 lb
xx
BB  
Thus,
80.0 lb
x
B , 60.0 lb
y
B www.elsolucionario.org

914. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.87 (Continued)

(b) Load applied at F. Free body: Member BCD:
Since BCD is a two-force member (with forces applied at B and C only),
the reaction at B must be directed along CB. We have, therefore,

0
x
B
The reaction at B is
0
x
B 60.0 lb
y
B 
From Eq. (1):
00 0
xx
AA 
The reaction at A is
0
x
A 40.0 lb
y
A 
www.elsolucionario.org

915. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.88
The 48-lb load can be moved along the line of action shown and applied at
A, D, or E. Determine the components of the reactions at B and F if the
48-lb load is applied (a) at A, (b) at D, (c) at E.

SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b) and (c) since the position of the load
along its line of action is immaterial.

0: (48lb)(8 in.) (12 in.) 0
Fx
MB  

32 lb 32 lb
xx
B B

0: 32 lb+ 0
xx
FF 

32 lb 32 lb
xx
F  F


0: 48 lb 0
yyy
FBF   
(1)

(a) Load applied at A.
Free body: Member CDB

CDB is a two-force member. Thus, the reaction at B must be directed along BC.

5in.
10 lb
32 lb 16 in.y
y
B
 B
From Eq. (1):
10 lb 48 lb 0
y
F 

38 lb 38 lb
yy
F F
Thus reactions are:

32.0 lb
x
B , 10.00 lb
y
B 

32.0 lb
x
F , 38.0 lb
y
F www.elsolucionario.org

916. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.88 (Continued)

(b) Load applied at D. Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be
directed along CF.

7in.
14 lb
32 lb 16 in.y
y
F

F
From Eq. (1):
14 lb 48 lb 0
y
B

34 lb 34 lb
yy
BB
Thus, reactions are:

32.0 lb
x
B , 34.0 lb
y
B 

32.0 lb
x
F , 14.00 lb
y
F 
(c) Load applied at E.
Free body: Member CDB

This is the same free body as in Part (a).
Reactions are same as ( a)


www.elsolucionario.org

917. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.89
The 48-lb load is removed and a 288-lb · in. clockwise couple is applied
successively at A, D, and E. Determine the components of the reactions at B
and F if the couple is applied (a) at A, (b) at D, (c) at E.

SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b), and (c), since the point of
application of the couple is immaterial.

0: 288 lb in. (12 in.) 0
Fx
MB   

24 lb 24 lb
xx
B  B

0: 24 lb+ 0
xx
FF  

24 lb 24 lb
xx
F F

0: 0
yyy
FBF  
(1)

(a) Couple applied at A.
Free body: Member CDB

CDB is a two-force member. Thus, reaction at B must be directed along BC.

5in.
7.5 lb
24 lb 16 in.y
y
B

B
From Eq. (1):
7.5 lb 0
y
F

7.5 lb 7.5 lb
yy
F F
Thus, reactions are:

24.0 lb
x
B , 7.50 lb
y
B 

24.0 lb
x
F ,7.50 lb
y
F www.elsolucionario.org

918. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.89 (Continued)

(b) Couple applied at D. Free body: Member ACF.

ACF is a two-force member. Thus, the reaction at F must be directed along CF.

7in.
10.5 lb
24 lb 16 in.y
y
F

F
From Eq. (1):
10.5 lb:
y
B

10.5 lb 10.5 lb
yy
B  B
Thus, reactions are:

24.0 lb
x
B , 10.50 lb
y
B 

24.0 lb
x
F , 10.50 lb
y
F 
(c) Couple applied at E.
Free body: Member CDB

This is the same free body as in Part (a).
Reactions are same as in ( a)


www.elsolucionario.org

919. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.90
(a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its
component parts can be obtained by removing the pulley and applying at A two forces equal and parallel
to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the
frame at a Point B, a force of magnitude equal to the tension in the cable should also be applied at B.

SOLUTION
First note that, when a cable or cord passes over a frictionless, motionless pulley,
the tension is unchanged.

12 12
0: 0
C
MrTrTTT  
(a) Replace each force with an equivalent force-couple.

(b) Cut the cable and replace the forces on pulley with equivalent pair of forces at A as above.

www.elsolucionario.org

920. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.91
Knowing that each pulley has a radius of 250 mm,
determine the components of the reactions at D and E.

SOLUTION
Free body: Entire assembly:


0: (4.8 kN)(4.25 m) (1.5 m) 0
Ex
MD  


13.60 kN
x
D

13.60 kN
x
D



0: 13.60 kN 0
xx
FE  


13.60 kN
x
E

13.60 kN
x
E



0: 4.8 kN 0
yyy
FDE   (1)
Free body: Member ACE:


0 : (4.8 kN)(2.25 m) (4 m) 0
Ay
ME  

2.70 kN
y
E 2.70 kN
y
E 
From Eq. (1):
2.70 kN 4.80 kN 0
y
D

7.50 kN
y
D 7.50 kN
y
D 
www.elsolucionario.org

921. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.92
Knowing that the pulley has a radius of 75 mm, determine the
components of the reactions at A and B.


SOLUTION
Free body: Entire frame and pulley:


0: (240 N)(75 mm) (600 mm) 0
By
MA   

30.0 N 30.0 N
yy
A  A 

0: 0 (1)
xxxxx
FBABA   

0: 30 N 240 N 0
yy
FB   

270 N
y
B 


Free body: Member ACD:


0 : (200) (30 N)(300) (240 N)(75) 0
Dx
MA   

45.0 N
x
A
45.0 N
x
A 
From Eq. (1):
45.0 N
Xx
AB 45.0 N
x
B 
Thus, reactions are

45.0 N
x
A , 30.0 N
y
A 

45.0 N
x
B , 270 N
y
B 
www.elsolucionario.org

922. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.93
A 3-ft-diameter pipe is supported every 16 ft by a small frame like
that shown. Knowing that the combined weight of the pipe and its
contents is 500 lb/ft and assuming frictionless surfaces, determine
the components (a) of the reaction at E, (b) of the force exerted at C
on member CDE.

SOLUTION
Free body: 16-ft length of pipe:
(500 lb/ft)(16 ft) 8 kipsW

Force Triangle





8 kips
35 4
BD



6 kips 10 kipsBD
 

Determination of CB  CD.
We note that horizontal projection of horizontal projection of BO OD CD
  


34
()
55
rr CD

Thus,
8
2(1.5 ft) 3 ft
4
CB CD r 

Free body: Member ABC:

0: (6 kips)( 3)
Ax
MCh h  


3
(6 kips)
x
h
C
h

 (1)

For 9 ft,h


93
(6 kips) 4 kips
9
x
C



www.elsolucionario.org

923. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.93 (Continued)

Free body: Member CDE: From above, we have

4.00 kips
x
C 
0: (10 kips)(7 ft) (4 kips)(6 ft) (8 ft) 0
Ey
MC   

5.75 kips,
y
C

5.75 kips
y
C



3
0: 4 kips (10 kips) 0
5
xx
FE    


2 kips,
x
E

2.00 kips
x
E 

4
0: 5.75 kips (10 kips) 0,
5
yy
FE   

2.25 kips
y
E



www.elsolucionario.org

924. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.94
Solve Problem 6.93 for a frame where h  6 ft.
PROBLEM 6.93 A 3-ft-diameter pipe is supported every 16 ft by
a small frame like that shown. Knowing that the combined weight
of the pipe and its contents is 500 lb/ft and assuming frictionless
surfaces, determine the components (a) of the reaction at E, (b) of
the force exerted at C on member CDE.

SOLUTION
See solution of Problem 6.91 for derivation of Eq. (1).
For
363
6 ft, C (6 kips) 3 kips
6
x
h
h
h

 

Free body: Member CDE:
From above, we have

3.00 kips
x
C 

0: (10 kips)(7 ft) (3 kips)(6 ft) (8 ft) 0
Ey
MC   


6.50 kips,
y
C

6.50 kips
y
C 

3
0: 3 kips (10 kips) 0
5
xx
FE    


3.00 kips,
x
E

3.00 kips
x
E 

4
0: 6.5 kips (10 kips) 0
5
yy
FE   


1.500 kips
y
E

1.500 kips
y
E 

www.elsolucionario.org

925. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.95
A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D.
Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the
additional load on each of the truck wheels due to the trailer.

SOLUTION
(a) Free body: Trailer:
(We shall denote by A, B, C the reaction at one wheel.)

0: (2400 lb)(2 ft) (11 ft) 0
A
MD   


436.36 lbD

0: 2 2400 lb 436.36 lb 0
y
FA   


981.82 lbA

982 lbA



Free body: Truck.

0: (436.36 lb)(3 ft) (2900 lb)(5 ft) 2 (9 ft) 0
B
MC   


732.83 lbC

733 lbC




0: 2 436.36lb 2900lb 2(732.83lb) 0
y
FB    


935.35 lbB

935 lbB



(b) Additional load on truck wheels.
Use free body diagram of truck without 2900 lb.

0: (436.36 lb)(3 ft) 2 (9 ft) 0
B
MC  


72.73 lbC

72.7 lbC





0: 2 436.36 lb 2(72.73 lb) 0
y
FB   


290.9 lbB

291lbB


www.elsolucionario.org

926. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.96
In order to obtain a better weight distribution over the
four wheels of the pickup truck of Problem 6.95, a
compensating hitch of the type shown is used to attach
the trailer to the truck. The hitch consists of two bar
springs (only one is shown in the figure) that fit into
bearings inside a support rigidly attached to the truck.
The springs are also connected by chains to the trailer
frame, and specially designed hooks make it possible to
place both chains in tension. (a) Determine the tension T
required in each of the two chains if the additional load
due to the trailer is to be evenly distributed over the four
wheels of the truck. (b) What are the resulting reactions at
each of the six wheels of the trailer-truck combination?
PROBLEM 6.95 A trailer weighing 2400 lb is attached
to a 2900-lb pickup truck by a ball-and-socket truck hitch
at D. Determine (a) the reactions at each of the six wheels
when the truck and trailer are at rest, (b) the additional
load on each of the truck wheels due to the trailer.

SOLUTION
(a) We small first find the additional reaction at each wheel due to the trailer.
Free body diagram: (Same
 at each truck wheel) 0: (2400 lb)(2 ft) 2 (14 ft) 2 (23 ft) 0
A
M    


64.86 lb

0: 2 2400 lb 4(64.86 lb) 0;
y
FA   


1070 lb;A

1070 lbA

Free body: Truck:
(Trailer loading only)

0: 2 (12 ft) 2 (3 ft) 2 (1.7 ft) 0
D
MT    


8.824
8.824(64.86 lb)
572.3 lb
T
T




572 lbT



Free body: Truck:
(Truck weight only)

0: (2900 lb)(5 ft) 2 (9 ft) 0
B
MC    

805.6 lbC 805.6 lbC
www.elsolucionario.org

927. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.96 (Continued)


0: 2 2900 lb 2(805.6 lb) 0
y
FB
   


644.4 lbB

644.4 lb
B


Actual reactions:
644.4 lb 64.86 709.2 lbBB   709 lbB


805.6 lb 64.86 870.46 lbCC    870 lb C


From part a: 1070 lb A


www.elsolucionario.org

928. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.97
The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind
the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is
located at G
m, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at
G
c and G l. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each
of the four wheels, (b) the forces exerted on the motor unit at C and D.


SOLUTION
(a) Free body: Entire machine:


AReaction at each front wheel

BReaction at each rear wheel

0: 75(3.2 m) 100(1.2 m) 2 (4.8 m) 300(5.6 m) 0
A
MB    


2 325 kNB

162.5 kNB 
0: 2 325 75 100 300 0
y
FA    


2150kNA

75.0 kNA 


www.elsolucionario.org

929. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.97 (Continued)

(b) Free body: Motor unit: 0: (1 m) 2 (2.8 m) 300(3.6 m) 0
D
MC B   

1080 5.6CB (1)

Recalling
162.5 kN, 1080 5.6(162.5) 170 kNBC

170.0 kNC 
0: 170 0
xx
FD  

170.0 kN
x
D 
 
0: 2(162.5) 300 0
yy
FD   

25.0 kN
y
D



www.elsolucionario.org

930. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.98
Solve Problem 6.97 assuming that the 75-kN load has been removed.
PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin
located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN
motor unit is located at G
m, while the centers of gravity of the 100-kN cab and 75-kN load are located,
respectively, at G
c and G l. Knowing that the machine is at rest with its brakes released, determine (a) the
reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.


SOLUTION
(a) Free body: Entire machine:
AReaction at each front wheel

BReaction at each rear wheel

0: 2 (4.8 m) 100(1.2 m) 300(5.6 m) 0
A
MB   


2375kNB

187.5 kNB 
0: 2 375 100 300 0
y
FA    


225kNA

12.50 kNA 
(b) Free body: Motor unit:
See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With 187.5 kN,B
we have

1080 5.6(187.5) 30 kNC 

30.0 kNC 

0: 30 0
xx
FD  

30.0 kN
x
D 


0: 2(187.5) 300 0
yy
FD    75.0 kN
y
D


www.elsolucionario.org

931. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.99
Knowing that P = 90 lb and Q = 60 lb, determine the
components of all forces acting on member BCDE of the
assembly shown.

SOLUTION
Free body: Entire assembly:



0: (12 in.) (60 lb)(4 in.) 90 lb 12 in. 0
C
MA   


110.0 lbA


110.0 lbA 

0: 60 lb 0
xx
FC  



60 lb
x
C

60.0 lb
x
C 

0: 110 lb 90 lb 0
yy
FC   


200 lb
y
C

200 lb
y
C 

Free body: Member BCDE:



 0: (10 in.) 200 lb (4 in.) 90 lb 8 in. 0
D
MB   


152.0 lbB

152.0 lbB 
www.elsolucionario.org

932. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.99 (Continued)


0: 60 lb 0
xx
FD  


60.0 lb
x
D

60.0 lbD 

0: 200 lb 152 lb 90 lb 0
yy
FD    


42.0 lb
y
D

42.0 lb
y
D 

www.elsolucionario.org

933. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.100
Knowing that P = 60 lb and Q = 90 lb, determine the
components of all forces acting on member BCDE of the
assembly shown.

SOLUTION
Free body: Entire assembly:


0: (12 in.) (60 lb)(12 in.) 90 lb 4 in. 0
C
MA   

90.0 lbA


90.0 lbA 
0: 90 lb 0
xx
FC  



90 lb
x
C

90.0 lb
x
C 

0: 60 lb 90 lb 0
yy
FC   


150 lb
y
C

150.0 lb
y
C 

Free body: Member BCDE:






www.elsolucionario.org

934. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.100 (Continued)


 0: (10 in.) 150 lb (4 in.) 60 lb 8 in. 0
D
MB   


108.0 lbB

108.0 lbB 


0: 90 lb 0
xx
FD  


90.0 lb
x
D

90.0 lbD 

0: 150 lb 108 lb 60 lb 0
yy
FD    


18.00 lb
y
D

18.00 lb
y
D 


www.elsolucionario.org

935. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.101
For the frame and loading shown, determine the components of all forces
acting on member ABE.

SOLUTION
FBD Frame:
0: (1.8 m) (2.1m)(12 kN) 0
Ey
MF  

14.00 kN
y
F

0: 14.00 kN 12 kN 0
yy
FE   


2kN
y
E

2.00 kN
y
E 
FBD member BCD:
0: (1.2 m) (12 kN)(1.8 m) 0 18.00 kN
By y
MC C   
But
C is  ACF, so 2 ; 36.0 kN
xyx
CCC

0: 0 36.0 kN
xxxxx
FBCBC    

36.0 kN
x
B on BCD

0: 18.00 kN 12 kN 0 6.00 kN
yy y
FB B      on BCD
On ABE:
36.0 kN
x
B 

6.00 kN
y
B 
FBD member ABE:
0: (1.2 m)(36.0 kN) (0.6 m)(6.00 kN)
(0.9 m)(2.00 kN) (1.8 m)( ) 0
A
x
M
E 


23.0 kN
x
E 

0: 23.0 kN 36.0 kN 0
xx
FA     13.00 kN
x
A 


0: 2.00 kN 6.00 kN 0
yy
FA    

4.00 kN
y
A 
www.elsolucionario.org

936. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.102
For the frame and loading shown, determine the components of all
forces acting on member ABE.


SOLUTION
FBD Frame:

0: (1.2 m)(2400 N) (4.8 m) 0
Fy
ME  

600 N
y
E 

FBD member
BC:

4.8 8
5.4 9
yxx
CCC

0: (2.4 m) (1.2 m)(2400 N) 0 1200 N
Cy y
MB B   
On ABE:
1200 N
y
B 

0: 1200 N 2400 N 0 3600 N
yyy
FCC     
so
9
4050 N
8
xyx
CCC

0: 0 4050 N
xxxx
FBCB     on BC
On ABE:
4050 N
x
B 

PROBLEM 6.102 (Continued) www.elsolucionario.org

937. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



FBD member
AB0E:
0: (4050 N) 2 0
Ax
Ma aE  

2025 N
x
E 2025 N
x
E 

0 : (4050 2025) N 0
xx
FA    2025 N
x
A 

0: 600 N 1200 N 0
yy
FA   

1800 N
y
A





www.elsolucionario.org

938. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.103
For the frame and loading shown, determine the components of
all forces acting on member ABD.

SOLUTION
Free body: Entire frame:







0: (12 in.) (360 lb)(15 in.) (240 lb)(33 in.) 0
A
ME   

1110 lbE

1110 lbE

0: 1110 lb 0
xx
FA  


1110 lb
x
A 1110 lb
x
A 

0: 360 lb 240 lb 0
yy
FA   

600 lb
y
A 600 lb
y
A 

Free body: Member CDE:



0: (1110 lb)(24 in.) (12 in.) 0
Cx
MD  

2220 lb
x
D

www.elsolucionario.org

939. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.103 (Continued)

Free body: Member ABD: From above: 2220 lb
x
D 

0: (18 in.) (600 lb)(6 in.) 0
By
MD  

200 lb
y
D 200 lb
y
D 

0: 2220 lb 1110 lb 0
xx
FB   

1110 lb
x
B 1110 lb
x
B 

0: 200 lb 600 lb 0
yy
FB   


800 lb
y
B

800 lb
y
B 
www.elsolucionario.org

940. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.104
Solve Problem 6.103 assuming that the 360-lb load has been
removed.
PROBLEM 6.103 For the frame and loading shown, determine
the components of all forces acting on member ABD.

SOLUTION
Free body diagram of entire frame.


0: (12 in.) (240 lb)(33 in.) 0
A
ME  

660 lbE

660 lbE

0: 660 lb 0
xx
FA  


660 lb
x
A 660 lb
x
A 

0: 240 lb 0
yy
FA  

240 lb
y
A 240 lb
y
A 

Free body: Member CDE:



0: (660 lb)(24 in.) (12 in.) 0
Cx
MD  

1320 lb
x
D

www.elsolucionario.org

941. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.104 (Continued)

Free body: Member ABD: From above: 1320 lb
x
D 

0: (18 in.) (240 lb)(6 in.) 0
By
MD  

80 lb
y
D 80.0 lb
y
D 

0: 1320 lb 660 lb 0
xx
FB   

660 lb
x
B 660 lb
x
B 

0: 80 lb 240 lb 0
yy
FB   


320 lb
y
B

320 lb
y
B 
www.elsolucionario.org

942. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.105
For the frame and loading shown, determine the components of
the forces acting on member DABC at B and D.

SOLUTION
Free body: Entire frame:
0: (0.6 m) (12 kN)(1 m) (6 kN)(0.5 m) 0
G
MH   

25 kN 25 kNH H

Free body: Member BEH:
0: (0.5 m) (25 kN)(0.2 m) 0
Fx
MB  

10 kN
x
B

Free body: Member DABC:
From above: 10.00 kN
x
B 

0: (0.8 m) (10 kN 12 kN)(0.5 m) 0
Dy
MB    

13.75 kN
y
B 13.75 kN
y
B 

0: 10 kN 12 kN 0
xx
FD    


22 kN
x
D 22.0 kN
x
D 

0: 13.75 kN 0
yy
FD   

13.75 kN
y
D 13.75 kN
y
D 
www.elsolucionario.org

943. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.106
Solve Problem 6.105 assuming that the 6-kN load has been
removed.
PROBLEM 6.105 For the frame and loading shown, determine
the components of the forces acting on member DABC at B and D.

SOLUTION
Free body: Entire frame
0 : (0.6 m) (12 kN)(1 m) 0
G
MH  

20 kN 20 kNH H

Free body: Member BEH
0: (0.5 m) (20 kN)(0.2 m) 0
Ex
MB  

8kN
x
B

Free body: Member DABC
From above: 8.00 kN
x
B 

0: (0.8 m) (8 kN 12 kN)(0.5 m) 0
Dy
MB   

12.5 kN
y
B 12.50 kN
y
B 

0: 8 kN 12 kN 0
xx
FD    


20 kN
x
D 20.0 kN
x
D 

0: 12.5 kN 0; 12.5 kN
yy y
FD D    

12.50 kN
y
D 

www.elsolucionario.org

944. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.107
The axis of the three-hinge arch ABC is a
parabola with the vertex at B. Knowing that
P  112 kN and Q  140 kN, determine (a) the
components of the reaction at A, (b) the
components of the force exerted at B on
segment AB.

SOLUTION

Free body: Segment AB:
0: (3.2 m) (8 m) (5 m) 0
Ax y
MB B P    (1)
0.75 (Eq. 1):
(2.4 m) (6 m) (3.75 m) 0
xy
BBP   (2)
Free body: Segment BC:
0: (1.8 m) (6 m) (3 m) 0
Cx y
MB B Q   
(3)

Add Eqs. (2) and (3):
4.2 3.75 3 0
x
BPQ

(3.75 3 )/4.2
x
BPQ (4)
From Eq. (1):
3.2
(3.75 3 ) 8 5 0
4.2
y
PQ B P

(9 9.6)/33.6
y
BPQ  (5)
given that
112 kN and 140 kN.PQ
(a) Reaction at A.
Considering again AB as a free body,

0: 0; 200 kN
xxxxx
FABAB    

200 kN
x
A 

0: 0
yyy
FAPB  

112 kN 10 kN 0
y
A

122 kN
y
A 122.0 kN
y
A 

www.elsolucionario.org

945. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.107 (Continued)

(b) Force exerted at B on AB. From Eq. (4): (3.75 112 3 140)/4.2 200 kN
x
B 

200 kN
x
B 
From Eq. (5):
( 9 112 9.6 140)/33.6 10 kN
y
B    

10.00 kN
y
B 
www.elsolucionario.org

946. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.108
The axis of the three-hinge arch ABC is a
parabola with the vertex at B. Knowing that
P  140 kN and Q  112 kN, determine (a) the
components of the reaction at A, (b) the
components of the force exerted at B on
segment AB.

SOLUTION

Free body: Segment AB:
0: (3.2 m) (8 m) (5 m) 0
Ax y
MB B P    (1)
0.75 (Eq. 1):
(2.4 m) (6 m) (3.75 m) 0
xy
BBP   (2)
Free body: Segment BC:
0: (1.8 m) (6 m) (3 m) 0
Cx y
MB B Q   
(3)

Add Eqs. (2) and (3):
4.2 3.75 3 0
x
BPQ

(3.75 3 )/4.2
x
BPQ (4)
From Eq. (1):
3.2
(3.75 3 ) 8 5 0
4.2
y
PQ B P

(9 9.6)/33.6
y
BPQ  (5)
given that
140 kN and 112 kN.PQ
(a) Reaction at A.
0: 0; 205 kN
xxx xx
FAB AB    

205 kN
x
A 

0: 0
yyy
FAPB  

140 kN ( 5.5 kN) 0
y
A

134.5 kN
y
A 134.5 kN
y
A 

www.elsolucionario.org

947. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.108 (Continued)

(b) Force exerted at B on AB. From Eq. (4): (3.75 140 3 112)/4.2 205 kN
x
B 

205 kN
x
B 
From Eq. (5):
( 9 140 9.6 112)/33.6 5.5 kN
y
B    

5.50 kN
y
B 
www.elsolucionario.org

948. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.109
Neglecting the effect of friction at the horizontal and vertical
surfaces, determine the forces exerted at B and C on member
BCE.

SOLUTION

Free body of Member ACD
0: (2 in.) (18 in.) 0 9
Hx y xy
MC C CC    (1)
Free body of Member BCE
0: (6 in.) (6 in.) (50 lb)(12 in.) 0
Bx y
MC C   
Substitute from (1):
9 (6) (6) 600 0
yy
CC

10 lb; 9 9(10) 90 lb
yxy
CCC   

90.6 lbC 6.34° 

0: 90 lb 0 90 lb
xx x
FB B   

0: 10 lb 50 lb 0 40 lb
yy y
FB B    

98.5 lbB 24.0° 
www.elsolucionario.org

949. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


.

PROBLEM 6.110
Neglecting the effect of friction at the horizontal and
vertical surfaces, determine the forces exerted at B and C
on member BCE.

SOLUTION

Free body of Member ACD
We note that AC is a two-force member.

or C 3
18 6
yx
xy
CC
C
(1)
On free body of Member BCE
0: (6 in.) (6 in.) (50.0 lb)(12 in.) 0
Hx y
MC C   
Substitute from (1):
3 (6) (6) 600 0
yy
CC

25.0 lb; 3 3(25.0) 75.0 lb
yxy
CCC   

79.1lbC 18.43° 


0: 25.0 lb 50.0 lb 0 25.0 lb
y
FB B

25.0 lbB 

www.elsolucionario.org

950. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.111
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each
link.

SOLUTION
Member FBDs:

I II
FBD I:
1
0: 0 2
2
B y AF AF y
MaCaFFC   
FBD II:
1
0: 0 2
2
D y EH EH y
MaCaFFC 
FBDs combined:
11 11
0: 0 2 2
22 22
GAFEHyy
MaPaFaFPC C     

2
y
P
C

2
so
2
AF
FPC 

2
2
EH
FPT 
FBD I:
11 1
0: 0 0
2222 2
yAFBGy BG
PP
FFFPC FP 


0
BG
F 
FBD II:
11 1
0: 0 0
2222 2
y y DG EH DG
PP
FCFF F      


2
DG
FPC 
www.elsolucionario.org

951. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.112
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each
link.

SOLUTION
Member FBDs:

I II
FBD I:
0: 2 0 2
Iyx yx
MaCaCaPCCP   
FBD II:

0: 2 0 2 0
Jyxyx
MaCaCCC   
Solving,
; as shown.
24
xy
PP
CC

FBD I:
1
0: 0 2
2
xBGxBGx
FFCFC    
2
BG
P
FC


12
0: 0
242
yAF
P
FFP P

   



4
AF
P
FC

FBD II:

1
0: 0 2
2
xxDGDGx
FCFFC    
2
DG
P
FC


12
0: 0
24242
y y EH EH
PP P
FC PFF

     



4
EH
P
FT

www.elsolucionario.org

952. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.113
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each
link.

SOLUTION
Member FBDs:

I II
From FBD I:

3
0: 0 3
222
Jxy xy
aaa
MCCPCCP     

FBD II:

3
0: 0 3 0
22
Kxyxy
aa
MCCCC    

Solving,
; as drawn.
26
xy
PP
CC

FBD I:
12
0: 0 2
62
ByAGAGy
MaCaFFCP    
2
6
AG
FPC 

11 22
0: 0 2
6222
x AG BF x BF AG x
FFFCFFCPP        

22
3
BF
FPC 
FBD II:
12
0: 0 2
62
DEHyEHy
MaFaCF C P    
2
6
EH
FPT 

11 22
0: 0 2
6222
x x DI EH DI EH x
FCF F FFC PP       

2
3
DI
FPC 
www.elsolucionario.org

953. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.114
Members ABC and CDE are pin-connected at C and supported by
the four links AF, BG, DG, and EH. For the loading shown,
determine the force in each link.

SOLUTION
We consider members ABC and CDE:

Free body: CDE:

0: (4 ) (2 ) 0
Jxy
MCaCa  

2
yx
CC (1)
Free body: ABC:

0: (2 ) (4 ) (3 ) 0
Kxy
MCaCaPa   
Substituting for C
y from Eq. (1): (2 ) 2 (4 ) (3 ) 0
xx
Ca CaPa

11
2
22
xy
CPC PP

   



1
0: 0
2
xBGx
FFC  


1
22 ,
2 2
BG x
P
FC P
   



2
BG
P
FT


www.elsolucionario.org

954. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.114 (Continued)


1
0: 0
2
yAFBGy
FFFPC    

1
222
AF
PP
FPP   

2
AF
P
FC

Free body: CDE:

1
0: 0
2
yDGy
FFC  

22
DG y
FCP 2
DG
FPT 

1
0: 0
2
xEHxDG
FFCF    


11
2
22 2
EH
P
FPP
   



2
EH
P
FC

www.elsolucionario.org

955. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.115
Solve Problem 6.112 assuming that the force P is replaced by a
clockwise couple of moment
M0 applied to member CDE at D.
PROBLEM 6.112 Members ABC and CDE are pin-connected at
C and supported by four links. For the loading shown, determine
the force in each link.

SOLUTION
Free body: Member ABC:
0: (2 ) ( ) 0
Jyx
MCaCa  

2
xy
CC
Free body: Member CDE:

0
0: (2 ) ( ) 0
Kyx
MCaCaM  
0
(2 ) ( 2 )( ) 0
yy
Ca CaM  
0
4
y
M
C
a

v

2:
xy
CC
0
2
x
M
C
a

v

0
0: 0; 0
222
xx
MDD
FC
a
    


0
2
M
D
a


0
2
DG
M
FT
a



0
0: ( ) ( ) 0
Dy
MEaCaM  


0
0
() () 0
4
M
Ea a M
a





03
4
M
E
a


03
4
EH
M
FC
a


Return to free body of ABC:

0
0: 0; 0
222
xx
MBB
FC
a
    


0
2
M
B
a


0
2
BG
M
FT
a



0
0: () (); () () 0
4
By
M
MAaCaAaa
a
   


0
4
M
A
a


0
4
AF
M
FC
a


www.elsolucionario.org

956. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.116
Solve Problem 6.114 assuming that the force P is replaced by a
clockwise couple of moment
M0 applied at the same point.
PROBLEM 6.114 Members ABC and CDE are pin-connected at C
and supported by the four links AF, BG, DG, and EH. For the
loading shown, determine the force in each link.

SOLUTION
Free body: CDE: (same as for Problem 6.114)

0: (4 ) (2 ) 0
Jxy
MCaCa  

2
yx
CC (1)
Free body: ABC:
0
0: (2 ) (4 ) 0
Kxy
MCaCaM  
Substituting for C
y from Eq. (1):
0
(2 ) 2 (4 ) 0
xx
Ca CaM

0
00
6
2
63
x
y
M
C
a
MM
C
aa


  




1
0: 0
2
xBGx
FFC  


0
2
2
6
BG x
M
FC
a
 

0
2
6
BG
M
FT
a



1
0: 0
2
yAFBGy
FFFC    

00 0
21
6362
AF
MM M
F
aa a
  

0
6
AF
M
FT
a






www.elsolucionario.org

957. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.116 (Continued)

Free body: CDE: (Use F.B. diagram of Problem 6.114.)

1
0: 0
2
yDGy
FFC  

0
2
2
3
DG y
M
FC
a

0
2
3
DG
M
FT
a
 

1
0: 0
2
xEHxDG
FFCF    

000
21
,
636 2
EH
MMM
F
aaa

     
 

0
6
EH
M
FC
a


www.elsolucionario.org

958. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.117
Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown.
Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D,
E, and H.

SOLUTION
Note that, if we assume P is applied to EG, each individual member FBD looks like
so
left right middle
22FFF


Labeling each interaction force with the letter corresponding to the joint of its application, we see that

22
22
22
2(4 8 16) 2
BAF
CBD
GCH
PF G C B F E



    

From
16 ,PF F
15
P
F
so
15
P
A

 2
15
P

D 

4
15
P

H 

8
15
P

E 
www.elsolucionario.org

959. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.118
Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached
to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces
are exerted at the connections, determine the vertical reactions at E, F, G, and H.

SOLUTION
We shall draw the free body of each member. Force P will be applied to member AFB. Starting with
member AED, we shall express all forces in terms of reaction E.
Member AED:
0: (3 ) ( ) 0
D
MAaEa  

3
E
A


0: (3 ) (2 ) 0
A
MDaEa   


2
3
E
D

Member DHC:

2
0: (3 ) ( ) 0
3
C
E
MaHa
   




2HE (1)

2
0: (2 ) ( ) 0
3
H
E
MaCa
   




4
3
E
C

Member CGB:

4
0: (3 ) ( ) 0
3
B
E
MaGa

   




4GE (2)

4
0: (2 ) ( ) 0
3
G
E
MaBa
   




8
3
E
B
www.elsolucionario.org

960. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.118 (Continued)

Member AFB: 0: 0
y
FFABP 

8
0
33EE
FP 
   
 
 


3FP E (3)

0: ( ) (3 ) 0
A
MFaBa  

8
(3)() (3)0
3E
PEa a
 




380;
5
P
PEE E 
5
P

E 
Substitute
5
P
E into Eqs. (1), (2), and (3).

22
5
P
HE
  



2
5
P
H

2
5
P

H 

44
5
P
GE

  


4
5
P
G

4
5
P

G 

33
5
P
FP EP
  



8
5
P
F

8
5
P

F 


www.elsolucionario.org

961. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.119
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each
frame, determine the reactions at the supports and indicate whether the frame is rigid.


SOLUTION
(a) Member FBDs:

I II
FBD II: 0: 0
yy
FB 
2
0: 0
B
MaF 
2
0F
FBD I:
2
0: 2 0,
A
MaFaP   but
2
0F
so
0P not rigid for 0P 
(b)
Member FBDs:

Note: Seven unknowns
12
(, , , ,, ,),
xyxy
AABBFFC but only six independent equations
System is statically indeterminate.


System is, however, rigid. 


www.elsolucionario.org

962. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.119 (Continued)

(c)
FBD whole: FBD right:

I II
FBD I: 0: 5 2 0
Ay
MaBaP  

2
5
y
PB 

2
0: 0
5
yy
FAPP  

3
5
y
PA
FBD II:
5
0: 0 5 2
22
Cxyxyx
aa
MBBBB P    
B
FBD I:
0: 0
xxxxx
FABAB    2
x
PA

2.09PA 16.70 

2.04PB 11.31 
System is rigid.

www.elsolucionario.org

963. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.120
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each
frame, determine the reactions at the supports and indicate whether the frame is rigid.

SOLUTION
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus,
the third force, at B, must be parallel to the link forces.
(a)
FBD whole:


4
0: 2 5 0 2.06
417 17
A
a
MaPBaBBP

     
2.06PB 14.04 

4
0: 0
17
xx
FA B   2
x
PA

1
0: 0
217
yy y
P
FAPB    A
2.06PA 14.04 
rigid

(b)
FBD whole:

Since
B passes through A, 2 0 only if 0.
A
MaP P  
no equilibrium if
0P not rigid www.elsolucionario.org

964. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.120 (Continued)

(c)
FBD whole:


134 17
0: 5 2 0
4417 17
A
a
MaB BaPBP    
1.031PB 14.04 

4
0: 0
17
xx x
FA BAP   

13
0: 0
4417
yy y
PP
FAPBAP   
1.250PA 36.9 
System is rigid.

www.elsolucionario.org

965. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.121
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each
frame, determine the reactions at the supports and indicate whether the frame is rigid.

SOLUTION
(a) Member FBDs:


I II
FBD I:
11
0: 2 0 2 ;
A
MaFaPFP    0: 0
yy y
FAP P   A
FBD II:
22
0: 0 0
B
MaFF  

11
0: 0, 2
xx x
FBFBFP      2
x
PB

0: 0
yy
FB  so 2PB 
FBD I:
12 21
0: 0 0 2
xx x
FAFFAFFP      2
x
PA
so
2.24PA 26.6 
Frame is rigid.

(b)
FBD left: FBD whole:


I II
FBD I:
5
0: 0 5
22 2
Exyxy
aa a
MPAAAAP     

FBD II:
0: 3 5 0 5 3
Bxyxy
MaPaAaAAAP     
This is impossible unless
0.P not rigid www.elsolucionario.org

966. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.121 (Continued)

(c)
Member FBDs:

I II
FBD I: 0: 0
y
FAP  PA



11
0: 2 0 2
D
MaFaAFP   

21 2
0: 0 2
x
FFFFP   
FBD II:
1
1
0: 2 0
2
B
F
MaCaFCP  
PC 

12
0: 0 0
xxx
FFFBBPP     

0: 0
xy y
FBCBCP      PB 
Frame is rigid.

www.elsolucionario.org

967. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.122
The shear shown is used to cut and trim electronic circuit board
laminates. For the position shown, determine (a) the vertical
component of the force exerted on the shearing blade at D, (b) the
reaction at C.

SOLUTION
We note that BD is a two-force member.
Free body: Member ABC: We have the components:

(400 N)sin30 200 N
x
P

(400 N)cos30 346.41 N
y
P


25
()
65
BD x BD
FF

60
()
65
BD y BD
FF

0:()(45)()(30) (45300sin30)
(30 300cos30 ) 0
CBDxBDyx
y
MF F P
P
    
 

3
25 60
(45) (30) (200)(195) (346.41)(289.81)
65 65
45 139.39 10
3097.6 N
BD BD
BD
BD
FF
F
F







(a) Vertical component of force exerted on shearing blade at D.

60 60
( ) (3097.6 N) 2859.3 N
65 65
BD y BD
FF  ( ) 2860 N
BD y
F 



www.elsolucionario.org

968. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.122 (Continued)

(b) Returning to FB diagram of member ABC,

0: ( ) 0
xBDxxx
FFPC  

25
()
65
25
(3097.6) 200
65
991.39
xBDxx BDx
x
CF P FP
C



991.39 N
x
C


0: ( ) 0
yBDyyy
FFPC  

60 60
( ) (3097.6) 346.41
65 65
yBDyy BDy
CF P FP 

2512.9 N
y
C 2512.9
y
C

2295 NC 2700 NC 68.5° 
www.elsolucionario.org

969. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.123
A 100-lb force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 in. long and that
a = 4 in., determine the horizontal force exerted on block E.

SOLUTION
Free body: Entire Frame

We have

4 in. sin15 1.03528 in.
1.03528 in.
sin = 9.9359
6 in.
BF






4cos15 6cos9.9359 9.7737 in.AD AG FD  



0: (100 lb)(10 in.)cos 15 ( )sin (9.7737 in.) 0
ABD
MF   



572.77 lb
BD
F
Horizontal force exerted on block.
( ) cos (572.77 lb)cos 9.9359
BD x BD
FF 



( ) 564 lb
BD x
F




www.elsolucionario.org

970. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.124
A 100-lb force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 in. long and that
a = 8 in., determine the horizontal force exerted on block E.

SOLUTION
Free body: Entire Frame

We have

8 in. sin15 2.0706 in.
2.0706 in.
sin = 20.188
6 in.
BF







8 in. cos15 7.7274 in.
6 in. cos20.188 5.6314 in.
13.359 in.
AF
FD
AD AF FD







0: (100 lb)(14 in.)cos 15 ( )sin (13.359 in.) 0
ABD
MF   



293.33 lb
BD
F
Horizontal force exerted on block.
( ) cos (293.33 lb)cos 20.188
BD x BD
FF 



()275lb
BD x
F



www.elsolucionario.org

971. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.125
The control rod CE passes through a
horizontal hole in the body of the toggle
system shown. Knowing that link BD is 250
mm long, determine the force
Q required to
hold the system in equilibrium when
  20.

SOLUTION
We note that BD is a two-force member.
Free body: Member ABC:

Since
250,BD
133.404
sin ; 7.679
250




0: ( sin )187.94 ( cos )68.404 (100 N)328.89 0
CBD BD
MF F    

[187.94sin 7.679 68.404cos7.679 ] 32,889
770.6 N
BD
BD
F
F
  



0: (770.6 N)cos7.679 0
xx
FC  


763.7 N
x
C
Member CQ:

0: 763.7 N
xx
FQC  

764 NQ 

Dimensions in mm www.elsolucionario.org

972. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.126
Solve Problem 6.125 when (a)   0,
(b)
  6.
PROBLEM 6.125 The control rod CE passes
through a horizontal hole in the body of the
toggle system shown. Knowing that link BD
is 250 mm long, determine the force
Q
required to hold the system in equilibrium
when
  20.

SOLUTION
We note that BD is a two-force member.
(a) 0: Free body: Member ABC
Since
35 mm
250 mm, sin ; 8.048
250 mm
BD


0: (100 N)(350 mm) sin (200 mm) 0
CBD
MF   

1250 N
BD
F

0: cos 0
xBD x
FF C   

(1250 N)(cos 8.048 ) 0 1237.7 N
xx
C   C
Member CE:
0: (1237.7 N) 0
x
FQ 

1237.7 NQ 1238 NQ 
(b)
6 Free body: Member ABC

Since
114.094 mm
250 mm, sin
250 mm
BD




3.232



Dimensions in mm www.elsolucionario.org

973. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.126 (Continued)


0: ( sin )198.90 ( cos )20.906 (100 N)348.08 0
CBD BD
MF F    

[198.90sin3.232 20.906cos3.232 ] 34808
1084.8 N
BD
BD
F
F
  



0: cos 0
xBD x
FF C   


(1084.8 N)cos3.232 0
x
C 


1083.1 N
x
C
Member DE:
0: 
xx
FQC
1083.1 NQ 1083 NQ 


www.elsolucionario.org

974. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.127
The press shown is used to emboss a small seal at E.
Knowing that P  250 N, determine (a) the vertical
component of the force exerted on the seal, (b) the reaction
at A.

SOLUTION
FBD Stamp D:
0: cos 20 0, cos 20
yBD BD
FEF EF    

FBD
ABC:


0: (0.2 m)(sin30 )( cos20 ) (0.2 m)(cos30 )( sin 20 )
ABD BD
MF F    

[(0.2 m)sin 30 (0.4 m)cos15 ](250 N) 0

793.64 N
BD
FC
and, from above,

(793.64 N)cos 20E
( a)
746 NE 

0: (793.64 N)sin 20 0
xx
FA  

271.44 N
x
A

0: (793.64 N)cos 20 250 N 0
yy
FA   

495.78 N
y
A
( b)
565 NA 61.3° 
www.elsolucionario.org

975. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.128
The press shown is used to emboss a small seal at E.
Knowing that the vertical component of the force exerted on
the seal must be 900 N, determine (a) the required vertical
force
P, (b) the corresponding reaction at A.

SOLUTION
FBD Stamp D:
0: 900 N cos 20 0, 957.76 N
yBDBD
FFFC   


(a)

FBD ABC:


0: [(0.2 m)(sin30 )](957.76 N)cos20 [(0.2 m)(cos30 )](957.76 N)sin 20
A
M    

[(0.2 m)sin 30 (0.4 m)cos15 ] 0P

301.70 N,P 302 NP 
(b)
0: (957.76 N)sin 20 0
xx
FA  

327.57 N
x
A

0: (957.76 N)cos20 301.70 N 0
yy
FA   

598.30 N
y
A

682 NA 61.3° 
www.elsolucionario.org

976. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.129
The pin at B is attached to member ABC and can slide freely along the
slot cut in the fixed plate. Neglecting the effect of friction, determine
the couple
M required to hold the system in equilibrium when 30 .

SOLUTION
Free body: Member ABC:
0: (25 lb)(13.856 in.) (3 in.) 0
C
MB  

115.47 lbB

0: 25 lb 0
yy
FC   


25 lb
y
C


0: 115.47 lb 0
xx
FC  


115.47 lb
x
C

Free body: Member CD:

15.196
sin ; 40.505
8





cos (8 in.)cos 40.505 6.083 in.CD


0: (25 lb)(5.196 in.) (115.47 lb)(6.083 in.) 0
D
MM   


832.3 lb in.M 


832 lb in.M



www.elsolucionario.org

977. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.130
The pin at B is attached to member ABC and can slide freely along the
slot cut in the fixed plate. Neglecting the effect of friction, determine
the couple
M required to hold the system in equilibrium when 60 .

SOLUTION
Free body: Member ABC:
0: (25 lb)(8 in.) (5.196 in.) 0
C
MB  


38.49 lbB


0: 38.49 lb 0
xx
FC  


38.49 lb
x
C


0: 25 lb 0
yy
FC   


25 lb
y
C

Free body: Member CD:

13
sin ; 22.024
8




cos (8 in.)cos22.024 7.416 in.CD

0: (25 lb)(3 in.) (38.49 lb)(7.416 in.) 0
D
MM   

360.4 lb in.M  360 lb in.M 
www.elsolucionario.org

978. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.131
Arm ABC is connected by pins to a collar at B and to crank CD at
C. Neglecting the effect of friction, determine the couple
M
required to hold the system in equilibrium when
0.

SOLUTION
Free body: Member ABC
0: 240 N 0
xx
FC  


240 N
x
C


0: (240 N)(500 mm) (160 mm) 0
C
MB  


750 NB


0: 750 N 0
yy
FC  


750 N
y
C


Free body: Member CD
0: (750 N)(300 mm) (240 N)(125 mm) 0
D
MM   


3
195 10 N mmM  


195.0 kN mM



www.elsolucionario.org

979. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.132
Arm ABC is connected by pins to a collar at B and to crank CD at
C. Neglecting the effect of friction, determine the couple
M
required to hold the system in equilibrium when
90 .

SOLUTION
Free body: Member ABC
0: 0
xx
FC 


0: (160 mm) (240 N)(90 mm) 0
By
MC  


135 N
y
C



Free body: Member CD

0: (135 N)(300 mm) 0
D
MM  


3
40.5 10 N mmM  


40.5 kN mM



www.elsolucionario.org

980. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.133
The Whitworth mechanism shown is used to produce a quick-return
motion of Point D. The block at B is pinned to the crank AB and is free to
slide in a slot cut in member CD. Determine the couple
M that must be
applied to the crank AB to hold the mechanism in equilibrium when
(a) α
 0, (b) α  30°.

SOLUTION
(a) Free body: Member CD:
0: (0.5 m) (1200 N)(0.7 m) 0
C
MB  

1680 NB
Free body: Crank AB:

0: (1680 N)(0.1 m) 0
A
MM  

168.0 N mM  168.0 N mM 
(b) Geometry:
100 mm, 30°AB  

50
tan 5.87
486.6

 

22
(50) (486.6)
489.16 mm
BC


Free body: Member CD:


0: (0.48916) (1200 N)(0.7)cos5.87 0
C
MB  

1708.2 NB


www.elsolucionario.org

981. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.133 (Continued)

Free body: Crank AB:

0: ( sin 65.87 )(0.1 m) 0
A
MMB   

(1708.2 N)(0.1 m)sin 65.87
155.90 N m
M
M
 

155.9 N mM 
www.elsolucionario.org

982. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.134
Solve Problem 6.133 when (a) α  60, (b) α  90.
PROBLEM 6.133 The Whitworth mechanism shown is used to produce
a quick-return motion of Point D. The block at B is pinned to the crank
AB and is free to slide in a slot cut in member CD. Determine the couple
M that must be applied to the crank AB to hold the mechanism in
equilibrium when (a) α
 0, (b) α  30°.

SOLUTION
(a) Geometry:
100 mm, 60°AB  

100cos60 86.60BE
 

22
400 100sin 60 450
86.60
tan 10.89
450
(86.60) (450) 458.26 mm
CE
BC






Free body: Member CD:




0: (0.45826 m) (1200 N)(0.7 m)cos10.89 0
C
MB  

1800.0 NB

Free body: Crank AB:

0: ( sin 40.89 )(0.1m) 0
A
MMB   

(1800.0 N)(0.1 m)sin 40.89M 
117.83 N mM  117.8 N m M 

www.elsolucionario.org

983. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.134 (Continued)

(b) Free body: Member CD:

22
0.1 m
tan
0.4 m
14.04
(0.1) (0.4) 0.41231 mBC






0: (0.41231) (1200 N)(0.7cos14.04 ) 0
C
MB  

1976.4 NB
Free body: Crank AB:


0: ( sin14.04 )(0.1 m) 0
A
MMB   

(1976.4 N)(0.1 m)sin14.04
47.948 N m
M
M
 

47.9 N mM 
www.elsolucionario.org

984. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.135
Two rods are connected by a frictionless collar B. Knowing that the magnitude
of the couple
MA is 500 lb · in., determine (a) the couple MC required for
equilibrium, (b) the corresponding components of the reaction at C.


SOLUTION

(a) Free body: Rod AB & collar:
0: ( cos )(6 in.) ( sin )(8 in.) 0
AA
MB B M    

(6cos 21.8 8sin 21.8 ) 500 0B

58.535 lbB
Free body: Rod BC:

0: 0
CC
MMB    

(58.535 lb)(21.541in.) 1260.9 lb in.
C
MB  



1261lb in.
C
M 
(b)
0: cos 0
xx
FCB   

cos (58.535 lb)cos 21.8 54.3 lb
x
CB   

54.3 lb
x
C 

0: sin 0
yy
FCB   

sin (58.535 lb)sin 21.8 21.7 lb
y
CB  

21.7 lb
y
C 
www.elsolucionario.org

985. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.136
Two rods are connected by a frictionless collar B. Knowing that the
magnitude of the couple
MA is 500 lb · in., determine (a) the couple MC
required for equilibrium, (b) the corresponding components of the reaction
at C.


SOLUTION




(a) Free body: Rod AB:
0: (10 in.) 500 lb in. 0
A
MB  

50.0B
Free body: Rod BC & collar:
0: (0.6 )(20 in.) (0.8 )(8 in.) 0
CC
MMB B   


(30 lb)(20 in.) (40 lb)(8 in.) 0
C
M

920 lb in.
C
M 920 lb in.
C
M 
(b)
0: 0.6 0
xx
FCB  

0.6 0.6(50.0 lb) 30.0 lb
x
CB  

30.0 lb
x
C 

0: 0.8 0
yy
FCB  

0.8 0.8(50.0 lb) 40.0 lb
y
CB 

40.0 lb
y
C  www.elsolucionario.org

986. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.137
Rod CD is attached to the collar D and passes through a collar
welded to end B of lever AB. Neglecting the effect of friction,
determine the couple
M required to hold the system in equilibrium
when
30 .

SOLUTION
FBD DC:

0: sin 30 (150 N)cos30 0
xy
FD
  

(150 N) ctn 30 259.81 N
y
D

FBD machine:
0: (0.100 m)(150 N) (259.81 N) 0
A
MdM  

0.040 mdb

0.030718 m
tan 30
b

so 0.053210 mb

0.0132100 m
18.4321 N m
d
M




18.43 N mM 

www.elsolucionario.org

987. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.138
Rod CD is attached to the collar D and passes through a
collar welded to end B of lever AB. Neglecting the effect
of friction, determine the couple
M required to hold the
system in equilibrium when
30 .

SOLUTION
Note: CDB
FBD DC: 0: sin 30 (300 N)cos30 0
xy
FD
  

300 N
519.62 N
tan 30
y
D

FBD machine:


0.200 m
0: 519.62 N 0
sin 30
A
MM 


207.85 N mM 208 N mM 
www.elsolucionario.org

988. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.139
Two hydraulic cylinders control the position
of the robotic arm ABC. Knowing that in the
position shown the cylinders are parallel,
determine the force exerted by each cylinder
when
160 NP and 80 N.Q

SOLUTION
Free body: Member ABC:


4
0: (150 mm) (160 N)(600 mm) 0
5
BAE
MF  


800 N
AE
F

800 N
AE
FT



3
0: (800 N) 80 N 0
5
xx
FB    


560 N
x
B


4
0: (800 N) 160 N 0
5
yy
FB    


800 N
y
B

Free body: Member BDF:

4
0: (560 N)(400 mm) (800 N)(300 mm) (200 mm) 0
5
FDG
MF   

100 N
DG
F 100.0 N
DG
FC 



www.elsolucionario.org

989. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.140
Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders
are parallel and both are in tension. Knowing the
600 N
AE
F and 50 N,
DG
F determine the forces P and
Q applied at C to arm ABC.


SOLUTION








Free body: Member ABC:

4
0: (600 N)(150 mm) (600 mm) 0
5
B
MP  


120 NP

120.0 NP



4
0: (600)(750 mm) (600 mm) 0
5
Cy
MB  


600 N
y
B

Free body: Member BDF:
0: (400 mm) (600 N)(300 mm)
Fx
MB 


4
(50 N)(200 mm) 0
5


470 N
x
B
Return to free body: Member ABC:

3
0: (600 N) 470 N 0
5
x
FQ   


110 NQ

110.0 NQ


www.elsolucionario.org

990. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.141
A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs
shown. Determine the forces exerted at D and F on tong BDF.

SOLUTION
Free body: Rail:


Free body: Upper link:


Free Body: Tong BDF:



(39ft)(44lb/ft) 1716lbW

By symmetry,
1
858 lb
2
yy
EF W 
By symmetry,
1
()() 858lb
2
AB y AC y
FF W
Since AB is a two-force member,

()() 9.6
( ) (858) 1372.8 lb
9.6 6 6
AB yAB x
AB x
FF
F



0: (Attach at .)
DAB
MFA


(8) ( ) (18) (0.8) 0
(8) (1372.8 lb)(18) (858 lb)(0.8) 0
xABx y
x
FF F
F




3174.6 lb
x
F

3290 lbF 15.12



0: ( ) 0
xxABxx
FDFF    


( ) 1372.8 3174.6 4547.4 lb
xABxx
DF F


0: ( ) 0
yyAByy
FDFF   


0
y
D

4550 lbD 

www.elsolucionario.org

991. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.142
A log weighing 800 lb is lifted by a pair of tongs as shown.
Determine the forces exerted at E and F on tong DEF.

SOLUTION
FBD AB:



FBD
DEF:


By symmetry: 400 lb
yy
AB
and
6
(400 lb)
5
480 lb
xx
AB


Note:
DB
so
480 lb
400 lb
x
y
D
D



(10.5 in.)(400 lb) (15.5 in.)(480 lb) (12 in.) 0
Fx
ME   

970 lb
x
E 970 lbE 

0: 480 lb 970 lb 0
490 lb
xx
x
FF
F
    


0: 400 lb 0
400 lb
yy
y
FF
F
  


633 lbF 39.2 

www.elsolucionario.org

992. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.143
The tongs shown are used to apply a total upward force of 45 kN on a pipe cap.
Determine the forces exerted at D and F on tong ADF.

SOLUTION
FBD whole:


FBD ADF:






By symmetry,
22.5 kNAB






0: (75 mm) (100 mm)(22.5 kN) 0
F
MD  


30.0 kND 
0: 0
xx
FFD 

30 kN
x
FD

0: 22.5 kN 0
yy
FF  


22.5 kN
y
F

so
37.5 kNF 36.9 
www.elsolucionario.org

993. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.144
If the toggle shown is added to the tongs of Problem 6.141 and a single
vertical force is applied at G, determine the forces exerted at D and F on
tong ADF.

SOLUTION
Free body: Toggle:
By symmetry,
1
(45 kN) 22.5 kN
2
y
A
AG is a two-force member.

22.5 kN
22 mm 55 mm
56.25 kN
x
x
A
A



Free body: Tong ADF:
0: 22.5 kN 0
yy
FF  


22.5 kN
y
F

0: (75mm) (22.5kN)(100mm) (56.25kN)(160mm) 0
F
MD   

150 kND 150.0 kND 


0: 56.25 kN 150 kN 0
xx
FF   


93.75 kN
x
F



96.4 kNF 13.50


www.elsolucionario.org

994. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.145
The pliers shown are used to grip a 0.3-in.-diameter rod.
Knowing that two 60-lb forces are applied to the handles,
determine (a) the magnitude of the forces exerted on the
rod, (b) the force exerted by the pin at A on portion AB of
the pliers.

SOLUTION
Free body: Portion AB:
(a) 0: (1.2 in.) (60 lb)(9.5 in.) 0
A
MQ  


475 lbQ



(b)
0: (sin30 ) 0
xx
FQ A  


(475 lb)(sin30 ) 0
x
A 


237.5 lb
x
A

237.5 lb
x
A

0: (cos30 ) 60 lb 0
yy
FQ A    


(475 lb)(cos30 ) 60 lb 0
y
A


471.4 lb
y
A

471.4 lb
y
A


528 lbA 63.3 
www.elsolucionario.org

995. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.146
Determine the magnitude of the gripping forces exerted
along line aa on the nut when two 50-lb forces are applied
to the handles as shown. Assume that pins A and D slide
freely in slots cut in the jaws.

SOLUTION
FBD jaw AB: 0: 0
xx
FB 

0: (0.5 in.) (1.5 in.) 0
B
MQA  

3
Q
A


0: 0
yy
FAQB  

4
3
y
Q
BAQ

FBD handle ACE: By symmetry and FBD jaw DE,

3
0
4
3
xx
yy
Q
DA
EB
Q
EB





4
0: (5.25 in.)(50 lb) (0.75 in.) (0.75 in.) 0
33
C
QQ
M   


350 lbQ 
www.elsolucionario.org

996. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.147
In using the bolt cutter shown, a worker applies two 300-N
forces to the handles. Determine the magnitude of the
forces exerted by the cutter on the bolt.

SOLUTION
FBD cutter AB: FBD handle BC:

I Dimensions in mm II
FBD I: 0: 0
xx
FB  FBD II: 0: (12 mm) (448 mm)300 N 0
Cy
MB  

11,200.0 N
y
B
Then FBD I: 0: (96 mm) (24 mm) 0
Ay
MBF   4
y
FB

44,800 N 44.8 kNF 
www.elsolucionario.org

997. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.148
Determine the magnitude of the gripping forces produced
when two 300-N forces are applied as shown.

SOLUTION
We note that AC is a two-force member.
FBD handle CD:

2.8
0: (126 mm)(300 N) (6 mm)
8.84
D
MA 

1
(30 mm) 0
8.84
A

 



Dimensions in mm
2863.6 8.84 NA
FBD handle AB:

1
0: (132 mm)(300 N) (120 mm) (2863.6 8.84 N)
8.84
(36 mm) 0
 

B
M
F

8.45 kNF 
Dimensions in mm
www.elsolucionario.org

998. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.149
Determine the force P that must be applied to the toggle CDE to
maintain bracket ABC in the position shown.

SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:

()()
()5()
30 150
CD yCD x
CD y CD x
FF
FF


Similarly,
()5()
DE y DE x
FF


0: ( ) ( )
yDEyCDy
FFF 

It follows that
()()
DE x CD x
FF


0: ( ) ( ) 0
xDExCDx
FPFF   


1
()()
2
DE x CD x
FF P

Also,

1
()()5 2.5
2
DE y CD y
FF PP

 



Free body: Member ABC:




1
0: (2.5 )(300) (450) (910 N)(150) 0
2
A
MPP

   




(750 225) (910 N)(150)P

140.0 NP 

Dimensions in mm www.elsolucionario.org

999. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.150
Determine the force P that must be applied to the toggle CDE to maintain
bracket ABC in the position shown.

SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:

()()
()5()
30 150
CD yCD x
CD y CD x
FF
FF

Similarly,
()5()
DE y DE x
FF

0: ( ) ( )
yDEyCDy
FFF 
It follows that
()()
DE x CD x
FF

0: ( ) ( ) 0
xDExCDx
FFFP  


1
()()
2
DE x CD x
FF P
Also,
1
()()5 2.5
2
DE y CD y
FF PP

 


Free body: Member ABC:





1
0: (2.5 )(300) (450) (910 N)(150) 0
2
A
MPP

   




(750 225) (910 N)(150)P

260 NP 

Dimensions
in mm www.elsolucionario.org

1000. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.151
Since the brace shown must remain in position even when the magnitude of P is
very small, a single safety spring is attached at D and E. The spring DE has a
constant of 50 lb/in. and an unstretched length of 7 in. Knowing that l = 10 in. and
that the magnitude of
P is 800 lb, determine the force Q required to release the
brace.

SOLUTION
Free body:
0 : 15 in. 35 in. 0
Ax
MQ C  



7
(1)
3
x
QC



0: 800 lb 0 or 800 lb
yy y
FC C   

Spring force.
Unstretched length:
0
7in.
Stretched length:
10 in.

50 lb/in.k

0
()
50(10 7) 150 lbFkxk 
 

Free body: Portion BC


0 : 150 lb 2 in. 800 lb 3 in. 20 in. 0
Bx
MC   

135.0 lb
x
C
Substituting into equation (1) gives
315 lbQ 
www.elsolucionario.org

1001. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.152
The specialized plumbing wrench shown is used in confined areas (e.g.,
under a basin or sink). It consists essentially of a jaw BC pinned at B to a
long rod. Knowing that the forces exerted on the nut are equivalent to a
clockwise (when viewed from above) couple of magnitude 135 lb  in.,
determine (a) the magnitude of the force exerted by pin B on jaw BC,
(b) the couple
M0 that is applied to the wrench.

SOLUTION











Free body: Jaw BC:
This is a two-force member.

5
8
2.4
1.5 in. in.
y x
yx
C C
CC


0:
xxx
FBC  (1)

0: 2.4
yyyx
FBCC   (2)
Free body: Nut:
0:
xxx
FCD 
135 lb in.M 


(1.125 in.) 135 lb in.
x
C 


120 lb
x
C

(a) Eq. (1): 120 lb
xx
BC
Eq. (2):
2.4(120 lb) 288 lb
yy
BC 


1/ 2
22 2 21/2
(120 288 )
xy
BBB   312 lbB 
(b) Free body: Rod:

0
0: (0.625 in.) (0.375 in.) 0
Dyx
MMB B  

0
(288)(0.625) (120)(0.375) 0M  

0
135.0 lb in.M 

www.elsolucionario.org

1002. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.153
The motion of the bucket of the front-end loader
shown is controlled by two arms and a linkage
that are pin-connected at D. The arms are located
symmetrically with respect to the central,
vertical, and longitudinal plane of the loader; one
arm AFJ and its control cylinder EF are shown.
The single linkage GHDB and its control cylinder
BC are located in the plane of symmetry. For the
position and loading shown, determine the force
exerted (a) by cylinder BC, (b) by cylinder EF.

SOLUTION
Free body: Bucket
0: (4500 lb)(20 in.) (22 in.) 0
JGH
MF  


4091 lb
GH
F
Free body: Arm BDH
0: (4091 lb)(24 in.) (20 in.) 0
DBC
MF  


4909 lb
BC
F



4.91 kips
BC
F C 
Free body: Entire mechanism
(Two arms and cylinders AFJE)

Note: Two arms thus
2
EF
F

18 in.
tan
65 in.
15.48




0: (4500 lb)(123 in.) (12 in.) 2 cos (24 in.) 0
ABCEF
MFF    


(4500 lb)(123 in.) (4909 lb)(12 in.) 2 cos15.48 (24 in.) 0
EF
F

10.690 lb
EF
F 10.69 kips
EF
FC 
www.elsolucionario.org

1003. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.154
The bucket of the front-end loader shown carries a
3200-lb load. The motion of the bucket is
controlled by two identical mechanisms, only
one of which is shown. Knowing that the
mechanism shown supports one-half of the
3200-lb load, determine the force exerted (a) by
cylinder CD, (b) by cylinder FH.

SOLUTION
Free body: Bucket: (one mechanism)
0: (1600 lb)(15 in.) (16 in.) 0
DAB
MF  

1500 lb
AB
F
Note: There are two identical support mechanisms.
Free body: One arm BCE:

8
tan
20
21.8





0: (1500 lb)(23 in.) cos 21.8 (15 in.) sin 21.8 (5 in.) 0
ECDCD
MFF     


2858 lb
CD
F 2.86 kips
CD
FC 
Free body: Arm DFG:


0: (1600 lb)(75 in.) sin 45 (24 in.) cos 45 (6 in.) 0
GFHFH
MFF     


9.428 kips
FH
F 9.43 kips
FH
FC 
www.elsolucionario.org

1004. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.155
The telescoping arm ABC is used to provide an elevated
platform for construction workers. The workers and the
platform together have a mass of 200 kg and have a
combined center of gravity located directly above C. For the
position when
  20, determine (a) the force exerted at B
by the single hydraulic cylinder BD, (b) the force exerted on
the supporting carriage at A.

SOLUTION
Geometry: (5 m)cos 20 4.6985 m
(2.4 m)cos 20 2.2553 m
(2.4 m)sin 20 0.8208 m
0.5 1.7553 m
0.9 1.7208 m
a
b
c
db
ec



 
 


1.7208
tan ; 44.43
1.7553
e
d
  

Free body: Arm ABC:
We note that BD is a two-force member.

2
(200 kg)(9.81 m/s ) 1.962 kNW
(a)
0: (1.962 kN)(4.6985 m) sin 44.43 (2.2553 m) cos 44.43(0.8208 m) 0
ABDBD
MFF    


9.2185 (0.9927) 0: 9.2867 kN
BD BD
FF

(b) 9.29 kN
BD
F 44.4 

0: cos 0
xxBD
FAF   


(9.2867 kN)cos 44.43 6.632 kN
x
A 6.632 kN
x
A

0: 1.962 kN sin 0
yy BD
FA F    


1.962 kN (9.2867 kN)sin 44.43 4.539 kN
y
A 


4.539 kN
y
A

8.04 kNA 34.4 
www.elsolucionario.org

1005. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.156
The telescoping arm ABC of Prob. 6.155 can be lowered
until end C is close to the ground, so that workers can easily
board the platform. For the position when
 = 20,
determine (a) the force exerted at B by the single hydraulic
cylinder BD, (b) the force exerted on the supporting carriage
at A.

SOLUTION
Geometry: (5 m)cos 20 4.6985 m
(2.4 m)cos 20 2.2552 m
(2.4 m)sin 20 0.8208 m
0.5 1.7553 m
0.9 0.0792 m
a
b
c
db
ec



 



0.0792
tan ; 2.584
1.7552
e
d
  
Free body: Arm ABC:
We note that BD is a two-force member.

2
(200 kg)(9.81 m/s )
1962 N 1.962 kN
W
W



(a)
0: (1.962 kN)(4.6985 m) sin 2.584 (2.2553 m) cos 2.584 (0.8208 m) 0
ABDBD
MFF     


9.2185 (0.9216) 0 10.003 kN
BD BD
FF

(b) 10.00 kN
BD
F 2.58 

0: cos 0
xxBD
FAF   


(10.003 kN)cos 2.583 9.993 kN
x
A 9.993 kN
x
A

0: 1.962 kN sin 0
yy BD
FA F    


1.962 kN (10.003 kN)sin 2.583 1.5112 kN
y
A 


1.5112 kN
y
A

10.11 kNA 8.60 
www.elsolucionario.org

1006. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.157
The motion of the backhoe bucket
shown is controlled by the hydraulic
cylinders AD, CG, and EF. As a result
of an attempt to dislodge a portion of a
slab, a 2-kip force
P is exerted on the
bucket teeth at J. Knowing that
  45,
determine the force exerted by each
cylinder.

SOLUTION
Free body: Bucket:
0:
H
M (Dimensions in inches)

43
(10) (10) cos (16) sin (8) 0
55
CG CG
FFP P  

(16 cos 8 sin )
14
CG
P
F
  (1)
Free body: Arm ABH and bucket: (Dimensions in inches)


43
0: (12) (10) cos (86) sin (42) 0
55
BADAD
MFFP P     


(86cos 42 sin )
15.6
AD
P
F
  (2)
Free body: Bucket and arms IEB  ABH:
Geometry of cylinder EF:
16 in.
tan
40 in.
21.801


 www.elsolucionario.org

1007. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.157 (Continued)



0: cos (18 in.) cos (28 in.) sin (120 in.) 0
IEF
MF P P    

(120 sin 28 cos )
cos 21.8 (18)
(120 sin 28 cos )
16.7126
EF
P
F
P 




(3)
For
2 kips,P

45
From Eq. (1):
2
(16 cos 45 8 sin 45 ) 2.42 kips
14
CG
F    2.42 kips
CG
FC 
From Eq. (2):
2
(86 cos 45 42 sin 45 ) 3.99 kips
15.6
AD
F    3.99 kips
AD
FC 
From Eq. (3):
2
(120 sin 45 28 cos 45 ) 7.79 kips
16.7126
EF
F 7.79 kips
EF
FT 
www.elsolucionario.org

1008. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.158
Solve Problem 6.157 assuming that the
2-kip force
P acts horizontally to the
right (
  0).
PROBLEM 6.157 The motion of the
backhoe bucket shown is controlled by
the hydraulic cylinders AD, CG, and
EF. As a result of an attempt to
dislodge a portion of a slab, a 2-kip
force
P is exerted on the bucket teeth at
J. Knowing that
  45, determine the
force exerted by each cylinder.

SOLUTION
Free body: Bucket:
0:
H
M (Dimensions in inches)

43
(10) (10) cos (16) sin (8) 0
55
CG CG
FFP P  

(16 cos 8 sin )
14
CG
P
F
  (1)
Free body: Arm ABH and bucket: (Dimensions in inches)


43
0: (12) (10) cos (86) sin (42) 0
55
BADAD
MFFP P     


(86cos 42 sin )
15.6
AD
P
F
  (2)
Free body: Bucket and arms IEB  ABH:
Geometry of cylinder EF:
16 in.
tan
40 in.
21.801



www.elsolucionario.org

1009. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.158 (Continued)



0: cos (18 in.) cos (28 in.) sin (120 in.) 0
IEF
MF P P    

(120 sin 28 cos )
cos 21.8 (18)
(120 sin 28 cos )
16.7126
EF
P
F
P 




(3)
For
2 kips,P

0
From Eq. (1):
2
(16 cos 0 8 sin 0) 2.29 kips
14
CG
F   2.29 kips
CG
FC 
From Eq. (2):
2
(86 cos 0 42 sin 0) 11.03 kips
15.6
AD
F   11.03 kips
AD
FC 
From Eq. (3):
2
(120 sin 0 28 cos 0) 3.35 kips
16.7126

EF
F 3.35 kips
EF
FC 
www.elsolucionario.org

1010. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.159
The gears D and G are rigidly attached
to shafts that are held by frictionless
bearings. If r
D  90 mm and r G  30
mm, determine (a) the couple
M0 that
must be applied for equilibrium, (b) the
reactions at A and B.

SOLUTION














(a) Projections on yz plane.
Free body: Gear G:
0: 30 N m (0.03 m) 0; 1000 N
G
MJJ   


Free body: Gear D:
0
0: (1000 N)(0.09 m) 0
D
MM  


0
90 N mM
0
(90.0 N m)Mi 
(b) Gear G and axle FH: 0: (0.3 m) (1000 N)(0.18 m) 0
F
MH  


600 NH


0: 600 1000 0
y
FF   

400 NF
Gear D and axle CE:
0: (1000 N)(0.18 m) (0.3 m) 0
C
ME  

600 NE

0: 1000 600 0
y
FC  


400 NC
www.elsolucionario.org

1011. Copyright © McGraw-Hill Education. Permission required for reproduction or display.








PROBLEM 6.159 (Continued)

Free body: Bracket AE:
0: 400 400 0
y
FA    0A 
0: (400 N)(0.32 m) (400 N)(0.2 m) 0
AA
MM   


48 N m
A
M  (48.0 N m)
A
 Mi 
Free body: Bracket BH: 0: 600 600 0
y
FB    0B 

0: (600 N)(0.32 m) (600 N)(0.2 m) 0
BB
MM   

72 N m
B
M  (72.0 N m)
B
 Mi 

www.elsolucionario.org

1012. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.160
In the planetary gear system shown, the radius of the central gear A is
a  18 mm, the radius of each planetary gear is b, and the radius of the
outer gear E is (a  2b). A clockwise couple of magnitude M
A  10 N  m
is applied to the central gear A and a counterclockwise couple of
magnitude M
S  50 N  m is applied to the spider BCD. If the system is
to be in equilibrium, determine (a) the required radius b of the
planetary gears, (b) the magnitude M
E of the couple that must be
applied to the outer gear E.

SOLUTION
FBD Central Gear:

FBD Gear C:

FBD Spider:

FBD Outer Gear:


By symmetry,
123
FFFF

10
0: 3( ) 10 N m 0, N m
3
AA
A
MrF F
r
    



44
0: ( ) 0,
CB
MrFF FF  

0: 0
xx
FC
 


0: 2 0, 2
yy y
FCF CF
    

Gears B and D are analogous, each having a central force of 2F.

0: 50 N m 3( )2 0
AAB
MrrF   


20
50 N m 3( ) N m 0
AB
A
rr
r
  

2.5 1 ,
AB B
AA
rr r
rr


1.5
BA
rr
Since
18 mm,
A
r
(a)
27.0 mm
B
r 

0: 3( 2 ) 0
AABE
MrrFM   


10 N m
3(18 mm 54 mm) 0
54 mm
E
M


(b)
40.0 N m
E
M


www.elsolucionario.org

1013. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.161*
Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The
bearings at B and D do not exert any axial force. A couple of magnitude 500 lb  in. (clockwise when
viewed from the positive x-axis) is applied to shaft CF at F. At a time when the arm of the crosspiece
attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft
AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted
on the crosspiece must be zero.)


SOLUTION
We recall from Figure 4.10 that a universal joint exerts on members it connects a force of unknown
direction and a couple about an axis perpendicular to the crosspiece.
Free body: Shaft DF:


0: cos30 500 lb in. 0
xC
MM  

577.35 lb in.
C
M
Free body: Shaft BC:
We use here ,,xyz with xalong.BC


0: (577.35 lb in.) ( 5 in.) ( ) 0
CR yz
MMi iiBjBk       
www.elsolucionario.org

1014. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.161* (Continued)

Equate coefficients of unit vectors to zero:

i: 577.35 lb in. 0
A
M 577.35 lb in.
A
M

j: 0
z
B 577 lb in.
A
M 

k: 0
y
B 0B
 0B

0: 0, since 0,BC B   F 0C
Return to free body of shaft DF.
0
D
M (Note that 0C and 577.35 lb in.
C
M )

(577.35 lb in.)(cos30 sin30 ) (500 lb in.)
    ij i

(6 in.) ( ) 0
xyz
EEE
  iijk
(500 lb in.) (288.68 lb in.) (500 lb in.)iji

(6 in.) (6 in.) 0
yz
EE
 kj
Equate coefficients of unit vectors to zero:

j: 288.68 lb in. (6 in.) 0
z
E
 48.1lb
z
E

k: 0
y
E

0: 0FCDE

0(48.1lb)0
yz x
DD E
 jki k

i: 0
x
E


j: 0
y
D


k: 48.1lb 0
z
D 48.1lb
z
D
Reactions are:
0
B 
(48.1lb)Dk 
(48.1lb)
Ek 
www.elsolucionario.org

1015. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.162*
Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical.
PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal
joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb · in.
(clockwise when viewed from the positive x-axis) is applied to shaft CF at F. At a time when the arm of
the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be
applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the
couples exerted on the crosspiece must be zero.)


SOLUTION
Free body: Shaft DF.


0: 500 lb in. 0
xC
MM  

500 lb in.
C
M
Free body: Shaft BC:

We resolve
(520 lb in.) i into components along x and y axes:

(500 lb in.)(cos30 sin30 )
C
   Mij

0: (500 lb in.)(cos30 sin 30 ) (5 in.) ( ) 0
CA yz
MBB

       Mi ijijk

(433 lb in.) (250 lb in.) (5 in.) (5 in.) 0
Ayz
MBB

   iijkj www.elsolucionario.org

1016. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.162* (Continued)

Equate to zero coefficients of unit vectors:

: 433 lb in. 0
A
M i 433 lb in.
A
M 

:250lbin.(5in.) 0
z
B j 50 lb
z
B

:0
y
kB
Reactions at B.
(50 lb)Bk
0: 0   FBC
(50 lb) 0 C
k (50 lb) Ck
Return to free body of shaft DF. 0: (6 in.) ( ) (4 in.) ( 50 lb)
Dxyz
EEE    Miijkik
(500 lb in.) (500 lb in.) 0  ii

(6 in.) (6 in.) (200 lb in.) 0
yz
EE
 kj j

:0
y
Ek

: (6 in.) 200 lb in. 0
z
Ej 33.3 lb
z
E

0: 0F   CDE

(50 lb) (33.3 lb) 0
yz x
DD E kjki k

:0
: 50 lb 33.3 lb 0
x
z
E
D

 
i
k


83.3 lb
z
D
Reactions are (50 lb)
Bk 
(83.3 lb)
Dk 
(33.3 lb)
Ek 
www.elsolucionario.org

1017. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.163*
The large mechanical tongs shown are used to grab and lift a
thick 7500-kg steel slab HJ. Knowing that slipping does not
occur between the tong grips and the slab at H and J, determine
the components of all forces acting on member EFH. (Hint:
Consider the symmetry of the tongs to establish relationships
between the components of the force acting at E on EFH and the
components of the force acting at D on DGJ.)

SOLUTION
Free body: Pin A:
2
(7500 kg)(9.81 m/s ) 73.575 kNTWmg  

0: ( ) ( )
1
0: ( ) ( )
2
xABxACx
yAByACy
FFF
FFFW 
  

Also,
()2()
AC x AC y
FFW

Free body: Member CDF:

1
0: (0.3) (2.3) (1.8) (0.5 m) 0
2
Dxy
MW WFF    
or
1.8 0.5 1.45
xy
FF W (1) www.elsolucionario.org

1018. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.163* (Continued)


0: 0
xxx
FDFW  
or
xx
EFW (2)

1
0: 0
2
yyy
FFDW   
or
1
2
yy
EF W (3)
Free body: Member EFH:

1
0: (1.8) (1.5) (2.3) (1.8 m) 0
2
Ex y x
MFFH W    
or
1.8 1.5 2.3 0.9 
xy x
FF H W (4)

0: 0
xxxx
FEFH   
or
xx x
EFH (5)
Subtract Eq. (2) from Eq. (5):
2
xx
FHW (6)
Subtract Eq. (4) from 3  (1):
3.6 5.25 2.3
xx
FWH (7)
Add Eq. (7) to 2.3  Eq. (6):
8.2 2.95
x
FW

0.35976
x
FW (8)
Substitute from Eq. (8) into Eq. (1):

(1.8)(0.35976 ) 0.5 1.45
y
WF W


0.5 1.45 0.64756 0.80244
y
FW W W 


1.6049
y
FW
(9)
Substitute from Eq. (8) into Eq. (2):
0.35976 ; 1.35976
xx
EWWEW
Substitute from Eq. (9) into Eq. (3):
1
1.6049 2.1049
2
yy
EWWE W 
From Eq. (5):
1.35976 0.35976 1.71952
xxx
HEF W W W  
Recall that
1
2
y
HW


www.elsolucionario.org

1019. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.163* (Continued)

Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams.
Substitute
73.575 kNW

100.0 kN
x
E 154.9 kN
y
E 
26.5 kN
x
F  118.1 kN
y
F 

126.5 kN
x
H 36.8 kN
y
H



www.elsolucionario.org

1020. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.164
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.

SOLUTION
Reactions:
0: 16 kN
Cx
M  A

0: 9 kN
yy
F A


0: 16 kN
x
F  C


Joint E:

3 kN
54 3
BE DE
FF




5.00 kN
BE
FT 


4.00 kN
DE
F C 

Joint B:

4
0: (5 kN) 0
5
xAB
FF  

4 kN
AB
F 4.00 kN
AB
FT 

3
0: 6 kN (5 kN) 0
5
yBD
FF    

9 kN
BD
F 9.00 kN
BD
FC 
Joint D:

3
0: 9 kN 0
5
yAD
FF   

15 kN
AD
F 15.00 kN
AD
FT 

4
0: 4 kN (15 kN) 0
5
xCD
FF    

16 kN
CD
F 16.00 kN
CD
FC 
www.elsolucionario.org

1021. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.165
Using the method of joints, determine the force in each
member of the double-pitch roof truss shown. State whether
each member is in tension or compression.

SOLUTION
Free body: Truss:


0: (18 m) (2 kN)(4 m) (2 kN)(8 m) (1.75 kN)(12 m)
(1.5 kN)(15 m) (0.75 kN)(18 m) 0
A
MH   
 


4.50 kNH


0: 0
0: 9 0
94.50
xx
yy
y
FA
FAH
A
 
 

4.50 kN
y
A
Free body: Joint A:


3.50 kN
215
7.8262 kN
ACAB
AB
FF
FC


7.83 kN
AB
FC 

7.00 kN
AC
FT 
www.elsolucionario.org

1022. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.165 (Continued)
Free body: Joint B:

22 1
0: (7.8262 kN) 0
55 2
xBD BC
FF F   

or
0.79057 7.8262 kN
BD BC
FF (1)

11 1
0: (7.8262 kN) 2 kN 0
55 2
yBD BC
FF F    

or
1.58114 3.3541
BD BC
FF (2)
Multiply Eq. (1) by 2 and add Eq. (2):

3 19.0065
6.3355 kN
BD
BD
F
F



6.34 kN
BD
FC


Subtract Eq. (2) from Eq. (1):

2.37111 4.4721
1.8861 kN
BC
BC
F
F


1.886 kN
BC
FC 
Free body: Joint C:

21
0: (1.8861kN) 0
52
yCD
FF  


1.4911 kN
CD
F 1.491 kN
CD
FT 

11
0: 7.00 kN (1.8861 kN) (1.4911 kN) 0
25
xCE
FF    


5.000 kN
CE
F

5.00 kN
CE
FT




Free body: Joint D:

212 1
0: (6.3355 kN) (1.4911 kN) 0
525 5
xDFDE
FFF    

or
0.79057 5.5900 kN
DF DE
FF (1)

111 2
0: (6.3355 kN) (1.4911 kN) 2 kN 0
525 5
yDFDE
FFF     
or
0.79057 1.1188 kN
DF DE
FF (2)
Add Eqs. (1) and (2):
2 6.7088 kN
DF
F

3.3544 kN
DF
F 3.35 kN
DF
FC 
Subtract Eq. (2) from Eq. (1):
1.58114 4.4712 kN
DE
F

2.8278 kN
DE
F 2.83 kN
DE
FC  www.elsolucionario.org

1023. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.165 (Continued)

Free body: Joint F:

12
0: (3.3544 kN) 0
25
xFG
FF  


4.243 kN
FG
F 4.24 kN
FG
FC 

11
0: 1.75 kN (3.3544 kN) ( 4.243 kN) 0
52
yEF
FF      


2.750 kN
EF
F

2.75 kN
EF
FT




Free body: Joint G:

111
0: (4.243 kN) 0
222
xGHEG
FFF   

or
4.243 kN
GH EG
FF (1)

111
0: (4.243 kN) 1.5 kN 0
222
yGHEG
FFF     
or
6.364 kN
GH EG
FF (2)
Add Eqs. (1) and (2):
2 10.607
GH
F

5.303
GH
F 5.30 kN
GH
FC 
Subtract Eq. (1) from Eq. (2):
2 2.121 kN
EG
F

1.0605 kN
EG
F 1.061 kN
EG
FC 
Free body: Joint H:


3.75 kN
11
EH
F

3.75 kN
EH
FT 
We can also write
3.75 kN
12
GH
F

5.30 kN (Checks)
GH
FC www.elsolucionario.org

1024. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.166
A stadium roof truss is loaded as shown. Determine the
force in members AB, AG, and FG.

SOLUTION
We pass a section through members AB, AG, and FG, and use the free body shown.


40
0: (6.3 ft) (1.8 kips)(14 ft)
41
(0.9 kips)(28 ft) 0
GAB
MF

 





8.20 kips
AB
F 8.20 kips
AB
FT 

3
0: (28 ft) (1.8 kips)(28 ft)
5
(1.8 kips)(14 ft) 0
DAG
MF

 




4.50 kips
AG
F 4.50 kips
AG
FT 

0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft)
(0.9 kips)(40 ft) 0
AFG
MF   


11.60 kips
FG
F 11.60 kips
FG
FC 
www.elsolucionario.org

1025. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.167
A stadium roof truss is loaded as shown. Determine the
force in members AE, EF, and FJ.

SOLUTION
We pass a section through members AE, EF, and FJ, and use the free body shown.


22
8
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
89
FAE
MF

  



17.46 kips
AE
F 17.46 kips
AE
F T 

0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
AEF
MF 

11.60 kips
EF
F 11.60 kips
EF
F C 

0: (8 ft) (0.9 kips)(8 ft) (1.8 kips)(20 ft) (1.8 kips)(34 ft) (0.9 kips)(48 ft) 0
EFJ
MF 

18.45 kips
FJ
F 18.45 kips
FJ
F C 
www.elsolucionario.org

1026. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.168
Determine the components of all forces acting on member
ABD of the frame shown.
SOLUTION
Free body: Entire frame:




0: (300 lb) (12 ft) (450 lb)(4 ft) (6 ft) 0
D
ME    

900 lbE 900 lbE 

0: 900 lb 0
xx
FD  

900 lb
x
D 

0: 300 lb 450 lb 0
yy
FD   

750 lb
y
D 

Free body: Member ABD:
We note that BC is a two-force member and that B is directed along BC.

0: (750 lb)(16 ft) (900 lb)(6 ft) (8 ft) 0
A
MB   

825 lbB 825 lbB 

0: 900 lb 0
xx
FA  

900 lb
x
A 900 lb
x
A 

0: 750 lb 825 lb 0
yy
FA   

75 lb
y
A 75.0 lb
y
A 
www.elsolucionario.org

1027. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.169
Determine the components of the reactions at A and E if the frame is loaded
by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.

SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since the point of application of the couple is
immaterial.

0: 36 N m (0.4 m) 0
Ex
MA   

90 N 90.0 N
xx
A  A

0: 90 + 0
xx
FE  

90 N 90.0 N
xx
E E

0: 0
yyy
FAE  
(1)

(a) Couple applied at B.
Free body: Member CE:
AC is a two-force member. Thus, the reaction at E must be directed along EC.

0.24 m
;45.0N
90 N 0.48 my
y
E

E
From Eq. (1):
45 N 0
y
A

45 N 45.0 N
yy
A  A
Thus, reactions are

90.0 N
x
A , 45.0 N
y
A 

90.0 N
x
E , 45.0 N
y
E 
(b) Couple applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.

0.16 m
;30N
90 N 0.48 my
y
A

A
www.elsolucionario.org

1028. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.169 (Continued)

From Eq. (1):
0
30 N 0
yy
y
AE
E



30 N 30 N
yy
E  E
Thus, reactions are
90.0 N
x
A , 30.0 N
y
A 

90.0 N
x
E , 30.0 N
y
E 
www.elsolucionario.org

1029. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.170
Knowing that the pulley has a radius of 50 mm, determine the
components of the reactions at B and E.

SOLUTION
Free body: Entire assembly:
0 : (300 N)(350 mm) (150 mm) 0
Ex
MB   


700 N
x
B

700 N
x
B

0: 700 0
xx
FNE   


700 N
x
E

700 N
x
E



0: 300 N 0
yyy
FBE   
(1)
Free body: Member ACE:


0: (700 N)(150 mm) (300 N)(50 mm) (180 mm) 0
Cy
ME   

500 N
y
E 500 N
y
E
From Eq. (1): 500 N 300 N 0
y
B

200 N
y
B 200 N
y
B
Thus, reactions are

700 N
x
B , 200 N
y
B 
700 N
x
E , 500 N
y
E 
www.elsolucionario.org

1030. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.171
For the frame and loading shown, determine the components of the
forces acting on member CFE at C and F.

SOLUTION
Free body: Entire frame:


0: (40 lb)(13 in.) (10 in.) 0
Dx
MA  


52 lb,
x
A

52 lb
x
A 
Free body: Member ABF:
0: (52 lb)(6 in.) (4 in.) 0
Bx
MF   


78 lb
x
F

Free body: Member CFE: From above:
78.0 lb
x
F 
0: (40 lb)(9 in.) (78 lb)(4 in.) (4 in.) 0
Cy
MF   


12 lb
y
F

12.00 lb
y
F



0: 78 lb 0
xx
FC  


78 lb
x
C

78.0 lb
x
C 
0: 40 lb 12 lb 0; 28 lb
yyy
FCC      28.0 lb
y
C


www.elsolucionario.org

1031. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.172
For the frame and loading shown, determine the reactions at A, B, D,
and E. Assume that the surface at each support is frictionless.

SOLUTION






Free body: Entire frame:
0: (1000 lb)sin30 0
x
FAB  

500 0AB  (1)

0: (1000 lb)cos30 0
y
FDE  


866.03 0DE 
(2)
Free body: Member ACE:

0: (6 in.) (8 in.) 0
C
MAE   


3
4
EA
(3)
Free body: Member BCD:

0: (8 in.) (6 in.) 0
C
MDB   


3
4
DB
(4)
Substitute E and D from Eqs. (3) and (4) into Eq. (2):

33
866.06 0
44
1154.71 0
AB
AB
 
 
(5)
From Eq. (1):
500 0AB  (6)
Eqs. (5)
(6): 2 654.71 0A 327.4 lbA 327 lbA 
Eqs. (5)
(6): 2 1654.71 0B 827.4 lbB 827 lbB 
From Eq. (4):
3
(827.4)
4
D
620.5 lbD 621lbD 
From Eq. (3):
3
(327.4)
4
E
245.5 lbE 246 lbE 

www.elsolucionario.org

1032. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.173
Water pressure in the supply system exerts a downward force of 135 N
on the vertical plug at A. Determine the tension in the fusible link DE
and the force exerted on member BCE at B.

SOLUTION
Free body: Entire linkage:
0: 135 0
y
FC   

135 NC

Free body: Member BCE:
0: 0
xx
FB 

0: (135 N)(6 mm) (10 mm) 0
BDE
MT  

81.0 N
DE
T 

0: 135 81 0
yy
FB  

216 N
y
B 216 NB



www.elsolucionario.org

1033. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.174
A couple M of magnitude 1.5 kN  m is applied to the crank of the engine system shown. For each of the
two positions shown, determine the force
P required to hold the system in equilibrium.


SOLUTION
(a) FBDs:

Note:
50 mm
tan
175 mm
2
7


FBD whole:
0: (0.250 m) 1.5 kN m 0 6.00 kN
Ay y
MC C   
FBD piston:
6.00 kN
0: sin 0
sin siny
y y BC BC
C
FCF F

    

0: cos 0
xBC
FF P  

6.00 kN
cos 7 kips
tan
BC
PF 


21.0 kNP 






Dimensions in mm www.elsolucionario.org

1034. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.174 (Continued)

(b)
FBDs:

Note:
2
tan as above
7

FBD whole:
0: (0.100 m) 1.5 kN m 0 15 kN
Ay y
MC C   

0: sin 0
sin
y
y y BC BC
C
FCF F

   

0: cos 0
xBC
FF P  

15 kN
cos
tan 2/7y
BC
C
PF

 52.5 kNP 

Dimensions in mm www.elsolucionario.org

1035. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.175
The compound-lever pruning shears shown can be
adjusted by placing pin A at various ratchet positions on
blade ACE. Knowing that 300-lb vertical forces are
required to complete the pruning of a small branch,
determine the magnitude P of the forces that must be
applied to the handles when the shears are adjusted as
shown.

SOLUTION
We note that AB is a two-force member.

()()
0.65 in. 0.55 in.
11
() ()
13
AB yAB x
ABy ABx
FF
FF

 (1)
Free body: Blade ACE:


0: (300 lb)(1.6 in.) ( ) (0.5 in.) ( ) (1.4 in.) 0
CABxABy
MFF   

Use Eq. (1):
11
()(0.5in.) ()(1.4in.)480lbin.
13
AB x AB x
FF
 


1.6846( ) 480
AB x
F

( ) 284.9 lb
AB x
F


11
( ) (284.9 lb)
13
AB y
F 

( ) 241.1lb
AB y
F 

Free body: Lower handle:


0: (241.1lb)(0.75 in.) (284.9 lb)(0.25 in.) (3.5 in.) 0
D
MP   

31.3 lbP 
www.elsolucionario.org

1036. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.F1
For the frame and loading shown, draw the free-body diagram(s) needed
to determine the force in member BD and the components of the reaction
at C.

SOLUTION
We note that BD is a two-force member.
Free body: ABC


www.elsolucionario.org

1037. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.F2
For the frame and loading shown, draw the free-body diagram(s)
needed to determine the components of all forces acting on
member ABC.

SOLUTION
We note that BE is a two-force member.
Free body: Frame


Note: Sum forces in y about A to determine
Ay.
Sum moments about E to determine
Ax.



Free body: ABC

Note: Sum moments about C to determine
FBE.
Sum forces in x to determine
Cx.
Sum forces in y to determine
Cy.

www.elsolucionario.org

1038. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 6.F3
Draw the free-body diagram(s) needed to determine all the forces
exerted on member AI if the frame is loaded by a clockwise
couple of magnitude 1200 lb·in. applied at point D.

SOLUTION
We note that AB, BC, and FG are two-force members.
Free body: Frame

Note: Sum moments about H to determine
Iy.

Free body: AI


www.elsolucionario.org

1039. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.F3 (Continued)

Note: Sum forces in y to determine FAB
Sum moments about G to determine
FBC
Sum forces in x to determine
FFG


www.elsolucionario.org

1040. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.F4
Knowing that the pulley has a radius of 0.5 m, draw the
free-body diagram(s) needed to determine the components
of the reactions at A and E.

SOLUTION
Free body: Frame


Note: Sum moments about E to determine
Ay.
Sum forces in y to determine
Ey.
Sum forces in x to determine
Ax and Ex.

Free body: ABC


Note: Sum moments about C to determine
Ax.
Use previously obtained relation between
Ax and Ex to determine Ex.

www.elsolucionario.org

1041. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.F5
An 84-lb force is applied to the toggle vise at C. Knowing that  = 90,
draw the free-body diagram(s) needed to determine the vertical force
exerted on the block at D.

SOLUTION
We note that BD is a two-force member.
Free body: ABC


Note: Sum moments about A to determine
FBD.
Free body: D

Note: Sum forces in y to determine
Dy.
www.elsolucionario.org

1042. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.F6
For the system and loading shown, draw the free-body diagram(s)
needed to determine the force
P required for equilibrium.

SOLUTION
We note that AB and BD are two-force members.
Free body: CDE


Note: Sum moments about C to determine
FBD.


Free body: Pin B


Note: Sum forces in x to determine
P.

www.elsolucionario.org

1043. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 6.F7
A small barrel weighing 60 lb is lifted by a pair of tongs as shown.
Knowing that a = 5 in., draw the free-body diagram(s) needed to
determine the forces exerted at B and D on tong ABD.

SOLUTION
We note that BC is a two-force member.
Free body: ABD

Note: Sum moments about D to determine
FBC.
Sum forces in x to determine
Dx.
Sum forces in y to determine
Dy.

www.elsolucionario.org

1044. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 6.F8
The position of member ABC is controlled by the
hydraulic cylinder CD. Knowing that θ  30°,
draw the free-body diagram(s) needed to
determine the force exerted by the hydraulic
cylinder on pin C, and the reaction at B.

SOLUTION
We note that CD is a two-force member.
Free body: ABC



Note: To find , consider geometry of triangle BCD:

Law of cosines:
222
( ) (0.5) (1.5) 2(0.5)(1.5)cos60CD
  


1.32288 mCD
Law of sines:
sin sin 60
0.5 m 1.32288 m
q 



19.1066
 

www.elsolucionario.org

1045. CHAPTER 7
www.elsolucionario.org

1046. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.1
Determine the internal forces (axial force, shearing force,
and bending moment) at Point J of the structure indicated.
Frame and loading of Problem 6.76.

SOLUTION
From Problem 6.76: 720 lb
x
C

140 lb
y
C

FBD of JC:


0: 720 lb 0
x
FF  


720 lbF

720 lbF 


0: 140 lb 0
y
FV  

140 lbV 140.0 lbV 


0: (140 lb)(8 in.) 0
J
MM  


1120 lb in.M 

1120 lb in.M 

www.elsolucionario.org

1047. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.2
Determine the internal forces (axial force, shearing force, and bending
moment) at Point J of the structure indicated.
Frame and loading of Problem 6.78.

SOLUTION
From Problem 6.78: 30 lbC
FBD of AE:


0: 120 lb 0
x
FF  


120 lbF

120.0 lbF 


0: 30 lb 0
y
FV  

30 lbV 30.0 lbV 


0: (30 lb)(4 in.) 120 lb 2 in. 0
J
MM   


120.0 lb in.M 


120.0 lb in.M 

www.elsolucionario.org

1048. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.3
Determine the internal forces at Point J when  90°.

SOLUTION
Reactions(90)



FBD BJ:




0: 0
Ay
M  B

12
0: 780 N 0
13
y
FA

  




845 N 845 NA A
5
0: (845 N) 0
13
xx
FB  

325 N 325 N
xx
B  B



0: 125 N 0FF   125.0 NF 67.4°

0: 300 N 0FV   300 NV 22.6° 

0: (325 N)(0.480 m) 0MM  

156 N mM  156.0 N mM 
www.elsolucionario.org

1049. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.4
Determine the internal forces at Point J when  0.

SOLUTION
Reactions(0)




FBD BJ:



Alternate



0: (780 N)(0.720 m) (0.3 m) 0
Ay
MB  


1872 N
y
B

1872 N
y
B

12
0: 1872 N 0
13
y
FA

  




2028 NA
5
0: (2028 N) 780 N 0
13
xx
FB

   



1560 N
x
B 1560 N
x
B



0: 1728 N 600 N 0FF   


2328 NF 2330 NF 67.4°

0: 720 N 1440 N 0FV   

720 NV 720 NV 22.6° 

22
480 200 520 mmBJ

0: (1440 N)(0.520 m) (720 N)(0.520 m) 0
J
MM  

374.4 N mM  374 N mM 

Computation of M using
:
xy
BB
0: (1560 N)(0.48 m) (1872 N)(0.2 m) 0
J
MM  

374.4 N mM  374 N mM 
www.elsolucionario.org

1050. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.5
For the frame and loading shown, determine the internal forces at
the point indicated:
Point J.

SOLUTION
From free body diagram, all reactions = 0



250
0(Force Triangle)
34 5
VF
F  


150 NV 150.0 NV 53.1°

200 NF 200 NF 36.9°


0: (250 N)(0.080 m) 0
J
MM  


20 N mM 

20.0 N mM 

www.elsolucionario.org

1051. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.6
For the frame and loading shown, determine the internal
forces at the point indicated:
Point K.

SOLUTION
From free body diagram, all reaction = 0
Free body KC:

60 3
tan
80 4



34
sin ; cos
55




4
0: (250 N)cos 250 N
5
x
FF 

  




200 NF 200 NF 36.9°

3
0: (250 N)sin 250 N
5
y
FV 

  




150 NV 150.0 NV 53.1°


0: (250 N)(0.060 m) 0
K
MM  


15.00 N mM 

15.00 N mM 

www.elsolucionario.org

1052. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.7
An archer aiming at a target is pulling with a 45-lb force on the
bowstring. Assuming that the shape of the bow can be approximated
by a parabola, determine the internal forces at Point J.

SOLUTION
FBD Point A:
By symmetry
12
TT


112
3
0: 2 45 lb 0 37.5 lb
5
x
FT TT

    


Curve CJB is parabolic:
2
xay
FBD BJ:
At
:8in.Bx
2
32 in.
8in. 1
128 in.(32 in.)
y
a



2
128
y
x

2
Slope of parabola tan
128 64
dx y y
dy

At J:
116
tan 14.036
64
J






So
14
tan 14.036 39.094
3




0: (37.5 lb)cos(39.094 ) 0
x
FV
  

29.1 lbV 14.04°



www.elsolucionario.org

1053. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.7 (Continued)


0: (37.5 lb)sin (39.094 ) 0
y
FF
  


23.647F 23.6 lbF 76.0° 

34
0: (16 in.) (37.5 lb) [(8 2) in.] (37.5 lb) 0
55
J
MM
 
   
 
 

540 lb in.M 


www.elsolucionario.org

1054. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.8
For the bow of Problem 7.7, determine the magnitude and location of
the maximum (a) axial force, (b) shearing force, (c) bending moment.
PROBLEM 7.7 An archer aiming at a target is pulling with a 45-lb
force on the bowstring. Assuming that the shape of the bow can be
approximated by a parabola, determine the internal forces at Point J.


SOLUTION
Free body: Point A


3
0: 2 45 lb 0
5
x
FT

  



37.5 lbT 

Free body: Portion of bow BC
0: 30 lb 0
yC
FF   30 lb
C
F 

0: 22.5 lb 0
xC
FV   22.5 lb
C
V 

0: (22.5 lb)(32 in.) (30 lb)(8 in.) 0
CC
MM   

960 lb in.
C
M 
Equation of parabola

2
xky
At B:
2 1
8 (32)
128
kk


Therefore, equation is
2
128
y
x (1)
The slope at J is obtained by differentiating (1):

2
,tan
128 64
x
ydy dx y
d
dy
 (2) www.elsolucionario.org

1055. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.8 (Continued)

(a) Maximum axial force
0: (30 lb)cos (22.5 lb)sin 0
V
FF    



Free body: Portion bow CK
30cos 22.5sinF
F is largest at
(0)C

30.0 lb at
m
FC 
(b) Maximum shearing force
0: (30 lb)sin (22.5 lb)cos 0
V
FV    

30sin 22.5cosV
V is largest at B (and D)
Where
1
max1
tan 26.56
2


   



30sin 26.56 22.5cos 26.56
m
V  33.5 lb
m
V at B and D 
(c) Maximum bending moment
0: 960 lb in. (30 lb) (22.5 lb) 0
K
MM x y    

960 30 22.5Mxy
M is largest at C, where
0.xy 960 lb in. at
m
MC 

www.elsolucionario.org

1056. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.9
A semicircular rod is loaded as shown. Determine the internal forces at Point J.

SOLUTION
FBD Rod:
0: (2 ) 0
Bx
MAr 

0
x
A

0: (120 N)cos60 0
x
FV
  

60.0 NV 

FBD AJ:
0: (120 N)sin 60 0
y
FF
  

103.923 NF

103.9 NF 

0: [(0.180 m)sin 60 ](120 N) 0
J
MM   

18.7061M

18.71M 
www.elsolucionario.org

1057. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.10
A semicircular rod is loaded as shown. Determine the internal forces at Point K.

SOLUTION
FBD Rod:
0: 120 N 0 120 N
yy y
FB    B

0: 2 0 0
Axx
MrB   B

0: (120 N)cos30 0
x
FV
  

103.923 NV

103.9 NV 
FBD BK:
0: (120 N)sin 30 0
y
FF
  

60 NF

60.0 NF 

0: [(0.180 m)sin 30 ](120 N) 0
K
MM   

10.80 N mM 
www.elsolucionario.org

1058. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.11
A semicircular rod is loaded as shown. Determine the internal
forces at Point Jknowing that
 30.

SOLUTION
FBD AB:
43
0: 2 (280 N) 0
55
A
MrCrCr

   



400 NC

4
0: (400 N) 0
5
xx
FA   

320 N
x
A

3
0: (400 N) 280 N 0
5
yy
FA   
FBD AJ:
40.0 N
y
A


0: (320 N)sin 30 (40.0 N)cos30 0
x
FF
   

194.641 NF

194.6 NF 60.0

0: (320 N)cos30 (40 N)sin30 0
y
FV
   

257.13 NV

257 NV 30.0
0
0: (0.160 m)(194.641 N) (0.160 m)(40.0 N) 0MM  

24.743M

24.7 N mM 
www.elsolucionario.org

1059. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.12
A semicircular rod is loaded as shown. Determine the
magnitude and location of the maximum bending moment in
the rod.


SOLUTION














Free body: Rod ACB

43
0: (0.16 m) (0.16 m)
55
(280 N)(0.32 m) 0
ACD CD
MF F
 
 
 
 


400 N
CD
F 

4
0: (400 N) 0
5
xx
FA  

320 N
x
A 320 N
x
A 
+
3
0: (400 N) 280 N 0
5
yy
FA   

40.0 N
y
A 40.0 N
y
A 
Free body: (For 90°)AJ 

0: (320 N)(0.16 m)sin (40.0 N)(0.16 m)(1 cos ) 0
J
MM    

51.2sin 6.4cos 6.4M (1)
For maximum value between A and C:

0: 51.2cos 6.4sin 0
dM
d 


51.2
tan 8
6.4
 82.87
Carrying into (1):

51.2sin82.87 6.4cos82.87 6.4 45.20 N mM

www.elsolucionario.org

1060. Copyright © McGraw-Hill Education. Permission required for reproduction or display.






PROBLEM 7.12 (Continued)

Free body: (For 90°)BJ 

0: (280 N)(0.16 m)(1 cos ) 0
J
MM    

(44.8 N m)(1 cos )M 
Largest value occurs for
90 , that is, at C, and is

44.8 N m
C
M 
We conclude that

max
45.2 N mM  for 82.9 
www.elsolucionario.org

1061. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.13
The axis of the curved member AB is a parabola with vertex at A. If a
verticalload
P of magnitude 450 lb is applied at A, determine the
internal forces at J when h 12 in., L 40 in., and a 24 in.


SOLUTION
Free body AB
0: 450 lb 0
yy
FB   
450 lb
y
B

0: (12 in.) (450 lb)(40 in.) 0
Ax
MB  

1500 lb
x
B

0: 1500 lb
x
F  A


Parabola:
2
ykx
At B:
2
12 in. (40 in.) 0.0075kk
Equation of parabola:
2
0.0075yx

slope 0.015
dy
x
dx

At J:
2
24 in. 0.0075(24) 4.32 in.
JJ
xy

slope 0.015(24) 0.36, tan 0.36, 19.8 

Free body AJ


0: (450 lb)(24 in.) (1500 lb)(4.32 in.) 0
J
MM  

4320 lb in.M 4320 lb in.M 

www.elsolucionario.org

1062. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.13 (Continued)


0: (450 lb)sin19.8 (1500 lb)cos19.8 0FF   

1563.8 lbF 1564 lbF 19.8° 


0: (450 lb)cos19.8 (1500 lb)sin19.8 0FV   

84.71lbV 84.7 lbV 70.2° 
www.elsolucionario.org

1063. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.14
Knowing that the axis of the curved member AB is a parabola with
vertex at A, determine the magnitude and location of the maximum
bending moment.

SOLUTION
Parabola
2
ykx
At B:
2
hkL

2
/khL
Equation of parabola
22
/
yhx L


0: ( ) ( ) 0
B
MPLAh   /
PLhA
Free body AJ At J:
22
/
JJ
xayhaL


22
0: ( / )( / ) 0
Ja
MPPLhhaLM  

2
a
MPa
L

 


For maximum:
2
10
dM a
P
da L






max
1
occurs at:
2
M aL 


2
max
(/2)
24
LLPL
MP
L




max
1
||
4
M PL
www.elsolucionario.org

1064. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.15
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point J of
the frame shown.


SOLUTION






FBD Frame with pulley and cord:


0: (1.8 m) (2.6 m)(360 N)
(0.2 m)(360 N) 0
Ax
MB 


560 N
x
B

FBD BE:
Note: Cord forces have been moved to pulley hub as per Problem 6.91.

0: (1.4 m)(360 N) (1.8 m)(560 N)
(2.4m) 0
E
y
M
B
 


630 N
y
B
FBD BJ:

3
0: 360 N (630 N 360 N)
5
4
(560 N) 0
5
x
FF
   


250 NF 36.9° 

43
0: (630 N 360 N) (560 N) 0
55
y
FV
    

120.0 NV 53.1° 

0: (0.6m)(360N) (1.2m)(560N)
(1.6 m)(630 N) 0
J
MM  


120.0 N mM 
www.elsolucionario.org

1065. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.16
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point K of
the frame shown.


SOLUTION

















Free body: frame and pulleys
0: (1.8 m) (360 N)(0.2 m)
(360 N)(2.6 m) 0
Ax
MB 


560 N
x
B 560 N
x
B 

0: 560 N 360 N 0
xx
FA   


920 N
x
A 920 N
x
A 

0: 360 N 0
yyy
FAB   


360 N
yy
AB (1)
Free body: member AE
We recall that the forces applied to a pulley may be applied
directly to the axle of the pulley.

0: (2.4 m) (360 N)(1.8 m) 0
Ey
MA  

270 N
y
A 270 N
y
A 
From (1): 360 N 270 N
y
B

630 N
y
B 630 N
y
B 





www.elsolucionario.org

1066. Copyright © McGraw-Hill Education. Permission required for reproduction or display.






PROBLEM 7.16 (Continued)

Free body: AK

0: 920N 360N 0
x
FF  


560 NF 560 NF 

0: 360 N 270 N 0
y
FV  


90.0 NV 90.0 NV 

0: (270 N)(1.6 m) (360 N)(1 m) 0
K
MM  

72.0 N mM  72.0 N mM 
www.elsolucionario.org

1067. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.17
A 5-in.-diameter pipe is supported every 9 ft by a small frame
consisting of two members as shown. Knowing that the combined
weight of the pipe and its contents is 10 lb/ft and neglecting the
effect of friction, determine the magnitude and location of the
maximum bending moment in member AC.

SOLUTION



















Free body: 10-ft section of pipe

4
0: (90 lb) 0
5
x
FD   72 lbD 

3
0: (90 lb) 0
5
y
FE   54 lbE 
Free body: Frame
0: (18.75 in.) (72 lb)(2.5 in.)
(54 lb)(8.75 in.) 0
By
MA  


34.8 lb
y
A 34.8 lb
y
A 

43
0: 34.8 lb (72 lb) (54 lb) 0
55
yy
FB    

55.2 lb
y
B 55.2 lb
y
B 

34
0: (72 lb) (54 lb) 0
55
xxx
FAB    


0
xx
AB (1)
Free body: Member AC
0: (72 lb)(2.5 in.) (34.8 lb)(12 in.) (9 in.) 0
Cx
MA   

26.4 lb
x
A 26.4 lb
x
A 
From (1):
26.4 lb
xx
BA 

26.4 lb
x
B 





www.elsolucionario.org

1068. Copyright © McGraw-Hill Education. Permission required for reproduction or display.






PROBLEM 7.17 (Continued)

Free body: Portion AJ
For 12.5 in. ( ) :xAJAD

34
0: (26.4 lb) (34.8 lb) 0
55
J
MxxM  

max
12
150 lb in. for 12.5 in.
Mx
Mx

 


max
150.0 lb in. atMD 
For 12.5 in.( ):xAJAD

34
0: (26.4 lb) (34.8 lb) (72 lb)( 12.5) 0
55
J
MxxxM    


max
900 60
150 lb in. for 12.5 in.
Mx
Mx

 

Thus:
max
150.0 lb in.M  at D 
www.elsolucionario.org

1069. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.18
For the frame of Problem 7.17, determine the magnitude and
location of the maximum bending moment in member BC.
PROBLEM 7.17 A 5-in.-diameter pipe is supported every 9 ft by
a small frame consisting of two members as shown. Knowing that
the combined weight of the pipe and its contents is 10 lb/ft and
neglecting the effect of friction, determine the magnitude and
location of the maximum bending moment in member AC.

SOLUTION

















Free body: 10-ft section of pipe

4
0: (90 lb) 0
5
x
FD   72 lbD 

3
0: (90 lb) 0
5
y
FE   54 lbE 
Free body: Frame
0: (18.75 in.) (72 lb)(2.5 in.)
(54 lb)(8.75 in.) 0
By
MA  


34.8 lb
y
A 34.8 lb
y
A 

43
0: 34.8 lb (72 lb) (54 lb) 0
55
yy
FB    

55.2 lb
y
B 55.2 lb
y
B 

34
0: (72 lb) (54 lb) 0
55
xxx
FAB    


0
xx
AB (1)
Free body: Member AC
0: (72 lb)(2.5 in.) (34.8 lb)(12 in.)
(9 in.) 0
C
x
M
A
 


26.4 lb
x
A 26.4 lb
x
A 
From (1):
26.4 lb
xx
BA 

26.4 lb
x
B 


www.elsolucionario.org

1070. Copyright © McGraw-Hill Education. Permission required for reproduction or display.






PROBLEM 7.18 (Continued)

Free body: Portion BK
For 8.75 in.( ):xBKBE

34
0: (55.2 lb) (26.4 lb) 0
55
K
MxxM  

max
12
105.0 lb in. for 8.75 in.
Mx
Mx

 

max
105.0 lb in. atME 
For 8.75 in.( ):xBKBE

34
0: (55.2 lb) (26.4 lb) (54 lb)( 8.75 in.) 0
55
K
MxxxM    


max
472.5 42
105.0 lb in. for 8.75 in.
Mx
Mx

 

Thus
max
105.0 lb in.M  at E

www.elsolucionario.org

1071. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.19
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point J of
the frame shown.

SOLUTION
Free body: Frame and pulleys
0: (1.8 m) (360 N)(2.6 m) 0
Ax
MB   

520 N
x
B 520 N
x
B 

0: 520 N 0
xx
FA  

520 N
x
A 520 N
x
A 

0: 360 N 0
yyy
FAB   

360 N
yy
AB (1)
Free body: Member AE
0: (2.4 m) (360 N)(1.6 m) 0
Ey
MA   

240 N
y
A 240 N
y
A 

From (1):
360 N 240 N
y
B

600 N
y
B 600 N
y
B 
Free body: BJ
We recall that the forces applied to a pulley may be applied directly to its axle.

34
0: (600 N) (520 N)
55
3
360 N (360 N) 0
5
y
F
F
 
 

200 NF 200 NF 36.9° 

www.elsolucionario.org

1072. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.19 (Continued)


434
0: (600 N) (520 N) (360 N) 0
555
x
FV   

120.0 NV 120.0 NV 53.1° 

0: (520 N)(1.2 m) (600 N)(1.6 m) (360 N)(0.6 m) 0
J
MM   

120.0 N mM  120.0 N mM 

www.elsolucionario.org

1073. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.20
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point K of
the frame shown.

SOLUTION
Free body: Frame and pulleys
0: (1.8 m) (360 N)(2.6 m) 0
Ax
MB   

520 N
x
B 520 N
x
B 

0: 520 N 0
xx
FA  

520 N
x
A 520 N
x
A 

0: 360 N 0
yyy
FAB   

360 N
yy
AB (1)
Free body: Member AE
0: (2.4 m) (360 N)(1.6 m) 0
Ey
MA   

240 N
y
A 240 N
y
A 

From (1):
360 N 240 N
y
B

600 N
y
B 600 N
y
B 
Free body: AK
0: 520 N 0
x
FF 

520 NF 520 NF 

0: 360 N 240 N 0
y
FV  

120.0 NV 120.0 NV 

0: (240 N)(1.6 m) (360 N)(0.8 m) 0
K
MM  

96.0 N mM  96.0 N mM 
www.elsolucionario.org

1074. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.21
A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.


SOLUTION
(a) FBD Rod:


FBD AJ:


(b)
FBD Rod:






0: 0
xx
FA 

0: 2 0
2
Dyy
P
MaPaA A   





0: 0
x
F  V 

0: 0
2
y
P
FF 

2
P

F 

0: 0
J
M  M 



43
0: 2 2 0
55
A
MaDaDaP
 
  
 
 
5
14
P
D


45
0: 0
514
xx
FA P  
2
7
x
P
A


35
0: 0
514
yy
FAPP  

11
14
y
P
A


www.elsolucionario.org

1075. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



FBD AJ:



(c) FBD Rod:


FBD AJ:

PROBLEM 7.21 (Continued)


0:
x
F
2
0
7
PV

2
7
P

V 

0:
y
F

11
0
14
P
F

11
14
P

F 

2
0: 0
7
J
P
MaM 

2
7
aP
M 



0:
A
M

4
0
25
aD
aP





5
2
P
D


0:
x
F
45
0
52
x
P
A
2
x
AP

35
0: 0
52
yy
P
FAP  

5
2
y
P
A




0:
x
F 20PV 2PV 

0:
y
F

5
0
2
P
F

5
2
P

F 

0: (2 ) 0
J
MaPM  2aPM  www.elsolucionario.org

1076. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.22
A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.


SOLUTION
(a) FBD Rod: 0: 2 0
D
MaPaA  

2
P

A

0: 0
2
x
P
FV 

2
P

V 




FBD AJ: 0:
y
F 0F


0: 0
2
J
P
MMa  

2
aP

M 


(b)
FBD Rod:

4
0: 0
25
D
a
MaPA

  




5
2
P

A




www.elsolucionario.org

1077. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.22 (Continued)



FBD AJ:
35
0: 0
52
x
P
FV 

3
2
P

V 


45
0: 0
52
y
P
FF 
2PF

3
2
aP
M 
(c) FBD Rod:

34
0: 2 2 0
55
D
MaPaAaA
 
   
 
 

5
14
P
A


35
0: 0
514
x
P
FV

  



3
14
P

V 

45
0: 0
514
y
P
FF 

2
7
P

F 

35
0: 0
514
J
P
MMa

  



3
14
aPM

www.elsolucionario.org

1078. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.23
A quarter-circular rod of weight W and uniform cross section is supported as
shown. Determine the bending moment at Point J when
 30.

SOLUTION
FBD Rod:

0: 0
xx
F  A

22
0: 0
Byy
rW
MWrA

   A





FBD AJ:
15 ,
30
weight of segment
90 3
W
W





12
sin sin15 0.9886
rr
rr




2
0: cos30 cos30 0
3
y
WW
FF

    

32 1
23
W





F


0
2
cos15 0
3
WW
MMrF r


   


0.0557WrM


www.elsolucionario.org

1079. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.24
For the rod of Problem 7.23, determine the magnitude and location of the
maximum bending moment.
PROBLEM 7.23 A quarter-circular rod of weight W and uniform cross section
is supported as shown. Determine the bending moment at Point J when
 30.

SOLUTION
FBD Rod: 0: 0
xx
FA 


22
0: 0
Byy
rW
MWrAA

  

,sin
2
r
r




Weight of segment
2
24
WW





42
0: cos2 cos2 0
x
W
FFW


   


22
(1 2 ) cos 2 (1 ) cos
WW
F
 
 
FBD AJ:
0
24
0: ( cos ) 0
W
MMFrr W 


   




24
(1 cos cos ) sin cos
WWr
Mr 
  

  
But,
11
sin cos sin 2 sin
22
  
so
2
(1 cos cos sin )
r
MW
  


2
(sin sin cos cos ) 0
dM rW
d
   

for
(1 ) sin 0

0for 0,1, ( 1,2,)
dM
nn
d 
 
Only 0 and 1 in valid range
At
00,at1radM 
at
57.3
max
0.1009MM Wr 

www.elsolucionario.org

1080. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.25
A semicircular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when
 60.

SOLUTION
FBD Rod:

2
0: 2 0
A
r
MWrB

  

W

B
0: sin 60 cos 60 0
3
y
WW
FF

   

0.12952FW
FBD BJ:


0
3
0: 0
23
WrW
MrF M


   




11
0.12952 0.28868
2
MWr Wr






On BJ
0.289
J
WrM



www.elsolucionario.org

1081. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.26
A semicircular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when
 150.

SOLUTION
FBD Rod:
0: 0
yy y
FAW W   A

2
0: 2 0
Bx
r
MWrA

  

x
W

A


FBD AJ:


5
0: cos 30 sin 30 0 0.69233
6
x
WW
FFFW

     

0
0: 0.25587 0
6
WW
MrrFM

  
   
  
  


0.25587 1
0.69233
6
MWr







(0.4166)MWr
On AJ
0.417WrM



www.elsolucionario.org

1082. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.27
A half section of pipe rests on a frictionless horizontal surface as shown. If
the half section of pipe has a mass of 9 kg and a diameter of 300 mm,
determine the bending moment at Point J when
 90°.

SOLUTION
For half section 9kgm

(9)(9.81) 88.29 NWmg 
Portion JC:

1
Weight 44.145 N
2
W

From Fig. 5.8B:

22(150)
95.49 mm
r
x
x

 


0: (44.145 N)(0.15 m) (44.145 N)(0.09549 m) 0
J
MM  

2.406 N mM
 2.41 N mM


www.elsolucionario.org

1083. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.28
A half section of pipe rests on a frictionless horizontal surface as shown. If
the half section of pipe has a mass of 9 kg and a diameter of 300 mm,
determine the bending moment at point J when
 90°.

SOLUTION
For half section 9kgm

2
(9 kg)(9.81 m/s ) 88.29 NWmg 
Free body JC Weight of portion JC,
1
44.145 N
2
W

150 mmr
From Fig. 5.8B:

2 2(150)
95.49 mm
r
x

 

0: (44.145 N)(0.09549 m) 0
J
MM  

4.2154 N mM
 4.22 N mM



www.elsolucionario.org

1084. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.29
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION
FBD beam:
(a) By symmetry:
1
()
22
y
L
AD w
4
y
wL
AD


Along AB:


0: 0
44
y
wL wL
FVV  


0: 0
4
(straight)
4
J
wL
MMx
wL
Mx
  


Along BC:

1
0: 0
4
y
wL
FwxV  


1
4
wL
Vwx

Straight with

1
0at
4
L
Vx

1
11
0: 0
244
k
x LwL
MMwx x

   




2
2
11
28 2
wL L
Mxx





Parabola with
2
13
at
32 4
L
MwLx

Section CD by symmetry
(b) From diagrams:
max
| | on and
4
wL
VABCD 

2
max
3
| | at center
32
wL
M 
 www.elsolucionario.org

1085. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.30
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION




2
00
11 1
22 2
xx
wx w x w
LL




By similar D’s







2
0
1
0: 0
2
y
x
FwV
L
  
2
0
1
2
x
Vw
L


2
0
1
0: 0
23
y
xx
FwM
L

  



3
0
1
6
x
Mw
L




max 0
1
||
2
VwL


2
max 01
||
6
MwL


www.elsolucionario.org

1086. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.31
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
(a) Reactions:

2
3
P
A

3
P
C

From A to B:

2
0:
3
y
P
FV 


1
2
0:
3
P
MM x 

From B to C:
0:
3
y
P
FV 


2
0: ( )
3
P
MMLx  

(b)
max
2
|| ;
3
P
V 
max
2
||
9
PL
M 

www.elsolucionario.org

1087. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.32
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION
(a)

From A to B:
0:
y
FVP 

1
0:MMPx 
From B to C:
0: 0 2
y
FPPVVP  

2
0: ( ) 0 2M Px P x a M M Px Pa    


(b)
max
|| 2;VP
max
|| 3MPa  www.elsolucionario.org

1088. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.33
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
(a) FBD Beam:
0
0: 0
Cy
MLAM 

0
y
M
L
A


0: 0
yy
FAC  

0
M
L
C

Along AB:

0
0
0: 0
y
M
FV
L
M
V
L
  



0
0
0: 0
J
M
MxM
L
M
Mx
L
 


Straight with
0
at
2
M
MB
Along BC:

00
0: 0
y
MM
FVV
LL
   


0
00
0: 0 1
K
M x
MMxM MM
LL

    



Straight with
0
at 0 at
2
M
MBMC
(b) From diagrams:
max 0
|| /VML 

0
max
|| at
2
M
MB 


www.elsolucionario.org

1089. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.34
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION
Free body: Portion AJ
0: 0
y
FPV 

VP 


0: 0 
Jx
MMPPL

()MPLx


(a) The V and M diagrams are obtained by plotting the functions V and M.

(b)
max
||VP 


max
||MPL 


www.elsolucionario.org

1090. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.35
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
(a)

Along AC:



0: 15 kN 0 15 kN
y
FVV   
 0: 15 kN 0 15 kN
J
MMx Mx   
15 kN m at MC 

Along CD:


0: 15 kN 25 kN 0 40 kN
y
FVV    
0: 1 m 25 kN 15 kN 0
K
MMx x   
25 kN m 40 kNMx  
27 kN m at 1.3 mMDx  
Along DE:




0: 15 kN 25 kN 30 kN 0
y
FV    
10 kNV www.elsolucionario.org

1091. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.35 (Continued)


 
11
0: 30 kN 0.3 m 25 kN
L
MMx x   


1
1.3 m 15 kN 0x 

1
27 kN m 10 kNMx  

1
31.0 kN m at 0.4 mMEx  
Along EB:


0: 15 kN 25 kN 30 kN 20 kN 0 30 kN
y
FVV     

 
22
0: 20 kN 0.7 m 25 kN
N
MMx x   


22
1.7 m 15 kN 0.4 m 30 kN 0xx   

2
31 kN m 30 kNMx  

2
55 kN m at 0.8 mMBx  
From diagrams:
max
40.0 kN on VCD 

max
55.0 kN m at MB 


www.elsolucionario.org

1092. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.36
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.

SOLUTION
Free body: Entire beam


0: (3.2 m) (40 kN)(0.6 m) (32 kN)(1.5 m) (16 kN)(3 m) 0
A
MB    


37.5 kNB

37.5 kNB 
0: 0
xx
FA 

0: 37.5 kN 40 kN 32 kN 16 kN 0
yy
FA     

50.5 kN
y
A 50.5 kNA 
(a) Shear and bending moment

Just to the right of A:
1
50.5 kNV
1
0M Just to the right of C:

2
0: 50.5 kN 40 kN 0
y
FV  
2
10.5 kNV 

22
0: (50.5 kN)(0.6 m) 0MM  
2
30.3 kN mM  

Just to the right of D:

3
0: 50.5 40 32 0
y
FV 
3
21.5 kNV 

33
0: (50.5)(1.5) (40)(0.9) 0MM   
3
39.8 kN mM  




www.elsolucionario.org

1093. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.36 (Continued)

Just to the right of E:

4
0: 37.5 0
y
FV  
4
37.5 kNV 

44
0: (37.5)(0.2) 0MM   
4
7.50 kN mM  
At B:
0
BB
VM 

(b)
max
|| 50.5kNV 










max
| | 39.8 kN mM  

www.elsolucionario.org

1094. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.37
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION
Free body: Entire beam


0: (6 ft) (6 kips)(2 ft) (12 kips)(4 ft) (4.5 kips)(8 ft) 0
A
ME    


16 kipsE

16 kipsE 
0: 0
xx
FA 

0: 16 kips 6 kips 12 kips 4.5 kips 0
yy
FA     

6.50 kips
y
A 6.50 kipsA 
(a) Shear and bending moment
Just to the right of A:

11
6.50 kips 0VM  

Just to the right of C:

2
0: 6.50 kips 6 kips 0
y
FV  
2
0.50 kipsV 

22
0: (6.50 kips)(2 ft) 0MM  
2
13 kip ftM 

Just to the right of D:

3
0: 6.50 6 12 0
y
FV 
3
11.5 kipsV 

33
0: (6.50)(4) (6)(2) 0MM   
3
14 kip ftM 



www.elsolucionario.org

1095. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.37 (Continued)

Just to the right of E:


4
0: 4.5 0
y
FV  
4
4.5 kipsV 

44
0: (4.5)2 0MM   
4
9kip ftM 
At B:
0
BB
VM 

(b)
max
|| 11.50kipsV 













max
| | 14.00 kip ftM  
www.elsolucionario.org

1096. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.38
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION
Free body: Entire beam


0: (120 lb)(10 in.) (300 lb)(25 in.) (45 in.) (120 lb)(60 in.) 0
C
ME    


300 lbE

300 lbE 
0: 0
xx
FC 

0: 300 lb 120 lb 300 lb 120 lb 0
yy
FC     

240 lb
y
C 240 lbC 
(a) Shear and bending moment
Just to the right of A:

1
0: 120 lb 0
y
FV  
11
120 lb, 0VM  

Just to the right of C:

2
0: 240 lb 120 lb 0
y
FV  
2
120 lbV

2
0: (120 lb)(10 in.) 0
C
MM  
2
1200 lb in.M  

Just to the right of D:

3
0: 240 120 300 0
y
FV   
3
180 lbV

33
0: (120)(35) (240)(25) 0,MM   
3
1800 lb in.M  


www.elsolucionario.org

1097. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.38 (Continued)

Just to the right of E:

4
0: 120 lb 0
y
FV   
4
120 lbV

44
0: (120 lb)(15 in.) 0MM   
4
1800 lb in.M  
At B:
0
BB
VM 

(b)
max
| | 180.0 lbV 











max
| | 1800 lb in.M  
www.elsolucionario.org

1098. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.39
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
Free body: Entire beam


0: (5 m) (60 kN)(2 m) (50 kN)(4 m) 0
A
MB   


64.0 kNB

64.0 kNB 
0: 0
xx
FA 

0: 64.0 kN 6.0 kN 50 kN 0
yy
FA    

46.0 kN
y
A 46.0 kNA 
(a) Shear and bending-moment diagrams.
From A to C:
0: 46 0
y
FV  46 kNV

0: 46 0
y
MMx  

(46 )kN mMx 

From C to D:
0: 46 60 0
y
FV  14 kNV

0: 46 60( 2) 0
j
MMxx   


(120 14 )kN mMx 
For
2m: 92.0kN m
C
xM 
For 3m: 78.0kN m
D
xM 




www.elsolucionario.org

1099. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.39 (Continued)

From D to B:
0: 64 25 0
y
FV   



(25 64)kNV

0: 64 (25 ) 0
2
j
MM


  




2
(64 12.5 )kN mM 

For 0: 64 kN
B
V 0
B
M 

(b)
max
| | 64.0 kNV  











max
| | 92.0 kN mM  



www.elsolucionario.org

1100. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.40
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
Free body: Entire beam


0: (50 kN)(2 m) (4 m) (40 kN)(5 m) 0
A
MD    

75 kND 
0: 75 50 40 0
y
FA 

15 kNA 15 kNA
From A to C:
0: 15 0
y
FV  15 kNV

1
0: (15)( ) 0MMx  

(15 ) kN mMx


From C to D:
0: 15 50 0
y
FV  35 kNV


2
0: (15)( ) 50( 2) 0MMxx   


(100 35 )kN mMx 

30 kN m at 2 m
40 kN m at 4 m
C
D
Mx
Mx
  
  


www.elsolucionario.org

1101. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.40 (Continued)

From D to B:
0: 20 0
y
FVx   20 kNVx

3
0: (20 ) 0
2
x
MMx

  



2
10 kN mMx 


40 kN, 40 kN m at 2 m
0 at 0
DD
BB
VM x
VM x

 

(b)
max
| | 40.0 kNV  



max
| | 40.0 kN mM  



www.elsolucionario.org

1102. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.41
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
(a) By symmetry:

1
8 kips (4 kips)(5 ft) 18 kips
2
yy
AB   AB


Along AC:


0 : 18 kips 0 18 kips
y
FVV  


0 : (18 kips) (18 kips)
J
MMx M x  


36 kip ft at ( 2 ft)MCx 

Along CD:


1
0: 18 kips 8 kips (4 kips/ft) 0
y
FxV   


1
10 kips (4 kips/ft)Vx


1
0 at 2.5 ft (at center)Vx


1
111
0: (4 kips/ft) (8 kips) (2 ft )(18 kips) 0
2
K
x
MM x x x     


2
11
36 kip ft (10 kips/ft) (2 kips/ft)Mxx 


1
48.5 kip ft at 2.5 ftMx

Complete diagram by symmetry
(b) From diagrams:
max
|| 18.00kipsV 


max
| | 48.5 kip ftM  

www.elsolucionario.org

1103. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.42
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
Free body: Entire beam





0: (10ft) (15kips)(3ft) (12kips)(6ft) 0
A
MB   


11.70 kipsB

11.70 kipsB 

0: 0
xx
FA 

0: 15 12 11.70 0
yy
FA  


15.30 kips
y
A

15.30 kipsA 
(a) Shear and bending-moment diagrams
From A to C:

0: 15.30 2.5 0
y
FxV  

(15.30 2.5 ) kipsVx

0: (2.5 ) 15.30 0
2
J
x
MMx x

   




2
15.30 1.25Mxx


For
0:x

15.30 kips
A
V

0
A
M 

For
6ft:x 0.300 kip
C
V 46.8 kip ft
C
M  





www.elsolucionario.org

1104. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.42 (Continued)


From C to B:

0: 11.70 0
y
FV  

11.70 kipsV 


0: 11.70 0
J
MM  


(11.70 ) kip ftM

For 4ft:
46.8 kip ft
C
M 

For
0: 0
B
M 
( b)
max
|| 15.30kipsV 








max
| | 46.8 kip ftM  



www.elsolucionario.org

1105. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.43
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed and knowing that P wa, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values
of the shear and bending moment.

SOLUTION




















Free body: Entire beam
0: (4 ) 2 0
yg
Fwawawa

3
4
g
ww 
(a) Shear and bending-moment diagrams
From A to C:


3
0: 0
4
1
4
y
FwxwxV
Vwx
  


3
0: ( ) 0
24 2
J
xx
M M wx wx

   




21
8
Mwx

For
0:x 0
AA
VM 
For
:xa
1
4
C
Vwa
21
8
C
Mwa 
From C to D:


3
0: 0
4
y
FwxwaV  

3
4
Vxaw






3
0: 0
24 2
J
ax
MMwax wx
 
   
 
 


23
82
a
Mwxwax




(1)


www.elsolucionario.org

1106. Copyright © McGraw-Hill Education. Permission required for reproduction or display.










PROBLEM 7.43 (Continued)

For
:xa
1
4
C
Vwa
21
8
C
Mwa 
For
2:xa
1
2
D
Vwa 0
D
M 
Because of the symmetry of the loading, we can deduce the values of V
and M for the right-hand half of the beam from the values obtained for
its left-hand half.
(b)
max
1
||
2
Vwa


To find
max
|| ,M we differentiate Eq. (1) and set
0:
dM
dx


2 2
2
34
0,
43
34 41
83 32 6
dM
wx wa x a
dx
wa
Mwawa
 
  

  
  


2
max1
||
6
Mwa

Bending-moment diagram consists of four distinct arcs of
parabola.

www.elsolucionario.org

1107. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.44
Solve Problem 7.43 knowing that P3wa.
PROBLEM 7.43 Assuming the upward reaction of the ground on beam
AB to be uniformly distributed and knowing that P wa, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.

SOLUTION














Free body: Entire beam

0: (4 ) 2 3 0
yg
Fwawawa   

5
4
g
ww 
(a) Shear and bending-moment diagrams
From A to C:


5
0: 0
4
1
4
y
FwxwxV
Vwx
  


5
0: ( ) 0
24 2
J
xx
MMwxwx

   




21
8
Mwx

For
0:x 0
AA
VM 
For
:xa
1
4
C
Vwa
21
8
C
Mwa 
From C to D:


5
0: 0
4
y
FwxwaV  

5
4
Vxaw






5
0: 0
24 2
J
ax
MMwax wx
 
   
 
 


25
82
a
Mwxwax




(1)
www.elsolucionario.org

1108. Copyright © McGraw-Hill Education. Permission required for reproduction or display.










PROBLEM 7.44 (Continued)

For
:xa
1
,
4
C
Vwa
21
8
C
Mwa 
For
2:xa
3
,
2
D
Vwa
2
D
Mwa 
Because of the symmetry of the loading, we can deduce the
values of V and M for the right-hand half of the beam from the
values obtained for its left-hand half.
(b)
max
3
||
2
Vwa

To find
max
|| ,M we differentiate Eq. (1) and set
0:
dM
dx


5
0
4
4
(outside portion )
5
dM
wx wa
dx
xaa CD



The maximum value of
||M occurs at D:

2
max
||Mwa 
Bending-moment diagram consists of four distinct arcs of
parabola.


www.elsolucionario.org

1109. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.45
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment diagrams,
(b) determine the maximum absolute values of the shear and bending
moment.

SOLUTION
(a) For equilibrium, 0: 6 12 6 (12 ft)=0
y
Fw   


2.0 kips/ftw

Along AD:

0: (2.0 kips/ft) 0
2
y
FxV
Vx
 




1
2
0: 2 0
2
x
MMx
Mx
  


Along DC:


0: 2 6 0
y
FxV 


26Vx



2
0: 6( 2) 2 0
2
x
MMx x   


2
612Mx x




www.elsolucionario.org

1110. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.45 (Continued)

Complete diagrams using symmetry.
(b)
max
| | 6.00 kipsV  

max
| | 12.00 kip ftM  

www.elsolucionario.org

1111. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.46
Solve Prob. 7.45 assuming that the 12-kip load has been
removed.
PROBLEM 7.45 Assuming the upward reaction of the ground on beam
AB to be uniformly distributed, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION
For equilibrium,

0: 12 6 6 0
yg
Fw 

1kip/ft
g
w 

(a) Shear and bending moment
From A to D:

0: 0
y
FxVVx  


2
1
0: ( ) 0,
22
xx
MMx M   



From D to C: 0: 6 0
y
FxV 


6Vx

2
0: 6( 2) ( ) 0
2
x
MMxx   


2
12 6
2
x
Mx





www.elsolucionario.org

1112. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.46 (Continued)


(b)
max
| | 4.00 kipsV  

max
| | 6.00 kip ftM  
www.elsolucionario.org

1113. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.47
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment
diagrams,(b) determine the maximum absolute values of the shear and
bending moment.

SOLUTION








For equilibrium,

0: (6 m) (8 kN/m)(3 m) 0
yg
Fw  

4 kN/m
g
w 
(a) Shear and bending-moment diagrams from A to C:
0: 4 0 (4 ) kN
y
FxVVx  
0: (4 )
2
J
x
MMx 

2
(2 ) kN mMx
For
0:x 0
AA
VM 
For
1.5 m: 6 kN,
C
xV 4.5 kN m
C
M 
From C to D:

0: 4 8( 1.5) 0
y
FxxV   

(12 4 ) kNVx

1.5
0: 8( 1.5) (4 ) 0
22
J
xx
MMx x

   


22
[2 4( 1.5) ]kN mMx x 

2
912 2Mxx  

For
3 m: 0,xV cL
9 kN mM 
For
4.5 m: 6 kN,
D
xV 4.5 kN m
D
M 
At B:
0
BB
VM 
(b)
max
| | 6.00 kNV  

max
| | 9.00 kN mM  
Bending-moment diagram consists or three distinct arcs of
parabola, all located above the x axis.
Thus:
0 everywhereM 
www.elsolucionario.org

1114. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.48
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment diagrams,
(b) determine the maximum absolute values of the shear and bending
moment.

SOLUTION
For equilibrium,



0: (6 m) (8 kN/m)(3 m) 0
yg
Fw  

4kN/m
g
w 

(a) Shear and bending-moment diagrams.
From A to C:

0: 8 4 0
(4)kN
y
FxxV
Vx
  



1
2
0: (8 ) (4 ) 0
22
(2 )kNm
xx
MMxx
Mx
   
 

For
0:x 0
AA
VM 
For
1.5 m:x

6kN
C
V 4.5 kN m
C
M  
From A to D:



0:412 0, (412)kN
y
FxVVx   


2
0: 12( 0.75) (4 ) 0
2
x
MMx x    


2
2129Mx x

www.elsolucionario.org

1115. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.48 (Continued)


For
3m:x

0,V

9.00 kN mM  


For
4.5 m:x

6 kN,
D
V

4.50 kN m
D
M  





At B:
0
BB
VM 


( b)
max
| | 6.00 kNV 



Bending-moment diagram consists of three distinct arcs of parabola.


max
| | 9.00 kN mM  


Since entire diagram is below the x axis:
0M everywhere 
www.elsolucionario.org

1116. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.49
Draw the shear and bending-moment diagrams for the beam AB, and
determine the maximum absolute values of the shear and bending moment.

SOLUTION
Reactions:
0: (0.4) (120)(0.2) 0
Ay
MB  


60 N
y
B


0:
x
F

0
x
B

0:
y
F

180 NA

Equivalent loading on straight part of beam AB.

From A to C:

0: 180 N
y
FV 


1
0: 180MM x 
From C to B: 0: 180 120 0
y
FV  

60 NV

0: (180 N)( ) 24 N m (120 N)( 0.2) 0
x
Mx xM     


60
M x






max
| | 180.0 NV 

max
|| 36.0NmM  
www.elsolucionario.org

1117. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.50
Draw the shear and bending-moment diagrams for the beam AB, and
determine the maximum absolute values of the shear and bending
moment.

SOLUTION
Free body: Entire beam


0: (0.9 m) (400 N)(0.3 m) (400 N)(0.6 m)
(400 N)(0.9 m) 0
A
MB  



800 NB

800 NB 

0: 0
xx
FA 


0: 800 N 3(400 N) 0
yy
FA   


400 N
y
A

400 NA 

We replace the loads by equivalent force-couple systems at C, D, and E.
We consider successively the following F-B diagrams.


1
1
400 N
0
V
M



5
5
400 N
180 N m
V
M

 


2
2
400 N
60 N m
V
M

 

6
6
400 N
60 N m
V
M

 


3
3
0
120 N m
V
M

 

7
7
800 N
120 N m
V
M

 


4
4
0
120 N m
V
M

 

8
8
800 N
0
V
M






www.elsolucionario.org

1118. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.50 (Continued)



( b)
max
|| 800NV 







max
| | 180.0 N mM  




www.elsolucionario.org

1119. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.51
Draw the shear and bending-moment diagrams for the
beam AB, and determine the maximum absolute values of
the shear and bending moment.

SOLUTION
Slope of cable CD is



710
y x
DD




7
0: (50 lb)(26 in.) (4 in.) (20 in.) (100 lb)(10 in.) (50 lb)(6 in.) 0
10
Hxx
MDD

     




2000 (4 14) 0
x
D 

200 lb
x
D


77
(200)
10 10
yx
DD

140 lb
y
D

0: 200 lb 0
xx
FH  

200 lb
x
H
0: 140 50 100 50 0
yy
FH   

60 lb
y
H
Equivalent loading on straight part of beam AB.

www.elsolucionario.org

1120. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.51 (Continued)

From A to E:

0: 50 lb
y
FV 


1
0: 50MMx 

From E to F: 0: 50 140 0
y
FV  

90 lbV


2
0: (50 lb) (140 lb)( 6) 800 lb in. 0Mxx M   


40 90
M x 

From F to G:
0: 50 140 100 0
y
FV   

10 lbV
3
0: 50 140( 6) 100( 16) 800 0Mxx x M    


1560 10Mx

From G to B: 0: 50 0
y
FV 

50 lbV

4
0: (50)(32 ) 0MM x  


1600 50Mx 



max max
| | 90.0 lb; | | 1400 lb in.VM 
www.elsolucionario.org

1121. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.52
Draw the shear and bending-moment diagrams for the
beam AB, and determine the maximum absolute values
of the shear and bending moment.

SOLUTION


0: (9 in.) (45 lb)(9 in.) (120 lb)(21in.) 0
G
FT    325 lbT

0: 325 lb 0
xx
FG   

325 lb
x
G


0: 45 lb 120 lb 0
yy
FG    165 lb
y
G
Equivalent loading on straight part of beam AB

From A to E:

0: 165 lb
y
FV 


1
0: 1625 lb in. (165 lb) 0MxM   


1625 165Mx 

From E to F:

0: 165 45 0
y
FV 


120 lbV


2
0 : 1625 lb in. (165 lb) (45 lb)( 9) 0MxxM    


1220 120Mx 


www.elsolucionario.org

1122. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.52 (Continued)

From F to B:
0 120 lb
y
FV 
3
0: (120)(21 ) 0MxM  


2520 120Mx 




max
| | 165.0 lbV 







max
| | 1625lb in.M  

www.elsolucionario.org

1123. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.53
Two small channel sections DF and EH have been
welded to the uniform beam AB of weight W 3 kN to
form the rigid structural member shown. This member is
being lifted by two cables attached at D and E. Knowing
that
 30 and neglecting the weight of the channel
sections, (a) draw the shear and bending-moment
diagrams for beam AB, (b) determine the maximum
absolute values of the shear and bending moment in the
beam.

SOLUTION
FBD Beam  channels:
(a) By symmetry:
12
TT T

0: 2 sin 60 3 kN 0
y
FT  

1
1
3
kN
3
3
23
3
kN
2
x
y
T
T
T



FBD Beam:
3
(0.5 m) kN
23
0.433 kN m
M


With cable force replaced by equivalent force-couple system at F and G
Shear Diagram: V is piecewise linear

0.6 kN/m with 1.5 kN
dV
dx




discontinuities at F and H.

(0.6 kN/m)(1.5 m) 0.9 kN
F
V 
V increases by 1.5 kN to
0.6 kN at F



0.6 kN (0.6 kN/m)(1 m) 0
G
V 
Finish by invoking symmetry


www.elsolucionario.org

1124. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.53 (Continued)

Moment diagram:
M is piecewise parabolic

decreasing with
dM
V
dx



with discontinuities of .433 kN at F and H.

1
(0.9 kN)(1.5 m)
2
0.675 kN m
F
M 
 

M increases by 0.433 kN m to –0.242 kN  m at F



1
0.242 kN m (0.6 kN)(1 m)
2
0.058 kN m
G
M  


Finish by invoking symmetry
(b)
max
|| 900NV 
at
F

and G



max
| | 675 N mM  
at F and G
www.elsolucionario.org

1125. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.54
Solve Problem 7.53 when 60 .
PROBLEM 7.53 Two small channel sections DF and
EH have been welded to the uniform beam AB of weight
W
3 kN to form the rigid structural member shown.
This member is being lifted by two cables attached at D
and E. Knowing that
30
 and neglecting the weight
of the channel sections, (a) draw the shear and bending-
moment diagrams for beam AB, (b) determine the
maximum absolute values of the shear and bending
moment in the beam.

SOLUTION
Free body: Beam and channels
From symmetry:

yy
ED

Thus:
tan
xxy
EDD  (1)


0: 3 kN 0
yyy
FDE   

1.5 kN
yy
DE 
From (1):
(1.5 kN) tan
x
D (1.5 kN) tanE 
We replace the forces at D and E by equivalent force-couple systems at F and H, where


0
(1.5 kN tan )(0.5 m) (750 N m) tanM  (2)

We note that the weight of the beam per unit length is

3 kN
0.6 kN/m 600 N/m
5m
W
w
L
  
(a) Shear and bending moment diagrams
From A to F:

0: 600 0 ( 600 ) N
y
FVxVx   

2
0: (600 ) 0, ( 300 )N m
2
J
x
MMx M x    

For
0:x 0
AA
VM 
For
1.5 m:x 900 N, 675 N m
FF
VM   
www.elsolucionario.org

1126. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.54 (Continued)

From F to H:


0: 1500 600 0
y
FxV   

(1500 600 ) NVx 

0
0: (600 ) 1500( 1.5) 0
2
J
x
MMx xM    


2
0
300 1500( 1.5) N mMM x x    
For 1.5 m:x
0
600 N, ( 675) N m
FF
VMM    
For
2.5 m:x
0
0, ( 375) N m
GG
VMM 
From G to B, The V and M diagrams will be obtained by symmetry,
(b)
max
|| 900 NV  

Making
60 in Eq. (2):
0
750 tan 60° 1299 N mM
Thus, just to the right of F:
1299 675 624 N mM  
and
1299 375 924 N m
G
M  


max
() | | 900 NbV  



max
|| 924 NmM  
www.elsolucionario.org

1127. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.55
For the structural member of Problem 7.53, determine
(a) the angle
 for which the maximum absolute value of
the bending moment in beam AB is as small as possible,
(b) the corresponding value of
max
||.M (Hint: Draw the
bending-moment diagram and then equate the absolute
values of the largest positive and negative bending
moments obtained.)
PROBLEM 7.53 Two small channel sections DF and EH
have been welded to the uniform beam AB of weight
W 3 kN to form the rigid structural member shown. This
member is being lifted by two cables attached at D and E.
Knowing that
 30 and neglecting the weight of the
channel sections, (a) draw the shear and bending-moment
diagrams for beam AB, (b) determine the maximum absolute
values of the shear and bending moment in the beam.

SOLUTION
See solution of Problem 7.50 for reduction of loading or beam AB to the following:

where
0
(750 N m) tanM  
[Equation (2)]
The largest negative bending moment occurs Just to the left of F:


11
1.5 m
0: (900 N) 0
2
MM

  


1
675 N mM  


The largest positive bending moment occurs
At G:

220
0: (1500 N)(1.25 m 1 m) 0MMM    

20
375 N mMM  

Equating
2
M and
1
:M

0
375 675M

0
1050 N mM www.elsolucionario.org

1128. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.55 (Continued)

(a) From Equation (2):
1050
tan 1.400
750
 54.5 

(b)
max
| | 675 N mM  
www.elsolucionario.org

1129. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.56
For the beam of Problem 7.43, determine (a) the ratio kP/wa for which
the maximum absolute value of the bending moment in the beam is as
small as possible, (b) the corresponding value of |M|
max. (See hint for
Problem 7.55.)
PROBLEM 7.43 Assuming the upward reaction of the ground on beam
AB to be uniformly distributed and knowing that P wa, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.

SOLUTION
Free body: Entire beam
0: (4 ) 2 0
yg
Fwawakwa   


(2 )
4
g
w
wk

Setting
g
w
w

(1)
We have
42k
(2)
Minimum value of B.M. For M to have negative values, we must have
.
g
ww
 We verify that M will then
be negative and keep decreasing in the portion AC of the beam. Therefore,
min
Mwill occur between C
and D.

From C to D:


0: 0
22
J
ax
MMwax wx

 
   
 
 


221
(2 )
2
Mwxaxa
 (3)
We differentiate and set
0:
dM
dx
 0xa
min
a
x

 (4)
Substituting in (3):

2
min
2
min112
1
2
1
2
Mwa
Mwa









(5)

www.elsolucionario.org

1130. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.56 (Continued)

Maximum value of bending moment occurs at D


3
0: (2 ) 0
2
DD
a
M M wa wa a


   




2
max 3
2
2
D
MMwa 

 


(6)

Equating
min
M and
max
:M

22
213
2
22
4210
wa wa


 






220
8



15
0.809
4



(a) Substitute in (2):
4(0.809) 2k 1.236k

(b) Substitute for
 in (5):

2
max min10.809
||
2(0.809)
MMwa

  2
max
| | 0.1180Mwa 
Substitute for
 in (4):
min
1.236
0.809
a
xa 
www.elsolucionario.org

1131. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.57
Determine (a) the distance a for which the maximum absolute
value of the bending moment in beam AB is as small as
possible, (b) the corresponding value of M
max. (See hint for
Prob. 7.55.)

SOLUTION
Tension in cords supporting load:

100 mm
0: 2 80 N 0;
50 mm
40 N
80 N
x
yy
y
y
x
T
FT
T
T
T
    




Translate cord tensions to straight part of beam ABat C and D:

(80 N) 0.04 m 3.2 N m
CD
MM  

FBD of beam AB:


From A to C:

0: 40 0 40 N
y
FVV  


11
0: 40 N 0 40MM x Mx   
From C to D:

www.elsolucionario.org

1132. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.57 (Continued)
0: 0
y
FV 

22
0: 3.2 N m 40 N 0 40 3.2MMaMa     
Equating the absolute values of
12
and MM :



40 40 3.2
40 40 3.2
40 40 3.2
80 3.2
0.04 m
aa
aa
aa
a
a

 
 



(a)
40.0 mma 
(b) Substituting:

max 2
40N .04 m 3.2 N mMM  
max
| | 1.600 N mM  


www.elsolucionario.org

1133. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.58
For the beam and loading shown, determine (a) the distance a
for which the maximum absolute value of the bending
moment in the beam is as small as possible, (b) the
corresponding value of |M|
max. (See hint for Problem 7.55.)

SOLUTION
Free body: Entire beam
0: 0
xx
FA 

0: (2.4) (3)(1.8) 3(1) (2) 0
Ey
MA a    


5
3.5 kN
6
y
Aa

5
3.5 kN
6
aA



Free body: AC



5
0: 3.5 (0.6 m) 0,
6
CC
MM a

   



2.1
2
C
a
M 


Free body: AD

5
0: 3.5 (1.4 m) (3 kN)(0.8 m) 0
6
DD
MM a

    



7
2.5
6
D
Ma  
Free body: EB 0: (2 kN) 0
EE
MM a   2
E
Ma 
We shall assume that
CD
MM and, thus, that
max
.
C
MM
We set
max min
||MM or ||2.1 2
2
CE
a
MM a
0.840 ma 

max
|| | |2 2(0.840)
CE
MMMa 
max
| | 1.680 N mM  
We must check our assumption.

7
2.5 (0.840) 1.520 N m
6
D
M  
Thus,
,
CD
MM O.K.
The answers are ( a)
0.840 ma 
( b)
max
| | 1.680 N mM  
www.elsolucionario.org

1134. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.59
A uniform beam is to be picked up by crane cables attached at A and
B. Determine the distance a from the ends of the beam to the points
where the cables should be attached if the maximum absolute value
of the bending moment in the beam is to be as small as possible.
(Hint: Draw the bending-moment diagram in terms of a, L, and the
weight w per unit length, and then equate the absolute values of the
largest positive and negative bending moments obtained.)

SOLUTION
w weight per unit length

To the left of A:
1
0: 0
2
x
MMwx

  



2
21
2
1
2
A
Mwx
Mwa



Between A and B:
2
11
0: ( ) ( ) 0
22
MMwLxawxx
 
   
 
 


2111
222
MwxwLxwLa  

At center C:
2
2
111
22 2 22
C
L
x
LL
Mw wL wLa

 
  
 
 





www.elsolucionario.org

1135. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.59 (Continued)

We set
22 22111111
||||:
282282
AC
M M wa wL wLa wa wL wLa

22
0.25 0aLa L



22
22
max11
()(21)
22
1
(0.207 ) 0.0214
2
 

aLLL L
MwL wL

0.207aL 
www.elsolucionario.org

1136. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.60
Knowing that P Q 150 lb, determine (a) the distance a for which the
maximum absolute value of the bending moment in beam AB is as small
as possible, (b) the corresponding value of
max
|| .M (See hint for Problem
7.55.)

SOLUTION











Free body: Entire beam
0: (150)(30) (150)(60) 0
A
MDa   


13,500
D
a


Free body: CB

13,500
0: (150)(30) ( 30) 0
CC
MM a
a
   


45
9000 1
C
M
a






Free body: DB
0: (150)(60 ) 0
DD
MM a 


150(60 )
D
Ma 


(a) We set

max min
45
| | or : 9000 1 150(60 )
CD
MM MM a
a





2700
60 60 a
a



2
2700 51.96 in.aa 52.0 in.a 
(b)
max
| | 150(60 51.96)
D
MM  

max
| | 1206 lb in.M  
www.elsolucionario.org

1137. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.61
Solve Problem 7.60 assuming that P  300 lb and Q 150lb.
PROBLEM 7.60 Knowing that P Q 150 lb, determine (a) the distance
a for which the maximum absolute value of the bending moment in beam
AB is as small as possible, (b) the corresponding value of
max
|| .M (See
hint for Problem 7.55.)

SOLUTION









Free body: Entire beam
0: (300)(30) (150)(60) 0
A
MDa   

18,000
D
a


Free body: CB

18,000
0: (150)(30) ( 30) 0
CC
MM a
a
    

40
13,500 1
C
M
a



 
Free body: DB
0: (150)(60 ) 0
DD
MM a 

150(60 )
D
Ma  
(a) We set

max min
40
| | or : 13,500 1 150(60 )
CD
MM MM a
a





3600
90 60 a
a



2
30 3600 0aa 

30 15.300
46.847
2
a



46.8 in.a 
(b)
max
| | 150(60 46.847)
D
MM  

max
| | 1973 lb in.M  
www.elsolucionario.org

1138. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.62*
In order to reduce the bending moment in the cantilever beam
AB, a cable and counterweight are permanently attached at end
B. Determine the magnitude of the counterweight for which the
maximum absolute value of the bending moment in the beam
is as small as possible and the corresponding value of
max
|| .M
Consider (a) the case when the distributed load is permanently
applied to the beam, (b) the more general case when the
distributed load may either be applied or removed.
SOLUTION
M due to distributed load:

2
0: 0
2
1
2
J
x
MMwx
Mwx
  


M due to counter weight:

0: 0
J
MMxw
Mwx
 

(a)
Both applied:

2
2
0at
x
w
MW x
dM W
Wwx x
dx w

  

And here
2
0
2
W
M
w
 so M max; Mmin must be at xL
So
2
min1
.
2
MWLwL
For minimum
max
||M set
max min
,MM
so
2
2
1
or
22
W
WL wL
w
 
222
20WwLWwL

22
2 (need )WwL wL   (2 1) 0.414WwLwL  





www.elsolucionario.org

1139. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.62* (Continued)

(b)
w may be removed

22
2
max
(2 1)
22
W
MwL
w

 2
max
0.0858MwL 

Without w,
max
at
MWx
MWLA



With w (see Part a)
2
2
max
2
min
2
at
2
1
at
2
w
MWx x
WW
Mx
ww
MWLwLxL


 

For minimum
max max min
, set (no ) (with )MMwM w 

211
24
WL WL wL W wL    

2
max1
4
MwL

With
1
4
WwL

www.elsolucionario.org

1140. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.63
Using the method of Section 7.6, solve Problem 7.29.
PROBLEM 7.29 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION








Reactions at A and D
Because of the symmetry of the supports and loading.

11
224
L
AD w wL

 




1
4
wLAD

Shear diagram
At A:
1
4
A
VwL
From B to C:
Oblique straight line
max
1
||
4
VwL

Bending-moment diagram
At A: 0
A
M
From B to C: ARC of parabola
2
max3
||
32
MwL

Since V has no discontinuity at BnorC, the slope of the parabola at
these points is the same as the slope of the adjoining straight line
segment.
www.elsolucionario.org

1141. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.64
Using the method of Section 7.6, solve Problem 7.30.
PROBLEM 7.30 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION






Free body: Entire beam

0
1
0: ( )( ) 0
2
y
FBwL  

0
1
2
wLB


Shear diagram
0
1
0: ( )( ) 0
23
BB
L
MwLM

  




2
01
6
B
wLM


Moment diagram
(b)
max 0
|| /2VwL 

2
max 0
|| /6MwL 
www.elsolucionario.org

1142. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.65
Using the method of Section 7.6, solve Problem 7.31.
PROBLEM 7.31 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION










Free body: Entire beam

2
0: ( ) 0
3
C
L
MPAL

  




2
3
PA


2
0: 0
3
Y
FPPC 


1
3
PC

Shear diagram
At A:
2
3
A
VP


max
2
||
3
VP


Bending-moment diagram
At A:
0
A
M

max
2
||
9
MPL

www.elsolucionario.org

1143. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.66
Using the method of Section 7.6, solve Problem 7.32.
PROBLEM 7.32 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION






Free body: Entire beam

0: 0
y
FCPP 


2PC


0: (2 ) ( ) 0
CC
MPaPaM   


3
C
PaM

Shear diagram
At A: A
VP


max
|| 2VP 

Bending-moment diagram At A:
0
A
M


max
|| 3MPa 



www.elsolucionario.org

1144. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.67
Using the method of Section 7.6, solve Problem 7.33.
PROBLEM 7.33 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values
of the shear and bending moment.


SOLUTION







Free body: Entire beam
0:
y
FAC 
0
0: 0
C
MAlM 

0
M
AC
L


Shear diagram
At A:
0
A
M
V
L



0
max
||
M
V
L
 
Bending-moment diagram
At A: 0
A
M
At B, M increases by M
0 on account of applied couple.

max 0
|| /2MM 
www.elsolucionario.org

1145. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.68
Using the method of Section 7.6, solve Problem 7.34.
PROBLEM 7.34 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION







Free body: Entire beam

0: 0
y
FBP 


PB
0
0: 0
BB
MMMPL 


0
B
M
Shear diagram
At A:
A
VP

max
||VP 
Bending-moment diagram
At A:
0A
MMPL

max
||MPL 
www.elsolucionario.org

1146. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.69
For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.

SOLUTION
Reactions: 00
xx
FA 
0: 24 kN m (8 kN)(3 m) 10 kN 6 m 8 kN 9 m (12 m) 0
A
ME     

11 kNE


0
y
F

15 kNA




(b)
max
|| 15.00kN;V
max
|| 42.0kNmM  
www.elsolucionario.org

1147. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.70
For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.

SOLUTION
Reactions 00
xx
FA 

0: 12 kN m (9 kN)(3.5 m) (18 kN)(2 m) 3 kN m (4.5 m) 0
D y
MA     

17.00 kN
y
A
0 17 kN 9 kN 18 kN 0
y
FD   
10.00kND





(b)
max
|| 17.00kN;V
max
|| 17.00kNmM  
www.elsolucionario.org

1148. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.71
Using the method of Section 7.6, solve Problem 7.39.
PROBLEM 7.39 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.


SOLUTION







Free body: Beam
0: 0
xx
FA 

0: (60 kN)(3 m) (50 kN)(1 m) (5 m) 0
By
MA   

46.0 kN
y
A

0: 46.0 kN 60 kN 50 kN 0
y
FB    

64.0 kNB
Shear diagram
At A: 46.0 kN
Ay
VA

max
|| 64.0kNV 
Bending-moment diagram
At A: 0
A
M

max
| | 92.0 kN mM  
Parabola from D to B. Its slope at D is same as that of straight-line
segment CD since V has no discontinuity at D.

www.elsolucionario.org

1149. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.72
Using the method of Section 7.6, solve Problem 7.40.
PROBLEM 7.40 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.


SOLUTION




Free body: Beam
0: 0
xx
FA 

0: (50 kN)(2 m) (40 kN)(1m) (4 m) 0
Dy
MA   

15.00 kN
y
A

0: 15 kN 40 kN 50 kN 0
y
FD    

75.0 kND
Shear diagram
At A: 15.00 kN
Ay
VA

max
| | 40.0 kNV 
Bending-moment diagram
At A: 0
A
M

max
| | 40.0 kN mM  
Parabola from D to B.
www.elsolucionario.org

1150. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.73
Using the method of Section 7.6, solve Problem 7.41.
PROBLEM 7.41 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION






Reactions at supports.
Because of the symmetry:

1
(8 8 4 5) kips
2
  AB


18 kipsAB 

Shear diagram
At A: 18 kips
A
V

max
|| 18.00kipsV 


Bending-moment diagram
At A: 0
A
M

max
| | 48.5 kip ftM  

Discontinuities in slope at C and D, due to the discontinuities of V.
www.elsolucionario.org

1151. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.74
Using the method of Section 7.6, solve Problem 7.42.
PROBLEM 7.42 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.

SOLUTION








Free body: Beam

0: 0
xx
FA 

0: (12 kips)(4ft) (15 kips)(7 ft) (10 ft) 0
By
MA   

15.3 kips
y
A 

0: 15.3 15 12 0
y
FB  

11.7 kipsB 

Shear diagram
At A: 15.3 kips
Ay
VA

max
|| 15.30kipsV 

Bending-moment diagram
At A: 0
A
M

max
| | 46.8 kip ftM  
www.elsolucionario.org

1152. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.75
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.

SOLUTION

Reactions

0: (16)(9) (45)(4) (8)(3) (12) 0
D
MA    

25 kipsA

25 kipsA


0: 25 16 45 8 0
yy
FD  

44,
y
D 44 kips
y
D


0: 0
xx
FD 



(b)
max
| | 36.0 kips;V
max
| | 120.0 kip ftM  


www.elsolucionario.org

1153. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.76
For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.

SOLUTION
Reactions
From symmetry, CG

0: 2(16 lb/in.)(10 in.) 100 lb 150 lb 100 lb 2 0
y
FC 

335 lbCG 335 lbCG




(b)
max
| | 175.0 lbs;V
max
| | 800 lb in.M  
www.elsolucionario.org

1154. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.77
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.

SOLUTION

Reactions

0: 45 kN m (90 kN)(3 m) (7.5 m) 0
A
MC   

30 kNC

30 kNC


0: 90 kN 30 kN 0
y
FA   

60 kNA




(b)
75.0 kN m,4.00 m from A 
www.elsolucionario.org

1155. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.78
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.

SOLUTION


0: (8.75)(1.75) (2.5) 0
A
MB  

6.125 kNB


0:
y
F

2.625 kNA



Similar
D’s


Add numerators
and denominators
1.05 m
2.5 2.5
2.625 3.625 6.25
x
xx



 


(b)
1.378 kN m,1.050 m from A 
www.elsolucionario.org

1156. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.79
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum
bending moment.


SOLUTION








Free body: Beam
0: 0
xx
FA 

0: (40 kN)(1.5 m) (3.75 m) 0
By
MA  

16.00 kN
y
A 
Shear diagram
16.00 kN
AC y
VV A
To determine Point E where
0,V we write

0 16 kN (20 kN/m)
EC
VV wu
u


0.800 mu 
We next compute all areas
Bending-moment diagram
At A: 0
A
M
Largest value occurs at E with

1.25 0.8 2.05 mAE
max
|| 26.4kNmM  

2.05 m from A 
From A to C and D to B: Straight line segments. From C to D: Parabola

www.elsolucionario.org

1157. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.80
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the magnitude and location of the
maximum bending moment.

SOLUTION
0: (8)(2) (4)(3.2) 4 0
A
MC  


7.2 kNC

0: 4.8 kN
y
F  A
(a) Shear diagram

Similar Triangles:
3.2 3.2
;2.4m
4.8 1.6 6.4

 
xx
x


Add num. & den.


Bending-moment diagram

(b)



max
| | 5.76 kN mM  
2.40from A 
www.elsolucionario.org

1158. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.81
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.

SOLUTION
Reactions
0: 0
xx
FA 

0: (800 lb/ft)(9 ft)(4.5 ft) (600 lb)(9 ft) (15 ft) 0
A
MC   

1800 lbC

1.800 kipsC


0: (800 lb/ft)(9 ft) 600 lb+1800 lb 0
y
FA   

4800 lbA 4.80 kipsA




Maximum moment occurs at zero shear,
4800 lb= 800 lb/ft 6.00 ftxx 

(b)
max
| | 14.40 kip ft, 6.00 ft from MA 
www.elsolucionario.org

1159. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.82
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.

SOLUTION
Reactions
0: 0
xx
FB 

0: (400 lb/ft)(20 ft)(10 ft) (3200 lb)(22.5 ft) (20 ft) 0
C
MB   

7600 lbB

7.60 kipsB


0: (400 lb/ft)(20 ft) 3200 lb 7600 lb 0
y
FC    

3600 lbC 3.60 kipsC




Maximum moment occurs at zero shear,
4400 lb
+2.5 ft 13.50 ft
400 lb/ft
xx

(b)
max
| | 16.20 kip ft, 13.50 ft from MA  www.elsolucionario.org

1160. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.83
(a) Draw the shear and bending-moment diagrams for beam AB,
(b) determine the magnitude and location of the maximum absolute value
of the bending moment.


SOLUTION












Reactions at supports
Because of symmetry of load

1
(300 8 300)
2
AB 


1350 lbAB 
Load diagram for AB
The 300-lb force at D is replaced by an equivalent force-couple system at
C.


Shear diagram
At A: 1350 lb
A
VA
To determine Point E where
0:V

0 1350 lb (300 lb/ft)
EA
VV wx
x




4.50 ftx 
We compute all areas
Bending-moment diagram
At A: 0
A
M
Note
600 lb ft drop at C due to couple

max
| | 3040 lb ftM  
 4.50 ftfrom A 
www.elsolucionario.org

1161. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.84
Solve Problem 7.83 assuming that the 300-lb force applied at D is
directed upward.
PROBLEM 7.83 (a) Draw the shear and bending-moment diagrams for
beam AB, (b) determine the magnitude and location of the maximum
absolute value of the bending moment.

SOLUTION












Reactions at supports
Because of symmetry of load:

1
(300 8 300)
2
AB 


1050 lbAB 

Load diagram
The 300-lb force at D is replaced by an equivalent force-couple
system at C.

Shear diagram
At A: 1050 lb
A
VA
To determine Point E where
0:V

0 1050 lb (300 lb/ft)
EA
VV wx
x




3.50 ftx 
We compute all areas
Bending-moment diagram
At A: 0
A
M
Note
600 lb ft increase at C due to couple

max
| | 1838 lb ftM  
 3.50 ftfrom A 
www.elsolucionario.org

1162. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.85
For the beam and loading shown, (a) write the equations of the shear
and bending-moment curves, (b) determine the magnitude and location
of the maximum bending moment
.

SOLUTION


0
01
cos
2
2
sin
2
dv x
ww
dx L
Lx
Vwdxw C
L



 

  



(1)

01
2
012
2
sin
2
2
cos
2
dM L x
Vw C
dx L
Lx
MVdxw CxC
L



 



 



(2)
Boundary conditions
At 0:x
11
00VC C 
At
0:x
2
02
2
20
2
cos(0) 0
2
L
Mw C
L
Cw



  







Eq. (1)
0
2
sin
2
Lx
Vw
L 






2
0
2
1cos
2
Lx
Mw
L 
 

 
 

max
M at :xL
2
2
max 0 0 2
24
|| |10|
L
Mw wL
 







www.elsolucionario.org

1163. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.86
For the beam and loading shown, (a) write the equations of the
shear and bending-moment curves, (b) determine the magnitude
and location of the maximum bending moment.

Solution

Eq. (7.1):
0
dV x
ww
dx L
 


2
001
1
2
xx
Vwdx w C
LL
  

(1)
Eq. (7.3):
2
01
1
2
dM x
VwC
dx L
 


23
01 0 12
11
26
xx
M wCdxw CxC
LL

    



(2)
(a) Boundary conditions

2
At 0: 0 0 0x MC
2
0C 

2
011
:0
6
xLM wLCL
10
1
6
CwL

Substituting C
1 and C 2 into (1) and (2):

2
00
11
26
x
Vw wL
L
 

2
0 2
11
23
x
VwL
L

 

 

3
00
11
66
x
M wwLx
L
 
3
2
0 3
1
6
xx
MwL
LL

 



(b) Max moment occurs when V 0:

2
2
13 0
xL

1
3
x
L

0.577
x L
www.elsolucionario.org

1164. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.86 (Continued)


3
2
max 0
111
6 33
MwL


 



2
max 0
0.0642MwL 
Note: At x 0,
0
11
23
A
AV wL




0
1
6
wLA

www.elsolucionario.org

1165. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.87
For the beam and loading shown, (a) write the equations of the shear
and bending-moment curves, (b) determine the magnitude and
location of the maximum bending moment.

SOLUTION
(a) We check that beam is in equilibrium 
00
1112
0
23223
L
wL wL L
   
 
   
    (ok)
Load:
0
00 3
() 1 1
22
w
x xx
wx w w
L LL
 
 
 
 

Shear: We use Eq. (7.1)

0
2
001
3
1
2
33
(1 )
24
dV x
ww
dx L
xx
Vw dxwx C
LL

  



     




Boundary Condition: at x=L, V=0 therefore
0110
31
0
44
wL L C C wL

   



Bending moment: Using Eq. (7.3)

2
00 0
3
2
00 02
31

44
11 1
()
24 4
dM x
VM Vdx wx w wLdx
dx L
x
Mx wx w wLx C
L

   


   


Boundary Condition: at x=0,
22
0 therefore 0MC C 


2
0
1
341
4
xx
Vx wL
LL
 
 
   
   



32
2
0
1
2
4
x xx
Mx wL
L LL
 
  
    
    




www.elsolucionario.org

1166. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.87 (Continued)

(b) Maximum bending moment
0
dM
V
dx


2
03 4 1
0 3 1 1 ; and
3
xx
LL
xx L
x xL
LL

 
 
 


   





2
max 01121
4 2793
MwL






2
max 01
, at
27 3
L
MwLx 
www.elsolucionario.org

1167. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.88
For the beam and loading shown, (a) write the equations of the shear
and bending-moment curves, (b) determine the magnitude and
location of the maximum bending moment.

SOLUTION

Eq. (7.1):
0
sin
dV x
ww
dx L

 

001
sin
cos
xLx
Vw dxw C
LL
 

  

(1)
Eq. (7.3):
01
cos
dM L x
Vw C
dx L


 

2
010 12 2
cos sin
Lx L x
M wCdxw CxC
LL

 




(2)
(a) Boundary conditions
At 0: 0xM therefore,
2
0C

:0xL M therefore,
1
0C



0
cos
Lx
Vx w
L








2
02
sin
Lx
Mw
L







(b) Max moment occurs when V 0 at
2
L
x
2
max 0 2
L
Mw

 
www.elsolucionario.org

1168. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.89
The beam AB is subjected to the uniformly distributed load shown and to
two unknown forces
P and Q. Knowing that it has been experimentally
determined that the bending moment is
800 N m
  at D and 1300 N m
at E, (a) determine
P and Q, (b) draw the shear and bending-moment
diagrams for the beam.

SOLUTION













(a) Free body: Portion AD
0: 0
xx
FC 

0: (0.3 m) 0.800 kN m (6 kN)(0.45 m) 0
Dy
MC   

11.667 kN
y
C 11.667 kNC 
Free body: Portion EB
0: (0.3 m) 1.300 kN m 0
E
MB  

4.333 kNB 
Free body: Entire beam
0: (6 kN)(0.45 m) (11.667 kN)(0.3 m)
(0.3 m) (4.333 kN)(0.6 m) 0
D
M
Q
 
 

6.00 kNQ 
0: 11.667 kN 4.333 kN
y
M 

6kN 6 kN 0P 

4.00 kNP 

Load diagram
(b) Shear diagram
At A: 0
A
V

max
|| 6 kNV  




www.elsolucionario.org

1169. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 7.89 (Continued)

Bending-moment diagram
At A: 0
A
M

max
| | 1300 N mM  
We check that

800 N m and 1300 N m
DE
MM   
As given:
At C:
900 N m
C
M  
www.elsolucionario.org

1170. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.90
Solve Problem 7.89 assuming that the bending moment was found to
be
650 N m at D and 1450 N m at E.
PROBLEM 7.89 The beam AB is subjected to the uniformly distributed
load shown and to two unknown forces
P and Q. Knowing that it has
been experimentally determined that the bending moment is
800 N m
at D and
1300 N m
  at E, (a) determine P and Q, (b) draw the shear
and bending-moment diagrams for the beam.

SOLUTION












(a) Free body: Portion AD
0: 0
xx
FC 

0: (0.3 m) 0.650 kN m (6 kN)(0.45 m) 0
D
MC   

11.167 kN
y
C 11.167 kNC 
Free body: Portion EB
0: (0.3 m) 1.450 kN m 0
E
MB  

4.833 kNB 
Free body: Entire beam
0: (6 kN)(0.45 m) (11.167 kN)(0.3 m)
(0.3 m) (4.833 kN)(0.6 m) 0
D
M
Q
 
 

7.50 kNQ 

0: 11.167 kN 4.833 kN
6kN 7.50 kN 0
y
M
P
 
 

2.50 kNP 
Load diagram

(b) Shear diagram
At A: 0
A
V

max
|| 6 kNV  



www.elsolucionario.org

1171. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.90 (Continued)

Bending-moment diagram
At A: 0
A
M

max
| | 1450 N mM  
We check that

650 N m and 1450 N m
DE
MM   
As given:
At C:
900 N m
C
M  
www.elsolucionario.org

1172. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.91*
The beam AB is subjected to the uniformly distributed load shown and to
two unknown forces
P and Q. Knowing that it has been experimentally
determined that the bending moment is
6.10 kip ft
  at D and
5.50 kip ft at E, (a) determine P and Q, (b) draw the shear and
bending-moment diagrams for the beam.

SOLUTION
(a) Free body: Portion DE
0: 5.50 kip ft 6.10 kip ft (1 kip)(2 ft) (4 ft) 0
ED
MV     

0.350 kip
D
V

0: 0.350 kip 1kip 0
yE
FV   

0.650 kip
E
V

Free body: Portion AD
0: 6.10 kip ft (2ft) (1 kip)(2 ft) (0.350 kip)(4 ft) 0
A
MP    

1.350 kipsP 

0: 0
xx
FA 

0: 1 kip 1.350 kip 0.350 kip 0
yy
FA    

2.70 kips
y
A 2.70 kipsA 

Free body: Portion EB

0: (0.650 kip)(4 ft) (1 kip)(2 ft) (2 ft) 5.50 kip ft 0
B
MQ    

0.450 kipQ 

0: 0.450 1 0.650 0
y
FB   

2.10 kipsB 





www.elsolucionario.org

1173. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.91* (Continued)
(b) Load diagram

Shear diagram
At A: 2.70 kips
A
VA

To determine Point G where
0,V we write


0 0.85 kips (0.25 kip/ft)
GC
VV w 




3.40 ft 
We next compute all areas
max
| | 2.70 kips at VA 

Bending-moment diagram
At A: 0
A
M
Largest value occurs at G with

2 3.40 5.40 ftAG 

max
| | 6.345 kip ftM  
5.40 ft from A

Bending-moment diagram consists of 3 distinct arcs of parabolas.

www.elsolucionario.org

1174. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.92*
Solve Problem 7.91 assuming that the bending moment was found to
be
5.96 kip ft at D and 6.84 kip ft at E.
PROBLEM 7.91* The beam AB is subjected to the uniformly
distributed load shown and to two unknown forces
P and Q. Knowing
that it has been experimentally determined that the bending moment is
6.10 kip ft at D and 5.50 kip ft
  at E, (a) determine Pand Q, (b) draw
the shear and bending-moment diagrams for the beam.

SOLUTION
(a) Free body: Portion DE
0: 6.84 kip ft 5.96 kip ft (1 kip)(2 ft) (4 ft) 0
ED
MV    

0.720 kip
D
V

0: 0.720 kip 1kip 0
yE
FV   

0.280 kip
E
V

Free body: Portion AD
0: 5.96 kip ft (2 ft) (1 kip)(2 ft) (0.720 kip)(4 ft) 0
A
MP    

0.540 kipP 

0: 0
xx
FA 


0: 1 kip 0.540 kip 0.720 kip 0
yy
FA    

2.26 kips
y
A 2.26 kipsA 

Free body: Portion EB
0: (0.280 kip)(4 ft) (1 kip)(2 ft) (2 ft) 6.84 kip ft 0
B
MQ    

1.860 kipsQ 

0: 1.860 1 0.280 0
y
FB   

3.14 kipsB 




www.elsolucionario.org

1175. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.92* (Continued)

(b) Load diagram


Shear diagram
At A: 2.26 kips
A
VA

To determine Point G where
0,V we write


GC
VV w 

0 (1.22 kips) (0.25 kip/ft)

4.88 ft 
We next compute all areas
max
| | 3.14 kips at VB 
Bending-moment diagram
At A: 0
A
M
Largest value occurs at G with


2 4.88 6.88 ftAG 

max
| | 6.997 kip ftM  
6.88 ft from A

Bending-moment diagram consists of 3 distinct arcs of parabolas.

www.elsolucionario.org

1176. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.93
Three loads are suspended as shown from the cable
ABCDE. Knowing that d
C 4 m, determine (a) the
components of the reaction at E, (b) the maximum tension
in the cable.

SOLUTION
(a)


0: (20 m) (2 kN)(5 m) (4 kN)(10 m) (6 kN)(15 m) 0
Ay
ME    

7kN
y
E 7.00 kN
y
E 
Portion CDE:



0: (7 kN)(10 m) (6 kN)(5 m) (4 m) 0
Cx
ME   

10 kN
x
E 10.00 kN
x
E 


(b) Maximum tension occurs in DE:

22 22
10 7
mxy
TEE  12.21 kN
m
T 
www.elsolucionario.org

1177. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.94
Knowing that the maximum tension in cable ABCDE is
25 kN, determine the distance d
C.

SOLUTION
Maximum T of 25 kN occurs in DE. See solution of Problem 7.93 for the determination of 7.00 kN
y
E


From force triangle
22 2
725
x
E
Portion CDE:
24 kN
x
E





0: (6 kN)(5 m) (24 kN) (7 kN)(10 m) 0
CC
Mh   

1.667 m
C
h 
www.elsolucionario.org

1178. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.95
If 8 ft,
C
d determine (a) the reaction at A, (b) the reaction at E.

SOLUTION
Free body: Portion ABC
0
C
M

216 300(8)0
xy
AA 

81200
xy
AA (1)
Free body: Entire cable


0: 6 32 (300 lb 200 lb 300 lb)16 ft 0
Exy
MAA      

3 16 6400 0
xy
AA
Substitute from Eq. (1):

3(8 1200) 16 6400 0
yy
AA 250 lb
y
A
Eq. (1)
8(250) 1200
x
A 800 lb
x
A


0: 0 800 lb 0
xxx x
FAE E       800 lb
x
E

0: 250 300 200 300 0
yy
FE      550 lb
y
E


(a) 838 lbA 17.35 
( b) 971 lbE 34.5 
www.elsolucionario.org

1179. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.96
If 4.5 ft,
C
d determine (a) the reaction at A, (b) the reaction at E.

SOLUTION
Free body: Portion ABC


0: 1.5 16 300 8 0
Cxy
MAA   

(2400 16 )
1.5
y
x
A
A


(1)



Free body: Entire cable
0: 6 32 (300 lb 200 lb 300 lb)16 ft 0
Exy
MAA      
3 16 6400 0
xy
AA
Substitute from Eq. (1):
3(2400 16 )
16 6400 0
1.5
100 lb



y
y
y
A
A
A

Thus
y
A acts downward 100 lb
y
A
Eq. (1)
(2400 16( 100))
2667 lb
1.5
x
A

 2667 lb
x
A

0: 0 2667 0
xxx x
FAE E       2667 lb
x
E

0: 300 200 300 0
yyy
FAE     

100 lb 800 lb 0
y
E  900 lb
y
E


(a) 2670 lbA 2.10
(b) 2810 lbE 18.65 
www.elsolucionario.org

1180. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.97
Knowing that 3m,
C
d determine (a) the distances
B
d and
D
d (b) the
reaction at E.

SOLUTION
Free body: Portion ABC
0: 3 4 (5 kN)(2 m) 0
Cxy
MAA   

410
33
xy
AA (1)

Free body: Entire cable


0: 4 10 (5 kN)(8 m) (5 kN)(6 m) (10 kN)(3 m) 0     
Exy
MAA

4101000
xy
AA
Substitute from Eq. (1):

410
4101000
33
18.571 kN
yy
y
AA
A

 



18.571 kN
y
A
Eq. (1)
410
(18.511) 21.429 kN
33
x
A 21.429 kN
x
A

0: 0 21.429 0
xxx x
FAE E       21.429 kN
x
E

0: 18.571 kN 5 kN 5 kN 10 kN 0
yy
FE   1.429 kN
y
E



(b) 21.5 kNE 3.81 

www.elsolucionario.org

1181. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.97 (Continued)

(a) Portion AB
0: (18.571 kN)(2 m) (21.429 kN) 0
BB
Md  

1.733 m
B
d 
Portion DE
Geometry

(3 m) tan 3.8°
0.199 m
4 m 0.199 m
D
h
d




 4.20 m
D
d 
www.elsolucionario.org

1182. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.98
Determine (a) distance d C for which portion DE of the cable is
horizontal, (b) the corresponding reactions at A and E.

SOLUTION
Free body: Entire cable

(b)
0: 5 kN 5 kN 10 kN 0
yy
FA  

20 kN
y
A


0: (4 m) (5 kN)(2 m) (5 kN)(4 m) (10 kN)(7 m) 0
A
ME    


25 kNE

25.0 kNE 

0: 25 kN 0
xx
FA    25 kN
x
A

32.0 kNA 38.7

(a) Free body: Portion ABC


0: (25 kN) (20 kN)(4 m) (5 kN)(2 m) 0
CC
Md   


25 70 0
C
d

2.80 m
C
d 
www.elsolucionario.org

1183. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.99
If dC 15 ft, determine (a) the distances d B and d D,
(b) the maximum tension in the cable.

SOLUTION
Free body: Entire cable


0: (30 ft) (7.5 ft) (2 kips)(6 ft) (2 kips)(15 ft) (2 kips)(21ft) 0
Ay x
ME E     

7.5 30 84 0
xy
EE (1)
Free body: Portion CDE
0: (15ft) (15ft) (2kips)(6ft) 0
Cy x
ME E   


15 15 12 0
xy
EE
(2)

1
Eq. (1) : 3.75 15 42 0
2
xy
EE (3)
(2) – (3):
11.25 30 0
x
E

2.6667 kips
x
E
Eq. (1):
7.5(2.6667) 30 84 0 3.4667 kips
yy
EE 

22 2 2
(2.6667) (3.4667)
mxy
TEE  4.37 kips
m
T 

www.elsolucionario.org

1184. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.99 (Continued)

Free body: Portion DE


0: (3.4667 kips)(9 ft) (2.6667 kips) 0
DD
Md  

11.70 ft
D
d 

Return to free body of entire cable (with 2.6667 kips, 3.4667 kips)
xy
EE
0: 3(2 kips) 3.4667 kips 0
yy
FA   

2.5333 kips
y
A


0: 2.6667 kips 0
xx
FA  

2.6667 kips
x
A

Free body: Portion AB


0:(7.5)(6)0
BxB y
MAd A   


(2.6667)( 7.5) (2.5333)(6) 0
B
d 

13.20 ft
B
d 
www.elsolucionario.org

1185. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.100
Determine (a) the distance d C for which portion BC of
the cable is horizontal, (b) the corresponding
components of the reaction at E.

SOLUTION
Free body: Portion CDE



0: 2(2 kips) 0 4 kips
yy y
FE E   


0: (4 kips)(15 ft) (2 kips)(6 ft) 0
CxC
MEd   



48 kip ft
xC
Ed
(1)


Free body: Entire cable


0: (4 kips)(30 ft) (7.5 ft) (2 kips)(6 ft) (2 kips)(15 ft) (2 kips)(21 ft) 0
Ax
ME     


4.8 kips
x
E

From Eq. (1):
48 48
4.8
C
x
d
E


10.00 ft
C
d 

Components of reaction at E:
4.80 kips
x
E ; 4.00 kips
y
E 


www.elsolucionario.org

1186. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.101
Knowing that m B 70 kg and m C 25 kg, determine the
magnitude of the force
P required to maintain equilibrium.

SOLUTION
Free body: Portion CD
0: (4 m) (3 m) 0
Cy x
MD D  


3
4
yx
DD

Free body: Entire cable


3
0: (14 m) (4 m) (10 m) (5 m) 0
4
AxBC
MDWWP    
(1)
Free body: Portion BCD

3
0: (10m) (5m) (6m) 0
4
BxxC
MDDW   


2.4
xC
DW
(2)

For
70 kg 25 kg
BC
mm

2
9.81 m/s :g

70 25
BC
WgWg

Eq. (2):
2.4 2.4(25 ) 60
xC
DW gg 

Eq. (1):
3
60 (14) 70 (4) 25 (10) 5 0
4
gggP 


100 5 0: 20gP P g 


20(9.81) 196.2 NP

196.2 NP 

www.elsolucionario.org

1187. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.102
Knowing that m B 18 kg and m C 10 kg, determine the
magnitude of the force
P required to maintain equilibrium.

SOLUTION
Free body: Portion CD
0: (4 m) (3 m) 0
Cy x
MD D  


3
4
yx
DD

Free body: Entire cable


3
0: (14 m) (4 m) (10 m) (5 m) 0
4
AxBC
MDWWP    
(1)
Free body: Portion BCD

3
0: (10m) (5m) (6m) 0
4
BxxC
MDDW   


2.4
xC
DW
(2)

For
18 kg 10 kg
BC
mm

2
9.81 m/s :g

18 10
BC
WgWg

Eq. (2):
2.4 2.4(10 ) 24
xC
DW gg 

Eq. (1):
3
24 (14) (18 )(4) (10 )(10) 5 0
4
gggP 


80 5 : 16gPP g


16(9.81) 156.96 NP

157.0 NP 
www.elsolucionario.org

1188. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.103
Cable ABC supports two loads as shown. Knowing that b 21 ft,
determine (a) the required magnitude of the horizontal force
P,
(b) the corresponding distance a.

SOLUTION
Free body: ABC


0: (12ft) (140lb) (180lb) 0
A
MP a b

(1)





Free body: BC



0: (9 ft) (180 lb)( ) 0
B
MP ba
 (2)



Data
21 ftb Eq. (1): 21 140 180(21) 0Pa
 

20
180
3
Pa
(3)

Eq. (2): 9 180(21 ) 0Pa 

20 420Pa (4)
Equate (3) and (4) through
20
: 180 20 420
3
Pa a
( b) 9.00 fta 

Eq. (3):
20
(9.00) 180
3
P
( a) 240 lbP 
www.elsolucionario.org

1189. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.104
Cable ABC supports two loads as shown. Determine the
distances a and b when a horizontal force
P of magnitude 200 lb
is applied at A.

SOLUTION
Free body: ABC


0: (12ft) (140lb) (180lb) 0
A
MP a b

(1)





Free body: BC



0: (9 ft) (180 lb)( ) 0
B
MP ba
 (2)



Data
200 lbP
Eq. (1): 200(21) 140 180 0ab

(3)
Eq. (2): 200(9) 180 180 0ab

(4)
(3) (4) : 2400 320 0a

7.50 fta 

Eq. (2): (200 lb)(9 ft) (180 lb)( 7.50 ft) 0b

17.50 ftb 
www.elsolucionario.org

1190. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.105
If a 3 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.

SOLUTION
Free body: Portion DE
0: (4 m) (5 m) 0
Dy x
ME E  


5
4
yx
EE

Free body: Portion CDE

5
0: (8 m) (7 m) (2 m) 0
4
Cxx
MEEP   


2
3
x
EP
(1)


Free body: Portion BCDE


5
0: (12m) (5m) (120kN)(4m) 0
4
Bxx
MEE   


10 480 0; 48 kN
xx
EE 

Eq. (1):
2
48 kN
3
P

72.0 kNP 




www.elsolucionario.org

1191. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.105 (Continued)

Free body: Entire cable


5
0: (16m) (2m) (3m) (4m) (120kN)(8m) 0
4
Ax x
MEEPQ     


(48 kN)(20 m 2 m) (72 kN)(3 m) (4 m) 960 kN m 0Q   


4120Q

30.0 kNQ 

www.elsolucionario.org

1192. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.106
If a = 4 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.

SOLUTION
Free body: Portion DE
0: (4 m) (6 m) 0
Dy x
ME E  


3
2
yx
EE

Free body: Portion CDE

3
0: (8 m) (8 m) (2 m) 0
2
Cxx
MEEP   


1
2
x
EP
(1)


Free body: Portion BCDE


3
0: (12m) (6m) (120kN)(4m) 0
2
Bxx
MEE   


12 480 40 kN
xx
EE

Eq (1):
11
;40kN
22
x
EP P

80.0 kNP 




www.elsolucionario.org

1193. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.106 (Continued)

Free body: Entire cable


3
0: (16m) (2m) (4m) (4m) (120kN)(8m) 0
2
Ax x
MEEPQ     


(40kN)(24m 2m) (80kN)(4m) (4m) 960kN m 0Q   


4 240Q

60.0 kNQ 

www.elsolucionario.org

1194. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.107
An electric wire having a mass per unit length of 0.6 kg/m is strung between two insulators at the same
elevation that are 60 m apart. Knowing that the sag of the cable is 1.5 m, determine (a) the maximum
tension in the wire, (b) the length of the wire.

SOLUTION

2
(0.6 kg/m)(9.81 m/s )
5.886 N/m
(5.886 N/m)(30 m)
176.580 N
w
W
W





(a)

0
0: (1.5 m) 176.580 N 15 m 0
B
MT   



0
222
1765.80 N
(176.580 N) (1765.80 N)
m
T
T


1775 N
m
T 
(b)
2
2
2
1
3
21.5m
30 m 1+
330 m
30.05 m
B
BB
B
y
sx
x


  












Length
2 2(30.05 m)
B
s Length 60.1 m 
www.elsolucionario.org

1195. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.108
The total mass of cable ACB is 20 kg. Assuming that the mass of the
cable is distributed uniformly along the horizontal, determine (a) the
sag h, (b) the slope of the cable at A.

SOLUTION
Free body: Entire frame






0: (4.5 m) (196.2 N)(4 m) (1471.5 N)(6 m) 0
Dx
MA   


2136.4 N
x
A

Free body: Entire cable

0: (8 m) (196.2 N)(4 m) 0
Dy
MA  


98.1 N
y
A



(a) Free body: Portion AC


0
0: 2136.4 N
xx
FTA  


0
0: (98.1 N)(2 m) 0
A
MTh
 


(2136.4 N) 196.2 N m 0h
 


0.09183 mh


91.8 mmh 

(b)
98.1 N
tan
2136.4 N
x
A
y
A
A



tan 0.045918
A



2.629
A

 

2.63
A
 

www.elsolucionario.org

1196. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.109
The center span of the George Washington Bridge, as originally constructed, consisted of a uniform
roadway suspended from four cables. The uniform load supported by each cable was
9.75 kips/ftw
along the horizontal. Knowing that the span L is 3500 ft and that the sag h is 316 ft, determine for the
original configuration (a) the maximum tension in each cable, (b) the length of each cable.


SOLUTION

(9.75 kips/ft)(1750 ft)
17,063 kips


B
Wwx
W


0
0
0: (316 ft) (17063 kips)(875 ft)
47,247 kips
B
MT c
T
  

(a)
22
0
22
(47,247 kips) (17,063 kips)
m
TTW


50,200 kips
m
T 
(b)
24
24
22
1
35
316 ft
0.18057
1750 ft
22
(1750 ft) 1+ (0.18057) (0.18057)
35
BB
BB
BB
B
B
yy
sx
xx
y
x

 
   
 










1787.3 ft; Length 2 3579.6 ft
BB
ss Length 3580 ft 
www.elsolucionario.org

1197. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.110
The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four
cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of
the center span to vary from h
w 386 ft in winter to 394 ft
s
h in summer. Knowing that the span is
L  4260 ft, determine the change in length of the cables due to extreme temperature changes.
SOLUTION
Eq. 7.10:
24
22
1
35
BB
BB
BB
yy
sx
xx
 
 
    
  
 


Winter:
1
386 ft, 2130 ft
2
BB
yh x L  

24
2386 2386
(2130) 1 2175.715 ft
3 2130 5 2130
B
s


  




Summer:
1
394 ft, 2130 ft
2
BB
yh x L  

24
2394 2394
(2130) 1 2177.59 ft
3 2130 5 2130
B
s


  





2( ) 2(2177.59 ft 2175.715 ft) 2(1.875 ft)
B
s    
Change in length
3.75 ft 
www.elsolucionario.org

1198. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.111
Each cable of the Golden Gate Bridge supports a load w 11.1 kips/ft along the horizontal. Knowing that
the span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the
length of each cable.

SOLUTION
Eq. (7.8) Page 386:
At B:
2
0
2
B
B
wx
y
T



2 2
0
(11.1 kip/ft)(2075 ft)
22(464ft)
B
B
wx
T
y

(a)
0
51.500 kipsT

(11.1 kips/ft)(2075 ft) 23.033 kips
B
Wwx 

22 2 2
0
(51.500 kips) (23.033 kips)
m
TTW 

56,400 kips
m
T 
(b)
24
22 464ft
1 0.22361
3 5 2075 ft
BB B
BB
BB B
yy y
sx
xy x

 
     
 




2422
(2075 ft) 1 (0.22361) (0.22361) 2142.1 ft
35
B
s

  



Length
2 2(2142.1 ft)
B
s Length 4284 ft 
www.elsolucionario.org

1199. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.112
Two cables of the same gauge are attached to a transmission tower
at B. Since the tower is slender, the horizontal component of the
resultant of the forces exerted by the cables at B is to be zero.
Knowing that the mass per unit length of the cables is 0.4 kg/m,
determine (a) the required sag h, (b) the maximum tension in each
cable.

SOLUTION

B
Wwx

0
0: ( ) 0
2
B
BBB
y
MTywx  

(a) Horiz. comp.
2
0
2
B
B
wx
T
y


Cable AB:
45 m
B
x

2
0
(45 m)
2
w
T
h

Cable BC: 2
0
30 m, 3 m
(30 m)
2(3 m)
BB
xy
w
T



Equate
00
TT
22
(45 m) (30 m)
22(3m)
ww
h

6.75 mh 
(b)
22 2
0
m
TTW
Cable AB:
(0.4 kg/m)(9.81 m/s) 3.924 N/mw
45 m, 6.75 m
BB
xyh

2 2
0
(3.924 N/m)(45 m)
588.6 N
22(6.75m)
B
B
wx
T
y
 

(3.924 N/m)(45 m) 176.58 N
B
Wwx 

22 2
(588.6 N) (176.58 N)
m
T
For AB:
615 N
m
T 




www.elsolucionario.org

1200. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.112 (Continued)

Cable BC:
30 m, 3 m
BB
xy

2 2
0
(3.924 N/m)(30 m)
588.6 N (Checks)
22(3m)
B
B
wx
T
y
 


(3.924 N/m)(30 m) 117.72 N
B
Wwx 

22 2
(588.6 N) (117.72 N)
m
T
For BC:
600 N
m
T 
www.elsolucionario.org

1201. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.113
A 76-m length of wire having a mass per unit length of 2.2 kg/m is used to span a horizontal distance of
75 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use
only the first two terms of Eq. (7.10).]

SOLUTION
First two terms of Eq. 7.10
(a)
2
1
(76 m) 38 m,
2
1
(75m) 37.5m
2
2
1
3
B
B
B
B
BB
B
s
x
yh
y
sx
x





 



2
2
2
38 m 37.5 m 1
3
0.02
0.141421
0.141421
37.5m
B
B
B
B
B
B
y
x
y
x
y
x
h


 









5.3033 mh 5.30 mh 
(b) Free body: Portion CB

(2.2 kg/m)(9.81m) 21.582 N/m
(38 m)(21.582 N/m)
820.12 N
B
w
Wsw
W




  

22
2
0
0
0
2.2 kg/m 9.81 m/s 37.5 m
:
2 2 5.3033 m
2861.6 N
2861.6 N
B
B
B
x
wx
yT
y
T
BT



www.elsolucionario.org

1202. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.113 (Continued)


 0: 820.12 N 0 820.12 N
yy y
FB B   


22 2 2
(2861.6 N) (820.12 N)
mxy
TBB 

2980 N
m
T 

www.elsolucionario.org

1203. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.114
A cable of length L +  is suspended between two points that are at the same elevation and a distance L
apart. (a) Assuming that  is small compared to L and that the cable is parabolic, determine the
approximate sag in terms of L and . (b) If L  100 ft and  4 ft, determine the approximate sag. [Hint:
Use only the first two terms of Eq. (7.10).

SOLUTION
Eq. 7.10 (First two terms)
(a)
2
2
1
3
B
BB
B
y
sx
x
 

  
 
 


/2
1
()
2
B
B
B
xL
sL
yh





2
2
12
()1
223
L
Lh
L
 

    
 

 


2
2
43
;;
23 8
h
hL
L



3
8
hL

(b)
3
100ft, 4ft. (100)(4);
8
Lhh
12.25 fth 
www.elsolucionario.org

1204. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.115
The total mass of cable AC is 25 kg. Assuming that the mass of the
cable is distributed uniformly along the horizontal, determine the
sag h and the slope of the cable at A and C.

SOLUTION
Cable: 25 kg
25 (9.81)
245.25 N
m
W



Block: 450 kg
4414.5 N
m
W


0: (245.25)(2.5) (4414.5)(3) (2.5) 0
5543 N
Bx
x
MC
C
   


0: 5543 N
xxx
FAC  

0: (5) (5543)(2.5) (245.25)(2.5) 0
Ay
MC   

2894 N
y
C

 0: 245.25 N 0
yyy
FCA   


2894 245.25 N 0
y
NA 

2649 N
y
A

Point A:
2649
tan 0.4779;
5543y
A
x
A
A
 

25.5
A
 

Point C:
2894
tan 0.5221;
5543y
C
x
C
C
 

27.6
A
 

Free body: Half cable
(12.5 kg) 122.6Wg
0
0: (122.6 N)(1.25 M) + (2649 N)(2.5 m) (5543 N) 0
1.2224 m; sag 1.25 m 1.2224 m
  

d
d
My
yh

0.0276 m 27.6 mmh 
www.elsolucionario.org

1205. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.116
Cable ACB supports a load uniformly distributed along the
horizontal as shown. The lowest Point C is located 9 m to the
right of A. Determine (a) the vertical distance a, (b) the length
of the cable, (c) the components of the reaction at A.

SOLUTION
Free body: Portion AC
0: 9 0
yy
FAw  

9
y
wA

0
0: (9 )(4.5 m) 0
A
MTaw   (1)
Free body: Portion CB


0: 6 0
yy
FBw  

6
y
wB


Free body: Entire cable


0
0: 15 (7.5 m) 6 (15 m) (2.25 m) 0
A
Mw wT   
(a)
0
10Tw
Eq. (1):
0
(9 )(4.5 m) 0Ta w

10 (9 )(4.5) 0wa w 4.05 ma 





www.elsolucionario.org

1206. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.116 (Continued)

(b) Length ACCB
Portion AC:
24
24
4.05
9 m, 4.05 m; 0.45
9
22
1
35
22
9 m 1+ 0.45 0.45 10.067 m
35
A
AA
A
AB
AC B
AA
AC
y
xya
x
yy
sx
xx
s
 

 
   
 








Portion CB:
24
6 m, 4.05 2.25 1.8 m; 0.3
22
6 m 1 0.3 0.3 6.341 m
35
B
BB
B
CB
y
xy
x
s
  






Total length 10.067 m 6.341 m Total length 16.41 m 
(c) Components of reaction at A.

2
2
0
9 9(60 kg/m)(9.81 m/s )
5297.4 N
10 10(60 kg/m)(9.81 m/s )
5886 N
y
x
Aw
AT w


 

5890 N
x
A 

5300 N
y
A 
www.elsolucionario.org

1207. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.117
Each cable of the side spans of the Golden Gate Bridge
supports a load w 10.2 kips/ft along the horizontal.
Knowing that for the side spans the maximum vertical
distance h from each cable to the chord AB is 30 ft and
occurs at midspan, determine (a) the maximum tension in
each cable, (b) the slope at B.

SOLUTION
FBD AB:
0: (1100 ft) (496 ft) (550 ft) 0
AByBx
MTTW    
11 4.96 5.5
By Bx
TTW (1)


FBD CB:


0: (550 ft) (278 ft) (275 ft) 0
2
CByBx
W
MTT   


11 5.56 2.75
By Bx
TTW (2)


Solving (1) and (2)
28,798 kips
By
T
Solving (1) and (2
51,425 kips
Bx
T 

22
max
tan
y
xy
x
B
BBB B
B
T
TTTT
T
   
So that ( a)
max
58,900 kipsT 
( b) 29.2
B
 
www.elsolucionario.org

1208. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.118
A steam pipe weighting 45 lb/ft that passes between two
buildings 40 ft apart is supported by a system of cables as
shown. Assuming that the weight of the cable system is
equivalent to a uniformly distributed loading of 5 lb/ft,
determine (a) the location of the lowest Point C of the
cable, (b) the maximum tension in the cable.

SOLUTION
Note:
40 ft
BA
xx

or
40 ft
AB
xx

(a) Use Eq. 7.8
Point A:
22
00
(40)
;9
22
AB
A
wx w x
y
TT


(1)
Point B:
22
00
;4
22
BB
B
wx wx
y
TT

(2)
Dividing (1) by (2):
2
2
( 40)9
;16ft
4
B
B
B
x
x
x



Point C is 16.00 ft to left of B

(b) Maximum slope and thus
max
Tis at A

40 16 40 24 ft
AB
xx

2 2
0
00
(50 lb/ft)( 24 ft)
; 9 ft ; 1600 lb
22
A
A
wx
yT
TT

 


(50 lb/ft)(24 ft) 1200 lb
AC
W


max
2000 lbT 
www.elsolucionario.org

1209. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.119*
A cable AB of span L and a simple beam ABof the same span are
subjected to identical vertical loadings as shown. Show that the
magnitude of the bending moment at a point C in the beam is equal
to the product T
0h, where T 0 is the magnitude of the horizontal
component of the tension force in the cable and h is the vertical
distance between Point C and the chord joining the points of support
A and B.

SOLUTION

0 loads
0: 0
BCy B
MLAaTM   (1)

FBD Cable:
(Where
loadsB
M includes all applied loads)

0 left
0: 0
CCy C
x
MxAhaTM
L

   


(2)

FBD AC:
(Where
leftC
M includes all loads left of C)

0 loads left
(1) (2) : 0
BC
xx
hT M M
LL

(3)
FBD Beam:

loads
0: 0
BByB
MLAM   (4)

left
0: 0
CByCC
MxAMM    (5)

FBD AC:


loads left
(4) (5): 0
BCC
xx
MMM
LL
  
(6)
Comparing (3) and (6)
0C
MhT Q.E.D.
www.elsolucionario.org

1210. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.120
Making use of the property established in Problem 7.119,
solve the problem indicated by first solving the
corresponding beam problem.
PROBLEM 7.94 Knowing that the maximum tension in
cable ABCDE is 25 kN, determine the distance d
C.

SOLUTION
Free body: beam AE

0: (20) 2(15) 4(10) 6(5) 0
E
MA     

5kNA


0:5246 0
y
FE 

7kNE




At E:
222 222
000
25 7 24 kN
m
TTE T T  
At C:
0
;40kNm(24kN)
CC C
MTh d
1.667 m
C
d 1.667 m
www.elsolucionario.org

1211. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.121
Making use of the property established in Problem 7.119, solve the
problem indicated by first solving the corresponding beam problem.
PROBLEM 7.97 (a) Knowing that 3m,
C
d determine the distances
d
Band
D
d.

SOLUTION


0: (10 m) (5 kN)(8 m) (5 kN)(6 m) (10 kN)(3 m) 0
B
MA    


10 kNA
Geometry:
1.6 m
3m 1.6m
1.4 m
CC
C
C
dh
h
h




Since MT
0h, h is proportional to M, thus

1.4 m
;
20 kN m 30 kN m 30 kN m
CBDB D
BC D
hhhh h
MMM
  



20
1.4 0.9333 m
30
B
h




30
1.4 1.4 m
30
D
h





0.8 m 0.9333 m
B
d 2.8 m 1.4 m
D
d

1.733 m
B
d  4.20 m
D
d 
www.elsolucionario.org

1212. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.122
Making use of the property established in Problem
7.119, solve the problem indicated by first solving the
corresponding beam problem.
PROBLEM 7.99 (a) If d C 15 ft, determine the
distances d
B and d D.

SOLUTION
Free body: Beam AE
0: (30) 2(24) 2(15) 2(9) 0
E
MA     

3.2 kipsA

0: 3.2 3(2) 0
y
FB    

2.8 kipsB






Geometry:
Given: 15 ft
C
d
Then,
3.75 ft 11.25 ft
CC
hd 
Since MT
0h, h is proportional to M. Thus,

11.25 ft
or,
19.2 30 25.2
CBD
BC D
BD
hhh
MMM
hh




or, 7.2 ft 9.45 ft
BD
hh
Then,
667.2
BB
dh  13.20 ft
B
d 

2.25 2.25 9.45
DD
dh 11.70 ft
D
d 

www.elsolucionario.org

1213. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.123
Making use of the property established in Problem
7.119, solve the problem indicated by first solving the
corresponding beam problem.
PROBLEM 7.100 (a) Determine the distance d C for
which portion BC of the cable is horizontal.


SOLUTION
Free body: Beam AE
0: (30) 2(24) 2(15) 2(9) 0
E
MA     

3.2 kipsA

0: 3.2 3(2) 0
y
FB  

2.8 kipsB






Geometry:
Given:
CB
dd
Then,
3.75 6
CB
hh

2.25
CB
hh (1)
Since MT
0h, h is proportional to M. Thus,

or,
19.2 30
CCBB
BC
hhhh
MM



0.64
BC
hh (2)
Substituting (2) into (1):

2.25 0.64 6.25 ft
CCC
hhh 
Then,
3.75 3.75 6.25
CC
dh 10.00 ft
C
d 

www.elsolucionario.org

1214. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.124*
Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential
equation d
2
y/dx
2
w(x)/T 0, where T 0is the tension at the lowest point.

SOLUTION
FBD Elemental segment:

0: ( ) ( ) ( ) 0
yy y
FTxxTxwxx   
So
000
()() ()
yy
Tx x Tx wx
x
TTT



But 0
y
Tdy
Tdx


So
0
()
xx x
dy dy
dx dx wx
xT





In
0
lim :
x

2
2
0
()dy wx
Tdx

Q.E.D.

www.elsolucionario.org

1215. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.125*
Using the property indicated in Problem 7.124, determine the curve assumed by a cable of span L and sag
h carrying a distributed load ww
0 cos (x/L), where x is measured from mid-span. Also determine the
maximum and minimum values of the tension in the cable.
PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined
by the differential equation d
2
y/dx
2
w(x)/T 0, where T 0is the tension at the lowest point.

SOLUTION


0
() cos
x
wx w
L



From Problem 7.124

2
0
2
00
()
cos
wdy wx x
TT Ldx 

So
0
0 0
sin using 0
WLdy x dy
dx T L dx


 



2
0
2
0
1 cos [using (0) 0]
wL x
yy
LT


 



But
22
00
022
0
1cos so
22
wL wLL
yh T
Th

  
  
  
  

And
0min
TT so
2
0
min 2
wL
T
h

 

0
max
00 /2
:
By
AB
xL
T wLdy
TTT
Tdx T


  

0
By
wL
T


2
22 0
max 0
1
By
wL L
TTT
h


 




www.elsolucionario.org

1216. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 7.126*
If the weight per unit length of the cable AB is w 0/cos
2
, prove that the
curve formed by the cable is a circular arc. (Hint: Use the property
indicated in Problem 7.124.)
PROBLEM 7.124 Show that the curve assumed by a cable that carries a
distributed load w(x) is defined by the differential equation
d
2
y/dx
2
w(x)/T 0, where T 0is the tension at the lowest point.

SOLUTION
Elemental Segment:
Load on segment*
0
2
()
cos
w
w x dx ds


But
0
3
cos , so ( )
cos
w
dx ds w x

From Problem 7.119
2
0
23
0 0
()
cos
wdy wx
Tdx T


In general
2
2
2
(tan ) sec
dy d dy d d
dx dx dx dxdx


 



So
00
32
00
coscos sec
wwd
dx TT


or
0
0
cos cos
T
ddxrd
w  
Giving
0
0
constant.
T
r
w

So curve is circular arc Q.E.D.
*For large sag, it is not appropriate to approximate ds by dx.

www.elsolucionario.org

1217. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.127
A 25-ft chain with a weight of 30 lb is suspended between two points at the same elevation. Knowing that
the sag is 10 ft, determine (a) the distance between the supports, (b) the maximum tension in the chain.

SOLUTION
(30 lb)/(25 ft) 1.2 lb/ftw

Eq. 7.17:

222 2 2 2
22
; (10 ft ) 12.5 ft
100 20 156.25
2.8125 ft
BB
ysc c c
cc c
c
   
 

Eq. 7.18: (1.2 lb/ft )(10 ft 2.8125 ft )
mB
Twy 
15.375 lb
m
T 15.38 lb
m
T 
Eq. 7.15: sinh ; 12.5 (2.8125 ft)sinh
sinh 4.444; 2.1972
BB
B
BB
xx
sc
cc
xx
cc



2.1972(2.8125 ft) 6.1796 ft; 2 2(6.1796 ft) 12.3592 ft
BB
xLx

12.36 ftL 
www.elsolucionario.org

1218. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.128
A 500-ft-long aerial tramway cable having a weight per unit length of 2.8 lb/ft is suspended between two
points at the same elevation. Knowing that the sag is 125 ft, find (a) the horizontal distance between the
supports, (b) the maximum tension in the cable.

SOLUTION

Given:
Length 500 ft
Unit mass 2.8 lb/ft
125 fth



Then,
22 2 2 22
2222
250 ft
125 ft
; (125 ) (250)
125 250 250
187.50 ft
B
B
BB
s
yhc c
ysc c c
cc c
c

 
   



sinh ; 250 187.50sinh
187.50
BB
B
xx
sc
c



206.06 ft
B
x

span 2 2(206.06 ft)
B
Lx  412 ftL 

(2.8 lb/ft)(125 ft 187.50 ft)
mB
Twy  875 lb
m
T 

www.elsolucionario.org

1219. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.129
A 40-m cable is strung as shown between two buildings. The
maximum tension is found to be 350 N, and the lowest point
of the cable is observed to be 6 m above the ground.
Determine (a) the horizontal distance between the buildings,
(b) the total mass of the cable.

SOLUTION


20 m
350 N
14 m 6 m 8 m
8m
B
m
B
s
T
h
yhc c



 

Eq. 7.17:
22 2 2 22
22
;(8 ) 20
64 16 400
21.0 m
BB
ysc c c
cc c
c
   
 

Eq. 7.15: sinh ; 20 (21.0)sinh
21.0
BB
B
x x
sc
c


(a) 17.7933 m; 2
BB
x Lx 35.6 mL 
(b) Eq. 7.18:
; 350 N (8 21.0)
12.0690 N/m
mB
Twy w
w



2
2 (40 m)(12.0690 N/m)
482.76 N
482.76 N
9.81 m/s
B
Wsw
W
m
g


 Total mass
49.2 kg 
www.elsolucionario.org

1220. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.130
A 50-m steel surveying tape has a mass of 1.6 kg. If the tape is stretched between two points at the same
elevation and pulled until the tension at each end is 60 N, determine the horizontal distance between the
ends of the tape. Neglect the elongation of the tape due to the tension.

SOLUTION


25 m
B
s


21.6 kg
9.81 m/s 0.31392 N/m 60 N
50 m
m
wT




Eq. 7.18:
; 60 N (0.31392 N/m) ; 191.131 m
mB BB
Twy y y 
Eq. 7.17:
22 2 2 22
; (191.131) (25) ; 189.489 m
BB
ysc cc   
Eq. 7.15: sinh ; 25 189.489 sinh
BB
B
xx
sc
cc


sinh 0.131933; 0.131553 m
BB
xx
cc


 0.131553 189.489 m 24.928 m
B
x

2 2(24.928 ft)
B
Lx 49.86 ftL 
www.elsolucionario.org

1221. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.131
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the force
P for which h 8 m, (b) the
corresponding span L.


SOLUTION
FBD Cable:


2
2222
20 m
20 m so 10 m
2
(0.2 kg/m)(9.81 m/s )
1.96200 N/m
8m
()
TB
B
BB B
ss
w
h
ych cs







  

So
22
22
2
(10 m) (8 m)
2(8 m)
2.250 m
BB
B
sh
c
h
c






Now
1
1
sinh sinh
10 m
(2.250 m)sinh
2.250 m
BB
BB
xs
sc xc
cc




 


4.9438 m
B
x

0
(1.96200 N/m)(2.250 m)PT wc  ( a) 4.41 NP 
22(4.9438m)
B
Lx ( b) 9.89 mL 
www.elsolucionario.org

1222. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.132
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Knowing that
the magnitude of the horizontal force applied to the collar is
P  20 N, determine (a) the sag h, (b) the span L.

SOLUTION
FBD Cable:


2
0
2222
22
22
20 m, (0.2 kg/m)(9.81 m/s ) 1.96200 N/m
20 N
10.1937 m
1.9620 N/m
()
20 m
20 10m
2
2(10.1937 m) 100 m 0
T
BB B
BB
sw
P
PT wcc
w
c
yhccs
hchs s
hh
 
 


  



4.0861 mh ( a) 4.09 mh 

11 10 m
sinh sinh (10.1937 m)sinh
10.1937 m
AB
BB
xs
sc xc
cc
 
 




8.8468 m

2 2(8.8468 m)
B
Lx ( b) 17.69 mL 
www.elsolucionario.org

1223. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.133
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the sag h for which L 15 m, (b) the
corresponding force
P.

SOLUTION
FBD Cable:


2
2
20 m
20 m 10 m
2
(0.2 kg/m)(9.81 m/s ) 1.96200 N/m
15 m
sinh sinh
7.5 m
10 m sinh
TB
L
B
B
ss
w
L
x
sc c
cc
c
c
 





Solving numerically:
5.5504 mc

7.5
cosh (5.5504)cosh
5.5504
11.4371 m
11.4371 m 5.5504 m
B
B
B
BB
x
yc
c
y
hyc
 




 

(a)
5.89 m
B
h 

(1.96200 N/m)(5.5504 m)Pwc ( b) 10.89 NP 
www.elsolucionario.org

1224. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.134
Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.

SOLUTION


30 ft
15 ft 20 ft
2
10 ft
2
sinh
10 ft
15 ft sinh
B
B
B
B
sL
L
x
x
sc
c
c
c
 




Solving numerically:
6.1647 ftc

cosh
10 ft
(6.1647 ft)cosh
6.1647 ft
16.2174 ft
16.2174 ft 6.1647 ft
B
B
BB
x
yc
c
hyc



 
10.05 ft
B
h 
www.elsolucionario.org

1225. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.135
A counterweight D is attached to a cable that passes over a small pulley at
A and is attached to a support at B. Knowing that L 45 ft and h 15 ft,
determine (a) the length of the cable from A to B, (b) the weight per unit
length of the cable. Neglect the weight of the cable from A to D.

SOLUTION
Given: 45 ft
15 ft
80 lb
22.5 ft
A
B
L
h
T
x




By symmetry: 80 lb
BAm
TTT
We have
22.5
cosh cosh
B
B
x
yc c
cc


and
15
B
yhc c 
Then
22.5
cosh 15cc
c


or
22.5 15
cosh 1
cc


Solve by trial for c:
18.9525 ftc
(a)
sinh
22.5
(18.9525 ft)sinh
18.9525
28.170 ft
B
B
x
sc
c




Length
2 2(28.170 ft) 56.3 ft
B
s  
(b)
()
mB
Twywhc

80 lb (15 ft 18.9525 ft)w 2.36 lb/ftw 

www.elsolucionario.org

1226. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.136
A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the
maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.


SOLUTION


45 m
B
s
Eq. 7.17:
sinh
30
45 sinh ; 18.494 m
B
B
x
sc
c
cc
c



Eq. 7.16: cosh
30
(18.494)cosh
18.494
48.652 m
48.652 18.494
B
B
B
B
B
x
yc
c
y
y
yhc
h






30.158 mh 30.2 mh 
Eq. 7.18:
300 N (48.652 m)
6.166 N/m
mB
Twy
w
w



Total weight of cable (Length)
(6.166 N/m)(90 m)
554.96 N
Ww


Total mass of cable
554.96 N
56.57 kg
9.81 m/s
W
m
g
 
56.6 kgm 
www.elsolucionario.org

1227. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.137
A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart.
Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.

SOLUTION

Eq. 7.18:
; 400 lb (2 lb/ft) ; 200 ft
mB BB
Twy y y 
Eq. 7.16: cosh
80 ft
200 ft cosh
B
B
x
yc
c
c
c



Solve for c: 182.148 ftc and 31.592 ftc

;
BB
yhchyc  
For
182.148 ft;c 200 182.147 17.852 fth  
For
31.592 ft;c 200 31.592 168.408 fth  
For
400 lb:
m
T smallest 17.85 fth 
www.elsolucionario.org

1228. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.138
A uniform cord 50 in. long passes over a pulley at B and is attached to a
pin support at A. Knowing that L 20 in. and neglecting the effect of
friction, determine the smaller of the two values of h for which the cord is
in equilibrium.

SOLUTION

Length of overhang:
50 in. 2
B
bs
Weight of overhang equals max. tension

(50 in. 2 )
mB B
TTwbw s  
Eq. 7.15:
sinh
B
B
x
sc
c


Eq. 7.16: cosh
B
B
x
yc
c


Eq. 7.18:
(50 in. 2 )
mB
BB
Twy
wswy


50 in. 2 sinh cosh
BB
xx
wc wc
cc






10 10
10: 50 2 sinh cosh
B
xcc
cc
 
Solve by trial and error:
5.549 in.c and 27.742 in.c
For c 5.549 in.
10 in.
(5.549 in.)cosh 17.277 in.
5.549 in.
; 17.277 in. 5.549 in.
B
B
y
yhc h

 
11.728 in.h 11.73 in.h 
For c 27.742 in.
10 in.
(27.742 in.)cosh 29.564 in.
27.742 in.
; 29.564 in. 27.742 in.
B
B
y
yhc h

 
1.8219 in.h 1.822 in.h 
www.elsolucionario.org

1229. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.139
A motor M is used to slowly reel in the cable shown. Knowing that the
mass per unit length of the cable is 0.4 kg/m, determine the maximum
tension in the cable when h 5 m.

SOLUTION


0.4 kg/m 10 m 5 m
cosh
cosh
2
5m
5m cosh 1
B
B
B
B
wLh
x
yc
c
L
hcc
c
c
c








Solving numerically:
max
2
3.0938 m
5 m 3.0938 m
8.0938 m
(0.4 kg/m)(9.81 m/s )(8.0938 m)
BB
BB
c
yhc
TTwy

 





max
31.8 NT 

www.elsolucionario.org

1230. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.140
A motor M is used to slowly reel in the cable shown. Knowing that the
mass per unit length of the cable is 0.4 kg/m, determine the maximum
tension in the cable when h 3 m.

SOLUTION


0.4 kg/m, 10 m, 3 m
cosh cosh
2
5m
3m cosh 1
B
B
BB
wLh
x L
yhcc c
cc
cc
c

 





Solving numerically:
max
2
4.5945 m
3 m 4.5945 m
7.5945 m
(0.4 kg/m)(9.81 m/s )(7.5945 m)
BB
BB
c
yhc
TTwy

 





max
29.8 NT 
www.elsolucionario.org

1231. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.141
The cable ACB has a mass per unit length of 0.45 kg/m.
Knowing that the lowest point of the cable is located at a
distance a 0.6 m below the support A, determine (a) the
location of the lowest Point C, (b) the maximum tension in
the cable.

SOLUTION
Note: 12 m
BA
xx
or,
12 m
AB
xx 
Point A:
12
cosh ; 0.6 cosh
AB
A
xx
yc c c
cc


(1)
Point B: cosh ; 2.4 cosh
BB
B
xx
yc c c
cc

(2)
From (1):
112 0.6
cosh
B
x c
cc c




(3)
From (2):
1 2.4
cosh
B
x c
cc




(4)
Add (3) + (4):
1112 0.6 2.4
cosh coshcc
cc c
 

 
 

Solve by trial and error:
13.6214 mc
Eq. (2):
13.6214 2.4 13.6214cosh
B
x
c



cosh 1.1762; 0.58523
BB
xx
cc



0.58523(13.6214 m) 7.9717 m
B
x
Point C is 7.97 m to left of B


2.4 13.6214 2.4 16.0214 m
B
yc   
Eq. 7.18:
2
(0.45 kg/m)(9.81 m/s )(16.0214 m)
mB
Twy
70.726 N
m
T 70.7 N
m
T 
www.elsolucionario.org

1232. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.142
The cable ACB has a mass per unit length of 0.45 kg/m.
Knowing that the lowest point of the cable is located at a
distance a 2 m below the support A, determine (a) the
location of the lowest Point C, (b) the maximum tension in
the cable.

SOLUTION
Note:

12 m
BA
xx
or

12 m
AB
xx 
Point A:
12
cosh ; 2 cosh
AB
A
xx
yc c c
cc


(1)
Point B: cosh ; 3.8 cosh
BB
B
xx
yc c c
cc

(2)
From (1):
112 2
cosh
B
x c
cc c




(3)
From (2):
1 3.8
cosh
B
x c
cc




(4)
Add (3)  (4):
1112 2 3.8
cosh coshcc
cc c
  

  
  

Solve by trial and error:
6.8154 mc
Eq. (2):
6.8154 m 3.8 m (6.8154 m)cosh
cosh 1.5576 1.0122
1.0122(6.8154 m) 6.899 m
B
BB
B
x
c
xx
cc
x




Point C is 6.90 m to left of B


3.8 6.8154 3.8 10.6154 m
B
yc   
Eq. (7.18):
2
(0.45 kg/m)(9.81 m/s )(10.6154 m)
mB
Twy

46.86 N
m
T 46.9 N
m
T 
www.elsolucionario.org

1233. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.143
A uniform cable weighing 3 lb/ft is held in the position shown by a
horizontal force
P applied at B. Knowing that P180lband A60,
determine (a) the location of Point B, (b) the length of the cable.

SOLUTION
Eq. 7.18:
0
180 lb
60 ft
3lb/ft
TPcw
P
cc
w

 

At A:
cos60
2
0.5
m
P
T
cw
cw




(a) Eq. 7.18: ()
2()
m
Twhc
cw w h c


2chchbc  60.0 ftb 
Eq. 7.16: cosh
cosh
(60ft 60ft) (60ft) cosh
60
cosh 2 1.3170
60 m 60 m
A
A
A
A
AA
x
yc
c
x
hcc
c
x
xx





79.02 ft
A
x 79.0 fta 
(b) Eq. 7.15:
79.02 ft
sinh (60 ft)sinh
60 ft
103.92 ft
B
A
A
x
sc
c
s



length
A
s 103.9 ft
A
s 
www.elsolucionario.org

1234. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.144
A uniform cable weighing 3 lb/ft is held in the position shown by a
horizontal force
P applied at B. Knowing that P 150lband A60,
determine (a) the location of Point B, (b) the length of the cable.

SOLUTION
Eq. 7.18:
0
150 lb
50 ft
3lb/ft
TPcw
P
c
w

 

At A:
cos60
2
0.5
m
P
T
cw
cw




(a) Eq. 7.18: ()
2()
m
Twhc
cw w h c


2chchcb  50.0 ftb 
Eq. 7.16: cosh
cosh
(50ft 50ft) (50ft)cosh
cosh 2 1.3170
A
A
A
A
AA
x
yc
c
x
hcc
c
x
c
xx
cc





1.3170(50 ft) 65.85 ft
A
x 65.8 fta 
(b) Eq. 7.15:
65.85 ft
sinh (50 ft)sinh
50 ft
A
A
x
sc
c

86.6 ft
A
s length 86.6 ft
A
s 
www.elsolucionario.org

1235. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.145
To the left of Point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the mass per unit length
of the cable is 2 kg/m, determine the force
F when a 3.6 m.

SOLUTION


3.6 m 4 m
cosh
cosh
3.6 m
4m cosh 1
D
D
D
xa h
x
yc
c
a
hcc
c
c
c
 







Solving numerically:
2.0712 mc
Then
4 m 2.0712 m 6.0712 m
B
yhc  

2
max
(2 kg/m)(9.81 m/s )(6.0712 m)
B
FT wy 119.1 NF 
www.elsolucionario.org

1236. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.146
To the left of Point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the mass per unit length
of the cable is 2 kg/m, determine the force
F when a 6 m.

SOLUTION


6m 4m
ycosh
cosh
6m
4m cosh 1
D
D
D
xa h
x
c
c
a
hcc
c
c
c
 







Solving numerically:
5.054 mc

2
4 m 5.054 m 9.054 m
(2 kg/m)(9.81 m/s )(9.054 m)
B
DD
yhc
FT wy
  
 
177.6 NF 
www.elsolucionario.org

1237. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.147*
The 10-ft cable AB is attached to two collars as shown. The
collar at A can slide freely along the rod; a stop attached to the
rod prevents the collar at B from moving on the rod. Neglecting
the effect of friction and the weight of the collars, determine the
distance a.


SOLUTION
Collar at A: Since 0, cable ' rod

Point A: cosh ; sinh
tan sinh
A
A
xdy x
yc
cdx c
xdy
cdx




sinh(tan (90 ))
sinh(tan (90 ))
A
A
x
c
xc



(1)
Length of cable 10 ft
10 ft
10 sinh sinh
10
sinh sinh
AB
BA
AC CB
xx
cc
cc
xx
cc c




110
sinh sinh
A
B
x
xc
cc




(2)

cosh cosh
AB
AB
xx
yc yc
cc

(3)
In ABD:
tan
BA
BA
yy
xx




(4) www.elsolucionario.org

1238. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.147* (Continued)

Method of solution:
For given value of , choose trial value of c and calculate:
From Eq. (1): x
A
Using value of x A and c, calculate:
From Eq. (2): x
B
From Eq. (3): y A and y B
Substitute values obtained for x
A, xB, yA, yBinto Eq. (4) and calculate 
Choose new trial value of
 and repeat above procedure until calculated value of  is equal to given value
of
.
For
30

Result of trial and error procedure:

1.803 ft
2.3745 ft
3.6937 ft
3.606 ft
7.109 ft
7.109 ft 3.606 ft
3.503 ft
A
B
A
B
BA
c
x
x
y
y
ay y








3.50 fta 
www.elsolucionario.org

1239. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.148*
Solve Problem 7.147 assuming that the angle  formed by the
rod and the horizontal is 45.
PROBLEM 7.147 The 10-ft cable AB is attached to two collars
as shown. The collar at A can slide freely along the rod; a stop
attached to the rod prevents the collar at B from moving on the
rod. Neglecting the effect of friction and the weight of the
collars, determine the distance a.


SOLUTION
Collar at A: Since 0, cable ' rod

Point A: cosh ; sinh
tan sinh
A
A
xdy x
yc
cdx c
xdy
cdx




sinh(tan (90 ))
sinh(tan (90 ))
A
A
x
c
xc


 (1)
Length of cable 10 ft
10 ft
10 sinh sinh
10
sinh sinh
AB
BA
AC CB
xx
cc
cc
xx
cc c





110
sinh sinh
A
B
x
xc
cc




(2)

cosh cosh
AB
AB
xx
yc yc
cc

(3)
www.elsolucionario.org

1240. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.148* (Continued)

In ABD: tan
BA
BA
yyxx



 (4)
Method of solution:
For given value of , choose trial value of c and calculate:
From Eq. (1): x
A
Using value of x A and c, calculate:
From Eq. (2): x
B
From Eq. (3): y A and y B
Substitute values obtained for x
A, xB, yA, yBinto Eq. (4) and calculate 
Choose new trial value of
 and repeat above procedure until calculated value of  is equal to given value
of
.
For
45

Result of trial and error procedure:

1.8652 ft
1.644 ft
4.064 ft
2.638 ft
8.346 ft
8.346 ft 2.638 ft
5.708 ft
A
B
A
B
BA
c
x
x
y
y
ay y

  




5.71fta 
www.elsolucionario.org

1241. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.149
Denoting by  the angle formed by a uniform cable and the horizontal, show that at any point (a) s  ctan
, (b) yc sec .

SOLUTION
(a) tan sinh
sinh tan Q.E.D.
dy x
dx c
x
sc c
c


(b) Also
222 2 2
(cosh sinh 1)ysc x x  
so
22 2 22
(tan 1) secyc c
and
sec Q.E.D.yc
www.elsolucionario.org

1242. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.150*
(a) Determine the maximum allowable horizontal span for a uniform cable of weight per unit length w if
the tension in the cable is not to exceed a given value T
m. (b) Using the result of part a, determine the
maximum span of a steel wire for which w 0.25 lb/ft and T
m 8000 lb.
SOLUTION

(a)
cosh
1
cosh




 



mB
B
B
B
B
Twy
x
wc
c
x
wx
x c
c
We shall find ratio

B
x
c
for when T m is minimum

2
11
sinh cosh 0
sinh
1
cosh
tanh
m BB
B
BBB
B
BB
B
B
dT xx
wx
xxx cc
d
ccc
x
c
xx
cc
xc
cx




 
 
 
 



Solve by trial and error for:
1.200
B
x
c

(1)

sinh sinh(1.200):
B
B
x
sc c
c


1.509
B
s
c


Eq. 7.17:
22 2
2
22 2 2
1 (1 1.509 )
1.810



 




BB
B
B
B
ysc
s
yc c
c
yc
www.elsolucionario.org

1243. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.150* (Continued)

Eq. 7.18:
1.810
1.810
mB
m
Twy
wc
T
c
w




Eq. (1): 1.200 1.200 0.6630
1.810
mm
B
TT
xc
ww
 

Span: 2 2(0.6630)
m
B
T
Lx
w


1.326
m
T
L
w


(b) For
0.25lb/ftw and 8000 lb
m
T

8000 lb
1.326
0.25lb/ft
42,432ft
L

8.04L miles 
www.elsolucionario.org

1244. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.151*
A cable has a mass per unit length of 3 kg/m and is
supported as shown. Knowing that the span L is 6 m,
determine the two values of the sag h for which the
maximum tension is 350 N.

SOLUTION


max
2
max max
max
max
max
cosh
2
(3 kg/m)(9.81 m/s ) 29.43 N/m
350 N
11.893 m
29.43 N/m
3m
cosh 11.893 m
L
yc hc
c
w
Twy
T
y
w
y
c
c







Solving numerically:
1
2
max
0.9241 m
11.499 m
c
c
hy c




1
11.893 m 0.9241 mh
1
10.97 mh 

2
11.893 m 11.499 mh
2
0.394 mh 
www.elsolucionario.org

1245. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.152*
Determine the sag-to-span ratio for which the maximum
tension in the cable is equal to the total weight of the entire
cable AB.


SOLUTION



max
1
2
2
2
cosh 2 sinh
22
1
tanh
22
1
tanh 0.549306
22
cosh 1 0.154701
2
0.5(0.154701)
0.14081
0.5493062
B
BB
BB
BB
h
cB
L
c
Twyws
ys
LL
cc
cc
L
c
L
c
hyc L
cc c
h
L







 
 
0.1408
B
h
L


www.elsolucionario.org

1246. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.153*
A cable of weight per unit length w is suspended between
two points at the same elevation that are a distance L apart.
Determine (a) the sag-to-span ratio for which the maximum
tension is as small as possible, (b) the corresponding values
of
Band T m.

SOLUTION

(a)
max
max
cosh
2
cosh sinh
22 2
B
L
Twywc
c
dT LL L
w
dc c c c






For
max
max
min , 0
dT
T
dc


2
tanh
2
Lc
cL

→
1.1997
2
L
c


cosh 1.8102
2
1 0.8102
1 2 0.8102
0.3375
22(1.1997)
B
B
y L
cc
yh
cc
hhc
LcL







0.338
h
L

(b)
max
0max
0
cosh cosh
22
B
T yLL
TwcT wc
cT cc
 

But
max
0max
0
cos sec
BB
T
TT
T

So
11
sec sec (1.8102) 56.46
B
B
y
c


 


56.5
B
 

max
2
(1.8102)
2 2(1.1997)
B
B
ycL L
Twyw w
cL

 



max
0.755TwL 
www.elsolucionario.org

1247. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.154
Knowing that the turnbuckle has been tightened until the tension in wire
AD is 850 N, determine the internal forces at point indicated:
Point J.

SOLUTION



160
0: (850 N) 0
340
x
FV

  



400 NV 400 NV 

300
0: (850 N) 0
340
y
FF

  



750 NF 750 NF 


300 160
0: (850 N)(120 mm) (850 N)(100 mm) 0
340 340
J
MM
 
   
 
 


130 N mM 

130 N mM 

www.elsolucionario.org

1248. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.155
Knowing that the turnbuckle has been tightened until the tension in wire
AD is 850 N, determine the internal forces at point indicated:
Point K.

SOLUTION
Free body AK:




22
160 300
340 mm
AD




On portion KBA:

160
0: (850 N) 0
340
x
FV

  



400 NV 400 NV 

300
0: (850 N) 0
340
y
FF

  



750 NF 750 NF 


300 160
0: (850 N)(120 mm) (850 N)(200 mm) 0
340 340
J
MM
 
   
 
 


170 N mM 

170.0 N mM 

Internal forces acting on KCD are equal and opposite

750 NF , 400 NV , 170.0 N mM 

www.elsolucionario.org

1249. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.156
Two members, each consisting of a straight and a
quarter- circular portion of rod, are connected as
shown and support a 75-lb load at A. Determine
the internal forces at Point J.

SOLUTION









Free body: Entire frame
0: (75lb)(12in.) (9in.) 0
C
MF  

100 lbF 

0: 0
xx
FC 

0: 75 lb 100 lb 0
yy
FC   

175 lb
y
C 175 lbC 
Free body: Member BEDF
0: (12 in.) (100 lb)(15 in.) 0
B
MD  

125 lbD 

0: 0
xx
FB 

0: 125 lb 100 lb 0
yy
FB   

25 lb
y
B 25 lbB 
Free body: BJ
0: (25 lb)sin 30 0
x
FF  

12.50 lbF 30.0°

0: (25 lb)cos30 0
y
FV  

21.7 lbV 60.0° 

0: (25 lb)(3 in.) 0
J
MM  

75.0 lb in.M 
www.elsolucionario.org

1250. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.157
Knowing that the radius of each pulley is 150 mm, that
 20, and neglecting friction, determine the internal
forces at (a) Point J, (b) Point K.

SOLUTION
Tension in cable  500 N. Replace cable tension by forces at pins A and B. Radius does not enter
computations: (cf. Problem 6.90)
(a) Free body: AJ


0: 500 N 0
x
FF 

500 NF 500 NF 

0: 500 N 0
y
FV  

500 NV 500 NV 

0: (500 N)(0.6 m) 0
J
M 

300 N mM 300 N mM 
(b) Free body: ABK


0: 500N 500N (500N)sin20 0
x
FV   

171.01 NV 171.0 NV 

www.elsolucionario.org

1251. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.157 (Continued)


0: 500 N (500 N)cos 20 0
y
FF   


969.8 NF

970 NF 

0: (500 N)(1.2 m) (500 N)sin 20 (0.9 m) 0
K
MM   

446.1 N mM

446 N mM 
www.elsolucionario.org

1252. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.158
For the beam shown, determine (a) the magnitude P of the two upward
forces for which the maximum absolute value of the bending moment in
the beam is as small as possible, (b) the corresponding value of
max
|| .M

SOLUTION
By symmetry: 60 kips
y
AB P 
Along AC:
0: (60 kips ) 0
(60 kips )
120kips ft (2ft) at 2ft
J
MMx P
MPx
MPx
  

 


Along CD:

0: ( 2 ft)(60 kips) (60 kips ) 0
120 kip ft
120 kip ft (4 ft) at 4 ft
K
MMx x P
MPx
MPx
   

 


Along DE:
0: ( 4 ft) ( 2 ft)(60 kips)
(60 kips ) 0
120 kip ft (4 ft) (const)
L
MMxPx
xP
MP
  



Complete diagram by symmetry
For minimum
max
|| ,M set
max min
MM

120 kip ft (2 ft) [120 kip ft (4 ft) ]PP  

( a)
40.0 kipsP 


min
120 kip ft (4 ft)MP (b)
max
| | 40.0 kip ftM  
www.elsolucionario.org

1253. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.159
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.


SOLUTION












Free body: Beam
0: (4 m) (100 kN)(2 m) 20 kN m 0
A
MB   

55 kNB

0: 0
xx
FA 

0: 55 100 0
yy
FA  

45 kN
y
A 
Shear diagram
At A: 45 kN
Ay
VA
To determine Point C where
0:V

0 45 kN (25 kN m)

 
CA
VV wx
x


1.8 mx
We compute all areas bending-moment
Bending-moment diagram
At A: 0
A
M
At B:
20 kN m
B
M 

max
|| 40.5kNmM  
 1.800 m from A 
Single arc of parabola
www.elsolucionario.org

1254. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.160
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.

SOLUTION
Reactions
0: 0
xx
FB 

0: (20 lb/in.)(9 in.)(40.5 in.) (125 in.)(24 in.) (125 in.)(12 in.) (36 in.) 0
E
MB    

327.5 lb
y
B

327.5 lb
y
B


0: (20 lb/in.)(9 in.) 125 lb 125 lb 327.5 lb 0
y
FE  

102.5 lbE 102.5 lbE




(b)
max
| | 180.0 lb;V
max
| | 1230 lb in.M  
www.elsolucionario.org

1255. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.161
For the beam shown, draw the shear and bending-moment diagrams,
and determine the magnitude and location of the maximum absolute
value of the bending moment, knowing that (a) M  0, (b) M  24
kip · ft.

SOLUTION
Free body: Beam
0: 0
xx
FA 

0: (16 kips)(6 ft) (8 ft) 0
By
MAM  

1
12 kips
8
y
AM (1) 

1
0: 12 16 0
8
y
FB M  

1
4 kips
8
BM
(2) 
(a)
0:M
Load diagram
Making M 0 in. (1) and (2).

12 kips
4 kips
y
A
B



Shear diagram
12 kips
Ay
VA
To determine Point D where 0:V
0 12 kips (4 kips/ft)
DA
VV wx
x


3 ftx

We compute all areas
B. M. Diagram At A: 0
A
M

max
| | 18.00 kip ft,M  

3.00 ft from A 
Parabola from A to C

www.elsolucionario.org

1256. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.161 (Continued)

(b) 24 kip ftM
Load diagram
Making 24 kip ftM in (1) and (2)

1
12 (24) 9 kips
8
1
4 (24) 7 kips
8
A
B
 
 

Shear diagram
9 kips
Ay
VA
To determine Point D where 0:V

0 9 kips (4 kips/ft)
DA
VV wx
x


2.25 ftx 





B. M. Diagram
At A: 24 kip ft
A
M 

max
| | 34.1 kip ft,M  

2.25 ft from
A
 Parabola from A to C.
www.elsolucionario.org

1257. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.162
The beam AB, which lies on the ground, supports the
parabolic load shown. Assuming the upward reaction of the
ground to be uniformly distributed, (a) write the equations of
the shear and bending-moment curves, (b) determine the
maximum bending moment.

SOLUTION
(a)
20
2
04
0: ( ) 0
L
ygw
FwL Lxxdx
L
   



230
02
04112
233
2
3
g
g
w
wL LL L wL
L
w
w






Define
so
xdx
d
LL net load
2
00
2
4
3
xx
ww w
LL







or
2
01
4
6
ww 






2
0
0
23
01
(0) 4
6
11 1
4
62 3
0
VV wL d
wL

 
 








 



23
02
(3 2)
3
VwL
  

223
00
00
223 4 2234
002
0(32)
3
21 11
(2 )
32 23
x
MM Vdx wL d
wL wL

 
     






(b) Max M occurs where
0V
2
13 2 0 
1
2


2
2 0
0
11 121
23 4816 48
wL
MwL

  
  
  
  


2
0
max
48
wL
M
at center of beam 
www.elsolucionario.org

1258. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.163
Two loads are suspended as shown from the cable ABCD.
Knowing that
1.8 m,
B
d determine (a) the distance ,
C
d
(b) the components of the reaction at D, (c) the maximum
tension in the cable.

SOLUTION
FBD Cable: 0: 0
xxxxx
FADAD   
0: (10 m) (6 m)(10 kN) (3 m)(6 kN) 0
Ay
MD   

7.8 kN
y
D

0: 6 kN 10 kN 7.8 kN 0
yy
FA    

8.2 kN
y
A
FDB AB:
0: (1.8 m) (3 m)(8.2 kN) 0
Bx
MA  

41
kN
3
x
A
From above
41
kN
3
xx
DA
FBD CD:

41
0: (4 m)(7.8 kN) kN 0
3
CC
Md

  


2.283 m
C
d

(a)
2.28 m
C
d 
(b)
13.67 kN
x
D 

7.80 kN
y
D 
Since
and , max is
xx yy AB
AB AB TT

2
22 2
41
kN (8.2 kN)
3
AB x y
TAA

 


(c)
max
15.94 kNT 
www.elsolucionario.org

1259. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.164
A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation
that are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum
tension in the wire.

SOLUTION

Eq. 7.16: cosh
60
30 cosh
B
B
x
yc
c
mcc
c



Solve by trial and error: 64.459 mc
Eq. 7.15: sin
60 m
(64.456 m) sinh
64.459 m
69.0478 m
B
B
B
B
x
sch
c
s
s




Length 2 2(69.0478 m) 138.0956 m
B
s  138.1 mL 
Eq. 7.18:
()
mB
Twywhc
2
(0.65 kg/m)(9.81 m/s )(30 m 64.459 m)

602.32 N
m
T 602 N
m
T 
www.elsolucionario.org

1260. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 7.165
A 10-ft rope is attached to two supports A and B as shown. Determine (a) the
span of the rope for which the span is equal to the sag, (b) the corresponding
angle θ
B.

SOLUTION


Note: Since
,Lh

22
B
Lh
x

Eq. 7.16:
cosh
/2
cosh
1
1cosh
2
B
B
x
y
c
h
hcc
c
hh
cc







Solve for h/c: 4.933
h
c

Eq. 7.16: cosh
sinh
x
yc
c
dy x
dx c


At B: tan sinh
B
B
B
xdy
dx c

Substitute
11
: tan sinh sinh 4.933
222
BB
hh
x
c

  

  
  


tan 5.848
B
 80.3
B
 
www.elsolucionario.org

1261. Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 7.165 (Continued)

Eq. 7.17:
1
sinh sinh
2
1
5ft sinh 4.933
2
5 ft (5.848)
0.855
B
B
x h
sc c
cc
c
c
c











Recall that 4.933
4.933(0.855) 4.218
h
c
h



4.22 fth 

www.elsolucionario.org

1262. CHAPTER 8
www.elsolucionario.org

1263. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.1
Determine whether the block shown is in equilibrium and find the magnitude
and direction of the friction force whenP 150 N.

SOLUTION
Assume equilibrium:


0: (500 N)sin 20 (150 N)cos 20 0
x
FF   

30.056 NF 30.056 NF

0: (500 N)cos 20 (150 N)sin 20 0
y
FN   
521.15 NN 521.15 NN

Maximum friction force:

0.30(521.15 N)
156.345 N
ms
FN



Since F is and
m
FF , block is in equilibrium 
30.1 NF 20.0 
www.elsolucionario.org

1264. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.2
Determine whether the block shown is in equilibrium and find the magnitude
and direction of the friction force when P 400 N.

SOLUTION
Assume equilibrium:


0: (500 N)sin 20° (400 N)cos 20° 0
x
FF   


204.87 NF 204.87 NF

0: (500 N)cos 20 (400 N)sin 20 0
y
FN   

606.65 NN 606.65 NN

Maximum friction force:
0.30(606.65 N)
181.995 N
ms
FN


Since
,
m
FF block moves up 
Actual friction force: 0.25(606.65 N)
kk
FF N  151.7 NF 20.0 
www.elsolucionario.org

1265. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.3
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when P  120 lb.

SOLUTION
Assume equilibrium:
0: (50 lb)sin 30 (120 lb)cos 40 0
x
FF   

66.925 lbF

0: (50 lb)cos 30 (120 lb)sin 40 0
y
FN   

120.436 lbN
Maximum friction force:

0.40(120.436 lb)
48.174 lb
ms
FN



We note that
.
m
FF Thus, block moves up 
Actual friction force:

0.30(120.436 lb) 36.131 lb,
kk
FF N    36.1 lbF 30.0 
www.elsolucionario.org

1266. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.4
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when P  80 lb.

SOLUTION
Assume equilibrium:
0: (50 lb)sin 30 (80 lb)cos 40 0
x
FF   

36.284 lbF

0: (50 lb)cos 30 (80 lb)sin 40 0
y
FN   

94.724 lbN
Maximum friction force:

0.40(94.724 lb)
37.890 lb
ms
FN



We note that
.
m
FF Thus, block is in equilibrium 
Thus 36.3 lbF 30.0 
Note:
0.30(94.724 lb) 28.417 lb,
kk
FN  .
K
FF If block is originally in motion, it will
keep moving with
28.4 lb.
K
F
www.elsolucionario.org

1267. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.5
Determine the smallest value of P required to (a) start the block up the
incline, (b) keep it moving up.

SOLUTION
(a) To start block up the incline:

1
0.40
tan 0.40 21.80
s
s







From force triangle:

50 lb
sin 51.80 sin 28.20
P


83.2 lbP 
(b) To keep block moving up:

1
0.30
tan 0.30 16.70
k
k









www.elsolucionario.org

1268. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.5 (Continued)


From force triangle:

50 lb
sin 46.70 sin 33.30°
P


66.3 lbP 
www.elsolucionario.org

1269. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.6
The 20-lb block A hangs from a cable as shown. Pulley C is connected by a
short link to block E, which rests on a horizontal rail. Knowing that the
coefficient of static friction between block E and the rail is s = 0.35 and
neglecting the weight of block E and the friction in the pulleys, determine
the maximum allowable value of θ if the system is to remain in equilibrium.

SOLUTION
Free-body diagrams:
1
0.35
tan 0.35 19.29
s
s








Force triangle for pulley C:
Law of sines:

sinsin
2
sin 2sin 2sin19.29
41.35
19.29 41.35
s
s
s
TT



 



 


60.6


Note: Answer is independent of .
A
W
www.elsolucionario.org

1270. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.7
The 10-kg block is attached to link AB and rests on a moving belt.
Knowing that
s = 0.30 and k = 0.25 and neglecting the weight of the
link, determine the magnitude of the horizontal force
P that should be
applied to the belt to maintain its motion (a) to the left as shown, (b) to
the right.

SOLUTION
Link AB is a two-force member.
(a)
Free-body diagram of block: Since there is motion between block and belt, use

1
0.25
tan 0.25 14.04
98.1 N
k
k
W






Free-body diagram of belt:

Force triangle:

98.1 N
sin 14.04 sin 40.96sin125
AB
F R





36.3 N
AB
FT 

122.586 NR
From free-body diagram of belt:
F 0 : (122.586 N)sin14.04 0
x
P 



29.7 NP







www.elsolucionario.org

1271. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.7 (Continued)

(b) Free-body diagram of block:Free-body diagram of belt:

Force triangle:

98.1 N
sin 14.04 sin 110.96sin 55
AB
F R





25.5 N
AB
FC 

86.053 NR
Free-body diagram of belt:
F 0 : (86.053 N)sin14.04 0
x
P  



20.9 NP


www.elsolucionario.org

1272. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.8
Considering only values of  less than 90, determine the smallest value of 
required to start the block moving to the right when (a)
75 lb,W

(b)
100 lb.W

SOLUTION
FBD block (Motion impending):


1
tan 14.036
30 lb
sin sin( )
sin 14.036
sin( )
30 lb



 




ss
ss
s
W
W

or
sin( 14.036 )
123.695 lb
W
(a)
175 lb
75 lb: 14.036 sin
123.695 lb
W


 51.4
 
(b)
1100 lb
100 lb: 14.036 sin
123.695 lb
W


 68.0
 
www.elsolucionario.org

1273. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.9
Knowing that θ = 40°, determine the smallest force P for which equilibrium of the
7.5-kg block is maintained.

SOLUTION
Free-body diagram of block and force triangle:
Since we seek smallest P downward motion impends.  
2
7.5 kg 9.81 m/s 73.575 NW



11
tan tan 0.45 24.23
73.575 N
sin 65.77 sin 64.23
ss
P


 




74.5 NP 
www.elsolucionario.org

1274. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.10
Knowing that P = 100 N, determine the range of values of θ for which equilibrium
of the 7.5-kg block is maintained.

SOLUTION
Free-body diagram of block and force triangle:
For motion impending downward,  
2
7.5 kg 9.81 m/s 73.575 NW






11
tan tan 0.45 24.23
100 N 73.575 N
sinsin 65.77
sin 24.23 0.67093
24.23 42.14
ss
s





 









17.91


For motion impending upward,



www.elsolucionario.org

1275. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.10 (Continued)




100 N 73.575 N
sinsin 114.23
sin 24.23 0.67093
24.23 42.14
s












66.37



17.91 66.4


www.elsolucionario.org

1276. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.11
The 50-lb block A and the 25-lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
static friction is 0.15 between the two blocks and zero between block
B and the incline, determine the value of
 for which motion is
impending.

SOLUTION
Since motion impends,
s
FN between AB
Free body: Block A
Impending motion:
1
0: 50cos 0
y
FN   
1
50cosN

11
0: 50sin 0
x
FT N    

1
50sin (50)cosT  (1)
Free body: Block B
Impending motion:
21
0: 25cos 0
y
FNN    

2
75cosN

11 2 2
: 25sin
x
FTN N 

 
12
50 cos 75 cos 25sinT (2)
Eq. (1)-Eq. (2):
 
12
0 25sin 100 cos 75 cos 0    
Substituting in for
12
0.15, 0, we have:

15
15cos 25sin : tan ;
25
 31.0 
www.elsolucionario.org

1277. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.12
The 50-lb block A and the 25-lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
static friction is 0.15 between all surfaces of contact, determine the
value of
 for which motion is impending.

SOLUTION
Since motion impends,
s
FN between AB
Free body: Block A
Impending motion:
1
0: 50cos 0
y
FN   
1
50cosN

11
0: 50sin 0
x
FT N    

1
50sin (50)cosT  (1)
Free body: Block B
Impending motion:
21
0: 25cos 0
y
FNN    

2
75cosN

11 2 2
: 25sin
x
FTN N 

 
12
50 cos 75 cos 25sinT (2)
Eq. (1)-Eq. (2):
 
12
0 25sin 100 cos 75 cos 0    
Substituting in for
12
0.15, we have:


26.25
175 0.15 cos 25sin : tan ;
25
 46.4 
www.elsolucionario.org

1278. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.13
Three 4-kg packagesA, B, and C are placed on a conveyor belt that
is at rest. Between the belt and both packages A and C the
coefficients of friction are
0.30
s
 and 0.20;
k
 between
package B and the belt the coefficients are
0.10
s
 and 0.08.
k

The packages are placed on the belt so that they are in contact with
each other and at rest. Determine which, if any, of the packages will
move and the friction force acting on each package.

SOLUTION
Consider C by itself: Assume equilibrium
0: cos15 0  
yC
FNW

cos 15 0.966
C
NW W


0: sin 15 0  
xC
FFW

sin 15 0.259
C
FW W
But
0.30(0.966 )
0.290
msC
FN
W
W


Thus,
Cm
FF Package C does not move 

2
0.259
0.259(4 kg)(9.81 m/s )
10.16 N
C
FW



10.16 N
C
F 
Consider B by itself: Assume equilibrium. We find,

0.259
0.966
B
B
FW
NW



But
0.10(0.966 )
0.0966
msB
FN
W
W


Thus,
.
Bm
FF Package B would move if alone 


www.elsolucionario.org

1279. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.13 (Continued)

Consider A and B together: Assume equilibrium

0.259
0.966
2(0.259 ) 0.518
( ) ( ) 0.3 0.1 0.386
AB
AB
AB
Am Bm A B
FF W
NN W
FF W W
FF NN W


 


Thus,
() ()
AB Am Bm
FF F F  A and B move 
 0.2(0.966)(4)(9.81)
AkA
FN 

7.58 N
A
F 

0.08(0.966)(4)(9.81)
BkB
FN

3.03 N
B
F 
www.elsolucionario.org

1280. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.14
Solve Problem 8.13 assuming that package B is placed to the right of
both packages A and C.
PROBLEM 8.13 Three 4-kg packagesA, B, and C are placed on a
conveyor belt that is at rest. Between the belt and both packages A
and C the coefficients of friction are
0.30
s
 and 0.20;
k

between package B and the belt the coefficients are
0.10
s
 and
0.08.
k
 The packages are placed on the belt so that they are in
contact with each other and at rest. Determine which, if any, of the
packages will move and the friction force acting on each package.
SOLUTION
Consider package B by itself: Assume equilibrium
0: cos15 0  
yB
FNW

cos 15 0.966
B
NW W

0: sin15 0  
xB
FFW

sin 15 0.259
B
FW W
But
0.10(0.966 )
0.0966
msB
FN
W
W


Thus,
.
Bm
FF. Package B would move if alone.
Consider all packages together: Assume equilibrium. In a manner similar to above, we find

0.966
0.259
3(0.259 )
0.777
ABC
ABC
ABC
NNN W
FFF W
FFF W
W





But
() ()
0.30(0.966 )
0.290
Am C m s
FF N
W
W 


and
( ) 0.10(0.966 )
0.0966
Bm
FW
W





www.elsolucionario.org

1281. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.14 (Continued)

Thus, ( ) ( ) ( ) 2(0.290 ) 0.0966
0.677
Am Cm Bm
FFF W W
W
 


and we note that
() () ()
ABC Am Cm Bm
FFF F F F  .
All packages move


2
2
0.20(0.966)(4 kg)(9.81 m/s )
7.58 N
0.08(0.966)(4 kg)(9.81 m/s )
3.03N
AC k
Bk
FF N
FN 








7.58 N
AC
FF ; 3.03 N
B
F 
www.elsolucionario.org

1282. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.15
A uniform crate with a massof 30 kg must be moved up along the 15°
incline without tipping. Knowing that force
P is horizontal, determine
(a) the largest allowable coefficient of static friction between the crate and
the incline, (b) the corresponding magnitude of force
P.

SOLUTION
(a) Free-body diagram
For tipping to be impending, reaction is at C,
Since the crate is a 3-force body.
C must pass through E where P and W intersect.
Geometry analysis:
tan
2
L
HJ 1tan
22
LL
HG HJ   cos 1 tan cos
2
L
GE HG

2
cos 1 tan cos
2
L
EK GE sin 1 tan cos sin
2
L
GK GE



2
2
1 cos sin cos
22
1 cos 15 sin15 cos15
2
0.84151
LL
EL EK
L
EL
EL L   
 







2
2
2
1tan cossin
22
sin
cos sin cos sin
22 cos
1cossin sin
2
1 cos15 sin15 sin 15
2
0.40849
L
LC GK
LL
LC
LL
LC
L
LC
L
LC
LC L


 

 

 

 


 
 

 

www.elsolucionario.org

1283. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.15 (Continued)
0.40849
tan 0.48543, 25.89
0.84151
ss
LC L
EL L
  


tan
ss
 0.485
s
 
(b) Force Triangle



  
2
tan
30 kg 9.81 m/s tan 15 25.89
s
PW
P 




255 NP 
www.elsolucionario.org

1284. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.16
A worker slowly moves a 50-kg crate to the left along a loading dock
by applying a force
P at corner B as shown. Knowing that the crate
starts to tip about the edge E of the loading dock when a 200 mm,
determine (a) the coefficient of kinetic friction between the crate and
the loading dock, (b) the corresponding magnitude P of the force.


SOLUTION
Free body: Crate Three-force body.
Reaction E must pass through K where P and W intersect.
Geometry:
(a) (0.6 m) tan15 0.16077 mHK


0.9 m 1.06077 m
0.4 m
tan 0.37708
1.06077 m



s
JK HK
tan 0.377
ss
 

20.66
s

Force triangle:
(b) (50 kg)(9.81 m/s)
490.5 N
W

Law of sines:

490.5 N
sin 20.66 sin 84.34°
173.91 N



P
P
173.9 NP 
www.elsolucionario.org

1285. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.17
A half-section of pipe weighing 200 lb is pulled by a cable as
shown. The coefficient of static friction between the pipe and the
floor is 0.40. If α = 30°, determine (a) the tension T required to
move the pipe, (b) whether the pipe will slide or tip.

SOLUTION
(a)FBD pipe:Note: assume that pipe slides
0; 200 sin30 0
yAB
FNN T    



200 0.5
AB
NN T 


AB sAsB sA B
FF N N N N   
0.4 200 0.5 (1)
AB
FF T 


0: cos30 0
0.866 80 0.2 0
xAB
FT FF
TT
   
 



75.0 lbT 

(b) Check that pipe does not tip about B:
 0: (200 lb) (2 ) cos30 sin 60 sin 30 cos60 0
BA
MrNrTrTrr     
  


2 200 lb (cos30 sin 60 sin 30 sin 30 sin 30 cos60 )
43.7 lb
A
A
NT
N 

   


Since 0, pipe does not tip.
A
N 

www.elsolucionario.org

1286. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.18
A 120-lb cabinet is mounted on casters that can be locked to prevent their
rotation. The coefficient of static friction between the floor and each caster
is 0.30. Assuming that the casters at both A and B are locked, determine
(a) the force
P required to move the cabinet to the right, (b) the largest
allowable value of h if the cabinet is not to tip over.

SOLUTION
FBD cabinet:
(a)
0: 0
yAB
AB
FNNW
NNW  

Impending slip:

AsA
BsB
FN
FN



So

AB s
FF W
0: 0
xAB
AB s
FPFF
PF F W

   
 120 lb
0.3
s
W




0.3(120 lb) 141.26 NP

36.0 lbP 
(b) For tipping,
0
AA
NF

0: (12 in.) 0
B
MhP W  


max
112in.
(12 in.) (12 in.)
0.3
s
W
h
P

 

max
40.0 in.h 
www.elsolucionario.org

1287. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.19
Wire is being drawn at a constant rate from a spool by applying a vertical
force
P to the wire as shown. The spool and the wire wrapped on the spool
have a combined weight of 20 lb. Knowing that the coefficients of friction at
both A and B are
s  0.40 and k  0.30, determine the required magnitude
of the force
P.

SOLUTION
Since spool is rotating

AkA BkB
FNFN

0: (3 in.) (6 in.) (6 in.) 0
GAB
MPF F   

36( )0
kA B
PNN (1)

0: 0
xAB
FFN  

BkA
NN (2)

0: 20 lb 0
20 0
yAB
AkB
FPNF
PN N

    
 
Substitute for
B
N from (2): 20 0
AkA
PN N 

2
20
1
A
k
P
N




(3)
Substitute from (2) into (1):
36( )0
kA kA
PNN

1
2(1 )
A
kk
P
N


 (4)
(3)(4):
22
20
12()
kkk
PP





Substitute
0.30:
k

2
20
1(0.3)P

2(0.3)(1.03)
P


20 1.3974 ; 2.3974 20;PP P  8.34 lbP 

www.elsolucionario.org

1288. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.20
Solve Problem 8.19 assuming that the coefficients of friction at B are zero.
PROBLEM 8.19 Wire is being drawn at a constant rate from a spool by
applying a vertical force
P to the wire as shown. The spool and the wire
wrapped on the spool have a combined weight of 20 lb. Knowing that the
coefficients of friction at both A and B are
s  0.40 and k  0.30, determine
the required magnitude of the force
P.

SOLUTION
Since spool is rotating

AkA
FN

0: (3 in.) (6 in.) 0
GA
MPF  

22
AkA
PF N  (1)


20: 20 lb 0
yA
FP N   

20
A
NP (2)
Substitute for
A
N from (2) into (1) 2(20 )
k
PP
Substitute
0.30: 2(0.3)(20 )
1.667 20
2.667 20
k
PP
PP
P



7.50 lbP 7.50 lbP 
www.elsolucionario.org

1289. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.21
The cylinder shown is of weight W and radius r. Express in terms W and r the
magnitude of the largest couple
M that can be applied to the cylinder if it is not to
rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B,
(b) 0.25 at A and 0.30 at B.

SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B

AAA
BBB
FN
FN




0: 0
xAB
AB BB
FNF
NF N

  



AAAABB
FN N

0: 0
(1 )
yBA
BAB
FNFW
NW

  

or
1
1
B
AB
NW



and
1
1
B
BBB
AB
AB
AABB
AB
FN W
FN W












0: ( ) 0
CAB
MMrFF   

1
1
A
B
AB
MWr






(a) For
0 and 0.30:
AB


0.300MWr 
(b) For
0.25 and 0.30:
AB


0.349MWr 
www.elsolucionario.org

1290. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.22
The cylinder shown is of weight W and radius r, and the coefficient of static
friction
s
 is the same at A and B. Determine the magnitude of the largest couple
M that can be applied to the cylinder if it is not to rotate.

SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B

AsA
BsB
FN
FN



0: 0
xAB
AB sB
FNF
NF N

  



2
AsAsB
FN N

2
0: 0
yBA
BsB
FNFW
NNW

  

or
2
1
B
s
W
N



and
2
2
2
1
1
s
B
s
s
A
W
F
W
F








0: ( ) 0
CAB
MMrFF   


2
2
1
ss
s
W
Mr




max 2
1
1
s
s
s
MWr







www.elsolucionario.org

1291. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.23
End A of a slender, uniform rod of length L and weight W bears on a
surface as shown, while end B is supported by a cord BC. Knowing that
the coefficients of friction are
0.40
s
 and 0.30,
k
 determine (a) the
largest value of
 for which motion is impending, (b) the corresponding
value of the tension in the cord.

SOLUTION
Free-body diagram
Three-force body. Line of action of R must pass through D, where T and R intersect.
Motion impends:

tan 0.4
21.80
s
s




(a) Since BGGA, it follows that BDDC and AD bisects
BAC

90
2
21.8 90
2
s







136.4 


(b) Force triangle (right triangle):

cos 21.8TW

0.928TW 
www.elsolucionario.org

1292. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.24
End A of a slender, uniform rod of length L and weight W bears on a surface as
shown, while end B is supported by a cord BC. Knowing that the coefficients
of friction are
0.40
s
 and 0.30,
k
 determine (a) the largest value of 
for which motion is impending, (b) the corresponding value of the tension in the
cord.

SOLUTION
Free-body diagram
Rod AB is a three-force body. Thus, line of action of R must pass through D,
where
W and T intersect.
Since AGGB, CDDB and the median AD of the isosceles
triangleABC bisects the angle
.
(a) Thus,
1
2
s

Since motion impends,

1
tan 0.40 21.80
2 2(21.8 )
s
s




 


43.6 
(b) Force triangle:
This is a right triangle.

sin
sin 21.8
s
TW
W 



0.371TW 

www.elsolucionario.org

1293. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.25
A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction
s
 is zero at B, determine the smallest value of
s
 at A for
which equilibrium is maintained.

SOLUTION
Free body: Ladder
Three-force body.
Line of action of
A must pass through D, where W and B intersect.


At A:
1.25 m
tan 0.2083
6m
ss
 

0.208
s
 
www.elsolucionario.org

1294. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.26
A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction
s
 is the same at A and B, determine the smallest value of
s

for which equilibrium is maintained.

SOLUTION
Free body: Ladder
Motion impending:

AsA
BsB
FN
FN




0: (1.25m) (6m) (2.5m) 0
ABsB
MW N N    

1.25
62.5
B
s
W
N


 (1)

0: 0
yAsB
FNNW   

1.25
62.5
AsB
s
A
s
NW N
W
NW 




(2)

0: 0
xsAB
FNN  
Substitute for
A
N and
B
N from Eqs. (1) and (2):

2
22
2
1.25 1.25
62.5 62.5
6 2.5 1.25 1.25
1.25 6 1.25 0
0.2


 




 


s
s
ss
ss s
ss
s
W W
W

and
5 (Discard)
s


0.200
s
 
www.elsolucionario.org

1295. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.27
The press shown is used to emboss a small seal at E.
Knowing that the coefficient of static friction between the
vertical guide and the embossing die D is 0.30, determine the
force exerted by the die on the seal.

SOLUTION
Free body: Member ABC
Dimensions in mm

0: cos 20 (100) sin 20 (173.205)
ABD BD
MF F   

(250 N)(100 386.37) 0
 

793.64 N
BD
F
Free body: Die
D

1
1
tan
tan 0.3
16.6992
s s
 



 


Force triangle:

793.64 N
sin 53.301 sin 106.6992°
664.35 N
D
D





On seal:
664 ND 
www.elsolucionario.org

1296. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.28
The machine base shown has a mass of 75 kg and is fitted with skids at A
and B. The coefficient of static friction between the skids and the floor is
0.30. If a force
P of magnitude 500 N is applied at corner C, determine the
range of values of θ for which the base will not move.

SOLUTION
Free-body: Machine base

2
(75 kg)(9.81 m/s ) 735.75 Nm

Assume sliding impends



AsA BsB
FN FN

0: sin 0
yAB
FNNWP    

()sin
AB
NN WP  (1)

0: cos 0
xAB
FFFP    

()cos0
sA B
NN P  (2)
Eq. (2)
:
Eq. (1)

cos
sin
s
P
WP





sin cos
ss
WP P 

0.30(735.75 N) 0.30(500 N) sin 500 cos
500cos 150sin 220.73
Solve for : 48.28






Assume tipping about B impends:
0
A
N

0: sin (0.4 m) cos (0.5 m) (0.2 m) 0
B
MP P W    

500 sin (0.4) 500cos (0.5) 735.75(0.2 m) 0
200sin 250cos 147.15
 

Solve for
: 78.70
Range for no motion:

48.3 78.7   

www.elsolucionario.org

1297. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.29
The 50-lb plate ABCD is attached at A and D to collars that can slide
on the vertical rod. Knowing that the coefficient of static friction is
0.40 between both collars and the rod, determine whether the plate is
in equilibrium in the position shown when the magnitude of the
vertical force applied at E is (a)
0,P (b) 20 lb.P

SOLUTION
(a) P 0 0: (2 ft) (50 lb)(3 ft) 0
DA
MN  
75 lb
A
N

0: 75 lb
xDA
FNN  

0: 50 lb 0
yAD
FFF   

50 lb
AD
FF
But: ( ) 0.40(75 lb) 30 lb
( ) 0.40(75 lb) 30 lb
Am s A
Dm s D
FN
FN
 
 

Thus:
() () 60lb
Am Dm
FF
and
() ()
Am Dm A D
FFFF Plate is in equilibrium 
(b) P 20 lb
0: (2 ft) (50 lb)(3 ft) (20 lb)(5 ft) 0
DA
MN   
25 lb
A
N

0: 25 lb
xDA
FNN  

0: 50 lb 20 lb 0
yAD
FFF    

30 lb
AD
FF
But: ( ) 0.4(25 lb) 10 lb
( ) 0.4(25 lb) 10 lb
Am s A
Dm s D
FN
FN
 
 

Thus:
() () 20lb
Am Dm
FF
and
() ()
AD Am Dm
FF F F  Plate moves downward 
www.elsolucionario.org

1298. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.30
In Problem 8.29, determine the range of values of the magnitude P of
the vertical force applied at E for which the plate will move
downward.
PROBLEM 8.29 The 50-lb plate ABCD is attached at A and D to
collars that can slide on the vertical rod. Knowing that the coefficient
of static friction is 0.40 between both collars and the rod, determine
whether the plate is in equilibrium in the position shown when the
magnitude of the vertical force applied at E is (a)
0,P

(b)
20 lb.P

SOLUTION
We shall consider the following two cases:
(1)
030 lbP

0: (2 ft) (50 lb)(3 ft) (5 ft) 0
DA
MN P   

75 lb 2.5
A
NP
( Note:
0
A
N and directed for 30 lbP as assumed here)

0:
xAD
FNN 

0: 50 0
50
yAD
AD
FFFP
FF P   

But:
0
() ()
0.40(75 2.5 )
30
Am m s A
FF N
P
P 


Plate moves if:
() ()
AD Am Dm
FF F F 
or 50 (30 ) (30 )PP P    10 lbP
(2) 30 lb
50 lbP

0: (2 ft) (50 lb)(3 ft) (5 ft) 0   
DA
MN P

2.5 75
A
NP
( Note:
A
N and directed for 30 lbP as assumed)

0:
xAD
FNN 

0: 50 0   
yAD
FFFP

50
AD
FF P
www.elsolucionario.org

1299. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.30 (Continued)

But:
() ()
0.40(2.5 75)
30 lb
Am Dm s A
FF N
P
P 


Plate moves if:
() ()
AD Am Dm
FF F F 
50 ( 30) ( 30) PP P 
110
36.7 lb
3
P

Thus, plate moves downward for:
10.00 lb 36.7 lbP 
(
Note: For 50 lb,Pplate is in equilibrium)


www.elsolucionario.org

1300. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.31
A window sash weighing 10 lb is normally supported by two 5-lb sash
weights. Knowing that the window remains open after one sash cord
has broken, determine the smallest possible value of the coefficient of
static friction. (Assume that the sash is slightly smaller than the frame
and will bind only at Points A and D.)

SOLUTION
FBD window: 5lbT
0: 0
xAD
AD
FNN
NN
  



Impending motion:
AsA
DsD
FN
FN




0: (18 in.) (27 in.) (36 in.) 0
DAA
MWNF   

10 lbW

3
2
2
2
34
AsA
A
s
WN N
W
N 





0: 0
yA D
FFWTF  

2
AD
W
FFWT

Now ( )
2
AD sA D
sA
FF N N
N 
 


Then
2
2
234
s
s
WW



or
0.750
s
 
www.elsolucionario.org

1301. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.32
A 500-N concrete block is to be lifted by the pair of tongs shown.
Determine the smallest allowable value of the coefficient of static
friction between the block and the tongs at F and G.

SOLUTION
Free body: Members CA, AB, BD
By symmetry:
1
(500) 250 N
2
yy
CD 

Since CA is a two-force member,

250 N
90 mm 75 mm 75 mm
300 Nyx
x
CC
C




0:
300 N
xxx
x
FDC
D
 


Free body: Tong DEF

0: (300 N)(105 mm) (250 N)(135 mm)
E
M 


(250 N)(157.5 mm) (360 mm) 0
x
F


290.625 N
x
F

Minimum value of :
s

250 N
290.625 Ny
s
x
F
F




0.860
s
 
www.elsolucionario.org

1302. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.33
A pipe of diameter 60 mm is gripped by the stillson wrench shown. Portions
AB and DE of the wrench are rigidly attached to each other, and portion CF is
connected by a pin at D. If the wrench is to grip the pipe and be self-locking,
determine the required minimum coefficients of friction at A and C.

SOLUTION
FBD ABD:

0: (15 mm) (110 mm) 0
DAA
MNF  

Impending motion:
AAA
FN

Then
15 110 0
A

or
0.13636
A
 0.1364
A
 

0: 0
xAxxA
FFDDF   
FBD pipe:

0: 0
yCA
FNN  

CA
NN
FBD DF:

0: (550 mm) (15 mm) (500 mm) 0
FCCx
MFND   

Impending motion:
CCC
FN
Then
550 15 500
A
C
C
F
N


But
and 0.13636
A
CA A
A
F
NN
N


So
550 15 500(0.13636)
C


0.1512
C


www.elsolucionario.org

1303. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.34
A safety device used by workers climbing ladders fixed to high
structures consists of a rail attached to the ladder and a sleeve that can
slide on the flange of the rail. A chain connects the worker’s belt to
the end of an eccentric cam that can be rotated about an axle attached
to the sleeve at C. Determine the smallest allowable common value of
the coefficient of static friction between the flange of the rail, the pins
at A and B, and the eccentric cam if the sleeve is not to slide down
when the chain is pulled vertically downward.


SOLUTION
Free body: Cam


0: (0.8 in.) (3in.) (6 in.) 0
CD sD
MN N P    

6
0.8 3
D
s
P
N


 (1)

Free body: Sleeve and cam


0: 0
xDAB
FNNN   

AB D
NNN (2)

0: 0
yABD
FFFFP   

or
()
sA B D
NNN P   (3)
Substitute from Eq. (2) into Eq. (3):
(2 )
2
 
sD D
s
P
NPN
(4)
Equate expressions for
D
N from Eq. (1) and Eq. (4):

6
;0.83 12
2 0.8 3
0.8
15
ss
ss
s
PP





0.0533
s
 
(Note: To verify that contact at pins A and B takes places as assumed, we shall check that
0
A
N and
0.)
B
N
www.elsolucionario.org

1304. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.34 (Continued)

From Eq. (4):
9.375
2 2(0.0533)
D
s
PP
NP

 
From free body of cam and sleeve:

0: (8 in.) (4 in.) (9 in.) 0
BA D
MN N P   

8 (9.375 )(4) 9
5.8125 0 OK
A
A
NPP
NP
 


From Eq. (2):
5.8125 9.375
3.5625 0 OK
AB D
B
B
NNN
PN P
NP




www.elsolucionario.org

1305. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.35
To be of practical use, the safety sleeve described in Problem 8.34
must be free to slide along the rail when pulled upward. Determine
the largest allowable value of the coefficient of static friction
between the flange of the rail and the pins at A and B if the sleeve is
to be free to slide when pulled as shown in the figure, assuming
(a)
60 , (b) 50 , (c) 40 .

SOLUTION
Note the cam is a two-force member.
Free body: Sleeve
We assume contact between rail and pins as shown.

0: (3 in.) (3 in.) (4 in.) (4 in.) 0    
CA B A B
MF F N N
But
AsA
BsB
FN
FN



We find
3( )4( )0
4
1.33333
3
sA B A B
s
NN NN
  

We now verify that our assumption was correct.

0: cos 0
xAB
F NNP    

cos
BA
NNP  (1)

0: sin 0
yAB
FFFP     

sin
sA sB
NNP 

sin
AB
s
P
NN

 (2)
Add Eqs. (1) and (2):
sin
2cos 0OK
B
s
NP









www.elsolucionario.org

1306. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.35 (Continued)

Subtract Eq. (1) from Eq. (2):
sin
2cos
A
s
NP






0
A
N only if
sin
cos 0
s




tan 1.33333
53.130
s




(a) For case (a): Condition is satisfied, contact takes place as shown. Answer is correct.

1.333
s
 
But for (b) and (c):
53.130 and our assumption is wrong,
A
N is directed to left.

0: cos 0
xAB
FNNP     

cos
AB
NNP  (3)

0: sin 0
yAA
FFFP    

()sin
sA B
NN P (4)

Divide Eq. (4) by Eq. (3):

tan
s
 (5)
(b) We make
50 in Eq. (5):

tan 50
s
 1.192
s

(c) We make
40 in Eq. (5):

tan 40
s
 0.839
s

www.elsolucionario.org

1307. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.36
Two 10-lb blocks A and B are connected by a slender rod of negligible weight. The
coefficient of static friction is 0.30 between all surfaces of contact, and the rod
forms an angle
30 .
 with the vertical. (a) Show that the system is in
equilibrium when
0.P
 (b) Determine the largest value of P for which
equilibrium is maintained.

SOLUTION
FBD blockB:
(a) Since 2.69 lbP to initiate motion, equilibrium exists with 0P

(b) For
max
,P motion impends at both surfaces:

Block B:
0: 10lb cos 30 0   
yB AB
FN F

3
10 lb
2
BAB
NF (1)
Impending motion:
0.3
BsB B
FN N

0: sin 30 0  
xBAB
FFF

20.6
AB B B
FF N (2)
Solving Eqs. (1) and (2):
3
10 lb (0.6 ) 20.8166 lb
2
BB
NN 
FBD blockA: Then 0.6 12.4900 lb
AB B
FN
Block A:
0: sin 30 0  
xAB A
FF N

11
(12.4900 lb) 6.2450 lb
22
AAB
NF 
Impending motion:
0.3(6.2450 lb) 1.8735 lb
AsA
FN 

0: cos 30 10 lb 0   
yAAB
FFF P

3
10 lb
2
3
1.8735 lb (12.4900 lb) 10 lb
2
2.69 lb
 
 

AAB
PF F
2.69 lbP 
www.elsolucionario.org

1308. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.37
A 1.2-m plank with a mass of 3 kg rests on two joists. Knowing
that the coefficient of static friction between the plank and the
joists is 0.30, determine the magnitude of the horizontal force
required to move the plank when (a) a = 750 mm, (b) a = 900 mm.

SOLUTION
Free body member AB:
In the vertical plane,
0: 0 hence (1)
22
AC C
LL
MNaW NW
a
   



2
0: 0 hence (2)
2
yAC A
aL
FNNW NW
a

   

In the horizontal plane,


0: 0 hence F (3)
AC C
L
MFaPL P
a
  



0: 0 hence (4)
ZAC A
La
FFPF FP
a

   

(a)
Substituting given data: 
 
2
0.75 m, 1.2 m, 0.30, 3 kg 9.81 m/s 29.430 N
s
aL W  
0.8 , 0.2 , 1.6 , 0.6
CACA
NWNWFPFP
To slip at A,
 or 0.6 0.30 0.2 0.1
AsA
FN P WPW
To slip at C,
 or 1.6 0.30 0.8 0.15
CsC
FN P WP W 
The plank will slip at A
2.94 NP 


www.elsolucionario.org

1309. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.37 (Continued)

(b) Substituting given data:
 
2
0.9 m, 1.2 m, 0.30, 3 kg 9.81 m/s 29.430 N
s
aL W   

2141
, , ,
3333
CACA
NWNWFPFP
To slip at A,
11
or 0.30 0.3
33
AsA
F NPWPW





To slip at C,
42
or 0.30 0.15
33
CsC
F NPWPW

 



The plank will slip at C for
0.15 29.43 NP 4.41 NP 
www.elsolucionario.org

1310. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.38
Two identical uniform boards, each of weight 40 lb, are temporarily
leaned against each other as shown. Knowing that the coefficient of
static friction between all surfaces is 0.40, determine (a) the largest
magnitude of the force
P for which equilibrium will be maintained,
(b) the surface at which motion will impend.

SOLUTION
Board FBDs:

Assume impending motion at C, so

0.4
CsC C
FN N
FBD II:
0: (6 ft) (8 ft) (3 ft)(40 lb) 0
BCC
MNF   
[6 ft 0.4(8 ft)] (3 ft)(40 lb)
C
N
or 42.857 lb
C
N
and
0.4 17.143 lb
CC
FN

0: 0
xBC
FNF  

17.143 lb
BC
NF

0: 40 lb 0
yB C
MF N  

40 lb 2.857 lb
BC
FN 
Check for motion at B:
2.857 lb
0.167 , OK, no motion.
17.143 lb
B
s
B
F
N

FBD I:
0: (8 ft) (6 ft) (3 ft)( 40 lb) 0
ABB
MNFP    

(8 ft)(17.143 lb) (6 ft)(2.857 lb)
40 lb
3 ft
11.429 lb
P



www.elsolucionario.org

1311. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.38 (Continued)

Check for slip at A (unlikely because of P):

0: 0 or 17.143 lb
xAB AB
FFN FN    

0: 40 lb 0
yA B
FNP F   
or
11.429 lb 40 lb 2.857 lb
48.572 lb
A
N

Then
17.143 lb
0.353
48.572 lb
A
s
A
F
N

OK,
no slip assumption is correct.
Therefore
(a)
max
11.43 lbP 
(b) Motion impends at C

www.elsolucionario.org

1312. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.39
Two rods are connected by a collar at B. A couple MA with a
magnitude of 15 N·m is applied to rod AB. Knowing that the
coefficient of static friction between the collar and the rod is
0.30, determine the largest couple
MC for which equilibrium
will be maintained.

SOLUTION

0.2
tan =63.43
0.1
0.2
tan 25.2
0.425






tan 0.3 16.7
ss s
  


For largest
MC, motion of B impends downward.
0.2
0.22362 m
sin 63.43
AB


0: cos 0
AsA
MBABM   

cos16.7 0.22362 m 15 N m 0 70.032 NBB  



0.2
0.46973 m
sin 25.2
BC


0: cos( ) 0
CC s
MMB BC     

 (70.032 N)cos 63.43 16.7 25.2 0.46973 m
C
M
 


30.6 N m
C
M


www.elsolucionario.org

1313. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.40
In Prob. 8.39, determine the smallest couple MC for
which equilibrium will be maintained.
PROBLEM 8.39
Two rods are connected by a collar at B. A couple MA
with a magnitude of 15 N·m is applied to rod
AB.
Knowing that the coefficient of static friction between the
collar and the rod is 0.30, determine the largest couple
MC for which equilibrium will be maintained.

SOLUTION

0.2
tan =63.43
0.1
0.2
tan 25.2
0.425






tan 0.3 16.7
ss s
  


For smallest
MC, motion of B impends upward.
0.2
0.22362 m
sin 63.43
AB


0: cos 0
AsA
MBABM   

cos16.7 0.22362 m 15 N m 0 70.032 NBB  



0.2
0.46973 m
sin 25.2
BC




www.elsolucionario.org

1314. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.40 (Continued)

0: cos( ) 0
CC s
MMB BC     

 (70.032 N)cos 63.43 16.7 25.2 0.46973 m
C
M
 


18.90 N m
C
M


www.elsolucionario.org

1315. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.41
A 10-ft beam, weighing 1200 lb, is to be moved to the left onto
the platform. A horizontal force
P is applied to the dolly, which
is mounted on frictionless wheels. The coefficients of friction
between all surfaces are
0.30
s
 and 0.25,
k
 and initially
2 ft.x
 Knowing that the top surface of the dolly is slightly
higher than the platform, determine the force
P required to start
moving the beam. (
Hint: The beam is supported at A and D.)
SOLUTION
FBD beam:


0: (8ft) (1200 lb)(5ft) 0  
AD
MN

750 lb
D
N

0: 1200 750 0
yA
FN   

450 lb
A
N

( ) 0.3(450) 135.0 lb
Am s A
FN 

( ) 0.3(750) 225 lb
Dm s D
FN 
Since
() (),
Am Dm
FF sliding first impends at A with

() 135 lb
AAm
FF
0: 0
135.0 lb
xAD
DA
FFF
FF  


FBD dolly:
From FBD of dolly:

0: 0
xD
FFP 

135.0 lb
D
PF 135.0 lbP 
www.elsolucionario.org

1316. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.42
(a) Show that the beam of Problem 8.41 cannot be moved if the
top surface of the dolly is slightly
lower than the platform.
(
b) Show that the beam can be moved if two 175-lb workers
stand on the beam at
B and determine how far to the left the beam
can be moved.
PROBLEM 8.41 A 10-ft beam, weighing 1200 lb, is to be
moved to the left onto the platform. A horizontal force
P is
applied to the dolly, which is mounted on frictionless wheels.
The coefficients of friction between all surfaces are
0.30
s

and
0.25,
k
 and initially 2 ft.x
 Knowing that the top
surface of the dolly is slightly higher than the platform,
determine the force
P required to start moving the beam. (Hint:
The beam is supported at
A and D.)

SOLUTION
(a) Beam alone
0: (8 ft) (1200 lb)(3 ft) 0
CB
MN  

450 lb
B
N

0: 450 1200 0
yC
FN   

750 lb
C
N

( ) 0.3(750) 225 lb
( ) 0.3(450) 135 lb
Cm s C
Bm s B
FN
FN
 
 

Since
() (),
Bm Cm
FF sliding first impends at B, and Beam cannot be moved 
(b) Beam with workers standing at B


0: (10 ) (1200)(5 ) 350(10 ) 0
CB
MNx x x    

9500 1550
10
B
x
N
x





0: (1200)(5) (10 ) 0
6000
10
BC
C
MNx
N
x
  


www.elsolucionario.org

1317. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.42 (Continued)

Check that beam starts moving for x 2 ft:
For x 2 ft:
9500 1550(2)
800 lb
10 2
6000
750 lb
10 2
( ) 0.3(750) 225 lb
( ) 0.3(800) 240 lb
B
C
Cm s C
Bm s B
N
N
FN
FN







 
 
Since
() (),
Cm Bm
FF sliding first impends at C, Beam moves 
How far does beam move?
Beam will stop moving when

()
CBm
FF
But
6000 1500
0.25
10 10
CkC
FN
x x
 
 

and
9500 1550 2850 465
( ) 0.30
10 10
Bm s B
x x
FN
xx

 
 


Setting
():
CBm
FF 1500 2850 465
 x 2.90ftx 
(Note: We have assumed that, once started, motion is continuous and uniform (no acceleration).)
www.elsolucionario.org

1318. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.43
Two 8-kg blocks A and B resting on shelves are connected by a rod of
negligible mass. Knowing that the magnitude of a horizontal force
P
applied at
C is slowly increased from zero, determine the value of P for
which motion occurs, and what that motion is, when the coefficient of
static friction between all surfaces is (
a) 0.40,
s
 (b) 0.50.
s


SOLUTION
(a) 0.40:
s
 Assume blocks slide to right.

2
(8 kg)(9.81 m/s ) 78.48 N
AsA
BsB
Wmg
FN
FN


 




0: 2 0
yAB
FNNW   

2
AB
NN W

0: 0
xAB
FPFF   
()(2)
0.40(2)(78.48 N) 62.78 N
AB sA B s
PF F N N W
P   

(1)

0: (0.1 m) ( )(0.09326 m) (0.2 m) 0
BA A
MP NW F    

(62.78)(0.1) ( 78.48)(0.09326) (0.4)( )(0.2) 0  
AB
NN

0.17326 1.041
A
N

6.01 N 0 OK
A
N
System slides:
62.8 NP 
(b) 0.50:
s
 See part a.
Eq. (1):
0.5(2)(78.48 N) 78.48 NP

0: (0.1 m) ( )(0.09326 m) (0.2 m) 0
BA A
MP NW F    


(78.48)(0.1) ( 78.48)(0.09326) (0.5) (0.2) 0  
AA
NN


0.19326 0.529
A
N

2.73 N 0
A
N  uplift, rotation about B
www.elsolucionario.org

1319. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.43 (Continued)

For
0:
A
N 0: (0.1 m) (0.09326 m) 0
B
MP W  

(78.48 N)(0.09326m)/(0.1) 73.19P
System rotates about
B: 73.2 NP 
www.elsolucionario.org

1320. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.44
A slender steel rod of length 225 mm is placed inside a pipe as
shown. Knowing that the coefficient of static friction between the rod
and the pipe is 0.20, determine the largest value of
 for which the
rod will not fall into the pipe.

SOLUTION
Motion of rod impends down at A and to left at B.

AsABsB
FNFN

0: sin cos 0   
xABB
FNN F

sin cos 0
AB sB
NN N   

(sin cos )
AB s
NN   (1)

0: cos sin 0
yAB B
FFN F W    

cos sin 0
sA B sB
NN N W  (2)
Substitute for
A
N from Eq. (1) into Eq. (2):

(sin cos ) cos sin 0
sB s B sB
NNNW  

22
(1 ) cos 2 sin (1 0.2 ) cos 2(0.2) sin
B
ss
WW
N
  

   (3)

75
0: (112.5cos ) 0
cos
AB
MN W 


  



Substitute for
B
N from Eq. (3), cancel W, and simplify to find

32
3
9.6cos 4sin cos 6.6667 0
cos (2.4 tan ) 1.6667



Solve by trial & error:
35.8 
www.elsolucionario.org

1321. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.45
In Problem 8.44, determine the smallest value of  for which the rod
will not fall out the pipe.
PROBLEM 8.44 A slender steel rod of length 225 mm is placed
inside a pipe as shown. Knowing that the coefficient of static friction
between the rod and the pipe is 0.20, determine the largest value of

for which the rod will not fall into the pipe.

SOLUTION
Motion of rod impends up at A and right at B.

AsABsB
FNFN

0: sin cos 0
xABB
FNN F    

sin cos 0
AB sB
NN N   

(sin cos )
AB s
NN   (1)

0: cos sin 0
yABB
FFNFW    

cos sin 0
sA B sB
NN N W   (2)
Substitute for
A
N from Eq. (1) into Eq. (2):

(sin cos ) cos sin 0
sB s B sB
NNNW 

22
(1 ) cos 2 sin (1 0.2 ) cos 2(0.2) sin
B
ss
WW
N
  

   (3)

75
0: (112.5cos ) 0
cos
AB
MN W 


  



Substitute for
B
N from Eq. (3), cancel W, and simplify to find

32
3
9.6 cos 4sin cos 6.6667 0
cos (2.4 tan ) 1.6667



Solve by trial + error:
20.5 
www.elsolucionario.org

1322. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.46
Two slender rods of negligible weight are pin-connected at C and
attached to blocks
A and B, each of weight W. Knowing that
80
 and that the coefficient of static friction between the
blocks and the horizontal surface is 0.30, determine the largest
value of
P for which equilibrium is maintained.

SOLUTION
FBD pinC:







sin 20 0.34202
cos 20 0.93969


AC
BC
FP P
FP P

0: sin 30 0  
yAAC
FNWF
or
0.34202 sin 30 0.171010 
A
NW P W P
FBD block
A: 0: cos 30 0  
xAAC
FFF
or 0.34202 cos 30 0.29620
A
FP P
For impending motion at
A:
AsA
FN
Then
0.29620
: 0.171010
0.3
A
A
s
F
NW P P

 
or
1.22500PW

0: cos 30 0  
yBBC
FNWF

0.93969 cos 30 0.81380 
B
NW P W P

0: sin 30 0  
xBC B
FF F
0.93969 sin 30 0.46985
B
FP P


www.elsolucionario.org

1323. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.46 (Continued)

FBD block B: For impending motion at B:
BsB
FN
Then
0.46985
: 0.81380
0.3
B
B
s
F P
NW P

 
or
1.32914PW
Thus, maximum
P for equilibrium
max
1.225PW 
www.elsolucionario.org

1324. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.47
Two slender rods of negligible weight are pin-connected at C and
attached to blocks A and B, each of weight W. Knowing that
P  1.260W and that the coefficient of static friction between the
blocks and the horizontal surface is 0.30, determine the range of
values of θ, between 0 and 180°, for which equilibrium is
maintained.

SOLUTION
AC and BC are two-force members
Free body: Joint C
Force triangle:

From Force triangle:
sin( 60 ) 1.26 sin( 60 )
AB
FP W    (1)

cos( 60 ) 1.26 cos( 60 )
BC
FP W 
   (2)
We shall, in turn, seek  corresponding to impending motion of each block
For motion of A impending to left
from solution of Prob. 8.46: 0.419
AC
F W
EQ (1):
0.419 1.26 sin( 60 )
sin( 60 ) 0.33254
60 19.423
AC
FWW 


 

 

79.42
  
For motion of B impending to right.
from solution of Prob. 8.46: 1.249
BC
F W
Eq. (2):
1.249 1.26 cos( 60 )
cos( 60 ) 0.99127
BC
FWW


 



60 7.58
 

60 7.58
  67.6 

60 7.58
  52.4 
www.elsolucionario.org

1325. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.47 (Continued)

For motion of A impending to right 180 60 16.7 103.3  
Law of sines:
sin16.7 sin103.3
0.29528
AB
AB
F W
FW





Note: Direction of
AB
F is kept same as in free body of Joint C.
Eq. (1):
0.29528 1.26 sin( 60 )
sin( 60 ) 0.23435
AB
FWW 
   


( 60 ) 13.553
  46.4 
Summary:

A moves
to right
No
motion
B moves
to right
No
motion
A moves
to left



46.4° 52.4° 67.6° 79.4°
No motion for: 46.4 52.4 and 67.6° 79.4°      
www.elsolucionario.org

1326. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.48
The machine part ABC is supported by a frictionless hinge at
B and a 10 wedge at C. Knowing that the coefficient of
static friction is 0.20 at both surfaces of the wedge,
determine (
a) the force P required to move the wedge to the
left, (
b) the components of the corresponding reaction at B.

SOLUTION

11
0.20
tan tan 0.20 11.3099
s
ss




 

Free body:
ABC


10 11.31 21.31  
0: ( cos 21.3099 )(250 mm) (600 N)(200 mm) 0
515.23 N
BC
C
MR
R   



Free body: Wedge
Force triangle:

www.elsolucionario.org

1327. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.48 (Continued)

Law of sines:

515.23 N
sin 32.6198 sin 78.690
P



(
a) 283 NP 
(
b) Returning to free body of ABC:

0: 600 (515.23)sin 21.3099 0
xx
FB   

412.76 N
x
B 413 N
x
B 

0: (515.23)cos 21.3099 0
yy
FB  

480.0 N
y
B 480 N
y
B 

www.elsolucionario.org

1328. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.49
Solve Prob. 8.48 assuming that the wedge is moved to the
right.
PROBLEM 8.48
The machine part ABC is supported by a frictionless hinge at
B and a 10 wedge at C. Knowing that the coefficient of
static friction is 0.20 at both surfaces of the wedge,
determine (
a) the force P required to move the wedge to the
left, (
b) the components of the corresponding reaction at B.

SOLUTION

11
0.20
tan tan 0.20 11.3099
s
ss




 

Free body:
ABC



10 11.31 21.31  
0: ( cos1.3099 )(250 mm) (600 N)(200 mm) 0
480.13 N
BC
C
MR
R   



Free body: Wedge



Force triangle:
www.elsolucionario.org

1329. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.49 (Continued)

Law of sines:

480.13 N
sin 12.6198 sin 78.690
P



(
a) 107.0 NP 
(
b) Returning to free body of ABC:

0: 600 (480.13)sin 1.3099 0
xx
FB   

610.98 N
x
B 611 N
x
B 

0: (480.13)cos1.3099 0
yy
FB  

480.0 N
y
B 480 N
y
B 
www.elsolucionario.org

1330. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.50
Two 8 wedges of negligible weight are used to move and position the
800-kg block. Knowing that the coefficient of static friction is 0.30 at all
surfaces of contact, determine the smallest force
P that should be applied
as shown to one of the wedges.

SOLUTION

1
0.30 tan 0.30 16.70
ss


 
Free body: 800-kg block and right-hand wedge

2
(800 kg)(9.81 m/s ) 7848 NW

Force triangle:
90 16.70 24.70
48.60  

Law of sines:
1
1 7848 N
sin 16.70 sin 48.60
3006.5 N



R
R

Free body: Left-hand wedge

Force triangle:
16.70 24.70
41.40

Law of sines:
1
sin 41.40 sin(90 16.70 )
RP




3006.5 N
sin 41.40 cos16.70°
P


2080 NP 

www.elsolucionario.org

1331. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.51
Two 8 wedges of negligible weight are used to move and position the
800-kg block. Knowing that the coefficient of static friction is 0.30 at all
surfaces of contact, determine the smallest force
P that should be applied
as shown to one of the wedges.

SOLUTION

1
0.30 tan 0.30 16.70
ss


 
Free body: 800-kg block

Force triangle:
2
(800 kg)(9.81 m/s ) 7848 N
90 2 90 2(16.70 )
56.60
s
W


  

Law of sines:
1
1 7848 N
sin 16.70 sin 56.60
2701 N



R
R

Free body: Right-hand wedge
Force triangle:


16.70 24.70
41.40

Law of sines:
1
sin 41.40 sin(90 24.70 )
RP




2701 N
sin 41.40 cos 24.70°
P


1966 NP 

www.elsolucionario.org

1332. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.52
The elevation of the end of the steel beam supported by a concrete
floor is adjusted by means of the steel wedges
E and F. The base
plate
CD has been welded to the lower flange of the beam, and the
end reaction of the beam is known to be 18 kips. The coefficient of
static friction is 0.30 between two steel surfaces and 0.60 between
steel and concrete. If the horizontal motion of the beam is
prevented by the force
Q, determine (a) the force P required to
raise the beam, (
b) the corresponding force Q.

SOLUTION
1
0.30 tan 0.30 16.70
ss


 
Free body: Beam and plate
CD

1
1
(18 kips)
cos16.7°
18.7926 kips
R
R


Free body: Wedge
E

(
a)
18.7926 kips
sin 45.4 cos28.7
P


15.26 kipsP 
(b)

(18 kips) tan16.7Q 5.40 kipsQ 
Free body: Wedge
F(To check that it does not move.)

www.elsolucionario.org

1333. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.52 (Continued)


Since wedge
F is a two-force body,
2
R and
3
R are colinear
Thus
28.7
But
1
concrete
tan 0.6 31.0

 OK
www.elsolucionario.org

1334. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.53
Solve Prob. 8.52 assuming that the end of the beam is to be
lowered.
PROBLEM 8.52
The elevation of the end of the steel beam supported by a concrete
floor is adjusted by means of the steel wedges
E and F. The base
plate
CD has been welded to the lower flange of the beam, and the
end reaction of the beam is known to be 18 kips. The coefficient of
static friction is 0.30 between two steel surfaces and 0.60 between
steel and concrete. If the horizontal motion of the beam is
prevented by the force
Q, determine (a) the force P required to
raise the beam, (
b) the corresponding force Q.

SOLUTION
Free body: Beam and plate
CD
1
0.30 tan 0.30 16.70
ss


 


1
1
(18 kips)
cos16.7°
18,792.6 lb
R
R


Free body: Wedge
E

(
a)
18.7926 kips
sin 21.4 sin85.3
P


6.88 kipsP 
www.elsolucionario.org

1335. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.53 (Continued)
(b)

(18 kips) tan16.7Q 5.40 kipsQ 
Free body: Wedge
F
(To check that it does not move.)
Since wedge
F is a two-force body,
2
R and
3
R are colinear
Thus
4.7
But
1
concrete
tan 0.6 31.0

 OK
www.elsolucionario.org

1336. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.54
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact is
0.25 and that
45 ,
 determine the smallest force P required to raise
block
A.

SOLUTION

11
tan tan 0.25 14.036
ss


 
FBD block A:


2
2 3kN
sin 104.036 sin 16.928°
10.0000 kN
R
R




FBD wedge B:


10.0000 kN
sin 73.072 sin 75.964°
9.8611 kN



P
P
9.86 kNP 
www.elsolucionario.org

1337. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.55
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact is
0.25 and that
45 ,
 determine the smallest force P for which
equilibrium is maintained.

SOLUTION

11
tan tan 0.25 14.036
ss


 
FBD block A:


2
2 3kN
sin(75.964 ) sin(73.072 )
3.0420 kN
R
R




FBD wedge B:


3.0420 kN
sin 16.928 sin 104.036°
0.91300 kN



P
P
913 NP 
www.elsolucionario.org

1338. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.56
Block A supports a pipe column and rests as shown on wedge B. The
coefficient of static friction at all surfaces of contact is 0.25. If
0,
P
determine (a) the angle
 for which sliding is impending, (b) the
corresponding force exerted on the block by the vertical wall.

SOLUTION
Free body: Wedge B
1
tan 0.25 14.04
s



(a) Since wedge is a two-force body,
2
R and
3
R must be equal and opposite.
Therefore, they form equal angles with vertical

s

and

2 2(14.04 )
ss
s
 

 


28.1 
Free body: Block A


1
(3 kN)sin 14.04 0.7278 kNR
(b) Force exerted by wall:
1
728 NR 14.04° 

www.elsolucionario.org

1339. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.57
A wedge A of negligible weight is to be driven between two 100-lb
blocks B and C resting on a horizontal surface. Knowing that the
coefficient of static friction between all surfaces of contact is 0.35,
determine the smallest force
P required to start moving the wedge
(a) if the blocks are equally free to move, (b) if block C is securely
bolted to the horizontal surface.

SOLUTION
Wedge angle
10.75 in.
tan 10.62
4 in.



(a) Free body: Block B

1
tan 0.35 19.29
s






1 100 lb
sin19.29 sin 40.8
R



1
50.557 lbR


Free body: Wedge From symmetry
31
RR


 
1
2sin
2 50.557 lb sin 10.62 19.29
s
PR
P 




50.4 lbP 
(b) Free body diagrams remain unchanged; therefore, result is the same.
50.4 lbP 
www.elsolucionario.org

1340. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.58
A 15 wedge is forced into a saw cut to prevent binding
of the circular saw. The coefficient of static friction
between the wedge and the wood is 0.25. Knowing that a
horizontal force
P with a magnitude of 30 lb was required
to insert the wedge, determine the magnitude of the forces
exerted on the board by the wedge after insertion.

SOLUTION
(a)Free body: Wedge being inserted,
0.25
s


1
tan 0.25 14.0362
s






Force Triangle:
Q = component of
R1 and R2perpendicular to P.

1
cot 21.536
2
15 lb cot 21.536
38.010 lb
QP
Q
Q






Free body: After P is removed only the components Q of
R1 and R2 remain;
38.0 lbQ 
Note: since 7.5 14.0362 ,

which is the angle of static friction, wedge remains in place
www.elsolucionario.org

1341. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.59
A 12 wedge is used to spread a split ring. The coefficient of static
friction between the wedge and the ring is 0.30. Knowing that a
force
P with a magnitude of 120 N was required to insert the wedge,
determine the magnitude of the forces exerted on the ring by the
wedge after insertion.

SOLUTION

11
tan tan 0.30 16.70
ss


 
FBD wedge:

From force triangle: Q = horizontal component of
R
120 N1
143.435 N
2tan 22.7
Q






FBD after wedge has been inserted: Wedge is now a two-force body with forces shown.


143.4 NQ 
Note: since 6.0 16.70 ,

which is the angle of static friction, wedge remains in place
www.elsolucionario.org

1342. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.60
The spring of the door latch has a constant of 1.8 lb/in. and in
the position shown exerts a 0.6-lb force on the bolt. The
coefficient of static friction between the bolt and the strike
plate is 0.40; all other surfaces are well lubricated and may be
assumed frictionless. Determine the magnitude of the force
P
required to start closing the door.

SOLUTION
Free body: Bolt


1
0.40
tan 0.40
21.801
s
s








Force triangle:

From force triangle,
(0.6 lb) tan 66.801
P
 

1.39997 lbP
 1.400 lbP 


www.elsolucionario.org

1343. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


––
PROBLEM 8.61
In Problem 8.60, determine the angle that the face of the bolt
should form with the line
BC if the force P required to close
the door is to be the same for both the position shown and the
position when
B is almost at the strike plate.
PROBLEM 8.60 The spring of the door latch has a constant
of 1.8 lb/in. and in the position shown exerts a 0.6-lb force on
the bolt. The coefficient of static friction between the bolt and
the strike plate is 0.40; all other surfaces are well lubricated
and may be assumed frictionless. Determine the magnitude of
the force
P required to start closing the door.

SOLUTION
For position shown in figure:
From Prob. 8.60: 1.400 lbP
For position when
B reaches strike plate: Free body: Bolt

1
0.40
tan 0.40 21.801
s
s




 



0.6 lb
3
0.6 lb (1.8 lb/in.) in.
8
1.275 lb
Fkx 








Force triangle:
From force triangle,

1.400 lb
tan( ) 1.09804
1.275 lb
47.675
s
s

 
 


47.675 21.801 25.874



180 90 90 90 25.874
  

64.1 


www.elsolucionario.org

1344. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.62
A 5 wedge is to be forced under a 1400-lb machine base at A.
Knowing that the coefficient of static friction at all surfaces is 0.20,
(a) determine the force
P required to move the wedge, (b) indicate
whether the machine base will move.

SOLUTION
Free body: Machine base
0: (1400 lb)(20 in) (70 in) 0
400 lb
By
y
MA
A
  



0: 1400 lb 0
yyy
FAB   


1400 400 1000 lb
y
B
Free body: Wedge
(Assume machine will not move)

0.20, tan 0.20 11.31
ss

We know that
400 lb
y
A
Force triangle:

(a)
(400 lb)(tan11.31 tan16.31 ) 197.0 lbP 197.0 lbP 
(b) Total maximum friction force at A and B:

0.20(1400 lb) 280 lb
mc
FW 
Since
:
m
PF Machine base will not move 
www.elsolucionario.org

1345. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.63
Solve Problem 8.62 assuming that the wedge is to be forced under the
machine base at
B instead of A.
PROBLEM 8.62 A 5 wedge is to be forced under a 1400-lb machine
base at
A. Knowing that the coefficient of static friction at all surfaces
is 0.20, (
a) determine the force P required to move the wedge,
(
b) indicate whether the machine base will move.

SOLUTION
See solution to Prob. 8.62 for F.B.D. of machine base and determination of 1000 lb.
y
B
Free body: Wedge (Assume machine will not move)

1
0.20 tan 0.20 11.31
ss




Force triangle:


(1000 lb)(tan 11.31 tan16.31 ) 493 lb
P
Total maximum friction force at
A and B:

0.20(1400 lb) 280 lb 493 lb
ms
FW  

(b) Since ,
m
PF machine base will move with wedge
(
a) We then have ,
m
PF

280 lbP 

www.elsolucionario.org

1346. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.64
A 15 wedge is forced under a 50-kg pipe as shown. The coefficient of static
friction at all surfaces is 0.20. (a) Show that slipping will occur between the
pipe and the vertical wall. (b) Determine the force
P required to move the
wedge.

SOLUTION
Free body: Pipe
0: sin (1 sin ) cos 0
BAA
MWrFr Nr     
Assume slipping at A:

cos (1 sin ) sin
sin
cos (1 sin )
sin 15
cos15 (0.20)(1 sin 15 )
0.36241

  

 





  

AsA
AsA
A
s
A
FN
NN W
W
N
W
N
W

0: sin sin cos 0
xB AA
FFWFN       

cos sin sin
0.36241 cos 15 0.20(0.36241 )sin 15 sin 15
0.072482   
 

BA sA
B
B
FN N W
FW WW
FW


0: cos cos sin 0
yB A A
FNWF N     

sin cos cos
(0.36241 )sin 15 0.20(0.36241 )cos15 cos15
1.12974   
 

BA sA
B
B
NN N W
NW WW
NW

Maximum available:
0.22595
BsB
FN W
(a) We note that
maxB
FF No slip at B 
(b) Free body: Wedge
2
0: cos sin 0
yBB
FNN F    

2
2
2
cos sin
(1.12974 )cos15 (0.07248 )sin 15
1.07249


BB
NN F
NW W
NW


www.elsolucionario.org

1347. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.64 (Continued)

2
0: cos sin 0   
xB B s
FF N NP

2
cos sin
(0.07248 )cos 15 (1.12974 )sin 15 0.2(1.07249 )
0.5769


BBs
PF N N
PW W W
PW


2
: 0.5769(50 kg)(9.81 m/s )Wmg P 283 NP


www.elsolucionario.org

1348. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.65
A 15 wedge is forced under a 50-kg pipe as shown. Knowing that the
coefficient of static friction at both surfaces of the wedge is 0.20, determine
the largest coefficient of static friction between the pipe and the vertical wall
for which slipping will occur at A.


SOLUTION
Free body: Pipe
0: cos ( sin ) 0
A B BB BB
MNr NrNrWr     

cos (1 sin )
cos15 0.2(1 sin15 )
1.4002
B
B
B
B
W
N
W
N
NW
 



  



0: sin cos 0
xAB BB
FNN N     


(sin cos )
(1.4002 )(sin15 0.2 cos15 )
0.63293 


AB B
A
NN
W
NW


0: cos sin 0
yABBB
FFWN N     

(cos sin )
(1.4002 )(cos15 0.2 sin )
0.28001
AB B
A
A
FN W
FW W
FW  




For slipping at A:
AAA
FN

0.28001
0.63293
A
A
A
F W
NW
 0.442
A
 
www.elsolucionario.org

1349. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.66*
A 200-N block rests as shown on a wedge of negligible weight. The
coefficient of static friction
s is the same at both surfaces of the wedge, and
friction between the block and the vertical wall may be neglected. For
P  100 N, determine the value of s for which motion is impending.
(
Hint: Solve the equation obtained by trial and error.)

SOLUTION
Free body: Wedge
Force triangle:

Law of sines:
2
sin(90 ) sin(15 2 ) 

 
s s
R P


2
sin(90 )
sin(15 2 )





s
s
RP (1)
Free body: Block
0
y
F
Vertical component of
R2 is 200 N
Return to force triangle of wedge. Note
100 NP


100 N (200 N) tan (200 N) tan(15 )
0.5 tan tan(15 )
 
 
 
s
s

Solve by trial and error
6.292
s


tan tan 6.292
ss

  0.1103
s
 
www.elsolucionario.org

1350. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.67*
Solve Problem 8.66 assuming that the rollers are removed and that s is the
coefficient of friction at all surfaces of contact.
PROBLEM 8.66* A 200-N block rests as shown on a wedge of negligible
weight. The coefficient of static friction
s is the same at both surfaces of
the wedge, and friction between the block and the vertical wall may be
neglected. For P 100 N, determine the value of
s for which motion is
impending. (Hint: Solve the equation obtained by trial and error.)


SOLUTION
Free body: Wedge
Force triangle:

Law of sines:
2
sin (90 ) sin (15 2 )
ss
R P


 

2
sin (90 )
sin (15 2 )
s
s
RP




(1)
Free body: Block (Rollers removed)
Force triangle:

Law of sines:
2
sin (90 ) sin (75 2 )
ss
R W


 

2
sin (90 )
sin (75 2 )
s
s
RW




(2)


www.elsolucionario.org

1351. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.67* (Continued)

Equate
R2 from Eq. (1) and Eq. (2):

sin (90 ) sin (90 )
sin (15 2 ) sin (75 2 )
100 lb
200 N
sin (90 )sin (15 2 )
0.5
sin(75 2 )sin(90 )
ss
ss
s s
s s
PW
P
W


 
 
 

 


 

 

Solve by trial and error:
5.784
s

 

tan tan 5.784
 
ss
0.1013
s
 
www.elsolucionario.org

1352. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.68
Derive the following formulas relating the load W and the force P exerted on the handle of the jack
discussed in Section 8.6. (a) P (Wr/a) tan (
s), to raise the load; (b) P (Wr/a) tan ( s), to lower the
load if the screw is self-locking; (c) P (Wr/a) tan (
s), to hold the load if the screw is not self-locking.

SOLUTION
FBD jack handle:
See Section 8.6. 0: 0 or
C
r
MaPrQ PQ
a
  

FBD block on incline:

(a) Raising load





tan ( )
s
QW 
tan ( )
s
r
PW
a
 
(b) Lowering load if screw is self-locking (i.e., if
)
s



tan ( )
s
QW 
tan ( )
s
r
PW
a
 
(c) Holding load is screw is not self-locking (i.e., if
)
s



tan ( )
s
QW 
tan ( )
s
r
PW
a
 
www.elsolucionario.org

1353. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.69
The square-threaded worm gear shown has a mean radius of 2 in. and a
lead of 0.5 in. The large gear is subjected to a constant clockwise couple of
9.6 kip · in. Knowing that the coefficient of static friction between the two
gears is 0.12, determine the couple that must be applied to shaft
AB in
order to rotate the large gear counterclockwise. Neglect friction in the
bearings at
A, B, and C.

SOLUTION
Free body: Large gear
0: (16 in.) 9.6 kip in. 0
0.6 kip 600 lb
C
MW
W  


Block-and-Incline analysis of worn gear

0.5 in.
tan 0.039789
2(2 in.)
2.28






1
0.12, tan 0.12 6.84
ss


 

(600 lb) tan9.12
96.32 lb
Torque (96.32 lb)(2 in.)
192.6 lb in.
r
Q
Q
 


 



Torque 16.05 lb ft 
www.elsolucionario.org

1354. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.70
In Problem 8.69, determine the couple that must be applied to shaft AB in
order to rotate the large gear clockwise.
PROBLEM 8.69 The square-threaded worm gear shown has a mean
radius of 2 in. and a lead of 0.5 in. The large gear is subjected to a constant
clockwise couple of 9.6 kip  in. Knowing that the coefficient of static
friction between the two gears is 0.12, determine the couple that must be
applied to shaft
AB in order to rotate the large gear counterclockwise.
Neglect friction in the bearings at
A, B, and C.

SOLUTION
Free body: Large gear
See solution to Prob. 8.69. We find 600 lbW
Block-and-incline analysis of worm gear
From Prob. 8.69 we have 2.28 and 6.84 .
s
 Since ,
s
gear is self-looking and a torque must
be applied to it to rotate large gear clockwise.


6.84 2.28 4.56
s
   





(600 lb) tan 4.56
47.85 lb
Q



Torque (47.85 lb)(2 in.)
95.70 lb in.
r
Q



Torque 7.98 lb ft 

www.elsolucionario.org

1355. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.71
High-strength bolts are used in the construction of many steel structures. For a 1-in.-
nominal-diameter bolt, the required minimum bolt tension is 51 kips. Assuming the
coefficient of friction to be 0.30, determine the required couple that should be applied to
the bolt and nut. The mean diameter of the thread is 0.94 in., and the lead is 0.125 in.
Neglect friction between the nut and washer, and assume the bolt to be square-threaded.

SOLUTION
Block-and-incline analysis of bolt and nut:


1
110.125 in.
tan
(0.94 in.)
2.4238
tan tan 0.30
16.6992
From force triangle:
(51,000 lb) tan (19.123 )
17,683.3 lb
Torque
2
0.94 in.
17,683.3 lb
2
8311.1lb in.
ss
s
Q
Q
d
Qr Q

















Torque
693 lb ft 
www.elsolucionario.org

1356. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.72
The position of the automobile jack shown is controlled by
a screw ABC that is single-threaded at each end (right-
handed thread at A, left-handed thread at C). Each thread
has a pitch of 2.5 mm and a mean diameter of 9 mm. If the
coefficient of static friction is 0.15, determine the
magnitude of the couple
M that must be applied to raise
the automobile.


SOLUTION
All members of jack are two-force members; analyze joint equilibrium at C and D.
Joint D:
Symmetry:
AD CD
FF


0: 2 sin 20 6000 N 0
yCD
FF  

8771.4 N
CD
F
Joint C:
Symmetry: 
CE CD
FF


0: 2 cos 25 0
xCD AC
FF F  


2(8771.4 N)cos20
16,484.8 N
AC
AC
F
F





www.elsolucionario.org

1357. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.72 (Continued)
Block-and-incline analysis of one screw:


1
2.5 mm
tan
(9 mm)
5.053
tan 0.15
8.531
16,484.8 N tan13.584
3983.3 N
s
Q
Q














But, we have two screws:
Torque 2 (3983.3 N) 0.009 mQr Qd Torque 35.8 N m 


www.elsolucionario.org

1358. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.73
For the jack of Prob. 8.72, determine the magnitude of the
couple
M that must be applied to lower the automobile.
PROBLEM 8.72
The position of the automobile jack shown is controlled by
a screw ABC that is single-threaded at each end (right-
handed thread at A, left-handed thread at C). Each thread
has a pitch of 2.5 mm and a mean diameter of 9 mm. If the
coefficient of static friction is 0.15, determine the
magnitude of the couple
M that must be applied to raise the
automobile.


SOLUTION All members of jack are two-force members; analyze joint equilibrium at C and D
Joint D:Symmetry:
AD CD
FF
0: 2 sin 20 6000 N 0
yCD
FF  
8771.4 N
CD
F
Joint C:
Symmetry: 
CE CD
FF 0: 2 cos 25 0
xCD AC
FF F  

2(8771.4 N)cos20
16,484.8 N
AC
AC
F
F







www.elsolucionario.org

1359. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.73 (Continued)

Block-and-incline analysis of one screw:


1
2.5 mm
tan
(9 mm)
5.053
tan 0.15
8.531
16,484.8 N tan 3.478
1001.87 N
s
Q
Q















But, we have two screws:
Torque 2 (1001.87 N) 0.009 mQr Qd Torque 9.02 N m 

www.elsolucionario.org

1360. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.74
The vise shown consists of two members connected by two double-threaded
screws with a mean radius of 0.25 in. and pitch of 0.08 in. The lower member
is threaded at A and B (
s = 0.35), but the upper member is not threaded. It is
desired to apply two equal and opposite forces of 120 lb on the blocks held
between the jaws. (a) What screw should be adjusted first? (b) What is the
maximum couple applied in tightening the second screw?

SOLUTION
Free body: Lower Jaw

0: 120 lb 10 in. 5 in. 0
B
MA  


240 lbA

0: 120 lb 0
y
FAB  


120 lbB

Since B < A, Screw A should be adjusted first.


Block-and-incline analysis of screw B:
For double-threaded screw: Lead 2 Pitch

2 0.08 in. 0.16 in.L

0.16 in.
tan 0.101859
2 2 0.25 in.
L
r

  5.816





www.elsolucionario.org

1361. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.74 (Continued)


11
tan tan 0.35
ss



19.29
s


25.11
s




tan 25.11
(120 lb) tan 25.11
56.238 lb
QW
Q
Q






Torque 56.238 lb 0.25 in.Qr Torque 14.06 lb in. 

www.elsolucionario.org

1362. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.75
The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius
6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The
coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of
the couple that must be applied to the sleeve in order to draw the rods closer together.


SOLUTION
To draw rods together:
Screw at A

1
2mm
tan
2 (6 mm)
3.037
tan 0.12
6.843
(2 kN) tan 9.88
348.3 N
Torque at
(348.3 N)(0.006 m)
2.0898 N m
s
Q
AQr















Same torque required at B Total torque
4.18N m 
www.elsolucionario.org

1363. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.76
Assuming that in Problem 8.75 a right-handed thread is used on both rods A and B, determine the
magnitude of the couple that must be applied to the sleeve in order to rotate it.
PROBLEM 8.75 The ends of two fixed rods A and B are each made in the form of a single-threaded
screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed
thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the
magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.


SOLUTION
From the solution to Problem 8.70,
Torque at
2.09 N mA
Screw at B: Loosening
3.037
6.843
(2 kN) tan 3.806
133.1 N




s
Q

Torque at
(133.1 N)(0.006 m)
0.79860 N m
BQr



Total torque
2.0848 N m 0.79860 N m  Total torque 2.89 N m 

www.elsolucionario.org

1364. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.77
A lever of negligible weight is loosely fitted onto a 75-mm-
diameter fixed shaft. It is observed that the lever will just start
rotating if a 3-kg mass is added at C. Determine the
coefficient of static friction between the shaft and the lever.

SOLUTION
0: (150) (100) 0   
OC D f
MW W Rr
But
2
2
(23 kg)(9.81m/s )
(30 kg)(9.81 m/s )
(53 kg)(9.81)
C
D
CD
W
W
RW W



Thus, after dividing by 9.81,

23(150) 30(100) 53 0
8.49 mm
f
f
r
r



But
8.49 mm
37.5 mm

f
s
r
r

0.226
s
 
www.elsolucionario.org

1365. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.78
A hot-metal ladle and its contents weigh 130 kips. Knowing that the
coefficient of static friction between the hooks and the pinion is 0.30,
determine the tension in cable
AB required to start tipping the ladle.

SOLUTION
Free body: Ladle


sin tan 0.30
sss


bearing
bearing
8 in.
sin (8 in.) (0.30) 2.4 in.
fs
r
rr


 


Ris tangent to friction circle at A.

0 : (64 in. ) (130 kips) 0
(64 2.4) (130)(2.4) 0
Aff
MT r r
T

 

4.70 kipsT

www.elsolucionario.org

1366. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.79
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the coefficient
of static friction between the shaft and the poorly lubricated
bearing is 0.40, determine the magnitude of the force
P required
to start raising the load.

SOLUTION
0: (45 ) (90 ) 0
Dff
MPrWr  

90
45
90 mm 4 mm
(196.2 N)
45 mm 4 mm
449.82 N
f
f
r
PW
r
P







450 NP 
www.elsolucionario.org

1367. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.80
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the coefficient
of static friction between the shaft and the poorly lubricated
bearing is 0.40, determine the magnitude of the force
P required
to start raising the load.

SOLUTION
Find P required to start raising load
0: (45 ) (90 ) 0
Dff
MPrWr  

90
45
90 mm 4 mm
(196.2 N)
45 mm 4 mm
411.54 N
f
f
r
PW
r
P







412 NP 
www.elsolucionario.org

1368. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.81
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the coefficient
of static friction between the shaft and the poorly lubricated
bearing is 0.40, determine the magnitude of the smallest force
P
required to maintain equilibrium.

SOLUTION
Find smallest P to maintain equilibrium
0: (45 ) (90 ) 0
Dff
MPrWr  

90
45
90 mm 4 mm
(196.2 N)
45 mm 4 mm
344.35 N
f
f
r
PW
r
P







344 NP 
www.elsolucionario.org

1369. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.82
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the coefficient
of static friction between the shaft and the poorly lubricated
bearing is 0.40, determine the magnitude of the smallest force
P
required to maintain equilibrium.

SOLUTION
Find smallest P to maintain equilibrium


0: (45 ) (90 ) 0
Dff
MPrWr  

90
45
90 mm 4 mm
(196.2 N)
45 mm 4 mm
376.38 N
f
f
r
PW
r
P







376 NP 
www.elsolucionario.org

1370. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.83
The block and tackle shown are used to raise a 150-lb load. Each of the 3-in.-
diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient
of static friction is 0.20, determine the tension in each portion of the rope as the
load is slowly raised.

SOLUTION
For each pulley: Axle diameter 0.5 in.

0.5 in.
sin 0.20 0.05 in.
2
fss
rr r

 




Pulley BC:
0: (3 in.) (150 lb)(1.5 in. ) 0
BCD f
MT r  

1
(150 lb)(1.5 in. 0.05 in.)
3
CD
T 77.5 lb
CD
T 

0: 77.5 lb 150 lb 0
yAB
FT    72.5 lb
AB
T 
Pulley DE:
0: (1.5 ) (1.5 ) 0  
BCDfEFf
MTrTr

1.5
1.5
1.5 in. 0.05 in.
(77.5 lb)
1.5 in. 0.05 in.
f
EF CD
f
r
TT
r







82.8 lb
EF
T 
www.elsolucionario.org

1371. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.84
The block and tackle shown are used to lower a 150-lb load. Each of the 3-in.-
diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient
of static friction is 0.20, determine the tension in each portion of the rope as the
load is slowly lowered.

SOLUTION
For each pulley:
0.5 in.
0.2 0.05 in.
2
fs
rr

 




Pulley BC:
0: (3 in.) (150 lb)(1.5 in. ) 0
BCD f
MT r  

(150 lb)(1.5 in. 0.05 in.)
3in.
CD
T

 72.5 lb
CD
T 

0: 72.5 lb 150 lb 0
yAB
FT    77.5 lb
AB
T 
Pulley DE:
(1.5 in. ) (1.5 in. ) 0
CD f EF f
TrTr  

1.5 in.
1.5 in.
1.5 in. 0.05 in.
(72.5 lb)
1.5 in. 0.05 in.
f
EF CD
f
r
TT
r







67.8 lb
EF
T 
www.elsolucionario.org

1372. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.85
A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the
coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the
required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.

SOLUTION


2
tan 0.02
100

Since a scooter rolls at constant speed, each wheel is in equilibrium. Thus,
W and R must have a common
line of action tangent to the friction circle.

(0.10)(12.5 mm)
1.25 mm
1.25 mm
tan tan 0.02
62.5 mm
fk
f
rr
rOB
OA






Diameter of wheel 2( ) 125.0 mmOA 
www.elsolucionario.org

1373. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.86
The link arrangement shown is frequently used in highway bridge
construction to allow for expansion due to changes in temperature. At each of
the 60-mm-diameter pins A and B the coefficient of static friction is 0.20.
Knowing that the vertical component of the force exerted by BC on the link is
200 kN, determine (a) the horizontal force that should be exerted on beam BC
to just move the link, (b) the angle that the resulting force exerted by beam
BC on the link will form with the vertical.

SOLUTION
Bearing:
30 mm
0.20(30 mm)
6mm
fs
r
rr







Resultant forces
R must be tangent to friction circles at Points C and D.
(a)

y
RVertical component  200 kN

tan
(200 kN) tan 1.375
4.80 kN



xy
RR


Horizontal force 4.80 kN 
(b)

6mm
sin
250 mm
sin 0.024



1.375 




www.elsolucionario.org

1374. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.87
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter
fixed shaft. Knowing that the coefficient of static friction between the
fixed shaft and the lever is 0.15, determine the force
P required to start
the lever rotating counterclockwise.

SOLUTION
0.15(1.25 in.)
0.1875 in.
fs
rr




0: (5 in. ) (50 lb) 0
Dff
MPr r  

50(0.1875)
5.1875
1.807 lb
P

1.807 lbP 
www.elsolucionario.org

1375. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.88
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed
shaft. Knowing that the coefficient of static friction between the fixed shaft
and the lever is 0.15, determine the force
P required to start the lever rotating
counterclockwise.

SOLUTION

0.15(1.25 in.)
0.1875 in.
2in.
tan
5in.
21.801
fs
f
rr
r







In EOD:

22
(2 in.) (5 in.)
5.3852 in.
sin
0.1875 in.
5.3852 in.
1.99531
f
OD
rOE
OD OD









Force triangle:

(50 lb) tan ( )
(50 lb) tan 23.796
22.049 lb
P 




22.0 lbP 
www.elsolucionario.org

1376. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.89
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter
fixed shaft. Knowing that the coefficient of static friction between the
fixed shaft and the lever is 0.15, determine the force
P required to start
the lever rotating clockwise.

SOLUTION
0.15(1.25 in.)
0.1875 in.
fs
f
rr
r




0: (5in. ) (50 lb) 0  
Dff
MPr r

50(0.1875)
5 0.1875
1.948 lb
P


1.948 lbP 
www.elsolucionario.org

1377. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.90
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed
shaft. Knowing that the coefficient of static friction between the fixed shaft
and the lever is 0.15, determine the force
P required to start the lever rotating
clockwise.

SOLUTION

0.15(1.25 in.)
0.1875 in.
5in.
tan
2in.
68.198
fs
rr







In EOD:

22
(2 in.) (5 in.)
5.3852 in.
0.1875 in.
sin
5.3852 in.
1.99531
OD
OD
OE
OD







Force triangle:


50 50 lb
tan ( ) tan 70.193
P


 18.01lbP 


www.elsolucionario.org

1378. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.91
A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-
mm-diameter axles. Knowing that the coefficients of friction are
0.020
s
 and 0.015,
k
 determine the
horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed.
Neglect rolling resistance between the wheels and the track.

SOLUTION

; 400 mm
sin tan
tan
62.5 mm
400 mm
0.15625
f
f
rrR
r r
RR
r
PW W
R
PW
W











For one wheel:
21
(30 mg)(9.81 m/s )
8
1
(294.3 kN)
8
W


For eight wheels of railroad car:
1
8(0.15625) (294.3 kN)
8
(45.984 ) kN
P




(a) To start motion: 0.020
s


(45.984)(0.020)
0.9197 kN
P

920 NP
(b) To maintain motion:
0.015
k


(45.984)(0.015)
0.6897 kN


P
690 NP
www.elsolucionario.org

1379. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.92
Knowing that a couple of magnitude 30 N m
 is required to start the
vertical shaft rotating, determine the coefficient of static friction between
the annular surfaces of contact.

SOLUTION
For annular contact regions, use Equation 8.8 with impending slipping:

33
21
22
21
2
3
s
RR
MN
RR





So,
33
22
2 (0.06 m) (0.025 m)
30 N m (4000 N)
3 (0.06 m) (0.025 m)
s






0.1670
s
 
www.elsolucionario.org

1380. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.93
A 50-lb electric floor polisher is operated on a surface
for which the coefficient of kinetic friction is 0.25.
Assuming that the normal force per unit area between
the disk and the floor is uniformly distributed, determine
the magnitude
Q of the horizontal forces required to
prevent motion of the machine.

SOLUTION
See Figure 8.12 and Eq. (8.9).
Using:

9in.
50 lbR
P


and

0.25
22
(0.25)(50 lb)(9 in.)
33
75 lb in.
k
k
MPR






0yields:
y
M (20 in.)
75 lb in. (20 in.)MQ
Q

3.75 lb
Q 
www.elsolucionario.org

1381. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.94*
The frictional resistance of a thrust bearing decreases as the shaft and bearing surfaces wear out. It is
generally assumed that the wear is directly proportional to the distance traveled by any given point of the
shaft and thus to the distance
r from the point to the axis of the shaft. Assuming, then, that the normal
force per unit area is inversely proportional to
r, show that the magnitude M of the couple required to
overcome the frictional resistance of a worn-out end bearing (with contact over the full circular area) is
equal to 75 percent of the value given by Eq. (8.9) for a new bearing.

SOLUTION
Using Figure 8.12, we assume
:
k
NAArr
r  

k
Nrrkr
r
 
We write
or
P NPdN 



2
00
2
2
00
2;
2
2
2
2 1
2222











 



 



R
kk
R
kk
k P
PkrRkk
R
Pr
N
R
Pr
MrFr Nr
R
PP R
M rdrd PR
RR

From Eq. (8.9) for a new bearing,

new
2
3
k
M PR
Thus,
1
2
2
new 3
3
4
M
M

new
0.75MM 
www.elsolucionario.org

1382. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.95*
Assuming that bearings wear out as indicated in Problem 8.94, show that the magnitude M of the couple
required to overcome the frictional resistance of a worn-out collar bearing is
1
122
()
k
MPRR
where P magnitude of the total axial force
R
1, R2 inner and outer radii of collar

SOLUTION
Let normal force on A be ,N and .



Nk
Ar
As in the text,
  FNMrF
The total normal force P is

2
1
2
0 0
lim









R
A Rk
PN rdrd
r


2
1
21
21
22()or
2( )
 


R
R P
PkdrkRRk
RR

Total couple:
2
1
2
worn
0 0
lim








R
A Rk
MMrrdrd
r




2
1
22
21
22
worn 2 1
21
2()
2( )

 
 



R
R PR R
MkrdrkRR
RR


worn 2 1
1
()
2
MPRR
 
www.elsolucionario.org

1383. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.96*
Assuming that the pressure between the surfaces of contact is uniform,
show that the magnitude M of the couple required to overcome frictional
resistance for the conical bearing shown is
33
21
22
21
2
3sin
k
PRR
M
RR
 




SOLUTION
Let normal force on A be N and .



N
k
A
So sin
r
NkA Ars s



  

where
 is the azimuthal angle around the symmetry axis of rotation.

sin
y
FN krr 
Total vertical force:
0
lim
y
A
PF



22
11
2
0
2
RR
RR
Pkrdrdkrdr





 




22
21
22
21
or
P
PkRR k
RR
 

Friction force:
FNkA 
Moment:
sin
r
MrFrkr



 

Total couple:
2
1
2
2
0 0
lim
sin
R
A Rk
MM rdrd













2
1
233
23
22
232
2
sin 3 sin
R
RkP
Mrdr RR
RR

 
 




33
21
22
21
2
3sin
RRP
M
RR
 



www.elsolucionario.org

1384. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.97
Solve Problem 8.93 assuming that the normal force per
unit area between the disk and the floor varies linearly
from a maximum at the center to zero at the
circumference of the disk.
PROBLEM 8.93 A 50-lb electric floor polisher is
operated on a surface for which the coefficient of
kinetic friction is 0.25. Assuming that the normal force
per unit area between the disk and the floor is
uniformly distributed, determine the magnitude
Q of the
horizontal forces required to prevent motion of the
machine.


SOLUTION
Let normal force on A be N and 1.
 


 
Nr
k
AR

11
rr
FNk Ak rr
RR  
 
    
 
 


2
0 00
lim 1
R
A r
PN krdrd
R




 





23
0
21 2
23
R rRR
Pk rdrk
RR


  
 



2
213
or
3
P
PkR k
R



2
0 00
3
2
0
34
3
3
2
lim 1
2
1
2
34 6
31
62



 




 
 
 


 









R
A
R r
MrF rkrdrd
R
r
kr dr
R
RR
kkR
R
P
RPR
R

where
0.25 9 in.
k
R 

www.elsolucionario.org

1385. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.97 (Continued)


50 lbPW

Then
1
(0.25)(50 lb)(9 in.)
2
56.250 lb in.


M

Finally
56.250 lb in.
20 in.
M
Q
d

 2.81 lbQ 
www.elsolucionario.org

1386. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.98
Determine the horizontal force required to move a 2500-lb automobile with 23-in.-diameter tires along a
horizontal road at a constant speed. Neglect all forms of friction except rolling resistance, and assume the
coefficient of rolling resistance to be 0.05 in.

SOLUTION
FBD wheel:


1
11.5 in.
0.05 in.
sin
r
b
b
r






1
tan tan sin
b
PW W
r




for each wheel, so for total

10.05
2500 lb tan sin
11.5
P




10.87 lbP  www.elsolucionario.org

1387. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.99
Knowing that a 6-in.-diameter disk rolls at a constant velocity down a 2 percent incline, determine the
coefficient of rolling resistance between the disk and the incline.

SOLUTION
FBD disk:


tan slope 0.02

tan (3 in.)(0.02)br 0.0600 in.b 
www.elsolucionario.org

1388. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.100
A 900-kg machine base is rolled along a concrete floor using a series of
steel pipes with outside diameters of 100 mm. Knowing that the
coefficient of rolling resistance is 0.5 mm between the pipes and the
base and 1.25 mm between the pipes and the concrete floor, determine
the magnitude of the force
P required to slowly move the base along the
floor.

SOLUTION
FBD pipe:



2
1
(900 kg)(9.81 m/s )
8829.0 N
.5 mm 1.25 mm
sin
100 mm
1.00273
Wmg








tanPW for each pipe, so also for total

(8829.0 N) tan (1.00273 )P 154.4 NP 
www.elsolucionario.org

1389. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.101
Solve Problem 8.85 including the effect of a coefficient of rolling resistance of 1.75 mm.
PROBLEM 8.85 A scooter is to be designed to roll down a 2 percent slope at a constant speed.
Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is
0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels
and the ground.


SOLUTION
Since the scooter rolls at a constant speed, each wheel is in equilibrium. Thus, W and R must have a
common line of action tangent to the friction circle.
Radius of wheela

2
tan 0.02
100

Since b and
f
rare small compared to a,

tan 0.02
f k
rb rb
aa
 
 

Data:
0.10, 1.75 mm, 12.5 mm
k
br 

(0.10)(12.5 mm) 1.75 mm
0.02
150 mm



a
a
Diameter 2 300 mma 
www.elsolucionario.org

1390. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.102
Solve Problem 8.91 including the effect of a coefficient of rolling resistance of 0.5 mm.
PROBLEM 8.91 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter
wheels with 125-mm-diameter axles. Knowing that the coefficients of friction are
0.020
s
 and 0.015,
k

determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant
speed. Neglect rolling resistance between the wheels and the track.


SOLUTION
For one wheel:

tan sin
tan
tan
88
f
f
rr
rb
a
rb
a
WWrb
Q
a













For eight wheels of car:
2
(30 Mg)(9.81 m/s ) 294.3 kN
400 mm, 62.5 mm, 0.5 mm


 
 
rb
PW
a
Wmg
ar b
(a) To start motion: 0.02
(0.020)(62.5 mm) 0.5 mm
(294.3 kN)
400 mm
s
P



1.288 kNP 
(b) To maintain constant speed 0.015
(0.015)(62.5 mm) 0.5 mm
(294.3 kN)
400 mm
k
P



1.058 kNP 
www.elsolucionario.org

1391. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.103
A rope having a weight per unit length of 0.4 lb/ft is wound 2
2
1 times
around a horizontal rod. Knowing that the coefficient of static friction
between the rope and the rod is 0.30, determine the minimum length x of
rope that should be left hanging if a 100-lb load is to be supported.


SOLUTION
Use eq. (8.14) with




1
2
0.4 lb/ft
100 lb 0.4 lb/ft 10 ft 104 lb
Tx
T

 


0.30 =2.5 2 5
s
 
Then,

2
1
T
e
T



0.30 5104
0.4
e
x



2.34 ftx 


www.elsolucionario.org

1392. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.104
A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the
hawser, a dockworker can resist a force of 5000 lb on the other end of the hawser. Determine (
a) the
coefficient of static friction between the hawser and the bollard, (
b) the number of times the hawser
should be wrapped around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force.

SOLUTION
(a)
12
2
1
2
1
2 turns 2(2 ) 4
80 lb, 5000 lb
ln
1 1 5000 lb
ln ln
480 lb
 






s
s
TT
T
T
T
T


14.1351
ln 62.5
44
s

 
 0.329
s


(b)
12
2
1
2
1
80 lb, 20,000 lb, 0.329
ln
1 1 20,000 lb
ln ln
0.329 80 lb
15.5215
ln(250) 16.783
0.329 0.329
s
s
TT
T
T
T
T 



 



Number of turns
16.783
2
 Number of turns 2.67 
www.elsolucionario.org

1393. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.105
Two cylinders are connected by a rope that passes over two fixed rods as
shown. Knowing that the coefficient of static friction between the rope
and the rods is 0.40, determine the range of the mass m of cylinder D for
which equilibrium is maintained.


SOLUTION
For each rod, , 0.4
2
s





For impending motion of A downward,


0.4 2
2
: 3.5136
s
AA
Qmg m
ee
mg Q m





  
 50 kg 3.5136 175.680 kgm
For impending motion of A upward,


3.1536
sA
m
e
m 

50 kg
14.2304 kg
5.5136
m

Range for equilibrium,
14.23 kg 175.7 kgm 

www.elsolucionario.org

1394. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.106
Two cylinders are connected by a rope that passes over two fixed rods as
shown. Knowing that for cylinder D upward motion impends when
m = 20 kg, determine (a) the coefficient of static friction between the rope
and the rods, (b) the corresponding tension in portion BC of the rope.

SOLUTION
For each rod,
2




(a)

:
s sBCAA
BC
Tmg m
ee
Tmg m 
  50 kg
20 kg
0.91629
0.29166
s
s
s
e








0.292
s
 
For impending motion of A upward,


3.1536
sA
m
e
m 

50 kg
14.2304 kg
5.5136
m

(b)

0.29166
2
20 kg
BC
T
e
g



310 N
BC
T 
www.elsolucionario.org

1395. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.107
Knowing that the coefficient of static friction is 0.25 between the rope and the
horizontal pipe and 0.20 between the rope and the vertical pipe, determine the
range of values of P for which equilibrium is maintained.

SOLUTION
Horizontal pipe
s h

Vertical pipe
s v


For motion of P to be impending downward:

(/2) (/2)
,,
400 N
hv h
PQS
ee e
QS
  
 
Multiply equation member by member:

(2 )(/2)
400
hvh
PQ S
e
QS


or:
()
400 N
hv
P
e

 (1)
For motion of
P to be impending upward, we find in a similar way

()
400 N
hv
P
e

 (2)
Given data:
0.25, 0.20
hv

From (1):
0.45
(400 N) 1644 NPe


From (2):
0.45
(400 N) 97.3 NPe


Range for equilibrium:
97.3 N 1644 NP 
www.elsolucionario.org

1396. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.108
Knowing that the coefficient of static friction is 0.30 between the rope and the
horizontal pipe and that the smallest value of P for which equilibrium is
maintained is 80 N, determine (a) the largest value of P for which equilibrium
is maintained, (b) the coefficient of static friction between the rope and the
vertical pipe.

SOLUTION
Horizontal pipe:
s h
Vertical pipe:
s v


For motion of P to be impending downward:

(/2) (/2)
,,
400 N
hv h
PQS
ee e
QS
  
 
Multiply equation member by member:

(2 )(/2)
400
hvh
PQ S
e
QS


or:
()
400 N
hv
P
e

 (1)
For motion of
P to be impending upward, we find in a similar way

()
400 N
hv
P
e

 (2)
Setting
max
PP in Eq. (1), 80 NP in Eq. (2) and 0.30
h
 in both, we get

(0.30) (0.30)max 80 N
(1 ); (2 )
400 N 400 N
vv
P
ee
 
 
(a)
Multiplying (1 ) by (2 ):

0max80
1
400 400
P
e
max
2000 NP 
(b)
From
400
(2 ) : (0.30 ) ln ln 5 1.60944
80
v
  

0.30 0.512
v
 0.212
v
 
www.elsolucionario.org

1397. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.109
A band belt is used to control the speed of a flywheel as shown.
Determine the magnitude of the couple being applied to the
flywheel, knowing that the coefficient of kinetic friction between
the belt and the flywheel is 0.25 and that the flywheel is rotating
clockwise at a constant speed. Show that the same result is
obtained if the flywheel rotates counterclockwise.

SOLUTION
Free body: Flywheel

For clockwise rotation of flywheel,
T1 and T2 are located as shown.

33
2 radians
42





3
0.25
22
1
3.2482
k
T
ee
T




 
21
3.2482TT
Free body: Handle

  
12
0: 0.08 m 100 N 0.32 m 0
C
MTT   

11
3.2482 400 NTT

12
94.157 N, 305.84 NTT
From free body: Flywheel
 
12
0: 0.15 m 0
E
MMTT  


94.157 N 305.84 N 0.15 m 0M  31.8 N mM 

www.elsolucionario.org

1398. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.110
The setup shown is used to measure the output of a small turbine. When the
flywheel is at rest, the reading of each spring scale is 14 lb. If a 105-lb · in.
couple must be applied to the flywheel to keep it rotating clockwise at
a constant speed, determine (a) the reading of each scale at that time, (b) the
coefficient of kinetic friction. Assume that the length of the belt does not
change.


SOLUTION
(a) Since the length of the belt is constant, the spring in scale B will increase in length by and the
spring in scale A will decrease by the same amount. Thus, the sum of the readings in scales A and B
remains constant:

14 lb 14 lb
AB
TT  28 lb
AB
TT (1)
On the other hand, the sum of the moments of T
A and T B about axle must be equal to moment of
couple:

( )(9.375 in.) 105 lb in.,
BA
TT 11.2 lb
BA
TT (2)
Solving (1) and (2) simultaneously

8.40 lb;
A
T 19.60 lb
B
T 
(b) Apply Eq. (8.13) with
21
, , 180 rad.
BA
TT TT 

2
1 19.60
ln : ln 0.84730
8.40
ss
T
T
  0.270
s
 
www.elsolucionario.org

1399. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.111
The setup shown is used to measure the output of a small turbine. The
coefficient of kinetic friction is 0.20 and the reading of each spring scale is
16 lb when the flywheel is at rest. Determine (a) the reading of each scale
when the flywheel is rotating clockwise at a constant speed, (b) the couple
that must be applied to the flywheel. Assume that the length of the belt does
not change.


SOLUTION
(a) Since the length of the belt is constant, the spring in scale B will increase in length by and the
spring in scale A will decrease by the same amount. Thus, the sum of the readings in scales A and B
remains constant:

16 lb 16 lb
AB
TT  32 lb
AB
TT (1)
We now apply Eq. (8.14) with

12
, , 0.20, 180 rad.
ABk
TT T T   

0.202
1
: 1.87446
k
B B
ATT
ee
TT
 


1.87446
BA
TT (2)
Substituting (2) into (1):

1.87446 32 lb
AA
TT

11.1325 lb
A
T 11.13 lb
A
T 
From (1):
32 lb 11.1325 lb
B
T

20.868 lb
B
T 20.9 lb
B
T 
(b) Couple applied to flywheel:

()
(20.868 lb 11.1325 lb)(9.375 in.)
BA
MTTr



91.3 lb in.M 
www.elsolucionario.org

1400. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.112
A flat belt is used to transmit a couple from drum B to drum A. Knowing
that the coefficient of static friction is 0.40 and that the allowable belt
tension is 450 N, determine the largest couple that can be exerted on drum A.

SOLUTION
FBD’s drums:

7
180 30
66
5
180 30
66
A
B









Since
,
BA
 slipping will impend first on B (friction coefficients being equal)
So
2max1
(0.4)5 /6
11
450 N or 157.914 N
sB
TT Te
Te T





12
0: (0.12 m)( ) 0
AA
MM TT  

(0.12 m)(450 N 157.914 N) 35.05 N m
A
M

35.1 N m
A
M 

www.elsolucionario.org

1401. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.113
A flat belt is used to transmit a couple from pulley A to
pulley B. The radius of each pulley is 60 mm, and a force
of magnitude P 900 N is applied as shown to the axle of
pulley A. Knowing that the coefficient of static friction is
0.35, determine (a) the largest couple that can be
transmitted, (b) the corresponding maximum value of the
tension in the belt.


SOLUTION
Drum A:

(0.35)2
1
21
3.0028
s
T
ee
T
TT
 




180 radians
(a) Torque:
0: (675.15 N)(0.06 m) (224.84 N)(0.06 m)
A
MM  
27.0 N mM 
(b)
12
0: 900N 0
x
FTT  


11
1
1
2
3.0028 900 N 0
4.00282 900
224.841 N
3.0028(224.841 N) 675.15 N
TT
T
T
T





max
675 NT 
www.elsolucionario.org

1402. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.114
Solve Problem 8.113 assuming that the belt is looped
around the pulleys in a figure eight.
PROBLEM 8.113 A flat belt is used to transmit a couple
from pulley A to pulley B. The radius of each pulley is 60
mm, and a force of magnitude P 900 N is applied as
shown to the axle of pulley A. Knowing that the coefficient
of static friction is 0.35, determine (a) the largest couple
that can be transmitted, (b) the corresponding maximum
value of the tension in the belt.


SOLUTION
Drum A:

4
240 240
180 3

 

60 1
sin
120 2
30





0.35(4/3 )2
1
21
4.3322
s
T
ee
T
TT
 



(a) Torque:
0: (844.3 N)(0.06 m) (194.9 N)(0.06 m) 0
B
MM   
39.0 N mM 
(b)
12
0: ( )cos30 900 N
x
FTT  


11
1
2
( 4.3322 )cos 30 900
194.90 N
4.3322(194.90 N) 844.3 NTT
T
T




max
844 NT 
www.elsolucionario.org

1403. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.115
The speed of the brake drum shown is controlled by a belt attached to the
control bar AD. A force
P of magnitude 25 lb is applied to the control bar
at A. Determine the magnitude of the couple being applied to the drum,
knowing that the coefficient of kinetic friction between the belt and the
drum is 0.25, that a 4 in., and that the drum is rotating at a constant
speed (a) counterclockwise, (b) clockwise.


SOLUTION
(a) Counterclockwise rotation
Free body: Drum

0.252
1
21
8 in. 180 radians
2.1933
2.1933
k
r
T
ee
T
TT
 

 


Free body: Control bar

12
0: (12 in.) (4 in.) (25 lb)(28 in.) 0
C
MT T   


11
1
2
(12) 2.1933 (4) 700 0
216.93 lb
2.1933(216.93 lb) 475.80 lbTT
T
T 




Return to free body of drum

12
0: (8 in.) (8 in.) 0
E
MMTT   

(216.96 lb)(8 in.) (475.80 lb)(8 in.) 0M

2070.9 lb in.M 2070 lb in.M 
(b) Clockwise rotation
0.252
1
21
8in. rad
2.1933
2.1933
k
r
T
ee
T
TT
 

 





www.elsolucionario.org

1404. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.115 (Continued)

Free body: Control rod
21
0: (12 in.) (4 in.) (25 lb)(28 in.) 0
C
MT T   

11
1
2
2
2.1933 (12) (4) 700 0
31.363 lb
2.1933(31.363 lb)
68.788 lb
TT
T
T
T





Return to free body of drum

12
0: (8 in.) (8 in.) 0
E
MMTT   

(31.363 lb)(8 in.) (68.788 lb)(8 in.) 0M

299.4 lb in.M 299 lb in.M 
www.elsolucionario.org

1405. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.116
The speed of the brake drum shown is controlled by a belt attached to the
control bar AD. Knowing that a 4 in., determine the maximum value of
the coefficient of static friction for which the brake is not self-locking
when the drum rotates counterclockwise.


SOLUTION

2
1
21
8 in., 180 radians
ss
s
r
T
ee
T
TeT
 





Free body: Control rod


12
0: (28 in.) (12 in.) (4 in.) 0
C
MP T T   

11
28 12 (4) 0PTeT

 
For self-locking brake:
0P

11
12 4
3
ln 3 1.0986
1.0986
0.3497









s
s
s
s
TTe
e
0.350
s
 
www.elsolucionario.org

1406. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.117
The speed of the brake drum shown is controlled by a belt attached to the
control bar AD. Knowing that the coefficient of static friction is 0.30 and
that the brake drum is rotating counterclockwise, determine the minimum
value of a for which the brake is not self-locking.


SOLUTION

0.302
1
21
8in., radians
2.5663
2.5663
s
r
T
ee
T
TT
 

 


Free body: Control rod


16 in.ba

12
0: (16 in. ) 0
C
MPbTbTa 
For brake to be self-locking,
0P

21 11
;2.5663 (16 )
2.5663 16


Ta Tb Ta T a
aa


3.5663 16a 4.49 in.a 
www.elsolucionario.org

1407. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.118
Bucket A and block C are connected by a cable that passes over drum B.
Knowing that drum B rotates slowly counterclockwise and that the
coefficients of friction at all surfaces are
s  0.35 and k  0.25,
determine the smallest combined mass m of the bucket and its contents
for which block C will (a) remain at rest, (b) start moving up the incline,
(c) continue moving up the incline at a constant speed.


SOLUTION
Free body: Drum

2
2 3
2/3
2
T
e
mg
Tmge



(1)

(a) Smallest m for block C to remain at rest
Cable slips on drum.
Eq. (1) with
2(0.25) /3
2
0.25; 1.6881

 
k
Tmge mg
Block C: At rest, motion impending
0: cos 30
cos 30
0.35 cos 30
100 kg
C
C
sC
C
FNmg
Nmg
FN mg
m

  

 


2
0: sin 30
C
FTFmg C   


1.6881 0.35 cos 30 sin 30 0
1.6881 0.19689
CC
C
mg mg mg
mm




0.11663 0.11663(100 kg);
C
mm

11.66 kgm 

(b) Smallest m to start block moving up
No slipping at both drum and block: 0.35
s

Eq. (1):
2(0.35) /3
2
2.0814Tmge mg






www.elsolucionario.org

1408. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.118 (Continued)

Block C:
Motion impending


100kg
C
m


0: cos30FNmg  


cos30
0.35 cos30
C
sC
Nmg
FN mg


 


2
0: sin30 0
C
FTFmg   

2.0814 0.35 cos30 sin30 0
2.0814 0.80311
0.38585 0.38585(100 kg)
CC
C
C
mg mg mg
mm
mm





38.6 kgm 
(c) Smallest m to keep block moving up drum: No slipping:
0.35
s

Eq. (1) with 0.35
s


2/3 2(0.35) /3
2
2
2.0814
s
Tmg mge
Tmg
 



Block C: Moving up plane, thus
0.25
k

Motion up

0: cos 30 0  
C
FNmg


cos 30
0.25 cos 30


 
C
kC
Nmg
FN mg


2
0: sin 30 0   
C
FTFmg

2.0814 0.25 cos 30 sin 30 0
2.0814 0.71651
0.34424 0.34424(100 kg)



CC
C
C
mg mg mg
mm
mm


34.4 kgm 
www.elsolucionario.org

1409. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.119
Solve Problem 8.118 assuming that drum B is frozen and cannot rotate.
PROBLEM 8.118 Bucket A and block C are connected by a cable that
passes over drum B. Knowing that drum B rotates slowly
counterclockwise and that the coefficients of friction at all surfaces are
0.35
s
 and 0.25,
k
 determine the smallest combined mass m of
the bucket and its contents for which block C will (a) remain at rest,
(b) start moving up the incline, (c) continue moving up the incline at a
constant speed.
SOLUTION
(a) Block C remains at rest: Motion impends
Drum:
0.35(2 /3)2
2
2.0814
 


k
T
ee
mg
Tmg

Block C: Motion impends
0: cos30 0
C
FNmg  

cos30
0.35 cos30
C
sC
Nmg
FN mg


 


2
0: sin30 0
C
FTFmg   

2.0814 0.35 cos30 sin30 0
2.0814 0.19689
0.09459 0.09459(100 kg)



CC
C
C
mg mg mg
mg m
mm
9.46 kgm 


(b) Block C: Starts moving up
0.35
s

Drum: Impending motion of cable

2
1
0.35(2/3 )
1
1
2.0814
0.48045






s
T
e
T
mg
e
T
mg
T
mg
www.elsolucionario.org

1410. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.119 (Continued)

Block C: Motion impends
0: cos30
C
FNmg  

cos30
0.35 cos30
C
sC
Nmg
FN mg


 


1
0: sin30 0
C
FTFmg   

0.48045 0.35 cos30 0.5 0
0.48045 0.80311
1.67158 1.67158(100 kg)



CC
C
C
mg m g m g
mm
mm
167.2 kgm 
(c) Smallest m to keep block moving
Drum: Motion of cable



0.25(2/3 )2
1
1
1
0.25
1.6881
0.59238
1.6881
 




k
k
T
ee
T
mg
T
mg
Tmg

Block C: Block moves
0: cos30 0
C
FNmg  

cos30
0.25 cos30
C
kC
Nmg
FN mg


 


1
0: sin30 0
C
FTFmg   

0.59238 0.25 cos30 0.5 0
0.59238 0.71651
1.20954 1.20954(100 kg)



CC
C
C
mg m g m g
mm
mm
121.0 kgm 
www.elsolucionario.org

1411. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.120
A cable is placed around three parallel pipes. Knowing that the
coefficients of friction are
0.25
s
 and 0.20,
k
 determine
(a) the smallest weight W for which equilibrium is maintained,
(b) the largest weight W that can be raised if pipe B is slowly
rotated counterclockwise while pipes A and C remain fixed.

SOLUTION
(a) 0.25
s
 at all pipes.


0.25 / 2 0.25 0.25 / 250 lb


AC BC
AB BC
TT
eee
TTW


/8 /4 /8 /8 /4 /8 /250 lb
4.8105
50 lb
4.8015; 10.394 lb
BCAB
AB BC
TT
eee e e
TTW
W
W
   
 

10.39 lbW 
(b) Pipe B rotated
;; ;
22

 
  
ks k



0.2 / 2 0.25 0.2 / 250 lb
BC BC
AB AB
TT
eee
TTW



/10 / 4 /10
/10 / 4 /10 / 2050 lb
0.85464
50 lb
0.85464
50 lb
58.504 lb
0.85464
BCAB
AB BC
TT
eee
TTW
ee
W
W

 
 
  



58.5 lbW 
www.elsolucionario.org

1412. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.121
A cable is placed around three parallel pipes. Two of the pipes are
fixed and do not rotate; the third pipe is slowly rotated. Knowing
that the coefficients of friction are
0.25
s
 and 0.20,
k

determine the largest weight W that can be raised (a) if only pipe
A is rotated counterclockwise, (b) if only pipe C is rotated
clockwise.

SOLUTION
(a) Pipe A rotates ;, ;
22
sk k

  
  


0.25 / 2 0.2 0.2 / 2
50 lb
BCAB AB
BC
TTT
eee
TW



/8 /5 /10
(1/ 8 1/ 5 1/10 ) 7 / 40
50 lb
0.57708
0.57708; 28.854 lb
50 lb


 
 


BCAB
AB BC
TT W
ee e
TT
ee
W
W
28.9 lbW 
(b) Pipe C rotates
;; ,
22
kk s

  
  


0.2 / 2 0.2 0.25 / 250 lb
AB
AB BC BC
T W
eee
TTT
 


/10 /5 /8 7 / 4050 lb
0.57708
50 lb
0.57708
28.854 lb
BCAB
AB BC
TT
eee e
TTW
W
W
  
   


28.9 lbW 
www.elsolucionario.org

1413. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.122
A cable is placed around three parallel pipes. Knowing that the
coefficients of friction are
0.25
s
 and 0.20,
k
 determine
(a) the smallest weight W for which equilibrium is maintained,
(b) the largest weight W that can be raised if pipe B is slowly
rotated counterclockwise while pipes A and C remain fixed.

SOLUTION
(a) Smallest W for equilibrium ,
s
B


0.25 0.25 0.25
50 lb
AC BC
AC BC
TT W
eee
TT



/4 /4 /4 3 /4
10.551
50 lb
10.551; 4.739 lb
50 lb
AC BC
AC BC
TT W
eee e
TT
W
W
 
 

4.74 lbW 
(b) Largest W which can be raised by pipe B rotated
,,,
kks
    


0.2 0.2 0.2550 lb
AC
AC BC BC
T W
eee
TTT



/5 /5 /4 (1/5 1/5 1/4)
3/2050 lb
1.602
50 lb 50 lb
1.602; 31.21 lb
1.602
  
 
 


AC BC
AC BC
TT
eee e
TTW
e
W
W

31.2 lbW 
www.elsolucionario.org

1414. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.123
A cable is placed around three parallel pipes. Two of the pipes are
fixed and do not rotate; the third pipe is slowly rotated. Knowing
that the coefficients of friction are
0.25
s
 and 0.20,
k

determine the largest weight W that can be raised (a) if only pipe A
is rotated counterclockwise, (b) if only pipe C is rotated clockwise.

SOLUTION
(a) Pipe A rotates
,,,
skk
    


0.25 0.2 0.2
50 lb
AC AC BC
BC
TTT
eee
TW



/4 /5 /5
(1/ 4 1/ 5 1/ 5) 3 / 20
50 lb
0.62423
0.62423; 31.21 lb
50 lb
AC BC
AC BC
TT W
ee e
TT
ee
W
W


 
 


31.2 lbW 
(b) Pipe C rotates
,,,
ksk
    


0.2 0.25 0.250 lb
BC BC
AC AC
TT
ee e
TT W

 

/5 /4 /5 (1/5 1/4 1/5) 3 /20
3/2050 lb
50 lb
1.602
50 lb
31.21 lb
1.602
AC BC
AC BC
TT
ee e e e
TTW
e
W
W
 
 
  


31.2 lbW 
www.elsolucionario.org

1415. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.124
A recording tape passes over the 20-mm-radius drive drum B and
under the idler drum C. Knowing that the coefficients of friction
between the tape and the drums are
0.40
s
 and 0.30
k
 and
that drum C is free to rotate, determine the smallest allowable value
of P if slipping of the tape on drum B is not to occur.

SOLUTION
FBD drive drum:


0: ( ) 0
BA
MrTTM 

300 N mm
15.0000 N
20 mm
A
M
TT
r

  

Impending slipping:
0.4
s
A
TTe Te
 

So
0.4
( 1) 15.0000 NTe


or
5.9676 NT
If C is free to rotate,
PT 5.97 NP 
www.elsolucionario.org

1416. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.125
Solve Problem 8.124 assuming that the idler drum C is frozen and
cannot rotate.
PROBLEM 8.124 A recording tape passes over the 20-mm-radius
drive drum B and under the idler drum C. Knowing that the
coefficients of friction between the tape and the drums are
0.40
s
 and 0.30
k
 and that drum C is free to rotate,
determine the smallest allowable value of P if slipping of the tape
on drum B is not to occur.

SOLUTION
FBD drive drum:


0: ( ) 0
BA
MrTTM 

300 N mm 15.0000 N
A
M
TT
r
   

Impending slipping:
0.4
s
A
TTe Te
 

So
0.4
( 1) 15.000 NeT


or
5.9676 NT
If C is fixed, the tape must slip
So
0.3 / 2
(5.9676 N) 9.5600 N
kC
PTe e
 
  9.56 NP 
www.elsolucionario.org

1417. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.126
The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe.
Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the
smallest value of
s
 for which the wrench will be self-locking when 200 mm, 30 mm,ar and
65 .


SOLUTION
For wrench to be self-locking (0),P the value of
s
 must prevent slipping of strap which is in contact
with the pipe from Point A to Point Band must be large enough so that at Point A the strap tension can
increase from zero to the minimum tension required to develop “belt friction” between strap and pipe.
Free body: Wrench handle

Geometry In CDH:
tan
sin
tan
sin







 
a
CH
a
CD
DE BH CH BC
a
DE r
a
AD CD CA r

On wrench handle 0: ( ) ( ) 0
DB
MTDEFAD  

sin
tan




B
a
r
TAD
aFDE
r
(1)

www.elsolucionario.org

1418. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.126 (Continued)

Free body: Strap at Point A


1
0: 2 0FTF  

1
2TF (2)
Pipe and strap
(2 ) radians
Eq. (8.13):
2
1
ln
s
T
T


1
ln
2
B
s
T
F

 (3)
Return to free body of wrench handle

0: sin cos 0
xB
FNFT    

sin cos
B
TN
FF

Since
,
s
FN we have
1
sin cos
B
s
T
F


or
sin
cos




s
B
T
F
(4)
(Note: For a given set of data, we seek the larger of the values of
s
 from Eqs. (3) and (4).)
For
200 mm, 30 mm, 65°ar 
Eq. (1):
200 mm
30 mm
sin 65
200 mm
30 mm
tan 65
190.676 mm
3.0141
63.262 mm
2 2 65 5.1487 radians
180








 

B
T
F
www.elsolucionario.org

1419. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.126 (Continued)

Eq. (3):
13.0141
ln
5.1487 rad 2
0.41015
5.1487


s

0.0797
 
Eq. (4):
sin 65
3.0141 cos 65
0.90631
2.1595
s



 



0.3497
 
We choose the larger value:
0.350
s
 
www.elsolucionario.org

1420. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.127
Solve Problem 8.126 assuming that 75 .
PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external
surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact,
determine the smallest value of
s
 for which the wrench will be self-locking when
200 mm, 30 mm,ar and 65 .


SOLUTION
For wrench to be self-locking (0),P the value of
s
 must prevent slipping of strap which is in contact
with the pipe from Point A to Point Band must be large enough so that at Point A the strap tension can
increase from zero to the minimum tension required to develop “belt friction” between strap and pipe.
Free body: Wrench handle

Geometry In CDH:
tan
sin
tan
sin
a
CH
a
CD
DE BH CH BC
a
DE r
a
AD CD CA r







 

On wrench handle 0: ( ) ( ) 0
DB
MTDEFAD  

sin
tan




B
a
r
TAD
aFDE
r
(1) www.elsolucionario.org

1421. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.127 (Continued)

Free body: Strap at Point A



1
0: 2 0FTF  

1
2TF (2)
Pipe and strap
(2 ) radians
Eq. (8.13):
2
1
ln
s
T
T


1
ln
2
B
s
T
F

 (3)
Return to free body of wrench handle

0: sin cos 0
xB
FNFT    

sin cos
B
TN
FF

Since
,
s
FN we have
1
sin cos
B
s
T
F


or
sin
cos




s
B
T
F
(4)
(Note: For a given set of data, we seek the larger of the values of
s
 from Eqs. (3) and (4).)
For
200 mm, 30 mm, 75°ar 
Eq. (1):
200 mm
30 mm
sin 75
200 mm
30 mm
tan 75
177.055 mm
7.5056
23.590 mm
2 2 75 4.9742
180








 

B
T
F
www.elsolucionario.org

1422. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.127 (Continued)

Eq. (3):
17.5056
ln
4.9742 rad 2
1.3225
4.9742
s



0.2659 
Eq. (4):
sin 75
7.5056 cos 75
0.96953
7.2468
s



 



0.1333 
We choose the larger value:
0.266
s
 
www.elsolucionario.org

1423. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.128
The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius
drum. Vertical motion of end E of the bar is prevented by the two stops
shown. Knowing that
0.30
s
 between the cable and the drum,
determine (a) the largest counterclockwise couple
0
M that can be applied
to the drum if slipping is not to occur, (b) the corresponding force exerted
on end E of the bar.

SOLUTION
Drum: Slipping impends 0.30
s


0.302
1
:2.5663
2.5663
D
B
DB
TT
ee
TT
TT
 


(a) Free-body: Drum and bar


0
0: (8 in.) 0
C
MME  

0
(3.78649 lb)(8 in.)
30.27 lb in.
M

0
30.3 lb in.M 
(b) Bar AE:
0: 10 lb 0
yBD
FTTE   

2.5663 10 lb 0
3.5663 10 lb 0
3.5663 10 lb

 

BB
B
B
TTE
TE
ET
(1)

0: (3 in.) (10 lb)(5 in.) (10 in.) 0
DB
ME T   

(3.5663 10 lb)(3 in.) 50 lb in. (10 in.) 0
BB
TT

20.699 80 3.8649 lb
BB
TT
Eq. (1):
3.5663(3.8649 lb) 10 lbE

3.78347 lbE 3.78 lbE 
www.elsolucionario.org

1424. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.129
Solve Problem 8.128 assuming that a clockwise couple
0
M is applied to
the drum.
PROBLEM 8.128 The 10-lb bar AE is suspended by a cable that passes
over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented
by the two stops shown. Knowing that
0.30
s
 between the cable and the
drum, determine (a) the largest counterclockwise couple
0
M that can be
applied to the drum if slipping is not to occur, (b) the corresponding force
exerted on end E of the bar.

SOLUTION
Drum: Slipping impends
2
1
0.30
0.30
2.5663
2.5663
s
B
D
BD
T
e
T
T
e
T
TT






(a) Free body: Drum and bar


0
0
0: (8 in.) 0
(2.1538 lb)(8 in.)
C
MME
M
  



0
17.23 lb in.M 
(b) Bar AE:

0: 10 lb 0
yBD
FTTE   

2.5663 10 lb
DD
TTE

3.5663 10 lb
D
ET  (1)

0: (10 in.) (10 lb)(5 in.) (13 in.) 0
BD
MT E   

(10 in.) 50 lb in. ( 3.5663 10 lb)(13 in.) 0
DD
TT   

36.362 80 lb in. 0; 2.200 lb
DD
TT
Eq. (1):
3.5633(2.200 lb) 10 lbE 

2.1538 lbE 2.15 lbE 
www.elsolucionario.org

1425. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.130
Prove that Eqs. (8.13) and (8.14) are valid for any shape of surface provided
that the coefficient of friction is the same at all points of contact.

SOLUTION
0: [ ( )]sin 0
2

   
n
FNTTT

or
(2 )sin
2

 NTT


0: [( ) ]cos 0
2

   
t
FTTT F

or
cos
2

FT

Impending slipping:
s
FN 
So
sin
cos 2 sin
222
ss
TTT


 

In limit as
 0:
or 
ss
dT
dT Td d
T

So
2
1
0
T
s
TdT
d
T




and
2
1
ln
s
T
T
 or
21
s
TTe

 
(Note: Nothing above depends on the shape of the surface, except it is assumed to be a smooth curve.)

www.elsolucionario.org

1426. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.131
Complete the derivation of Eq. (8.15), which relates the tension in both parts of a V belt.

SOLUTION
Small belt section:
Side view: End view:


0: 2 sin [ ( )]sin 0
22 2
y
N
FTTT
  


0: [( ) ]cos 0
2
x
FTTT F

  

Impending slipping:
2
cos sin
22
sin
2


  
ss
TT
FNT

In limit as
 0:
or
sin sin
22
 



ss
Td dT
dT d
T

so
2
1
0
sin
2





T
s
TdT
d
T

or
2
1
ln
sin
2



sT
T

or
2
/sin
21
s
TTe


 
www.elsolucionario.org

1427. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.132
Solve Problem 8.112 assuming that the flat belt and drums are replaced by a
V belt and V pulleys with
36 . (The angle  is as shown in Figure 8.15a.)
PROBLEM 8.112 A flat belt is used to transmit a couple from drum B to
drum A. Knowing that the coefficient of static friction is 0.40 and that the
allowable belt tension is 450 N, determine the largest couple that can be
exerted on drum A.

SOLUTION
Since  is smaller for pulley B. The belt will slip first at B.

2
5
6
/sin2
1
(0.4) /sin18 3.389
1
1
1
rad 5
150 rad
180° 6
450 N
450 N
29.63; 15.187 N
s
T
e
T
ee
T
T
T 






 






Torque on pulley A:


max 1
0: ( )(0.12 m) 0
(450 N 15.187 N)(0.12 m) 0
B
MMTT
M
   
 


52.18 N mM 52.2 N mM 
www.elsolucionario.org

1428. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.133
Solve Problem 8.113 assuming that the flat belt and pulleys
are replaced by a V belt and V pulleys with
36 .
 (The
angle
 is as shown in Figure 8.15a.)
PROBLEM 8.113 A flat belt is used to transmit a couple
from pulley A to pulley B. The radius of each pulley is 60
mm, and a force of magnitude
900 NP
 is applied as
shown to the axle of pulley A. Knowing that the coefficient
of static friction is 0.35, determine (a) the largest couple
that can be transmitted, (b) the corresponding maximum
value of the tension in the belt.

SOLUTION
Pulley A: rad

2
/sin2
1
0.35 / sin182
1
3.5582
1
21
35.1
35.1
s
T
e
T
T
e
T
T
e
T
TT 








12
0: 900 N 0
x
FTT  

11
1
2
35.1 900 N 0
24.93 N
35.1(24.93 N) 875.03 N



TT
T
T

21
0: (0.06 m) (0.06 m) 0
A
MMT T   

(875.03 N)(0.06 m) (24.93 N)(0.06 m) 0M

51.0 N mM 

max 2
TT
max
875 NT 
www.elsolucionario.org

1429. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.134
The coefficients of friction are 0.40
s
 and 0.30
k
between all
surfaces of contact. Determine the smallest force
P required to start the
30-kg block moving if cable AB (a) is attached as shown, (b) is removed.

SOLUTION
(a) Free body: 20-kg block

2
1
11
(20 kg)(9.81 m/s ) 196.2 N
0.4(196.2 N) 78.48 N
s
W
FN


 


11
0: 0 78.48 N    FTFTF
Free body: 30-kg block

2
2
2
22
N 294.3 N 490.5 N
(30 kg)(9.81 m/s ) 294.3 N
196.2
0.4(490.5 N) 196.2 N
s
W
N
FN




 


12
0: 0FPFFT   


78.48 N 196.2 N 78.48 N 353.2 NP

353 NP 
(b) Free body: Both blocks
Blocks move together

2
(50 kg)(9.81 m/s ) 490.5 NW

0: 0FPF  


0.4(490.5 N) 196.2 N
s
PN 

196.2 NP 
www.elsolucionario.org

1430. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.135
The coefficients of friction are 0.40
s
 and 0.30
k
between all
surfaces of contact. Determine the smallest force
P required to start
the 30-kg block moving if cable AB (a) is attached as shown, (b) is
removed.

SOLUTION
(a) Free body: 20-kg block

2
1
11
(20 kg)(9.81 m/s ) 196.2 N
0.4(196.2 N) 78.48 N
s
W
FN


 


11
0: 0 78.48 NFTFTF    
Free body: 30-kg block

2
2
2
22
(30 kg)(9.81 m/s ) 294.3 N
196.2 N 294.3 N 490.5 N
0.4(490.5 N) 196.2 N
s
W
N
FN



 


12
0: 0FPFF   


78.48 N 196.2 N 274.7 NP

275 NP 
(b) Free body: Both blocks
Blocks move together

2
(50 kg)(9.81 m/s )
490.5 N
W



0: 0
0.4(490.5 N) 196.2 N
s
FPF
PN

  
 

196.2 NP 
www.elsolucionario.org

1431. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.136
A 120-lb cabinet is mounted on casters that can be locked to prevent their
rotation. The coefficient of static friction between the floor and each caster
is 0.30. If
32 in.,h determine the magnitude of the force P required to
move the cabinet to the right (a) if all casters are locked, (b) if the casters at
B are locked and the casters at A are free to rotate, (c) if the casters at A are
locked and the casters at B are free to rotate.

SOLUTION
FBD cabinet:Note: for tipping,

0
AA
NF

tip
0: (12 in.) (32 in.) 0
B
MWP  


tip
2.66667P
(a) All casters locked. Impending slip:

AsA
BsB
FN
FN




0: 0  
yAB
FNNW


AB
NNW 120 lbW
So
AB s
FF W 0.3
s

0: 0
xAB
AB s
FPFF
PF F W

   


0.3(120 lb)P or 36.0 lbP 

tip
( 0.3 OK)PWP
(b) Casters at A free, so 0
A
F
Impending slip:

BsB
FN

0: 0
xB
FPF  

BsB B
s
P
PF N N

 

www.elsolucionario.org

1432. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 8.136 (Continued)


0: (32 in.) (12 in.) (24 in.) 0
AB
MPWN   

8 3 6 0 0.25
0.3
P
PW P W  

tip
( 0.25 OK)PWP

0.25(120 lb)P or 30.0 lbP 
(c) Casters at B free, so
0
B
F
Impending slip:
AsA
FN

0: 0
xAAsA
FPFPFN     

0.3
A
s
PP
N



0: (12 in.) (32 in.) (24 in.) 0
BA
MWPN   

386 0
0.3
0.107143 12.8572
P
WP
PW
 



tip
(OK)PP

 12.86 lbP 

www.elsolucionario.org

1433. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.137
A slender rod of length L is lodged between peg C and the vertical wall and
supports a load
P at end A. Knowing that the coefficient of static friction
between the peg and the rod is 0.15 and neglecting friction at the roller,
determine the range of values of the ratio L/a for which equilibrium is
maintained.

SOLUTION
FBD rod:
Free-body diagram: For motion of B impending upward:
0: sin 0
sin
BC
a
MPLN



  




2
sin
C
PL
N
a
 (1)

0: sin cos 0
yC sC
FN N P    

(sin cos )
C
NP 

Substitute for
C
N from Eq. (1), and solve for a/L.

2
sin (sin cos )
s
a
L
  (2)
For
30 and 0.15:
s

2
sin 30 (sin 30 0.15cos 30 )
0.092524 10.808
 

a
L
aL
La
For motion of B impending downward, reverse sense of friction force
.
C
F To do this we make
0.15 in. Eq. (2).
s

Eq. (2):
2
sin 30 (sin 30 ( 0.15)cos 30 )
0.15748 6.350
 

a
L
aL
La
Range of values of L/a for equilibrium:
6.35 10.81
L
a
 
www.elsolucionario.org

1434. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.138
The hydraulic cylinder shown exerts a force of 3 kN directed
to the right on Point B and to the left on Point E. Determine
the magnitude of the couple
M required to rotate the drum
clockwise at a constant speed.

SOLUTION
Free body: Drum
0: (0.25 m)( ) 0
CLR
MM FF   

(0.25 m)( )
LR
MFF (1)
Since drum is rotating
0.3
0.3
LkL L
RkR R
FN N
FN N



Free body: Left arm ABL
0: (3 kN)(0.15 m) (0.15 m) (0.45 m) 0
ALL
MFN   

0.45 kN m (0.3 )(0.15 m) (0.45 m) 0
0.405 0.45
  

LL
L
NN
N


1.111 kN
0.3 0.3(1.111 kN)
0.3333 kN
L
LL
N
FN



(2)
Free body: Right arm DER
0: (3 kN)(0.15 m) (0.15 m) (0.45 m) 0
DRR
MFN   
0.45 kN m (0.3 )(0.15 m) (0.45 m) 0
RR
NN  

0.495 0.45
R
N

0.9091 kN
0.3(0.9091 kN)
0.2727 kN
R
RkR
N
FN




(3)
Substitute for
L
Fand
R
Finto (1): (0.25 m)(0.333 kN 0.2727 kN)M

0.1515 kN mM

151.5 N mM 
www.elsolucionario.org

1435. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.139
A rod DE and a small cylinder are placed between two guides as shown. The
rod is not to slip downward, however large the force
P may be; i.e., the
arrangement is said to be self-locking. Neglecting the weight of the cylinder,
determine the minimum allowable coefficients of static friction at A, B, and C.

SOLUTION
Free body: Cylinder
Since cylinder is a two-force body,
B
Rand
C
R have the same line of action. Thus :
BC

From triangle OBC:

BC

Thus:
2
BC


For no sliding, we must have
tan ( ) , tan ( )
BBs CCs
 
Therefore:
() tan,
2
Bs

 () tan
2
cs

 
We also note that R B and R C are indeterminate.
Free body: Rod
Since R B is indeterminate, it may be as large as necessary to satisfy equation 0,
y
F
no matter how large
P is or how small
A
 is. Therefore

()
As
 may have any value 
www.elsolucionario.org

1436. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.140
Bar AB is attached to collars that can slide on the inclined rods
shown. A force
P is applied at Point D located at a distance a from
end A. Knowing that the coefficient of static friction
s between
each collar and the rod upon which it slides is 0.30 and neglecting
the weights of the bar and of the collars, determine the smallest
value of the ratio a/L for which equilibrium is maintained.

SOLUTION
FBD bar and collars:
Impending motion:
1
1
tan
tan 0.3
16.6992
ss






Neglect weights: 3-force FBD and
90ACB
so
cos(45 )
sin (45 )
sin (45 16.6992 )cos(45 16.6992 )
s
s
a
AC
l
a
l





    

0.225
a
l

www.elsolucionario.org

1437. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.141
Two 10 wedges of negligible weight are used to move and position the 400-
lb block. Knowing that the coefficient of static friction is 0.25 at all surfaces
of contact, determine the smallest force
P that should be applied as shown to
one of the wedges.

SOLUTION
Free body: Block and top wedge
1
tan 0.25 14.04
s


 
Force triangle

Law of sines:
2
2 400 lb
sin104.04 sin 51.92
493 lb
R
R




Free body: Lower wedge Force triangle

Law of sine:
493 lb
sin(14.04 24.04 ) sin 75.96
P

  

313.4 NP 313 lbP 
www.elsolucionario.org

1438. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.142
A 10 wedge is used to split a section of a log. The coefficient of static
friction between the wedge and the log is 0.35. Knowing that a force
P of
magnitude 600 lb was required to insert the wedge, determine the magnitude
of the forces exerted on the wood by the wedge after insertion.

SOLUTION
FBD wedge (impending motion ):

1
1
tan
tan 0.35
19.29
ss







By symmetry:
12
RR

1
0: 2 sin(5 ) 600 lb 0   
ys
FR
or
12
300 lb
729.30 lb
sin (5°+19.29°)
 RR

When P is removed, the vertical components of
1
R and
2
R vanish, leaving the horizontal components

12
1
cos(5 )
(729.30 lb)cos(5 19.29 )
xx
s
RR
R





12
665 lb
xx
RR 
(Note that
5,
s
 so wedge is self-locking.)
www.elsolucionario.org

1439. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.143
In the gear-pulling assembly shown the square-threaded screw AB has a mean radius
of 15 mm and a lead of 4 mm. Knowing that the coefficient of static friction is 0.10,
determine the couple that must be applied to the screw in order to produce a force of
3 kN on the gear. Neglect friction at end A of the screw.

SOLUTION
Block/Incline:

14mm
tan
30 mm
2.4302







1
1
tan
tan (0.10)
5.7106





ss


(3000 N) tan (8.1408 )
429.14 N
Q



Couple
(0.015 m)(429.14 N)
6.4371 N m



rQ

6.44 N mM 
www.elsolucionario.org

1440. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.144
A lever of negligible weight is loosely fitted onto a 30-mm-
radius fixed shaft as shown. Knowing that a force
P of
magnitude 275 N will just start the lever rotating clockwise,
determine (a) the coefficient of static friction between the
shaft and the lever, (b) the smallest force
P for which the
lever does not start rotating counterclockwise.

SOLUTION
(a) Impending motion

2
(40 kg)(9.81 m/s ) 392.4 NW

0: (160 ) (100 ) 0
Dff
MPrWr  

160 100
(160 mm)(275 N) (100 mm)(392.4 N)
275 N 392.4 N
7.132 mm
sin
7.132 mm
0.2377
30 mm
f
f
f
fss
f
s
PW
r
PW
r
r
rr r
r
r










 
0.238
s
 
(b) Impending motion
sin
(30 mm)(0.2377)
7.132 mm
fss
f
rr r
r 


0: (160 ) (100 ) 0
Dff
MPrWr  

100
160
100 mm 7.132 mm
(392.4 N)
160 mm 7.132 mm
218.04 N
f
f
r
PW
r
P
P







218 NP 
www.elsolucionario.org

1441. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.145
In the pivoted motor mount shown the weight W of the 175-
lb motor is used to maintain tension in the drive belt.
Knowing that the coefficient of static friction between the
flat belt and drums A and B is 0.40, and neglecting the
weight of platform CD, determine the largest couple that can
be transmitted to drum B when the drive drum A is rotating
clockwise.

SOLUTION
FBD motor and mount:

Impending belt slip: cw rotation

0.40
21 1 1
3.5136
s
TTe Te T
 
 

21
0: (12in.)(175lb) (7in.) (13in.) 0
D
MTT   


1
121
2100 lb [(7 in.)(3.5136) 13 in.]
55.858 lb, 3.5136 196.263 lb
T
TTT



FBD drum at B:

0: (3 in.)(196.263 lb 55.858 lb) 0
BB
MM   

3in.r

421 lb in.
B
M 
www.elsolucionario.org

1442. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.F1
Knowing that the coefficient of friction between the 25-kg block and the
incline is
s = 0.25, draw the free-body diagram needed to determine both the
smallest value of P required to start the block moving up the incline and the
corresponding value of
.

SOLUTION
Free body: Block
2
(25 kg)(9.81 m/s ) 245.25 NW

11
tan tan 0.25 14.04
ss


 


Force Triangle:


Note: For minimum P,
P must be perpendicular to R; thus, 14.04
s





www.elsolucionario.org

1443. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 8.F2
Two blocks A and B are connected by a cable as
shown. Knowing that the coefficient of static friction
at all surfaces of contact is 0.30 and neglecting the
friction of the pulleys, draw the free-body diagrams
needed to determine the smallest force
P required to
move the blocks.

SOLUTION

Free body: Block A
Motion impends to left, with


0.30
A SA A
F NN








Free body: Block B



Motion impends to left, with

0.30
BsB B
F NN



www.elsolucionario.org

1444. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.F3
A cord is attached to and partially wound around a cylinder with a weightof
W and radius r that rests on an incline as shown. Knowing that θ = 30°, draw
the free-body diagram needed to determine both the tension in the cord and
the smallest allowable value of the coefficient of static friction between the
cylinder and the incline for which equilibrium is maintained.

SOLUTION

Free body: Cylinder
For minimum
s
, motion impends at A with
AsA
FN



www.elsolucionario.org

1445. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 8.F4
A 40-kg packing crate must be moved to the left along the floor without
tipping. Knowing that the coefficient of static friction between the crate
and the floor is 0.35, draw the free-body diagram needed to determine
both the largest allowable value of α and the corresponding magnitude of
the force
P.

SOLUTION


2
(40 kg) (9.81 m/s ) 392.4 NWmg 
Free body diagram:

If the crate is about to tip about C, contact between crate and ground is at C only, and the reaction
R is
applied at C. As the crate is about to slide,
R must form with the vertical an angle

11
tan tan 0.35 19.29 .
ss


 


Since the crate is a 3-force body,
P must pass through point E where Rand W intersect. Free body of the
crate is then:



Note: After the crate starts moving,
s
 should be replaced with the lower value .
k
 This will yield a
larger value of .

www.elsolucionario.org

1446. CHAPTER 9
www.elsolucionario.org

1447. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

SOLUTION

At

2
0, : 0xybbka 
2
b
k
a

then

2
2
;
b
yxadAydx
a
 



22
22
2
0
4322
2
0
5423
2
0
2
11 1
52 3
y
a
y
a
a
dI x dA x y dx
b
Ixxadx
a
b
xaxaxdx
a
b
xaxax
a

















31
30
y
Iab 
www.elsolucionario.org

1448. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

SOLUTION

1/3
ykx
For
:
xa
1/ 3
bka

1/3
/kba
Thus:
1/ 3
1/3b
yx
a



22
y
dI x dA x ydx

21/3 7/3
1/3 1/3
7/3 10/3
1/3 1/3
0
3
10
y
a
yy
bb
dI x x dx x dx
aa
bb
IdI xdx a
aa


 




33
10
y
I ab

www.elsolucionario.org

1449. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

SOLUTION
By observation
h
yx
b

Now
22
2
[( ) ]
1
y
dI x dA x h y dx
x
hx dx
b





Then
2
0
1
b
yy x
IdI hx dx
b

 





4
3
0
1
34
b
x
hx
b




31
or
12
y
Ibh 
www.elsolucionario.org

1450. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

SOLUTION

At
3
0, : xybbka 
3
b
k
a
 then
3
3
x;
b
ydAydx
a



22
3
2
3
0
6
3
3
0
1
1
3 6
y
a
y
a
dI x dA x b y dx
x
Ibx dx
a
x
bx
a











31
6
y
Iab 
www.elsolucionario.org

1451. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.

SOLUTION

At

2
0, : 0xybbka 
2
b
k
a
 then
2
2b
yxa
a
 or
1 , take the negative root so that when 0.
y
xa yb x
b

  



22
252
12
0
372
12
0
1
121
37
x
b
x
b
dI y dA y x dy
Iay ydy
b
ay y
b





 

 




31
21
x
Iab 
www.elsolucionario.org

1452. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.6
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.

SOLUTION

1/ 3
1/3b
yx
a



3 3
31/3
1/3
332
0
11 1
33 3
11
332
x
a
xx
bb
dI y dx x dx xdx
aa
bba
IdI xdx
aa

 


 




31
6
x
I ab
www.elsolucionario.org

1453. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.7
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.

SOLUTION
By observation
h
yx
b

or
b
xy
h

Now
22
2
3
()
x
dI y dA y x dy
b
yydy
h
b
ydy
h






Then
3
0
h
xxbI dI y dy
h




4
01
4
h
b
y
h





31
or
4
x
Ibh 
www.elsolucionario.org

1454. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.

SOLUTION

At
3
, : xayb bka 
3
b
k
a
 then
313
313
;
ba
yxx y
ab


22 13
13
73
13
0
10 3
13
0
3
10
x
b
x
b
a
dI y dA y y dy
b
a
Iydy
b
a
y
b















33
10
x
Iab 
www.elsolucionario.org

1455. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.9
Determine by direct integration the moment of inertia of the
shaded area with respect to the x axis.

SOLUTION
At 3, :xayb
3
(3 )bkaa
or
3
8
b
k
a

Then
3
3
()
8
b
yxa
a

Now
3
3
3
3
3
9
91
3
1
()
38
()
1536
x
dI y dx
b
xa dx
a
b
xadx
a






Then
333
3
910
99
3
10
9
1
() ()
101536 1536
[(3 ) 0]
15,360
a
a
xx
a
a
bb
IdI xadx xa
aa
b
aa
a

   






31
or
15
x
Iab 

www.elsolucionario.org

1456. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.10
Determine by direct integration the moment of inertia of the
shaded area with respect to the x axis.

SOLUTION

At
0, :xyb (1 0)bc orcb
At
,0:xay
1/ 2
0(1 )bka
1/ 2
1
ork
a

Then
1/ 2
1/ 2
1
x
yb
a




Now
3
1/ 2
3
1/ 2
11
1
33
x
x
dI y dx b dx
a

 



or
1/ 2 3/ 2
3
1/ 2 3/ 2
1
13 3
3
x
xxx
dI b dx
aaa





Then
1/ 2 3/ 2
3
1/ 2 3/ 2
0
1
2133
3
a
xx xxx
IdI b dx
aaa

   





3/2 2 5/2
3
1/ 2 3/ 2
0
232
2
325
a
xxx
bx
aaa




31
or
15
x
Iab 


www.elsolucionario.org

1457. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.11
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.

SOLUTION

At
,:xayb
/
or
aa b
bke k
e


Then
//1xa xab
ye be
e


Now
3/13
33(/ 1)11
()
33
1
3
xa
x
xa
dI y dx be dx
be dx




Then
3
33(/ 1) 3(/ 1)
0
0
1
333
a
a
xa xa
xx
ba
IdI be dx e
 
 





331
(1 )
9
ab e


3
or 0.1056
x
Iab 
www.elsolucionario.org

1458. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.12
Determine by direct integration the moment of inertia of the
shaded area with respect to the y axis.

SOLUTION
At 3, :xayb
3
(3 )bkaa
or
3
8
b
k
a

Then
3
3
()
8
b
yxa
a

Now
22 2 3
3
23 2 2 3
3
() ( )
8
(3 3 )
8
y
b
dI x dA x y dx x x a dx
a
b
xx xa xa adx
a

  




Then
3
54 3232
3
3
652433
3
652433
3
652433
(3 3 )
8
13 3 1
65 4 38
13 3 1
(3 ) (3 ) (3 ) (3 )
65 4 38
13 3 1
() () () ()
65 4 3
a
yy
a
a
ab
IdI xxaxaaxdx
a
b
xaxaxax
a
b
aaaaaaa
a
aaaaaaa
  








  
 


or
3
3.43
y
Iab 

www.elsolucionario.org

1459. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.13
Determine by direct integration the moment of inertia of the
shaded area with respect to the y axis.

SOLUTION
At 0, :xyb (1 0)bc orcb

,0:xa y
1/ 2
0(1 )cka
1/ 2
1
ork
a

1/ 2
1/ 2
1
x
yb
a





12
22 2
12
1
y
x
dI x dA x ydx x b dx
a

   


12
2
12
0
21
a
yy x
IdI bx dx
a

  



72
3
12
0
12
2
37
a
y
x
Ibx
a






Then
32
21
y
Iab 

www.elsolucionario.org

1460. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.14
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

SOLUTION
At ,:xa yb
/aa
bke
or
b
k
e


Then
//1xa xab
ye be
e


Now
22
2/1
()
()
y
xa
dI x dA x ydx
xbe dx



Then
2/1
0
a
xa
yy
IdI bxedx




Now use integration by parts with

2/1
/1
2
xa
xa
ux dve dx
du xdx v ae





Then
2 /1 2 /1 /1
000
3/1
0
()2
2
aa a
xa xa xa
a
xa
x e dx x ae ae xdx
aaxedx
 







Using integration by parts with

/1
/1xa
xa
ux dve dx
du dx v ae





Then

3/1 /1
0
0
322/1
0
2( )| ( )
2()|
a
xa a xa
y
xa a
Iba axae ae dx
ba aa ae

 
 


 





32221
2( )ba aa a ae

 

3
or 0.264
y
Iab 
www.elsolucionario.org

1461. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.15
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.

SOLUTION

21/2
11 2 2
ykx ykx
For
12
0andx yyb

21/2
12
12 21/2
bka bka
bb
kk
aa



Thus,

21/2
12 21/2bb
yxy x
aa


21
1/ 2 2
1/ 2 2
0
3/2 3
1/ 2 2
()
2
3 3
1
3
a
dA y y dx
bb
Axxdx
aa
ba ba
A
aa
Aab










33
21
33
3/2 6
3/2 611
33
11
33
x
dI y dx y dx
bb
xdx xdx
aa





33
3/2 6
3/2 6
00
35/2 37
3
3/2 65
2
33
21
7152133
aa
xxbb
IdI xdx xdx
aa
ba ba
ab
aa
 




 

33
35
x
Iab 


33
352 x
x
ab
b
ab
I
k
A


9
35
x
kb 
www.elsolucionario.org

1462. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.16
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

SOLUTION
At 2, :xayh 
2
2hka
or
2
4
h
k
a

Then
2
2
4
h
yx
a

Now
2
2
4
h
dA ydx x dx
a

2
2
2
4
a
ah
AdA xdx
a



2
333
22
87
3331244
a
a
hx h aa
Aah
aa
   
   
   

36
611
33 64
x
h
dI y dx x dx
a





Then
3
2
6
6
0
192
a
xxh
IdI xdx
a



2
37 3 77
66
128
77192 192
a
x
a
hx h aa
I
aa
   
  
  


 
33127
0.094494
7 192
x
Iahah

3
0.0945
x
Iah 

3
22
0.094494
0.161983
7
12
x
x
I ah
kh
A
ah
 


0.402
x
kh 
www.elsolucionario.org

1463. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.17
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.

SOLUTION
See figure of solution on Problem 9.15.

22
21
21/22 5/2 4
1/ 2 2 1/ 2 2
0001
()
3
y
aaa
y
Aab dIxdAxyydx
bb b b
I xx xdx xdx xdx
aa a a








7/2 5
3
1/ 2 27
2
21
575
y
bb ba
I ab
aa




33
35
y
I ab


33
352
3
y
y
ab
abI
k
A


9
35
y
ka 


www.elsolucionario.org

1464. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.18
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.

SOLUTION
See figure of solution on Problem 9.16

2222
22
2
5
2
4
22
0 7

1244
544
y
a
a
y
a
hh
y x A ah dA ydx dI x dA x x dx
aa
hhx
Ixdx
aa 




 





55
3
2
32 31
55204
y
haa
I ha
a




331
20
y
I ha 


331
2022
7
12
93
35y
y
haI
ka
Aah
 

93
35
y
ka 


www.elsolucionario.org

1465. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.19
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.

SOLUTION
1
:y At 2, 0:xay 0sin(2)cka
2or
2
ak k
a



At
,:xa yh
sin ( )
2
hc a
a

 or ch
2
:y At ,2:xa y h 2hmab
At
2, 0:xay 0(2)ma b
Solving yields
2
,4
h
mbh
a
 
Then
12
2
sin 4
2
2
(2)
h
yh x y x h
aa
h
xa
a


Now
21
2
() (2)sin
2
h
dAyydx xah xdx
aa 
  



Then
22
(2)sin
2
a
a
AdA h xa xdx
aa

 





2
2
12
(2) cos
2
a
a
a
hxa x
aa

 




221 2
(2) 1
0.36338
a
haaah
a
ah

  
   
  
  






www.elsolucionario.org

1466. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.19 (Continued)

Find:
and
xx
Ik
We have
33
21
3
33
3
11 12
(2) sin
33 3 2
8
(2)sin
32
x
h
dI y y dx x a h x dx
aa
h
xa x
aa 

 
  
      
   





Then
3
2
33
3
8
(2)sin
32
a
xx
ah
I dI x a x dx
aa

 
 
 
 


Now
32 2
sin sin (1 cos ) sin sin cos
 
Then
3
2
32
3
23
43
3
3
4
3
3
8
(2)sin sincos
3222
222
(2) cos cos
3232
22 2
(2)
33
22
1
33
a
x
a
a
ah
I xa x x xdx
aaaa
haa
xa x x
aaa
haa
aa
a
ah





 

 
 

  













3
0.52520
x
I ah
3
or 0.525
x
I ah
and
3
2
0.52520
0.36338
x
x
I ah
k
A ah
 or
1.202
x
kh 
www.elsolucionario.org

1467. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.20
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.

SOLUTION
1
:y At 2, 0:xay 0sin(2)cka
2or
2
ak k
a


At ,:xayh sin ( )
2
hc a
a

 or ch
2
:y At ,2:xay h 2hmab
At 2, 0:xay 0(2)ma b
Solving yields
2
,4
h
mbh
a
 

Then
12
2
sin 4
2
2
(2)
h
yh x y x h
aa
h
xa
a



Now
21
2
() (2)sin
2
h
dAyydx xah xdx
aa 
 
  
 
 

Then
22
(2)sin
2
a
a
AdA h x a x dx
aa

 
 
 
 



2
2
12
(2) cos
2
a
a
a
hxa x
aa

 
 
 
 


221 2
(2) 1
0.36338
a
haaah
a
ah
 
  
   
  
  






www.elsolucionario.org

1468. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.20 (Continued)

Find:
and
y y
I k
We have
22
322 2
(2)sin
2
2
(2)sin
2
y
h
dI x dA x x a h x dx
aa
hxaxx xdx
aa 

 
  

 
 
 
 

Then
2
3222
(2)sin
2
a
yy
a
I dI h x ax x x dx
aa

 
  
 
 


Now using integration by parts with

2
sin
2
ux dv xdx
a



2
2cos
2
a
du x dx v x
a



Then
22 22
sin cos cos (2 )
222
aa
x xdx x x x xdx
aaa

 
 




Now let
cos
2
ux dv xdx
a



2
sin
2
a
du dx v x
a



Then
22 242 2
sin cos sin sin
2222
aaa a
x xdx x x x x x dx
aaaa

 
 
  
 
 



43
2
23
2
2321 2
43
2816
cos sin cos
222
y
a
a
Ih x ax
a
aa a
xxxx x
aaa

 





   

 


3
43 2
3
2
43
2
3
21 2 2 16
(2 ) (2 ) (2 )
43
21 2 8
() () ()
43
0.61345
aa
haaa a
a
a
haaa a
a
ah
 


   


  



or
3
0.613
y
Iah


and
3
2
0.61345
0.36338y
y
I ah
k
A ah
 or
1.299
y
ka 
www.elsolucionario.org

1469. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.21
Determine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point P.


SOLUTION

We have
22
21
[( ) ]
x
dI y dA y x x dy 
Then
2
22
2
2
33
0
333
4
2(2) (20)
11
22
33
12
2() (2)()
33
10
aa
x
a
aa
a
Iyaadyyady
ay ay
aa aa a
a





 
 
 

 




Also
22
21
[( ) ]
y
dI x dA x y y dx 
Then
2
22
0
4
2(2) (20)
10
aa
y
a
Ixaadxxadx
a







Now
44
10 10
Pxy
JII a a  or
4
20
P
Ja 
and
4
2
20
(4 )(2 ) (2 )( )
P
P
J a
k
Aaaaa




210
3
a or
1.826
P
ka 

www.elsolucionario.org

1470. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.22
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown with respect to Point P.


SOLUTION
By observation
b
yx
a


First note
(2)
(2)
dA y b dx
b
xadx
a



Now
33
top bottom
3
3
3
33
311
33
1
(2)
3
1
(8)
3
x
dI y y dx
b
x bdx
a
b
xadx
a












Then
3
23
3
3
43
3
3
43 43 3
3
1
(8)
3
11
8
34
11 1 16
() 8 () ( ) 8 ( )
34 4 3
a
xx
a
a
ab
IdI xadx
a
b
xax
a
b
aaa aaa ab
a


 




 
 
 


Also
22
(2)
y
b
dI x dA x x a dx
a
 
 
 
 

Then
2
43
43 4 33
(2)
12
43
12 1 2 4
() () () ()
43 4 3 3
a
yy
a
a
ab
IdI xxadx
a
b
xax
a
b
aaa aaa ab
a


 




 
 
 


www.elsolucionario.org

1471. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.22 (Continued)

Now
3316 4
33
Pxy
JII ab ab 

224
(4)
3
P
Jabab

and
224
2 3
1
2
(4)
(2 )(3 ) (2 )(2 )
P
P
ab a bJ
k
A ab ab





221
(4)
3
ab
or
22
4
3
P
ab
k


 www.elsolucionario.org

1472. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.23
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown with respect to Point P.

SOLUTION
1
:y At 2, 2:xaya

2
11 1
2(2)or
2
ak a k
a


2
:y At 0, :xya ac
At
2, 2:xaya

2
2
2(2)aak a or
2
1
4
k
a

Then
22
12
2211
24
1
(4 )
4
yxyax
aa
ax
a



Now
22 2
21
2211
() (4)
42
1
(4 )
4
dA y y dx a x x dx
aa
axdx
a

  




Then
2
2
22 2 3 2
0
0
1118
2(4) 4
4233
a
a
AdA axdx axx a
aa

    




Now
33
33 22 2
21
642246 6
33
642246
3
11 11 1
(4 )
33 34 2
11 1
(64 48 12 )
364 8
1
(64 48 12 7 )
192
x
dI y y dx a x x dx
aa
aaxaxx xdx
aa
aaxaxxdx
a

 
    
  







www.elsolucionario.org

1473. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.23 (Continued)

Then
2
642246
3
01
2 (64 48 12 7 )
192
a
xx
I dI a a x a x x dx
a
   



2
643257
3
0
643257
3
44
112
64 16
596
112
64 (2 ) 16 (2 ) (2 ) (2 )
596
11232
128 128 32 128
96 5 15
a
ax ax ax x
a
aa aa aa a
a
aa




 

 
 





Also
22 22 1
(4 )
4
y
dI x dA x a x dx
a
 
 
 
 

Then
2
2
222 23 5
0
0
23 5 4 4
1141
2(4)
4235
14 1 32 11 32
(2 ) (2 )
23 5 2 35 15
a
a
yy
IdI xaxdx axx
aa
aa a a a
a
 
   
 
 






Now
4432 32
15 15
Pxy
JII a a  or
464
15
P
Ja 
and
464
22 15
28
3
8
5
P
P
aJ
ka
A a
 
or 1.265
P
ka 
www.elsolucionario.org

1474. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.24
Determine the polar moment of inertia and the polar radius of gyration of
the shaded area shown with respect to Point P.

SOLUTION
The equation of the circle is

222
xyr
So that
22
xry
Now
22
dA xdy r y dy
Then
22
/2
2
r
r
AdA rydy

 


Let
sin ; cosyr dyr d
Then
/2
22
/6
/2
/2
22 2
/6
/6
22 632
2
2(sin)cos
sin 2
2cos 2
24
sin 3
22
22 4 38
2.5274
Arrrd
rdr
rr
r



















 
  
 
 




Now
 
2222
x
dI y dA y r y dy 
Then
22 2
/2
2
r
xx
r
IdI yrydy

 


Let
sin ; cosyr dyr d
Then
/2
22 2
/6
/2
22
/6
2(sin) (sin)cos
2 sin (cos)cos
x
Irrrrd
rrrd













Now
22 1
sin 2 2sin cos sin cos sin 2
4
   

www.elsolucionario.org

1475. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.24 (Continued)

Then
/24
/2
42
/6
/6
24
632
4
1sin4
2sin2
4228
sin
22 2 8
3
23 16
x
r
Ir d
r
r










 

 
 
  
 

  






Also
 
3
322
11
33
y
dI x dy r y dy
Then
223/2
/21
2( )
3
r
yy
r
I dI r y dy

 


Let sin ; cosyr dyr d  
Then
/2
223/2
/62
[(sin)]cos
3
y
I rr r d


 





/2
33
/62
( cos ) cos
3
y
I rrd







Now
42 2 2 2 1
cos cos (1 sin ) cos sin 2
4
  
Then
/2
42 2
/6
/2
4
/621
cos sin 2
34
2sin21sin4
324 428
y
Ir d
r



















2
4 636 322
4
4
sin sin21 1
3242 2 442 8
21313
3 4 16 12 4 2 48 32 2
293
3464
r
r
r
 
 

     
    
 
    
  
  
 
  








www.elsolucionario.org

1476. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.24 (Continued)

Now
4
4
32 93
2 3 16 3 4 64
Pxy
r
JII r 
    
 
 


44 3
1.15545
316
rr




4
or 1.155
P
J r
and
4
2
2
1.15545
2.5274
P
P
J r
k
A r

or 0.676
P
kr 
www.elsolucionario.org

1477. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.25
(a) Determine by direct integration the polar moment of inertia of the annular
area shown with respect to Point O. (b) Using the result of Part a, determine
the moment of inertia of the given area with respect to the x axis.


SOLUTION
(a) 2dA u du

22
3
(2 )
2
O
dJ u dA u u du
udu 




2
1
2
1
3
4
2
1
2
4
R
OO
R
R
R
JdJ udu
u 




44
21
2
O
JRR


 

(b) From Eq. (9.4): (Note by symmetry.)
xy
II

44
21
2
1
24
Oxy x
xO
JII I
IJ RR


 


44
21
4
x
IRR


 
www.elsolucionario.org

1478. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.26
(a) Show that the polar radius of gyration k O of the annular area shown is
approximately equal to the mean radius
12
()2
m
RRR for small values of
the thickness
21
.tR R (b) Determine the percentage error introduced by
using R
m in place of k O for the following values of t/R m: 1,
1
2
, and
1
10
.

SOLUTION
(a) From Problem 9.23:

44
21
2
O
JRR




44
2122
22
21
O
O
RR
J
k
A RR








222
211
2
O
kRR

Thickness:
21
tR R
Mean radius:
12
1
()
2
m
RRR
Thus,
12
11
and
22
mm
RR t RR t 

22
222
11 1 1
22 2 4
Om m m
kRtRtRt






For
t small compared to :
m
R
22
Om
kR
Om
kR 
(
b)
(exact value) (approximation value)
Percentage error 100
(exact value)






2
1221
4
4
22 21
1
4
4
11
P.E. (100) 100
1
m
m
t
R
mm
t
m
R
RtR
Rt


 






www.elsolucionario.org

1479. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.26 (Continued)

For
1:
m
t
R

1
4
1
4
11
P.E. (100) 10.56%
1





For
1
:
2
m
t
R



2
11
42
2
11
42
11
P.E. (100) 2.99%
1

 

 For
1
:
10
m
t
R



2
11
410
2
11
410
11
P.E. (100) 0.1250%
1




www.elsolucionario.org

1480. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.27
Determine the polar moment of inertia and the polar radius of gyration of
the shaded area shown with respect to Point O.


SOLUTION
At ,2:Ra 2()aak 
or
a
k


Then
1
a
Ra a



  



Now
()( )dA dr r d
rdr d


Then
(1 / )
(1 / )
2
00 0
0
1
2
a
a
AdA rdrd r d

 




 


 


23
22
0
0
3
232
11
11
223
17
1(1)
66
Aa da
aa









 
  
 
  









Now
22
()
O
dJ r dA r rdr d
Then
(1 / )
3
00a
OO
JdJ rdrd







(1 / ) 4
44
00
0
55
445
0
11
1
44
11
11(1)
45 20
a
rda d
aa






 



  


  
  
 
   
 
   
  


431
or
20
O
Ja 
and
431
22 20
27
6
93
70
O
O
aJ
ka
A a

  or 1.153
O
ka 
www.elsolucionario.org

1481. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.28
Determine the polar moment of inertia and the polar radius of gyration of
the isosceles triangle shown with respect to Point O.

SOLUTION
By observation:
2
b
h
yx

or
2
b
xy
h

Now
2
b
dA xdy y dy
h





and
23
2
x
b
dI y dA y dy
h

Then
3
0
2
2
h
xxb
IdI ydy
h




4
3
0
1
44
h
by
bh
h


From above:
2h
yx
b

Now
2
()
h
dA h y dx h x dx
b

  




(2)
h
bxdx
b

and
22
(2)
y
h
dI x dA x b x dx
b
 



www.elsolucionario.org

1482. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.28 (Continued)

Then
/2
2
0
2(2)
b
yyh
I dI x b x dx
b
 



/2
34
0
11
2
32
b
h
bx x
b
 

 
 


34
3
11
2
32 22 48
hbb b
bh
b

 
 
 


Now
3311
448
Oxy
JII bh bh 
22
or (12 )
48
O
bh
Jhb

and
22
222 48
1
2
(12 ) 1
(12 )
24
bh
O
O
hbJ
khb
Abh

  
or
22
12
24
O
hb
k


 www.elsolucionario.org

1483. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.29*
Using the polar moment of inertia of the isosceles triangle of Problem
9.28, show that the centroidal polar moment of inertia of a circular area of
radius r is
4
/2.r (Hint: As a circular area is divided into an increasing
number of equal circular sectors, what is the approximate shape of each
circular sector?)
PROBLEM 9.28 Determine the polar moment of inertia and the polar
radius of gyration of the isosceles triangle shown with respect to Point O.


SOLUTION
First the circular area is divided into an increasing number of identical circular sectors. The sectors can be
approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of
the isosceles triangles are as shown.
For an isosceles triangle (see Problem 9.28):

22
(12 )
48
O
bh
Jhb
Then with
andbr hr 

22
sector1
() ()()[12()]
48
O
Jrrrr  

421
(12 )
48
r
 


Now
sector sector 42
00 1
lim lim [12 ( ) ]
48OO
dJ J
r
d



 
 
 
 


41
4
r
Then 
2
244
circle sector
0
011
()
44
OO
JdJ rdr


 


or
4
circle
()
2
O
Jr

 
www.elsolucionario.org

1484. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.30*
Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than
2
/2 .A
 (Hint:
Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same
area and the same centroid.)

SOLUTION
From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is

4
cir
()
2
C
Jr


The area of the circle is

2
cir
Ar
So that
2
cir
[()]
2
C
A
JA


Two methods of solution will be presented. However, both methods depend upon the observation that as a
given element of area dA is moved closer to some Point C. The value of
C
J will be decreased
2
(;
C
J rdA as r decreases, so must ).
C
J

Solution 1
Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so
that its area is equal to A. To minimize the value of
(),
CA
Jthe area would have to be distributed as
closely as possible about C. This is accomplished by winding the strip into a tightly wound roll with C as
its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus
increasing the value of
().
CA
J (The process is analogous to rewinding a length of tape back into a roll.)
Since the shape of the roll is circular, with the centroid of its area at C, it follows that

2
( ) Q.E.D.
2
CA
A
J

 



where the equality applies when the original area is circular.







www.elsolucionario.org

1485. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.30* (Continued)

Solution 2
Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid)
at Point
C. Without loss of generality, assume that

12 34
AAAA 
It then follows that

cir 1 2 3 4
() () [() () () ()]
CA C C C C C
J JJAJAJAJA   
Now observe that

12
() ( )0
CC
JA JA
 

34
() ()0
CC
JA JA
 
Since as a given area is moved farther away from
C, its polar moment of inertia with respect to C must
increase.

cir
() ()
CA C
JJ
2
or ( ) Q.E.D.
2
CA
A
J

 
www.elsolucionario.org

1486. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.

SOLUTION
First note that

123
2
2
2
[(24)(6) (8)(48) (48)(6)] mm
(144 384 288) mm
816 mm
AA A A




Now
123
() () ()
xx x x
II I I
where

322
1
4
41
( ) (24 mm)(6 mm) (144 mm )(27 mm)
12
(432 104,976) mm
105,408 mm
x
I




34
2
322
3
441
( ) (8 mm)(48 mm) 73,728 mm
12
1
( ) (48 mm)(6 mm) (288 mm )(27 mm)
12
(864 209,952) mm 210,816 mm
x
x
I
I

 
Then
4
(105,408 73,728 210,816) mm
x
I

4
389,952 mm
34
or 390 10 mm
x
I 
and
4
2
2
389,952 mm
816 mm
x
x
I
k
A

or 21.9 mm
x
k 
www.elsolucionario.org

1487. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.32
Determine the moment of inertia and the radius of gyration of
the shaded area with respect to the x axis.


SOLUTION
First note that

123
2
2
2
[(5)(6) (4)(2) (4)(1)] in
(30 8 4) in
18 in
AA A A




Now
123
() () ()
xx x x
II I I
where
34
11
( ) (5 in.)(6 in.) 90 in
12
x
I

322
2
41
( ) (4 in.)(2 in.) (8 in )(2 in.)
12
2
34 in
3
x
I


2
32
3
4
13
( ) (4 in.)(1in.) (4 in ) in.
12 2
1
9 in
3
x
I






Then
421
90 34 9 in
33
x
I

 


4
or 46.0 in
x
I 
and
4
2
4
46.0 in
18 in
x
x
I
k
A

or 1.599 in.
x
k 
www.elsolucionario.org

1488. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.33
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.

SOLUTION
First note that

123
2
2
2
[(24 6) (8)(48) (48)(6)] mm
(144 384 288) mm
816 mm
AA A A
 



Now
123
() () ()
yy y y
II I I
where

34
1
34
2
34
31
( ) (6 mm)(24 mm) 6912 mm
12
1
( ) (48 mm)(8 mm) 2048 mm
12
1
( ) (6 mm)(48 mm) 55,296 mm
12
y
y
y
I
I
I




Then
44
(6912 2048 55,296) mm 64,256 mm
y
I 
or
34
64.3 10 mm
y
I 
and
4
2
2
64,256 mm
816 mmy
y
I
k
A

or 8.87 mm
y
k 
www.elsolucionario.org

1489. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.34
Determine the moment of inertia and the radius of gyration
of the shaded area with respect to the y axis.

SOLUTION
First note that

123
2
2
2
[(5)(6) (4)(2) (4)(1)] in
(30 8 4) in
18 in
AA A A




Now
123
() () ()
yy y y
II I I
where

34
11
( ) (6 in.)(5 in.) 62.5 in
12
y
I

34
212
( ) (2 in.)(4 in.) 10 in
12 3
y
I

34
311
( ) (1 in.)(4 in.) 5 in
12 3
y
I
Then
421
62.5 10 5 in
33
y
I




4
or 46.5 in
y
I 
and
4
2
2
46.5 in
18 iny
y
I
k
A

or 1.607 in.
y
k 
www.elsolucionario.org

1490. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.35
Determine the moments of inertia of the shaded area shown with respect to the x and
y axes.

SOLUTION We have 
12
3
() ()
xx x x
II I I where
3
1
41
() (2)(2a)
12
4
3
x
Ia
a



4
2
3
()
8
AA BB
II a





2
2
2
2
AA x
IIAd

 
23
2
42 4
2 3
48
823 89
xx
a
II aa a 
 
 
 
 



2
2
242 4
2
2
8445
89 2 3 3 8
xx
a
IIAd a aa a 

  





3
2
242 4
3
3
8445
89 2 3 3 8
xx
a
IIAd a aa a 
 
  
 
 

Finally,
44 4445 45
338 38
x
Ia a a
 
 
 
  4
4
x
Ia 
Also
12
3
() ()
yy y y
II I I
where
 
322 4
1116
() 2 2 2
12 3
y
Iaaaaa
but
2
3
()
yy
II
Then
416
3
y
Ia 

www.elsolucionario.org

1491. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.36
Determine the moments of inertia of the shaded area shown with respect to the
x and y axes.

SOLUTION
We have 
12
3
() ()
xx x x
II I I
where 
223
1
41
() (4)(4a) 4 2
12
256
3
x
Iaaa
a




4
2
3
()
8
AA BB
II a





2
2
2
2
AA x
IIAd

 
23
2
42 4
2 3
48
823 89
xx
a
II aa a 
 
 
 
 



2
2
242 4
2
2
8541013
89 2 2 3 3 4
xx
a
IIAd a aa a 
  
  
  
  



3
2
242 4
3
3
8345
2
89 2 2 3 4
xx
a
IIAd a aa a 
  
  
  
  

Finally,
444256 10 13 5
2
334 4
x
Ia a a

 

 4
69.9
x
Ia 
Also
12
3
() ()
yy y y
II I I

www.elsolucionario.org

1492. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.36 (Continued)


 
322 4
11 256
() 4 4 4 2
12 3
y
I aa a a a



242 4
2
317
() 2
82 8
yy
I Iaaa a



 



Then

44256 17
2
38
y
I aa 




4
72.0
y
I a
www.elsolucionario.org

1493. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.37
The centroidal polar moment of inertia
CJ of the 24-in
2
shaded
area is 600 in
4
. Determine the polar moments of inertia JB and JD
of the shaded area knowing that
JD = 2JB and d = 5 in.

SOLUTION
Given:
2
24 inA
4
2600 in
DBC
JJJ
2
( ) (1)
BC CB
JJAd

2
( ) (2)
DC CD
JJAd
  
22
2
CCD CCB
JAd JAd

  
22 222 2
and 2
CB CD
dad d ad

 

22 22
22
22
42
1
2
1600
5
224
5 in.
C
C
Aad J Aad
J
ad
A
a
a
 









Then from Eqn. (1)  
22
600 24 5 5
B
J 
4
1800 in
B
J 
Then from Eqn. (2)  
22
600 24 10 5
D
J 
4
3600 in
D
J 
www.elsolucionario.org

1494. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.38
Determine the centroidal polar moment of inertia
CJ of the 25-
in
2
shaded area knowing that the polar moments of inertia of the
area with respect to points A, B, and D are, respectively, J
A = 281
in
4
, JB = 810 in
4
, and J D = 1578 in
4
.

SOLUTION

 
22 2

A C CA CA
JJAd d a 
2
( ) (1)
BC
JJAa
 
22 22

BC CB CB
JJAd d ad 

22
() (2)
BC
JJAad 
 
22 22
4
D C CD CD
JJAd d ad  

22
4 (3)
DC
JJAad 
Eqn. (3) – Eqn. (2):
2
3 (4)
DB
JJ Aa
Eqn. (4) -3[Eqn. (1)]
33
D BAC
JJJJ 

1
281 1578 810
3
C
J 
4
25.0 in
C
J 

www.elsolucionario.org

1495. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.39
Determine the shaded area and its moment of inertia with respect to the
centroidal axis parallel to AA knowing that d
1 = 25 mm and d 2 = 10 mm
and that its moments of inertia with respect to AA and BB are 2.2  10
6

mm
4
and 4  10
6
mm
4
, respectively.

SOLUTION


64 2
2.2 10 mm (25 mm)
AA
IIA
  (1)

64 2
410mm (35mm)
BB
IIA
  

622
(4 2.2) 10 (35 25 )
BB AA
II A
 

6
1.8 10 (600)A
2
3000 mmA 
Eq. (1):
62
2.2 10 (3000)(25)I
34
325 10 mmI 
www.elsolucionario.org

1496. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.40
Knowing that the shaded area is equal to 6000 mm
2
and that its moment of
inertia with respect to AA is 18  10
6
mm
4
, determine its moment of inertia
with respect to BB for d
1 = 50 mm and d 2 = 10 mm.

SOLUTION
Given:
2
6000 mmA

262 22
1
; 18 10 mm (6000 mm )(50 mm)
AA
IIAd I
  

64
310mmI

264 2 2
6264
3 10 mm (6000 mm )(50 mm 10 mm)
3 10 6000(60) 24.6 10 mm
BB
IIAd
   
   


64
24.6 10 mm
BB
I
 
www.elsolucionario.org

1497. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.41
Determine the moments of inertia
x
I and
y
I of the area shown with
respect to centroidal axes respectively parallel and perpendicular to side
AB.

SOLUTION
First locate centroid C of the area.
Symmetry implies 0X


2
,mmA
,mmy
3
,mmyA
Flng. 180 40 7200 20 144,000
Web 80 60 4800 80 384,000
 12,000 528,000
Then
23
: (12,000 mm ) 528,000 mmYA yA X 
or 44.0 mmY (measured from top of flange)
Now () ()
x xF xW
II I
where

 

232 64
32 2641
( ) (180 mm)(40 mm) 7200 mm 44 mm 20 mm 5.170 10 mm
12
1
( ) (60mm)(80mm) 4800mm (80mm 44 mm) 8.78 10 mm
12
xF
xW
I
I






Then
64
(5.107 8.78) 10 mm
x
I
or
64
13.89 10 mm
x
I 
Also () ()
yyFyW
II I
Then   
33 411
40 180 80 60 mm
12 12
y
I





or
64
20.9 10 mm
y
I 
www.elsolucionario.org

1498. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.42
Determine the moments of inertia
x
Iand
y
I of the area shown with respect to
centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION
First locate C of the area.
Symmetry implies 18 mm.X

2
,mmA
,mmy
3
,mmyA
1 36 28 1008 14 14,112
2
1
36 42 756
2
  42 31,752
 1764 45,864


Then
23
: (1764 mm ) 45,864 mmYA yA Y 
or 26.0 mmY
Now
12
() ()
xx x
II I
where
32 2
1
434
32 2
2
4341
( ) (36 mm)(28 mm) (1008 mm )[(26 14)mm]
12
(65,856 145,152)mm 211.008 10 mm
1
( ) (36 mm)(42 mm) (756 mm )[(42 26)mm]
36
(74,088 193,536)mm 267.624 10 mm
x
x
I
I

 

 
Then
34
(211.008 267.624) 10 mm
x
I
or
34
479 10 mm
x
I 

Dimensions in mm www.elsolucionario.org

1499. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.42 (Continued)

Also
12
() ()
yy y
II I
where
334
1
32
2
4341
( ) (28 mm)(36 mm) 108.864 10 mm
12
11
( ) 2 (42 mm)(18 mm) 18 mm 42 mm (6 mm)
36 2
2(6804 13,608)mm 40.824 10 mm
y
y
I
I

 

 
 
 

2
[( )
y
I is obtained by dividing
2
Ainto ]
Then
34
(108.864 40.824 10 mm
y
I or
34
149.7 10 mm
y
I 
www.elsolucionario.org

1500. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.43
Determine the moments of inertia
x
I and
y
I of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION

First locate centroid C of the area.

2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1 58 40 2.5 4 100 160
2 25 10  1.9 4.3 –19 –43
 30 81 117
Then
23
:(30in)81inXA xA X 
or 2.70 in.X
and
23
: (30 in ) 117 inYA yA Y 
or 3.90 in.Y
Now
12
() ()
xx x
II I
where
32 2
1
44
32 2
2
41
( ) (5 in.)(8 in.) (40 in )[(4 3.9) in.]
12
(213.33 0.4) in 213.73 in
1
( ) (2 in.)(5 in.) (10 in )[(4.3 3.9) in.]
12
(20.83 1.60) 22.43 in
x
x
I
I




www.elsolucionario.org

1501. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.43 (Continued)

Then
4
(213.73 22.43) in
x
I or
4
191.3 in
x
I 
Also
12
() ()
yy y
II I
where
32 2
1
44
32 2
2
441
( ) (8 in.)(5 in.) (40 in )[(2.7 2.5) in.]
12
(83.333 1.6) in 84.933 in
1
( ) (5 in.)(2 in.) (10 in )[(2.7 1.9) in.]
12
(3.333 6.4) in 9.733 in
y
y
I
I



 
Then
4
(84.933 9.733) in
y
I or
4
75.2 in
y
I 
www.elsolucionario.org

1502. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.44
Determine the moments of inertia
x
I and
y
I of the area shown with
respect to centroidal axes respectively parallel and perpendicular to side
AB.

SOLUTION

First locate centroid C of the area.

2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1 3.6 0.5 1.8 1.8 0.25 3.24 0.45
2 0.5 3.8 1.9 0.25 2.4 0.475 4.56
3 1.3 1 1.3 0.65 4.8 0.845 6.24
 5.0 4.560 11.25
Then
23
:(5in)4.560inXA xA X 
or 0.912 in.X
and
23
: (5 in ) 11.25 inYA yA Y 
or 2.25 in.Y


www.elsolucionario.org

1503. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.44 (Continued)

Now
123
() () ()
x xx x
IIII
where
32 2
1
44
32 2
2
441
( ) (3.6 in.)(0.5 in.) (1.8 in )[(2.25 0.25) in.]
12
(0.0375 7.20) in 7.2375 in
1
( ) (0.5 in.)(3.8 in.) (1.9 in )[(2.4 2.25) in.]
12
(2.2863 0.0428) in 2.3291in
x
x
I
I

 

 

32 2
3
441
( ) (1.3 in.)(1in.) (1.3 in )[(4.8 2.25 in.)]
12
(0.1083 8.4533) in 8.5616 in
x
I
 

Then
44
(7.2375 2.3291 8.5616) in 18.1282 in
x
I 
or
4
18.13 in
x
I 
Also
123
() () ()
yy y y
II I I
where
32 2
1
441
( ) (0.5 in.)(3.6 in.) (1.8 in )[(1.8 0.912) in.]
12
(1.9440 1.4194) in 3.3634 in
y
I
 


32 2
2
44
32 2
3
441
( ) (3.8 in.)(0.5 in.) (1.9 in )[(0.912 0.25) in.]
12
(0.0396 0.8327) in 0.8723 in
1
( ) (1in.)(1.3 in.) (1.3 in )[(0.912 0.65) in.]
12
(0.1831 0.0892) in 0.2723 in
y
y
I
I

 

 
Then
44
(3.3634 0.8723 0.2723)in 4.5080 in
y
I 
or
4
4.51in
y
I 
www.elsolucionario.org

1504. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.45
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.

SOLUTION
First locate centroid C of the figure.



2
,inA
, in.y
3
,inyA
1
2
(6) 56.5487
2


2.5465


 144
2
1
(12)(4.5) 27
2
 1.5 –40.5
 29.5487 103.5
Then
23
: (29.5487 in ) 103.5 inYA yA Y 
or 3.5027 in.Y
(a)
12
()()
OO O
JJ J
where
44
1
222
( ) (6 in.) 107.876 in
4
() () ()
O
Ox y
J
JII




Now
34
21
( ) (12 in.)(4.5 in.) 91.125 in
12
x
I

and
34
21
( ) 2 (4.5 in.)(6 in.) 162.0 in
12
y
I






[Note:
2
()
y
I
 is obtained using ]


www.elsolucionario.org

1505. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.45 (Continued)

Then
4
2
4
( ) (91.125 162.0) in
253.125 in
O
J


Finally
44
(1017.876 253.125) in 764.751in
O
J 
or
4
765 in
O
J 
(b)
2
OC
JJAY
or
422
764.751in. (29.5487 in )(3.5027 in.)
C
J
or
4
402 in
C
J 
www.elsolucionario.org

1506. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.46
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.

SOLUTION
Determination of centroid C of entire section:

Section Area, in
2
,in.x
3
,inxA
1
2
(4) 4
4


16
3
21.333
2
1
(4)(4) 8
2


4
3
 10.6667
3
1
(4)(4) 8
2


4
3
10.6667
 28.566 21.333


23
: (28.566 in ) 21.333 in
0.74680 in.; by symmetry, 0.74680 in.
XA xA X
XYX
 


(a) For section
:
44 411 1
(4) 50.265 in
44 16
x
Ir 





For sections
,:
33 411
(4)(4) 21.333 in
12 12
x
Ibh 
For total area,
4
50.265 2(21.333) 92.931in
x
I 
By symmetry,
4
;185.862in
yx Oxy
II JII
4
185.9 in
O
J 
(b) Parallel axis theorem:
222
() ( )
CO O
JJAOC JAXY  

22
185.862 (28.566)(0.74680 0.74680 )
C
J 
4
154.0 in
C
J 
www.elsolucionario.org

1507. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.47
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.

SOLUTION



Determination of centroid C of entire section

233
(4000 mm ) 122.67 10 mm
30.667 mm
YA yA
Y
Y



Area mm
2
,mmy
3
,mmyA
1
1
(160)(80) 6400
2


80
3

3
170.67 10
2
1
(80)(60) 2400
2

20
3
48 10
 4000
3
122.67 10
(a) Polar moment of inertia :
O
J
Section
:
3641
(160)(80) 6.8267 10 mm
12
x
I
For I
y consider the following two triangles

364
6641
2 (80)(80) 6.8267 10 mm
12
(6.8267 6.8267)10 13.653 10 mm
y
Oxy
I
JII




   


Dimensions in mm Symmetry: 0X www.elsolucionario.org

1508. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.47 (Continued)

(b) Section
:
3641
(80)(60) 1.44 10 mm
12
x
I
For I
y consider the following two triangles

364
6641
2 (60)(40) 0.640 10 mm
12
(1.44 0.640)10 2.08 10 mm
y
Oxy
I
JII




   

Entire section
66
12
( ) ( ) 13.653 10 2.08 10
OO O
JJ J 

64
11.573 10 mm
O
J
64
11.57 10 mm
O
J 
(c)
Polar moment of inertia of intire area
O
J

2
62
64
11.573 10 (4000)(30.667)
7.811 10 mm
OC
C
C
JJAY
J
J





64
7.81 10 mm
C
J 
www.elsolucionario.org

1509. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.48
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.

SOLUTION
First locate centroid C of the figure.
Note that symmetry implies 0.Y


Dimensions in mm
2
,mmA
,mmX
3
,mmXA
1 (84)(42) 5541.77
2


112
35.6507

 197,568
2
2
(42) 2770.88
2


56
17.8254

 49,392
3 (54)(27) 2290.22
2


72
22.9183

 52,488
4
2
(27) 1145.11
2


36
11.4592

 –13,122
 4877.32 –108,810
Then
23
: (4877.32 mm ) 108,810 mmXA xA X 
or 22.3094X
(a)
1234
()() () ()
OO O O O
JJ J J J
where
22
1
64
( ) (84 mm)(42 mm)[(84 mm) (42 mm) ]
8
12.21960 10 mm
O
J



www.elsolucionario.org

1510. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.48 (Continued)


4
2
64
22
3
64
4
4
64
() (42mm)
4
2.44392 10 mm
( ) (54 mm)(27 mm)[(54 mm) (27 mm) ]
8
2.08696 10 mm
() (27mm)
4
0.41739 10 mm
O
O
O
J
J
J










Then
64
64
(12.21960 2.44392 2.08696 0.41739) 10 mm
12.15917 10 mm


O
J
or
64
12.16 10 mm
O
J 
(b)
2
OC
JJAX
or
64 2 2
12.15917 10 mm (4877.32 mm )( 22.3094 mm)
C
J 
or
6
9.73 10 mm


C
J 

www.elsolucionario.org

1511. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.49
Two channels and two plates are used to form the column
section shown. For b  200 mm, determine the moments of
inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.

SOLUTION


For C250  22.8

2
64
64
2890 mm
28.0 10 mm
0.945 10 mm
x
y
A
I
I




Total area
432
2[2890 mm (10 mm)(375 mm)] 13.28 10 mmA 
Given b 200 mm:

64 3 2 1
2[28.0 10 mm ] 2 (375 mm)(10 mm) (375 mm)(10 mm)(132 mm)
12
x
I

  




6
186.743 10
64
186.7 10 mm
x
I 

6
232
3
186.743 10
14.0620 10 mm
13.28 10
x
x
I
k
A

  


118.6 mm
x
k 



From Figure 9.13B www.elsolucionario.org

1512. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.49 (Continued)

Channel

Plate

22 3
2
64 2 64
66 6 1
( ) 2[ ] 2 (10 mm)(375 mm)
12
200 mm
2 0.945 10 mm (2890 mm ) 16.10 mm 87.891 10 mm
2
2[0.945 10 38.955 10 ] 87.891 10
yy y
IIAd IAd

    




   






64
167.691 10 mm
64
167.7 10 mm
y
I 

6
23
3
167.691 10
12.6273 10
13.28 10y
y
I
k
A

  


112.4 mm
y
k 
www.elsolucionario.org

1513. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.50
Two L6  4 
2
1-in. angles are welded together to form the section shown.
Determine the moments of inertia and the radii of gyration of the combined
section with respect to the centroidal x and y axes.

SOLUTION

From Figure 9.13A: Area A 4.75 in
2

4
4
17.3 in
6.22 in
x
y
I
I





2422 4
2[ ] 2[17.3 in (4.75 in )(1.02 in.) ] 44.484 in
xxO
IIAy
   

4
44.5 in
x
I 
Total area  2(4.75)  9.50 in
2

4
22
2
44.484 in
4.6825 in
area9.50 in
x
x
I
k 

2.16 in.
x
k 

2422
2[ ] 2[6.22 in (4.75 in )(1.269 in.) ]
yyO
IIAx
  

4
27.738 in
4
27.7 in
y
I 
Total area  2(4.75)  9.50 in
2

4
2
2
27.738 in
2.9198
area9.50 iny
y
I
k 

1.709 in
y
k 

From Figure 9.13A www.elsolucionario.org

1514. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.51
Two L6  4 
2
1-in. angles are welded together to form the section
shown. Determine the moments of inertia and the radii of gyration
of the combined section with respect to the centroidal x and y axes.

SOLUTION
W section:
24 4
9.12 in 110 in 37.1in
xy
AI I
Angle:
244
1.44 in 1.23 in 1.23 in
xy
AI I

total W A
2
4
9.12 4(1.44) 14.880 in
AAA

 

Now
WA
() 4()
xx x
II I
where
A
2
A
42 2 4
()
1.23 in (1.44 in )(4.00 in. 0.836 in.) 34.907 in
xx
IIAd
  

Then
44
(110 4 34.907) in 249.63 in
x
I 
4
250 in
x
I 
and
4
2
2
total
249.63 in
14.880 in
x
x
I
k
A


4.10 in.
x
k 
Also
WA
() 4()
yy y
III
where


A
2
A
2 44
(1.23 1.44 5 0.836 ) in 26.198 in
yy
IIAd
  


44
(37.1 4 26.198) in 141.892 in
y
I 
4
141.9 in
y
I 

4
2
2
total
141.892 in
14.880 iny
y
I
k
A


3.09 in.
y
k 

www.elsolucionario.org

1515. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.52
Two 20-mm steel plates are welded to a rolled S section as shown. Determine
the moments of inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.

SOLUTION
S section:
2
64
64
6010 mm
90.3 10 mm
3.88 10 mm
x
y
A
I
I



Note:
total S plate
2
2
2
6010 mm 2(160 mm)(20 mm)
12,410 mm
AAA


Now
Splate
() 2()
xx x
II I
where
plate
2
plate
32 2
64
()
1
(160 mm)(20 mm) (3200 mm )[(152.5 10) mm]
12
84.6067 10 mm
xx
IIAd


Then
64
(90.3 2 84.6067) 10 mm
x
I 

64
259.5134 10 mm or
64
260 10 mm
x
I 
and
64
2
2
total
259.5134 10 mm
12410 mm
x
x
I
k
A

 or
144.6 mm
x
k 
Also
Splate
() 2()
yy y
II I

64 3
64 1
3.88 10 mm 2 (20 mm)(160 mm)
12
17.5333 10 mm

 


 or
6
17.53 10 mm


y
I 
and
64
2
2
total
17.5333 10 mm
12,410 mmy
y
I
k
A


or
37.6 mm
y
k 
www.elsolucionario.org

1516. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.53
A channel and a plate are welded together as shown to form a
section that is symmetrical with respect to the y axis.
Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.

SOLUTION

From Figure 9.13B
2
3.37 inB
For C8  11.5:
4
1.31in
y
I
(Note change of axes)
4
32.5 in
y
I
Location of centroid
:YA yA
22
[3.7 in (12 in.)(0.5 in.)] (3.37 in )(1.938 in.)
0.69702 in.
Y
Y


Moment of inertia with respect to x axis.
Plate:
23 2
41
(12 in.)(0.5 in.) (12 in.)(0.5 in.)(0.69702 in.)
12
0.125 2.9150 3.0400 in
xx
IIAY  
 

Channel:
24 2
0
4
1.31in (3.37 in.)(1.938 in. 0.69702 in.)
1.31 5.1899 6.4999 in
xx
II AY
   
 
Entire section:
444
3.0400in 6.4999in 9.5399in
x
I
4
9.54 in
x
I 
Moment of inertia with respect to y axis.
Plate:
341
(0.5in.)(12in.) 72in
12
y
I
Channel:
4
32.5 in
y
I
Entire section:
44 4
72 in 32.5 in 104.5 in
y
I 
4
104.5 in
y
I 
www.elsolucionario.org

1517. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.54
The strength of the rolled W section shown is increased by welding a channel to
its upper flange. Determine the moments of inertia of the combined section with
respect to its centroidal
x and y axes.

SOLUTION
W section:
2
64
64
14,400 mm
554 10 mm
63.3 10 mm
x
y
A
I
I



Channel:
2
64
64
2890 mm
0.945 10 mm
28.0 10 mm
x
y
A
I
I



First locate centroid
C of the section.

A, mm
2
,mmy
3
,mmyA
W Section 14,400 231 33,26,400
Channel 2,890 49.9 1,44,211
 17,290 31,82,189

Then
23
: (17,290 mm ) 3,182,189 mmYA yA Y 
or 184.047 mmY
Now
WC
() ()
xx x
II I
where
2
W
64 2 22
64
2
C
64 2 22
64
()
554 10 mm (14,400 mm )(231 184.047) mm
585.75 10 mm
()
0.945 10 mm (2,890 mm )(49.9 184.047) mm
159.12 10 mm
xx
xx
IIAd
IIAd

  


  

www.elsolucionario.org

1518. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.54 (Continued)

Then
64
(585.75 159.12) 10 mm
x
I
or
64
745 10 mm
x
I 
also
W
64
() ()
(63.3 28.0) 10 mm
yy yC
II I

or
64
91.3 10 mm
y
I 

www.elsolucionario.org

1519. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.55
Two L76  76  6.4-mm angles are welded to a C250  22.8 channel.
Determine the moments of inertia of the combined section with respect to
centroidal axes respectively parallel and perpendicular to the web of the
channel.


SOLUTION
Angle:
2
64
929 mm
0.512 10 mm

 
xy
A
II
Channel:
2
64
64
2890 mm
0.945 10 mm
28.0 10 mm
x
y
A
I
I



First locate centroid
C of the section


2
,mmA
,mmy
3
,mmyA
Angle 2(929) 1858 21.2 39,389.6
Channel 2890 16.1 46,529
 4748 7139.4

Then
23
: (4748 mm ) 7139.4 mmYA yAY 
or 1.50366 mmY
Now 2( ) ( )
xxLxC
II I
where
64 2 2
64
264 2 2
64
( ) 0.512 10 mm (929 mm )[(21.2 1.50366) mm]
0.990859 10 mm
( ) 0.949 10 mm (2890 mm )[(16.1 1.50366) mm]
1.56472 10 mm
xL x
xC x
IIAd
IIAd

    

    


Then
64
[2(0.990859) 1.56472 10 ] mm
x
I
or
64
3.55 10 mm
x
I 







www.elsolucionario.org

1520. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.55 (Continued)

Also
2( ) ( )
yyLyC
I II
where
2642 2
64
( ) 0.512 10 mm (929 mm )[(127 21.2) mm]
10.9109 10 mm
()
yL y
yC y
IIAd
II
    


Then
64
[2(10.9109) 28.0] 10 mm
y
I
or
64
49.8 10 mm
y
I 

www.elsolucionario.org

1521. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.56
Two steel plates are welded to a rolled W section as indicated.
Knowing that the centroidal moments of inertia
x
Iand
y
I of
the combined section are equal, determine (a) the distance a,
(b) the moments of inertia with respect to the centroidal x and y
axes.

SOLUTION

For W-section:
24 4
11.2 in , 385 in , 26.7 in
xy
AI I
Section Area, in
2
in.x
3
,inxA
Plate 2(1)(26) 52 a 52a
W-section 11.2  
 63.2 a

 
23
: 63.2 in 52 inXA xA X a 
0.82279 in.xa
Entire section:
  
2
32 3
4
( 2 );
1
26 in. 1 in. 26 in 7.05 in. 0.5 in.
12
=1484.232 in
WPL PL
xx x x x
II I I IAd 



 
4
385 2 1484.232 3353.46 in
x
I 
4
3350 in
x
I 

x y
II
4
3350 in
y
I 
   

2
32 2
24
( 2 );
1
1 in. 26 in. 26 in 0.82279 in.
12
= 1464.667 0.81649 in
WPL PL
yy y y y
II I I IAd
aa
a
 



www.elsolucionario.org

1522. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.56 (Continued)


 

2
242
24
26.7 in 11.2 in 0.82279 in.
26.7 7.5822 in
WW
yy
IIAd
a
a




  
22
26.7 7.5822 2 1464.667 0.81649 3353.46
y
Ia a   

6.57 in.a 

www.elsolucionario.org

1523. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.57
The panel shown forms the end of a trough that is filled with water to the
line
AA. Referring to section 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel (the
center of pressure).

SOLUTION
From section 9.2:
2
,
AA
R ydA M y dA



Let
yP distance of center of pressure from .AA We must have:

2
:
AA AA
PAAP
ydA
MI
Ry M y
RyA ydA






(1)
For triangular panel:

3111
12 3 2
AA
IahyhAah



31
12
11
32
AA
P
ahI
y
yA hah


1
2
P
yh 

www.elsolucionario.org

1524. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.58
The panel shown forms the end of a trough that is filled with water
to the line
AA. Referring to Section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).

SOLUTION
Using the equation developed on page 506 of the text:

AA
P
I
y
yA


Now
2
41
(2 ) 2
232
7
3
YA yA
h
bh h bh
bh


 




and
12
()()
AA AA AA
II I
 
where

33
1
2
23
2
312
() (2)()
33
114
() (2)() 2
36 2 3
11
6



   



AA
AA x
Ibhbh
IIAd bh bhh
bh

Then
3332115
362
AA
Ibhbhbh
 
Finally,
3
25
2
7
3
P
bh
y
bh


or
15
14
P
yh 

www.elsolucionario.org

1525. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.59
The panel shown forms the end of a trough that is filled with water to the line
AA. Referring to section 9.2, determine the depth of the point of application of
the resultant of the hydrostatic forces acting on the panel (the center of
pressure).

SOLUTION
Using the equation developed on page 506 of the text have:

AA
P
I
y
yA


For a quarter circle:
4
2
16
4
34
AA
Ir
r
yAr





Then
4
16
24
34
P
r
r
y
r




 or or
3
16
P
yr

 


www.elsolucionario.org

1526. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.60*
The panel shown forms the end of a trough that is filled with water
to the line AA. Referring to section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).


SOLUTION
Using the equation developed on page 506 of the text:

AA
P
I
y
yA


For a parabola:
24
53
yhAah
Now
31
()
3
AA
dI h y dx

By observation
2
2h
yx
a

So that
3 3
2223
26
642246
6
11
()
33
1
(3 3 )
3
AA
hh
dI h x dx a x dx
aa
h
aaxaxxdx
a


  




Then
3
642246
6
0
3
643 25 7
6
0
3
1
2(33 )
3
231
357
32
105
a
AA
ah
I a ax ax x dx
a
h
ax ax ax x
a
ah








Finally,
332
105
24
53
p
ah
y
hah



or
4
7
P
yh 

www.elsolucionario.org

1527. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.61
A vertical trapezoidal gate that is used as an automatic valve is held
shut by two springs attached to hinges located along edge AB.
Knowing that each spring exerts a couple of magnitude 1470 N · m,
determine the depth d of water for which the gate will open.


SOLUTION
From section 9.2:

SS
P
I
RyA y
yA



where R is the resultant of the hydrostatic forces acting on the gate
and y
P is the depth to the point of application of R. Now

3
11
[( 0.17) m] 1.2 m 0.51 m [( 0.34) m] 0.84 0.51 m
22
(0.5202 0.124848) m
yA yA h h
h

        




Recalling that
,py we have

33 2 3
(10 kg/m 9.81 m/s )(0.5202 0.124848) m
5103.162( 0.24) N
Rh
h
 


Also
12
()()
SS SS SS
II I
 
where
2
1
32
24
24
()
11
(1.2 m)(0.51 m) 1.2 m 0.51 m [( 0.17) m]
36 2
[0.0044217 0.306( 0.17) ] m
(0.306 0.10404 0.0132651) m
SS x
IIAd
h
h
hh






 

2
2
2
32
24
24
()
11
(0.84 m)(0.51 m) 0.84 m 0.51 m [( 0.34) m]
36 2
[0.0030952 0.2142( 0.34) ] m
(0.2142 0.145656 0.0278567) m
SS X
IIAd
h
h
hh






 
www.elsolucionario.org

1528. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.61 (Continued)

Then
12
24
()()
(0.5202 0.249696 0.0411218) m
SS SS SS
II I
hh
 
 
and
24
3
2
(0.5202 0.244696 0.0411218) m
(0.5202 0.124848) m
0.48 0.07905
m
0.24
P
hh
y
h
hh
h







For the gate to open, require that

open
:()
AB P
M MyhR
Substituting
2
0.48 0.07905
2940 N m m 5103.162( 0.24) N
0.24
hh
hh
h

   




or 5103.162(0.24 0.07905) 2940h
or
2.0711 mh

Then
(2.0711 0.79) md
 
or 2.86 md

www.elsolucionario.org

1529. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.62
The cover for a 0.5-m-diameter access hole in a water storage
tank is attached to the tank with four equally spaced bolts as
shown. Determine the additional force on each bolt due to the
water pressure when the center of the cover is located 1.4 m
below the water surface.


SOLUTION
From section 9.2:

AA
P
I
RyAy
yA



where R is the resultant of the hydrostatic forces acting on the cover
and y
P is the depth to the point of application of R.
Recalling that
,py we have

33 2 2
(10 kg/m 9.81 m/s )(1.4 m)[ (0.25 m) ]
2696.67 N
R 


Also
2422
4
(0.25 m) [ (0.25 m) ](1.4 m)
4
0.387913 m


  

AA x
IIAy
Then
4
2
.387913 m
1.41116 m
(1.4 m)[ (0.25 m) ]
P
y




Now note that symmetry implies

AB CD
FF FF
Next consider the free-body of the cover.
We have
0: [2(0.32 m) sin 45 ](2 )
[0.32 sin 45 (1.41116 1.4)] m
(2696.67 N) 0
CD A
MF 



or
640.92 N
A
F
Then
0: 2(640.92 N) 2 2696.67 N 0
zC
FF   
or
707.42 N
C
F

641 N
AB
FF 
and
707 N
CD
FF 
www.elsolucionario.org

1530. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.63*
Determine the x coordinate of the centroid of the volume shown.
(Hint: The height y of the volume is proportional to the x coordinate;
consider an analogy between this height and the water pressure on a
submerged surface.)

SOLUTION

First note that
h
yx
a

Now
EL
xdV xdV


where
EL
h
xxdVydA xdA
a





Then

2
()
()
h
a
zA
h
A
a
xxdA xdA
I
x
xAxdA xdA




where
()
zA
I and
()
AA
x pertain to area.
OABC:
()
zA
I is the moment of inertia of the area with respect to the z axis,
A
x is the x coordinate
of the centroid of the area, and A is the area of OABC. Then

12
33
3
() () ()
11
()() ()()
312
5
12
zA zA zA
III
ba ba
ab



and
2
()
1
()
232
2
3
A
xA xA
aa
ab ab
ab

   
 
   
   










www.elsolucionario.org

1531. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.63* (Continued)

Finally,
35
12
22
3
ab
x
ab


or
5
8
xa
Analogy with hydrostatic pressure on a submerged plate:
Recalling that ,Py it follows that the following analogies can be established.
Height ~yP

~
()~
dV ydA pdA dF
xdV x ydA ydF dM
 


Recalling that
x
P
dM
M
y
R dF









It can then be concluded that
~
P
xy

www.elsolucionario.org

1532. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.64*
Determine the x coordinate of the centroid of the volume shown;
this volume was obtained by intersecting an elliptic cylinder with
an oblique plane. (Hint: The height y of the volume is
proportional to the x coordinate; consider an analogy between this
height and the water pressure on a submerged surface.)


SOLUTION
Following the “Hint,” it can be shown that (see solution to Problem 9.63)

()
()
zAA
I
x
xA
 x
where
()
zA
I and
()
A
xA are the moment of inertia and the first moment of the area, respectively, of the
elliptical area of the base of the volume with respect to the z axis. Then

2
32
64
()
(39 mm)(64 mm) [ (64 mm)(39 mm)](64 mm)
4
12.779520 10 mm
zA z
IIAd







63
( ) (64 mm)[ (64 mm)(39 mm)]
0.159744 10 mm
A
xA 


Finally
64
64
12.779520 10 mm
0.159744 10 mm
x 




or
80.0 mmx 

www.elsolucionario.org

1533. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.65*
Show that the system of hydrostatic forces acting on a
submerged plane area A can be reduced to a force
P at the
centroid C of the area and two couples. The force
P is
perpendicular to the area and is of magnitude P
sin ,Ay
where
is the specific weight of the liquid, and the couples are (sin)
xx
I
Mi and (sin),
yxy
I 
M jwhere
xy
I xy dA

 (see section 9.8). Note that the couples are
independent of the depth at which the area is submerged.


SOLUTION
The pressure p at an arbitrary depth (sin)yis
(sin)py
so that the hydrostatic force dF exerted on an infinitesimal area dA is

(sin)dF y dA

Equivalence of the force
P and the system of infinitesimal forces dF requires
:sinsinFP dF y dA ydA
  


or sinPAy 
Equivalence of the force and couple
(, )
 
x y
PM M and the system of infinitesimal hydrostatic
forces requires
:()
xx
M yP M ydF



Now
2
( sin ) sinydF y y dA y dA  
 


(sin)
x
I

Then (sin)
x x
yP M I
 
or (sin) ( sin)
xx
MIyAy


2
sin ( )
x
IAy
or sin
xx
MI
 
:
yy
M xP M x dF



Now ( sin ) sinxdF x y dA xy dA 
 


( sin )
xy
I (Equation 9.12)
www.elsolucionario.org

1534. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.65* (Continued)

Then
(sin)
y xy
xPM I

or (sin) ( sin)
yxy
MIxAy

sin ( )
xy
IAx y
or sin
yxy
MI 
 

www.elsolucionario.org

1535. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.66*
Show that the resultant of the hydrostatic forces acting on a
submerged plane area A is a force
P perpendicular to the area
and of magnitude P
sin ,AypA  where is the specific
weight of the liquid and p is the pressure at the centroid C
of the area. Show that
P is applied at a Point C P, called the
center of pressure, whose coordinates are
/
Pxy
xIAy and
/,
Px
yIAy where
xy
I xydA (see section 9.8). Show also
that the difference of ordinates
P
yy is equal to
2
/
x
ky
 and
thus depends upon the depth at which the area is submerged.

SOLUTION
The pressure P at an arbitrary depth (sin)yis
(sin)Py
so that the hydrostatic force dP exerted on an infinitesimal area dA is
(sin)dP y dA
The magnitude
P of the resultant force acting on the plane area is then

sin sin
sin ( )
PdP y dA ydA
yA 
 



Now sinpy PpA
Next observe that the resultant
P is equivalent to the system of infinitesimal forces d P. Equivalence then
requires
:
xP
M yP ydP


Now
2
( sin ) sinydP y y dA y dA 
 


(sin)
x
I

Then
(sin)
Px
yP I

or
(sin)
sin ( )
x
P
I
y
yA



or
x
P
I
y
Ay





www.elsolucionario.org

1536. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.66* (Continued)

:
yP
M xP xdP


Now (sin) sinxdP x y dA xydA 
 


(sin)
xy
I (Equation 9.12)
Then
(sin)
Pxy
xPI
or
(sin)
sin ( )
xy
P
I
x
yA


or
xy
P
I
x Ay


Now
2
xx
IIAy

From above ()
xP
IAyy
By definition
2
x
x
IkA

Substituting
22
()
Px
Ayy k A Ay

Rearranging yields
2
x
P
k
yy
y

 
Although
x
k
is not a function of the depth of the area (it depends only on the shape of A), yis dependent
on the depth.
()(depth)
P
yyf 

www.elsolucionario.org

1537. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.67
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.


SOLUTION
First note
2
2
1
4
x
ya
a


221
4
2
ax
We have
xy x y EL EL
dI dI x y dA

where 0(symmetry)
xy EL
dI x x


2211
4
24
EL
yy ax 

221
4
2
dA ydx a x dx 
Then
2
22 22
011
44
42
a
xy xy
I dI x ax axdx

  





2
2
23 22 4
0
0
111
(4 ) 2
884
a
a
ax x dx ax x







4
24
1
2(2) (2)
84
a




or
41
2
xy
Ia 

www.elsolucionario.org

1538. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.68
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.

SOLUTION

xy x y EL EL
dI dI x y dA

But
1
0 (by symmetry)
2
x y EL EL
x
dI x x y h
dA hd




2
00
2
211
22
1
22
bb
xy xy
I dI x h hdx h xdx
b
h

 



 


221
4
xy
I bh

www.elsolucionario.org

1539. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.69
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.

SOLUTION

For
12
12
,or
b
xa bka k
a
 
Thus,
12
12b
yx
a



, But 0 (by symmetry)
xy x y EL EL x y
dI dI x y dA dI
  
with
2
EL EL
y
xxy dAydx

2
12
12
00
223
2
0
0
1
22
1
223
aa
xy xy
a
ayb
I dI x ydx x x dx
a
bbx
xdx
aa
  
 
  
  

 

 



22
/6
xy
Iab 

www.elsolucionario.org

1540. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.70
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.

SOLUTION

For
2
, ; or
k
xay a a k a
a
 
Thus,
2
a
y
x

, But 0 (by symmetry)
xy x y EL EL x y
dI dI x y dA dI
  
with
2
1
22
EL EL
ya
xxy dAydx
x

  




22 4
2
4
2
4
11
22
ln
2
2
ln
2
aa
xy xy
aa
a
a aa a
I dI x dx dx
xxx
a
x
aa
a

  



 


4
0.347
xy
I a

www.elsolucionario.org

1541. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.71
Using the parallel-axis theorem, determine the product of
inertia of the area shown with respect to the centroidal x
and y axes.


SOLUTION
Dimensions in mm
We have
123
()() ()
xy xy xy xy
II I I
Now symmetry implies

2
() 0
xy
I
and for the other rectangles

xy x y
II xyA

where
0 (symmetry)
xy
I

Thus
13
()()
xy
IxyAxyA

2
, mmA
, mmx , mmy
4
, mmxyA
1 10 80 800 55 20 880,000
3 10 80 800 55 20 880,000
 1,760,000

64
1.760 10 mm 
xy
I 

www.elsolucionario.org

1542. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.72
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.

SOLUTION

Given area  Rectangle  (Two triangles)
For rectangle
0,
xy
I by symmetry
For each triangle:

Compare these triangles with the triangle of sample Problem 9.6, where
221
.
72
xy
Ibh

For orientation of axes of this problem,

22 2 2 3 411
(60mm) (40mm) 80 10 mm
72 72
xy
Ibh
   


www.elsolucionario.org

1543. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.72 (Continued)


2
Area, mm
, mmx , mmy
4
, mmxyA
1 1
(60)(40) 1200
2


40
 26.67
6
1.280 10
2 1
(60)(40) 1200
2

40
 26.67
6
1.280 10



64
2.56 10 mmxyA
For two triangles:
36
64
( ) 2(80 10 ) 2.56 10
2.40 10 mm
xy x y
IIxyA
     
 

Since we must subtract triangles,
64
2.40 10 mm
xy
I  

www.elsolucionario.org

1544. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.73
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.


SOLUTION
We have
12
()()
xy xy xy
II I
For each semicircle
xy x y
II xyA

and
0 (symmetry)
xy
I

Thus
xy
IxyA

2
, inA
, in.x , in.y xyA
1
2
(6) 18
2
  3
8

 432
2
2
(6) 18
2
  3
8

432
 864

4
864 in
xy
I 

www.elsolucionario.org

1545. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.74
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.


SOLUTION

We have
12
()()
xy xy xy
II I
For each rectangle

xy x y
II xyA

and 0 (symmetry)
xy
I

Thus
xy
IxyA

2
, inA
, in.x , in.y
4
, inxyA
1 30.25 0.75 0.520 0.362 0.141180
2 0.25 1.75 0.4375 0.855 0.638 0.238652
 0.379832

4
0.380 in
xy
I 

www.elsolucionario.org

1546. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.75
Using the parallel-axis theorem, determine the product of
inertia of the area shown with respect to the centroidal x and
y axes.

SOLUTION
We have
123
()() ()
xy xy xy xy
II I I
Now symmetry implies
1
() 0
xy
I
and for the other rectangles

xy x y
II xyA

where 0

xy
I (symmetry)
Thus
23
()()
xy
IxyAxyA

2
,mmA
,mmX ,mmy
4
,mmxyA
2 8 32 256 –46 –20 235,520
3 8 32 256 46 20 235,520
 471,040

34
471 10 mm
xy
I 

www.elsolucionario.org

1547. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.76
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.

SOLUTION
We have
123
()() ()
xy xy xy xy
II I I


Now, symmetry implies
1
() 0
xy
I

and for each triangle 
xy x y
II xyA

where, using the results of Sample Problem 9.6,
122
72
xy
Ibh
 for both triangles. Note that the sign of
xy
Iis unchanged because the angles of rotation are 0and 180 for triangles 2 and 3, respectively.

Now

2
,inA

,in.x

,in.y

4
,inxyA

2

1
(9)(15) 67.5
2


–9
7

–4252.5
3

1
(9)(15) 67.5
2


9

–7 –4252.5


–8505
Then
22 4
41
2 (9 in.) (15 in.) 8505 in
72
9011.25 in
xy
I

 




or
4
9010 in
xy
I 

www.elsolucionario.org

1548. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.77
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal
x and y axes.

SOLUTION
We have
123
()() ()
xy xy xy xy
II I I
For each rectangle
xy x y
II xyA

and 0

xy
I (symmetry)
Thus
xy
IxyA


2
,inA
,in.x ,in.y
4
,inxyA
1 3.6 0.5 1.8 0.888 –2.00 –3.196
2 0.5 3.8 1.9 –0.662 0.15 –0.18867
3 1.3 1.0 1.3 –0.262 2.55 –0.86853
 –4.25320

4
4.25 in
xy
I 

www.elsolucionario.org

1549. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.78
Using the parallel-axis theorem, determine the product of
inertia of the area shown with respect to the centroidal
x
and
y axes.

SOLUTION
We have
12
()()
xy xy xy
II I
For each rectangle
xy x y
II xyA

and 0

xy
I (symmetry)
Thus
xy
IxyA


2
,mmA
,mmx ,mmy
4
,mmxyA
1 139.3 12.7 1769.11 +32.05 +18.55
3
1051.78 10
2 12.7 (102) 1295.4 -43.95 -26.1
3
1485.95 10

3
2537.7 10

64
2.54 10 mm
xy
I 


www.elsolucionario.org

1550. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.79
Determine for the quarter ellipse of Problem 9.67 the moments of
inertia and the product of inertia with respect to new axes obtained by
rotating the
x and y axes about O (a) through 45 counterclockwise,
(
b) through 30 clockwise.

SOLUTION
From Figure 9.12:
3
4
3
4
(2 )( )
16
8
(2 ) ( )
16
2








x
y
Iaa
a
Iaa
a

From Problem 9.67:
41
2

xy
Ia
First note
44 4
44 411 5
()
228216
11 3
()
228216
xy
xy
II a a a
II a a a




  



  



Now use Equations (9.18), (9.19), and (9.20).
Equation (9.18):
11
()()cos2sin2
22

 
xxy xy xy
III II I

44 453 1
cos 2 sin 2
16 16 2
  aa a
Equation (9.19):
11
()()cos2sin2
22

 
yxy xy xy
IIIII I

44 453 1
cos 2 sin 2
16 16 2
  aa a
Equation (9.20):
1
()sin2cos2
2

 
xy x y xy
III I

4431
sin 2 cos 2
16 2
  aa

www.elsolucionario.org

1551. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.79 (Continued)

(a)
45 :
 
44 453 1
cos90 sin 90
16 16 2
x
Iaa a

 
or
4
0.482

x
I a

4453 1
cos90
16 16 2

 
y
I aa
or
4
1.482

y
I a

4431
sin 90 cos90
16 2



xy
Ia a

4
or 0.589

xy
I a
(b)
30 :
 
44 453 1
cos( 60 ) sin( 60 )
16 16 2
x
Iaa a

 
or
4
1.120

x
I a

44 453 1
cos( 60 ) sin( 60 )
16 16 2


 
y
Iaa a
or
4
0.843

y
I a

4431
sin( 60 ) cos( 60 )
16 2



xy
Ia a
or
4
0.760

xy
I a

www.elsolucionario.org

1552. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.80
Determine the moments of inertia and the product of inertia of the
area of Problem 9.72 with respect to new centroidal axes obtained
by rotating the x and y axes 30 counterclockwise.

SOLUTION
From Problem 9.72:
64
2.40 10 mm
xy
I

Now, with two rectangles and two triangles:

33
6411
2 (60 mm)(40 mm) 2 (60 mm)(40 mm)
312
3.20 10 mm
x
x
I
I
 

 
 



336411
2 (40 mm)(60 mm) 2 (40 mm)(60 mm) 7.20 10 mm
312
y
I
 
 
 
 

Now
664
66411
( ) (3.20 7.20) 10 5.20 10 mm
22
11
( ) (3.20 7.20) 10 2.00 10 mm
22
xy
xy
II
II   
   
Using Eqs. (9.18), (9.19), (9.20):
Eq. (9.18):
64
11
()()cos2sin2
22
[5.20 ( 2.00)cos60 (2.40)sin 60 ] 10 mm
xxy xy xy
III II I 
 
  
or
64
2.12 10 mm
x
I
 

www.elsolucionario.org

1553. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.80 (Continued)

Eq. (9.19): 64
11
()()cos2sin2
22
[5.20 ( 2.00)cos60 (2.40)sin 60 ] 10 mm
yxy xy xy
IIIII I 
 
  

or
64
8.28 10 mm
y
I
 
Eq. (9.20):
64
1
( )sin 2 sin 2
2
[( 2.00)sin 60 (2.40)cos60 ] 10 mm
xy x y xy
III I 
 
  

or
64
0.532 10 mm
xy
I
  

www.elsolucionario.org

1554. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.81
Determine the moments of inertia and the product of inertia of
the area of Problem 9.73 with respect to new centroidal axes
obtained by rotating the x and y axes 60 counterclockwise.

SOLUTION
From Problem 9.73:

4
864 in
xy
I
Now
12
() ()
xx x
II I
where
4
12
4
() () (6in.)
8
162 in
xx
II




Then
44
2(162 in ) 324 in
x
I 
Also
12
() ()
yy y
II I
where
4224
12
( ) ( ) (6 in.) (6 in.) (3 in.) 324 in
82
yy
II

 
  



Then
44
2(324 in ) 648 in
y
I 
Now
4
411
( ) (324 648 ) 486 in
22
11
( ) (324 648 ) 162 in
22
xy
xy
II
II  
   
  
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
4
11
()()cos2sin2
22
[486 ( 162 )cos120 864sin120 ] in
xxy xy xy
III II I 

 
  
or
4
1033 in
x
I
 




www.elsolucionario.org

1555. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.81 (Continued)

Eq. (9.19): 4
11
()()cos2sin2
22
[486 ( 162 )cos120 864sin120 ] in
yxy xy xy
IIIII I
 

 
  

or
4
2020 in
y
I
 
Eq. (9.20):
4
1
()sin2cos2
2
[( 162 )sin120 864cos120 ] in
xy x y xy
III I 

 
  

or
4
873 in
xy
I
 

www.elsolucionario.org

1556. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.82
Determine the moments of inertia and the product of inertia of the
area of Problem 9.75 with respect to new centroidal axes obtained
by rotating the x and y axes 45 clockwise.

SOLUTION
From Problem 9.75:

4
471,040 mm
xy
I
Now
123
() () ()
xx x x
II I I
where
3
1
41
( ) (100 mm)(8 mm)
12
4266.67 mm


x
I

32
231
( ) ( ) (8 mm)(32 mm) [(8 mm)(32 mm)](20 mm)
12
124,245.33
xx
II 


Then
44
[4266.67 2(124,245.33)] mm 252,757 mm
x
I 
Also
123
() () ()
yy y y
II I I
where
34
1
32
23
41
( ) (8 mm)(100 mm) 666,666.7 mm
12
1
( ) ( ) (32 mm)(8 mm) [(8 mm)(32 mm)](46 mm)
12
543,061.3 mm
y
yy
I
II

 


Then
44
[666,666.7 2(543,061.3)] mm 1,752,789 mm
y
I 
Now
44
4411
( ) (252,757 1,752,789) mm 1,002,773 mm
22
11
( ) (252,757 1,752,789) mm 750,016 mm
22
xy
xy
II
II
  
  
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
11
()()cos2sin2
22
[1,002,773 ( 750,016)cos( 90 ) 471,040sin( 90 )]
xxy xy xy
III II I 
 
  
or
64
1.474 10 mm
x
I
 

www.elsolucionario.org

1557. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.82 (Continued)

Eq. (9.19):
11
()()cos2sin2
22
[1,002,773 ( 750,016)cos( 90 ) 471,040sin( 90 )]
yxy xy xy
IIIII I 
 
  

or
64
0.532 10 mm
y
I
 
Eq. (9.20):
1
()sin2cos2
2
[( 750,016)sin( 90 ) 471,040cos( 90 )]
xy x y xy
III I 
 
    

or
64
0.750 10 mm
xy
I
 

www.elsolucionario.org

1558. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.83
Determine the moments of inertia and the product of inertia of the
1
4
L3 2 -in. angle cross section of Problem 9.74 with respect to
new centroidal axes obtained by rotating the x and y axes 30
clockwise.

SOLUTION
From Figure 9.13:

4
4
0.390 in
1.09 in
x
y
I
I



From Problem 9.74:

4
0.37983 in
xy
I
Now
44
4411
( ) (0.390 1.09) in 0.740 in
22
11
( ) (0.390 1.09) in 0.350 in
22
xy
xy
II
II
  
  
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
11
()()cos2sin2
22
[0.740 ( 0.350)cos( 60 ) ( 0.37983)sin( 60 )]
xxy xy xy
III II I 
 
  
or
4
0.236 in
x
I
 
Eq. (9.19):
11
()()cos2sin2
22
[0.740 ( 0.350)cos( 60 ) ( 0.37983)sin( 60 )]

 
  
yxy xy xy
IIIII I
or
4
1.244 in
y
I
 
Eq. (9.20):
1
()sin2cos2
2
[( 0.350)sin( 60 ) ( 0.37983)cos( 60 )]

 
  
xy x y xy
III I
or
4
0.1132 in
xy
I
 

www.elsolucionario.org

1559. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.84
Determine the moments of inertia and the product of inertia of the
L152  102  12.7-mm angle cross section of Prob. 9.78 with
respect to new centroidal axes obtained by rotating the x and y axes
30 clockwise.

SOLUTION
From Figure 9.13: Note the axes are reversed in sketch.
64
64
2.59 10 mm
7.20 10 mm
x
y
I
I



From Problem 9.78:

64
2.54 10 mm
xy
I
Now
64
64
64 6411
( ) (2.59 7.20) 10 mm
22
4.895 10 mm
11
( ) (2.59 7.20) 10 mm 2.305 10 mm
22
xy
xy
II
II
  

    
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
64
11
()()cos2sin2
22
[4.895 2.305cos( 60 ) 2.54sin( 60 )] 10 mm
xxy xy xy
III II I 
 
  
or
64
5.94 10 mm
x
I
 
Eq. (9.19):
64
11
()()cos2sin2
22
[4.895 2.305cos( 60 ) 2.54sin( 60 )] 10 mm
yxy xy xy
IIIII I 
 
  
or
64
3.85 10 mm
y
I
 
Eq. (9.20):
64
1
()sin2cos2
2
[( 2.305sin( 60 ) 2.54cos( 60 )] 10 mm
xy x y xy
III I 
 
    
or
64
3.27 10 mm
xy
I
 
www.elsolucionario.org

1560. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.85
For the quarter ellipse of Problem 9.67, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.

SOLUTION
From Problem 9.79:
44
82
xy
IaIa


Problem 9.67:
41
2

xy
Ia
Now Eq. (9.25):

41
2
44
82
22
tan2
8
0.84883
3


 
 

xy
m
xy
aI
II aa

Then
2 40.326 and 220.326 
m

or
20.2 and 110.2
m
  
Also Eq. (9.27):
2
2
max, min
44
2 2
44 4
4
22
1
28 2
11
28 2 2
(0.981,748 0.772,644)
xy xy
xy
II II
II
aa
aa a
a


 
  











or
4
max
1.754Ia 
and
4
min
0.209Ia 
By inspection, the a axis corresponds to
min
I and the b axis corresponds to
max
.I

www.elsolucionario.org

1561. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.86
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments of
inertia.
Area of Problem 9.72.

SOLUTION
From Problem 9.80:
64
64
3.20 10 mm
7.20 10 mm
x
y
I
I



From Problem 9.72:
64
2.40 10 mm
xy
I
Now Eq. (9.25):
6
6
2 2(2.40 10 )
tan2 1.200
(3.20 7.20) 10
xy
m
xy
I
II


  
 

Then
2 50.194 and 230.194
m

 
or
25.1 and 115.1
m
  
Also Eq. (9.27):
2
2
max, min
22
xy xy
xy
II II
I I

  



Then
2
264
max, min
64
3.20 7.20 3.20 7.20
(2.40) 10 mm
22
(5.20 3.1241) 10 mm
I

 
 

 

 

or
64
max
8.32 10 mmI 
and
64
min
2.08 10 mmI 
By inspection, the a axis corresponds to
min
I and the b axis corresponds to
max
.I


www.elsolucionario.org

1562. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.87
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments of
inertia.
Area of Problem 9.73.

SOLUTION
From Problem 9.81:
44
324 in 648 in
xy
II
Problem 9.73:
4
864 in
xy
I
Now Eq. (9.25):
2 2(864)
tan2
324 648
1.69765
xy
m
xy
I
II

 



Then
2 59.500 and 239.500 
m

or
29.7 and 119.7
m
  
Also Eq. (9.27):
2
2
max, min
22

  


xy xy
xy
II II
II

Then
2
2
max, min
4
324 648 324 648
864
22
(1526.81 1002.75) in
I  
 




or
4
max
2530 inI 
and
4
min
524 inI 
By inspection, the a axis corresponds to
min
I and the b axis corresponds to
max
.I


www.elsolucionario.org

1563. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.88
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments
of inertia.
Area of Problem 9.75.

SOLUTION
From Problem 9.82:
4
252,757 mm
x
I

4
1,752,789 mm
y
I
Problem 9.75:
4
471,040 mm
xy
I
Now Eq. (9.25):
2
tan 2
2(471,040)
252,757 1,752,789
0.62804
xy
m
xy
I
II





Then
2 32.130 and 212.130°
m

or
16.07 and 106.1 
m

Also Eq. (9.27):
2
2
max, min
22

  


xy xy
xy
II II
II

Then
2
2
max, min
4
252,757 1,752,789 252,757 1,752,789
471040
22
(1,002,773 885,665) mm
I
 
 




or
64
max
1.888 10 mmI 
and
64
min
0.1171 10 mmI 
By inspection, the a axis corresponds to
min
Iand the b axis corresponds to
max
.I

www.elsolucionario.org

1564. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.89
For the angle cross section indicated, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
The
1
4
L3 2 -in. angle cross section of Problem 9.74.

SOLUTION
From Problem 9.83:
44
0.390 in 1.09 in
xy
II
Problem 9.74:
4
0.37983 in
xy
I
Now Eq. (9.25):
2 2( 0.37983)
tan 2
0.390 1.09
1.08523


 

xy
m
xy
I
II

Then
2 47.341 and 132.659°
m
 
or
23.7 and 66.3°
m
  
Also Eq. (9.27):
2
2
max, min
22

  


xy xy
xy
II II
II

Then
2
2
max, min
24
0.390 1.09 0.390 1.09
( 0.37983)
22
(0.740 0.51650) in
I
 
 




or
4
max
1.257 inI 
and
4
min
0.224 inI 
By inspection, the a axis corresponds to
min
I and the b axis corresponds to
max
.I


www.elsolucionario.org

1565. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.90
For the angle cross section indicated, determine the
orientation of the principal axes at the origin and the
corresponding values of the moments of inertia.
The
L152 102 12.7-mm  angle cross section of
Problem 9.78.

SOLUTION
From Figure 9.13B:
64
2.59 10 mm
x
I

64
7.20 10 mm
y
I
Problem 9.78:
64
2.54 10 mm
xy
I
Now Eq. (9.25):
6
6
2 2(2.54 10 )
tan 2
(2.59 7.20) 10
1.10195
xy
m
xy
I
II


 
 


Then
2 47.777 and 227.777°
m

or
23.9 and 113.9°
m
 
Also Eq. (9.27):
2
2
max, min
22

  


xy xy
xy
II II
II

Then
2
264
max, min
64
2.59 7.20 2.59 7.20
2.54 10 mm
22
(4.895 3.4300) 10 mm
I

 
 

 

 

or
64
max
8.33 10 mmI 
and
64
min
1.465 10 mmI 

www.elsolucionario.org

1566. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.91
Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67
the moments of inertia and the product of inertia with respect to new
axes obtained by rotating the x and y axes about O (a) through 45
counterclockwise, (b) through 30 clockwise.

SOLUTION
From Problem 9.79:
4
8


x
Ia

4
2


y
Ia
Problem 9.67:
41
2

xy
Ia
The Mohr’s circle is defined by the diameter XY, where

44 4 411
,and ,
82 2 2
Xaa Ya a 

 
 

Now
44 4 4
ave11 5
( ) 0.98175
228216 

   


xy
III aa a a
and
2 2 2
2444
11
22822 
 

xy
xy
II
RIaaa

4
0.77264 a
The Mohr’s circle is then drawn as shown.

2
tan 2


xy
m
xy
I
II



41
2
44
82
2



a
aa

0.84883
or
240.326
m





www.elsolucionario.org

1567. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.91 (Continued)

Then
90 40.326
49.674
  


180 (40.326 60 )
79.674
  
(a)
45 :
 
44
ave
cos 0.98175 0.77264 cos49.674
   
x
II R a a
or
4
0.482

x
I a

44
ave
cos 0.98175 0.77264 cos 49.674
   
y
II R a a
or
4
1.482

y
I a

4
sin 0.77264 sin 49.674
  
xy
IR a
or
4
0.589

xy
I a
(b) 30 : 
44
ave
cos 0.98175 0.77264 cos79.674
   
x
II R a a
or
4
1.120

x
I a

44
ave
cos 0.98175 0.77264 cos79.674
   
y
II R a a
or
4
0.843

y
I a

4
sin 0.77264 sin 79.674
 
xy
IR a
or
4
0.760

xy
I a

www.elsolucionario.org

1568. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.92
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the area of Problem 9.72 with respect to new
centroidal axes obtained by rotating the x and y axes 30
counterclockwise.


SOLUTION
From Problem 9.80:
64
3.20 10 mm
x
I

64
7.20 10 mm
y
I
From Problem 9.72:
64
2.40 10 mm
xy
I
The Mohr’s circle is defined by the diameter XY, where
66
(3.20 10 , 2.40 10 )X 
and
66
(7.20 10 , 2.40 10 ).Y 
Now
664
ave11
( ) (3.20 7.20) 10 5.20 10 mm
22
xy
III 
and
22
2264
64
11
( ) (3.20 7.20) (2.40) 10 mm
22
3.1241 10 mm
xy xy
RIII

  
     
   



The Mohr’s circle is then drawn as shown.

6
6
2
tan 2
2(2.40 10 )
(3.20 7.20) 10
1.200
xy
m
xy
I
II







or
2 50.1944
m

Then
60 50.1944 9.8056
  




www.elsolucionario.org

1569. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.92 (Continued)

Then
6
ave
cos (5.20 3.1241cos9.8056 ) 10
x
II R 
   
or
64
2.12 10 mm
x
I
 

6
ave
cos (5.20 3.1241cos9.8056 ) 10
y
II R 
   
or
64
8.28 10 mm
y
I
 

6
sin (3.1241 10 )sin 9.8056
xy
IR 
   
or
64
0.532 10 mm
xy
I
  

www.elsolucionario.org

1570. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.93
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the area of Problem 9.73 with respect to new
centroidal axes obtained by rotating the x and y axes 60
counterclockwise.

SOLUTION
From Problem 9.81:
4
324 in
x
I 

4
648 in
y
I 
Problem 9.73:
4
864 in
xy
I
The Mohr’s circle is defined by the diameter XY, where
(324 , 864)X and(648 , 864).Y
Now
4
ave11
( ) (324 648 ) 1526.81 in
22
xy
III   
and
2
22 2
4
11
( ) (324 648 ) 864
22
1002.75 in
xy xy
RIII 

  




The Mohr’s circle is then drawn as shown.

2
tan 2
2(864)
324 648
1.69765
xy
m
xy
I
II







or
2 59.500
m

Then 120 59.500
60.500
 



www.elsolucionario.org

1571. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.93 (Continued)

Then
ave
cos 1526.81 1002.75cos60.500
   
x
II R
or
4
1033 in
x
I
 

ave
cos 1526.81 1002.75cos60.500   
y
II R 
or
4
2020 in
y
I
 
sin 1002.75sin 60.500
  
xy
IR
or
4
873 in
xy
I
 

www.elsolucionario.org

1572. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.94
Using Mohr’s circle, determine the moments of inertia and
the product of inertia of the area of Problem 9.75 with
respect to new centroidal axes obtained by rotating the x and
y axes 45 clockwise.

SOLUTION
From Problem 9.82:
4
252,757 mm
x
I

4
1, 752, 789 mm
y
I
Problem 9.75:
4
471,040 mm
xy
I
The Mohr’s circle is defined by the diameter XY, where X (252,757;
471,040) and Y (1,752,789;
471,040).
Now
ave
4
11
( ) (252,757 1,752,789)
22
1,002,773 mm
xy
III 


and
2
2
22
4
1
()
2
1
(252,757 1,752,789) 471,040
2
885,665 mm
xy xy
RIII









The Mohr’s circle is then drawn as shown.

2
tan 2
2(471,040)
252,757 1,752,789
0.62804
xy
m
xy
I
II






or
2 32.130
m

Then 180 (32.130 90 )
57.870
 



www.elsolucionario.org

1573. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.94 (Continued)

Then
ave
cos 1,002,773 885,665cos57.870
x
II R 
   
or
64
1.474 10 mm
x
I
 

ave
cos 1,002,773 885,665cos57.870
y
II R 
   
or
64
0.532 10 mm

y
I 
sin 885,665sin 57.870
xy
IR 
 
or
64
0.750 10 mm

xy
I 

www.elsolucionario.org

1574. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.95
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the
1
4
L3 2 -in. angle cross section of
Problem 9.74 with respect to new centroidal axes obtained by
rotating the x and y axes 30 clockwise.

SOLUTION
From Problem 9.83:
4
0.390 in
x
I

4
1.09 in
y
I
Problem 9.74:

4
0.37983 in
xy
I
The Mohr’s circle is defined by the diameter XY, where
(0.390, 0.37983)X  and (1.09, 0.37983).Y
Now
ave
4
1
()
2
1
(0.390 1.09)
2
0.740 in
xy
III


and
2
2
2
2
4
1
()
2
1
(0.390 1.09) ( 0.37983)
2
0.51650 in
xy xy
RIII









The Mohr’s circle is then drawn as shown.

2
tan 2
2( 0.37983)
0.390 1.09
1.08523






xy
m
xy
I
II

or
2 47.341 
m

Then
60 47.341 12.659  
www.elsolucionario.org

1575. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.95 (Continued)

Then
ave
cos 0.740 0.51650 cos12.659
x
II R 
 
or
4
0.236 in
x
I
 

ave
cos 0.740 0.51650cos 12.659
  
y
II R
or
4
1.244 in
y
I 
sin 0.51650 sin 12.659
xy
IR 
 
or
4
0.1132 in
xy
I
 

www.elsolucionario.org

1576. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.96
Using Mohr’s circle, determine the moments of inertia and
the product of inertia of the
L152 102 12.7-mm angle
cross section of Problem 9.78 with respect to new
centroidal axes obtained by rotating the x and y axes 45
counterclockwise.


SOLUTION
From Problem 9.84:
64
2.59 10 mm
x
I

64
7.20 10 mm
y
I
Problem 9.78:
64
2.54 10 mm
xy
I
The Mohr’s circle is defined by the diameter XY, where
66
(2.59 10 , 2.54 10 ),X 
6
(7.20 10 ,Y 
6
2.54 10 ).
Now
664
ave11
( ) (2.59 7.20) 10 4.895 10 mm
22
xy
III  
and


2
2
2 264
64
4.895 2.54 2.54 10 mm
3.43 10 mm
avg x xy
RII I


The Mohr’s circle is then drawn as shown.

6
6
tan
2.54 10
2.305 10
1.10195
DX
DC





or
42.777

www.elsolucionario.org

1577. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.96 (Continued)

Then 180 60 47.777
72.223




and
6
ave
cos (4.895 3.43cos72.223 ) 10
x
II R 
   
or
64
5.94 10 mm
x
I
 

6
ave
cos (4.895 3.43cos72.223 ) 10
y
II R 
   
or
64
3.85 10 mm
y
I
 

6
sin (3.43 10 )sin 72.223
xy
IR 
 
or
64
3.27 10 mm
xy
I
 

www.elsolucionario.org

1578. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.97
For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine
the orientation of the principal axes at the origin and the corresponding
values of the moments of inertia.

SOLUTION
From Problem 9.79:
44
82
xy
IaIa


Problem 9.67:
41
2

xy
Ia
The Mohr’s circle is defined by the diameter XY, where

44 4 411
, and ,
82 2 2
Xa a Ya a 

 
 

Now
44 4
ave11
( ) 0.98175
2282 
  


xy
III aa a
and

2
2
2 2
44 4
4
1
()
2
11
28 2 2
0.77264










xy xy
RIII
aa a
a

The Mohr’s circle is then drawn as shown.


41
2
44
82
2
tan 2
2
0.84883






xy
m
xy
I
II
a
aa

or
240.326
m

and
20.2
m


www.elsolucionario.org

1579. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.97 (Continued)

The principal axes are obtained by rotating the xy axes through

20.2 counterclockwise 

about O.

Now
44
max, min ave
0.98175 0.77264 IIR a a
or
4
max
1.754Ia 
and
4
min
0.209Ia 
From the Mohr’s circle it is seen that the a axis corresponds to
min
Iand the b axis corresponds to
max
.I

www.elsolucionario.org

1580. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.98
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes and the
corresponding values of the moments of inertia.
Area of Problem 9.78.

SOLUTION
From Fig. 9.13:
64
2.59 10 mm
x
I

64
7.20 10 mm
y
I
From Problem 9.78:
64
2.54 10 mm
xy
I
The Mohr’s circle is defined by the diameter XY, where
66
(2.59 10 , 2.54 10 )X 
and
66
(7.20 10 , 2.54 10 ).Y 
Now
ave
64
64
1
()
2
1
(2.59 7.20) 10 mm
2
4.895 10 mm
xy
III


and
2
2
2
264
64
1
()
2
1
(2.59 7.20) (2.54) 10 mm
2
3.430 10 mm
xy xy
RIII










The Mohr’s circle is then drawn as shown.

6
6
2
tan 2
2(2.54 10 )
1.10195
(2.59 7.20) 10
xy
m
xy
I
II



 

or
2 47.777
m

and
23.9
m



www.elsolucionario.org

1581. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.98 (Continued)

The principal axes are obtained by rotating the xy axes through
23.9° counterclockwise

about C.

Now
6
max,min
(4.895 3.43) 10
ave
IIR  
or
64
max
8.33 10 mmI 
and
64
min
1.465 10 mmI 
From the Mohr’s circle it is seen that the x
’
axis corresponds to
min
Iand the y
’
axis corresponds to
max
.I

www.elsolucionario.org

1582. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.99
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes and the corresponding
values of the moments of inertia.
Area of Problem 9.76.

SOLUTION
From Problem 9.76:

4
9011.25 in
xy
I
Now
123
() () ()
xx x x
II I I
where
34
11
( ) (24 in.)(4 in.) 128 in
12
x
I

32
23
411
( ) ( ) (9 in.)(15 in.) (9 in.)(15 in.) (7 in.)
36 2
4151.25 in
xx
II

 



Then
4
4
[128 2(4151.25)] in
8430.5 in
x
I

Also
123
() () ()
yy y y
II I I
where
34
11
( ) (4 in.)(24 in.) 4608 in
12
y
I

32
23
411
( ) ( ) (15 in.)(9 in.) (9 in.)(15 in.) (9 in.)
36 2
5771.25 in
yy
II

 



Then
44
[4608 2(5771.25)] in 16150.5 in
y
I 



www.elsolucionario.org

1583. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.99 (Continued)

The Mohr’s circle is defined by the diameter XY, where
(8430.5, 9011.25)X  and
(16150.5, 9011.25).Y
Now
4
ave11
( ) (8430.5 16150.5) 12290.5 in
22
xy
III  
and
2
2
2
2
4
1
()
2
1
(8430.5 16150.5) ( 9011.25)
2
9803.17 in
xy xy
RIII









The Mohr’s circle is then drawn as shown.

2
tan 2
2( 9011.25)
8430.5 16150.5
2.33452






xy
m
xy
I
II

or
2 66.812 
m

and
33.4 
m

The principal axes are obtained by rotating the xy axes through

33.4 clockwise 

about C.



Now
max, min ave
12290.5 9803.17 IIR
or
34
max
22.1 10 inI 
and
4
min
2490 inI 
From the Mohr’s circle it is seen that the a axis corresponds to
min
Iand the b axis corresponds to
max
.I

www.elsolucionario.org

1584. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.100
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes and the corresponding
values of the moments of inertia.
Area of Problem 9.73

SOLUTION
From Problem 9.81:
44
324 in 648 in
xy
II
Problem 9.73:
4
864 in
xy
I
The Mohr’s circle is defined by the diameter XY, where
(324 , 864)
X and (648 , 864).Y
Now
4
ave11
( ) (324 648 ) 1526.81 in
22
xy
III   
and
22
2
2
41
()
2
1
(324 648 ) 864
2
1002.75 in
xy xy
RIII











The Mohr’s circle is then drawn as shown.

2
tan 2
2(864)
324 648
1.69765







xy
m
xy
I
II

or 2 59.4998

 
m

and 29.7


m

The principal axes are obtained by rotating the xy axes through
29.7° counterclockwise

about C.


www.elsolucionario.org

1585. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.100 (Continued)

Now
max, min ave
1526.81 1002.75IIR 
or
4
max
2530 inI 
and
4
min
524 inI 

From the Mohr’s circle it is seen that the a axis corresponds to
min
Iand the b axis corresponds to
max
.I

www.elsolucionario.org

1586. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.101
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes and the corresponding
values of the moments of inertia.
Area of Problem 9.74.

SOLUTION
From Problem 9.83:
4
4
0.390 in
1.09 in
x
y
I
I


Problem 9.74:
4
0.37983 in
xy
I
The Mohr’s circle is defined by the diameter XY, where
(0.390, 0.37983) and (1.09, 0.37983).XY
Now
4
ave11
( ) (0.390 1.09) 0.740 in
22
xy
III 
and
2
2
2
2
4
1
()
2
1
(0.390 1.09) ( 0.37983)
2
0.51650 in
xy xy
RIII









The Mohr’s circle is then drawn as shown.

2
tan 2
2( 0.37983)
0.390 1.09
1.08523






xy
m
xy
I
II

Then
247.341 
m

and
23.7 
m

The principal axes are obtained by rotating the xy axes through
23.7° clockwise

about C.

www.elsolucionario.org

1587. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.101 (Continued)

Now
max, min ave
0.740 0.51650 IIR
or
4
max
1.257 inI 
and
4
min
0.224 inI 

From the Mohr’s circle it is seen that the a axis corresponds to
min
I and the b axis corresponds to
max
.I

www.elsolucionario.org

1588. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.102
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.77
(The moments of inertia
x
I and
y
I of the area of Problem 9.102
were determined in Problem 9.44).

SOLUTION
From Problem 9.44:
4
4
18.1282 in
4.5080 in
x
y
I
I


Problem 9.77:
4
4.25320 in
xy
I
The Mohr’s circle is defined by the diameter XY, where
(18.1282, 4.25320) and (4.5080, 4.25320).XY
Now
4
ave11
( ) (18.1282 4.5080) 11.3181 in
22
xy
III  
and
2
2
2
2
4
1
()
2
1
(18.1282 4.5080) ( 4.25320)
2
8.02915 in
xy xy
RIII









The Mohr’s circle is then drawn as shown.

2
tan 2
2( 4.25320)
18.1282 4.5080
0.62454






xy
m
xy
I
II

or
2 31.986
m

and
15.99
m

The principal axes are obtained by rotating the xy axes through
15.99° counterclockwise

about C.
www.elsolucionario.org

1589. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.102 (Continued)

Now
max, min ave
11.3181 8.02915 IIR
or
4
max
19.35 inI 
and
4
min
3.29 inI 

From the Mohr’s circle it is seen that the a axis corresponds to
max
I and the b axis corresponds to
min
.I

www.elsolucionario.org

1590. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.103
The moments and product of inertia of an L4  3 
1
4
-in. angle cross section with respect to two
rectangular axes x and y through C are, respectively,
x
I 1.33 in
4
,
y
I 2.75 in
4
, and 0,xyI with the
minimum value of the moment of inertia of the area with respect to any axis through C being minI 0.692
in
4
. Using Mohr’s circle, determine (a) the product of inertia
xyIof the area, (b) the orientation of the
principal axes, (c) the value of max.I

SOLUTION
(Note: A review of a table of rolled-steel shapes reveals that the given values of
x
I and
y
I are obtained
when the 4-in. leg of the angle is parallel to the x axis. Further, for 0,
xy
I the angle must be oriented as
shown.)
Now
4
ave11
()(1.332.75)2.040 in
22
xy
III 
and
min ave
or 2.040 0.692  IIR R

4
1.348 in
Using
ave
I and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, (1.33, )
xy
XI
and (2.75, | |).
xy
YI
(a) We have
2
22
1
()
2




xy xy
RIII
or
2
22
1
1.348 (1.33 2.75)
2
xy
I

 


Solving for
xy
I and taking the negative root (since 0)
xy
I yields
4
1.14586 in .
xy
I

4
1.146 in
xy
I 
(b) We have
2 2( 1.14586)
tan 2
1.33 2.75
1.61389


 

xy
m
xy
I
II

or
2 58.217 29.1   
mm

The principal axes are obtained by rotating the xy axes through
29.1° clockwise

about C.
(c) We have
max ave
2.040 1.348 IIR
or
4
max
3.39 inI 
www.elsolucionario.org

1591. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.104
Using Mohr’s circle, determine for the cross section of the
rolled-steel angle shown the orientation of the principal
centroidal axes and the corresponding values of the moments of
inertia. (Properties of the cross sections are given in Figure 9.13.)

SOLUTION
From Figure 9.13B:
64
64
0.162 10 mm
0.454 10 mm


x
y
I
I

We have
12
()()
xy xy xy
II I
For each rectangle
xy x y
II xyA

and 0

xy
I (symmetry)
xy
IxyA
2
,mmA ,mmx ,mmy
4
, mmxyA
1 76 6.4 486.4 13.1 9.2 58620.93
2 6.4 (51 6.4) 285.44  21.7 –16.3 –100962.98
 –159583.91

4
159584 mm
xy
I
The Mohr’s circle is defined by the diameter XY where
66
(0.162 10 , 0.159584 10 ) X
and
66
(0.454 10 , 0.159584 10 )Y
Now
6
ave
6411
( ) (0.162 0.454) 10
22
0.3080 10 mm
  

xy
III
www.elsolucionario.org

1592. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.104 (Continued)

and
22
226
64
11
( ) (0.162 0.454) ( 0.159584) 10
22
0.21629 10 mm
xy xy
RIII

  
      
   


The Mohr’s circle is then drawn as shown.

6
6
2
tan 2
2( 0.159584 10 )
(0.162 0.454) 10
1.09304






xy
m
xy
I
II

or
247.545
m

and
23.8 
m

The principal axes are obtained by rotating the xy axes through
23.8° clockwise

About C.
Now
6
max, min ave
(0.3080 0.21629) 10  IIR
or
64
max
0.524 10 mmI 
and
64
min
0.0917 10 mmI 

From the Mohr’s circle it is seen that the a axis corresponds to
min
I and the b axis corresponds to
max
.I

www.elsolucionario.org

1593. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.105
Using Mohr’s circle, determine for the cross section of the rolled-
steel angle shown the orientation of the principal centroidal axes
and the corresponding values of the moments of inertia.
(Properties of the cross sections are given in Figure 9.13.)

SOLUTION
From Figure 9.13B:
64
64
3.93 10 mm
1.06 10 mm
x
y
I
I


Determine
xy
I:


12
xy xy xy
II I and for each rectangle,
'' ''
and by symmetry 0 thus
xy x y x y xy
I I xyA I I xyA  

2
, mmA
, mmx , mmy
4
, mmxyA
1 76 12.7 965.2 19.1 -37.85 -697,777
2 12.7 127 12.7 1451.61 -12.55 25.65 -467,284
 -1,165,061


www.elsolucionario.org

1594. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.105 (Continued)
64
1.16506 10 mm
xy
I 
The Mohr’s circle is defined by the diameter
XY, where
66
(3.93 10 , 1.165061 10 )X  
and
66
(1.06 10 ,1.165061 10 ).Y 
Now
ave
6
64
1
()
2
1
(3.93 1.06) 10
2
2.495 10 mm
xy
III


and
2
2
2
26 64
1
()
2
1
(3.93 1.06) ( 1.165061) 10 1.84840 10 mm
2
xy xy
RIII






 


The Mohr’s circle is then drawn as shown.

6
6
2 2( 1.165061 10 )
tan 2
(3.93 1.06) 10
0.81189
xy
m
xy
I
II


 
 


or
2 39.073
m
and
19.54
m
The principal axes are obtained by rotating the
xy axes through
19.54° counterclockwise

about
C.
Now
6
max, min ave
(2.495 1.84840) 10  IIR

or
64
max
4.34 10 mmI

and
64
min
0.647 10 mmI


From the Mohr’s circle it is seen that the
a axis corresponds to
max
I and the b axis corresponds to min
.I www.elsolucionario.org

1595. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.106*
For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are
x
I 1200 in
4
and
y
I 300 in
4
, respectively. Knowing that after rotating the x and y axes about the
centroid 30
 counterclockwise, the moment of inertia relative to the rotated x axis is 1450 in
4
, use Mohr’s
circle to determine (
a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.
SOLUTION
We have
4
ave11
( ) (1200 300) 750 in.
22
 
xy
III
Now observe that
ave
,
x
II ,
xx
II
 and 260.  This is possible only if 0.
xy
I Therefore, assume
0
xy
I and (for convenience) 0.
xy
I
 Mohr’s circle is then drawn as shown.
We have
260
m

Now using
ABD:
ave
41200 750
cos 2 cos 2
450
(in )
cos 2
x
mm
m
II
R


 



Using
AEF:
ave
41450 750
cos cos
700
(in )
cos
x
II
R


 



Then
450 700
60 2
cos 2 cos


m
m

or
9cos(60 2 ) 14 cos 2
mm
 
Expanding:
9(cos 60 cos 2 sin 60 sin 2 ) 14 cos 2
mmm

or
14 9 cos 60
tan 2 1.21885
9sin60




m

or
2 50.633 and 25.3 
mm

(
Note:260
m
 implies assumption
0
xy
I
 is correct.)
Finally,
4450
709.46 in
cos 50.633
R


(
a) From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given
centroidal
x and y axes through 
m
about the centroid C or
25.3° counterclockwise



www.elsolucionario.org

1596. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.106* (Continued)


(
b) We have
max, min ave
750 709.46IIR
or
4
max
1459 inI 
and
4
min
40.5 inI 
From the Mohr’s circle it is seen that the
a axis corresponds to
max
I and the b axis corresponds to
min
.I

www.elsolucionario.org

1597. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.107
It is known that for a given area
y
I 48  10
6
mm
4
and xyI –20  10
6
mm
4
, where the x and y axes are
rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by
rotating the
x axis 67.5 counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia
x
I of the area, (b) the principal centroidal moments of inertia.

SOLUTION
First assume
x y
IIand then draw the Mohr’s circle as shown. (Note: Assuming
x y
IIis not consistent
with the requirement that the axis corresponding to max
()
xy
I is obtained after rotating the x axis through 67.5°
CCW.)

From the Mohr’s circle we have

2 2(67.5 ) 90 45
m
(
a) From the Mohr’s circle we have

6
6
|| 20 10
248102
tan 2 tan 45
xy
xy
m
I
II


  


or
64
88.0 10 mm
x
I 
(
b) We have
6
ave
6411
()(88.048)10
22
68.0 10 mm
xy
III 

and
6
64
|| 20 10
28.284 10 mm
sin 2 sin 45


 
xy
m
I
R

Now
6
max, min ave
(68.0 28.284) 10IIR  
or
64
max
96.3 10 mmI 
and
64
min
39.7 10 mmI 

www.elsolucionario.org

1598. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.108
Using Mohr’s circle, show that for any regular polygon (such as a pentagon) ( a) the moment of inertia
with respect to every axis through the centroid is the same, (
b) the product of inertia with respect to every
pair of rectangular axes through the centroid is zero.

SOLUTION
Consider the regular pentagon shown, with centroidal axes x and y.

Because the
y axis is an axis of symmetry, it follows that
0.
xy
I Since 0,
xy
I the x and y axes must be
principal axes. Assuming
max min
and ,
xy
II II the Mohr’s circle is then drawn as shown.

Now rotate the coordinate axes through an angle
 as shown; the resulting moments of inertia,
x
I and ,
y
I
and product of inertia, ,
xy
I are indicated on the Mohr’s circle. However, the x axis is an axis of
symmetry, which implies 0.

xy
I For this to be possible on the Mohr’s circle, the radius R must be equal
to zero (thus, the circle degenerates into a point). With
0,R it immediately follows that
(
a)
ave 
xyx y
III I I (for all moments of inertia with respect to an axis through C) 
(
b)
0

xy x y
II (for all products of inertia with respect to all pairs of rectangular axes with
origin at
C) 

www.elsolucionario.org

1599. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.109
Using Mohr’s circle, prove that the expression
2
xyxy
III
 is independent of the orientation of the x and
y axes, where Ix, Iy, and Ixy represent the moments and product of inertia, respectively, of a given area
with respect to a pair of rectangular axes
x and y through a given Point O. Also show that the given
expression is equal to the square of the length of the tangent drawn from the origin of the coordinate
system to Mohr’s circle.

SOLUTION
First observe that for a given area A and origin O of a rectangular coordinate system, the values of
ave
I
and
R are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the
moments of inertia,

x
I and ,
y
I and the product of inertia, ,
xy
I having been computed for an arbitrary
orientation of the

xy axes.

From the Mohr’s circle

ave
ave
cos 2
cos 2
sin 2







x
y
xy
II R
II R
IR

Then, forming the expression
2
 
xy xy
II I


22
ave ave
222 22
ave
22
ave
( cos2)( cos2) ( sin2)
cos 2 ( sin 2 )
which is a constant

    
 

xy xy
II I I R I R R
IR R
IR


2
 
xy xy
II I is independent of the orientation of the coordinate axes Q.E.D. 
Shown is a Mohr’s circle, with line
,OA of length L, the required tangent.

Noting that
OAC is a right angle, it follows that

22 2
ave
LI R
or
22
 
xyxy
LII I Q.E.D. 

www.elsolucionario.org

1600. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.110
Using the invariance property established in the preceding problem, express
the product of inertia
Ixy of an area A with respect to a pair of rectangular axes
through
O in terms of the moments of inertia Ix and Iy of A and the principal
moments of inertia
Imin and Imax of A about O. Use the formula obtained to
calculate the product of inertia
Ixy of the L3  2 
1
4
-in. angle cross section
shown in Figure 9.13A, knowing that its maximum moment of inertia is 1.257
in
4
.

SOLUTION
Consider the following two sets of moments and products of inertia, which correspond to two different
orientations of the coordinate axes whole origin is at Point
O.
Case 1:
,,
xxyyxyxy
IIIII I
 
Case 2:
max min
,,0
xyxy
II II I
 
The invariance property then requires

2
max minxy xy
II I I I or
max minxy x y
IIIII  
From Figure 9.13A:
4
4
1.09 in
0.390 in
x
y
I
I


Using Eq. (9.21):
max min
 
xy
III I
Substituting
min
1.09 0.390 1.257 I
or
4
min
0.223 inI
Then
4
(1.09)(0.390) (1.257)(0.223)
0.381 in
xy
I

The two roots correspond to the following two orientations of the cross section.
For
4
0.381 in
xy
I 



and for
4
0.381 in
xy
I 

www.elsolucionario.org

1601. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.111
A thin plate of mass m is cut in the shape of an equilateral triangle of side a.
Determine the mass moment of inertia of the plate with respect to (a) the
centroidal axes AA and BB, (b) the centroidal axis CC that is perpendicular
to the plate.

SOLUTION

213 3
Area
22 4
aa a







mass area area
MassmV tA
m
I tI I
A





(a) Axis AA:
3
4
,area
13 3
2
12 2 2 96
AA
a
I aa

 
 



324 2
, mass , area31
496 24
m
AA AA
m
I Iaama
A


  



Axis BB:
3
4
,area
13 3
36 2 96
BB
I aa a





(we check that
AABB
II)

324 2
, mass , area31
496 24
m
BB BB
m
I Iaama
A


  



(b) Eq. (9.38):
21
2
24
CC AA BB
I II ma
 





21
12
CC
I ma
 

www.elsolucionario.org

1602. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.112
Determine the mass moment of inertia of a ring of mass m, cut from a
thin uniform plate, with respect to (
a) the axis AA, (b) the centroidal axis
CC that is perpendicular to the plane of the ring.

SOLUTION

mass area area
MassmV tA
m
ItI I
A 
 


We first determine



22 22
21 21
44 44
,area 2 1 2 1
444
AA
Ar r rr
Irrrr


 
 

(
a)

 
44 22
, mass , area 2 1 2 1
22
21 1
44

 

AA AA
mm
II rrmrr
A rr



22
121
4
AA
Imrr
 
(
b) By symmetry:
BB AA
II

Eq. (9.38):
2
CC AA BB AA
III I
   

22
121
2

CC
Imrr 
www.elsolucionario.org

1603. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.113
A thin, semielliptical plate has a mass m. Determine the mass moment of
inertia of the plate with respect to (
a) the centroidal axis BB, (b) the
centroidal axis
CC that is perpendicular to the plate.

SOLUTION

mass area area
mass
2
mtA
m
ItI I
ab 




Area:
21
2
Aa

2
,area ',area
2
3
',area
3
',area 2
4
823
64
1
8 9
xBB
BB
BB
II Ay
b
Iabab
Iab














(
a)
3
,mass 2264
1
8 9
BB
m
Iab
ab
 

 

 
 


2
2164
1
4 9
mb






2
0.0699
BB
Imb
 
(
b) Fig. (9.12):
332
', ', 21
, then
884
AA area AA mass
m
IabI abma
ab
  

 
 


22
,,, 211 64
1
44 9
CC mass AA mass BB mass
III mamb



 




221
m( 0.279 )
4
CC
Iab
 

www.elsolucionario.org

1604. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.114
The parabolic spandrel shown was cut from a thin, uniform plate.
Denoting the mass of the spandrel by
m, determine its mass
moment of inertia with respect to (
a) the axis BB, (b) the axis
DD that is perpendicular to the spandrel. (Hint: See Sample
Problem 9.3.)

SOLUTION
First note mass
2
mV tA
tab 

 




Also
mass area
area
2
ItI
m
I
ab



(
a) We have
or
2
, area ,area
2
3
,area
4
823
xBB
BB
II Ay
b
Iabab









(
a) From Sample Problem 9.3:
Then
3
,area
3
,mass1
21
31
21
BB
BB
Iab
m
Iab
ab




or
21
7
BB
Imb
 
(
b) From Sample Problem 9.3:
Now
3
,area
2
, area 11 ,area1
5
3
4
AA
AA
Iab
IIAa









www.elsolucionario.org

1605. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.114 (Continued)

and
2
22 ,area 11 , area
1
4
I IAa






Then
22
22 , area , area
322
3
13
44
1119
5 3 16 16
1
30
AA
II Aaa
ab ab a a
ab

 

   

 
 

 




and
3
22 , mass
231
30
1
10
m
I ab
ab
ma


Finally,
,mass ,mass 22 ,massDD BB
III


2211
710
mb ma
or
221
(7 10 )
70
DD
I ma b
 
www.elsolucionario.org

1606. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.115
A piece of thin, uniform sheet metal is cut to form the machine
component shown. Denoting the mass of the component by
m,
determine its mass moment of inertia with respect to (
a) the x axis,
(
b) the y axis.

SOLUTION
First note
2
mass
1
(2 )( ) (2 )
22
3
2
mV tA
a
taa a
ta 

 
 

 



Also
mass area
area2
2
3
ItI
m
I
a


(a) Now
,area 1,area 2,area
33
4
() 2()
11
()(2) 2 ()
12 12 2
7
12
xx x
II I
a
aa a
a






Then
4
,mass 227
123
x
m
Ia
a

or
27
18
x
Ima 
(b) We have
,area 1,area 2,area
3
3
4
() 2()
11
(2 )( ) 2 ( )
3122
31
48
zz z
II I
a
aa a
a



 





Then
4
,mass 2
2231
483
31
72
z
m
Ia
a
ma



www.elsolucionario.org

1607. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.115 (Continued)

Finally,
, mass , mass , mass
22
2
731
18 72
59
72
yxz
III
ma ma
ma



or
2
0.819
y
Ima 

www.elsolucionario.org

1608. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.116
A piece of thin, uniform sheet metal is cut to form the machine
component shown. Denoting the mass of the component by m,
determine its mass moment of inertia with respect to (a) the axis
AA, (b) the axis BB, where the AA and BB axes are parallel to the
x axis and lie in a plane parallel to and at a distance a above the xz
plane.


SOLUTION
First note that the x axis is a centroidal axis so that

2
,massx
II md
and that from the solution to Problem 9.115,

2
,mass7
18
x
I ma
(a) We have
22
,mass7
()
18
AA
I ma m a

or
2
1.389
AA
I ma
 
(b) We have
22
,mass7
(2)
18
BB
Imama

or
2
2.39
BB
I ma
 

www.elsolucionario.org

1609. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.117
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the x
axis, (b) the y axis.


SOLUTION
First note
2
mass
1
(2 )( ) (2 )( ) 3
2
mV tA
taa aa ta
 




Also
mass area area 2
3
m
ItI I
a

(a) Now , area 1,area 2,area
33
4
() ()
11
(2 )( ) (2 )( )
312
5
6
xx x
II I
aa aa
a



Then
4
,mass 25
63
x
m
Ia
a

2
, mass5
or
18
x
Ima 
(b) We have

, area 1,area 2, area
2
33 4
() ()
1111
( )(2 ) ( )(2 ) (2 )( ) 2 2 10
3362 3
zz z
II I
aa aa aa a a a


  
  

   


Then
42
,mass 210
10
33
x
m
Iama
a
 
Finally,
, mass , mass , mass
22
510
18 3
yxz
III
ma ma



265
18
ma 2
,mass
or 3.61
y
Ima 
www.elsolucionario.org

1610. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.118
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the
centroidal axis CC that is perpendicular to the plate, (b) the axis
AA that is parallel to the x axis and is located at a distance 1.5a
from the plate.


SOLUTION
First locate the centroid C.

22 2 2 1
:(2 )(2)2 2()
3
XA xA Xa a aa a aa

    



or
14
9
Xa


22 2 2 11
:(2 ) (2) ()
23
ZAzAZaa aa aa
 
   
 
 

or
4
9
Z a
(a) We have
22
,mass ,mass
()

yCC
II mXZ
From the solution to Problem 9.117:

2
,mass65
18
y
I ma
Then
22
2
,mass
65 14 4
18 9 9
cc
Imamaa

 

   

 
 

or
2
0.994

cc
I ma
(b) We have
2
,mass ,mass
()
xBB
II mZ

and
2
,mass ,mass
(1.5 )

AA BB
IIma
Then
2
2
,mass ,mass
4
(1.5 )
9

 

   

 
 
AA x
IImaa

www.elsolucionario.org

1611. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.118 (Continued)

From the solution to Problem 9.193:

2
,mass5
18
x
Ima
Then
2
22
,mass
54
(1.5 )
18 9
AA
Imamaa



  




or
2
2.33

AA
Ima 
www.elsolucionario.org

1612. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.119
Determine by direct integration the mass moment of inertia
with respect to the z axis of the right circular cylinder shown,
assuming that it has a uniform density and a mass m.

SOLUTION
For the cylinder:
2
mV aL
For the element shown:
2
dm a dx
m
dx
L

and
2
22
1
4
zz
dI dI x dm
adm xdm


Then
22
0
23
01
4
11
43
L
zz
L m
IdI a x dx
L
m
ax x
L

 









221
(3 4 )
12
z
I ma L 

www.elsolucionario.org

1613. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.120
The area shown is revolved about the x axis to form a
homogeneous solid of revolution of mass m. Using direct
integration, express the mass moment of inertia of the solid
with respect to the x axis in terms of m and h.

SOLUTION
We have
2

hh
yxh
a

so that
()
h
rxa
a

For the element shown:

22
2 1
2
()
x
dm r dx dI r dm
h
xa dx
a






Then
22
23
022
0
2
33 2
2
1
() [()]
3
17
(8 )
33
a
ahh
mdm xadx xa
aa
h
aa ah
a
 
    



Now
4
22
0
44
555
044
4
11
() ()
22
11 1
[( ) ] (32 )
25 10
31
10
a
xx
ah
IdI rrdx xadx
a
hh
xa a a
aa
ah
 
 


  


  

 

From above:
23
7
ah m

Then
2231 3 93
10 7 70
x
Imhmh




or
2
1.329
x
Imh 

www.elsolucionario.org

1614. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.121
The area shown is revolved about the x axis to form a
homogeneous solid of revolution of mass m. Determine by
direct integration the mass moment of inertia of the solid with
respect to (a) the x axis, (b) the y axis. Express your answers in
terms of m and the dimensions of the solid.


SOLUTION
We have at (, ):
k
ah h
a

or
kah
For the element shown:

2
2
4r
dm r dx
ah
dx
x








Then
2
3
3
22
22 2
1
112
33
a
a
a
a ah
mdm dx
x
ah
x
ah ah
aa


 








 

 



(a) For the element:
2411
22

x
dI r dm r dx
Then
3
4
3
44
3
33
44 4
22 2
1111
223
111126
63 627
1 2 13 13
63 9 54
a
a
xx
a
a
ah
IdI dx ah
x x
ah ah
aa
ah h mh
 
 

  
  
 
  


   



  


or
2
0.241
x
Imh 

www.elsolucionario.org

1615. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.121 (Continued)

(b) For the element:
2
22
1
4
yy
dI dI x dm
rdm xdm


Then
22
3
2
3
22 22
3
22 22
43
22 22
33
1
4
11
1
412
31 1
3
212 12 (3 ) ( )
a
yy
a
a
a
a
aah ah
IdI x dx
xx
ah ah
ah dx ah x
xx
ah ah
ma a a
aa

 

 
  
 
 

  
 
 

  


 




2
22
3126 13
23
21227 108
 
  

h
ma a m h a
a

or
22
(3 0.1204 )
y
Ima h 

www.elsolucionario.org

1616. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.122
Determine by direct integration the mass moment of inertia with
respect to the x axis of the tetrahedron shown, assuming that it has
a uniform density and a mass m.


SOLUTION
We have 1




ay
xyaa
hh
and
1




by
zybb
hh
For the element shown:
2
11
1
22
y
dm xz dy ab dy
h

 

 
 

Then
2
0
0
3
33
1
1
2
1
1
23
1
(1) (1 1)
6
1
6
h
h
y
mdm ab dy
h
hy
ab
h
abh
abh







 




 








Now, for the element:
4
33
,area
11
1
36 36
AA
y
Ixzab
h


 



Then
4
3
, mass ,area
1
() 1
3
AA AA
y
dI tI dy ab
h




  








www.elsolucionario.org

1617. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.122 (Continued)

Now

2
2
,mass
4
3
2 2
2
4 34
32
2
1
3
1
1
36
11
11
32
11
12
12 2
xAA
dI dI y z dm
y
ab dy
h
yy
yb ab dy
hh
yyy
ab dy ab y dy
hh h






 







   
    
   

  
 

Now
1
6
m abh
Then
4234
2
2
13
12
2
 
  
 
x
bym yy
dI m y dy
hh h h h

and
4 34
0
22
2
0
5 45
23
2
25 3 4 5
2
162
2
11
16
25 32 5
1111
(1) 6 () () ()
25 3 2 5
xx
h
h
my yy
IdI b y dy
hh h h
mhy yy
by
hh h h
m
bh h h h
hh h


 
    
 
 
   
 
  
  
 


or
221
()
10

x
I mb h 
www.elsolucionario.org

1618. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.123
Determine by direct integration the mass moment of inertia with
respect to the y axis of the tetrahedron shown, assuming that it has
a uniform density and a mass m.

SOLUTION
We have 1




ay
xyaa
hh
and
1




by
zybb
hh
For the element shown:
2
11
1
22
y
dm xz dy ab dy
h

 

 
 

Then
2
0
0
3
33
1
1
2
1
1
23
1
(1) (1 1)
6
1
6
h
h
y
mdm ab dy
h
hy
ab
h
abh
abh







 




 








Also
33
, area , area11

12 12
BB DD
IxzI zx

Then, using
mass area
ItI
we have
33
,mass ,mass11
() ()
12 12
BB DD
dI dy xz dI dy zx

 

 
 




www.elsolucionario.org

1619. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.123 (Continued)

Now

,mass ,mass
22
22
22
1
()
12
1
1()1
12




 
 
  
 
  
 
yBB DD
dI dI dI
xz x z dy
yy
ab a b dy
hh

We have
4
22
1
()1
62
y
my
m abh dI a b dy
hh






Then



4
0
22
0
5
22
22 5 5
()1
2
1
25
(1) (1 1)
10
yy
h
h
my
I dI a b dy
hh
mhy
ab
hh
m
ab



  


 

  

 
 
  
 


or
221
()
10

y
I ma b 
www.elsolucionario.org

1620. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.124
Determine by direct integration the mass moment of inertia and
the radius of gyration with respect to the x axis of the paraboloid
shown, assuming that it has a uniform density and a mass m.

SOLUTION
222
:ryzkx at
2
2
,; ;
a
xhra a kh k
h
  
Thus:
2
2
a
rx
h



2
2
2
2
00
1
;
2
hh
a
dm r dx xdy
h
a
mdm xdxm ah
h
 
 
 



 
2
22 4
00 0
2
24
2
2
001

2
11
22
1
2 2
x
hh h
xx
hh
dI r dm
I dI r r dx r dx
aa
xdx xdx
h h  






 

 


 



41
6
x
I ah 
Recall:
21
;
2
mah

2211
23
x
I ah a






21
3
x
I ma


2
2 21
13
3
x
ma
I
ka
mm


 




3
x
a
k



www.elsolucionario.org

1621. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.125
A thin rectangular plate of mass m is welded to a vertical shaft AB as
shown. Knowing that the plate forms an angle
 with the y axis,
determine by direct integration the mass moment of inertia of the
plate with respect to (a) the y axis, (b) the z axis.

SOLUTION






Projection on yz plane





Mass of plate: mtab

(a) For element shown:
dm btdv


2222
222
222
0
33
222
0
2221
( sin ) sin
12
1
sin
12
1
sin
12
11
sin sin
12 3 12 3
1
()(4sin)
12
yy
a
yy
a
dI dI v dm b dm v dm
bv btdv
IdIbt bv dv
va
bt b v bt ab
tab b a 


  

  





 



  





www.elsolucionario.org

1622. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.125 (Continued)


2221
(4sin)
12
y
Imba
 

(b)
2222
222
222
0 1
(cos) cos
12
1
cos
12
1
cos
12
zz
a
zz
dI dI v dm b dm v dm
b v btdv
IdIbt bv dv


  





 





33
22 22
0
222
11
cos cos
12 3 12 3
1
()(4cos)
12
a
va
bt b v bt ab
tab b a
  


  





2221
(4cos)
12
z
Imba
 

www.elsolucionario.org

1623. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.126*
A thin steel wire is bent into the shape shown. Denoting the mass per
unit length of the wire by m, determine by direct integration the mass
moment of inertia of the wire with respect to each of the coordinate
axes.

SOLUTION
First note
1/ 3 2/3 2/ 3 1/ 2
()
dy
xa x
dx

 
Then
2
2/3 2/3 2/3
2/3
11()
dy
xa x
dx
a
x

 







For the element shown:
2
1/3
1
dy
dm m dL m dx
dx
a
mdx
x









Then
1/3
1/3 2/3
1/3
00
33
22
a aa
mdm m dx max ma
x
   



Now
1/3
22/32/33
1/3
0
2
1/3 4/3 1/3 2/3 5/3
1/3
0
1/3 2 2/3 4/3 4/3 2/3 2 8/3
0
33
()
33
39 33
24 28
3933 3
2428 8
a
x
a
a a
Iydm ax mdx
x
a
ma ax axx dx
x
ma a x a x a x x
ma ma

  



 













or
21
4

x
Ima 
Symmetry implies
21
4

y
Ima 
 www.elsolucionario.org

1624. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.126* (Continued)

Alternative solution:

1/3
22 1/35/3
1/3
00
1/3 8/3 3
0
2
33
88
1
4
aa
y
a a
I xdm x m dx ma x dx
x
ma x ma
ma

  


 


 

Also
22
()
zyx
IxydmII 


or
21
2

z
I ma

www.elsolucionario.org

1625. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.127
Shown is the cross section of an idler roller. Determine its
mass moment of inertia and its radius of gyration with respect
to the axis AA. (The specific weight of bronze is 0.310 lb/in
3
;
of aluminum, 0.100 lb/in
3
; and of neoprene, 0.0452 lb/in
3
.)

SOLUTION
First note for the cylindrical ring shown that
 
22 22
21 21
44
mV t dd tdd

 
  
and, using Figure 9.28, that



22
21
21
22 22
22 11
44
21
2222
21 21
22
12
11
2222
1
84 4
1
84
1
84
1
8
AA
dd
Im m
tdd tdd
td d
td d d d
md d







 

 
 

 












Now treat the roller as three concentric rings and, working from the bronze outward, we have

22
23 2
22
32
22
32
11331
ft/s (0.310 lb/in ) in. in
432.2 16 8 4
11 1 3
(0.100 lb/in ) in. in
16 2 8
11 1 1
(0.0452 lb/in ) in. 1 in
16 8 2
m
 
 
   
 


 





  
  


62
32
(479.96 183.41 769.80) 10 lb s /ft
1.4332 10 lb s /ft






www.elsolucionario.org

1626. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.127 (Continued)

and
22 22
22 2
62 2
2
113 31
(479.96) (183.41)
848 82
11 1ft
(769.80) 1 10 lb s /ft in
82 144 in
AA
I


 
  
     
    


 
  


92
62
(84.628 62.191 1012.78) 10 lb ft s
1.15960 10 lb ft s


 


or
62
1.160 10 lb ft s


 
AA
I 
Now
62
262
32
1.15960 10 lb ft s
809.09 10 ft
1.4332 10 lb s /ft





  

AA
AA
I
k
m

Then
3
28.445 10 ft



AA
k
or 0.341in.

AA
k 

www.elsolucionario.org

1627. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.128
Shown is the cross section of a molded flat-belt
pulley. Determine its mass moment of inertia and its
radius of gyration with respect to the axis AA. (The
density of brass is 8650 kg/m
3
and the density of the
fiber-reinforced polycarbonate used is 1250 kg/m
3
.)

SOLUTION
First note for the cylindrical ring shown that

22
21
4


 mV t dd
and, using Figure 9.28, that




22
21
21
22 22
22 11
44
21
2222
21 21
22
12
11
2222
1
84 4
1
84
1
84
1
8







 

 
 

 











AA
dd
Im m
tdd tdd
td d
td d d d
md d

Now treat the pulley as four concentric rings and, working from the brass outward, we have


3222
3222
222
222
8650 kg/m (0.0175 m) (0.011 0.005 ) m
4
1250 kg/m [(0.0175 m) (0.017 0.011 ) m
(0.002 m) (0.022 0.017 ) m
(0.0095 m) (0.028 0.022 ) m ]}
m







3
3
(11.4134 2.8863 0.38288 2.7980) 10 kg
17.4806 10 kg





www.elsolucionario.org

1628. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.128 (Continued)

and
22 2 2
22
22321
[(11.4134)(0.005 0.011 ) (2.8863)(0.011 0.017 )
8
(0.38288)(0.017 0.022 )
(2.7980)(0.022 0.028 )] 10 kg m
AA
I





92
92
(208.29 147.92 37.00 443.48) 10 kg m
836.69 10 kg m





or
92
837 10 kg m


 
AA
I 
Now
92
262
3
836.69 10 kg m
47.864 10 m
17.4806 10 kg





  

AA
AA
I
k
m

or 6.92 mm

AA
k 

www.elsolucionario.org

1629. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.129
The machine part shown is formed by machining a conical
surface into a circular cylinder. For
1
2
,bh determine the
mass moment of inertia and the radius of gyration of the
machine part with respect to the y axis.

SOLUTION

Mass:
222
cyl cone 11
326
h
mahm a ah
  

22
cyl cyl cone cone
4413
:
210
11
220
y
II ma I ma
ah ah
 



For entire machine part:

22 2
cyl cone
44 4
cyl cone15
66
11 9
22020
y
mm m ah ah ah
II I ah ah ah  
     
   

or
22569
6520





y
Iah a
227
50
y
Ima 
Then
22 27
50
y
I
ka
m

0.735
y
ka 

www.elsolucionario.org

1630. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.130
Knowing that the thin hemispherical shell shown has a mass m and thickness t,
determine the mass moment of inertia and the radius of gyration of the shell with
respect to the x axis. (Hint: Consider the shell as formed by removing a hemisphere
of radius r from a hemisphere of radius r + t; then neglect the terms containing t
2

and t
3
and keep those terms containing t.)

SOLUTION
Consider shell to be formed by removing hemisphere of radius r from hemisphere of radius .rt

For a hemisphere,


222
3321
; 4 2
52
14 2
23 3
Imr A r r
mV r r
 
 

 



Then
32
522
53
4
15
I rr
r







Now following the hint, for a hemispherical shell,

5 5
54 32 54
15
4
510 ...
15
IrtrI rrtrt r






Neglect terms with powers of
1.t

42242
2
33
I rt r rt 
Mass of shell,
 
22
22mV tAtr rt   

Substituting with expression for m to simplify,

22
3
Imr
Radius of gyration,
22 2
3
I
kr
m



0.816kr 
www.elsolucionario.org

1631. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.131
A square hole is centered in and extends through the aluminum
machine component shown. Determine (a) the value of a for
which the mass moment of inertia of the component with respect
to the axis AA, which bisects the top surface of the hole, is
maximum, (b) the corresponding values of the mass moment of
inertia and the radius of gyration with respect to the axis AA.
(The specific weight of aluminum is 0.100 lb/in
3
.)

SOLUTION
First note
2
11
mVbL
and
2
22
mVaL
(a) Using Figure 9.28 and the parallel-axis theorem, we have

12
2
22
11
2
22
22
222 2 2
4224
()()
1
()
12 2
1
()
12 2
11 5
() ()
64 12
(2 3 5 )
12
AA AA AA
II I
a
mb b m
a
ma a m
bL b a aL a
L
bbaa


 


















Then
23
(6 20 ) 0
12



AA
dI L
ba a
da

or
3
0and
10
aab

Also
2
22 22
2
1
(6 60 ) ( 10 )
12 2



AA
dI L
ba Lba
da

Now for
0,a
2
2
0
AA
dI
da


and for
3
,
10
ab

2
2
0
AA
dI
da




www.elsolucionario.org

1632. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.131 (Continued)


max
()
AA
I occurs when
3
10
ab
Then
3
(4.2 in.)
10
a
or 2.30 in.a

(b) From part a:

24
42 4
max
2
3
44
2
3349
() 2 3 5
12 10 10 240
49 49 0.100 lb/in 1ft
(15 in.)(4.2 in.)
240 240 12 in. 32.2 ft/s
AA
AL
L
Ibbbb Lb
Lb
g





   





 


or
32
max
( ) 20.6 10 lb ft s


 
AA
I 
Now
2 max()
AA
AA
I
k
m



where
22
12
()mm m Lb a 

2
2
2
3
10
7
10


 

  

 

 

Lb b
Lb
Then
449
222240
27
10
77
(4.2 in.)
24 24


AA
Lb
kb
Lb

or 2.27 in.

AA
k 

www.elsolucionario.org

1633. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.132
The cups and the arms of an anemometer are fabricated from
a material of density
. Knowing that the mass moment of
inertia of a thin, hemispherical shell of mass m and thickness
t with respect to its centroidal axis GG is
2
5/12,ma
determine (a) the mass moment of inertia of the anemometer
with respect to the axis AA, (b) the ratio of a to l for which
the centroidal moment of inertia of the cups is equal to 1
percent of the moment of inertia of the cups with respect to
the axis AA.

SOLUTION
(a) First note
2
arm arm
4


mV dl
and
cup cup
[(2 cos )( )( )]

  

dm dV
atad
Then
/2
2
cup cup
0
2/2
0
2
2cos
2[sin]
2
mdm atd
at
at


  
 





Now
anem cups arms
() () ()
AA AA AA
III

Using the parallel-axis theorem and assuming the arms are slender rods, we have

arm
2
anem cup cup arm arm
22
22 2
cup cup arm arm
222
cup arm
22 2 2
() 3() 3
51
3()3
12 2 2 2
5
32
3
5
3(2 ) 2
34
AA GG AG AG
IImdImd
al
ma m la ml m
malalml
at a la l d

 

 

  
  
    
    








2
()ll




or
22
22
anem 2
5
() 6 21
34






AA
aa dl
Ilat
ll





www.elsolucionario.org

1634. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.132 (Continued)

(b) We have
cup
cup
()
0.01
()



GG
AA
I
I

or
222
cup cup55
0.01 2 (from part )
12 3
ma m a lal a





Now let
.
a
l
Then
22 5
50.12 21
3


 



or
2
40 2 1 0

Then
2
2 ( 2) 4(40)( 1)
2(40)

 


or 0.1851 and = 0.1351
  
0.1851
a
l 
www.elsolucionario.org

1635. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.133
After a period of use, one of the blades of a shredder has been worn to
the shape shown and is of mass 0.18 kg. Knowing that the mass
moments of inertia of the blade with respect to the AA and BB axes are
2
0.320 g m
 and
2
0.680 g m , respectively, determine (a) the location
of the centroidal axis GG, (b) the radius of gyration with respect to axis
GG.

SOLUTION
(a) We have (0.08 ) m
BA
dd
and, using the parallel axis

2
2
AA GG A
BB GG B
IImd
IImd




Then

22
22
(0.08 )
(0.0064 0.16 )
BB AA B A
AA
A
IImdd
mdd
md
 



Substituting:
32 2
(0.68 0.32) 10 kg m 0.18 kg(0.0064 0.16 ) m
A
d

  
or
27.5 mm
A
d 
to the right of A
(b) We have
2


AA GG A
IImd
or
32 2
32
0.32 10 kg m 0.18 kg (0.0275 m)
0.183875 10 kg m



  

GG
I
Then
32
232
0.183875 10 kg m
1.02153 10 m
0.18 kg




  
GG
GG
I
k
m

or
32.0 mm

GG
k 

www.elsolucionario.org

1636. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.134
Determine the mass moment of inertia of the 0.9-lb machine component shown
with respect to the axis AA.

SOLUTION
First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger
cone as indicated.

Now
2.4
or 1.2 in.
0.4 1.2


hh
h
The weight of the body is given by

123 123
()()   Wmggmm m gVVV
or
2
3
2223
0.9 lb 32.2 ft/s
1 ft
(0.8) (2.4) (0.2) (1.2) (0.6) (3.6) in
33 12 in.



 



233
32.2 ft/s (2.79253 0.02909 0.78540) 10 ft

   
or
24
13.7266 lb s /ft
Then
32
1
32
2
32
3
(13.7266)(2.79253 10 ) 0.038332 lb s /ft
(13.7266)(0.02909 10 ) 0.000399 lb s /ft
(13.7266)(0.78540 10 ) 0.010781 lb s /ft





m
m
m
www.elsolucionario.org

1637. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.134 (Continued)

Finally, using Figure 9.28, we have

123
222
11 2 2 3 3
()()()
13 3
21010
  
 
AA AA AA AA
II I I
ma ma ma


2
22222
13 3 1 ft
(0.038332)(0.8) (0.000399)(0.2) (0.010781)(0.6) (lb s /ft) in
21010 12 in.

 




62
(85.1822 0.0333 8.0858) 10 lb ft s


or
62
77.1 10 lb ft s


 
AA
I 

www.elsolucionario.org

1638. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


To the instructor:
The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are
derived at this time for use in the solutions of Problems 9.135 through 9.140.
Thin rectangular plate

2
22
22
22
() ( )
1
()
12 2 2
1
()
3


 
  
 
 


xm x m
IImd
bh
mb h m
mb h


2
2
22
() ( )
11
12 2 3





ym y m
IImd
b
mb m mb

2
2
22
()
11
12 2 3





zzm
II md
h
mh m mh
Thin triangular plate
We have
1
2





mV bht
and
3
,area1
36
z
Ibh
Then
, mass ,area
3
2
1
36
1
18
zz
ItI
tbh
mh



Similarly,
2
,mass1
18
y
Imb
Now
22
, mass , mass , mass1
()
18
xyz
III mbh 
Thin semicircular plate
We have
2
2





mV at
and
4
, area ,area
8
yz
II a


www.elsolucionario.org

1639. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


(Continued)

Then
,mass ,mass ,area
4
2
8
1
4
yz y
II tI
ta
ma 




Now
2
, mass , mass , mass1
2
xyz
III ma
Also
22
,mass ,mass ,mass 2 116
or
29
xx x
IImy Im a



 



and
22
,mass ,mass ,mass 2 116
or
49
zz z
IImy Im a



 




4
3


a
yz
Thin Quarter-Circular Plate
We have
2
4





mV at
and
4
, area ,area
16
yz
II a


Then
,mass ,mass ,area
4
2
16
1
4
yz y
II tI
ta
ma 




Now
2
,mass ,mass ,mass1
2
xyz
III ma
Also
22
,mass ,mass
()
xx
IImyz

or
2
,mass 2132
29
x
Im a







and
2
,mass ,massyy
IImz

or
2
,mass 2116
49
y
Im a







www.elsolucionario.org

1640. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.135
A 2-mm thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m
3
,
determine the mass moment of inertia of the component with
respect to each of the coordinate axes.

SOLUTION


3
1
3
2
7850 kg/m (0.120 m)(0.200 m)(0.002 m) 0.3768 kg
7850 kg/m (0.120 m)(0.100 m)(0.002 m) 0.1884 kg
mV
mV
 
 

Panel 1:
  
22221
0.3768 kg 0.12 m 0.3768 kg 0.1 m 0.06 m
12
xx
IImd
    
 

32
5.577 10 kg m
x
I

 

22221
0.3768 kg 0.2 m 0.3768 kg 0.1 m 0.1 m
12
yy
IImd
    
 

32
8.792 10 kg m
y
I

 

22 2221
0.3768 kg 0.2 m 0.12 m 0.3768 kg 0.1 m 0.06 m
12
zz
IImd     

32
6.833 10 kg m
z
I

 

www.elsolucionario.org

1641. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.135 (Continued)

Panel 2:

22 2 221
0.1884 kg 0.12 m 0.1 m 0.1884 kg 0.05 m 0.06 m
12
xx
IImd      
 
32
1.532 10 kg m
x
I

 

22221
0.1884 kg 0.1 m 0.1884 kg 0.05 m 0.2 m
12
yy
IImd
    
 

32
8.164 10 kg m
y
I

 

22221
0.1884 kg 0.12 m 0.1884 kg 0.06 m 0.2 m
12
zz
IImd
    
 

32
8.440 10 kg m
z
I

 

Total Mass Moments of Inertia:
12T
III
32
7.11 10 kg m
x
I

  

32
16.96 10 kg m
y
I

  

32
15.27 10 kg m
z
I

  
www.elsolucionario.org

1642. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.136
A 2-mm thick piece of sheet steel is cut and bent into the
machine component shown. Knowing that the density of steel is
7850 kg/m
3
, determine the mass moment of inertia of the
component with respect to each of the coordinate axes.

SOLUTION
First compute the mass of each component. We have

ST ST
mV tA
Then

32
1
(7850 kg/m )(0.002 m)(0.35 0.39) m
2.14305 kg


m

322
2
(7850 kg/m )(0.002 m) 0.195 m 0.93775 kg
2
m






32
3 1
(7850 kg/m )(0.002 m) 0.39 0.15 m 0.45923 kg
2
m




Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have

123
2
2
2
222
2
() () ()
10.39
(2.14305 kg)(0.39 m) (2.14305 kg) m
12 2
1 16 4 0.195
(0.93775 kg)(0.195 m) (0.93775 kg) (0.195) m
239
xx x x
II I I








 
   
      
     

22
222 2
1 0.39 0.15
(0.45923 kg)[(0.39) (0.15) ] m (0.45923 kg) m
18 3 3
 
 
  
 


2
[(0.027163 0.081489) (0.011406 0.042081) (0.004455 0.008909)] kg m 

2
(0.108652 0.053487 0.013364) kg m 

2
0.175503 kg m
or
32
175.5 10 kg m
x
I

  
www.elsolucionario.org

1643. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.136 (Continued)


123
22
222 2
() () ()
10.350.39
(2.14305 kg)[(0.35) (0.39) ] m (2.14305 kg) m
12 2 2
yy y y
II I I
 
 
  
  

22
2
2221
(0.93775 kg)(0.195 m) (0.93775 kg)(0.195 m)
4
10.39
(0.45923 kg)(0.39 m) (0.45923 kg) (0.35) m
18 3
[(0.049040 0.147120) (0.008914 0.035658)
(0.003880 0.064017)] kg m




 
 
  
 
 
 
2


2
(0.196160 0.044572 0.067897) kg m 

2
0.308629 kg m
or
32
309 10 kg m

 
y
I 

123
2
2
() () ()
10.35
(2.14305 kg)(0.35 m) (2.14305 kg) m
12 2







zz z z
II I I


2
2
222
21
(0.93775 kg)(0.195 m)
4
10.15
(0.45923 kg)(0.15 m) (0.45923 kg) (0.35) m
18 3
[(0.021877 0.065631) 0.008914) (0.000574 0.057404)] kg m




 
 
  
 
 


2
2
(0.087508 0.008914 0.057978)kg m
0.154400 kg m
 


or
32
154.4 10 kg m
z
I

  

www.elsolucionario.org

1644. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.137
A subassembly for a model airplane is fabricated from three pieces of 1.5-
mm plywood. Neglecting the mass of the adhesive used to assemble the
three pieces, determine the mass moment of inertia of the subassembly
with respect to each of the coordinate axes. (The density of the plywood
is
3
780 kg/m .)

SOLUTION
First compute the mass of each component. We have

mV tA
Then

32
12
3 1
(780 kg/m )(0.0015 m) 0.3 0.12 m
2
21.0600 10 kg
mm


 





3223
3
(780 kg/m )(0.0015 m) 0.12 m 13.2324 10 kg
4





m
Using the equations derived above and the parallel-axis theorem, we have

12
123
2
32 3
32
62
() ()
() () ()
10.12
2 (21.0600 10 kg)(0.12 m) (21.0600 10 kg) m
18 3
1
(13.2324 10 kg)(0.12 m)
2
[2(16.8480 33.6960) (95.2733)] 10 kg m
[2(50.5440) (95.2733)] 10
xx
xx x x
II
II I I







 









62
kg m



or
62
196.4 10 kg m

 
x
I 





www.elsolucionario.org

1645. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.137 (Continued)


123
2
32
() () ()
10.3
(21.0600 10 kg)(0.3 m) (21.0600 kg) m
18 3




 



yy y y
II I I 

3222
22
321
(21.0600 10 kg)[(0.3) (0.12) ] m
18
0.3 0.12
(21.0600 10 kg) m
33


 


 
    
   


321
(13.2324 10 kg)(0.12 m)
4

 

 
 


62
[(105.300 210.600) (122.148 244.296)
(47.637)] 10 kg m


 

62
(315.900 366.444 47.637) 10 kg m

 
or
62
730 10 kg m

 
y
I 
Symmetry implies

y z
II

62
730 10 kg m

 
z
I 


www.elsolucionario.org

1646. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.138
A section of sheet steel 0.03 in. thick is cut and bent into the
sheet metal machine component shown. Determine the mass
moment of inertia of the component with respect to each of the
coordinate axes. (The specific weight of steel is 490 lb/ft
3
.)

SOLUTION
First consider one triangular plate with local axes u-v. We have

3
32
2
490 lb/ft 1 4.5 in. 0.03 in.
1 ft 7.1332 10 lb s /ft
2 12 in./ft 12 in./ft32.2 ft/s
mV tA
g


 
     


mass area area
m
I tI I
A
Then
3211
1 36 18
2
u
m
I bh mh
bh




 
 ,
21
18
v
I mb

2
362
14.5 in.
7.1332 10 55.728 10 lb ft s
18 12 in./ft
u
I

 


 
23621
7.1332 10 1 ft 396.29 10 lb ft s
18
v
I

 
For composite area:
2
263
1.5 in.
2 2 55.728 10 7.1332 10
12 in./ft
xu
IImd

 

     
  


62
334 10 lb ft s
x
I

  

For ,
y
I consider plates separately.
Plate 1 in
x-y plane:
1
2
263 62
2 in.
396.29 10 7.1332 10 594.43 10 lb ft s
12 in./ft
yv
IImd
  
       





www.elsolucionario.org

1647. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.138 (Continued)

Plate 2 in
x-z plane:
Note: By Eqn. 9.38,

66
'
62
'
55.728 10 396.29 10
452.02 10 lb ft s
yuv
y
III
I


   


Now applying the Parallel Axis Theorem,


2
2
2
2
'
22
63
6 2
21.5
452.02 10 7.1332 10
12 12
761.62 10 lb ft s
yy
y
y
IImd
I
I




  
     
  



Finally, total for both plates 1 and 2 gives,


12
66
594.43 10 761.62 10
yy y
II I
 
   


32
1.356 10 lb ft s
y
I

  
By symmetry,
y z
II
32
1.356 10 lb ft s
z
I

  

www.elsolucionario.org

1648. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.139
A framing anchor is formed of 0.05-in.-thick galvanized steel. Determine the
mass moment of inertia of the anchor with respect to each of the coordinate
axes. (The specific weight of galvanized steel is
3
470 lb/ft .)

SOLUTION
First compute the mass of each component. We have

G.S.

mV tA
g
Then

33
2
1 2
62
470 lb/ft 1 ft
0.05 in. (2.25 3.5) in
12 in.32.2 ft/s
3325.97 10 lb s /ft
m








33
2
2 2
62
470 lb/ft 1 ft
0.05 in. (2.25 1) in
12 in.32.2 ft/s
950.28 10 lb s /ft
m








33
2
3 2
62
470 lb/ft 1 1 ft
0.05 in. 2 4.75 in
212 in.32.2 ft/s
2006.14 10 lb s /ft
m







Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have

123
62 2
2
62 2
() () ()
1
(3325.97 10 lb s /ft)(3.5 in.)
12
3.5 1 ft
(3325.97 10 lb s /ft) in
2 12 in.
xx x x
II I I










 


62 2
22
62 2 21
(950.28 10 lb s /ft)(1 in.)
12
11 ft
(950.28 10 lb s /ft) (3.5) in
212 in.





  
     
    

www.elsolucionario.org

1649. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.139 (Continued)


62 2 22
22 2
62 21
(2006.14 10 lb s /ft)[(4.75) (2) ] in
18
21 1 ft
(2006.14 10 lb s /ft) 4.75 2 in
33 12 in.


 



   
  


62
[(23.578 70.735) (0.550 82.490)
(20.559 145.894)] 10 lb ft s


 


62
(94.313 83.040 166.453) 10 lb ft s

 
or
62
344 10 lb ft s

 
x
I 

123
62 2
22
62
() () ()
1
(3325.97 10 lb s /ft)(2.25 in.)
12
2.25 1 ft
(3325.97 10 lb s /ft) in.
212 in.
yy y y
II I I









 

 



62 2 22
22 2
62 21
(950.28 10 lb s /ft)[(2.25) (1) ] in
12
2.25 1 1 ft
(950.28 10 lb s /ft) in
22 12 in.


 


  
     
    


62 2
22
62 2 21
(2006.14 10 lb s /ft)(2 in.)
18
11 ft
(2006.14 10 lb s /ft) (2.25) 2 in
3 12 in.





  
     
    


62
[(9.744 29.232) (3.334 10.002)
(3.096 76.720)] 10 lb ft s

 
 

62
(38.976 13.336 79.816) 10 lb ft s


or
62
132.1 10 lb ft s

 
y
I 





www.elsolucionario.org

1650. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.139 (Continued)


123
62 2 22
22 2
62 2
() () ()
1
(3325.97 10 lb s /ft)[(2.25) (3.5) ] in
12
2.25 3.5 1 ft
(3325.97 10 lb s /ft) in
22 12 in.
zz z z
II I I







  
     
    


62 2
22
62 221
(950.28 10 lb s /ft)(2.25 in.)
12
2.25 1 ft
(950.28 10 lb s /ft) (3.5) in
212 in.





  
     
    


62 2
22
62 2 21
(2006.14 10 lb s /ft)(4.75 in.)
18
21 ft
(2006.14 10 lb s /ft) (2.25) 4.75 in
312 in.






   
  


62
[(33.322 99.967) (2.784 89.192)
(17.463 210.231)] 10 lb ft s


 


62
(133.289 91.976 227.694) 10 lb ft s


or
62
453 10 lb ft s

 
z
I 

www.elsolucionario.org

1651. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.140*
A farmer constructs a trough by welding a rectangular piece
of 2-mm-thick sheet steel to half of a steel drum. Knowing
that the density of steel is
3
7850 kg/mand that the thickness
of the walls of the drum is 1.8 mm, determine the mass
moment of inertia of the trough with respect to each of the
coordinate axes. Neglect the mass of the welds.
SOLUTION
First compute the mass of each component. We have

ST ST
mV tA
Then

32
1
(7850 kg/m )(0.002 m)(0.84 0.21) m
2.76948 kg
m


32
2
(7850 kg/m )(0.0018 m)( 0.285 0.84) m
10.62713 kg
m 


322
34
(7850 kg/m )(0.0018 m) 0.285 m
2
1.80282 kg
mm

 




Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have

34
1234
2
22
() ()
() () () ()
10.21
(2.76948 kg)(0.21 m) (2.76948 kg) 0.285 m
12 2
xx
xx x x x
II
II I I I










22
2
2 1
[(10.62713 kg)(0.285 m) ] 2 (1.80282 kg)(0.285 m)
2
[(0.01018 0.08973) (0.86319) 2(0.07322)] kg m
[(0.09991 0.86319 2(0.07322)] kg m




 
 

or
2
1.110 kg m
x
I 



www.elsolucionario.org

1652. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.140* (Continued)


1234
222
22
2
() () () ()
1
(2.76948 kg)[(0.84) (0.21) ] m
12
0.84 0.21
(2.76948 kg) 0.285 m
22
yy y y y
II I I I




 
   
   


2
222
10.84
(10.62713 kg)[(0.84) 6(0.285) ] m (10.62713 kg) m
12 2

 
 



21
(1.80282 kg)(0.285 m)
4






221
(1.80282 kg)(0.285 m) (1.80282 kg)(0.84 m)
4






2
[(0.17302 0.57827) (1.05647 1.87463)
(0.03661) (0.03661 1.27207)] kg m

 


2
(0.75129 2.93110 0.03661 1.30868) kg m 
or
2
5.03 kg m
y
I 

1234
2
2
2
222
() () () ()
10.84
(2.76948 kg)(0.84 m) (2.76948 kg) m
12 2
10.84
(10.62713 kg)[(0.84) 6(0.285) ] m (10.62713 kg) m
12 2
zz z z z
II I I I







 
 


2
221
(1.80282 kg)(0.285 m)
4
1
(1.80282 kg)(0.285 m) (1.80282 kg)(0.84 m)
4










2
[(0.16285 0.48854) (1.05647 1.87463)
(0.03661) (0.03661 1.27207)] kg m

 

2
(0.65139 2.93110 0.03661 1.30868) kg m 
or
2
4.93 kg m
z
I 

www.elsolucionario.org

1653. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.141
The machine element shown is fabricated from steel.
Determine the mass moment of inertia of the assembly with
respect to (a) the x axis, (b) the y axis, (c) the z axis. (The
density of steel is
3
7850 kg/m .)

SOLUTION
First compute the mass of each component. We have

ST
mV
Then

32
1
(7850 kg/m )( (0.08 m) (0.04 m)]
6.31334 kg

m

32
2
(7850 kg/m )[ (0.02 m) (0.06 m)] 0.59188 kgm 

32
3
(7850 kg/m )[ (0.02 m) (0.04 m)] 0.39458 kgm
Using Figure 9.28 and the parallel-axis theorem, we have
(a)
123
222 2
() () ()
1
(6.31334 kg)[3(0.08) (0.04) ] m (6.31334 kg)(0.02 m)
12




xx x x
II I I

222 21
(0.59188 kg)[3(0.02) (0.06) ] m (0.59188 kg)(0.03 m)
12





222 21
(0.39458 kg)[3(0.02) (0.04) ] m (0.39458 kg)(0.02 m)
12





32
[(10.94312 2.52534) (0.23675 0.53269)
(0.09207 0.15783)] 10 kg m


 


32
(13.46846 0.76944 0.24990) 10 kg m



32
13.98800 10 kg m


or
32
13.99 10 kg m
x
I

  

www.elsolucionario.org

1654. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.141 (Continued)

(b)
123
2
22
22
() () ()
1
(6.31334 kg)(0.08 m)
2
1
(0.59188 kg)(0.02 m) (0.59188 kg)(0.04 m)
2
1
(0.39458 kg)(0.02 m ) (0.39458 kg)(0.04 m)
2
yy y y
II I I














32
[(20.20269) (0.11838 0.94701)
(0.07892 0.63133)] 10 kg m


 


32
(20.20269 1.06539 0.71025) 10 kg m



32
20.55783 10 kg m


or
32
20.6 10 kg m
y
I

  
(c)
123
222 2
() () ()
1
(6.31334 kg)[3(0.08) (0.04) ] m (6.31334 kg)(0.02 m)
12



zz z z
II I I

222 2221
(0.59188 kg)[3(0.02) (0.06) ] m (0.59188 kg)[(0.04) (0.03) ] m
12





222 2221
(0.39458 kg)[3(0.02) (0.04) ] m (0.03958 kg)[(0.04) (0.02) ] m
12





32
[(10.94312 2.52534) (0.23675 1.47970)
(0.09207 0.78916)] 10 kg m


 

32
(13.46846 1.71645 0.88123) 10 kg m



32
14.30368 10 kg m


or
32
14.30 10 kg m
z
I

  

www.elsolucionario.org

1655. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



To the Instructor:
The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the
solutions of Problems 9.142 through 9.145.
From Figure 9.28:
Cylinder
2
cyl cyl1
()
2

x
Ima

22
cyl cyl cyl1
() () (3 )
12
 
yz
II maL
Symmetry and the definition of the mass moment of inertia
2
()Irdm imply

semicylinder cylinder
1
() ()
2
II

2
sc cyl11
()
22




x
Ima
and
22
sc sc cyl11
() () (3 )
212

 


yz
II maL
However,
sc cyl
1
2
mm
Thus,
2
sc sc1
()
2

x
Ima
and
22
sc sc sc1
() () (3 )
12
 
yz
II maL
Also, using the parallel axis theorem find

2
sc 2116
29






x
Im a

22
sc 2116 1
4129



 


z
Im a L
where
x and z are centroidal axes.
www.elsolucionario.org

1656. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.142
Determine the mass moments of inertia and the radii of
gyration of the steel machine element shown with respect
to the x and y axes. (The density of steel is 7850 kg/m
3
.)

SOLUTION

First compute the mass of each component. We have

ST
mV
Then
33
1
323
23
32 3
45
(7850 kg/m )(0.24 0.04 0.14) m
10.5504 kg
(7850 kg/m ) (0.07) 0.04 m 2.41683 kg
2
(7850 kg/m )[ (0.044) (0.04)] m 1.90979 kg
m
mm
mm





  


  

Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to
Problem 9.144) for a semicylinder, we have

12345 2 3
222 2 2
22 2
() () () () () where() ()
11
(10.5504 kg)(0.04 0.14 ) m 2 (2.41683 kg)[3(0.07 m) (0.04 m) ]
12 12
1
(1.90979 kg)[3(0.044 m) (0.04 m) ] (1.90979 kg)(0.04 m)
12
1
1
xx x x x x x x
II I I I I I I 
 
  
 




22
2
2
(1.90979 kg)[3(0.044 m) (0.04 m) ]
2
[(0.0186390) 2(0.0032829) (0.0011790 0.0030557) (0.0011790)] kg m
0.0282605 kg m



  


or
32
28.3 10 kg m
x
I

  
www.elsolucionario.org

1657. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.142 (Continued)


12345
() () () () ()
y yy y y y
IIIIII
where
234 5
() () () |()|
yyy y
III I
Then
222
2
22
2
2
21
(10.5504 kg)(0.24 0.14 ) m
12
1 16 4 0.07
2 (2.41683 kg) (0.07 m ) (2.41683 kg) 0.12 m
239
[(0.0678742) 2(0.0037881 0.0541678)] kg m
0.1837860 kg m
y
I





   

   
    
 


or
32
183.8 10 kg m
y
I

  
Also
12345 2 34 5
where , | |
(10.5504 2 2.41683) kg 15.38406 kg
mm m m m m m m m m  
 
Then
2
2
0.0282605 kg m
15.38406 kg
x
x
I
k
m


or 42.9 mm
x
k 
and
2
2
0.1837860 kg m
15.38406 kgy
y
I
k
m


or
109.3 mm
y
k 

www.elsolucionario.org

1658. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.143
Determine the mass moment of inertia of the steel machine
element shown with respect to the x axis. (The density of
steel is
3
490 lb/ft .)

SOLUTION
First compute the mass of each component. We have

2
490 lb s
;
32.2 ft
mV
g
 
 

Then
3
1 3
33.6 1.2 6
ft
12
15 10
m










2
3
2
3
1.8 1.6
ft
212 12
4.7124 10
m












2
33
3
1.5 1.2
ft 4.9087 10
12 12
m
 








Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to
Problem 9.142) for a semicylinder, we have

 
1
1
123
22 2
33
3
() () ()
11.26 3
15 10 15 10
12 12 12 12
1.26250 10
xx x x
x
x
II I I
I
I





 
 
   
 

  



Half-cylinder, centroidal,

2"
223
22 5
4.7124 10 1.8 1.6
3 3 3.3489 10
12 12 12 12
x
m
IrL 



 
  






www.elsolucionario.org

1659. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.143 (Continued)
Half-cylinder about base,

2" 2'
2'
2'
2
2
53
5
0.7639
3.3489 10 4.7124 10
12
1.43925 10
xx
x
x
IImd
I
I






 





22'
2
2
2
22
53
4
5.2 0.6 0.7639
1.43925 10 4.7124 10
12 12
9.6015 10
xx
x
x
IImd
I
I





 
 
   
 
  
 




3
3
22 2
22 23
3
3
12
4.9087 10 1.5 1.2 2.4
34.908710
12 12 12 12
x
x
m
IrLmd
I






  
 
 
 



3
4
2.1961 10
x
I



Now combining all components and substituting for
,


344490
1.26250 10 9.6015 10 2.1961 10
32.2
x
I


or
32
30.5 10 lb ft s
x
I

  


www.elsolucionario.org

1660. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.144
Determine the mass moment of inertia of the steel machine
element shown with respect to the yaxis. (The density of
steel is
3
490 lb/ft .)

SOLUTION
First compute the mass of each component. We have

2
490 lb s
;
32.2 ft
mV
g
 
 

Then
3
1 3
33.6 1.2 6
ft
12
15 10
m










2
3
2
3
1.8 1.6
ft
212 12
4.7124 10
m












2
33
3
1.5 1.2
ft 4.9087 10
12 12
m
 








Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to
Problem 9.142) for a semicylinder, we have

 
1
1
123
22 2
33
3
() () ()
13.66 3
15 10 15 10
12 12 12 12
1.36250 10
yy y y
y
y
II I I
I
I





 
  
   
  

   






www.elsolucionario.org

1661. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.144 (Continued)

22'
2
2
2
22 23
3
4
4.7124 10 1.8 1.6 5.2
3 4.7124 10
12 12 12 12
9.1837 10
yy
y
y
IImd
I
I








  

 

 




3
223
3
4.9087 10 1.5 2.4
4.9087 10
212 12
y
I




  

 
 


3
4
2.3470 10
y
I



Now combining all components and substituting for
,


344490
1.36250 10 9.1837 10 2.3470 10
32.2
y
I


or
32
31.1 10 lb ft s
y
I

  



www.elsolucionario.org

1662. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.145
Determine the mass moment of inertia of the steel fixture
shown with respect to (a) the x axis, (b) the y axis, (c) the
z axis. (The density of steel is 7850 kg/m
3
.)

SOLUTION
First compute the mass of each component. We have

ST
mV
Then
33
1
7850 kg/m (0.08 0.05 0.160) m
5.02400 kg
m


33
2
323
3
7850 kg/m (0.08 0.038 0.07) m 1.67048 kg
7850 kg/m 0.024 0.04 m 0.28410 kg
2
m
m







Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
(a)
123
22
222 2
() () ()
10.050.16
(5.02400 kg)[(0.05) (0.16) ] m (5.02400 kg) m
12 2 2
xx x x
II I I
 
 
  
 


22
222 2
1 0.038 0.07
(1.67048 kg)[(0.038) (0.07) ] m (1.67048 kg) 0.05 0.05 m
12 2 2
 
 
  
  


222
2116 1
(0.28410 kg) (0.024) (0.04) m
4129

 
 
 


22
2
4 0.024 0.04
(0.28410 kg) 0.05 0.16 m
32


 
  
  


32
32
[(11.7645 35.2936) (0.8831 13.6745) (0.0493 6.0187)] 10 kg m
(47.0581 14.5576 6.0680) 10 kg m






32
26.4325 10 kg m


32
or 26.4 10 kg m
x
I

  
www.elsolucionario.org

1663. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.145 (Continued)

(b)
123
22
222 2
() () ()
10.080.16
(5.02400 kg)[(0.08) (0.16) ] m (5.02400 kg) m
12 2 2
yy y y
II I I
 
 
  
 

22
222 2
10.080.07
(1.67048 kg)[(0.08) (0.07) ] m (1.67048 kg) 0.05 m
12 2 2
 
  
   
   


22
222 2
10.080.04
(0.28410 kg)[3(0.024) (0.04) ] m (0.28410 kg) 0.16 m
12 2 2
 
  
   
   


32
32
[(13.3973 40.1920) (1.5730 14.7420) (0.0788 6.0229)] 10 kg m
(53.5893 16.3150 6.1017) 10 kg m






32
31.1726 10 kg m


32
or 31.2 10 kg m
y
I

  
(c)
123
22
222 2
() () ()
1 0.08 0.05
(5.02400 kg)[(0.08) (0.05) ] m (5.02400 kg) m
12 2 2
zz z z
II I I
 
 
  
 

22
222 2
1 0.08 0.038
(1.67048 kg)[(0.08) (0.038) ] m (1.67048 kg) 0.05 m
12 2 2
 
  
   
   


22
22
2
1 16 0.08 4 0.024
(0.28410 kg) (0.024 m) (0.28410 kg) 0.05 m
2239

 
   
     
     


32
32
[(3.7261 11.1784) (1.0919 4.2781) (0.0523 0.9049)] 10 kg m
(14.9045 5.3700 0.9572) 10 kg m


 



32
8.5773 10 kg m

 
32
or 8.58 10 kg m
z
I

  

www.elsolucionario.org

1664. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


To the instructor:
The following formulas for the mass moment of inertia of wires are derived or summarized at this time
for use in the solutions of Problems 9.146 through 9.148.
Slender Rod

21
0 (Figure 9.28)
12
xyz
IIImL


21
(Sample Problem 9.9)
3
yz
II mL
Circle
We have
22
y
Irdmma


Now
yxz
III
And symmetry implies
xz
II

21
2
xz
II ma
Semicircle
Following the above arguments for a circle, We have

221
2
xz y
II ma Ima 
Using the parallel-axis theorem

2 2
zz
a
IImx x

 
or
2
214
2
z
Im a







www.elsolucionario.org

1665. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.146
Aluminum wire with a weight per unit length of 0.033 lb/ft is used to
form the circle and the straight members of the figure shown.
Determine the mass moment of inertia of the assembly with respect to
each of the coordinate axes.

SOLUTION
First compute the mass of each component. We have

AL
1WW
mL
ggL





Then
1 2
32
2345 2
2
11 ft
0.033 lb/ft (2 16 in.)
12 in.32.2 ft/s
8.5857 10 lb s /ft
11 ft
0.033 lb/ft 8 in.
12 in.32.2 ft/s
0.6832 lb s /ft
m
mmmm


 

   

Using the equations given above and the parallel-axis theorem, we have

12345
2
32 2
() () () () ()
11 ft
(8.5857 10 lb s /ft)(16 in.)
212 in.
xxxxxx
II I I I I



 




2
22
2
22
2
22 2222
2
11 ft
(0.6832 lb s /ft)(8 in.)
3 12 in.
1 ft
[0 (0.6832 lb s /ft)(8 in.) ]
12 in.
11 ft
(0.6832 lb s /ft)(8 in.) (0.6832 lb s /ft)(4 16 ) in
12 12 in.
1
(0.6832 lb s /
12

 



  







2
22222
1 ft
ft)(8 in.) (0.6832 lb s /ft)(8 12 ) in
12 in.






32
[(7.6315) (0.1012) (0.3036) (0.0253 1.2905) (0.0253 0.9868)] 10 lb ft s

 

32
(7.6315 0.1012 0.3036) 1.3158 1.0121) 10 lb ft s



32
10.3642 10 lb ft s


32
or 10.36 10 lb ft s
x
I

  
www.elsolucionario.org

1666. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.146 (Continued)


24 35
12345
2
32 2
() (), () ()
() () () () ()
1 ft
[(8.5857 10 lb s /ft)(16 in.) ]
12 in.
yy yy
yy y y y y
II II
II I I I I




 




2
32 2
2
32 2 2 2
1 ft
2[0 (0.6832 10 lb s /ft)(16 in.) ]
12 in.
11 ft
2 (0.6832 10 lb s /ft)(8 in.) (0.6832 lb s /ft)(12 in.)
12 12 in.

 
   



 




32
32
[(15.2635) 2(1.2146) 2(0.0253 0.6832)] 10 lb ft s
[15.2635 2(1.2146) 2(0.7085)] 10 lb ft s



  


32
19.1097 10 lb ft s


32
or 19.11 10 lb ft s
y
I

  
Symmetry implies
x z
II
32
10.36 10 lb ft s
z
I

  

www.elsolucionario.org

1667. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.147
The figure shown is formed of
1
8
-in.-diametersteel wire. Knowing that the
specific weight of the steel is 490 lb/ft
3
, determine the mass moment of inertia of
the wire with respect to each of the coordinate axes.

SOLUTION
First compute the mass of each component. We have

ST
ST
mV AL
g


Then

233
12 2
490 lb/ft 1 1 ft
in. ( 18 in.)
48 12 in.32.2 ft/s
mm 


  
    
  
  



32
6.1112 10 lb s /ft



233
34 2
490 lb/ft 1 1 ft
in. 18 in.
48 12 in.32.2 ft/s
mm 

  
    
  
  



32
1.9453 10 lb s /ft


Using the equations given above and the parallel-axis theorem, we have

34
1234
2
32 2
2
32 2 32 2
3
() ()
() () () ()
11 ft
(6.1112 10 lb s /ft)(18 in.)
212 in.
11 ft
(6.1112 10 lb s /ft)(18 in.) (6.1112 10 lb s /ft)(18 in.)
212 in.
1
2 (1.9453 10 l
12
xx
xx x x x
II
II I I I






 



  



2
22 32222
32
32
1 ft
b s /ft)(18 in.) (6.1112 10 lb s /ft)(9 18 ) in
12 in.
[(6.8751) (6.8751 13.1502) 2(0.3647 5.4712)] 10 lb ft s
[6.8751 20.6252 2(5.8359)] 10 lb ft s







  


32
39.1721 10 lb ft s


2
or 0.0392 lb ft s
x
I 
www.elsolucionario.org

1668. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.147 (Continued)


12 34
1234
2
32 2
2
32 2
32
() (),() ()
() () () ()
1 ft
2[(6.1112 10 lb s /ft)(18 in.) ]
12 in.
1 ft
2[(0 1.9453 10 lb s /ft)(18 in.) ]
12 in.
[2(13.7502) 2(4.3769)] 10 lb ft s
yy y y
yy y y y
II II
II I I I






 



   





32
36.2542 10 lb ft s


2
or 0.0363 lb ft s
y
I 

34
1234
2
32 2
32 2
2
2
32 22
() ()
() () () ()
11 ft
(6.1112 10 lb s /ft)(18 in.)
212 in.
14
(6.1112 10 lb s /ft) (18 in.)
2
218 1
(6.1112 10 lb s /ft) (18) in
zz
zz z z z
II
II I I I








 


 
 


 
    
  
2
2
32 2
32
32
ft
12 in.
11 ft
2 (1.9453 10 lb s /ft)(18 in.)
3 12 in.
[(6.8751) (1.3024 19.3229) 2(1.4590)] 10 lb ft s
[6.8751 20.6253 2(1.4590)] 10 lb ft s







 


 
  


32
30.4184 10 lb ft s


2
or 0.0304 lb ft s
z
I 

www.elsolucionario.org

1669. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.148
A homogeneous wire with a mass per unit length of 0.056 kg/m is
used to form the figure shown. Determine the mass moment of
inertia of the wire with respect to each of the coordinate axes.

SOLUTION
First compute the mass m of each component. We have

(/)
0.056 kg/m 1.2 m
0.0672 kg
mmLL



Using the equations given above and the parallel-axis theorem, we have

123456
() () () () () ()
xxxxxxx
II I I I I I
Now
134 6 25
() () () () and() ()
xxxx xx
IIII II 
Then
22
21
4 (0.0672 kg)(1.2 m) 2[0 (0.0672 kg)(1.2 m) ]
3
[4(0.03226) 2(0.09677)] kg m
x
I




 

2
0.32258 kg m or
2
0.323 kg m
x
I 

123456
() () () () () ()
yyyyyyy
II I I I I I
Now
126 45
() 0,() (),and() ()
yyy yy
III II 
Then
22
2222
2
21
2 (0.0672 kg)(1.2 m) [0 (0.0672 kg)(1.2 m) ]
3
1
2 (0.0672 kg)(1.2 m) (0.0672 kg)(1.2 0.6 ) m
12
[2(0.03226) (0.09677) 2(0.00806 0.12096)] kg m
[2(0.03226) (0.09677) 2(0.12902)] kg m
y
I








 
 

2
0.41933 kg m or
2
0.419 kg m
y
I 
Symmetry implies
yz
II
2
0.419 kg m
z
I 
www.elsolucionario.org

1670. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.149
Determine the mass products of inertia I xy, Iyz, and I zx of the
steel fixture shown. (The density of steel is 7850 kg/m
3
.)

SOLUTION
First compute the mass of each component. We have

mV
Then
33
1
33
2
323
3
7850 kg/m (0.08 0.05 0.16) m 5.02400 kg
7850 kg/m (0.08 0.038 0.07) m 1.67048 kg
7850 kg/m 0.024 0.04 m 0.28410 kg
2
m
m
m







Now observe that the centroidal products of inertia,
,,and,
xy yz zx
II I
   of each component are zero
because of symmetry. Now

2
3
0.038
0.05 m 0.031 m
2
40.024
0.05 m 0.039814 m
3
y
y


 



 



and then
m, kg
,mx ,my ,mz
2
, kg mmx y
2
, kg mmy z
2
, kg mmz x
1 5.02400 0.04 0.025 0.08
3
5.0240 10


3
10.0480 10


3
16.0768 10


2 1.67048 0.04 0.031 0.085
3
2.0714 10


3
4.4017 10


3
5.6796 10


3 0.28410 0.04 0.039814 0.14
3
0.4524 10


3
1.5836 10


3
1.5910 10



Finally,

3
123
( ) ( ) ( ) [(0 5.0240) (0 2.0714) (0 0.4524)] 10
xy xy xy xy
II I I

   

32
2.5002 10 kg m

 or
32
2.50 10 kg m
xy
I

  
www.elsolucionario.org

1671. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.149 (Continued)


3
123
( ) ( ) ( ) [(0 10.0480) (0 4.4017) (0 1.5836)] 10
yz yz yz yz
II I I

   

32
4.0627 10 kg m


or
32
4.06 10 kg m
yz
I

  

3
123
( ) ( ) ( ) [(0 16.0768) (0 5.6796) (0 1.5910)] 10
zx zx zx zx
II I I

   

32
8.8062 10 kg m


or
32
8.81 10 kg m
zx
I

  

www.elsolucionario.org

1672. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.150
Determine the mass products of inertia I xy, Iyz, and I zx
of the steel machine element shown. (The density of
steel is 7850 kg/m
3
.)

SOLUTION
Since the machine element is symmetrical with respect to the xy plane, I yz = Izx= 0 
Also,
0
xy
I
for all components except the cylinder, since these are symmetrical with respect to the xz plane.



For the cylinder

32
62
(7850 kg/m ) (0.022 m) (0.02 m) 0.2387 kg
0 (0.2387 kg)(0.06 m)(0.02 m) 286.4 10 kg m
xy x y
mV
II mxy


 
    


62
286 10 kg m
xy
I

  




www.elsolucionario.org

1673. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.151
Determine the mass products of inertia I xy, Iyz, and I zx of the cast
aluminum machine component shown. (The specific weight of
aluminum is 0.100 lb/in
3
)

SOLUTION
First compute the mass of each component. We have

AL
AL
mV V
g



Then
3
3
1 2
32
3
32
2 2
32
0.100 lb/in
(5.9 1.8 2.2) in
32.2 ft/s
72.5590 10 lb s /ft
0.100 lb/in
in(1.1) 1.8
232.2 ft/s
10.6248 10 lb s /ft
m
m






 




3
3
3 2
32
0.100 lb/in
(1.4 1.1 1.2) in
32.2 ft/s
5.7391 10 lb s /ft
m




Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  of each component are zero
because of symmetry. Now

2
3
41.1
5.9 in.
3
6.36685 in.
1.1
1.8 in.
2
1.25 in.
x
y


 













www.elsolucionario.org

1674. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.151 (Continued)

and then

2
,lb s /ftm
,ftx ,fty ,ftz
2
,lb ft smx y
2
,lb ft smy z
2
,lb ft smz x
1
3
72.5590 10


2.95
12


0.9
12

1.1
12
3
1.33781 10


3
0.49884 10


3
1.63510 10


2
3
10.6248 10


6.36685
12


0.9
12

1.1
12
3
0.42279 10


3
0.07305 10


3
0.51674 10


3
3
5.7391 10


0.7
12


1.25
12

1.3
12
3
0.03487 10


3
0.06476 10


3
0.03627 10



Finally,
123
3
()() ()
[(0 1.33781) (0 0.42279) (0 0.03487)] 10
xy xy xy xy
II I I


   
or
32
1.726 10 lb ft s
xy
I

  

123
3
()() ()
[(0 0.49884) (0 0.07305) (0 0.06476)] 10
yz yz yz yz
II I I


   
or
32
0.507 10 lb ft s
yz
I

  

123
3
()() ()
[(0 1.63510) (0 0.51674) (0 0.3627)] 10
zx zx zx zx
II I I

    
or
32
2.12 10 lb ft s
zx
I

  
www.elsolucionario.org

1675. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.152
Determine the mass products of inertia I xy, Iyz, and I zx of the
cast aluminum machine component shown. (The specific
weight of aluminum is 0.100 lb/in
3
)

SOLUTION
First compute the mass of each component. We have

AL
AL
mV V
g




Then
3
3
1 2
32
3
3
2 2
32
0.100 lb/in
(7.8 0.8 1.6) in
32.2 ft/s
31.0062 10 lb s /ft
0.100 lb/in
(2.4 0.8 2) in
32.2 ft/s
11.9255 10 lb s /ft
m
m







3
23 2
3 2
3
23 2
4 2
0.100 lb/in
(0.8) 0.8 in 2.4977 lb s /ft
232.2 ft/s
0.100 lb/in
[ (0.55) 0.6] in 1.7708 lb s /ft
32.2 ft/s
m
m 






Now observe that the centroidal products of inertia,
,,and,
xy yz zx
II I
   of each component are zero
because of symmetry. Now

3
40.8
7.8 in. 8.13953 in.
3
x


  





www.elsolucionario.org

1676. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.152 (Continued)

and then


2
,lb s /ftm
,ftx ,fty ,ftz
2
,lb ft smx y
2
,lb ft smy z
2
,lb ft smz x
1
3
31.0062 10


3.9
12


0.4
12
0.8
12

6
335.901 10


6
68.903 10


6
671.801 10


2
3
11.9255 10


1.2
12


0.4
12
2.6
12

6
39.752 10


6
86.129 10


6
258.386 10


3
3
2.4977 10


8.13953
12

0.4
12
0.8
12

6
56.473 10


6
5.550 10


6
112.945 10


4
3
1.7708 10


7.8
12


1.1
12

0.8
12

6
105.511 10


6
10.822 10


6
76.735 10



6
537.637 10


6
171.404 10


6
1119.867 10



Then ()
xy x y
IImxy
  or
62
538 10 lb ft s
xy
I

  
()
yz y z
IImyz
  or
62
171.4 10 lb ft s
yz
I

  
()
zx z x
I Imzx
  or
62
1120 10 lb ft s
zx
I

  


0
0
0 www.elsolucionario.org

1677. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.153
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of steel
is 7850 kg/m
3
, determine the mass products of inertia I xy, Iyz,
and I
zx of the component.

SOLUTION

First compute the mass of each component.
We have
ST ST
mV tA
Then
33
1
33
2
(7850 kg/m )[(0.002)(0.4)(0.45)]m
2.8260 kg
1
7850 kg/m (0.002) 0.45 0.18 m
2
0.63585 kg
m
m


 

 



Now observe that

111
22
()()()0
()()0xy yz zx
yz zxIII
II  
 


From Sample Problem 9.6:
22
2,area 2 21
()
72
xyIbh 
Then
22
2ST 2,areaST 22 22211
() ()
72 36
xy xy
ItI tbhmbh
 





Also
112 2
0.45
0 0.225 m 0.075 m
3
xyz x

   



www.elsolucionario.org

1678. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.153 (Continued)

Finally,
332
1
( ) (0 0) (0.63585 kg)(0.45 m)(0.18 m)
36
0.18 m
(0.63585 kg)( 0.075 m)
3
( 1.43066 10 2.8613 10 ) kg m
xy xy
IImxy


     





 
    

or
32
4.29 10 kg m
xy
I

  
and ( ) (0 0) (0 0) 0
yz y z
IImyz
      
or
0
yz
I

( ) (0 0) (0 0) 0
zx z x
IImzx
  
or 0
zx
I

www.elsolucionario.org

1679. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.154
A section of sheet steel 2 mm thick is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m
3
,
determine the mass products of inertia I
xy, Iyz, and I zx of the
component.

SOLUTION


3
1
3
2
7850 kg/m (0.120 m)(0.200 m)(0.002 m) 0.3768 kg
7850 kg/m (0.120 m)(0.100 m)(0.002 m) 0.1884 kg
mV
mV
 
 

For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes.

m, kg
,xm ,ym ,zm mxy myz mzx
 0.3768 0.1 0.06 0.1 3
2.261 10


3
2.261 10


3
3.768 10


 0.1884 0.2 0.06 0.05
3
2.261 10


3
0.565 10


3
1.884 10


 3
4.522 10


3
2.826 10


3
5.652 10




32
4.52 10 kg m
xy
I

  

32
2.83 10 kg m
yz
I

  

32
5.65 10 kg m
zx
I

  
www.elsolucionario.org

1680. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.155
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of
steel is 7850 kg/m
3
, determine the mass products of inertia
I
xy, Iyz, and I zx of the component.

SOLUTION
First compute the mass of each component. We have

ST ST
mV tA
Then
32
1
322
2
32
3
(7850 kg/m )(0.002 m)(0.35 0.39) m
2.14305 kg
(7850 kg/m )(0.002 m) 0.195 m
2
0.93775 kg
1
(7850 kg/m )(0.002 m) 0.39 0.15 m
2
0.45923 kg
m
m
m













Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero
and

33
()()0
xy zx
II
 
Also
3, mass ST 3, area
() ()
yz yz
ItI 
 
Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to
a 90° rotation, we have

22
3,area 3 31
()
72
yz
Ibh
 
Then
22
3ST 33 33311
()
72 36
yz
Itbhmbh






Also
12
2
0
40.195
m 0.082761 m
3
yx
y




www.elsolucionario.org

1681. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.155 (Continued)

Finally,

()
0.15
(0 0) (0 0) 0 (0.45923 kg)(0.35 m) m
3
xy x y
IImxy
 
 
 
 
 


32
8.0365 10 kg m

  
32
or 8.04 10 kg m
xy
I

  

32
32
()
(0 0) [0 (0.93775 kg)(0.082761 m)(0.195 m)]
1 0.15 0.39
(0.45923 kg)(0.39 m)(0.15 m) (0.45923 kg) m m
36 3 3
[(15.1338) (0.7462 2.9850)] 10 kg m
(15.1338 2.2388) 10 kg m
yz y z
IImyz


 

 

 





32
12.8950 10 kg m


32
or 12.90 10 kg m
yz
I

  

32
()
[0 (2.14305 kg)(0.175 m)(0.195 m)] (0 0)
0.39
0 (0.45923 kg) m (0.35 m)
3
(73.1316 20.8950) 10 kg m
zx z x
IImzx

 
 
 

 




32
94.0266 10 kg m


32
or 94.0 10 kg m
zx
I

  
www.elsolucionario.org

1682. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.156
A section of sheet steel 2 mm thick is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m
3
,
determine the mass products of inertia I
xy, Iyz, and I zx of the
component.

SOLUTION
First compute the mass of each component. We have

ST ST
mV tA
Then

32
13
322
2
322
4 1
(7850 kg/m )(0.002 m) 0.225 0.135 m 0.23844 kg
2
(7850 kg/m )(0.002 m) 0.225 m 0.62424 kg
4
(7850 kg/m )(0.002 m) 0.135 m 0.22473 kg
4
mm
m
m



   











Now observe that the following centroidal products of inertia are zero because of symmetry.

11 2 2
33 4
()()0 ()()0
()()0 ()()0
xy yz xy yz
xy zx xy zx
II II
II II
   
    
 

Also
12
0yy
34
0xx
Now ()
xy x y
IImxy
 
so that
0
xy
I 
Using the results of Sample Problem 9.6, we have

22
,area1
24
uv
Ibh
Now
, mass ST , area
22
ST
1
24
1
12
uv uv
ItI
tbh
mbh






Thus,
1111
1
()
12
zx
Imbh
www.elsolucionario.org

1683. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.156 (Continued)

While
3333
1
()
12
yz
Imbh
because of a 90° rotation of the coordinate axes.
To determine
uv
I for a quarter circle, we have


EL ELuv u v
dI d I u v dm

Where
22
EL EL
22
ST ST ST11
22
uuv v au
dm t dA tv du t a u du
 

  
Then

22
22
ST
0
22
ST
0
22 4 4 4
ST ST
01
()
2
1
()
2
11 1 1 1
22 4 8 2
a
uv uv
a
a
IdI uau ta udu
tua udu
tau u ta ma 




  









Thus
2
2221
()
2
zx
Ima


because of a 90° rotation of the coordinate axes. Also

2
4441
()
2
yz
Ima


Finally,

111 2 22 3 4
2
32
()[( ) ][( ) ]() ()
11
(0.23844 kg)(0.225 m)(0.135 m) (0.22473 kg)(0.135 m)
12 2
( 0.60355 0.65185) 10 kg m
yz yz y z y z yz yz
IIImyzImyzII

 
      

 


   
or
62
48.3 10 kg m
yz
I

  

12 3333 4444
2
32
()()() [( ) ][( ) ]
11
(0.23844 kg)(0.225 m)(0.135 m) (0.62424 kg)(0.225 m)
12 2
(0.60355 5.02964) 10 kg m
zx zx zx zx z x z x
IIII ImzxImzx

 
      






or
32
4.43 10 kg m
zx
I

   

0
0 0
0 0 www.elsolucionario.org

1684. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.157
The figure shown is formed of 1.5-mm-diameter
aluminum wire. Knowing that the density of aluminum is
2800 kg/m
3
, determine the mass products of inertia I xy, Iyz,
and I
zx of the wire figure.

SOLUTION

First compute the mass of each component. We have

AL AL
mV AL
Then

32
14
3
32
25
3
32
36
3
(2800 kg/m ) (0.0015 m) (0.25 m)
4
1.23700 10 kg
(2800 kg/m ) (0.0015 m) (0.18 m)
4
0.89064 10 kg
(2800 kg/m ) (0.0015 m) (0.3 m)
4
1.48440 10 kg
mm
mm
mm

























www.elsolucionario.org

1685. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.157 (Continued)

Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  of each component are zero
because of symmetry.
m, kg ,mx ,my ,mz
2
,kg mmx y
2
,kg mmy z
2
,kg mmz x
1
3
1.23700 10

 0.18 0.125 0
6
27.8325 10

 0 0
2
3
0.89064 10

 0.09 0.25 0
6
20.0394 10

 0 0
3
3
1.48440 10

 0 0.25 0.15 0
6
55.6650 10

 0
4
3
1.23700 10

 0 0.125 0.3 0
6
46.3875 10

 0
5
3
0.89064 10

 0.09 0 0.3 0 0
6
24.0473 10


6
3
1.48440 10

 0.18 0 0.15 0 0
6
40.0788 10



6
47.8719 10


6
102.0525 10


6
64.1261 10



Then ()
xy x y
IImxy
  or
62
47.9 10 kg m
xy
I

  
()
yz y z
IImyz
  or
62
102.1 10 kg m
yz
I

  
()
zx z x
I Imzx
  or
62
64.1 10 kg m
zx
I

  

0
0
0 www.elsolucionario.org

1686. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.158
Thin aluminum wire of uniform diameter is used to form the figure shown.
Denoting by m the mass per unit length of the wire, determine the mass
products of inertia I
xy, Iyz, and I zxof the wire figure.

SOLUTION
First compute the mass of each component. We have

m
mLmL
L





Then
15 1 1
24 21
322
22
()
22
mmm R mR
mmmRR
mm R mR


 


 





Now observe that because of symmetry the centroidal products of inertia,
,,
xyyz
II
  and ,
zx
I
 of
components 2 and 4 are zero and

11 3 3
55
()()0 ()()0
()()0
xy zx xy yz
yz zx
II II
II
   
  


Also
12 2 3 4 45
000xx y y y z z  
Using the parallel-axis theorem [Equations (9.47)], it follows that
xy yz zx
III for components 2 and 4.
To determine
uv
Ifor one quarter of a circular arc, we have
uv
dI uvdm
where cos sinua va
and [( )]dm dV A ad 
where A is the cross-sectional area of the wire. Now

22
mm a A a

 

 
 

so that
dm m ad
and
3
(cos)(sin)( )
sin cos
uv
dI a a m ad
ma d 
 

www.elsolucionario.org

1687. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.158 (Continued)

Then
/2
3
0
/2
32 3
0
sin cos
11
sin
22
uv uv
I dI m a d
ma ma










Thus,
3
111
()
2
yz
I mR
and
33
325111
() ()
22
zx xy
I mR I mR  
because of
90 rotations of the coordinate axes. Finally,


1111 3 333 5
()[( ) ][( ) ]()
xy xy xy xy xy
IIImxyImxyI
  
or
3
11
2
xy
I mR 

13333 5555
()()[( ) ][( ) ]
yz yz yz y z y z
III ImyzImyz
      
or
3
11
2
yz
I mR 

1111 3 5 555
()[( ) ]() [( ) ]
zx zx z x zx z x
I IImzxI Imzx
      
or
3
21
2
zx
I mR 

0 0
0 0
0 0 www.elsolucionario.org

1688. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.159
Brass wire with a weight per unit length w is used to form the figure shown.
Determine the mass products of inertia I
xy, Iyz, and I zx of the wire figure.

SOLUTION
First compute the mass of each component. We have

1W
mwL
gg


Then
1
2
3
4
(2 ) 2
()
(2 ) 2
3
23
2
Ww
maa
gg
ww
maa
gg
ww
maa
gg
ww
maa
gg








Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  of each component are zero
because of symmetry.
m x y z mx y my z mz x
1 2
w
a
g 2a a a
3
4
w
a
g
3
2
w
a
g
3
4
w
a
g
2
w
a
g
2a
1
2
a 0
3w
a
g
0 0
3 2
w
a
g 2a 0 a 0 0
3
4
w
a
g
4 3
w
a
g 2a
3
2
a 2a
3
9
w
a
g
3
9
w
a
g
3
12
w
a
g





3
(1 5 )
w
a
g
3
11
w
a
g
3
4(12)
w
a
g 

www.elsolucionario.org

1689. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.159 (Continued)

Then
()
xy x y
IImxy
  or
3
(1 5 )
xy
w
Ia
g
 
()
yz y z
IImyz
  or
3
11
yz
w
Ia
g
 
()
zx z x
I Imzx
  or
3
4(12)
zx
w
Ia
g
 

0
0
0 www.elsolucionario.org

1690. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.160
Brass wire with a weight per unit length w is used to form the figure
shown. Determine the mass products of inertia I
xy, Iyz, and I zx of the wire
figure.


SOLUTION
First compute the mass of each component. We have

1W
mwL
gg


Then
1
2
3
33
22
(3 ) 3
()
Ww
maa
gg
ww
maa
gg
ww
maa
gg









Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  of each component are zero
because of symmetry.
m x y z mx y my z mz x
1
3
2
w
a
g

23
2
a





2a
1
2
a

3
9
w
a
g

33
2
w
a
g

39
4
w
a
g


2 3
w
a
g 0
1
2
a 2a 0
3
3
w
a
g 0
3
w
a
g

2
()a

a a
3
2
w
a
g

3w
a
g

3
2
w
a
g



3
11
w
a
g

3
3
2
w
a
g





31
4
w
a
g





www.elsolucionario.org

1691. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.160 (Continued)

Then
()
xy x y
IImxy
  or
3
11
xy
w
I a
g
 
()
yz y z
IImyz
  or
31
(6)
2
yz
w
Ia
g
 

()
zx z x
I Imzx
  or
31
4
zx
w
I a
g
 


0
0
0 www.elsolucionario.org

1692. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.161
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of
inertia.


SOLUTION
We have
xy yz zx
I xydm I yzdm I zxdm
 
(9.45)
and xxxyyyzzz    (9.31)
Consider
xy
I xydm


Substituting for x and for y

()()
xy
Ixxyydm
xydm y xdm x ydm xy dm
 
  

  

By definition
xy
I xydm




and xdm mx
ydm my





However, the origin of the primed coordinate system coincides with the mass center G, so that

0xy

xy x y
II mxy
 Q.E.D. 
The expressions for
yz
I and
zx
I are obtained in a similar manner.
www.elsolucionario.org

1693. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.162
For the homogeneous tetrahedron of mass m shown, (a) determine by
direct integration the mass product of inertia I
zx, (b) deduce I yz and Ixy from
the result obtained in part a.


SOLUTION
(a) First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown.
Now 1
az
xzaa
cc

   


and
1
bz
yzbb
cc

   


The mass dm of the slab is

2
11
1
22
z
dm dV xydz ab dz
c
 
 
  
 
 

Then
2
0
3
0
1
1
2
11
1
23 6
c
c z
mdm ab dz
c
cz
ab abc
c



 



 

 
 



Now
EL ELzx z x
dI dI z x dm

where 0(symmetry)
zx
dI

and
EL EL
11
1
33
z
zzx xa
c





Then
2
0
234
2
23
0
345
2
23
0
11
11
32
1
33
6
131
245
c
zx zx
c
c zz
IdI za ab dz
cc
zzz
ab z dz
ccc
mzzz
az
cc cc


 
    
  

 








or
1
20
zx
Imac 
www.elsolucionario.org

1694. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.162 (Continued)

(b) Because of the symmetry of the body,
xy
I and
yz
I can be deduced by considering the circular
permutation of
(, ,)xyz and (,,)abc.
Thus,

1
20
xy
Imab 

1
20
yz
Imbc 
Alternative solution for part a:
First divide the tetrahedron into a series of thin horizontal slices of thickness
dy as shown.
Now
1
ay
xyaa
bb

   


and
1
cy
zycc
bb

   


The mass
dm of the slab is

2
11
1
22
y
dm dV xzdy ac dy
b
 
 
  
 
 

Now
,areazx zx
dI tdI
where
tdy
and
22
,area1
24
zx
dI x z from the results of Sample Problem 9.6.
Then
22
44
22
1
() 1 1
24
11
11
24 4
yy
dIzx dy a c
bb
ymy
a c dy ac dy
bbb


 
  

 
  
 

Finally,
4
0
1
1
4
b
zx zxmy
IdI ac dy
bb

  




5
0
1
1
45
b
mby
ac
bb

 
  
 
or
1
20
zx
Imac 


www.elsolucionario.org

1695. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.162 (Continued)

Alternative solution for part a:
The equation of the included face of the tetrahedron is

1
xyz
abc

so that 1
xz
yb
ac





For an infinitesimal element of sides ,,dx dy and :dz
dm dV dydxdz 
From part a 1
z
xa
c





Now

(1 / ) (1 / / )
00 0
()
ca zc b xazc
zx
I zxdm zx dydxdz 






(1 / )
00
(1 / )
3
22
0
0
23 2
23 2
0
3
2
0
234
2
23
0
1
11 1
23 2
111
11 1
23 2
1
1
6
1
33
6
ca zc
xz
ac
azc
c
c
c
zx b dxdz
xz
bz x x dz
ac
zzzz
bz a a a dz
ca c c c
z
baz dz
c
zzz
ab z dz
ccc









 

 
  
 
  
   
 





 






345
2
23
0
131
245
c
c
mzzz
az
cc cc





or
1
20
zx
I mac
www.elsolucionario.org

1696. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.163
The homogeneous circular cone shown has a mass m.
Determine the mass moment of inertia of the cone with respect
to the line joining the origin O and Point A.

SOLUTION
First note that
2
22
39
(3) (3)
22
OA
daaaa




Then
9
2
13 1
33 (22)
23
OA
aaa
a




ijk ijk
For a rectangular coordinate system with origin at Point A and axes aligned with the given
,,
xyz axes,
we have (using Figure 9.28)

22 231 3
(3 )
54 10
xz y
IImaa I ma

  




2111
20
ma

Also, symmetry implies
0
xy yz zx
III

With the mass products of inertia equal to zero, Equation (9.46) reduces to

222
OA x x y y z z
II I I


222
22 2
111 1 3 2 111 2
20 3 10 3 20 3
ma ma ma
   

   
   


2193
60
ma

2
or 3.22
OA
I ma
www.elsolucionario.org

1697. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.164
The homogeneous circular cylinder shown has a mass m. Determine the mass
moment of inertia of the cylinder with respect to the line joining the origin O and
Point A that is located on the perimeter of the top surface of the cylinder.

SOLUTION
From Figure 9.28:
21
2
y
Ima
and using the parallel-axis theorem

2
22 2 2
11
(3 ) (3 4 )
12 2 12
xz
h
II mah m ma h

    



Symmetry implies
0
xy yz zx
III
For convenience, let Point A lie in the yz plane. Then

22
1
()
OA
ha
ha
 jk
With the mass products of inertia equal to zero, Equation (9.46) reduces to


222
OA x x y y z z
II I I

22
222
22 22
11
(3 4 )
212
ha
ma m a h
ha ha
 
 
 
 

or
22
2
22
1103
12
OA
ha
Ima
ha




Note: For Point A located at an arbitrary point on the perimeter of the top surface,
OA
 is given by

22
1
(cos sin )
OA
aha
ha
 ij k
which results in the same expression for
.
OA
I

0www.elsolucionario.org

1698. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.165
Shown is the machine element of Problem 9.141. Determine its
mass moment of inertia with respect to the line joining the
origin O and Point A.

SOLUTION

First compute the mass of each component.
We have
3
23
ST 2
0.284 lb/in
(0.008819 lb s /ft in )
32.2 ft/s
mV V V
  
Then

32
1
32
2
32
3
(7850 kg/m )[ (0.08 m) (0.04 m)] 6.31334 kg
(7850 kg/m )[ (0.02 m) (0.06 m)] 0.59188 kg
(7850 kg/m )[ (0.02 m) (0.04 m)] 0.39458 kg
m
m
m 





Symmetry implies
1
0()0
yz zx xy
II I 
and
23
()()0
xy xy
II
 
Now
22 2 333
32
32
()
[0.59188 kg (0.04 m)(0.03 m)] [0.39458 kg ( 0.04 m)( 0.02 m)]
(0.71026 0.31566) 10 kg m
0.39460 10 kg m
xy x y
IImxymxymxy


   






www.elsolucionario.org

1699. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.165 (Continued)

From the solution to Problem 9.141, we have

32
32
32
13.98800 10 kg m
20.55783 10 kg m
14.30368 10 kg m
x
y
z
I
I
I



 
 
 

By observation
1
(2 3 )
13
OA
λ i
j
Substituting into Eq. (9.46)

222
222
OA x x y y z z xy x y yz y z zx z x
II I I I I I
     

22
32
32
23
(13.98800) (20.55783)
13 13
22
2(0.39460) 10 kg m
13 13
(4.30400 14.23234 0.36425) 10 kg m



 
  
  


 
 
 


32
or 18.17 10 kg m
OA
I

  


0 0 0 www.elsolucionario.org

1700. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.166
Determine the mass moment of inertia of the steel fixture of
Problems 9.145 and 9.149 with respect to the axis through
the origin that forms equal angles with the x, y, and z axes.

SOLUTION
From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
32
32
32
26.4325 10 kg m
31.1726 10 kg m
8.5773 10 kg m
x
y
z
I
I
I



 
 


Problem 9.149:
32
32
32
2.5002 10 kg m
4.0627 10 kg m
8.8062 10 kg m
xy
yz
zx
I
I
I



 
 
 

From the problem statement it follows that

x yz

Now
222 2
13 1
xyz x
  
or
1
3
xyz

Substituting into Eq. (9.46)

222
222
OL x x y y z z xy x y yz y z zx z x
II I I I I I
     
Noting that
222 1
3
xyzxyyzzx

   
We have
32
1
[26.4325 31.1726 8.5773
3
2(2.5002 4.0627 8.8062)] 10 kg m
OL
I


  


32
or 11.81 10 kg m
OL
I

  
www.elsolucionario.org

1701. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.167
The thin bent plate shown is of uniform density and weight W. Determine its
mass moment of inertia with respect to the line joining the origin O and
Point A.

SOLUTION
First note that
12
1
2
W
mm
g


and that
1
()
3
OA
ijk
Using Figure 9.28 and the parallel-axis theorem, we have

12
2
2
22
22
222
() ()
11 1
12 2 2 2
11 1
()
12 2 2 2 2
111 11 1
2 124 62 2
xx x
II I
WWa
a
gg
WWaa
aa
gg
WW
aaa
gg

 
 

  
   
    






12
22
22
2
22
222
() ()
11 1
()
12 2 2 2 2
11 1
()
12 2 2 2
111 15
262 124
yy y
II I
WWaa
aa
gg
WW a
aa
gg
WW
aaa
gg

   
   
  
 
  
   
 

 
 


12
2
2
2
22
222
() ()
11 1
12 2 2 2
11 1
()
12 2 2 2
111 15 5
2124 124 6
zz z
II I
WWa
a
gg
WW a
aa
gg
WW
aaa
gg

 
 

 
  
   





www.elsolucionario.org

1702. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.167 (Continued)

Now observe that the centroidal products of inertia,
,,
xyyz
II
 and ,
zx
I
 of both components are zero
because of symmetry. Also,
1
0y
Then
2
22 211
() ()
224
xy x y
Wa W
I Imxymxy a a
g g


    




2
22211
()
2228
yz y z
Wa a W
I Imyzmyz a
gg


    




11 1 2 2 2
()
zx z x
I Imzxmzxmzx
   

2113
()
22222 8
Wa a Wa W
aa
g gg
 

 
 

Substituting into Equation (9.46)

222
222
OA x x y y z z xy x y yz y z zx z x
II I I I I I
     
Noting that

222 1
3
xyzxyyzzx
   
We have

22 2 2 2 2
211 5 1 1 3
2
32 6 4 8 8
114 3
2
36 4
OA
WWW WWW
Iaaaaaa
gg g g g g
W
a
g
 
 

 

 


25
or
18
OA
W
I a
g
 

0
0
0www.elsolucionario.org

1703. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.168
A piece of sheet steel of thickness t and specific weight  is cut and
bent into the machine component shown. Determine the mass
moment of inertia of the component with respect to the joining the
origin O and Point A.

SOLUTION
First note that

1
(2 )
6
OA
λ ijk
Next compute the mass of each component. We have

()mV tA
g



Then
2
1
22
2
(2 2 ) 4
22
t
mtaa a
gg
t
mta a
gg

 






Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem
9.134) for a semicircular plate for component 2, we have

12
22 2 222
22 2 2 2
2222
4
() ()
1
4[(2)(2)]4( )
12
1
[(2 ) ( ) ]
42 2
21
42 5
324
18.91335
xx x
II I
tt
aa a aaa
gg
tt
aa a a a
gg
tt
aaaa
gg
t
a
g

 









 
 
 
 



www.elsolucionario.org

1704. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.168 (Continued)


12
22 22
2
222 2
2
222 2
22
4
() ()
1
4(2)4()
12
116 4
()
22 2 3 9
111616
41 1
322 99
7.68953
yy y
II I
tt
aa aa
gg
tta
aaaa
gg
tt
aaa a
gg
t
a
g

 








  
  
     
     
  
 
  
  



12
22 22
2
222 2
2
222 2
22
4
() ()
1
4(2)4()
12
116 4
(2 )
24 2 3 9
1 1 16 16
41 4
324 99
12.00922
zz z
II I
tt
aa aa
gg
tta
aaa a
gg
tt
aaa a
gg
t
a
g

 








  
  
     
     
  
 
  
  


Now observe that the centroidal products of inertia, ,,
xyyz
II
 and ,
zx
I
 of both components are zero
because of symmetry. Also
1
0.x
Then
2
22 2
44
() (2)
23
1.33333
xy x y
ta
I Imxymxy a a
g
t
a
g





   






111 2 22
22
4
()
4()() (2)()
2
7.14159
yz y z
I Imyzmyzmyz
tt
aaa a aa
gg
t
a
g


   




2
222
4 4
() ()
23
0.66667
zx z x
ta
IImzxmzx aa
g
t
a
g




   





0
0
0 www.elsolucionario.org

1705. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.168 (Continued)

Substituting into Eq. (9.46)

222
22
44
222
12
18.91335 7.68953
66
OA x x y y z z xy x y yz y z zx z x
II I I I I I
tt
aa
gg
   
  
 
  
 


2
44
44
4
112
12.00922 2 1.33333
666
22 11
2 7.14159 2 0.66667
66 66
(3.15223 5.12635 2.00154 0.88889 4.76106 0.22222)
tt
aa
gg
tt
aa
gg
t
a
g

  
  
 
   
   
  


or
4
4.41
OA
t
I a
g

 
www.elsolucionario.org

1706. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.169
Determine the mass moment of inertia of the machine
component of Problems 9.136 and 9.155 with respect to
the axis through the origin characterized by the unit vector
(4 8 )/9.
λ ijk

SOLUTION
From the solutions to Problems 9.136 and 9.155. We have
Problem 9.136:
32
32
32
175.503 10 kg m
308.629 10 kg m
154.400 10 kg m
x
y
z
I
I
I



 
 
 

Problem 9.155:
32
32
32
8.0365 10 kg m
12.8950 10 kg m
94.0266 10 kg m
xy
yz
zx
I
I
I



 
 
 

Substituting into Eq. (9.46)

222
222
OL x x y y z z xy x y yz y z zx z x
II I I I I I
     

222
32
3
481
175.503 308.629 154.400
999
48 81
2( 8.0365) 2(12.8950)
99 99
14
2(94.0266) 10 kg m
99
(34.6673 243.855 1.906 6.350
2.547 9.287) 10 k



  
 
  
  

  
  
  
  
 

 
 

 
2
gm

or
32
281 10 kg m
OL
I

  
www.elsolucionario.org

1707. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.170
For the wire figure of Problem 9.148, determine the mass
moment of inertia of the figure with respect to the axis
through the origin characterized by the unit vector
(3 6 2)/7.  ijk

SOLUTION
First compute the mass of each component. We have

0.056 kg/m 1.2 m
0.0672 kg
m
mL
L

 




Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  for each
component are zero because of symmetry.
Also

16 4 5 6 123
000xx y y y zz z  
Then
222 333
()
xy x y
IImxymxymxy
   

2
(0.0672 kg)(0.6 m)(1.2 m) (0.0672 kg)(1.2 m)(0.6 m)
0.096768 kg m





()0
yz y z
IImyz
  


444 555
2
()
(0.0672 kg)(0.6 m)(1.2 m) (0.0672 kg)(1.2 m)(0.6 m)
0.096768 kg m
zx z x
IImzxmzxmzx
   



From the solution to Problem 9.148, we have

2
2
0.32258 kg m
0.41933 kg m
x
yz
I
II

 



0
0
0www.elsolucionario.org

1708. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.170 (Continued)

Substituting into Eq. (9.46)

222
222
2
222
362
0.32258 0.41933 0.41933
777
36 23
2(0.096768) 2(0.096768) kg m
77 77
OL x x y y z z xy x y yz y z zx z x
II I I I I I     

  
   
  

 
   
 
  


2
(0.059249 0.30808 0.034231 0.071095 0.023698) kg m
OL
I 
or
2
0.354 kg m
OL
I 

0 www.elsolucionario.org

1709. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.171
For the wire figure of Problem 9.147, determine the mass moment of inertia of
the figure with respect to the axis through the origin characterized by the unit
vector
(3 6 2)/7.  ijk

SOLUTION
First compute the mass of each component. We have

ST
ST
mV AL
g



Then

323
12 2
32
490 lb/ft 1 1 ft
in. ( 18 in.)
48 12in.32.2 ft/s
6.1112 10 lb s /ft
mm 


 
     
 



323
34 2
2
490 lb/ft 1 1 ft
in. 18 in.
48 12in.32.2 ft/s
1.9453 lb s /ft
mm 
 
     
 


Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  for each component are zero
because of symmetry.
Also
34 1 12
00 0xx y zz  

Then
222
()
xy x y
IImxymxy
  

2
32
32
218 1ft
(6.1112 10 lb s /ft) in. (18 in.)
12 in.
8.75480 10 lb ft s


 
 
 
   


333 444
()
yz y z
IImyzmyzmyz
   
Now
34 34 4 3
,, 0
yz
mm yy z z I
()or0
zx z x zx
IImzx I
  


0
0
0www.elsolucionario.org

1710. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.171 (Continued)

From the solution to Problem 9.147, we have

32
32
32
39.1721 10 lb ft s
36.2542 10 lb ft s
30.4184 10 lb ft s
x
y
z
I
I
I







Substituting into Eq. (9.46)

222
222
32
32
222
362
39.1721 36.2542 30.4184
777
36
2( 8.75480) 10 lb ft s
77
(7.19488 26.6357 2.48313 6.43210) 10 lb ft s
OL x x y y z z xy x y yz y z zx z x
II I I I I I   


  

  

  
  


    

 


or
2
0.0427 lb ft s
OL
I 

0 0 www.elsolucionario.org

1711. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.172
For the wire figure of Problem 9.146, determine the mass moment
of inertia of the figure with respect to the axis through the origin
characterized by the unit vector
(3 6 2)/7.  ijk

SOLUTION
First compute the mass of each component. We have

AL
1
(/)
W
mWLL
gg


Then

1 2
32
11ft
(0.033 lb/ft)(2 16 in.)
12 in.32.2 ft/s
8.5857 10 lb s /ft
m






2345 2
32
11ft
(0.033 lb/ft)(8 in.)
12 in.32.2 ft/s
0.6832 10 lb ft/s
mmmm

 


Now observe that the centroidal products of inertia,
,,and,
xyyz zx
II I
  of each component are zero
because of symmetry. Also

145 1 123
00 0xxx y zz z  
Then
22 2 333
()
xy x y
IImxymxymxy
   

32
32
32
32 16 4
0.6832 10 lb s /ft ft ft
12 12
12 8
0.6832 10 lb s /ft ft ft
12 12
(0.30364 0.45547) 10 lb ft s
0.75911 10 lb ft s



 










Symmetry implies
32
0.75911 10 lb ft s
yz xy yz
II I



()0
zx z x
IImzx
  



0
0www.elsolucionario.org

1712. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.172 (Continued)

From the solution to Problem 9.146, we have

32
32
10.3642 10 lb ft s
19.1097 10 lb ft s
xz
y
II
I


  


Substituting into Eq. (9.46)

222
222
32
222
362
10.3642 19.1097 10.3642
777
36 62
2(0.75911) 2(0.75911) 10 lb ft s
77 77
(1.90663 14.03978 0.8
OL x x y y z z xy x y yz y z zx z x
II I I I I I   

  

  

  
  

 

 
  
 
32
4606 0.55771 0.37181) 10 lb ft s

 

or
32
16.61 10 lb ft s
OL
I

  

0 www.elsolucionario.org

1713. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.173
For the homogeneous circular cylinder shown, of radius a and length L,
determine the value of the ratio a/L for which the ellipsoid of inertia of
the cylinder is a sphere when computed (a) at the centroid of the cylinder,
(b) at Point A.

SOLUTION
(a) From Figure 9.28:

2
221
2
1
(3 )
12
x
yz
Ima
II maL

 

Now observe that symmetry implies
0
xy yz zx
III
Using Eq. (9.48), the equation of the ellipsoid of inertia is then

222 22 222 22 11 1
1: (3 ) (3 ) 1
212 12
xyz
Ix Iy Iz max m a L y m a L    
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore

22211
(3 )
212
ma m a L or
1
3
a
L


(b) Using Figure 9.28 and the parallel-axis theorem, we have

2
2
22 2 21
2
117
(3 )
12 4 4 48
x
yz
Ima
L
II maLm ma L


  
    
  
  

Now observe that symmetry implies

0
xy yz zx
III
  
From Part a it then immediately follows that

222117
2448
ma m a L




or
7
12
a
L


www.elsolucionario.org

1714. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.174
For the rectangular prism shown, determine the values of the
ratios b/a and c/a so that the ellipsoid of inertia of the prism is a
sphere when computed (a) at Point A, (b) at Point B.

SOLUTION
(a) Using Figure 9.28 and the parallel-axis theorem, we have
at Point A

22
2
22
22
2
22 221
()
12
1
()
12 2
1
(4 )
12
11
() (4)
12 2 12
x
y
z
Imbc
a
Imacm
ma c
a
I mab m mab









 


Now observe that symmetry implies

0
xy yz zx
III
  
Using Eq. (9.48), the equation of the ellipsoid of inertia is then

222 222 222 222 11 1
1: ( ) (4 ) (4 ) 1
12 12 12
xyz
Ix Iy Iz mb cx ma cy ma bz
 
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore

22 22
2211
()(4)
12 12
1
(4 )
12
mb c m a c
ma b
 


Then
22 22
4bc ac 
or 2
b
a
 
and
22 22
4bc ab 
or 2
c
a
 

www.elsolucionario.org

1715. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.174 (Continued)

(b) Using Figure 9.28 and the parallel-axis theorem, we have at Point B

2
22 2 2
2
22 2 2
22
11
() (4)
12 12 2
11
() (4)
12 2 12
1
()
12
x
y
z
c
I mb c m mb c
c
I ma c m ma c
Imab













Now observe that symmetry implies

0
xy yz zx
III
     
 
From part a it then immediately follows that

22 22 22111
(4)(4)()
12 12 12
mb c ma c ma b  

Then
2222
44bcac or
1
b
a

and
2222
4bcab
 or
1
2
c
a

www.elsolucionario.org

1716. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.175
For the right circular cone of Sample Problem 9.11, determine the
value of the ratio a/h for which the ellipsoid of inertia of the cone is a
sphere when computed (a) at the apex of the cone, (b) at the center of
the base of the cone.

SOLUTION
(a) From Sample Problem 9.11, we have at the apex A

2
223
10
31
54
x
yz
Ima
II mah


 



Now observe that symmetry implies
0
xy yz zx
III
Using Eq. (9.48), the equation of the ellipsoid of inertia is then


222 22 222 222 331 31
1: 1
10 54 54
xyz
IxIyIz max mahy mahz

     


For the ellipsoid to be a sphere, the coefficients must be equal. Therefore,

222331
10 5 4
ma m a h




or
2
a
h

(b) From Sample Problem 9.11, we have

23
10
x
Ima

and at the centroid C
2231
20 4
y
Imah






Then
2
22 22
31 1
(3 2 )
20 4 4 20
yz
h
II ma h m ma h


    



Now observe that symmetry implies

0
xy yz zx
III
  
From Part a it then immediately follows that

22231
(3 2 )
10 20
ma m a h or
2
3
a
h


www.elsolucionario.org

1717. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.176
Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of
the body with respect to any one of the three axes cannot be larger than the sum of the mass moments of
inertia of the body with respect to the other two axes. That is, prove that the inequality
x yz
IIIand
the two similar inequalities are satisfied. Further, prove that
1
2yx
II if the body is a homogeneous solid
of revolution, where x is the axis of revolution and y is a transverse axis.

SOLUTION
(i) To prove
y zx
III
By definition
22
22
()
()
y
z
I zxdm
I xydm





Then
22 2 2
()()
yz
I I z x dm x y dm   



22 2
()2yzdm xdm 
 

Now
22 2
( ) and 0
x
y z dm I x dm 



Q.E.D.
yzx
III
The proofs of the other two inequalities follow similar steps.
(ii) If the x axis is the axis of revolution, then

y z
II
and from part (i)
y zx
III
or
2
y x
II
or
1
Q.E.D.
2
yx
II 
www.elsolucionario.org

1718. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.177
Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a
sphere, and use this property to determine the moment of inertia of the cube with respect to one of its
diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of
revolution, and determine the principal moments of inertia of the cube at that point.

SOLUTION
(a) At the center of the cube have (using Figure 9.28)

22 211
()
12 6
xyz
III maa ma  
Now observe that symmetry implies

0
xy yz zx
III
Using Equation (9.48), the equation of the ellipsoid of inertia is

22 2 2 22111
1
666
ma x ma y ma z





or
222 2
2 6
()xyz R
ma
 

which is the equation of a sphere.
Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through
the center O of the cube must always
be the same
1
.
OL
R
I




21
6
OL
Ima 
(b) The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes
through corner A. For a rectangular coordinate system at A and with one of the coordinate axes
aligned with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then
rotated 120 about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid
will also remain unchanged after it is rotated. As noted at the end of Section 9.17, this is possible
only if the ellipsoid is an ellipsoid of revolution, where the diagonal is both the axis of revolution
and a principal axis.

It then follows that
21
6
xOL
II ma
 
In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal
and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallel-
axis theorem between the center of the cube and corner A for any perpendicular axis

2
2
13
62
yz
IImama


  



or
211
12
yz
II ma
 
(Note: Part b can also be solved using the method of Section 9.18.) www.elsolucionario.org

1719. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.177 (Continued)

First note that at corner A

2
22
3
1
4
xyz
xy yz zx
III ma
III ma



Substituting into Equation (9.56) yields

322 26 39 55 121
20
48 864
kmak mak ma  

For which the roots are

2
1
2
231
6
11
12
kma
kk ma



www.elsolucionario.org

1720. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.178
Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with
origin at O, prove that the sum I
x + Iy + Izof the mass moments of inertia of the body cannot be smaller
than the similar sum computed for a sphere of the same mass and the same material centered at O.
Further, using the result of Problem 9.176, prove that if the body is a solid of revolution, where x is the
axis of revolution, its mass moment of inertia I
yabout a transverse axis y cannot be smaller than 3ma
2
/10,
where a is the radius of the sphere of the same mass and the same material.

SOLUTION
(i) Using Equation (9.30), we have

22 22 2 2
()()()
xyz
III yzdm zxdm xydm     



222
2( )xyzdm



2
2rdm


where r is the distance from the origin O to the element of mass dm. Now assume that the given
body can be formed by adding and subtracting appropriate volumes
1
V and
2
V from a sphere of
mass m and radius a which is centered at O; it then follows that

1 2 body sphere
()mmm m m
Then

1
body sphere
()()()
xyz xyz xyzV
III III III    

2
()
xyzV
III
or
12
22
body sphere
()()22
xyz xyz
mm
III III rdm rdm    


Now,
12
mm and
12
rr for all elements of mass dm in volumes 1 and 2.

12
22
0
mm
rdm rdm


so that
body sphere
( ) ( ) Q.E.D.
xyz xyz
III III  






www.elsolucionario.org

1721. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.178 (Continued)

(ii) First note from Figure 9.28 that for a sphere

22
5
xyz
III ma
Thus
2
sphere6
()
5
xyz
III ma 
For a solid of revolution, where the x axis is the axis of revolution, we have

y z
II
Then, using the results of part (i)

2
body6
(2)
5
xy
IIma
From Problem 9.178 we have
1
2
yx
I I
or
body
(2 ) 0
yx
II
 
Adding the last two inequalities yields

2
body6
(4 )
5
y
I ma
or
2
body3
( ) Q.E.D.
10
y
Ima
www.elsolucionario.org

1722. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.179*
The homogeneous circular cylinder shown has a mass m, and
the diameter OB of its top surface forms 45 angles with the x
and z axes. (a) Determine the principal mass moments of
inertia of the cylinder at the origin O. (b) Compute the angles
that the principal axes of inertia at O form with the coordinate
axes. (c) Sketch the cylinder, and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.

SOLUTION
(a) First compute the moments of inertia using Figure 9.28 and the
parallel-axis theorem.

2 2
22 2
22 2
113
(3 )
12 2 12 2
13
()
22
xz
y
aa
II maa m ma
Imama ma

 
    
 
 



Next observe that the centroidal products of inertia are zero because of symmetry. Then


21
2222
xy x y
aa
I I mx y m ma


  




21
2222
yz y z
aa
II myzm ma


  




21
222
zx z x
aa
I I mz x m ma


  



Substituting into Equation (9.56)

32213 3 13
12 2 12
KmaK






22 2
22
13 3 3 13 13 13 1 1 1
()
12 2 2 12 12 12 2 22 22
ma K

   

  
   



2 2
2
23
13 3 13 13 1 3 1
12 2 12 12 2 2 22
13 1 1 1 1
2()0
12 222 22 22
ma
 
  

 
 
    
  


0
0
0 www.elsolucionario.org

1723. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.179* (Continued)

Simplifying and letting
2
Kma
 yields

3211 565 95
0
3 144 96
  
Solving yields

123
19
0.363383 1.71995
12

The principal moments of inertia are then

2
1
0.363Kma 

2
2
1.583Kma 

2
3
1.720Kma 
(b) To determine the direction cosines
, ,
xyz
 of each principal axis, we use two of the equations
of Equations (9.54) and (9.57).
Thus,

() 0
xxxyyzxz
IK I I (9.54a)

()0
zx x yz y z z
II IK    (9.54c)

222
1
xyz

 (9.57)
(Note: Since
,
xyyz
II Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of
y
 during the solution process.)
Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)

22213 1 1
0
12 2 22
xyz
ma K ma ma
 
   
 
  

2221113
0
212 22
xy z
ma ma ma K 
  
 
  
   

or
13 1 1
0
12 2 22
xyz
  

 


(i)
and
11 13
0
21222
xy z
 

  


(ii)
Observe that these equations will be identical, so that one will need to be replaced, if

13 1 19
or
12 2 12
  www.elsolucionario.org

1724. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.179* (Continued)

Thus, a third independent equation will be needed when the direction cosines associated with
2
K
are determined. Then for
1
K and
3
K
Eq. (i) through Eq. (ii):
13 1 1 13
0
12 2 2 12
xz
 
   
    
   
  
or
zx

Substituting into Eq. (i):
13 1 1
0
12 2 22
xyx
  

 



or
7
22
12
yx
 





Substituting into Equation (9.57):

2
22
7
22 ( ) 1
12
xxx






or
2
2
7
28 1
12
x



 


 (iii)

1
:K Substituting the value of
1
 into Eq. (iii):

2
2
1
7
2 8 0.363383 ( ) 1
12
x








or
11
( ) ( ) 0.647249
xz

and then
1
7
( ) 2 2 0.363383 (0.647249)
12
y







0.402662


11 1
( ) ( ) 49.7 ( ) 113.7
xz y
 
    

3
:K Substituting the value of
3
 into Eq. (iii):

2
2
3
7
2 8 1.71995 ( ) 1
12
x



 




or
33
( ) ( ) 0.284726
xz

and then
3
7
( ) 2 2 1.71995 (0.284726)
12
y







0.915348


33 3
( ) ( ) 73.5 ( ) 23.7
xz y
 
   www.elsolucionario.org

1725. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.179* (Continued)

K
2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b),
and (9.57).
Now

() 0
xy x y y yz z
IIKI   (9.54b)
Substituting for the moments and products of inertia.

22 213 1
0
222 22
xy z
ma ma K ma 
  
    
  
 

or
13 1
0
222 22
xyz


  


(iv)
Substituting the value of
2
 into Eqs. (i) and (iv):

222
222
13 19 1 1
() () () 0
12 12 2 22
13191
() () () 0
21222 22
xyz
xyz



 



  



or
222
1
() () () 0
2
xyz
  
and
222
2
() () () 0
6
xyz

Adding yields
2
() 0
y

and then
22
() ()
yx

Substituting into Equation (9.57)

22 2
22 2
() () ( ) 1
xy x
 
or
22
11
() and()
22
xz


222
() 45.0() 90.0()135.0
xyz
    
(c) Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B.
Principal axis 2 lies in the xz plane.

0 www.elsolucionario.org

1726. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.180
For the component described in Problem 9.165, determine
(a) the principal mass moments of inertia at the origin, (b) the
principal axes of inertia at the origin. Sketch the body and show
the orientation of the principal axes of inertia relative to the x, y,
and z axes.

SOLUTION
(a) From the solutions to Problems 9.141 and 9.165 we have
Problem 9.141:
32
32
32
13.98800 10 kg m
20.55783 10 kg m
14.30368 10 kg m
x
y
z
I
I
I



 
 
 

Problem 9.165:
32
0
0.39460 10 kg m
yz zx
xy
II
I


 

Eq. (9.55) then becomes

2
0
00or()()()()0
00
xxy
xy y x y z z xy
z
IK I
IIK IKIKIKIKI
IK

  


Thus
22
0() 0
zxyxyxy
IK II IIKKI

Substituting:
32
1
14.30368 10 kg mK

 
or
32
1
14.30 10 kg mK

  
and

33 32 32
(13.98800 10 )(20.55783 10 ) (13.98800 20.55783)(10 ) (0.39460 10 ) 0KK
  
 
or
23 6
(34.54583 10 ) 287.4072 10 0KK

  
Solving yields
32
2
13.96438 10 kg mK


or
32
2
13.96 10 kg mK

  
and
32
3
20.58145 10 kg mK

 or
32
3
20.6 10 kg mK

  

www.elsolucionario.org

1727. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.180 (Continued)

(b) To determine the direction cosines
,,
xyz
 of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
K
1: Begin with Eqs. (9.54a) and (9.54b) with 0.
yz zx
II


11 1
111
()()()0
() ( )() 0
xxxyy
xy x y y
IK I
IIK
 

 
  

Substituting:

33
11
33
11
[(13.98800 14.30368) 10 ]( ) (0.39460 10 )( ) 0
(0.39460 10 )( ) [(20.55783 14.30368) 10 ]( ) 0
xy
xy




 
    

Adding yields
11
() () 0
xy

Then using Eq. (9.57)
222
111
() () () 1
xyz

or
1
() 1
z



111
()90.0()90.0()0
xyz

 
K
2: Begin with Eqs. (9.54b) and (9.54c) with 0.
yz zx
II


222
22
() ( )() 0
()()0
xy x y y
zz
IIK
IK
 
 
(i)
Now
22
() 0
zz
IK 

Substituting into Eq. (i):

33
2
(0.39460 10 )( ) [(20.55783 13.96438) 10 ]( ) 0
xz y


    
or
22
( ) 0.059847( )
y x
 
Using Eq. (9.57):
222
222
( ) [0.05984]( ) ] ( ) 1
xxz

 
or
2
( ) 0.998214
x

and
2
( ) 0.059740
y


22 2
( ) 3.4 ( ) 86.6 ( ) 90.0
xy z
 
 
K
3: Begin with Eqs. (9.54b) and (9.54c) with 0.
yz zx
II


333
33
() ( )() 0
()()0
xy x y y
zz
IIK
IK
 
 
(ii)

0 0
0 www.elsolucionario.org

1728. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.180 (Continued)

Now
33
() 0
zz
IK  
Substituting into Eq. (ii):

33
33
(0.39460 10 )( ) [(20.55783 20.58145) 10 ]( ) 0
xy


    
or
33
( ) 16.70618( )
yx

Using Eq. (9.57):
222
333
( ) [ 16.70618( ) ] ( ) 1
xxz
  
or
3
( ) 0.059751 (axes right-handed set "_")
x
 
and
3
( ) 0.998211
y


333
( ) 93.4 ( ) 3.43 ( ) 90.0
xyz
   
Note: Since the principal axes are orthogonal and
23
() () 90,
zz
 it follows that

23 2 3
|( ) | |( ) | |( ) | | ( ) |
xy y z
  
The differences in the above values are due to round-off errors.
(c) Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.




0 www.elsolucionario.org

1729. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.181*
For the component described in Problems 9.145 and 9.149,
determine (a) the principal mass moments of inertia at the
origin, (b) the principal axes of inertia at the origin. Sketch
the body and show the orientation of the principal axes of
inertia relative to the x, y, and z axes.

SOLUTION
(a) From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
32
26.4325 10 kg m
x
I

 Problem 9.149:
32
2.5002 10 kg m
xy
I

 

32
31.1726 10 kg m
y
I


32
4.0627 10 kg m
yz
I

 

32
8.5773 10 kg m
z
I

 
32
8.8062 10 kg m
zx
I

 
Substituting into Eq. (9.56):

332
2226
2
2
[(26.4325 31.1726 8.5773)(10 )]
[(26.4325)(31.1726) (31.1726)(8.5773) (8.5773)(26.4325)
(2.5002) (4.0627) (8.8062) ](10 )
[(26.4325)(31.1726)(8.5773) (26.4325)(4.0627)
(31.1726)(8.8062) (8
KK
K







2
9
.5773)(2.5002)
2(2.5002)(4.0627)(8.8062)](10 ) 0



or
3326 9
(66.1824 10 ) (1217.76 10 ) (3981.23 10 ) 0KKK


Solving yields

32
1
4.1443 10 kg mK


32
1
or 4.14 10 kg mK

  

32
2
29.7840 10 kg mK


32
2
or 29.8 10 kg mK

  

32
3
32.2541 10 kg mK

 
32
3
or 32.3 10 kg mK

  



www.elsolucionario.org

1730. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.181* (Continued)

(b) To determine the direction cosines
, ,
xyz
 of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then

1
:K Begin with Eqs. (9.54a) and (9.54b).

11 1 1
1111
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 
 

Substituting:

333
111
[(26.4325 4.1443)(10 )]( ) (2.5002 10 )( ) (8.8062 10 )( ) 0
xyz




333
111
(2.5002 10 )( ) [(31.1726 4.1443)(10 )]( ) (4.0627 10 )( ) 0
xyz


     
Simplifying

11 1
11 1
8.9146( ) ( ) 3.5222( ) 0
0.0925( ) ( ) 0.1503( ) 0
xy z
xy z

 
 
  

Adding and solving for
1
():
z


11
( ) 2.4022( )
zx
 
and then
11
1
( ) [8.9146 3.5222(2.4022)]( )
0.45357( )
y x
x
 


Now substitute into Eq. (9.57):

222
11 1
( ) [0.45357( ) ] [2.4022( ) ] 1
xx x

or
1
( ) 0.37861
x

and
1
( ) 0.17173
y

1
( ) 0.90950
z


111
( ) 67.8 ( ) 80.1 ( ) 24.6
xyz

 

2
:K Begin with Eqs. (9.54a) and (9.54b).

22 2 2
2222
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 

 
   

Substituting:

333
222
[(26.4325 29.7840)(10 )]( ) (2.5002 10 )( ) (8.8062 10 )( ) 0
xyz




333
222
(2.5002 10 )( ) [(31.1726 29.7840)(10 )]( ) (4.0627 10 ) ) 0
xyz


    
www.elsolucionario.org

1731. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.181* (Continued)

Simplifying

22 2
22 2
1.3405( ) ( ) 3.5222( ) 0
1.8005( ) ( ) 2.9258( ) 0
xy z
xy z

 
  
  

Adding and solving for
2
():
z


22
( ) 0.48713( )
zx
 
and then
22
2
( ) [ 1.3405 3.5222( 0.48713)]( )
0.37527( )
yx
x
 
  

Now substitute into Eq. (9.57):

22 2
22 2
( ) [0.37527( ) ] [ 0.48713( ) ] 1
xx x
 
  
or
2
( ) 0.85184
x

and
2
( ) 0.31967
y

2
( ) 0.41496
z


122
( ) 31.6 ( ) 71.4 ( ) 114.5
xy z

 

3
:K Begin with Eqs. (9.54a) and (9.54b).

33 3 3
3333
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 

 
   

Substituting:

333
333
[(26.4325 32.2541)(10 )]( ) (2.5002 10 )( ) (8.8062 10 )( ) 0
xyz




333
333
(2.5002 10 )( ) [(31.1726 32.2541)(10 )]( ) (4.0627 10 )( ) 0
xyz


     
Simplifying

33 3
33 3
2.3285( ) ( ) 3.5222( ) 0
2.3118( ) ( ) 3.7565( ) 0
xy z
xy z

 
  
 

Adding and solving for
3
():
z


33
( ) 0.071276( )
zx
 
and then
33
3
( ) [ 2.3285 3.5222(0.071276)]( )
2.5795( )
y x
x
 
 


www.elsolucionario.org

1732. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.181* (Continued)

Now substitute into Eq. (9.57):

22 2
33 3
( ) [ 2.5795( ) ] [0.071276( ) ] 1
xx x
    (i)
or
3
( ) 0.36134
x

and
3
( ) 0.93208
y

3
( ) 0.025755
z


33 3
( ) 68.8 ( ) 158.8 ( ) 88.5
xy z
     
(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
that is,
3
( ) 0.36134.
x


Then
333
( ) 111.2 ( ) 21.2 ( ) 91.5
xyz
   
www.elsolucionario.org

1733. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.182*
For the component described in Problem 9.167, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes of
inertia at the origin. Sketch the body and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.

SOLUTION
(a) From the solution of Problem 9.167, we have

22
22
2211
24
1
8
53
68
xxy
yyz
zzx
WW
I aI a
gg
WW
I aIa
gg
WW
I aIa
gg




Substituting into Eq. (9.56):

322
2222
2
22 215
1
26
1551113
(1) (1)
2662488
1511 3 51 1
(1) (1) 2
2628 8 64 4
W
KaK
g
W
aK
g
 
 
 
    
     
    
    

   
    
3
2
13
0
88
W
a
g
   
 
   

Simplifying and letting
2W
K aK
g
 yields

32
2.33333 1.53125 0.192708 0KKK
 
Solving yields

123
0.163917 1.05402 1.11539KKK
The principal moments of inertia are then
2
1
0.1639
WK a
g


2
2
1.054
WK a
g


2
3
1.115
WK a
g
 
www.elsolucionario.org

1734. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.182* (Continued)
(b) To determine the direction cosines , ,
xyz
 of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then

1
:K Begin with Eqs. (9.54a) and (9.54b).

11 1 1
1211
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 

 
  

Substituting

222
111113
0.163917 ( ) ( ) ( ) 0
248
xyz
WWW
aaa
ggg

     
     
     


222
11111
( ) (1 0.163917) ( ) ( ) 0
48
xyz
WWW
aaa
ggg

  
    
   

Simplifying

11 1
1.34433( ) ( ) 1.5( ) 0
xy z
 
 

11 1
0.299013( ) ( ) 0.149507( ) 0
xy z
  
Adding and solving for
1
():
z


11
( ) 0.633715( )
zx
 
and then
11
1
( ) [1.34433 1.5(0.633715)]( )
0.393758( )
y x
x
 


Now substitute into Eq. (9.57):

222
11 1
( ) [0.393758( ) ] (0.633715( ) ] 1
xx x
 
 
or
1
( ) 0.801504
x
 
and
1
( ) 0.315599
y

1
( ) 0.507925
z


111
( ) 36.7 ( ) 71.6 ( ) 59.5
xyz

 

2
:K Begin with Eqs. (9.54a) and (9.54b):

22 2 2
222 2
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
Ik I I
IIkI
 

 
   



www.elsolucionario.org

1735. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.182* (Continued)

Substituting

222
222
222
222113
1.05402 ( ) ( ) ( ) 0
248
11
( ) (1 1.05402) ( ) ( ) 0
48
xyz
xyz
WWW
aaa
ggg
WWW
aaa
ggg


     
     
     
  
    
   

Simplifying

22 2
2.21608( ) ( ) 1.5( ) 0
xy z
 
  

22 2
4.62792( ) ( ) 2.31396( ) 0
xy z
 
 
Adding and solving for
2
()
z


22
( ) 2.96309( )
zx
 
and then
22
2
( ) [ 2.21608 1.5( 2.96309)]( )
2.22856( )
yx
x
 
  

Now substitute into Eq. (9.57):

22 2
22 2
( ) [2.22856( ) ] [ 2.96309( ) ] 1
xx x
 
or
2
( ) 0.260410
x

and
2
( ) 0.580339
y

2
( ) 0.771618
z


222
( ) 74.9 ( ) 54.5 ( ) 140.5
xyz

 

3
:K Begin with Eqs. (9.54a) and (9.54b):

33 3 3
3333
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 

 
  

Substituting

222
333
222
333113
1.11539 ( ) ( ) ( ) 0
248
11
( ) (1 1.11539) ( ) ( ) 0
48
xyz
xyz
WWW
aaa
ggg
WWW
aaa
ggg


     
     
     
  
    
   

Simplifying

33 3
33 3
2.46156( ) ( ) 1.5( ) 0
2.16657( ) ( ) 1.08328( ) 0
xy z
xy z

   
www.elsolucionario.org

1736. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.182* (Continued)

Adding and solving for
3
()
z


33
( ) 0.707885( )
zx

and then
33
3
( ) [ 2.46156 1.5( 0.707885)]( )
1.39973( )
yx
x

  

Now substitute into Eq. (9.57):

22 2
33 3
( ) [ 1.39973( ) ] [ 0.707885( ) ] 1
xx x
    (i)
or
3
( ) 0.537577
x

and
33
( ) 0.752463 ( ) 0.380543
yz
 

33 3
( ) 57.5 ( ) 138.8 ( ) 112.4
xy z
      
(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
that is,
3
( ) 0.537577.
x


Then
333
( ) 122.5 ( ) 41.2 ( ) 67.6
xyz
   
www.elsolucionario.org

1737. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.183*
For the component described in Problem 9.168, determine (a) the
principal mass moments of inertia at the origin, (b) the principal
axes of inertia at the origin. Sketch the body and show the
orientation of the principal axes of inertia relative to the x, y, and z
axes.

SOLUTION
(a) From the solution to Problem 9.168, we have

4
4
4
18.91335
7.68953
12.00922
x
y
z
t
I a
g
t
I a
g
t
I a
g







4
4
4
1.33333
7.14159
0.66667
xy
yz
zx
t
I a
g
t
I a
g
t
I a
g







Substituting into Eq. (9.56):

342
2
2224
(18.91335 7.68953 12.00922)
[(18.91335)(7.68953) (7.68953)(12.00922) (12.00922)(18.91335)
(1.33333) (7.14159) (0.66667) ]
t
KaK
g
t
aK
g

 
 



 



2
22
3
4
[(18.91335)(7.68953)(12.00922) (18.91335)(7.14159)
(7.68953)(0.66667) (12.00922)(1.33333)
2(1.33333)(7.14159)(0.66667)] 0
t
a
g




 


Simplifying and letting

4
yields
t
KaK
g



32
38.61210 411.69009 744.47027 0KK K
www.elsolucionario.org

1738. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.183* (Continued)

Solving yields

12 3
2.25890 17.27274 19.08046KK K 
The principal moments of inertia are then
4
1
2.26
tK a
g

 

4
2
17.27
tK a
g

 

4
3
19.08
tK a
g

 
(b) To determine the direction cosines
, ,
xyz
 of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then

1
:K Begin with Eqs. (9.54a) and (9.54b):

11 1 1
1211
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 

 
   

Substituting

444
11 1
(18.91335 2.25890) ( ) 1.33333 ( ) 0.66667 ( ) 0
xy z
ttt
aaa
gag
     
    
  


444
111
1.33333 ( ) (7.68953 2.25890) ( ) 7.14159 ( ) 0
xyz
ttt
aaa
ggg
  
    
  

Simplifying

11 1
11 1
12.49087( ) ( ) 0.5( ) 0
0.24552( ) ( ) 1.31506( ) 0
xy z
xy z

 
 
  

Adding and solving for
1
()
z


11
( ) 6.74653( )
zx
 
and then
11
1
( ) [12.49087 (0.5)(6.74653)]( )
9.11761( )
y x
x
 


Now substitute into Eq. (9.57):
22 2
11 1
( ) [9.11761( ) ] [6.74653( ) ] 1
xx x
 
 
or
1
( ) 0.087825
x

and
11
( ) 0.80075 ( ) 0.59251
yz


111
( ) 85.0 ( ) 36.8 ( ) 53.7
xyz

 
www.elsolucionario.org

1739. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.183* (Continued)


2
:K Begin with Eqs. (9.54a) and (9.54b):

22 2 2
2222
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
 

 
   

Substituting

44 4
22 2
444
222
[(18.91335 17.27274) ( ) 1.33333 ( ) 0.66667 ( ) 0
1.33333 ( ) (7.68953 17.27274) ( ) 7.14159 ( ) 0
xy z
xyz
tt t
aa a
gg g
ttt
aaa
ggg 
 

    
     
    
 
    
  

Simplifying

22 2
22 2
1.23046( ) ( ) 0.5( ) 0
0.13913( ) ( ) 0.74522( ) 0
xy z
xy z

 
 
 

Adding and solving for
2
()
z


22
( ) 5.58515( )
zx
 
and then
22
2
( ) [1.23046 (0.5)( 5.58515)]( )
4.02304 ( )
yx
x
 


Now substitute into Eq. (9.57):

22 2
22 2
( ) [4.02304( ) ] [ 5.58515( ) ] 1
xx x
 
  
or
2
( ) 0.14377
x

and
22
( ) 0.57839 ( ) 0.80298
yz


222
( ) 81.7 ( ) 54.7 ( ) 143.4
xyz

 

3
:K Begin with Eqs. (9.54a) and (9.54b):

33 3 3
3333
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz y
IK I I
IIKI
 

 
   

Substituting

44 4
33 3
[(18.91335 19.08046) ( ) 1.33333 ( ) 0.66667 ( ) 0
xy z
tt t
aa a
gg g 
     
     
    


444
333
1.33333 ( ) (7.68953 19.08046) ( ) 7.14159 ( ) 0
xyz
ttt
aaa
ggg
  
    
  

www.elsolucionario.org

1740. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.183* (Continued)

Simplifying

33 3
33 3
0.12533( ) ( ) 0.5( ) 0
0.11705( ) ( ) 0.62695( ) 0
xy z
xy z
 
 
 
Adding and solving for
3
()
z


33
( ) 0.06522( )
zx

and then
33
3
( ) [ 0.12533 (0.5)(0.06522)]( )
0.15794( )
yx
x

 

Now substitute into Eq. (9.57):

222
33 3
( ) [ 0.15794( ) ] [0.06522( ] ] 1
xx x
   (i)
or
3
( ) 0.98571
x

and
33
( ) 0.15568 ( ) 0.06429
yz
 

33 3
( ) 9.7 ( ) 99.0 ( ) 86.3
xy z
      
(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
that is,
3
( ) 0.98571.
x


Then
333
( ) 170.3 ( ) 81.0 ( ) 93.7
xyz
   
www.elsolucionario.org

1741. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.184*
For the component described in Problems 9.148 and 9.170,
determine (a) the principal mass moments of inertia at the origin,
(b) the principal axes of inertia at the origin. Sketch the body and
show the orientation of the principal axes of inertia relative to the
x, y, and z axes.

SOLUTION
(a) From the solutions to Problems 9.148 and 9.170. We have

22
2
0.32258 kg m 0.41933 kg m
0.096768 kg m 0
xyz
xy zx yz
III
II I 
  

Substituting into Eq. (9.56) and using

0
y z xy zx yz
II I I I 

32
22
22
[0.32258 2(0.41933)]
[2(0.32258)(0.41933) (0.41933) 2(0.096768) ]
[(0.32258)(0.41933) 2(0.41933)(0.096768) ] 0
KK
K


 

Simplifying

32
1.16124 0.42764 0.048869 0KK K
Solving yields

2
1
0.22583 kg mK
2
1
or 0.226 kg mK 

2
2
0.41920 kg mK
2
2
or 0.419 kg mK 

2
3
0.51621 kg mK
2
3
or 0.516 kg mK 
(
b) To determine the direction cosines , ,
xyz
 of each principal axis, use two of the equations of
Eqs. (9.54) and (9.57). Then

1
:
K Begin with Eqs. (9.54b) and (9.54c):

1113
() ( )() () 0
xy x y y yz z
IIKI
   

1113
() () ( )() 0
zx x yz y z z
II IK 
  



0
0 www.elsolucionario.org

1742. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.184* (Continued)

Substituting

11
11
(0.096768)( ) (0.41933 0.22583)( ) 0
(0.096768)( ) (0.41933 0.22583)( ) 0
xy
xz



Simplifying yields

11 1
( ) ( ) 0.50009( )
y zx

Now substitute into Eq. (9.54):

22
1
( ) 2[0.50009( ) ] 1
xx

 
or
1
( ) 0.81645
x

and
11
( ) ( ) 0.40830
yz


111
( ) 35.3 ( ) ( ) 65.9
xyz

 

2
:K Begin with Eqs. (9.54a) and (9.54b)

22 2 2
()()()()0
xxxyyzxz
IK I I
 

2222
() ( )() () 0
xy x y y yz z
IIKI
   
Substituting

222
(0.32258 0.41920)( ) (0.096768)( ) (0.096768)( ) 0
xyz
   (i)

22
(0.96768)( ) (0.41933 0.41920)( ) 0
xy

   (ii)
Eq. (ii)

2
() 0
x

and then Eq. (i)

22
() ()
zy
 
Now substitute into Eq. (9.57):

22 2
22 2
() () [()] 1
xy y
 
  
or
2
1
()
2
y

and
2
1
()
2
z


222
( ) 90.0 ( ) 45.0 ( ) 135.0
xyz

 



0
0 www.elsolucionario.org

1743. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.184* (Continued)


3
:K Begin with Eqs. (9.54b) and (9.54c):

3333
() ( )() () 0
xy x y y yz z
IIKI  

3333
() () ( )() 0
zx x yz y z z
II IK   
Substituting

33
(0.096768)( ) (0.41933 0.51621)( ) 0
xy


33
(0.096768)( ) (0.41933 0.51621)( ) 0
xz

Simplifying yields

33 3
() () ()
yz x
 
Now substitute into Eq. (9.57):

22
33
() 2[()] 1
xx
  (i)
or
3
1
()
3
x

and
3
1
() ()
3
yz


333
( ) 54.7 ( ) ( ) 125.3
xyz
    
(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
That is,
3
1
()
3
x

Then
3
( ) 125.3
x

33
() () 54.7
yz
 

0
0 www.elsolucionario.org

1744. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.185
Determine by direct integration the moments of inertia of the
shaded area with respect to the
x and y axes.

SOLUTION

2
2
3
2
3
2
33456
3456
0
34 5 6 7
3456
0
33
4
11
4
33
64
33
3
64 1 3 1 1
34 5 2 7
64 1311 64 3
34527 3420
x
a
xx
a
xx
yh
aa
xx
dI y dx h dx
aa
hxxxx
I dI dx
aaaa
hx x x x
aaaa
hh
aa





  



  










316
105
x
I ah


2
2
4
xx
yh
aa






2
22
2
34
2
0
4
4
y
a
y
dA ydx
xx
dI x dA hx dx
aa
xx
Ih dx
aa


  









45 33
2
0
44
4455
a
y
xxaa
Ih h
aa
  
 

  

31
5
y
I ha


www.elsolucionario.org

1745. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.186
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the
y axis.

SOLUTION

22
22
2
2
22
1
1
2
2
y
xy
ab
x
yb
a
dA ydx
dI x dA x ydx






2
22
2
00
221
aa
yy x
I dI x ydx b x dx
a
  
 

Set:
sin cos
xadxad 

/2
22 2
0
/2 /2
322 3 2
00
/2
/2
33
0
0
2 sin 1 sin cos
1
2sincos 2 sin2
4
11 11
(1 cos 4 ) sin 4
22 44
y
Iba a d
ab d ab d
ab d ab





 









331
0
42 8
ab ab





31
8
y
I ab 
From solution of Problem 9.16:
1
2
A ab
Thus:

31
22 2 8
1
2
1
4y
yy y
I ab
IkA k a
Aab



1
2
y
ka 
www.elsolucionario.org

1746. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.187
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the
x axis.

SOLUTION
At
11 2
,:xayyb
2
1 2
2
2 2
:or
:2 or
b
ybka k
a
b
ybbca c
a

 

Then
2
2
12 22
2
bx
yxyb
aa

 


Now
2
222
21 22 2
2
() 2 ()
xb b
dA y y dx b x dx a x dx
aa a

     



Then
22 2 3
22
0
02214
()
33
a
a
bb
AdA axdx axx ab
aa

    




Now
3
32
33 2
21 22
3
6422466
6
3
642246
6
11 1
2
33 3
1
(8 12 6 )
3
2
(4 6 3 )
3
x
xb
dI y y dx b x dx
aa
b
aaxaxxxdx
a
b
aaxaxxdx
a
 
  
      
   




Then
3
642246
6
0
3
643657
6
0
2
(4 6 3 )
3
231
42
357
a
xx
ab
IdI a a x a x x dx
a
b
ax ax ax x
a
   







3172
105
ab 3
or 1.638
x
I ab
and
3172
22 105
4
3
43
35
x
x
abI
kb
Aab
 


or 1.108
x
kb 
www.elsolucionario.org

1747. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.188
Determine the moments of inertia
x
I and
y
I of the area shown
with respect to centroidal axes respectively parallel and
perpendicular to side AB.

SOLUTION


Dimensions in mm
First locate centroid C of the area.
Symmetry implies
30 mm.Y


2
,mmA
,mmx
3
,mmxA
1 108 60 6480 54 349,920
2
1
72 36 1296
2
   46 –59,616
 5184 290,304
Then
23
: (5184 mm ) 290,304 mmXA xA X 
or 56.0 mmX
Now
12
() ()
xx x
II I
where
364
1
32
2
4341
( ) (108 mm)(60 mm) 1.944 10 mm
12
11
( ) 2 (72 mm)(18 mm) 72 mm 18 mm (6 mm)
36 2
2(11,664 23,328) mm 69.984 10 mm
x
x
I
I

 

 

 
2
[( )
x
I is obtained by dividing
2
Ainto ]
Then
64
(1.944 0.069984) 10 mm
x
I 
or
64
1.874 10 mm
x
I 
www.elsolucionario.org

1748. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.188 (Continued)

Also
12
() ()
yy y
II I
where
32 2
1
4641
( ) (60 mm)(108 mm) (6480 mm )[(56.54) mm]
12
(6,298,560 25,920) mm 6.324 10 mm
y
I
 


32 2
2
4641
( ) (36 mm)(72 mm) (1296 mm )[(56 46) mm]
36
(373,248 129,600) mm 0.502 10 mm
y
I
 

Then
64
(6.324 0.502)10 mm
y
I
or
64
5.82 10 mm
y
I 
www.elsolucionario.org

1749. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.189
Determine the polar moment of inertia of the area shown with
respect to (a) Point O, (b) the centroid of the area.


SOLUTION
First locate centroid C of the area.


2
,mmA
,mmy
3
,mmyA
1 (54)(36) 3053.6
2


48
15.2789

 46,656
2
2
(18) 508.9
2


24
7.6394

 3888
 2544.7 42,768

Then
23
: (2544.7 mm ) 42,768 mmYA yA Y 
or 16.8067 mmY 
(a)
12
22 4
664
64
()()
(54 mm)(36 mm)[(54 mm) (36 mm) ] (18 mm)
84
(3.2155 10 0.0824 10 ) mm
3.1331 10 mm
OO O
JJ J





or
64
3.13 10 mm
O
J 
(b)
2
()
OC
JJAY
or
64 2 2
3.1331 10 mm (2544.7 mm )(16.8067 mm)
C
J 
or
64
2.41 10 mm
C
J 
www.elsolucionario.org

1750. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.190
Two L4  4 
1
2
-in. angles are welded to a steel plate as shown. Determine
the moments of inertia of the combined section with respect to centroidal
axes respectively parallel and perpendicular to the plate.

SOLUTION

For
1
44 -in.
2
 angle:
24
23
3.75 in , 5.52 in
(12.5 in ) 33.85 in
2.708 in.
xy
AII
YA y A
Y
Y





Section Area, in
2
in.y
3
,inyA
Plate (0.5)(10) 5 5 25
Two angles 2(3.75) 7.5  
 12.5 




www.elsolucionario.org

1751. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.190 (Continued)

Entire section:

23 2 21
( ) (0.5)(10) (0.5)(10)(2.292) 2[5.52 (3.75)(1.528) ]
12
xx
IIAd


     




4
41.667 26.266 1604 17.511 96.48 in
4
96.5 in
x
I 

321
(10)(0.5) 2[5.52 (3.75)(1.43) ]
12
y
I

4
0.104 11.04 15.367 26.51in 
4
26.5 in
y
I 

www.elsolucionario.org

1752. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 9.191
Using the parallel-axis theorem, determine the product of inertia
of the L5  3 
1
2
-in. angle cross section shown with respect to the
centroidal x and y axes.


SOLUTION
We have

12
()()
xy xy xy
II I
For each rectangle:

xy x y
II xyA

and 0
xy
I
 (symmetry)
Thus
xy
IxyA

2
, inA
,in.x ,in.y
4
,inxyA
1
1
31.5
2
 0.754
1.49 1.68519
2
1
4.5 2.25
2

0.496 1.01 1.12716
 2.81235


4
2.81in
xy
I 
www.elsolucionario.org

1753. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.192
For the L5  3 
1
2
-in. angle cross section shown, use Mohr’s circle
to determine (a) the moments of inertia and the product of inertia
with respect to new centroidal axes obtained by rotating the x and y
axes 30 clockwise, (b) the orientation of the principal axes through
the centroid and the corresponding values of the moments of inertia.

SOLUTION
From Figure 9.13a:

44
9.43 in 2.55 in
xy
II
From the solution to Problem 9.191

4
2.81235 in
xy
I
The Mohr’s circle is defined by the diameter XY, where

(9.43 2.81235), (2.55, 2.81235)XY
Now
4
ave11
( ) (9.43 2.55) 5.99 in
22
xy
III 
and
22
22
4
11
( ) (9.43 2.55) ( 2.81235)
22
4.4433 in
xy xy
RIII
  
 
  
  

The Mohr’s circle is then drawn as shown.

2
tan 2
2( 2.81235)
9.43 2.55
0.81754
xy
m
xy
I
II







or
2 39.267
m

and
19.6335
m

Then 180 (39.267 60 )
80.733
 
 www.elsolucionario.org

1754. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.192 (Continued)

(a) We have
ave
cos 5.99 4.4433cos80.733
x
II R 
   
or
4
5.27 in
x
I
 

ave
cos 5.99 4.4433cos80.733
y
II R 
   
or
4
6.71in
y
I
 
sin 4.4433sin80.733
xy
IR 
  
or
4
4.39 in
xy
I
 
(b) First observe that the principal axes are obtained by rotating the xy axes through
19.63° counterclockwise

about C.
Now
max, min ave
5.99 4.4433IIR 
or
4
max
10.43 inI 

4
min
1.547 inI 

From the Mohr’s circle it is seen that the a axis corresponds to
max
I and the b axis corresponds to
min
.I
www.elsolucionario.org

1755. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.193
A thin plate of mass m was cut in the shape of a parallelogram as
shown. Determine the mass moment of inertia of the plate with
respect to (a) the x axis, (b) the axis BB, which is perpendicular to
the plate.


SOLUTION

mass area area
mass 


mtA
m
ItI I
A



(a) Consider parallelogram as made of horizontal strips and slide strips to form a square since distance
from each strip to x axis is unchanged.


4
,area1
3
x
Ia

4
, mass ,area 21
3
xx
mm
II a
A a 




21
3

x
Ima 
(b) For centroidal axis y:

44
,area
42
, mass , area 2
222211
2
12 6
11
66
17
66
y
yy
yy
Iaa
mm
II ama
A a
I I ma ma ma ma


 





 


   

For axis
BBto plate, Eq. (9.38):

2217
36
  
BB x y
III mama
23
2

BB
Ima 
www.elsolucionario.org

1756. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.194
A thin plate of mass m was cut in the shape of a parallelogram as
shown. Determine the mass moment of inertia of the plate with
respect to (a) the y axis, (b) the axis AA, which is perpendicular to
the plate.

SOLUTION
See sketch of solution of Problem 9.117.
(a) From part b of solution of Problem 9.117:

21
6

y
I ma

2221
6
yy
IIma mama
  
27
6

y
I ma
(b) From solution of Problem 9.115:

2211
and
63

yx
I ma I ma
Eq. (9.38):

AA y x
III

2211
63
ma ma

21
2

AA
I ma

www.elsolucionario.org

1757. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.195
A 2-mm thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m
3
,
determine the mass moment of inertia of the component with respect
to each of the coordinate axes.

SOLUTION
First compute the mass of each component. We have

ST ST
mV tA
Then
32
1
32
2
322
3
(7850 kg/m )(0.002 m)(0.76 0.48) m
5.72736 kg
1
(7850 kg/m )(0.002 m) 0.76 0.48 m
2
2.86368 kg
(7850 kg/m )(0.002 m) 0.48 m
4
2.84101 kg
m
m
m













Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have

123
2
2
2
2 2
() () ()
10.48
(5.72736 kg)(0.48 m) (5.72736 kg) m
12 2
10.481
(2.86368 kg)(0.48 m) (2.86368 kg) m (2.84101 kg)(0.48 m)
18 3 2
[(0.109965 0.329896) (0.036655 0.073310
xx x x
II I I







 


 
 


2
2
) (0.327284)] kg m
(0.439861 0.109965 0.327284) kg m
 

or
2
0.877 kg m
x
I 





www.elsolucionario.org

1758. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.195 (Continued)


123
22
222 2
2
2 2
() () ()
10.760.48
(5.72736 kg)(0.76 0.48 ) m (5.72736 kg) m
12 2 2
10.761
(2.86368 kg)(0.76 m) (2.86368 kg) m (2.84101 kg)(0.48 m)
18 3 4
[(0.3856
yy y y
II I I
 
 
  
 

 


 
 


2
2
42 1.156927) (0.091892 0.183785) (0.163642)] kg m
(1.542590 0.275677 0.163642) kg m
 
 

or
2
1.982 kg m
y
I 

123
2
2
22
222 2
2
() () ()
10.76
(5.72736 kg)(0.76 m) (5.72736 kg) m
12 2
1 0.76 0.48
(2.86368 kg)(0.76 0.48 ) m (2.86368 kg) m
18 3 3
1
(2.84101 kg)(0.48 m)
4
zz z z
II I I






 
 
  
 






2
2
[(0.275677 0.827031) (0.128548 0.257095)
(0.163642)] kg m
(1.102708 0.385643 0.163642) kg m
z
I

 

or
2
1.652 kg m
z
I 

www.elsolucionario.org

1759. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 9.196
Determine the mass moment of inertia of the steel
machine element shown with respect to the z axis.
(The specific weight of steel is
3
490 lb/ft .)

SOLUTION
First compute the mass of each component. We have

ST
ST
mV V
g



Then

33
3
1 2
32
490 lb/ft 1 ft
(2.7 3.7 9) in
12 in.32.2 ft/s
791.780 10 lb s /ft
m








33
23
2 2
32
490 lb/ft 1 ft
1.35 1.4 in
212 in.32.2 ft/s
35.295 10 lb s /ft
m 








33
23
3 2
32
490 lb/ft 1 ft
( 0.6 1.2) in
12 in.32.2 ft/s
11.952 10 lb s /ft
m









33
332
4 2
490 lb/ft 1 ft
(0.9 0.6 9) in 42.799 10 lb s /ft
12 in.32.2 ft/s
m





The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the
equations derived above (component 2).

www.elsolucionario.org

1760. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.196 (Continued)

Find:
z
I
We have
1234
32 2 22
22 2
22
() () () ()
1
(791.780 10 lb s /ft)[(2.7) (3.7) ] in
12
2.7 3.7 1 ft
(791.780 lb s /ft) in
22 12 in.
zz z z z
II I I I






  
      
    

32 2
2
22
32 2 2 116
(35.295 10 lb s /ft) (1.35 in.)
29
4 1.35 1 ft
(35.295 10 lb s /ft) (1.35) 3.7 in
312 in.





 


 
    
  


32 2
2
32 2 221
(11.952 10 lb s /ft)(0.6 in.)
12
1 ft
(11.952 10 lb s /ft)[(1.35) (3.7) ] in
12 in.





   



32 2 22
22
32 2 21
(42.799 10 lb s /ft)[(0.9) (0.6) ] in
12
0.6 1 ft
(42.799 10 lb s /ft) (1.35) in
212 in.






  
      
    


32
[(9.6132 28.8395) (0.1429 4.9219)
(0.0149 1.2875) (0.0290 0.5684)] 10 lb ft s
z
I

 
 


32
(38.4527 5.0648 1.3024 0.5974) 10 lb ft s


or
2
0.0442 lb ft s
z
I 
www.elsolucionario.org

1761. CHAPTER 10
www.elsolucionario.org

1762. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.1
Determine the vertical force P that must be applied at C to
maintain the equilibrium of the linkage.

SOLUTION
Assume
A
y

2
CA
yy

DA
yy

11
22
EDA
yyy

1
2
GE A
yy y 
Also
A
y
a

 0.3 ma
Virtual Work: We apply both P and M to member ABC.

80 60 20 40 0
11
80 (2 ) 60 20 40 0
22
80 2 60 10 20 0
AC D E G
A
AA A A A
UyPyM y y y
y
yPy M y y y
a
M
P
a

   
 
   





2 130 N
M
P
a

(1)
Now from Eq. (1) for
0M 2 130 NP 65.0 NP


www.elsolucionario.org

1763. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.2
Determine the horizontal force P which must be applied at A so
that the linkage is in equilibrium for the position shown.

SOLUTION
Assume 


10 in.
A
x


4in.
C
y


4in.
DC
yy 


2
63
D
y
 



2
15 15 10 in.
3
G
x

 




Virtual Work: Assume that force P is applied at A.

0:U

30 60 240 80 0
AC D G
UPx y y x      


 
2
10 in. 30 lb 4 in. 60 lb 4 240 lb in.
3
P
   

  




 80 lb 10 in. 0


10 120 240 160 800 0P 


10 1320P 132.0 lbP 
www.elsolucionario.org

1764. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.3
Determine the couple M that must be applied to member ABC to
maintain the equilibrium of the linkage.

SOLUTION
Assume
A
y

DA
yy

11
22
EDA
yyy

1
2
GE A
yy y 
Also
0.3
A
y


Virtual Work: We apply both P and M to member ABC.

80 60 20 40 0
11
80 (2 ) 60 20 40 0
0.3 m 2 2
80 2 60 10 20 0
0.3 m
AC D E G
A
AA A A A
UyPyM y y y
y
yPy M y y y
M
P   

   
 
   


 


2 130 N
0.3 m
M
P
(1)
Now from Eq. (1) for
0P
130 N
0.3 m
M

39.0 N mM




www.elsolucionario.org

1765. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.4
Determine the couple M that must be applied to member ABC to
maintain the equilibrium of the linkage.

SOLUTION
Assume 

10
A
x

4
C
y

4
DC
yy 

2
63
D
y
 


15
2
15
3
G
x







10
Virtual Work: We shall assume that a force
P and a couple M are applied to member ABC as shown.
30 40 180 80 0
ACD G
UPxM y y x       

2
(10 ) 30(4 ) 60(4 ) 240 80(10 ) 0
3
PM
     

  




10 120 240 160 800 0PM

(10 in.) 1320 lb in.PM  (1)
Now from Eq. (1) for
0P 1320 lb in.M 

www.elsolucionario.org

1766. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.5
A spring of constant 15 kN/m connects Points C and F of the linkage
shown. Neglecting the weight of the spring and linkage, determine the force
in the spring and the vertical motion of Point G when a vertical downward
120-N force is applied (a) at Point C, (b) at Points C and H.

SOLUTION

4
44
33
22
GC
HCH C
FC F C
ECE C
yy
yyy y
yyy y
yyy y








For spring:

FC
yy 
Q  Force in spring (assumed in tension)

()(3)2
FC CC C
Qkkyy kyy ky   (1)
(a)
120 N, 0CEFH
Virtual Work:
0: (120 N) 0
CCF
U y Qy Qy   

120 (3 ) 0
CC C
yQyQy  

60 NQ 60.0 N CQ 
Eq. (1):
2,60N2(15kN/m), 2mm
CCC
Qky y y 
At Point G:
4 4( 2 mm) 8 mm
GC
yy  8.00 mm
G
y 
(b)
120 N, 0 CH EF
Virtual Work:
0: (120 N) (120 N) 0
CHCF
UyyQyQy     

120 120(4 ) (3 ) 0
CCCC
yyQyQy  

300 NQ 300 N CQ 
Eq. (1):
2300N2(15kN/m),10mm
CCC
Qky yy 
At Point G:
44(10mm)40mm
GC
yy  40.0 mm
G
y 
www.elsolucionario.org

1767. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.6
A spring of constant 15 kN/m connects Points C and F of the linkage
shown. Neglecting the weight of the spring and linkage, determine the force
in the spring and the vertical motion of Point G when a vertical downward
120-N force is applied (a) at Point E, (b) at Points E and F.

SOLUTION

4
44
33
22
GC
HCH C
FC F C
ECE C
yy
yyy y
yyy y
yyy y








For spring:
FC
yy 
Q  Force in spring (assumed in tension)

()(3)2
FC CC C
kkyy kyy ky  Q (1)
(a)
120 N, 0ECFH
Virtual Work:
0: (120 N) 0
ECF
UyQyQy   
120(2 ) (3 ) 0
CC C
yQyQy 

120 NQ 120.0 N CQ 
Eq. (1):
2 , 120 N 2(15 kN/m) , 4 mm
CCC
Qky y y 
At Point G:
44(4mm)16mm
GC
yy  16.00 mm
G
y 
(b)
120 N, 0  EF CH
Virtual Work:
0 : (120 N) (120 N) 0
EFCF
UyyQyQy    

120(2 ) 120(3 ) (3 ) 0
CCCC
yyQyQy

300 NQ 300 N CQ 
Eq. (1):
2 , 300 N 2(15 kN/m) , 10 mm
CCC
Qky y y 
At Point G:
44(10mm)40mm
GC
yy  40.0 mm
G
y 
www.elsolucionario.org

1768. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.7
The two-bar linkage shown is supported by a pin and bracket at B and
a collar at D that slides freely on a vertical rod. Determine the force
P
required to maintain the equilibrium of the linkage.

SOLUTION

Assume
A
y:

8in. 1
;
16 in. 2
CACA
yyyy
Since bar CD move in translation

EFC
yyy
or
1
2
EF A
yy y
Virtual Work:
0: (100 lb) (150 lb) 0
AEF
UPy y y     

11
100 150 0
22
AA A
Py y y 

  




125 lbP

125.0 lbP 
www.elsolucionario.org

1769. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.8
Knowing that the maximum friction force exerted by the bottle on the cork is
60 lb, determine (a) the force
P that must be applied to the corkscrew to open the
bottle, (b) the maximum force exerted by the base of the corkscrew on the top of
the bottle.

SOLUTION
From sketch 4
AC
yy
Thus
4
AC
yy
(a) Virtual Work:
0: 0
AC
UPyFy
(4 ) 0
CC
Py Fy

1
4
PF

1
60 lb: (60 lb) 15 lb
4
FP

15.00 lbP 
(b) Free body: Corkscrew
00;15lb60lb0
y
FRPFR    
On corkscrew:
45 lbR
,

On bottle:
45.0 lbR 
www.elsolucionario.org

1770. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.9
Rod AD is acted upon by a vertical force P at end A, and by two equal and
opposite horizontal forces of magnitude Q at points B and C. Derive an
expression for the magnitude Q of the horizontal forces required for
equilibrium.

SOLUTION
We have sin
cos
2sin
2cos
3cos
3sin
C
C
B
B
A
A
xa
xa
xa
xa
ya
ya 
 

 

 








Virtual Work: We note that
P tends to increase y A and Q tends to increase x B, while Q tends to decrease
xC. Therefore

0
(3sin ) (2cos ) (cos ) 0
ABC
UPy Qx Qx
Pa Qa Qa

  
   


cos 3 sinQP
 

3tanQP
 
www.elsolucionario.org

1771. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.10
The slender rod AB is attached to a collar A and rests on a small wheel at C.
Neglecting the radius of the wheel and the effect of friction, derive an
expression for the magnitude of the force
Q required to maintain the
equilibrium of the rod.

SOLUTION
For :AA C

2
tan
() tan
cos
A
A
AC a
yACa
a
y 



 


For
:CC B

2
sin
sin tan
sin tan
cos
cos
B
B
BC l A C
la
yBCl a
a
yl 


  


 


Virtual Work:
22
22
0: 0
cos 0
cos cos
cos
cos cos
AB
UQyPy
aa
QPl
aa
QPl
  



 
   
 
 
 

 
 


3
cos 1
l
QP
a





www.elsolucionario.org

1772. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.11
The slender rod AB is attached to a collar A and rests on a small wheel at C.
Neglecting the radius of the wheel and the effect of friction, derive an
expression for the magnitude of the force
Q required to maintain the
equilibrium of the rod.

SOLUTION
For :AA C

2
tan
() tan
cos
A
A
AC a
yACa
a
y 



 


For
:BB C

sin
sin tan
sin tan
tan tan
cos
sin
B
B
BC l AC
la
BC l a
BB
xBBl a
xl 






 

 


Virtual Work:
0: 0
BA
UPxQy

2
(sin ) 0
cos
a
Pl Q 


 



or
2
sin cosPl Qa 
2
sin cos
l
QP
a
 
www.elsolucionario.org

1773. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.12
Knowing that the line of action of the force Q passes through Point
C, derive an expression for the magnitude of
Q required to maintain
equilibrium.

SOLUTION

We have
2cos ; 2sin
AA
yl y l 

2sin ; ( ) cos
22CD l CD l



Virtual Work:
0: ( ) 0
A
UPyQCD  
(2sin ) cos 0
2
Pl Ql

 
  



sin
2
cos( /2)
QP

  www.elsolucionario.org

1774. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.13
Solve Problem 10.12 assuming that the force P applied at Point A
acts horizontally to the left.
PROBLEM 10.12 Knowing that the line of action of the force Q
passes through Point C, derive an expression for the magnitude of
Q
required to maintain equilibrium.

SOLUTION

We have
2cos ; 2cos
AA
xl xl 

2sin ; ( ) cos
22CD l CD l



Virtual Work:
0: ( ) 0
A
UPxQCD
(2 cos ) cos 0
2
Pl Ql

 




cos
2
cos( /2)
QP

  www.elsolucionario.org

1775. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.14
The mechanism shown is acted upon by the force P; derive an expression for the
magnitude of the force
Q required to maintain equilibrium.

SOLUTION
Virtual Work:
We have 2sin
2cos
A
A
xl
xl 


and
3cos
F
yl 

3sin
F
yl
Virtual Work:
0: 0
AF
UQxPy
(2 cos ) ( 3 sin ) 0Ql P l  
3
tan
2
QP  
www.elsolucionario.org

1776. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.15
Derive an expression for the magnitude of the couple M required to
maintain the equilibrium of the linkage shown.

SOLUTION

We have
cos
B
xl 

sin
B
xl (1)

sin
C
yl 

cos
C
yl

Now
1
2
B
xl
Substituting from Equation (1)

1
sin
2
ll 
or
2sin 
Virtual Work: 0:U 0
C
MPy 
(2sin ) (cos ) 0MPl  
or
1cos
2sin
MPl



2tan
Pl
M

 
www.elsolucionario.org

1777. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.16
Derive an expression for the magnitude of the couple M required to maintain the
equilibrium of the linkage shown.

SOLUTION
We have

sin
cos
cos
sin
B
B
A
A
xl
xl
yl
yl







Virtual Work: 0: 0
BA
UMPxPy

(cos ) ( sin ) 0MPl Pl  

(sin cos )MPl  
www.elsolucionario.org

1778. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.17
A uniform rod AB of length l and weight W is suspended from two cords
AC and BC of equal length. Derive an expression for the magnitude of
the couple
M required to maintain equilibrium of the rod in the position
shown.

SOLUTION


1
cos tan cos
2
1
tan sin
2
G
G
yh l
yl 



Virtual Work:
0
G
UWy M 

1
tan sin 0
2
1
tan sin
2
Wl M
MWl
  








www.elsolucionario.org

1779. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.18
The pin at C is attached to member BCD and can slide along a slot cut in
the fixed plate shown. Neglecting the effect of friction, derive an
expression for the magnitude of the couple
M required to maintain
equilibrium when the force
P that acts at D is directed (a) as shown,
(b) vertically downward, (c) horizontally to the right.

SOLUTION
We have cos
D
xl 

sin
D
xl

3sin
D
yl 

3cos
D
yl
Virtual Work:
0: ( cos ) ( sin ) 0
DD
UMP xP y  
( cos )( sin ) ( sin )(3 cos ) 0MP l P l     

(3sin cos cos sin )MPl  (1)
(a) For
P directed along BCD, 
Equation (1):
(3sin cos cos sin )MPl 

(2sin cos )MPl sin 2MPl 
(b) For P directed , 90
Equation (1):
(3sin 90 cos cos90 sin )MPl  3cosMPl 
(c) For P directed , 180
Equation (1):
(3sin180 cos cos180 sin )MPl  sinMPl 

www.elsolucionario.org

1780. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.19
For the linkage shown, determine the couple M required for
equilibrium when
1.8 ft,l 40 lb,Q and 65 .

SOLUTION


1
2
cos
l
C




Virtual Work:
0:U 0MQC 

1
0
2cos
l
MQ
 







1
2cosQl
M


Data:
1 (40 lb)(1.8 ft)
85.18 lb ft
2cos65°
M
85.2 lb ftM



www.elsolucionario.org

1781. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.20
For the linkage shown, determine the force Q required for
equilibrium when
18 in.,l 600 lb in.,M and 70 .

SOLUTION


1
2cosl
C



Virtual Work:
0:U 0MQC 

1
0
2cos
l
MQ
 







2cosM
Q
l

Data:
2(600 lb in.)cos70°
22.801 lb
18 in.
Q

 22.8 lbQ 70.0 
www.elsolucionario.org

1782. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.21
A 4-kN force P is applied as shown to the piston of the engine
system. Knowing that
= 50 mmAB and BC 200 mm,
determine the couple
M required to maintain the equilibrium
of the system when (a)
30, (b) 150 .

SOLUTION
Analysis of the geometry:

Law of sines

sin sin
AB BC


sin sin
AB
BC
 (1)
Now

cos cos
C
xAB BC

sin sin
C
xAB BC  (2)
Now, from Equation (1)
cos cos
AB
BC
 
or
cos
cos
AB
BC
 

 (3)
From Equation (2)

cos
sin sin
cos
C
AB
xAB BC
BC


  


or
(sin cos sin cos )
cos
C
AB
x

 
Then
sin( )
cos
C
AB
x



www.elsolucionario.org

1783. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.21 (Continued)

Virtual Work:
0: 0
C
UPxM  

sin( )
0
cos
AB
PM
 
 



Thus,
sin( )
cos
MAB P



 (4)
(a)
4 kN, 30°P

Eq. (1):
50 mm
sin sin 30
200 mm
 7.181
 
Eq. (4):
sin (30°+7.181°)
(0.05 m) (4 kN)
cos7.818
M

121.8 N mM



(b)
4 kN, 150°P

Eq. (1):
50 mm
sin sin160 7.181
200 mm

Eq. (4):
sin (150°+7.181°)
(0.05 m) (4 kN)
cos 7.181°
M
78.2 N mM


www.elsolucionario.org

1784. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.22
A couple M of magnitude 100 N  m is applied as shown to the
crank of the engine system. Knowing that AB
50 mm and
200 mm,BC
 determine the force P required to maintain the
equilibrium of the system when (a)
60 , (b) 120 .

SOLUTION
Analysis of the geometry:

Law of sines

sin sin
AB BC


sin sin
AB
BC
 (1)
Now

cos cos
C
xAB BC

sin sin
C
xAB BC  (2)
Now, from Equation (1)
cos cos
AB
BC
 
or
cos
cos
AB
BC
 

 (3)
From Equation (2)

cos
sin sin
cos
C
AB
xAB BC
BC


  


or
(sin cos sin cos )
cos
C
AB
x

 
Then
sin( )
cos
C
AB
x



www.elsolucionario.org

1785. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.22 (Continued)

Virtual Work:
0: 0
C
UPxM  

sin( )
0
cos
AB
PM
 
 
  



Thus,
sin( )
cos
MAB P

 (4)
(a)
100 N m, 60M 
Eq. (1):
50 mm
sin sin 60 12.504
200 mm

Eq. (4):
sin (60 12.504 )
100 N m (0.05 m)
cos 12.504
P
 




2047 NP 2.05 kNP 
(b)
100 N m, 120M 
Eq. (1):
50 mm
sin sin120 12.504
200 mm

Eq. (4):
sin (120° 12.504 )
100 N m (0.05 m)
cos12.504
P




2649 NP 2.65 kNP 
www.elsolucionario.org

1786. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.23
Rod AB is attached to a block at A that can slide freely in the vertical
slot shown. Neglecting the effect of friction and the weights of the
rods, determine the value of
 corresponding to equilibrium.

SOLUTION



400sin
400cos
A
A
y
y 




100cos
100sin
D
D
x
x 



Virtual Work:
(160 N) (800 N) 0
AD
Uyx  
(160)(400 cos ) (800)( 100sin ) 0 

sin
0.8; tan 0.8
cos


 38.7 

Dimensions in mm. www.elsolucionario.org

1787. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.24
Solve Problem 10.23 assuming that the 800-N force is replaced by a
24-N m clockwise couple applied at D.
PROBLEM 10.23 Rod AB is attached to a block at A that can slide
freely in the vertical slot shown. Neglecting the effect of friction and
the weights of the rods, determine the value of
 corresponding to
equilibrium.

SOLUTION


(0.4 m)sin
A
y 

(0.4 m)cos
A
y
Virtual Work:
(160 N) (24 N m) 0
A
Uy    
(160 N)(0.4 m)cos (24 N m) 0 

cos 0.375 68.0 
www.elsolucionario.org

1788. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.25
Determine the value of  corresponding to the equilibrium position of the rod
of Problem 10.10 when
30 in.,l 5 in.,a 25 lb,P and 40 lb.Q
PROBLEM 10.10 The slender rod AB is attached to a collar A and rests on a
small wheel at C. Neglecting the radius of the wheel and the effect of friction,
derive an expression for the magnitude of the force
Q required to maintain the
equilibrium of the rod.


SOLUTION
For triangle :AA C tanAC a

() tan
A
yACa  

2
cos
A
a
y


For
triangle :CC B

sinBC l A C

sin tanla

sin tan
B
yBCl a  

2
cos
cos
B
a
yl
  

Virtual Work:
0: 0
AB
UQyPy  

22
cos 0
cos cos
aa
QPl  

 
   
 
 


22
cos
cos cos
aa
QPl

 

 
 

or
3
cos 1
l
QP
a







www.elsolucionario.org

1789. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.25 (Continued)

with
30 in., 5 in., 25 lb, and 40 lblaP Q
  

330 in.
(40 lb) (25 lb) cos 1
5 in.


 


or
3
cos 0.4333 40.8
 

www.elsolucionario.org

1790. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.26
Determine the values of  corresponding to the equilibrium position of the rod
of Prob. 10.11 when l = 24 in., a = 4 in., P = 10 lb, and Q = 18 lb.
PROBLEM 10.11 The slender rod AB is attached to a collar A and rests on a
small wheel at C. Neglecting the radius of the wheel and the effect of friction,
derive an expression for the magnitude of the force
Q required to maintain the
equilibrium of the rod.

SOLUTION
For triangle :AA C tanAC a

() tan
A
yACa  

2
cos
A
a
y


For
triangle :BB C

sin
sin tan
BC l AC
la




sin tan
tan tanBC l a
BB 
 



cos
B
xBBl a  

sin
B
xl
Virtual Work:
0: 0
BA
UPxQy

2
(sin ) 0
cos
a
Pl Q 


 




2
sin cosPl Qa 
or
2
sin cos
l
QP
a




www.elsolucionario.org

1791. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.26 (Continued)

with
24 in., 4 in., 10 lb, and 18 lblaP Q
  

224 in.
18 lb (10 lb) sin cos
4 in.

or
2
sin cos 0.300 
Solving numerically
19.81
 and 51.9 



www.elsolucionario.org

1792. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.27
Determine the value of  corresponding to the equilibrium position
of the mechanism of Problem 10.12 when
= 80 NP and
100 N.Q
PROBLEM 10.12 Knowing that the line of action of the force Q
passes through Point C, derive an expression for the magnitude of
Q
required to maintain equilibrium.

SOLUTION
From geometry

2cos , 2sin
2sin , ( ) cos
22
AA
yl y l
CD l CD l 




Virtual Work:
0:U ()0
A
Py Q CD 
(2sin ) cos 0
2
Pl Ql

 
  



or

2
sin
2
cos
QP



with
80 N, 100 NPQ


2
sin
(100 N) 2(80 N)
cos





2
sin
0.625
cos



or


22
2
2sin cos
0.625
cos




36.42 36.4 
(Additional solutions discarded as not applicable are 180 ) 

www.elsolucionario.org

1793. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.28
Determine the value of  corresponding to the equilibrium position of the
mechanism of Prob. 10.14 when P = 270 N and Q = 960 N.


SOLUTION
Virtual Work:

2 sin , 2 cos
3cos , 3sin
AA
FF
xl xl
yl y l 
 



0:U 0
AF
Qx Py

0:U 0
AF
Qx Py 

(2 cos ) ( 3 sin ) 0Ql P l  

3
tan
2
QP 
Data:
270 N, 960 NPQ

3
(960 N) (270 N) tan
2
 67.1 

www.elsolucionario.org

1794. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.29
Two rods AC and CE are connected by a pin at C and by a
spring AE. The constant of the spring is k, and the spring is
unstretched when
30 .
 For the loading shown, derive an
equation in P,
, l, and k that must be satisfied when the system
is in equilibrium.


SOLUTION
cos
E
yl 

sin
E
yl
Spring:
Unstretched length 2l

2(2 sin ) 4 sinxl l

4cosxl

(2)Fkx l

(4 sin 2 )Fkl l
Virtual Work:
0:U 0
E
Py Fx

( sin ) (4 sin 2 )(4 cos ) 0Pl kl l l   

sin 8 (2sin 1)cos 0Pkl  
or
cos
(1 2 sin )
8sinP
kl 



12sin
8tanP
kl 




www.elsolucionario.org

1795. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.30
Two rods AC and CE are connected by a pin at C and by a spring
AE. The constant of the spring is 1.5 lb/in., and the spring is
unstretched when
30 .
 Knowing that 10 in.l and
neglecting the weight of the rods, determine the value of

corresponding to equilibrium when
40 lb.P


SOLUTION
From the analysis of Problem 10.29
12sin
8tan
P
kl




Then with
160 N, 0.2 m, and 300 N/mPl k
 

160 N 1 2sin
8(300 N/m)(0.2 m) tan




or
12sin 1
0.3333
tan 3




Solving numerically
25.0
 

www.elsolucionario.org

1796. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.31
Solve Problem 10.30 assuming that force P is moved to C and
acts vertically downward.
PROBLEM 10.30 Two rods AC and CE are connected by a pin
at C and by a spring AE. The constant of the spring is 1.5 lb/in.,
and the spring is unstretched when
30 . Knowing that
l
10 in. and neglecting the weight of the rods, determine the
value of
 corresponding to equilibrium when 40 lb.P

SOLUTION
cos , sin
CC
yl y l 
Spring:

Unstretched length 2
2(2 sin ) 4 sin
4cos
(2)
(4 sin 2 )
l
xl l
xl
Fkx l
Fkl l









Virtual Work:
0:
C
UPyFx 

(sin ) (4sin 2)(4cos )0Pl kl l l     

sin 8 (2sin 1)cos 0Pkl
or
cos
(2sin 1)
8sinP
kl 



with
200 mm, 300 N/m, and 160 Nlk P 

(160 N) cos
(2sin 1)
8(300 N/m)(0.2) sin



or
cos 1
(2sin 1)
sin 3



Solving numerically
39.65
and
68.96 39.7
and 69.0 
www.elsolucionario.org

1797. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.32
Two bars AD and DG are connected by a pin at D and by a spring
AG. Knowing that the spring is 300 mm long when unstretched and
that the constant of the spring is 5 kN/m, determine the value of x
corresponding to equilibrium when a 900-N load is applied at E as
shown.

SOLUTION



1
36 2 2
EE
xxx
yyx
  




Linear Spring:
 5000 N/m 0.30 m
SP
Fks x 
Virtual Work:
0U

0
SP E
UFxP  

 
1
5000 N/m 0.30 m 900 N 0
2
xx x


 




5000 1500 450 0x

0.390 mx or 390 mmx 


www.elsolucionario.org

1798. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.33
Solve Prob. 10.32 assuming that the 900-N load is applied at C
instead of E.
Problem 10.32: Two bars AD and DG are connected by a pin at D
and by a spring AG. Knowing that the spring is 300 mm long when
unstretched and that the constant of the spring is 5 kN/m, determine
the value of x corresponding to equilibrium when 900-N load is
applied at E as shown.


SOLUTION



1
66
CC
x
yyx
 




Linear Spring:

 5000 N/m 0.30 m
SP
Fks x 
Virtual Work:
0U


0
SP C
UFxPy


 
1
5000 N/m 0.30 m 900 N 0
6
xx x


 




5000 1500 150 0x

0.330 mx or 330 mmx 
www.elsolucionario.org

1799. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.34
Two 5-kg bars AB and BC are connected by a pin at B and by a
spring DE. Knowing that the spring is 150 mm long when
unstretched and that the constant of the spring is 1 kN/m, determine
the value of x corresponding to equilibrium.

SOLUTION
First note:
2
bar
(5 kg)(9.81 m/s ) 49.05 NW

During the virtual displacement, Points D and E move apart a distance
2
3
()
DEx  and the total work
done by the forces exerted at D and E is 
2
3
Fx

213
0: 49.05 N 49.05 N 0
344
73.575 N
UFx x x
F

  
   
  
  


For
73.575 N,F elongation of spring is

373.575 N
73.575 10 m
1000 N/m
73.575 mm
F
k




Since undeformed length of spring is 150 mm, total length is

2
150 mm 73.575 mm
3
DE x  335 mmx 
www.elsolucionario.org

1800. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.35
A vertical force P of magnitude 150 N is applied to end E of cable
CDE, which passes over a small pulley D and is attached to the
mechanism at C. The constant of the spring is k  4 kN/m, and the
spring is unstretched when
  0. Neglecting the weight of the
mechanism and the radius of the pulley, determine the value of

corresponding to equilibrium.

SOLUTION

0.2 m
0.1 m
lBC
r



90CBD
2sin 45
2
cos 45
2
vl
vl
sr sr
Fkskr



 





 





Virtual Work:
0: 0UPvFs 
cos 45 ( ) 0
2
Pl krr

  
   






2
2
22
2
cos 45
(150 N)(0.2 m)
0.75
(4000 N/m)(0.1 m)
0.75
cos 45
Pl
kr
Pl
kr











0.67623 rad 38.745 38.7 

www.elsolucionario.org

1801. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.36
A load W of magnitude 72 lb is applied to the mechanism at C.
Neglecting the weight of the mechanism, determine the value of

corresponding to equilibrium. The constant of the spring is
20 lb/in.,k and the spring is unstretched when 0.

SOLUTION

sin
cos
C
C
sr
sr
Fkskr
yl
yl









Virtual Work:
0
C
UFsWy  
() (cos )0kr r W l  

22
(72 lb)(15 in.)
1.500
cos (20 lb/in.)(6 in.)
CWl
kr

 
Solving by trial and error:
0.91486 rad 52.4 

www.elsolucionario.org

1802. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.37
Knowing that the constant of spring CD is k and that the spring is unstretched
when rod ABC is horizontal, determine the value of
 corresponding to
equilibrium for the data indicated.
300 N, 400 mm, 5 kN/m.Pl k 

SOLUTION
sin
A
yl 

cos
A
yl
Spring:
vCD
Unstretched when
0
so that
0
2vl
For
:
90
2sin
2
cos 45
2
vl
vl










Stretched length:
0
2 sin 45 2
2
svv l l

   



Then
2sin 45 2
2
Fkskl

  



Virtual Work: 0: 0
A
UPyFv
cos 2sin 45 2 cos 45 0
22
Pl kl l

  
 
 
 

or
1
2sin45 cos45 2cos45
cos 2 2 2
P
kl  
 

 
 


1
2sin 45 cos 45 cos 2 cos 45
cos 2 2 2 

 

 
 



2
cos 45
12
cos






www.elsolucionario.org

1803. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.37 (Continued)

Now, with
300 N, 400 mm, and 5 kN/mPl k 

 
2
cos 45(300 N)
12
(5000 N/m)(0.4 m) cos





or

2
cos 45
0.60104
cos





Solving numerically
22.6
 

www.elsolucionario.org

1804. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.38
Knowing that the constant of spring CD is k and that the spring is
unstretched when rod ABC is horizontal, determine the value of

corresponding to equilibrium for the data indicated.
75 lb, 15 in., 20 lb/in.Pl k
 

SOLUTION
From the analysis of Problem 10.37, we have

 
2
cos 45
12
cos
P
kl





with
75 lb, 15 in. and 20 lb/in.Pl k 

 
2
cos 45(75 lb)
12
(20 lb/in.)(15 in.) cos





or
 
2
cos 45
0.53033
cos





Solving numerically
51.1
 

www.elsolucionario.org

1805. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.39
The lever AB is attached to the horizontal shaft BC that passes
through a bearing and is welded to a fixed support at C. The
torsional spring constant of the shaft BC is K; that is, a couple of
magnitude K is required to rotate end B through 1 rad. Knowing
that the shaft is untwisted when AB is horizontal, determine the
value of
 corresponding to the position of equilibrium when
100 N, 250 mm,Pl and 12.5 N · m/rad.K

SOLUTION
We have sin
A
yl 

cos
A
yl
Virtual Work:
0: 0
A
UPyM

cos 0Pl K 
or
cos
Pl
K

 (1)
with
100 N, 250 mm and 12.5 N m/radPl K 

(100 N)(0.250 m)
cos 12.5 N m/rad



or
2.0000
cos



Solving numerically
59.0 

www.elsolucionario.org

1806. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.40
Solve Problem 10.39 assuming that 350 N,P 250 mm,l and
12.5 N · m/rad.K Obtain answers in each of the following
quadrants:
0 90 , 270 360°, 360 450 .     
PROBLEM 10.39 The lever AB is attached to the horizontal shaft
BC that passes through a bearing and is welded to a fixed support
at C. The torsional spring constant of the shaft BC is K; that is, a
couple of magnitude K is required to rotate end B through 1 rad.
Knowing that the shaft is untwisted when AB is horizontal,
determine the value of
 corresponding to the position of
equilibrium when
100PN, l 250 mm, and
12.5 N · m/rad.K

SOLUTION
Using Equation (1) of Problem 10.39 and

350 N, 250 mm and 12.5 N m/radPl K 
We have
(350 N)(0.250 m)
cos 12.5 N m/rad



or
7or 7cos
cos



 (1)
The solutions to this equation can be shown graphically using any appropriate graphing tool, such as
Maple, with the command:
plot ({theta, 7 * cos(theta)}, 0.5* Pi/2);t
Thus, we plot
and 7cos in the rangeyy

5
0
2





www.elsolucionario.org

1807. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.40 (Continued)

We observe that there are three points of intersection, which implies that Equation (1) has three roots in
the specified range of
.

0 90 ; 1.37333 rad, 78.69
2

 
    


78.7
 

3
270 360 2 ; 5.65222 rad, 323.85
2
 
     


 324
 

5
360 450 2 ; 6.61597 rad, 379.07
2
  
    


 379
 

www.elsolucionario.org

1808. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.41
The position of boom ABC is controlled by the hydraulic cylinder
BD. For the loading shown, determine the force exerted by the
hydraulic cylinder on pin B when
 = 70.

SOLUTION
We have

(4.5ft)cos 4.5sin
AA
yy  

222
22
( ) ( ) ( ) 2( )( )cos
(3) (2) 2(3)(2)cos
BD BC CD BC CD 




2
( ) 13 12cosBD  (1)

Differentiating:
2( ) ( ) 12sinBD BD

6
() sin
BD
BD (2)
Virtual Work: Noting that
P tends to decrease
A
y and
BD
F tends to increase BD, write
()0
ABD
UPyF BD   

6
8 kips( 4.5sin ) sin 0
BD
F
BD 

  





36 sin
6sin
BD
BD
F


or,
6 kips( )
BD
FBD (3)
Making
70 in Eq. (1), we have

2
( ) 13 12cos70 8.8958
2.9826 ft
BD
BD
 


Carrying into Eq. (3):
6(2.9826) 17.8956 kips
BD
F 17.90 kips
BD
F 
www.elsolucionario.org

1809. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.42
The position of boom ABC is controlled by the hydraulic
cylinder BD. For the loading shown, determine the largest
allowable value of the angle
 if the maximum force that
the cylinder can exert on pin B is 25 kips.

SOLUTION
(a) See solution of Problem 10.41 for the derivation of Eq. (3):

(6 kips)( )
BD
F BD 
(b) For
max
( ) 25kips,
BD
F  we have

25 kips (6 kips)( )
4.1667 ft
BD
BD



Carrying this value into Eq. (1) of Problem 10.41, write

2
2
( ) 13 12cos
(4.1667) 13 12cos
cos 0.36343
BD






111.3
 

www.elsolucionario.org

1810. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.43
The position of member ABC is controlled by
the hydraulic cylinder CD. For the loading
shown, determine the force exerted by the
hydraulic cylinder on pin C when
55 .

SOLUTION


222
222
2
(0.8 m)sin
0.8cos
2( )( )cos(90 )
0.5 1.5 2(0.5)(1.5)sin
2.5 1.5sin
A
A
y
y
CD BC BD BC BD
CD
CD 





 

 (1)

3cos
2( )( ) 1.5cos
4
CD CD
CD
CD


 
Virtual Work:
0: (10 kN) 0
ACDCD
UyF  

3cos
10(0.8cos ) 0
4
CD
F
CD

 





32
3
CD
FCD (2)
For
55 :
Eq. (1):
2
2.5 1.5sin 55 1.2713; 1.1275 mCD CD  
Eq. (2):
32 32
(1.1275) 12.027 kN
33
CD
FCD  12.03 kN
CD
F 

www.elsolucionario.org

1811. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.44
The position of member ABC is controlled by
the hydraulic cylinder CD. Determine the angle

knowing that the hydraulic cylinder exerts a 15-kN
force on pin C.

SOLUTION


222
222
2
(0.8 m)sin
0.8cos
2( )( )cos(90 )
0.5 1.5 2(0.5)(1.5)sin
2.5 1.5sin
A
A
y
y
CD BC BD BC BD
CD
CD 





 

 (1)

3cos
2( )( ) 1.5cos ;
4
CD CD
CD
CD


 
Virtual Work:
0: (10 kN) 0
ACDCD
UyF  

3cos
10(0.8cos ) 0
4
CD
F
CD

 





32
3
CD
FCD (2)
For
15 kN:
CD
F
Eq. (2):
32
15 kN :
3
CD
45
1.40625 m
32
CD
Eq. (1):
2
(1.40625) 2.5 1.5sin : sin 0.34831 

20.38 20.4 

www.elsolucionario.org

1812. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.45
The telescoping arm ABC is used to provide an elevated
platform for construction workers. The workers and the
platform together weigh 500 lb and their combined center of
gravity is located directly above C. For the position when
20 ,
 determine the force exerted on pin B by the single
hydraulic cylinder BD.

SOLUTION

In
:ADE
2.7 ft
tan
1.5 ft
60.945
2.7 ft
3.0887 m
sin 60.945





AE
DE
AD

From the geometry:
(15 ft ) sin
(15 ft ) cos


C
C
y
y
Then, in triangle BAD: Angle
BAD
Law of cosines:

222
2( )( )cos( )BD AB AD AB AD 
or
22 2
(7.2 ft) (3.0887 ft) 2(7.2 ft)(3.0887 ft)cos( )BD   

22 2
61.380 ft (44.477cos( )) ftBD   (1)
www.elsolucionario.org

1813. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.45 (Continued)

And then

2( )( ) (44.477sin( ))
44.477sin( )
2( )
BD BD
BD
BD





Virtual Work:
0: 0
CBD
UPyFBD  
Substituting
2
(44.477 ft )sin( )
(500 lb)(15 ft)cos 0
2( )
BD
F
BD

  
  


or
cos
337.25 lb/ft
sin( )
BD
FBD





(2)
Now, with
20 and 60.945
Equation (1):

2
2
61.380 44.477cos(60.945 20 )
54.380
7.3743 ft
BD
BD
BD
 



Equation (2):

cos 20
337.25 (7.3743 ft) lb/ft
sin(60.945 20 )
BD
F
 


 
or
2366 lb
BD
F 2370 lb
BD
F 

www.elsolucionario.org

1814. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.46
Solve Problem 10.45 assuming that the workers are lowered
to a point near the ground so that
20 . 
PROBLEM 10.45 The telescoping arm ABC is used to
provide an elevated platform for construction workers. The
workers and the platform together weigh 500 lb and their
combined center of gravity is located directly above C. For
the position when
20 ,
 determine the force exerted on
pin B by the single hydraulic cylinder BD.

SOLUTION
Using the figure and analysis of Problem 10.45, including Equations (1) and (2), and with 20 ,  we
have
Equation (1):
2
2
61.380 44.477cos(60.945 20 )
27.785
5.2711 ft
BD
BD
BD
 


Equation (2):
cos( 20 )
337.25 (5.2711)
sin(60.945 20 )
BD
F


 


2549 lb
BD
F or 2550 lb
BD
F 

www.elsolucionario.org

1815. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.47
Denoting by
s
 the coefficient of static friction between collar C and the
vertical rod, derive an expression for the magnitude of the largest couple
M
for which equilibrium is maintained in the position shown. Explain what
happens if
tan .
s


SOLUTION


Member BC: We have

cos
B
xl 

sin
B
xl (1)
and

sin
C
yl 

cos
C
yl (2)
Member AB: We have

1
2
B
xl
Substituting from Equation (1),

1
sin
2
ll
 
or
2sin  (3)
Free body of rod BC
For
max
,M motion of collar C impends upward

0: sin ( )( cos ) 0
Bs
MNlPNl    

tan
tan
s
s
NNP
P
N





www.elsolucionario.org

1816. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.47 (Continued)

Virtual Work: 0: ( ) 0
sC
UMPNy 

tan
max
(2sin ) ( )cos 0
()
2tan 2tan

  




 



s
s
P
s
s
MPNl
P
PN
M ll
or
max
2(tan )
s
Pl
M


 
If
tan , ,
s
M system becomes self-locking. 

www.elsolucionario.org

1817. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.48
Knowing that the coefficient of static friction between collar C and the
vertical rod is 0.40, determine the magnitude of the largest and smallest
couple
M for which equilibrium is maintained in the position shown, when
  35, l  600 mm, and300 N.P


SOLUTION
From the analysis of Problem 10.50, we have

max
2(tan )
s
Pl
M



With
35 , 0.6 m, 300 NlP
 

max
(300 N)(0.6 m)
2(tan35 0.4)
299.80 N m



M

max
300 N mM  
For
min
,Mmotion of C impends downward and F acts upward. The equations of Problem 10.50 can still
be used if we replace
s
by .
s
 Then

min
2(tan )
s
Pl
M



Substituting,

min
(300 N)(0.6 m)
2(tan35 0.4)
81.803 N m



M

min
81.8 N mM 

www.elsolucionario.org

1818. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.49
A block of weight W is pulled up a plane forming an angle  with the horizontal by a force P directed
along the plane. If
 is the coefficient of friction between the block and the plane, derive an expression for
the mechanical efficiency of the system. Show that the mechanical efficiency cannot exceed
1
2
if the
block is to remain in place when the force
P is removed.

SOLUTION

Input work
Output work ( sin )


Px
Wx


Efficiency:

sin sin
or
Wx W
Px P 


 (1)

0: sin 0 or sin
x
F PFW PW F      (2)

0: cos 0 or cos
y
FNW NW    

cosFNW 
Equation (2):
sin cos (sin cos )PW W W     
Equation (1):
sin 1
or
(sin cos ) 1 cot
W
W

   



If block is to remain in place when0,Pwe know (see Chapter 8) that
s
 or, since

tan , tan
s

Multiply by
cot : cot tan cot 1 
Add 1 to each side:
1cot 2
Recalling the expression for
, we find
1
2
 

www.elsolucionario.org

1819. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.50
Derive an expression for the mechanical efficiency of the jack discussed in Section 8.6. Show that if the
jack is to be self-locking, the mechanical efficiency cannot exceed
1
2
.

SOLUTION
Recall Figure 8.9a. Draw force triangle


tan( )
tan so that tan
Input work tan( )
Output work ( ) tan
s
s
QW
yx y x
Qx W x
Wy W x




 


Efficiency:
tan
;
tan( )
s
Wx
Wx




tan
tan( )
s




 
From Page 432, we know the jack is self-locking if


Then
2
s
 
so that
tan( ) tan 2
s
  
From above
tan
tan( )
s





It then follows that
tan
tan 2



But
2
2tan
tan 2
1tan




Then
22
tan (1 tan ) 1 tan
2tan 2 



1
2
 

www.elsolucionario.org

1820. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.51
Denoting by
s
 the coefficient of static friction between the block attached to
rod ACE and the horizontal surface, derive expressions in terms of P,
,
s
and 
for the largest and smallest magnitude of the force
Q for which equilibrium is
maintained.

SOLUTION
For the linkage:

0: 0 or
22
   
A
BA
x P
MxP
A
Then:
1
22
ss s
P
FA P
  
Now
2sin
2cos
A
A
xl
xl 


and
3cos
3sin
F
F
yl
yl 



Virtual Work:
max
0: ( ) 0
AF
UQFxPy

max
1
(2 cos ) ( 3 sin ) 0
2
s
QPl Pl 





or
max
31
tan
22
s
QP P 

max
(3tan )
2
s
P
Q
 
For
min
,Qmotion of A impends to the right and F acts to the left. We change
s
to
s
and find

min
(3tan )
2
s
P
Q
 

www.elsolucionario.org

1821. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.52
Knowing that the coefficient of static friction between the block attached to
rod ACE and the horizontal surface is 0.15, determine the magnitude of the
largest and smallest force
Q for which equilibrium is maintained when
30 , 0.2 m,l
 and P  40 N.

SOLUTION
Using the results of Problem 10.48 with

30
0.2 m
40 N, and 0.15
s
l
P






We have
max
(3tan )
2
(40 N)
(3tan 30 0.15)
2
37.64 N


s
P
Q

max
37.6 NQ 
and
min
(3tan )
2
(40 N)
(3tan 30 0.15)
2
31.64 N


s
P
Q

min
31.6 NQ 


www.elsolucionario.org

1822. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.53
Using the method of virtual work, determine separately the
force and couple representing the reaction at A.

SOLUTION
Force at A. We give a vertical virtual displacement
A
yto Point A, keeping member AB horizontal.

From the geometry of the diagram:

51
;
82
551 5
662 12
AB
CBEB
FE B B
yy
yyyy
yy y y

  


 


Virtual Work:

 
0;
(800 N) (600 N) 0
55
800 600 0
812
AC F
BBB
U
Ay y y
Ay y y


  
  
  
  
  

250 NA

250 NA


Couple at A. We rotate beam AB through
about Point A.


www.elsolucionario.org

1823. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.53 (Continued)

From the geometry of the diagram:

1.8
11
(1.8 ) 0.9
55
55
(1.8 ) 1.125
88
55
(0.9 ) 0.75
66
B
EB
CB
FE
y
yy
yy
yy
 
 
 

 
 
 

Virtual Work:
0; (800 N) (600 N) 0
2 800(1.125 ) 600(0.75 ) 0
ACF
A
UM y y
M
  
  
 


450 N m
A
M
  450 N m
A
M



www.elsolucionario.org

1824. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.54
Using the method of virtual work, determine the reaction at D.

SOLUTION
We release the support at D and assume a virtual displacement
D
yfor Point D.

From similar triangles:

3
8
3
2
553 5
662 4
CD
ED
FE D D
yy
yy
yy y y

  


 



Virtual Work:
0; y (800N) (600N) 0
DC F
UD y y     

35
800 600 0
84
DD D
Dy y y 

  




1050 ND 1050 ND 

www.elsolucionario.org

1825. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.55
Referring to Problem 10.43 and using the value
found for the force exerted by the hydraulic cylinder
CD, determine the change in the length of CD
required to raise the 10-kN load by 15 mm.
PROBLEM 10.43 The position of member ABC is
controlled by the hydraulic cylinder CD. For the
loading shown, determine the force exerted by the
hydraulic cylinder on pin C when
55 .

SOLUTION

Virtual Work: Assume both
A
yand
CD
increase
0: (10 kN) 0
ACDCD
UyF  
Substitute:
15 mm and 12.03 kN
ACD
yF (10 kN)(15 mm) (12.03 kN) 0
CD


12.47 mm
CD

The negative sign indicates that CD shortened
12.47 mm shorter
CD
 
www.elsolucionario.org

1826. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.56
Referring to Problem 10.45 and using the value found for the
force exerted by the hydraulic cylinder BD, determine the
change in the length of BD required to raise the platform
attached at C by 2.5 in.
PROBLEM 10.45 The telescoping arm ABC is used to
provide an elevated platform for construction workers. The
workers and the platform together weigh 500 lb and their
combined center of gravity is located directly above C. For
the position when
20 ,
 determine the force exerted on
pin B by the single hydraulic cylinder BD.


SOLUTION

Virtual Work: Assume both
C
yand
BD
increase
0: (500 lb) 0
CBDBD
UyF  
Substitute:
2.5 in. and 2370 lb
CBD
yF (500 lb)(2.5 in.) (2370 lb) 0
BD


0.527 in.
BD

The positive sign indicates that cylinder BD increases in length
0.527 in. longer
BD
 

www.elsolucionario.org

1827. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.57
Determine the vertical movement of joint D if the length of
member BF is increased by 1.5 in. (Hint: Apply a vertical
load at joint D, and, using the methods of Chapter 6,
compute the force exerted by member BF on joints B and
F. Then apply the method of virtual work for a virtual
displacement resulting in the specified increase in length of
member BF. This method should be used only for small
changes in the lengths of members.)

SOLUTION
Apply vertical load P at D.


0: (40 ft) (120 ft) 0
H
MPE  

3
P

E
3
0: 0
53
yBF
P
FF 


5
9
BF
FP

Virtual Work:
We remove member BF and replace it with forces
BF
Fand
BF
Fat pins F and B, respectively. Denoting
the virtual displacements of Points B and F as
B
r and ,
F
rrespectively, and noting that P and D

have
the same direction, we have
Virtual Work:
0: ( ) 0
BF F BF B
UPD Fr F r

cos cos 0
BF F F BF B B
PDFr Fr

(cos cos)0
BF B B F F
PD F r r
where
(cos cos) ,
BBFFBF
rr which is the change in
length of member BF. Thus,

0
5
(1.5 in.) 0
9
0.833 in.







BF BF
PD F
PD P
D
0.833 in.D


www.elsolucionario.org

1828. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.58
Determine the horizontal movement of joint D if the length
of member BF is increased by 1.5 in. (See the hint for
Problem 10.57.)


SOLUTION
Apply horizontal load P at D.


0: (30 ft) (120 ft) 0
Hy
MPE  

4

y
P
E
3
0: 0
54
yBF
P
FF 


5
12
BF
FP

We remove member BF and replace it with forces
BF
Fand
BF
Fat pins F and B, respectively. Denoting
the virtual displacements of Points B and F as
B
r and ,
F
rrespectively, and noting that P and D

have
the same direction, we have
Virtual Work:
0: ( ) 0
BF F BF B
UPD Fr F r

cos cos 0
BF F F BF B B
PDFr Fr

(cos cos)0
BF B B F F
PD F r r
where
(cos cos) ,
BBFFBF
rr which is the change in
length of member BF. Thus,

0
5
(1.5 in.) 0
9
0.625 in.







BF BF
PD F
PD P
D
0.625 in.D 
www.elsolucionario.org

1829. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.59
Using the method of Section 10.8, solve Problem 10.29.
PROBLEM 10.29 Two rods AC and CE are connected by a pin
at C and by a spring AE. The constant of the spring is k, and the
spring is unstretched when
30 .
 For the loading shown,
derive an equation in P,
, l, and k that must be satisfied when
the system is in equilibrium.


SOLUTION
Spring:
2(2 sin ) 4 sinAE x l l 

Unstretched length:
0
4 sin 30 2xl l

Deflection of spring


0
2
2
2
2(2sin 1)
1
2
1
(cos)2(2sin 1)
2
4(2sin 1)2cos sin 0
E
sxx
sl
VksPy
Vk Pll
dV
kl Pl
d







 



cos
(2sin 1) 0
sin 8
P
kl




12sin
8tan
P
kl




www.elsolucionario.org

1830. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.60
Using the method of Section 10.8, solve Problem 10.30.
PROBLEM 10.30 Two rods AC and CE are connected by a pin
at C and by a spring AE. The constant of the spring is 1.5 lb/in.,
and the spring is unstretched when
30 .
 Knowing that
10 in.l and neglecting the weight of the rods, determine the
value of
 corresponding to equilibrium when 40 lb.P

SOLUTION
Using the result of Problem 10.59, with

40 lb
10 in. and 1.5 lb/in.
12sin
8tan






P
lk
P
kl
or
12sin 40 lb
tan 8(1.5 lb/in.)(10 in.)
1
3






Solving numerically,
25.0
 

www.elsolucionario.org

1831. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.61
Using the method of Section 10.8, solve Problem 10.31.
PROBLEM 10.31 Solve Problem 10.30 assuming that force P is
moved to C and acts vertically downward.


SOLUTION
Spring: 2(2 sin ) 4 sinAE x l l 
Unstretched length:
0
4 sin 30 2xl l
Deflection of spring
0
2
2
2(2sin 1)
1
2
1
[2 (2sin 1)] ( cos )
2





C
sxx
sl
VksPy
kl Pl

22
2
2(2sin 1) cos
4(2sin 1)2cos sin 0
cos
(1 2 sin ) 0
sin 8
2sin 1
8tan











Vkl Pl
dV
kl Pl
d
P
kl
P
kl

with
40 lb, 10 in. and 1.5 lb/in. Pl k
We have
(40 lb) 2sin 1
8(1.5 lb/in.)(10 in.) tan



or
2sin 1 1
tan 3




Solving numerically
39.65 and 68.96° 39.7

69.0 


www.elsolucionario.org

1832. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.62
Using the method of Sec. 10.8, solve Prob. 10.30.
Problem 10.30: Two bars AD and DG are connected by a pin at
D and by a spring AG. Knowing that the spring is 300 mm long
when unstretched and that the constant of the spring is 5 kN/m,
determine the value of x corresponding to equilibrium when a
900-N load is applied at E as shown.

SOLUTION
Spring: 0.3 msx

36 2
E
xx x
y  






Potential Energy:

21
2
E
VksWy


21
0.3
22
x
kx W





For equilibrium:


1
0.3 0
2
dV
kx W
dx
 




1
5000 N/m 0.3 m 900 N 0
2
x 

Solving
0.390 mx 390 mmx 



www.elsolucionario.org

1833. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.63
Using the method of Section 10.8, solve Problem 10.33.
PROBLEM 10.33 Two 5-kg bars AB and BC are connected by a pin
at B and by a spring DE. Knowing that the spring is 150 mm long
when unstretched and that the constant of the spring is 1 kN/m,
determine the value of x corresponding to equilibrium.

SOLUTION
First note:
2
bar
(5 kg)(9.81 m/s ) 49.05 NW

Since unstretched length of spring is 150 mm, or 0.15 m, we have

2
2
2
0.15
3
13
(49.05 N) (49.05 N)
244
12
(1000 N/m) 0.15 12.2625 36.7875
23
22
1000 0.15 12.2625 36.7875 0
33
S
S
x
xx
k
Vxxx
dV
x
dx
 
 








V

0.335 mx 335 mmx 

www.elsolucionario.org

1834. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.64
Using the method of Section 10.8, solve Problem 10.35.
PROBLEM 10.35 A vertical force P of magnitude 150 N is applied to
end E of cable CDE, which passes over a small pulley D and is
attached to the mechanism at C. The constant of the spring is k  4
kN/m, and the spring is unstretched when
  0. Neglecting the weight
of the mechanism and the radius of the pulley, determine the value of

corresponding to equilibrium.

SOLUTION

1
(90 ) 45
22
2sin 2sin 45
2
BC BD l
CD l l 






 



For
0
0: ( ) 2 sin45 2CD l l

0
()
22sin45
2
E
yCDCD
ll

 

 



Potential energy:

2
2
21
2
1
() 2 2sin45
22
1
2 cos 45 0
22
E
VksPy
Vkr Pll
dV
kr Pl
d






 

 
 

 





2
2
cos 45
Pl
kr






22
2
(150 N)(0.2 m)
0.75
(4000 N/m)(0.1 m)
0.75
cos 45
Pl
kr






Solve by trial and error:
0.67623 rad


38.745
 38.7 
www.elsolucionario.org

1835. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.65
Using the method of Section 10.8, solve Problem 10.37.
PROBLEM 10.37 and 10.38 Knowing that the constant of spring CD is
k and that the spring is unstretched when rod ABC is horizontal, determine
the value of
 corresponding to equilibrium for the data indicated.
PROBLEM 10.37 300 N, 400 mm, 5 kN/m.Pl k 

SOLUTION
Spring
90
2sin
2
2sin 45
2








vl
vl

Unstretched
(0)
0
2sin45 2vl l
Deflection of spring
0
2
22
2
2sin 45 2
2
11
2sin 45 2 ( sin )
22 2
2sin 45 2 cos 45 cos 0
22
A
svv l l
VksPy kl Pl
dV
kl Pl
d






   



 


 

 
 


2sin45 cos45 2cos45 cos
22 2
cos 2 cos 45 cos
2
P
kl
P
kl
 


 
    
 
 





Divide each member by cos


2
cos 45
12
cos
P
kl




Then with
300 N, 0.4 mPl and 5000 N/mk


2
cos 45 300 N
12
cos (5000 N/m)(0.4 m)
0.15






or

2
cos 45
0.60104
cos





Solving numerically
22.6 
www.elsolucionario.org

1836. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.66
Using the method of Section 10.8, solve Problem 10.38.
PROBLEM 10.37 and 10.38 Knowing that the constant of spring CD is k
and that the spring is unstretched when rod ABC is horizontal, determine
the value of
 corresponding to equilibrium for the data indicated.
PROBLEM 10.38 75 lb, 15 in., 20 lb/in.Pl k
 

SOLUTION
Using the results of Problem 10.65 with 75 lb, 15 in. and 20 lb/in., we have Pl k

 
2
cos 45
12
cos
75 lb
(20 lb/in.)(15 in.)
0.25
P
kl







or
 
2
cos 45
0.53033
cos





Solving numerically
51.058
  51.1 

www.elsolucionario.org

1837. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.67
Show that equilibrium is neutral in Problem 10.1.
PROBLEM 10.1 Determine the vertical force P that must be
applied at C to maintain the equilibrium of the linkage.

SOLUTION







Designate vertical distance between link and as .AC DG b
We have
11
,2,, ,
22
AC DEG
ybuybuy uy uy u    



(80 N) ( )+ 60 N 20 N 40 N
11
80(b ) 2 60 20 40
22
80 2 60 10 20 0
65 N
AC D E G
VyPy y y y
VuPbuu u u
dV
P
du
P
  




     


Substituting
270 NP in the expression for V, we have
 80 65 80 130 60 10 20
145
Vb u
Vb
 


Thus V is constant
and equilibrium is neutral


www.elsolucionario.org

1838. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.68
Show that equilibrium is neutral in Problem 10.7.
PROBLEM 10.7 The two-bar linkage shown is supported by a pin
and bracket at B and a collar at D that slides freely on a vertical rod.
Determine the force
P required to maintain the equilibrium of the
linkage.


SOLUTION

2
(100 lb) (150 lb)
(2 ) (100 lb)( ) (150 lb)( )
2 100 150 0
125 lb
A
E
F
AEF
yu
yu
yu
VPy y y
VPu u u
dV
P
du
P



 
  
  


Now, substitute
125 lbP in expression for V, making 0.V Thus, V is constant and
equilibrium is neutral.


www.elsolucionario.org

1839. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.69
Two uniform rods, each of mass m, are attached to gears of
equal radii as shown. Determine the positions of equilibrium of
the system and state in each case whether the equilibrium is
stable, unstable, or neutral.


SOLUTION
Potential energy

2
2
sin cos
22
(cos sin )
2
(sin cos)
2
(sin cos )
2






  



 

ll
VW W Wmg
l
W
dV Wl
d
dV Wl
d

For equilibrium:
0: sin cos

dV
d
or
tan 1
Thus
45.0 and 135.0   
Stability:
At
45.0 : 
2
2
[sin( 45) cos45]
2
22
0
222
dV Wl
d
Wl



 



45.0 , Unstable  
At
135.0 :
2
2
(sin135 cos135 )
2
22
0
22 2
dV Wl
d
Wl







135.0 , Stable 
www.elsolucionario.org

1840. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.70
Two uniform rods, AB and CD, are attached to gears of
equal radii as shown. Knowing that
8 lb
AB
W and
4 lb,
CD
W determine the positions of equilibrium of the
system and state in each case whether the equilibrium is
stable, unstable, or neutral.

SOLUTION
Potential energy

22
(3.5 kg 9.81 m/s ) sin (1.75 kg 9.81 m/s ) cos
22
(8.5838 N) ( 2sin cos )

 
   
 
 

ll
V
l


2
2
(8.5838 N) ( 2cos sin )
(8.5838 N) (2sin cos )




dV
l
d
dV
l
d

Equilibrium:
0: 2cos sin 0
  
dV
d
or
tan 2
Thus
63.4 and 116.6  
Stability:
At
63.4 : 
2
2
(8.5838 N) [2sin( 63.4 ) cos( 63.4 )]
(8.5838 N) ( 1.788 0.448) 0
dV
l
d
l




63.4 , Unstable  
At
116.6 :
2
2
(8.5838 N) [2sin(116.6 ) cos(116.6 )]
(8.5838 N) (1.788 0.447) 0
dV
l
d
l




116.6 , Stable 

www.elsolucionario.org

1841. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.71
Two uniform rods AB and CD, of the same length l, are attached
to gears as shown. Knowing that rod AB weighs 3 lb and that rod
CD weighs 2 lb, determine the positions of equilibrium of the
system and state in each case whether the equilibrium is stable,
unstable, or neutral.


SOLUTION

sin cos 2
22
1
cos sin 2
2


 
 
 
 
 
AB CD
AB CD
ll
VW W
dV
Wl Wl
d

Equilibrium:
0
1
cos sin 2 0
2
cos 4 sin cos 0
cos 1 sin 0
AB CD
AB CD
CD
AB
AB
dV
d
Wl Wl
WW
W
W
W






 

 


1
cos 0 and sin
4
90 , 270 , sin
4
 


   
AB
CD
AB
CD
W
W
W
W
(1)
For
3 lb and 2 lb, We have
AB CD
WW

13
90 , 270 , sin
8
90 , 270°, 22.0°, 158.0°
 
  

   
   

Stability:
2
2
1
sin 2 cos 2 )
2






AB CD
dV
WW l
d
(2)

2
2
1
270 : (3)sin(270 ) 2(2)cos(540 ) 5.5 0, unstable
2
dV
ll
d



   




2
2
1
22.0 : (3)sin 220 2(2)cos44.0 0 Stable
2
dV
l
d



   




www.elsolucionario.org

1842. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.71 (Continued)


2
2
1
90 : (3)sin 90 2(2)cos180 2.5 0, Unstable
2
dV
ll
d



    




2
2
1
158.0 : (3)sin158 2(2)cos316 3.44 0, Stable
2
dV
ll
d



    



www.elsolucionario.org

1843. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.72
Two uniform rods, each of mass m and length l, are attached to drums that are
connected by a belt as shown. Assuming that no slipping occurs between the
belt and the drums, determine the positions of equilibrium of the system and
state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

2
2
cos 2 cos
22
( 2sin 2 sin )
2
1
(4cos2 cos)
2









 
 
Wmg
ll
VW W
dV l
W
d
dV
W
d
Equilibrium:
0: ( 2sin 2 sin ) 0
2 

dV Wl
d
or
sin ( 4cos 1) 0
Solving
0, 75.5°, 180°, and 284°
Stability:
2
2
( 4cos 2 cos )
2 
 
dV l
W
d
At
0:
2
2
(4 1) 0
2
dV l
W
d

 0, Unstable 
At
75.5 :
2
2
( 4( .874) .25) 0
2
dV l
W
d

 75.5 , Stable 
At
180 :
2
2
(4 1) 0
2
dV l
W
d

 180.0 , Unstable 
At
284 :
2
2
( 4( .874) .25) 0
2
dV l
W
d

 284 , Stable 
www.elsolucionario.org

1844. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.73
Using the method of Section 10.8, solve Problem 10.39.
Determine whether the equilibrium is stable, unstable, or neutral.
(Hint: The potential energy corresponding to the couple exerted
by a torsion spring is
21
2
,
K where K is the torsional spring
constant and
 is the angle of twist.)
PROBLEM 10.39
The lever AB is attached to the horizontal shaft
BC that passes through a bearing and is welded to a fixed support
at C. The torsional spring constant of the shaft BC is K; that is, a
couple of magnitude K is required to rotate end B through 1 rad.
Knowing that the shaft is untwisted when AB is horizontal,
determine the value of
 corresponding to the position of
equilibrium when P  100 N, l  250 mm, and
12.5 N m/rad.K

SOLUTION
Potential energy
2
2
21
sin
2
cos
sin







VKPl
dV
KPl
d
dV
KPl
d

Equilibrium:
0: cos

dV K
dPl
For
100 N, 0.25 m., 12.5 N m/rad Pl K

12.5 N m/rad
cos
(100)(0.25 m)
0.500






Solving numerically, we obtain

1.02967 rad 59.000 59.0 
Stability

2
2
(12.5 N m/rad) (100 N)(0.25 m)sin 59.0 0
dV
d

  Stable 

www.elsolucionario.org

1845. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.74
In Problem 10.40, determine whether each of the positions of
equilibrium is stable, unstable, or neutral. (See hint for Problem 10.73.)
PROBLEM 10.40 Solve Problem 10.39 assuming that 350 N,P
250 mm,l and 12.5 N m/rad.K Obtain answers in each of the
following quadrants:
0 90°, 270° 360°, 360° 450°.   

SOLUTION
Potential energy
2
2
21
sin
2
cos
sin






VKPl
dV
KPl
d
dV
KPl
d
Equilibrium

0: cos

dV K
dPl
For
350 N, 0.250 m and 12.5 N m/rad
12.5 N m/rad
cos
(350 N)(0.250 m)
Pl K

 


or
cos
7



Solving numerically
1.37333 rad, 5.652 rad, and 6.616 rad
or 78.7 , 323.8 , 379.1 
Stability at 78.7 :
2
2
(12.5 N m/rad) (350 N)(0.250 m)sin 78.7

 
dV
d

98.304 0 78.7 , Stable 
At 323.8 :
2
2
(12.5 N m/rad) (350 N)(0.250 m)sin323.8

 
dV
d

39.178 N m 0   324 , Unstable 
At At 379.1 :
2
2
(12.5 N m/rad) (350 N)(0.250 m)sin379.1

 
dV
d

44.132 N m 0 379 , Stable 

www.elsolucionario.org

1846. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.75
A load W of magnitude 100 lb is applied to the mechanism at C.
Knowing that the spring is unstretched when
15 , determine that
value of
 corresponding to equilibrium and check that the
equilibrium is stable.

SOLUTION
We have
2
00
22
0
2
0
cos
1
[( )] 15 rad
212
1
() cos
2
()sin

 
 
 

 


C
C
yl
Vkr Wy
kr Wl
dV
kr Wl
d
Equilibrium
2
0
0: ( ) sin 0 

dV
kr wl
d (1)
with
100 lb, 50 lb./in., 20 in., and 5 in.WR l r  

2
(50 lb./in.)(25 in. ) (100 lb)(20 in.)sin 0
12


 



or
0.625 sin 0.16362
Solving numerically
1.8145 rad 103.97°

104.0 
Stability
2
2
2
cos

dV
kr Wl
d (2)
or
1250 2000cos
For
104.0 : 1734 in. lb 0 Stable 

www.elsolucionario.org

1847. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.76
A load W of magnitude 100 lb is applied to the mechanism at C.
Knowing that the spring is unstretched when
30 ,
 determine that
value of
 corresponding to equilibrium and check that the
equilibrium is stable.

SOLUTION
Using the solution of Problem 10.75, particularly Equation (1), with 15° replaced by 30 rad :
6





For equilibrium
2
sin 0
6


 


kr Wl

With
50 lb/in., 100 lb, 5 in., and 20 in.kWr l 

2
(50 lb/in.)(25 in. ) (100 lb)(20 in.)sin 0
6


 



or
1250 654.5 200sin 0
  
Solving numerically,
1.9870 rad 113.8°
 

113.8
 
Stability: Equation (2), Problem 75:

2
2
2
cos



dV
kr Wl
d
or
1250 2000cos
 
For
113.8 :
 2057 in. lb 0  Stable 

www.elsolucionario.org

1848. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.77
A slender rod AB, of weight W, is attached to two blocks
A and B that can move freely in the guides shown.
Knowing that the spring is unstretched when
0,
y
determine the value of
y corresponding to equilibrium
when
80 N,W 500 mm,l and 600 N/m.k

SOLUTION
Deflection of spring  s, where
22
22



slyl
ds y
dy
ly
Potential energy:
 
2
22
22
221
22
1
2
1
2
1
1
2


 


 



y
VksW
dV ds
ks W
dy dy
dV y
kl y l W
dy
ly
l
kyW
ly

Equilibrium
22
1
0: 1
2

 



dV l W
y
dy k
ly

Now
80 N, 0.500 m, and 600 N/mWl k 
Then
22
0.500 m 1 (80 N)
1
2 (600 N/m)
(0.500)





y
y

or
2
0.500
1 0.066667
0.25





y
y

Solving numerically,
0.357 my 357 mmy 

www.elsolucionario.org

1849. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.78
A slender rod AB, of weight W, is attached to two blocks
A and B that can move freely in the guides shown.

Knowing that both springs are unstretched when 0,
y
determine the value of y corresponding to equilibrium
when
80 N,W l  550 mm, and 600 N/m.k

SOLUTION
Spring deflections

22
22
 

AD
BC
Slly
Slyl


   
   
22
22
22 22
22 22
22 2211
22 2
11
22 2
2
AD BC
y
VkS kSW
y
Vklly klylW
dV y y W
kl l y k l y l
dy
ly ly

    
 
    
 

 


22 22
0: 1 1
2





dV l l W
y
dy k
ly ly

Data:
80 N, 0.5 m, 600 N/m Wlk

22 22
0.5 0.5 80
0.066667
2(1200)
(0.5) (0.5)
y
yy

 
 


Solve by trial and error:
0.252 my 252 mmy 

www.elsolucionario.org

1850. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.79
A slender rod AB, of weight W, is attached to two blocks A and B that
can move freely in the guides shown. The constant of the spring is
k,
and the spring is unstretched when
AB is horizontal. Neglecting the
weight of the blocks, derive an equation in
, W, l, and k that must be
satisfied when the rod is in equilibrium.


SOLUTION

Elongation of spring:
sin cos
(sin cos 1)


sl l l
sl
Potential energy:
2
2211
sin
22
1
(sin cos 1) sin
22

  

VksW Wmg
l
kl mg

2 1
(sin cos 1)(cos sin ) cos
2
  

dV
kl mgl
d (1)
Equilibrium:
0: (sin cos 1)(cos sin ) cos 0
2   

dV mg
dkl
or
cos (sin cos 1)(1 tan ) 0
2
mg
kl  

 




www.elsolucionario.org

1851. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.80
A slender rod AB, of weight W, is attached to two blocks A and B that
can move freely in the guides shown. Knowing that the spring is
unstretched when AB is horizontal, determine three values of

corresponding to equilibrium when
300 lb,
W 16 in.,l and
75 lb/in.k State in each case whether the equilibrium is stable,
unstable, or neutral.


SOLUTION
Using the results of Problem 10.79, particularly the condition of equilibrium
cos (sin cos 1)(1 tan ) 0
2  
   


mg
kl
Now, with
300 lb, 16 in., and 75 lb/in.Wl k 

300 lb
0.25
2 (16 in.)(75 lb/in.)

W
kl

Thus:

cos (sin cos 1)(1 tan ) 0.25 0   

cos 0 and (sin cos 1)(1 tan ) 0.25
  
First equation yields
90 .
 Solving the second equation by trial, we find 9.39 and 34.16°
Values of  for equilibrium are

9.39 , 34.2°, and 90.0°
 
Stability: we differentiate Eq. (1).

2
2
2
22 2 2 2
2
2
2
2
1
[(cos sin )(cos sin ) (sin cos 1)( sin cos )] sin
2
cos sin 2cos sin sin cos 2cos sin sin cos sin
2
1sincos2sin2
2
(1.25 sin c
   
 
 


 
 
  
 
 

 



dy
kl wl
ds
W
kl
kl
W
kl
kl
dV
kl
d
os 2sin 2 )


9.39 :

2
2
2
(1.25sin 9.4 cos9.4 2sin18.8)


dV
kl
d

2
(0.55) 0kl  Stable 
www.elsolucionario.org

1852. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.80 (Continued)

34.2 :

2
2
2
(1.25sin 34.2 cos34.3 2sin 68.4 )

   
dv
kl
d

2
(0.33) 0kl
 Unstable 
90.0 :
2
2
2
(1.25sin90 cos90 2sin180 )
dV
kl
d

   

2
(1.25) 0kl
  Stable 

www.elsolucionario.org

1853. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.81
A spring AB of constant k is attached to two identical gears as
shown. Knowing that the spring is undeformed when
0,

determine two values of the angle
 corresponding to
equilibrium when
P  30 lb, a  4 in., b  3 in., r  6 in., and
k  5 lb/in. State in each case whether the equilibrium is
stable, unstable, or neutral.

SOLUTION
Elongation of spring

2
2
2
2
2( sin ) 2 sin
1
2
1
(2 sin )
2
4sincos
2sin2
sa a
VksPb
ka Pb
dV
ka Pb
d
ka Pb








 (1)
Equilibrium
2
0: sin2
2 

dV Pb
d ka

2
(30 lb)(3 in.)
sin 2 ; sin 2 0.5625
2(5 lb/in.)(4 in.)


2 34.229 and 145.771 

17.11 and 72.9  
Stability: We differentiate Eq. (1)

2
2
2
4cos2
dV
ka
d 


2
22
2
17.11 : 4 cos34.2 4 (0.83) 0
dv
ka ka
d
    Stable 

2
22
2
72.9 : 4 cos145.8 4 ( 0.83) 0
dv
ka ka
d
     Unstable 

www.elsolucionario.org

1854. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.82
A spring AB of constant k is attached to two identical gears as
shown. Knowing that the spring is undeformed when
0,

and given that
60 mm, 45 mm, 90 mm,abr and
6 kN/m,k determine (a) the range of values of P for which
a position of equilibrium exists, (b) two values of

corresponding to equilibrium if the value of P is equal to half
the upper limit of the range found in part a.

SOLUTION
Elongation of spring

2( sin ) 2 sinsa a
Potential energy

22
2211
(2 sin )
22
2sin

 

VksPb ka Pb
Vka Pb

Equilibrium

2
2
0: 4 sin cos
2sin2 0
dV dV
ka Pb
dd
ka Pb 

 

(1)

max
max22
sin 2 ; For ; 1
22
PbPb
P
ka ka
(a)
max
2
(0.045 m)
1
2(6000 N/m)(0.06 m)

P

max
960 NP 
(b) For
max
1
,
2
PP

1
sin 2 ; 2 30 and 150
2
 

15.00 and 75.0  

www.elsolucionario.org

1855. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.83
A slender rod AB is attached to two collars A and B that can move
freely along the guide rods shown. Knowing that
30 and
P  Q  400 N, determine the value of the angle
 corresponding
to equilibrium.

SOLUTION

Law of Sines
sin(90 ) sin(90 )
cos( ) cos 
 

  


A
A
y L
y L

or
cos( )
cos


A
yL
From the figure:
cos( )
cos
cos



B
yL L
Potential Energy:
cos( ) cos( )
cos
cos cos
BA
VPyQy PL L QL
 

 
    




sin( ) sin( )
sin
cos cos
sin( )
() sin
cos 

 


 
   



 
dV
PL QL
d
LP Q PL


www.elsolucionario.org

1856. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.83 (Continued)

Equilibrium
sin( )
0: ( ) sin 0
cos




 
dV
LP Q PL
d
or
()sin()sincos
PQ P

( )(sin cos cos sin ) sin cos
 PQ P
or
()cossin sincos0
  PQ Q

sin sin
0
cos cos



 
PQ
Q

tan tan 


PQ
Q (2)
With
400 N, 30
PQ

800 N
tan tan 30 1.1547
400 N
 49.1
 

www.elsolucionario.org

1857. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.84
A slender rod AB is attached to two collars A and B that can move
freely along the guide rods shown. Knowing that
30 , 100 N,P
 and 25 N,Q determine the value of the
angle
 corresponding to equilibrium.

SOLUTION
Using Equation (2) of Problem 10.83, with 100 N, 25 N, and 30 , PQ we have

(100 N)(25 N)
tan tan30
(25 N)
57.735
89.007


 


89.0
 

www.elsolucionario.org

1858. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.85
Cart B, which weighs 75 kN, rolls along a sloping track that forms an
angle
 with the horizontal. The spring constant is 5 kN/m, and the
spring is unstretched when
0.
x Determine the distance x
corresponding to equilibrium for the angle
 indicated.
Angle = 30 . 

SOLUTION


(4 m)tanx (1)

sin 4tan sin
(4 m)cos


B
yx
AC

For
0,x
0
()4mAC
Stretch of spring.
0
2
2
41
() 44 1
cos cos
1
(75 kN)
2
11
(5 kN/m)16 1 (75 kN)4 tan sin
2cos
B
sAC AC
Vks y




   









22
1sin sin
80 1 300
cos cos cos
 



dV
d

Equilibrium
1
0: 1 sin 3.75sin
cos






dV
d
(2)
Given: 30 , sin 0.5 
Eq. (2):
1
1 sin 3.75(0.5) 1.875
cos



 



Solve by trial and error:
70.46
Eq. (1):
(4 m) tan 70.46x 11.27 mx 
www.elsolucionario.org

1859. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.86
Cart B, which weighs 75 kN, rolls along a sloping track that forms an
angle
 with the horizontal. The spring constant is 5 kN/m, and the
spring is unstretched when
0.
x Determine the distance x
corresponding to equilibrium for the angle
 indicated.
Angle = 60°.

SOLUTION


(4 m)tanx (1)

sin 4tan sin
(4 m)cos


B
yx
AC

For
0,x
0
()4mAC
Stretch of spring:
0
2
2
41
() 44 1
cos cos
1
(75 kN)
2
11
(5 kN/m)16 1 (75 kN)4 tan sin
2cos
B
sAC AC
Vks y




   









22
1sin sin
80 1 300
cos cos cos
 



dV
d

Equilibrium
1
0: 1 sin 3.75sin
cos






dV
d
(2)
Given: 60 , sin 0.86603 
Eq. (2):
1
1 sin 3.75(0.86603) 3.2476
cos



 



Solve by trial and error:
76.67
Eq. (1):
(4 m)tan 26.67x 16.88 mx 
www.elsolucionario.org

1860. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.87
Collar A can slide freely on the semicircular rod shown. Knowing
that the constant of the spring is k and that the unstretched length of
the spring is equal to the radius r, determine the value of

corresponding to equilibrium when
50 lb, 9 in.,Wr and
15 lb/in.k

SOLUTION
Stretch of spring
2( cos )
(2cos 1)




sABr
sr r
sr
Potential energy:
2
22
21
sin 2
2
1
(2cos 1) sin 2
2
(2cos 1)2sin 2 cos2


 
 

  
VksWr Wmg
Vkr Wr
dV
kr Wr
d
Equilibrium
2
0: (2cos 1)sin cos 2 0 
   
dV
kr Wr
d

(2cos 1)sin
cos 2


W
kr

Now
(50 lb)
0.37037
(15 lb/in.)(9 in.)

W
kr

Then
(2cos 1)sin
0.37037
cos 2



Solving numerically,
0.95637 rad 54.8 54.8 

www.elsolucionario.org

1861. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.88
Collar A can slide freely on the semicircular rod shown. Knowing that the constant
of the spring is k and that the unstretched length of the spring is equal to the radius
r, determine the value of
 corresponding to equilibrium when 50 lb, 9 in.,Wr
and k  15 lb/in.

SOLUTION
Stretch of spring

2
22
2
2( cos )
(2cos 1)
1
cos 2
2
1
(2cos 1) cos2
2
(2cos 1)2sin 2 sin 2



 
 



  
sABr r r
sr
VksWr
kr Wr
dV
kr Wr
d

Equilibrium

2
0: (2cos 1)sin sin2 0 
   
dV
kr Wr
d

2
(2cos 1)sin (2sin cos ) 0  kr Wr
or
(2cos 1)sin
2cos


W
kr

Now
(50 lb)
0.37037
(15 lb/in.)(9 in.)

W
kr

Then
2cos 1
0.37037
2cos



Solving
37.4 

www.elsolucionario.org

1862. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.89
Two bars AB and BC of negligible weight are attached to a
single spring of constant k that is unstretched when the bars
are horizontal. Determine the range of values of the
magnitude P of two equal and opposite forces
P and  P for
which the equilibrium of the system is stable in the position
shown.

SOLUTION


cos
sin
xl
yl




2
221
2
2
1
2cos sin
2
VPxky
vPl kl





2
2
2
2
2
2sin sincos
1
2sin sin2
2
2cos cos2
dV
Pl kl
d
Pl kl
dV
Pl kl
d



 
 
 
(1)
For equilibrium position
0 to be stable

2
2
2
20
dV
Pl kl
d

   (2)

1
2
Pkl









www.elsolucionario.org

1863. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.89 (Continued)

Note: For
1
2
,
Pkl we have
2
2
0
dV
d

 and we must determine which is the first derivative to be  0.
Differentiating Eq. (1):

3
2
3
4
22
4
2 sin 2 sin 2 0 for 0
2 cos 4 cos2 2 4 for 0
dV
Pl kl
d
dV
Pl kl Pl kl
d

 
   
 

But
1
2
.
Pkl Thus
4
4
22
40
dV
d
kl kl

  and we conclude that the equilibrium is unstable for
1
2
.
Pkl
The sign < in Eq. (2) is thus correct.


www.elsolucionario.org

1864. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.90
A vertical bar AD is attached to two springs of constant k and is in
equilibrium in the position shown. Determine the range of values of the
magnitude P of two equal and opposite vertical forces
P and  P for which
the equilibrium position is stable if (a)
,AB CD (b) 2.AB CD

SOLUTION
For both (a) and (b): Since P and P are vertical, they form a couple of moment

sin
P
MPl
The forces
F and F exerted by springs must, therefore, also form a couple, with moment

cos
F
MFa
We have
(sin cos)
PF
dU M d M d
Pl Fa d


but
1
sin
2





Fksk a

Thus,
21
sin sin cos
2





dU Pl ka d

From Equation (10.19), page 580, we have

21
sin sin 2
4
   dV dU Pl d ka d
or
21
sin sin 2
4

 
dV
Pl ka
d
and
2
2
2
1
cos cos 2
2

 
dV
Pl ka
d (1)
For
0:
2
2
2
1
2

 
dV
Pl ka
d


www.elsolucionario.org

1865. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.90 (Continued)

For Stability:

2
2
2
1
0, 0
2
dV
Pl ka
d

 
or (for Parts a and b)
2
2
ka
P
l


Note: To check that equilibrium is unstable for
2
2
,
ka
l
P we differentiate (1) twice:

3
2
3
4
2
4
sin sin 2 0, for 0,
cos 2 cos 2
dV
Pl ka
d
dV
Pl ka
d 



  


For
0

42
22
4
220
2
dV ka
Pl ka ka
d


Thus, equilibrium is unstable when
2
2

ka
P
l


www.elsolucionario.org

1866. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.91
Rod AB is attached to a hinge at A and to two springs, each of
constant k. If
25 in., 12 in.,hd and 80 lb,W determine the
range of values of k for which the equilibrium of the rod is stable
in the position shown. Each spring can act in either tension or
compression.

SOLUTION
We have sin cos
CB
xd yh
Potential Energy:
2
221
2
2
sin cos






CB
VkxWy
kd Wh
Then
2
2
2sincos sin
sin 2 sin 



dV
kd Wh
d
kd Wh

and
2
2
2
2cos2 cos

dV
kd Wh
d (1)
For equilibrium position
0 to be stable, we must have

2
2
2
20
dV
kd Wh
d


or
21
2
kd Wh
(2)
Note: For
21
2
,kd Wh we have
2
2
0,


dV
d
so that we must determine which is the first derivative that is not
equal to zero. Differentiating Equation (1), we write

3
2
3
4
2
2
4 sin 2 sin 0 for 0
8 cos2 cos
dV
kd Wh
d
dV
kd Wh
d  


   
 

For
0:
4
2
4
8

 
dV
kd Wh
d






www.elsolucionario.org

1867. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.91 (Continued)

Since
4
4
21
2
,4 0,
dV
d
kd Wh Wh Wh

 we conclude that the equilibrium is unstable for
21
2
kd Wh and
the
sign in Equation (2) is correct.
With
80 lb, 25 in., and 12 in.Wh d
 
Equation (2) gives
21
(12 in.) (80 lb)(25 in.)
2
k 

or
6.944 lb/in.k 6.94 lb/in.k 

www.elsolucionario.org

1868. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.92
Rod AB is attached to a hinge at A and to two springs, each of
constant
k. If 45 in., 6 lb/in.,
 hk and 60 lb,W determine the
smallest distance
d for which the equilibrium of the rod is stable in
the position shown. Each spring can act in either tension or
compression.

SOLUTION
Using Equation (2) of Problem 10.91 with

45 in., 6 lb/in., and 60 lbhk W
 

21
(6 lb/in.) (60 lb)(45 in.)
2
d

or
22
225 in.d

15.0000 in.d smallest 15.00 in.d 

www.elsolucionario.org

1869. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.93
Two bars are attached to a single spring of constant k
that is unstretched when the bars are vertical.
Determine the range of values of P for which the
equilibrium of the system is stable in the position
shown.

SOLUTION

2
sin sin
33
LL
s

For small values of
 and 

2
2
2
2
2
2
2
2
21
cos cos
33 2
12
(cos2 2cos ) sin
323
2
( 2sin 2 2sin ) sin cos
39
2
(2sin 2 2sin ) sin 2
39
4
(4cos2 2cos ) cos2
39
LL
VP ks
PL L
Vk
dV PL
kL
d
PL
kL
dV PL
kL
d

 
 

 
 









  
  
  

when
0:
2
2
2
64
39
dV PL
kL
d

 
For stability:
2
2
2
4
0, 2 0
9
dV
PL kL
d

 
2
9
PkL


www.elsolucionario.org

1870. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.94
Two bars are attached to a single spring of constant k
that is unstretched when the bars are vertical.
Determine the range of values of P for which the
equilibrium of the system is stable in the position
shown.

SOLUTION

2
sin sin
33

LL
a
For small values of
 and 

2

sin
3
L
s

2
2
2
2
22
221
cos cos
33 2
1
(2cos cos2 ) sin
323
(2sin 2sin2) sin cos
39
2
(sin sin 2 ) sin 2
318
2
(cos 2cos2 ) cos2
39
LL
VP ks
PL L
k
dV PL kL
d
dV PL kL
d
dV PL kL
d

 
 

 

 









  
  
  

when
0:
22
2
2
9

 
dV kL
PL
d
For stability:
22
2
0, 2 0
9
dV kL
PL
d


1
18
PkL


www.elsolucionario.org

1871. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.95
The horizontal bar BEH is connected to three vertical bars. The collar at
E can slide freely on bar DF. Determine the range of values of Q for
which the equilibrium of the system is stable in the position shown when
a  24 in., b  20 in., and P  150 lb.


SOLUTION

First note
sin sinAa b
For small values of
and : ab
or
a
b


22
22
()cos2()cos
()cos 2cos
() sin 2sin
( ) cos 2 cos
VPab Qab
a
abP Q
b
dV a a
ab P Q
dbb
dV a a
ab P Q
bdb 




  

 


 
  
 

 
   


when
0:
22
22
() 2
dV a
ab P Q
db


  




www.elsolucionario.org

1872. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.95 (Continued)

Stability:
22
22
0: 2 0
dV a
PQ
db

 

2
2
2
b
P Q
a
 (1) 
or
2
2
2
a
QP
b

(2) 
with 150 lb, 24 in., and 20 in.Pa b 
Equation (1):
2
2
(24 in.)
(150 lb) 108.000 lb
2(20 in.)
Q

For stability
108.0 lbQ 

www.elsolucionario.org

1873. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.96
The horizontal bar BEH is connected to three vertical bars. The collar
at
E can slide freely on bar DF. Determine the range of values of P for
which the equilibrium of the system is stable in the position shown when
a  150 mm, b  200 mm, and Q  45 N.

SOLUTION
Using Equation (2) of Problem 10.95 with

45 N, 150 mm, and 200 mmQa b 
Equation (2)
2
2
(200 mm)
2(45N)
(150 mm)
160.000 N
P


For stability
160.0 NP 

www.elsolucionario.org

1874. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.97*
Bars AB and BC, each of length l and of negligible weight, are attached to two
springs, each of constant k. The springs are undeformed, and the system is in
equilibrium when
1  2  0. Determine the range of values of P for which
the equilibrium position is stable.


SOLUTION
We have
12
12
22
sin
sin sin
cos cos
11
22
B
C
C
CBC
xl
xl l
yl l
VPy kx kx




 
or
2 22
12 1121
(cos cos ) sin (sin sin )
2
VPl kl
   

For small values of
1
and
2
:

22
11 2 2 1 1 2 211
sin , sin , cos 1 , cos 1
22
        
Then

22
22212
112
1
11
222
VPl kl


 
 


and
2
1112
1
2
212
2
22
22
22
12
2
2
12
[( )]
()
2
V
Pl kl
V
Pl kl
VV
Pl kl Pl kl
V
kl







 


   
  
   

Stability: Conditions for stability (see Page 583).


www.elsolucionario.org

1875. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.97* (Continued)

For
12
12
0: 0 (condition satisfied)
VV


 
  

2
222
22
12 12
0
VVV

  
 


Substituting
22 2
24 22 3 24
222
()( 2)( )0
320
30
kl Pl kl Pl kl
kl Pl Pkl kl
PklPkl
  
 
 

Solving
35 35
or
22
P kl P kl

<
or
0.382 or 2.62
P kl P kl<

2
2
2
1
0: 2 0
V
Pl kl


 
or
1
2
Pkl

2
2
2
2
0: 0
V
Pl kl



or
Pkl

Therefore, all conditions for stable equilibrium are satisfied when
0 0.382Pkl 
www.elsolucionario.org

1876. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.98*
Solve Problem 10.97 knowing that l  800 mm and k  2.5 kN/m.
PROBLEM 10.97* Bars AB and BC, each of length l and of negligible
weight, are attached to two springs, each of constant
k. The springs are
undeformed, and the system is in equilibrium when
1  2  0. Determine
the range of values of
P for which the equilibrium position is stable.

SOLUTION
From the analysis of Problem 10.97 with

800 mm and 2.5 kN/m
0.382 0.382(2500 N/m)(0.8 m) 764 N
lk
Pkl
 
 
764 NP 

www.elsolucionario.org

1877. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.99*
Two rods of negligible weight are attached to drums of radius r
that are connected by a belt and spring of constant k. Knowing
that the spring is undeformed when the rods are vertical,
determine the range of values of P for which the equilibrium
position
1  2  0 is stable.

SOLUTION

Left end of spring moves from a to b. Right end of spring moves from
ato.b Elongation of spring

12 12
2
12
22
12 1 2
2
12 1
1
2
12 2
2
2
2
12
1
2
2
22
2
2
2
12
()
1
cos cos
2
1
()cos cos
2
()sin
()sin
cos
cos
sab abr r r
Vkspl wl
kr pl wl
v
kr pl
v
kr wl
v
kr pl
v
kr wl
v
kr 

  

 


 








 
 


  




www.elsolucionario.org

1878. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.99* (Continued)

For
22 2
22 2
12 22
212
0: , ,
,
vv rv
kr pl kr wl kr


  
     
Conditions for stability (see Page 583)

2
222
22
12 12
22 2 2
22
0
()( )( )0
() 0
vvv
kr kr pl kr wl
pl kr wl kr wl

  
 


 



2
22
2
;
kr
l
wkr kr W
PP
lkr wl w


 



22
2
2
1
:0;
vkr
kr pl P
l


 
We choose:
2
2
kr
l
kr W
P
l W







www.elsolucionario.org

1879. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.100*
Solve Problem 10.99 knowing that k  20 lb/in., r  3 in., l  6
in., and (
a) W  15 lb, (b) W  60 lb.
PROBLEM 10.99* Two rods of negligible weight are attached
to drums of radius
r that are connected by a belt and spring of
constant
k. Knowing that the spring is undeformed when the rods
are vertical, determine the range of values of
P for which the
equilibrium position
1  2  0 is stable.

SOLUTION

22
20 lb/in.
3in.
6in.
(20 lb/in.)(3 in.)
30 lb
6in.
k
r
l
kr
l





(
a)
15 lb
15 lb: (30 lb)
(30 lb) (15 lb)
WP

10.00 lbP 
(
b)
60 lb
60 lb: (30 lb)
(30lb) (60lb)
WP

20.0 lbP 

www.elsolucionario.org

1880. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.101
Determine the vertical force P that must be applied at G to
maintain the equilibrium of the linkage.


SOLUTION


Assuming
A
y
it follows

12
1.5
8
CAA
yyy

1.5
EC A
yy y 

18
3(1.5 ) 4.5
6
DA A A
yy y y   

10 10
(1.5 ) 2.5
66
GA A A
yy y y   
Then, by virtual work

0: (300 lb) (100 lb) 0
ADG
UyyPy

300 100(4.5 ) (2.5 ) 0
300 450 2.5 0
AAA
yyPy
P
 


60.0 lbP


60.0 lbP 
www.elsolucionario.org

1881. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.102
Determine the couple M that must be applied to member DEFG to
maintain the equilibrium of the linkage.


SOLUTION

Assume
A
y:
12
1.5
8
CAA
yyy
,
1.5
EC A
yy y 


18
3(1.5 ) 4.5
6
DE A A
yy y y   


1.5 1
664
EA
A
yy
y
 
 
Virtual Work:
0:U (300 lb) (100 lb) 0
1
300 100(1.5 ) 0
4
1
300 450 0
4
AD
AAA
yyM
yyMy
M





 

600 lb in.M  600 lb in.M


www.elsolucionario.org

1882. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.103
Determine the force P required to maintain the equilibrium of the
linkage shown. All members are of the same length and the wheels at
A and B roll freely on the horizontal rod.

SOLUTION
Using yC as independent variable:

2
D C
yy 2
D C
yy 

3
FC
yy 3
FC
yy


4
4
GH C
GH C
yy y
yy y

 


Virtual Work:

(400 N) (100 N) (75 N) (150 N) 0
400 100(2 ) (3 ) (75 150)(4 ) 0
CDFGH
CCC C
Uy yPyy y
yyPy y  
 
  

3 400 200 900P 500 NP


500 NP


www.elsolucionario.org

1883. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.104
Derive an expression for the magnitude of the force Q
required to maintain the equilibrium of the mechanism
shown.

SOLUTION

We have
2 cos so that 2 sin
2
DD
xl x l
Al
Bl  




Virtual Work:
0: 0
D
UQxPAPB   

(2sin ) (2 ) ( ) 0
2sin 3 0
Ql Pl Pl
Ql Pl  
   


3
2sin
P
Q

 

www.elsolucionario.org

1884. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.105
Derive an expression for the magnitude of the couple M required
to maintain the equilibrium of the linkage shown.

SOLUTION


3sin 3cos
4sin 4cos
EE
FF
ya ya
ya ya 
 


Virtual Work:
0:U 0
EF
MPyPy    

(3 cos ) (4 cos ) 0MPa Pa     7cosMPa 
www.elsolucionario.org

1885. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.106
A vertical load W is applied to the linkage at B. The
constant of the spring is k, and the spring is unstretched
when AB and BC are horizontal. Neglecting the weight of
the linkage, derive an equation in
, W, l, and k that must
be satisfied when the linkage is in equilibrium.

SOLUTION


2cos 2sin
sin cos
(2 ) 2 (1 cos )
CC
BB
C
xl x l
yl yl
Fksklx kl 
 


   

Virtual Work:
0: 0
CB
UFxWy
2(1cos)(2sin ) (cos ) 0kl l W l   

2
4(1cos)sin coskl Wl 
or

(1 cos ) tan
4
W
kl
From above
(1 cos ) tan
4
W
kl 

www.elsolucionario.org

1886. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.107
A force P of magnitude 240 N is applied to end E of cable CDE,
which passes under pulley D and is attached to the mechanism at
C. Neglecting the weight of the mechanism and the radius of the
pulley, determine the value of
 corresponding to equilibrium.
The constant of the spring is
4 kN/m,k and the spring is
unstretched when
90 .

SOLUTION

2
2
2sin
2
sr
sr
Fkskr
CD l











 





1
()2cos
22
cos
2
CD l
l








Virtual Work:
Since F tends to decrease s and P tends to decrease CD, we have

()0UFsPCD  

() cos 0
22
kr r P l

   
 
  
  


2
22
2 (240 N)(0.3 m)
1.25
cos (4000 N/m)(0.12 m)
pl
kr



 

Solving by trial and error:
0.33868 rad 19.40 

www.elsolucionario.org

1887. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.108
Two identical rods ABC and DBE are connected by a pin at B
and by a spring CE. Knowing that the spring is 4 in. long when
unstretched and that the constant of the spring is 8 lb/in.,
determine the distance x corresponding to equilibrium when a
24-lb load is applied at E as shown.


SOLUTION

Deformation of spring

2
2
2
4 in. 4
3
112
(24 lb) (8 lb/in.) 4 4
2623
22
844
33
x
sEC
xx
Vks x
dV x
dx
 

  







Equilibrium:
16 2
0440
33
dV x
dx






23
44
316
23
4
34
x
x





7.13 in.x 

www.elsolucionario.org

1888. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.109
Solve Problem 10.108 assuming that the 24-lb load is applied at
C instead of E.
PROBLEM 10.108 Two identical rods ABC and DBE are
connected by a pin at B and by a spring CE. Knowing that the
spring is 4 in. long when unstretched and that the constant of the
spring is 8 lb/in., determine the distance x corresponding to
equilibrium when a 24-lb load is applied at E as shown.


SOLUTION

Deformation of spring

2
2
2
4 in. 4
3
1512
(24 lb) (8 lb/in.) 4 20
2623
22
8420
33
x
sEC
xx
Vks x
dV x
dx
 

  







Equilibrium:
16 2
04200
33
dV x
dx






23
420
316
2
43.75
3
x
x





11.63 in.x 

www.elsolucionario.org

1889. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.110
Two uniform rods, each of mass m and length l, are attached to gears
as shown. For the range
0 180 ,
 determine the positions of
equilibrium of the system and state in each case whether the
equilibrium is stable, unstable, or neutral.


SOLUTION

Potential energy cos1.5 cos
22
ll
VW W Wmg

 




2
2
( 1.5sin1.5 ) ( sin )
22
(1.5 sin 1.5 sin )
2
(2.25cos1.5 cos )
2
dV Wl Wl
d
Wl
dV Wl
d 



 
 
 

For equilibrium
0: 1.5sin1.5 sin 0
dV
d 

Solutions: One solution, by inspection, is
0, and a second angle less than 180° can be found
numerically:

2.4042 rad 137.8
Now
2
2
(2.25cos1.5 cos )
2
dV Wl
d 
 
At
0:
2
2
(2.25cos0 cos0 )
2
dV Wl
d

  

(3.25)( 0)
2
Wl
  0, Unstable 
www.elsolucionario.org

1890. Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 10.110 (Continued)

At
137.8 :

2
2
[2.25cos(1.5 137.8 ) cos137.8 ]
2
dV Wl
d


(2.75)( 0)
2
Wl
  137.8 , Stable 

www.elsolucionario.org

1891. Copyright © McGraw-Hill Education. Permission required for reproduction or display.




PROBLEM 10.111
A homogeneous hemisphere of radius r is placed on an incline as shown.
Assuming that friction is sufficient to prevent slipping between the
hemisphere and the incline, determine the angle
 corresponding to
equilibrium when
10 .

SOLUTION

Detail

3
8
(sin()cos)
3
sin sin
8
CG r
VWr CG
dV
Wr r
d
 



 

 


Equilibrium:
3
0, sin sin 0
8
dV
d

  

3
sin sin
8
 (1)
For
10
3
sin10 sin
8
 sin 0.46306, 27.6 

www.elsolucionario.org

1892. Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 10.112
A homogeneous hemisphere of radius r is placed on an incline as shown.
Assuming that friction is sufficient to prevent slipping between the
hemisphere and the incline, determine (a) the largest angle
 for which a
position of equilibrium exists, (b) the angle
 corresponding to equilibrium
when the angle
 is equal to half the value found in part a.

SOLUTION

Detail

3
8
(sin()cos)
3
sin sin
8
CG r
VWr CG
dV
Wr r
d
 



 

 


Equilibrium:
3
0, sin sin 0
8
dV
d

  

3
sin sin
8
 (1)
(a) For
max
, 90° 
Eq. (1)
max max
33
sin sin 90 , sin 22.02
88
 
max
22.0 
(b) When
1
max2
11.01
Eq. (1)
3
sin11.01 sin ; sin 0.5093
8
  30.6 
Note: We can also use CGDand law of sines to derive Eq. (1).

sin sin 3
; sin sin ; sin sin
8
CG
CG CD CD

 

www.elsolucionario.org