Lecture Notes on Engineering Statics.

Lecture notes on Engineering Statics for undergrads.

1. Engineering Mechanics
Statics
Supported with MATLAB Codes
Dr. Ahmed Momtaz Hosny
PhD in Aircraft Dynamics and Control, BUAA
Lecturer at KMA
Lecture Notes & Solved Examples with MATLAB Applications

2. List of Contents

Chapter 1 Introduction to Mechanics
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Statics of Particles
Statics of Rigid Bodies, Equivalent Systems of Forces
Equilibrium of Rigid Bodies
Distributed Forces, Centroids and Center of Gravity
Analysis of Structures, Simple Truss
(Pin Joints & Two-Forces Members)
Chapter 7 Shear and Bending Moment of Beams
Chapter 8

Friction, Pulleys, Tension Belts and Journal Bearings

3. Chapter 1
Introduction to Mechanics

1.1 WHAT IS MECHANICS?
Mechanics can be defined as that science which describes and predicts the
conditions of rest or motion of bodies under the action of forces. It is divided into
three parts: mechanics of rigid bodies, mechanics of deformable bodies, and
mechanics of fluids . The mechanics of rigid bodies is subdivided into statics and
dynamics, the former dealing with bodies at rest, the latter with bodies in motion.
In this part of the study of mechanics, bodies are assumed to be perfectly rigid.
Actual structures and machines, however, are never absolutely rigid and deform
under the loads to which they are subjected. But these deformations are usually
small and do not appreciably affect the conditions of equilibrium or motion of the
structure under consideration. They are important, though, as far as the resistance
of the structure to failure is concerned and are studied in mechanics of materials,
which is a part of the mechanics of deformable bodies. The third division of
mechanics, the mechanics of fluids, is subdivided into the study of incompressible
fluids and of compressible fluids. An important subdivision of the study of
incompressible fluids is hydraulics, which deals with problems involving water.
Mechanics is a physical science, since it deals with the study of physical
phenomena. However, some associate mechanics with mathematics, while many
consider it as an engineering subject. Both these views are justified in part.
Mechanics is the foundation of most engineering sciences and is an indispensable
prerequisite to their study. However, it does not have the empiricism found in some
engineering sciences, i.e., it does not rely on experience or observation alone; by
its rigor and the emphasis it places on deductive reasoning it resembles
mathematics. But, again, it is not an abstract or even a pure science; mechanics is
an applied science. The purpose of mechanics is to explain and predict physical
phenomena and thus to lay the foundations for engineering applications.

4. 1.2 SYSTEMS OF UNITS
With the four fundamental concepts introduced in the preceding section are
associated the so-called kinetic units, i.e., the units of length, time, mass, and force.

International System of Units (SI Units). In this system, which will be in
universal use after the United States has completed its conversion to SI units, the
base units are the units of length, mass, and time, and they are called, respectively,
the meter (m), the kilogram (kg), and the second (s). All three are arbitrarily
defined. The unit of force is a derived unit. It is called the newton (N) and is
defined as the force which gives an acceleration of 1 m/s 2 to a mass of 1 kg.

1 N = (1 kg)(1 m/s
2

)
Multiples and submultiples of the fundamental SI units may be obtained through
the use of the prefixes defined in Table 1.1 . The multiples and submultiples of the
units of length, mass, and force most frequently used in engineering are,
respectively, the kilometer (km) and the millimeter (mm); the megagram (Mg) and
the gram (g); and the kilonewton (kN). According to Table 1.1 ,

5. Other derived SI units used to measure the moment of a force, the work of a force,
etc., are shown in Table 1.2 . While these units will be introduced in later chapters
as they are needed.




U.S. Customary Units. Most practicing American engineers still commonly use a
system in which the base units are the units of length, force, and time. These units
are, respectively, the foot (ft), the pound (lb), and the second (s).

The U.S. customary units most frequently used in mechanics are listed in Table
1.3 with their SI equivalents.

7. Chapter 2
Statics of Particles


2.1 INTRODUCTION
In this chapter you will study the effect of forces acting on particles. First you will
learn how to replace two or more forces acting on a given particle by a single force
having the same effect as the original forces. This single equivalent force is the
resultant of the original forces acting on the particle. Later the relations which
exist among the various forces acting on a particle in a state of equilibrium will be
derived and used to determine some of the forces acting on the particle. The use of
the word “particle” does not imply that our study will be limited to that of small
corpuscles. What it means is that the size and shape of the bodies under
consideration will not significantly affect the solution of the problems treated in
this chapter and that all the forces acting on a given body will be assumed to be
applied at the same point. Since such an assumption is verified in many practical
applications, you will be able to solve a number of engineering problems in this
chapter. The first part of the chapter is devoted to the study of forces contained in a
single plane, and the second part to the analysis of forces in three-dimensional
space.

2.2 FORCE ON A PARTICLE. RESULTANT OF TWO FORCES
A force represents the action of one body on another and is generally characterized
by its point of application, its magnitude, and its direction. Forces acting on a
given particle, however, have the same point of application. Each force considered
in this chapter will thus be completely defined by its magnitude and direction. The
magnitude of a force is characterized by a certain number of units. As indicated in
Chap. 1, the SI units used by engineers to measure the magnitude of a force are the
newton (N) and its multiple the kilonewton (kN), equal to 1000 N, while the U.S.
customary units used for the same purpose are the pound (lb) and its multiple the
kilopound (kip), equal to 1000 lb. The direction of a force is defined by the line of
action and the sense of the force. The line of action is the infinite straight line
along which the force acts; it is characterized by the angle it forms with some fixed

8. axis (Fig. 2.1). The force itself is represented by a segment of that line; through the
use of an appropriate scale, the length of this segment may be chosen to represent
the magnitude of the force.
Fig. 2.1


Finally, the sense of the force should be indicated by an arrowhead. It is important
in defining a force to indicate its sense. Two forces having the same magnitude and
the same line of action but different sense, such as the forces shown in Fig. 2.1a
and b, will have directly opposite effects on a particle. Experimental evidence
shows that two forces P and Q acting on a particle A (Fig. 2.2a) can be replaced by
a single force R which has the same effect on the particle (Fig. 2.2c). This force is
called the resultant of the forces P and Q and can be obtained, as shown in Fig.
2.2b, by constructing a parallelogram, using P and Q as two adjacent sides of the
parallelogram. The diagonal that passes through A represents the resultant. This
method for finding the resultant is known as the parallelogram law for the addition
of two forces. This law is based on experimental evidence; it cannot be proved or
derived mathematically.




Fig. 2.2

9. 2.3 VECTORS
It appears from the above that forces do not obey the rules of addition defined in
ordinary arithmetic or algebra. For example, two forces acting at a right angle to
each other, one of 4 lb and the other of 3 lb, add up to a force of 5 lb, not to a force
of 7 lb. Forces are not the only quantities which follow the parallelogram law of
addition. As you will see later, displacements, velocities, accelerations, and
momenta are other examples of physical quantities possessing magnitude and
direction that are added according to the parallelogram law. All these quantities
can be represented mathematically by vectors, while those physical quantities
which have magnitude but not direction, such as volume, mass, or energy, are
represented by plain numbers or scalars. Vectors are defined as mathematical
expressions possessing magnitude and direction, which add according to the
parallelogram law. Vectors are represented by arrows in the illustrations and will
be distinguished from scalar quantities in this text through the use of boldface type
(P). In longhand writing, a vector may be denoted by drawing a short arrow above
the letter used to represent it (????????????��⃗) or by underlining the letter (P)
Two vectors which have the same magnitude and the same direction are said to be
equal, whether or not they also have the same point of application (Fig. 2.4); equal
vectors may be denoted by the
. The last method
may be preferred since underlining can also be used on a typewriter or computer.
The magnitude of a vector defines the length of the arrow used to represent the
vector. In this text, italic type will be used to denote the magnitude of a vector.
Thus, the magnitude of the vector P will be denoted by P. A vector used to
represent a force acting on a given particle has a well- defined point of application,
namely, the particle itself. Such a vector is said to be a fixed, or bound, vector and
cannot be moved without modifying the conditions of the problem. Other physical
quantities, however, such as couples (see Chap. 3), are represented by vectors
which may be freely moved in space; these vectors are called free vectors. Still
other physical quantities, such as forces acting on a rigid body (see Chap. 3 ), are
represented by vectors which can be moved, or slid, along their lines of action;
they are known as sliding vectors.
same letter.

10. The negative vector of a given vector P is defined as a vector having the same
magnitude as P and a direction opposite to that of P (Fig. 2.5); the negative of the
vector P is denoted by -P. The vectors P and -P are commonly referred to as equal
and opposite vectors. Clearly, we have

P + (-P) = 0


2.4 ADDITION OF VECTORS
We saw in the preceding section that, by definition, vectors add according to the
parallelogram law. Thus, the sum of two vectors P and Q is obtained by attaching
the two vectors to the same point A and constructing a parallelogram, using P and
Q as two sides of the parallelogram (Fig. 2.6). The diagonal that passes through A
represents the sum of the vectors P and Q, consequently we can write

R = P + Q = Q + P

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3

PROBLEM 2.1
Two forces are applied at point B of beam AB. Determine
graphically the magnitude and direction of their resultant using
(a) the parallelogram law, (b) the triangle rule.

SOLUTION
(a) Parallelogram law:

(b) Triangle rule:

We measure:
3.30 kN, 66.6R α==° 3.30 kN=R
66.6° 

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4


PROBLEM 2.2
The cable stays AB and AD help support pole AC . Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.

SOLUTION

We measure:
51.3
59.0α
β=°
=°
(a) Parallelogram law:

(b) Triangle rule:

We measure:
139.1 lb,R= 67.0
γ=° 139.1lbR= 67.0° 


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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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5


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° 

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6


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° 

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7


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 25P
=
°° 101.4 NP= 
(b)
30 25 180
180 25 30
125
β
β
°+ + °= °
=°−°−° =°

120 N
sin30 sin125
=
°° R
196.6 N=R 

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8


PROBLEM 2.6
A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that
α = 25°, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical. (b) What is the corresponding magnitude of the
resultant?

SOLUTION

Using the triangle rule and the law of sines:
(a)
1600 N
sin 25° sin 75P
=
° 3660 NP= 
(b)
25 75 180
180 25 75
80
β
β
°+ + °= °
=°−°−°
=°

1600 N
sin 25° sin80
R
=
° 3730 NR= 

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13

PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing that
α = 20°, determine by trigonometry (a) the required magnitude
of the force P if the resultant R of the two forces applied at A is
to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines:
(a)
50 60 180
180 50 60
70
β
β
+°+°= °
=°−°−°
=°

425 lb
sin 70 sin 60
P
=
°°
392 lbP= 
(b)
425 lb
sin 70 sin 50
R
=
°° 346 lbR= 

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14


PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing that
the magnitude of P is 500 lb, determine by trigonometry (a) the
required angle
α if the resultant R of the two forces applied at A
is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines:
(a) ( 30 ) 60 180
180 ( 30 ) 60
90
sin (90 ) sin 60
425 lb 500 lbα
β
βα
βα
α
+°+°+= °
=°−+°−°
=°−
°− °
=

90 47.402α°− = ° 42.6α=° 
(b)
500 lb
sin (42.598 30 ) sin 60
R
=
°+ ° °
551 lbR= 

19. %%%%%%%%%Problem 2.12 How to calculate angle Alfa for different values of P
%%%%%%%%%Considering Vertical Resultant R and Constant Force of 425 with
%%%%%%%%%angle 30 degree

%% Calculating the regular problem

%% Beta=180 - (Alfa+30)- 60 ie Beta= 90- Alfa ;

%% Substituting in Sines Law sin(Beta)/425 = sin(pi/3)/500;

B =(sin(pi/3)/500)*425;

Beta = asin(B)

Beta = Beta * 57.3 %%%% in Radians

Alfa=90 - Beta

%%%%%%%%% Calculating the resultant Force

R=(500/sin(pi/3))* sin((Alfa+30)/57.3)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Performing a loop calculation for P=[500 550 600 650 700 750 800 850 900]

P=[500 550 600 650 700 750 800 850 900]

for I=0:1:8

I=I+1

B =(sin(pi/3)/P(I))*425;

Beta = asin(B)

Beta = Beta * 57.3 %%%% in Degrees

Beta1(I)=Beta

Alfa(I)=90 - Beta


R(I)=(P(I)/sin(pi/3))* sin((Alfa(I)+30)/57.3)


end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Plotting Alfa and Beta for Different P

20. figure(1);
plot(P,Alfa);
xlabel('P in LB'),grid on;;ylabel('Alfa in Degrees'),grid on;;
figure(2);
plot(P,Beta1);
xlabel('P in LB'),grid on;;ylabel('Beta in Degrees'),grid on;;
figure(3);
plot(P,R);
xlabel('P in LB'),grid on;;ylabel('R Resultant'),grid on;;
figure(4);
plot(P,Alfa,'+',P,Beta1,'o');
xlabel('P in LB'),grid on;;ylabel('Alfa Beta in Degrees'),grid on;;

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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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17

PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM 2.2 The cable stays AB and AD help support pole
AC. Knowing that the tension is 120 lb in AB and 40 lb in AD,
determine graphically the magnitude and direction of the
resultant of the forces exerted by the stays at A using (a) the
parallelogram law, (b) the triangle rule.

SOLUTION


8
tan
10
38.66
6
tan
10
30.96
α
α
β
β=
=°
=
=°

Using the triangle rule:
180
38.66 30.96 180
110.38α
βψ
ψ
ψ
++= °
°+ °+ = °
=°
Using the law of cosines:
222
(120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38
139.08 lbR
R
=+− °
=
Using the law of sines:
sin sin110.38
40 lb 139.08 lb
γ °
=


15.64
(90 )
(90 38.66 ) 15.64
66.98
γ
φαγ
φ
φ
=°
=°−+
=°− °+ °
=°
139.1 lb=R
67.0° 

23. %%%%%%%%%Problem 2.15 How to calculate Resultant R and its angle from the
%%%%%%%%%horizontal plane

%% Calculating the regular problem

%% tan(Alfa)=8/10 & tan(Beta)=6/10 then we can get Alfa & Beta;

Alfa=atan(8/10)*57.3
Beta=atan(6/10)*57.3

Epsai = 180 - Alfa - Beta

%%%%% Calculating the resultant from the Law of Cosines

R = ((120^2) + (40^2) - 2*40*120*cos(Epsai/57.3))^0.5

%%%%% Calculating the angle Gama to obtain angle Fai as shown in the figure
%%%%% sin(Gama)/40=sin(Epsai)/R consequently we can obtain sin(Gama) in
%%%%% form of G

G=(sin(Epsai/57.3)/R)*40

Gama=asin(G)*57.3 %%%%%%% in Degrees

Fai=90- Alfa + Gama



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Performing a loop calculation for P=[500 550 600 650 700 750 800 850 900]

T1=[100 120 140 160 180 200 220 240 260];
T2=[30 40 50 60 70 80 90 100 110];

for I=0:1:8

I=I+1;

%%%%% Calculating the resultant from the Law of Cosines

R(I) = ((T1(I)^2) + (T2(I)^2) - 2*T1(I)*T2(I)*cos(Epsai/57.3))^0.5;

%%%%% Calculating the angle Gama to obtain angle Fai as shown in the figure
%%%%% sin(Gama)/40=sin(Epsai)/R consequently we can obtain sin(Gama) in
%%%%% form of G

G=(sin(Epsai/57.3)/R(I))*40;

Gama=asin(G)*57.3; %%%%%%% in Degrees

Fai(I)=90- Alfa + Gama;

24. end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Plotting Resultant R and the Corresponding Angle Fai for different T1 & T2


figure(1);
plot3(T1,T2,R);
xlabel('T1 in LB'), grid on;ylabel('T2 in LB'), grid on;zlabel('R in LB'),
grid on;

figure(2);
plot(R,Fai);
xlabel('R in LB'), grid on;ylabel('Fai in Degrees'), grid on;

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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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18


PROBLEM 2.16
Solve Problem 2.4 by trigonometry.
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
Using the force triangle and the laws of cosines and sines:
We have:
180 (50 25 )
105
γ=°−°+° =°



Then
222
2
(4 kips) (6 kips) 2(4 kips)(6 kips)cos105
64.423 kips
8.0264 kipsR
R=+− °
=
=
And
4 kips 8.0264 kips
sin(25 ) sin105
sin(25 ) 0.48137
25 28.775
3.775
α
α
α
α
=
°+ °
°+ =
°+ = °
=°

8.03 kips=R
3.8° 

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31


PROBLEM 2.29
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= 
(b)
35
12
35
(720 N)
12
2100 N
yx
PP=
=
=

2.10 kN=
y
P 

27. %%%%%%%%%Problem 2.29 How to calculate Resultant P by knowing one component
%%%%%%%%%Px = 720 N as shown in figure

Px = 720;


%% Calculating the regular problem

%% Taking into consideration that Px = P*sin(theta)where theta = the angle
%% between Py and P -----> sin(theta)= 2.4/(2.4^2 + 7^2)^0.5

s= 2.4/(2.4^2 + 7^2)^0.5;

P = Px/s

%% Taking into consideration that Py = P*cos(theta)where theta = the angle
%% between Py and P -----> cos(theta) = 7/(2.4^2 + 7^2)^0.5

Py = P*(7/(2.4^2 + 7^2)^0.5)



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Performing a loop calculation with Length L (from point D to point C
%% varies from L=[1 2.4 4 6 8 10 12 14 16 18]) therefore calculating the
%% corresponding P & Py

L=[1 2.4 4 6 8 10 12 14 16 20 30 40];

for I=0:1:11

I=I+1;

%% Taking into consideration that Px = P*sin(theta)where theta = the angle
%% between Py and P -----> sin(theta)= L(I)/(L(I)^2 + 7^2)^0.5

s= L(I)/(L(I)^2 + 7^2)^0.5;

P(I) = Px/s;

%% Taking into consideration that Py = P*cos(theta)where theta = the angle
%% between Py and P -----> cos(theta) = 7/(L(I)^2 + 7^2)^0.5

Py(I) = P(I)*(7/(L(I)^2 + 7^2)^0.5);



end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Plotting Resultant P & Px & Py with respect to length L

28. Px = [720 720 720 720 720 720 720 720 720 720 720 720];

figure(1);
plot(L,P,'o',L,Px,'*',L,Py,'+');
xlabel('L in m '), grid on;ylabel('P & Px & Py in N'), grid on;

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33


PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.23.
PROBLEM 2.23 Determine the x and y components of each of
the forces shown.

SOLUTION
Components of the forces were determined in Problem 2.23:
Force x Comp. (N) y Comp. (N)
800 lb +640 +480
424 lb –224 –360
408 lb +192 –360
608
x
R=+ 240
y
R=−

(608 lb) ( 240 lb)
tan
240
608
21.541
240 N
sin(21.541°)
653.65 N
xy
y
x
RR
R
R
R
α
α
=+
=+−
=
=
=°
=
=
Rij
ij
654 N=R
21.5° 

30. %%%%%%%%%Problem 2.31 How to calculate Resultant R by knowing all the
%%%%%%%%%components of the applied forces as shown in figure


%%%Calculating the regular problem as it is stated in the book

%%Calculating the X & Y Components for each applied force as following:-
%%%First force T1 = 800 N & x = 800 mm & y = 600 mm
T1 = 800
T1x = 800*(800/(800^2 + 600^2)^0.5)
T1y = 800*(600/(800^2 + 600^2)^0.5)
%%%Second force T2 = 408 N & x = 480 mm & y = - 900 mm
T2 = 408
T2x = 408*(480/(480^2 + 900^2)^0.5)
T2y = 408*(- 900/(480^2 + 900^2)^0.5)
%%%Third force T3 = 424 N & x = - 560 mm & y = - 900 mm
T3 = 424
T3x = 424*(- 560/(560^2 + 900^2)^0.5)
T3y = 424*(- 900/(560^2 + 900^2)^0.5)
%%%%Calculating the Resultant R as a summation of X components and Y
%%%%components as following:-
Rx = T1x + T2x + T3x
Ry = T1y + T2y + T3y

R=(Rx^2 + Ry^2)^0.5

%%%%Calculating the angle of the resultant R as following:-
theta = atan(Ry/Rx)*57.3 %%%% in Degrees

%%In case of several acting forces or even enormous number of acting forces
%%A Successive Loop should be constructed as following:-

%%T = [T1 T2 T3 T4 T5 .........]
%%x = [x1 x2 x3 x4 x5 .........]
%%y = [y1 y2 y3 y4 y5 .........]

T = [800 408 424]
x = [800 480 - 560]
y = [600 - 900 -900]

Rx=0;
Ry=0;

for H=0:1:2

H=H+1
Tx(H)= T(H)*(x(H)/(x(H)^2 + y(H)^2)^0.5)
Ty(H)= T(H)*(y(H)/(x(H)^2 + y(H)^2)^0.5)
Rx = Rx + Tx(H)
Ry = Ry + Ty(H)

end

R=(Rx^2 + Ry^2)^0.5

31. %%%%Calculating the angle of the resultant R as following:-
theta = atan(Ry/Rx)*57.3 %%%% in Degrees

32. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
34

PROBLEM 2.32
Determine the resultant of the three forces of Problem 2.21.
PROBLEM 2.21 Determine the x and y components of each of the
forces shown.

SOLUTION
Components of the forces were determined in Problem 2.21:
Force x Comp. (N) y Comp. (N)
80 N +61.3 +51.4
120 N +41.0 +112.8
150 N –122.9 +86.0
20.6
x
R=− 250.2
y
R=+

( 20.6 N) (250.2 N)
tan
250.2 N
tan
20.6 N
tan 12.1456
85.293
250.2 N
sin85.293
xy
y
x
RR
R
R
R
α
α
α
α
=+
=− +
=
=
=
=°
=
°Rij
ij
251 N=R
85.3° 

33. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
36


PROBLEM 2.34
Determine the resultant of the three forces of Problem 2.24.
PROBLEM 2.24 Determine the x and y components of each of the
forces shown.

SOLUTION
Components of the forces were determined in Problem 2.24:
Force x Comp. (lb) y Comp. (lb)
102 lb −48.0 +90.0
106 lb +56.0 +90.0
200 lb −160.0 −120.0
152.0
x
R=− 60.0
y
R=

( 152 lb) (60.0 lb)
tan
60.0 lb
tan
152.0 lb
tan 0.39474
21.541
α
α
α
α
=+
=− +
=
=
=
=°
xy
y
x
RR
R
RRij
ij


60.0 lb
sin 21.541
R=
°
163.4 lb=R
21.5° 

34. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
38


PROBLEM 2.36
Knowing that the tension in rope AC is 365 N, determine the
resultant of the three forces exerted at point C of post BC.

SOLUTION
Determine force components:
Cable force AC:
960
(365 N) 240 N
1460
1100
(365 N) 275 N
1460
=− =−
=− =−
x
y
F
F
500-N Force:
24
(500 N) 480 N
25
7
(500 N) 140 N
25
x
y
F
F
== ==

200-N Force:
4
(200 N) 160 N
5
3
(200 N) 120 N
5
x
y
F
F
==
=− =−
and
22
22
240N 480N 160N 400N
275 N 140 N 120 N 255 N
(400 N) ( 255 N)
474.37 N
=Σ =− + + =
=Σ =− + − =−
=+
=+−
=
xx
yy
xy
RF
RF
RRR
Further:
255
tan
400
32.5
α
α=
=°

474 N=R
32.5° 

35. %%%%%%%%%Problem 2.36 How to calculate Resultant P by knowing all the
%%%%%%%%%components of the applied forces as shown in figure


%%%Calculating the regular problem as it is stated in the book

%%Calculating the X & Y Components for each applied force as following:-
%%%First force T1 = 500 N & x = 24 m & y = 7 m
T1 = 500
T1x = 500*(24/(24^2 + 7^2)^0.5)
T1y = 500*(7/(24^2 + 7^2)^0.5)
%%%Second force T2 = 200 N & x = 4 m & y = - 3 m
T2 = 200
T2x = 200*(4/(4^2 + 3^2)^0.5)
T2y = 200*(- 3/(4^2 + 3^2)^0.5)
%%%Third force T3 = 365 N & x = - 960 mm & y = - 1100 mm
T3 = 365
T3x = 365*(- 960/(960^2 + 1100^2)^0.5)
T3y = 365*(- 1100/(960^2 + 1100^2)^0.5)
%%%%Calculating the Resultant R as a summation of X components and Y
%%%%components as following:-
Rx = T1x + T2x + T3x
Ry = T1y + T2y + T3y

R=(Rx^2 + Ry^2)^0.5

%%%%Calculating the angle of the resultant R as following:-
theta = atan(Ry/Rx)*57.3 %%%% in Degrees

%%In case of several acting forces or even enormous number of acting forces
%%A Successive Loop should be constructed as following:-

%%T = [T1 T2 T3 T4 T5 .........]
%%x = [x1 x2 x3 x4 x5 .........]
%%y = [y1 y2 y3 y4 y5 .........]

for I=1:1:50

%%P is the highest force of 500 N, a range of P will be proposed
%%from 0 : 500 N in the following loop to discuss the effect on the
%%resultant

P(I)=I*10;

T = [P(I) 200 365];
x = [24 4 - 960];
y = [7 - 3 -1100];

Rx=0;
Ry=0;

for H=0:1:2

H=H+1;

36. Tx(H)= T(H)*(x(H)/(x(H)^2 + y(H)^2)^0.5);
Ty(H)= T(H)*(y(H)/(x(H)^2 + y(H)^2)^0.5);
Rx = Rx + Tx(H);
Ry = Ry + Ty(H);

end

R(I)=(Rx^2 + Ry^2)^0.5;

%%%%Calculating the angle of the resultant R as following:-
theta(I) = atan(Ry/Rx)*57.3; %%%% in Degrees


end

figure(1);
plot(P,theta,'o');
xlabel('Required P (The Highest Force)'), grid on;ylabel('theta in Degrees'),
grid on;


figure(2);
plot(P,R,'+');
xlabel('Required P (The Highest Force)'), grid on;ylabel('Resultant in N'),
grid on;

38. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
39


PROBLEM 2.37
Knowing that α = 40°, determine the resultant of the three forces
shown.

SOLUTION
60-lb Force: (60 lb)cos 20 56.382 lb
(60 lb)sin 20 20.521lb
x
y
F
F
=°=
=°=
80-lb Force:
(80 lb)cos60 40.000 lb
(80lb)sin60 69.282lb
x
y
F
F
=°=
=°=
120-lb Force:
(120 lb)cos30 103.923 lb
(120 lb)sin30 60.000 lb
x
y
F
F
=°=
=− °=−
and
22
200.305 lb
29.803 lb
(200.305 lb) (29.803 lb)
202.510 lb
xx
yy
RF
RF
R
=Σ =
=Σ =
=+
=
Further:
29.803
tan
200.305
α=

129.803
tan
200.305
8.46
α
−
=
=°
203 lb=R
8.46° 

39. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
40


PROBLEM 2.38
Knowing that α = 75°, determine the resultant of the three forces
shown.

SOLUTION
60-lb Force: (60 lb) cos 20 56.382 lb
(60 lb)sin 20 20.521 lb
x
y
F
F=° =
=° =
80-lb Force:
(80 lb) cos 95 6.9725 lb
(80 lb)sin 95 79.696 lb
x
y
F
F=° =−
=° =
120-lb Force:
(120 lb) cos 5 119.543 lb
(120 lb)sin 5 10.459 lb
x
y
F
F=° =
=° =
Then
168.953 lb
110.676 lb
xx
yy
RF
RF=Σ =
=Σ =
and
22
(168.953 lb) (110.676 lb)
201.976 lbR=+
=

110.676
tan
168.953
tan 0.65507
33.228
α
α
α=
=
=°
202 lb=R
33.2° 

40. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
42


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= 
(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= 

41. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
45


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
1100
tan
960
48.888
400
tan
960
22.620
α
α
β
β=
=°
=
=°



Force Triangle

Law of sines:
15.696 kN
sin 22.620 sin 48.888 sin108.492
AC BC
TT
==
°° °




(a)
15.696 kN
(sin 22.620 )
sin108.492
AC
T=°
° 6.37 kN
AC
T= 
(b)
15.696 kN
(sin 48.888 )
sin108.492
BC
T=°
° 12.47 kN
BC
T= 

42. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
46

PROBLEM 2.44
Two cables are tied together at C and are loaded as shown. Determine
the tension (a) in cable AC, (b) in cable BC.

SOLUTION

3
tan
2.25
53.130
1.4
tan
2.25
31.891
α
α
β
β=
=°
=
=° Free-Body Diagram





Law of sines: Force-Triangle

660 N
sin 31.891 sin 53.130 sin 94.979
AC BC
TT
==
°°°




(a)
660 N
(sin 31.891 )
sin94.979
AC
T=°
° 350 N
AC
T= 
(b)
660 N
(sin 53.130 )
sin94.979
BC
T=°
° 530 N
BC
T= 

43. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
47

PROBLEM 2.45
Knowing that 20 ,α=° determine the tension (a) in cable AC,
(b) in rope BC .

SOLUTION
Free-Body Diagram Force Triangle

Law of sines:
1200 lb
sin 110 sin 5 sin 65
AC BC
TT
==
°° °
(a)
1200 lb
sin 110
sin 65
AC
T=°
° 1244 lb
AC
T= 
(b)
1200 lb
sin 5
sin 65
BC
T=°
° 115.4 lb
BC
T= 

44. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
48


PROBLEM 2.46
Knowing that 55α=° and that boom AC exerts on pin C a force directed
along line AC , determine (a) the magnitude of that force, (b) the tension in
cable BC.

SOLUTION
Free-Body Diagram Force Triangle

Law of sines:
300 lb
sin 35 sin 50 sin 95
AC BC
FT
==
°°°
(a)
300 lb
sin 35
sin 95
AC
F=°
° 172.7 lb
AC
F= 
(b)
300 lb
sin 50
sin 95
BC
T=°
° 231 lb
BC
T= 

45. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
54


PROBLEM 2.52
Two cables are tied together at C and loaded as shown.
Determine the range of values of P for which both cables
remain taut.

SOLUTION
Free Body: C

12 4
0: 0
13 5
xA C
TΣ= − + =FP

13
15
AC
TP= (1)

53
0: 480 N 0
13 5
yA CB C
TT PΣ= + + − =F
Substitute for
AC
T from (1):
513 3
480 N 0
13 15 5
BC
PT P

++− =




14
480 N
15
BC
TP=− (2)
From (1),
0
AC
TΣ requires 0.PΣ
From (2),
0
BC
TΣ requires
14
480 N, 514.29 N
15
PPαα
Allowable range:
0514 NPαα 

46. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
55

PROBLEM 2.53
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that
30α=° and 10
β=° and that the combined
weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the
traction cable CD.

SOLUTION
Free-Body Diagram


0: cos 10 cos 30 cos 30 0
xACB ACB CD
FT T TΣ= °− °− °=


0.137158
CD ACB
TT=
(1)

0: sin 10 sin 30 sin 30 900 0
yACB A CB CD
FT T TΣ= °+ °+ °− =

0.67365 0.5 900
ACB CD
TT+=

(2)
(a) Substitute (1) into (2):
0.67365 0.5(0.137158 ) 900
ACB ACB
TT+=

1212.56 N
ACB
T= 1213 N
ACB
T= 
(b) From (1):
0.137158(1212.56 N)
CD
T= 166.3 N
CD
T= 

47. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
56


PROBLEM 2.54
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD .
Knowing that
25α=° and 15
β=° and that the tension in
cable CD is 80 N, determine (a) the combined weight of the
boatswain’s chair and the sailor, (b) in tension in the support
cable ACB.

SOLUTION
Free-Body Diagram

0: cos 15 cos 25 (80 N)cos 25 0
xACB ACB
FT TΣ= °− °− °=

1216.15 N
ACB
T=

0: (1216.15 N)sin 15 (1216.15 N)sin 25
y
FΣ= °+ °


(80 N)sin 25 0
862.54 N
W
W
+° −=
=

( a)
863 NW= 

(b)
1216 N
ACB
T= 

48. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
57

PROBLEM 2.55
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium and
that
500P=lb and 650Q=lb, determine the magnitudes of
the forces exerted on the rods A and B .

SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:

0
AB
=++ + =RPQF F
Substituting components:
(500 lb) [(650 lb)cos50 ]
[(650 lb)sin 50 ]
(cos50)(sin50)0
BA A
FF F
=− + °
−°
+− °+ °=
Rj i
j
iij
In the y -direction (one unknown force):

500 lb (650 lb)sin 50 sin 50 0
A
F−− °+ °=
Thus,
500 lb (650 lb)sin50
sin 50
A
F
+°
=
°

1302.70 lb= 1303 lb
A
F= 
In the x -direction:
(650 lb)cos50 cos50 0
BA
FF°+ − °=
Thus,
cos50 (650 lb)cos50
(1302.70 lb)cos50 (650 lb)cos50
BA
FF=°− °
=° −°

419.55 lb= 420 lb
B
F= 

49. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
58


PROBLEM 2.56
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium
and that the magnitudes of the forces exerted on rods A and B
are
750
A
F= lb and 400
B
F= lb, determine the magnitudes of
P and Q .

SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:

0
AB
=++ + =RPQF F
Substituting components:
cos 50 sin 50
[(750 lb)cos 50 ]
[(750 lb)sin 50 ] (400 lb)
PQ Q=− + ° − °
−°
+°+
Rj i j
i
j i

In the x -direction (one unknown force):

cos 50 [(750 lb)cos 50 ] 400 lb 0Q °− ° + =

(750 lb)cos 50 400 lb
cos 50
127.710 lb
Q
°−
=
°
=

In the y -direction:
sin 50 (750 lb)sin 50 0PQ−− °+ °=

sin 50 (750 lb)sin 50
(127.710 lb)sin 50 (750 lb)sin 50
476.70 lb
PQ=− °+ °
=− °+ °
=
477 lb; 127.7 lbPQ== 

50. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
61


PROBLEM 2.59
For the situation described in Figure P2.45, determine (a) the
value of
α for which the tension in rope BC is as small as
possible, (b) the corresponding value of the tension.
PROBLEM 2.45 Knowing that
20 ,α=° determine the tension
(a) in cable AC, (b) in rope BC .

SOLUTION
Free-Body Diagram Force Triangle

To be smallest,
BC
T must be perpendicular to the direction of .
AC
T
(a) Thus,
5α=° 5.00α=°

(b)
(1200 lb)sin 5
BC
T=° 104.6 lb
BC
T= 

51. %%%%%%%%%Problem 2.59 How to calculate value of alfa for which the tension
%%%%%%%%%in rope BC is as small as possible.

%%Measuring alfa from vertical line upward
%%Considering the Law of Sines consequently (Tac/sin(90+90- alfa)) =
%%(Tbc/sin(90+90- 5)) = (1200/sin(180 -5-alfa).

for I=1:1:180
alfa(I)=I
Tbc(I) = (1200/sin((5+alfa(I))/57.3))*sin((180- 5)/57.3);
Tac(I) = (1200/sin((5+alfa(I))/57.3))*sin((180- alfa(I))/57.3);


end

figure(1);
plot(alfa,Tbc,'o');
xlabel('alfa in Degrees '),grid on;ylabel('Tbc in LB'),grid on;
xlim([0 180])
ylim([0 2000])

figure(2);
plot(alfa,Tac,'+');
xlabel('alfa in Degrees '),grid on;ylabel('Tac in LB'),grid on;
xlim([0 180])
ylim([0 2000])

figure(3);
plot(alfa,Tac,'+',alfa,Tbc,'o');
xlabel('alfa in Degrees '),grid on;ylabel('Tac in LB & Tbc in LB'),grid on;
xlim([0 180])
ylim([0 2000])

53. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
62


PROBLEM 2.60
For the structure and loading of Problem 2.46, determine (a) the value of α for
which the tension in cable BC is as small as possible, (b) the corresponding
value of the tension.

SOLUTION
BC
Tmust be perpendicular to
AC
F to be as small as possible.
Free-Body Diagram: C Force Triangle is a right triangle

To be a minimum,
BC
Tmust be perpendicular to .
AC
F
(a) We observe:
90 30α=°−° 60.0α=° 
(b)
(300 lb)sin 50
BC
T=°
or
229.81lb
BC
T= 230 lb
BC
T= 

54. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
63


PROBLEM 2.61
For the cables of Problem 2.48, it is known that the maximum
allowable tension is 600 N in cable AC and 750 N in cable BC.
Determine (a) the maximum force P that can be applied at C,
(b) the corresponding value of
α.

SOLUTION
Free-Body Diagram Force Triangle

(a) Law of cosines
222
(600) (750) 2(600)(750)cos(25 45 )P=+− °+°

784.02 NP= 784 NP= 
(b) Law of sines

sin sin (25 45 )
600 N 784.02 N
β °+ °
=


46.0
β=° 46.0 25α∴= °+° 71.0α=° 

55. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
64


PROBLEM 2.62
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:
20.625m=× 1.250 m 

56. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
65


PROBLEM 2.63
Collar A is connected as shown to a 50-lb load and can slide on
a frictionless horizontal rod. Determine the magnitude of the
force P required to maintain the equilibrium of the collar when
(a)
4.5 in.,x= (b) 15 in.x=

SOLUTION
(a) Free Body: Collar A Force Triangle

50 lb
4.5 20.5P
= 10.98 lbP= 



(b) Free Body: Collar A Force Triangle

50 lb
15 25P
= 30.0 lbP= 



57. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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66


PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the
collar is in equilibrium when P = 48 lb.

SOLUTION
Free Body: Collar A Force Triangle


222
(50) (48) 196
14.00 lb
N
N
=−=
=


Similar Triangles

48 lb
20 in. 14 lb
x
=


68.6 in.x= 

58. %%%%%%%%%Problem 2.64 How to calculate value of alfa while the tension in
%%%%%%%%%rope AB is varing from T = 0 LB to 250 LB
%%%%%%%%%

%%Measuring alfa from horizontal line to the right at Point A
%%Considering the Law of Sines consequently (T/sin(90)) = (48/sin(90- alfa))
%%= (N/sin(alfa) consequently Beta = sin(90- alfa)=48/T

for I = 1:1:50

T(I)=I*5
Beta(I) = 48/T(I);
alfa(I) = - asin(Beta(I))*57.3 + 90;
X(I) = 20/tan(alfa(I)/57.3)
N(I) = (T(I))*sin(alfa(I)/57.3)


end

figure(1);
plot(T,X,'o');
xlabel('T in LB '),grid on;ylabel('X in in'),grid on;
xlim([0 250])
ylim([0 100])

figure(2);
plot(T,alfa,'*');
xlabel('T in LB '),grid on;ylabel('alfa in Degrees'),grid on;
xlim([0 250])
ylim([0 100])

figure(3);
plot(T,N,'+');
xlabel('T in LB '),grid on;ylabel('N in LB'),grid on;
xlim([0 250])
ylim([0 200])

60. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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67

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 N
Q=°
=

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
27.4α
α°+ =
=°
or
55 82.375
55 360 82.375
222.6α
α
α°+ =− °
°+ = °− °
=°
For
600 lb:R< 27.4 222.6α°< < 

61. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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68


PROBLEM 2.66
A 200-kg crate is to be supported by the rope-and-pulley arrangement shown.
Determine the magnitude and direction of the force P that must be exerted on the free
end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on
each side of a simple pulley. This can be proved by the methods of Ch. 4.)

SOLUTION
Free-Body Diagram: Pulley A





5
0: 2 cos 0
281
cos 0.59655
53.377
x
FPP α
α
α

Σ= − + = 

=
=± °

For
53.377 :α=+ °
16
0: 2 sin53.377 1962 N 0
281
y
FP P

Σ= + °− = 


724 N=P
53.4° 
For
53.377 :α=− ° 16
0: 2 sin( 53.377 ) 1962 N 0
281
y
FP P

Σ= + − °− = 

1773=P
53.4° 

62. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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73


PROBLEM 2.71
Determine (a) the x , y, and z components of the 900-N force, (b) the
angles
θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

cos 65
(900 N)cos 65
380.36 N
h
h
FF
F
=°
=°
=


(a)
sin 20
(380.36 N)sin 20°
xh
FF=°
=


130.091 N,=−
x
F 130.1 N
x
F=− 

sin 65
(900 N)sin 65°
815.68 N,
y
y
FF
F
=°
=
=+
816 N
y
F=+ 

cos 20
(380.36 N)cos 20
357.42 N
=°
=°
=+
zh
z
FF
F
357 N
z
F=+ 
(b)
130.091 N
cos
900 N
x
x
F
F
θ
−
==
98.3
x
θ=° 

815.68 N
cos
900 Ny
y
F
F
θ
+
==
25.0
y
θ=° 

357.42 N
cos
900 N
z
z
F
F
θ
+
==
66.6
z
θ=° 

63. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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95


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.
(40 in.) (45 in.) (60 in.)
(425 lb)
85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
=++=
=−+
=++=
−+
===
ijk
ijk
ijk
T λ



(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


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

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

544.8 lb
cos 0.59677
912.92 lb
z
θ== 53.4
z
θ=° 

64. %%%%%%%%%Problem 2.93 How to calculate the resultant at point A from
%%different acting forces using the unit vector & vector summation concept,
%%considering the tension in cable AB = 425 LB and in cable AC = 510 LB

%% Calculating each force in vector form by calculating the unit vector for
%% each

AB=[40 -45 60] %%% 40i - 45j + 60k
AC=[100 -45 60] %%% 100i - 45j + 60k

lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5

Tab=lamdaAB.*425

lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5

Tac=lamdaAC.*510

R = Tab + Tac

Rm=(R(1)^2 + R(2)^2 + R(3)^2)^0.5

%%Calculating theta angles with respect to X, Y, Z

thetaX = acos(R(1)/Rm)*57.3

thetaY = acos(R(2)/Rm)*57.3

thetaZ = acos(R(3)/Rm)*57.3

65. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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112


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




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



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= 

66. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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113


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
QΣ= + ++ =FTTj i

(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
AB AB
AC AC
=− − + =
=− − − =
ijk
ijk




48 12 19
53 53 53
12 3 4
13 13 13
AB AB AB AB AB
AC AC AC AC AC
AB
TT T
AB
AC
TT T
AC

===−−+



== =−−−


T λ ijk
T λ ijk



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

48 12
: 1200 N 0
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 NQΣφ 
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=−

67. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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114


PROBLEM 2.109
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AC is 60 N, determine the
weight of the plate.

SOLUTION
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.
Free Body A:

0: 0
AB AC AD
FPΣ= + + + =TTT j
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




Thus:

()
812 9
17 17 17
0.6 0.64 0.48
59.67.2
13 13 13
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD

===−−+


== =−+

== =−−


T λ ijk
T λ ijk
T λ ijk




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

85
0.6
17 13
12 9.6
0.64
17 13
97 .2
0.48 0
17 13
AB AC AD
AB AC AD
AB AC AD
TTT
TTTP
TTT

−+ +



+− − − +



++− =


i
j
k



Dimensions in mm

68. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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115
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
(544 N) 0.64(60 N) (572 N)
17 13
844.8 N
P=+ +
=
Weight of plate 845 NP== 

69. %%%%%%%%%Problem 2.109 How to calculate weight of a plate supported by
%%tensions at point A in cables AD & AB & AC knowing that tension in AC is
%%60 N


AB=[- 320 -480 360] %%% -320i -450j + 360k
AC=[450 -480 360] %%% 450i - 450j + 360k
AD=[250 -480 -360] %%% 250i - 450j + 360k

%%Tab=lamdaAB.*Tab
lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5


%%Tac=lamdaAC.*Tac
lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5
Tac=lamdaAC.*60

%%Tad=lamdaAD.*Tad
lamdaAD=AD./(AD(1)^2 + AD(2)^2 + AD(3)^2)^0.5




%%%%Consider the weight force = Wj as it is upward

%%%%As long the plate is in equilibrium state therefore the summation of
%%%%forces is equal to zero in all directions X, Y, Z
%%%Calculating the weight of the plate in matrix calculations
%%%[lamdaAB(1) lamdaAD(1) 0] [Tab] [- Tac(1)]
%%%[lamdaAB(2) lamdaAD(2) 1] * [Tad] = [- Tac(2)]
%%%[lamdaAB(3) lamdaAD(3) 0] [ W ] [- Tac(3)]

A = [lamdaAB(1) lamdaAD(1) 0;lamdaAB(2) lamdaAD(2) 1;lamdaAB(3)
lamdaAD(3) 0]

C = [- Tac(1) -Tac(2) -Tac(3)]'
X = inv(A)*C

Tab = X(1)
Tad = X(2)
W = X(3)

70. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
116

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== 

71. %%%%%%%%%Problem 2.110 How to calculate weight of a plate supported by
%%tensions at point A in cables AD & AB & AC knowing that tension in AD is
%%520 N


AB=[- 320 -480 360] %%% -320i -450j + 360k
AC=[450 -480 360] %%% 450i - 450j + 360k
AD=[250 -480 -360] %%% 250i - 450j + 360k

lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5

%%Tab=lamdaAB.*Tab

lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5

%%Tac=lamdaAC.*Tac

lamdaAD=AD./(AD(1)^2 + AD(2)^2 + AD(3)^2)^0.5

Tad=lamdaAD.*520


%%%%Consider the weight force = - Wj as it is downward

%%%%As long the plate is in equilibrium state therefore the summation of
%%%%forces is equal to zero in all directions X, Y, Z
%%%Calculating the weight of the plate in matrix calculations
%%%[lamdaAB(1) lamdaAC(1) 0] [Tab] [- Tad(1)]
%%%[lamdaAB(2) lamdaAC(2) - 1] * [Tac] = [- Tad(2)]
%%%[lamdaAB(3) lamdaAC(3) 0] [ W ] [- Tad(3)]

A = [lamdaAB(1) lamdaAC(1) 0;lamdaAB(2) lamdaAC(2) 1;lamdaAB(3)
lamdaAC(3) 0]

C = [- Tad(1) -Tad(2) -Tad(3)]'
X = inv(A)*C

Tab = X(1)
Tac = X(2)
W = X(3)

72. Chapter 3
Statics of Rigid Bodies
Equivalent Systems of Forces

3.1 INTRODUCTION
In the preceding chapter it was assumed that each of the bodies considered could
be treated as a single particle. Such a view, however, is not always possible, and a
body, in general, should be treated as a combination of a large number of particles.
The size of the body will have to be taken into consideration, as well as the fact
that forces will act on different particles and thus will have different points of
application. Most of the bodies considered in elementary mechanics are assumed to
be rigid, a rigid body being defined as one which does not deform. Actual
structures and machines, however, are never absolutely rigid and deform under the
loads to which they are subjected. But these deformations are usually small and do
not appreciably affect the conditions of equilibrium or motion of the structure
under consideration. They are important, though, as far as the resistance of the
structure to failure is concerned and are considered in the study of mechanics of
materials. In this chapter you will study the effect of forces exerted on a rigid body,
and you will learn how to replace a given system of forces by a simpler equivalent
system. This analysis will rest on the fundamental assumption that the effect of a
given force on a rigid body remains unchanged if that force is moved along its line
of action (principle of transmissibility).

The basic system is called a force -couple system. In the case of concurrent,
coplanar, or parallel forces, the equivalent force-couple system can be further
reduced to a single force, called the resultant of the system, or to a single
couple, called the resultant couple of the system.

3.2 EXTERNAL AND INTERNAL FORCES
Forces acting on rigid bodies can be separated into two groups: (1) external forces
and (2) internal forces.

73. 1. The external forces represent the action of other bodies on the rigid body under
consideration. They are entirely responsible for the external behavior of the rigid
body. They will either cause it to move or ensure that it remains at rest. We shall
be concerned only with external forces in this chapter and in Chaps. 4 and 5.

2 . The internal forces are the forces which hold together the particles forming the
rigid body. If the rigid body is structurally composed of several parts, the forces
holding the component parts together are also defined as internal forces. Internal
forces will be considered in Chaps. 6 and 7.

As an example of external forces, let us consider the forces acting on a disabled
truck that three people are pulling forward by means of a rope attached to the front
bumper (Fig. 3.1). The external forces acting on the truck are shown in a free -body
diagram (Fig. 3.2). Let us first consider the weight of the truck. Although it
embodies the effect of the earth’s pull on each of the particles forming the truck,
the weight can be represented by the single force W. The point of application of
this force, i.e., the point at which the force acts, is defined as the center of gravity
of the truck. It will be seen in Chap. 5 how centers of gravity can be determined.
The weight W tends to make the truck move vertically downward. In fact, it would
actually cause the truck to move downward, i.e., to fall, if it were not for the
presence of the ground. The ground opposes the downward motion of the truck by
means of the reactions R 1 and R 2. These forces are exerted by the ground on the
truck and must therefore be included among the external forces acting on the truck.
The people pulling on the rope exert the force F . The point of application of F is
on the front bumper. The force F tends to make the truck move forward in a
straight line and does actually make it move, since no external force opposes this
motion. (Rolling resistance has been neglected here for simplicity.) This forward
motion of the truck, during which each straight line keeps its original orientation
(the floor of the truck remains horizontal, and the walls remain vertical), is known
as a translation. Other forces might cause the truck to move differently. For
example, the force exerted by a jack placed under the front axle would cause the
truck to pivot about its rear axle. Such a motion is a rotation . It can be concluded,
therefore, that each of the external forces acting on a rigid body can, if unopposed,
impart to the rigid body a motion of translation or rotation, or both.

74. 3.3 VECTOR PRODUCT OF TWO VECTORS
In order to gain a better understanding of the effect of a force on a rigid body, a
new concept, the concept of a moment of a force about a point, will be introduced
at this time. This concept will be more clearly understood, and applied more
effectively, if we first add to the mathematical tools at our disposal the vector
product of two vectors. The vector product of two vectors P and Q is defined as
the vector V which satisfies the following conditions.

1. The line of action of V is perpendicular to the plane containing P and Q (Fig.
3.3 a).
2. The magnitude of V is the product of the magnitudes of P and Q and of the sine
of the angle u formed by P and Q (the measure of which will always be 180° or
less); we thus have
V = PQ sin θ
3. The direction of V is obtained from the right -hand rule. Close your right hand
and hold it so that your fingers are curled in the same sense as the rotation through
u which brings the vector P in line with the vector Q ; your thumb will then
indicate the direction of the vector V (Fig. 3.3 b ). Note that if P and Q do not have
a common point of application, they should first be redrawn from the same point.
The three vectors P , Q, and V —taken in that order—are said to form a right-
handed rule.
As stated above, the vector V satisfying these three conditions (which define it
uniquely) is referred to as the vector product of P and Q ; it is represented by the
mathematical expression in vector form as:
V = P x Q

75. Fig. 3.3
We can now easily express the vector product V of two given vectors P and Q in
terms of the rectangular components of these vectors. Resolving P and Q into
components, we first write:

V = P x Q = (P xi + P yj + P zk) x (Q xi + Q yj + Q z
V = (P
k) yQz - PzQy)i + (P zQx - PxQz)j + (P xQy - PyQx
The rectangular components of the vector product V are thus found to be:
)k

Which is more easily memorized by the following determinate:

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161


PROBLEM 3.1
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 sin 25 )
(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



77. %%%%%%%%%Problem 3.1 How to calculate the moment of force about point B as
%%%%%%%%%shown in figure.

%%Calulating the acting force of 20 LB in vector form as following:-
%%F = (Fx)i + (Fy)j
%%F = 20*cos(25/57.3)i - 20*sin(25/57.3)j

F=[ 20*cos(25/57.3) - 20*sin(25/57.3)]
%%% F = 20*cos(25/57.3)i - 20*sin(25/57.3)j


%%Calulating the moment arm BA in vector form as shown in figure.
%% BA = (x)i + (y)j
%% BA = - 9*cos(65/57.3)i + 9*sin(65/57.3)j

BA=[- 9*cos(65/57.3) 9*sin(65/57.3)]
%%% BA = -9*cos(65/57.3)i + 9*sin(65/57.3)j


%%Calculating the moment acting about point B by force of 20 LB as
%%Mb = BA x F by the cross product method as following:-
%% | i j k |
%%determinate of | BA(1) BA(2) 0 |
%% | F(1) F(2) 0 |
%%
Mb = BA(1)*F(2) - F(1)*BA(2) %%%%Mb= (Mb)k

%%Negative sign of the moment magnitude represents a clock wise acting
%%moment

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166


PROBLEM 3.6
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)cos40
0.153209 m
(0.2 m)sin 40
0.128558 mx
y=°
=
=°
=

/
(0.153209 m) (0.128558 m)
AO
∴= +rij




(a)
(300 N)sin30
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= 

79. %%%%%%%%%Problem 3.6 How to calculate the moment of force about point O as
%%%%%%%%%shown in figure.

%%Calulating the acting force of 300 N in vector form as following:-
%%F = (Fx)i + (Fy)j
%%F = 300*sin(30/57.3)i + 300*cos(30/57.3)j

F=[ 300*sin(30/57.3) 300*cos(30/57.3)]
%%% F = 300*sin(30/57.3)i + 300*cos(30/57.3)j


%%Calulating the moment arm OA in vector form as shown in figure.
%% OA = (x)i + (y)j
%% OA = 200*cos(40/57.3)i + 200*sin(40/57.3)j

OA=[200*cos(40/57.3) 200*sin(40/57.3)]
%%% OA = 200*cos(40/57.3)i + 200*sin(40/57.3)j


%%Calculating the moment acting about point O by force of 300 N as
%%Mo = OA x F by evaluating the cross product as following:-
%% | i j k |
%%Mo =determinate of | OA(1) OA(2) 0 |
%% | F(1) F(2) 0 |
%%
Mo = OA(1)*F(2) - F(1)*OA(2) %%%%Mo= (Mo)k in N.mm

Mo = Mo/1000 %%%%Mo in N.m

%%Negative sign of the moment magnitude represents a clock wise acting

%%Calculating the perpendicular distance from point O to the line of the
%%acting force of the 300 N as following:-

D = Mo/300 %%%% D in meter

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171

PROBLEM 3.11
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
MTyTx=+

(960 N)(0) (400 N)(1.90 m)
760 N m
=+
=⋅
( b)

or
760 N m
D
=⋅M


81. %%%%%%%%%Problem 3.11 How to calculate the moment of force about point D as
%%%%%%%%%shown in figure using unit vector method to resolve the tension in
%%%%%%%%%cable BC into the horizontal and vertical components

%%unit vector of CB = unit vector of CE ---> lamdaCB = lamdaCE therefore
%%lamdaCE can be easily calculated from the geometry
%%CE = ((- d)i + 0j) - ((o.2)i + (0.857)j)
d=1.9 %%%%d = 1.9 meter
CE = [(- d - 0.2) (0 - 0.875)]
lamdaCE =CE./(CE(1)^2 + CE(2)^2)^0.5


%%Calulating the acting tension of 1040 N in vector form as following:-
%%T = (Fx)i + (Fy)j
%%T = 1040*lamdaCE
T = 1040*lamdaCE

%%Calulating the moment arm DC in vector form as shown in figure.
%% DC = (x)i + (y)j
%% DC = (0.2)i + (0.875)j
DC = [(0.2) (0.875)]

%%Calculating the moment acting about point D by force of 1040 N as
%%MD = DC x T by evaluating the cross product as following:-
%% | i j k |
%%MD =determinate of | DC(1) DC(2) 0 |
%% | T(1) T(2) 0 |
%%
MD1 = DC(1)*T(2) - T(1)*DC(2) %%%%MD1 = (MD1)k in N.m
%%Negative sign of the moment magnitude represents a clock wise acting

%%Calulating the moment arm DE in vector form as shown in figure.
%% DE = (x)i + (y)j
%% DE = (-1.9)i (0)j
DE = [(- 1.9) (0)]
%%Calculating the moment acting about point D by force of 1040 N as
%%MD = DE x T by evaluating the cross product as following:-
%% | i j k |
%%MD =determinate of | DE(1) DE(2) 0 |
%% | T(1) T(2) 0 |
%%
MD2 = DE(1)*T(2) - T(1)*DE(2) %%%%MD2= (MD2)k in N.m
%%Negative sign of the moment magnitude represents a clock wise acting

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172

PROBLEM 3.12
It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If
d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the
required moment about Point D.


SOLUTION

Slope of line:
0.875 m 7
2.80 m 0.2 m 24
EC==
+
Then
24
25
ABx AB
TT=
and
7
25
ABy AB
TT=
We have
() ()
D ABx ABy
MTyTx=+

24 7
960 N m (0) (2.80 m)
25 25
1224 N
AB AB
AB
TT
T⋅= +
=
or 1224 N
AB
T= 

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173

PROBLEM 3.13
It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D .


SOLUTION

The minimum value of d can be found based on the equation relating the moment of the force
AB
Tabout D :

max
()()
DABy
MT d=
where
960 N m
D
M=⋅

max max
( ) sin (2400 N)sin
AB y AB
TT θθ==
Now
22
22
0.875 m
sin
( 0.20) (0.875) m
0.875
960N m 2400N ( )
( 0.20) (0.875)
d
d
d
θ=
++
 
 ⋅=
 ++
 

or
22
( 0.20) (0.875) 2.1875dd++ =
or
22 2
( 0.20) (0.875) 4.7852dd++ =
or
2
3.7852 0.40 0.8056 0dd−− =
Using the quadratic equation, the minimum values of d are 0.51719 m and
0.41151 m.−
Since only the positive value applies here,
0.51719 md=
or
517 mmd= 

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175


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
α
β αβ αβ=++−

SOLUTION
Note: (cos sin )
(cos sin )
(cos sin )
B
B
Cββ
ββ
αα
=+
′=−
=+Bij
Bij
Cij
By definition,
||sin()BC α
β×= −BC (1)

||sin()BC α
β′×= +BC (2)
Now
(cos sin ) (cos sin )BC
ββ αα×= + × +BC i j i j

(cos sin sin cos )BC
βαβα=− k (3)
and
(cos sin ) (cos sin )BC
ββ αα′×= − × +BC i j i j

(cos sin sin cos )BC
βαβα=+ k (4)
Equating the magnitudes of
×BCfrom Equations (1) and (3) yields:

sin( ) (cos sin sin cos )BC BCα
ββ αβα−= − (5)
Similarly, equating the magnitudes of
′×BC from Equations (2) and (4) yields:

sin( ) (cos sin sin cos )BC BCα
ββ αβα+= + (6)
Adding Equations (5) and (6) gives:

sin( ) sin( ) 2cos sinα
β αββ α−+ +=
or
11
sin cos sin( ) sin( )
22
α
β αβ αβ=++− 

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176

PROBLEM 3.16
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.

SOLUTION
(a) We have ||A=×PQ
where
733=− + −Pijk

225=++Qijk
Then
73 3
225
[(15 6) ( 6 35) ( 14 6) ]
(21) (29) ( 20)
×=− −
=++−+ +−−
=+ −
ijk
PQ
ijk
ijk

22 2
(20) (29) ( 20)A=++− or 41.0A= 
(b) We have
||A=×PQ
where
652=−−Pijk

251=− + −Qijk
Then
652
25 1
[(5 10) (4 6) (30 10) ]
(15) (10) (20)
×= − −
−−
=+ +++ − =++
ijk
PQ
ij k
ijk

22 2
(15) (10) (20)A=++ or 26.9A= 

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177

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) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.

SOLUTION
(a) We have
||
×
=
×
AB
λ
AB

where
125=+ −Ai jk

475=−−Bijk
Then
125
475
(1035) (205) (78)
15(3 1 1 )
×= + −
−−
=− − + + +−−
=−−
ijk
AB
ijk
ijk
and
222
| | 15(3) (1) (1) 1511×= − +− +− =AB

15( 3 1 1 )
15 11
−−−
=
ijk
λ or
1
(3 )
11
=−−−
λ ijk 
(b) We have ||
×
=
×
AB
λ
AB

where
332=−+Aijk

264=− + −Bijk
Then
332
26 4
(12 12) ( 4 12) (18 6) (8 12 )
×= −
−−
=− +−+ +− =+
ijk
AB
ijk
jk
and
22
| | 4 (2) (3) 4 13×= + =AB

4(2 3 )
413
+
=
jk
λ or
1
(2 3 )
13
=+
λ jk 

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179

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
=−
−
ijk
M

(15 12) (16 10) (6 12)=− +−− +−−ijk 32618
O
=− −Mijk 
(b)
86 10
435
O
=− −
−
ijk
M

(30 30) ( 40 40) (24 24)=− +−+ +−ijk 0
O
=M 
(c)
865
435
O
=−
−
ijk
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.

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180

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.

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213


PROBLEM 3.53
A single force P acts at C in a direction perpendicular to the
handle BC of the crank shown. Knowing that M
x = +20 N ·
m and M y = −8.75 N · m, and M z = −30 N · m, determine the
magnitude of P and the values of
φ and θ.

SOLUTION

(0.25 m) (0.2 m)sin (0.2 m)cos
sin cos
0.25 0.2sin 0.2cos
0sincos
C
OC
PP
PP
θθ
φφ
θθ
φφ=+ +
=− +
=×=
−
ri j k
Pjk
ij k
MrP

Expanding the determinant, we find

(0.2) (sin cos cos sin )
x
MP θ
φθφ=+

(0.2) sin( )
x
MP θ
φ=+ (1)

(0.25) cos
y
MP
φ=− (2)

(0.25) sin
z
MP
φ=− (3)
Dividing Eq. (3) by Eq. (2) gives: tan
z
y
M
M
φ= (4)

30 N m
tan
8.75 N m
φ
−⋅
=
−⋅


73.740
φ= 73.7φ=° 
Squaring Eqs. (2) and (3) and adding gives:

22 22 22
(0.25) or 4
yz yz
MM P P MM+= = + (5)

22
4 (8.75) (30)
125.0 N
P=+
=
125.0 NP= 
Substituting data into Eq. (1):

( 20 N m) 0.2 m(125.0 N)sin( )
( ) 53.130 and ( ) 126.87
20.6 and 53.1 θ
φ
θφ θφ
θθ
+⋅= +
+= ° += °
=− ° = °


53.1Q=° 

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217


PROBLEM 3.57
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

λ
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)
47DG
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
−−
=⋅ ×=
−
Mrλ

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 

91. %%%%%%%%%Problem 3.57 How to calculate the moment of force
%%%%%%%%%about a definite straight line AB
%%%%%%%%%shown in figure using unit vector method to resolve the
%%%%%%%%%force acting P into horizontal and vertical component and resolve
%%%%%%%%%the rod AB into horizontal and vertical compnents of unity

%%unit vector of AB where AB = 32i - 30j - 24j
AB = [32 - 30 -24]
lamdaAB =AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5

%%unit vector of DG where DG = 16i - 38j + 18j
DG = [21 - 38 18]
lamdaDG =DG./(DG(1)^2 + DG(2)^2 + DG(3)^2)^0.5

%%Calulating the acting force P of 235 LB in vector form as following:-
%%P = (Fx)i + (Fy)j + (Fz)K
%%P = 235*lamdaDG
P = 235*lamdaDG

%%Calulating the moment arm CD in vector form as shown in figure.
%% CD = (x)i + (y)j + (z)k
%% CD= 0i + 23j + 0k
CD = 23 %%%% DC = 23j

%%Calculating the moment acting about point C by force of 235 LB as
%%MC = CD x P by evaluating the cross product as following:-
%% | i j k |
%%MC =determinate of | 0 CD 0 |
%% | P(1) P(2) P(3)|
%%

MC = [CD*P(3) - CD*P(1)] %%%%MC = (MCx)i + (MCy)k in LB.in
%%Negative sign of the moment magnitude represents a clock wise acting


%%Calculating the moment of force P about AB Rod
Mab=lamdaAB(1)*MC(1) + lamdaAB(3)*MC(2)%%%%%%Scalar value of acting
%%moment by force P about the rod AB in LB.in

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218


PROBLEM 3.58
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

λ
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)
29DH 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
λ

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 

93. %%%%%%%%%Problem 3.58 How to calculate the moment of force
%%%%%%%%%about a definite straight line AB
%%%%%%%%%shown in figure using unit vector method to resolve the
%%%%%%%%%force acting Q into horizontal and vertical component and resolve
%%%%%%%%%the rod AB into horizontal and vertical compnents of unity

%%unit vector of AB where AB = 32i - 30j - 24j
AB = [32 - 30 -24]
lamdaAB =AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5

%%unit vector of DH where DH = - 16i - 21j - 12j
DH = [- 16 -21 -12]
lamdaDH =DH./(DH(1)^2 + DH(2)^2 + DH(3)^2)^0.5

%%Calulating the acting force Q of 174 LB in vector form as following:-
%%Q = (Fx)i + (Fy)j + (Fz)K
%%Q = 174*lamdaDH
Q = 174*lamdaDH

%%Calulating the moment arm CD in vector form as shown in figure.
%% CD = (x)i + (y)j + (z)k
%% CD= 0i + 23j + 0k
CD = 23 %%%% DC = 23j

%%Calculating the moment acting about point C by force of 174 LB as
%%MC = CD x Q by evaluating the cross product as following:-
%% | i j k |
%%MC =determinate of | 0 CD 0 |
%% | Q(1) Q(2) Q(3)|
%%

MC = [CD*Q(3) - CD*Q(1)] %%%%MC = (MCx)i + (MCy)k in LB.in
%%Negative sign of the moment magnitude represents a clock wise acting


%%Calculating the moment of force Q about AB Rod
Mab=lamdaAB(1)*MC(1) + lamdaAB(3)*MC(2) %%%%%%Scalar value of acting
%%moment by force Q about the rod AB in LB.in

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271

PROBLEM 3.107
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
MWΣ= =
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
MWΣ= =
Then
(84 lb)(6 ft) 52 lb( ) 64 lb(6 ft) 0d−− =

2.31 ft to the right of dC= 

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272


PROBLEM 3.108
A couple of magnitude M = 54 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)(30cos60)
30 sin 60 ( 45 )
(30 lb) (15.9808 lb)
Σ=−+ °
+°+−
=− +
FR j i
ji
ij
or
34.0 lb=R
28.0° 
(b) First reduce the given forces and couple to an equivalent force-couple system
(, )
B
RM at B.
We have
: (54 lb in) (12 in.)(10 lb) (8 in.)(45 lb)
186 lb in.
BB
MMΣ=⋅+ −
=− ⋅
Then with R at D,
: 186 lb in (15.9808 lb)
B
MaΣ−⋅=
or
11.64 in.a=
and with R at E,
: 186 lb in (30 lb)
B
MCΣ−⋅=
or
6.2 in.C=
The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in.
below B . 

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278


PROBLEM 3.114
Four ropes are attached to a crate and exert the forces shown.
If the forces are to be replaced with a single equivalent force
applied at a point on line AB, determine (a) the equivalent
force and the distance from A to the point of application of
the force when
30 ,α=° (b) the value of α so that the single
equivalent force is applied at Point B.


SOLUTION
We have

(a) For equivalence,
: 100 cos30 400 cos65 90 cos65
xx
FRΣ− °+ °+ °=
or
120.480 lb
x
R=

: 100 sin 160 400 sin 65 90 sin 65
yy
FR αΣ++° +° =
or
(604.09 100sin ) lb
y
R α=+ (1)
With
30 ,α=° 654.09 lb
y
R=
Then
22 654.09
(120.480) (654.09) tan
120.480
665 lb or 79.6
R θ
θ=+ =
==°
Also
: (46 in.)(160 lb) (66 in.)(400 lb)sin 65
(26 in.)(400 lb)cos65 (66 in.)(90 lb)sin 65
(36 in.)(90 lb)cos65 (654.09 lb)
A
M
dΣ+°
+°+°
+°=

or
42,435 lb in. and 64.9 in.
A
MdΣ= ⋅ = 665 lbR=
79.6° 
and R is applied 64.9 in. to the right of A. 
(b) We have
66 in.d=
Then
: 42,435 lb in (66 in.)
Ay
MRΣ⋅ =
or
642.95 lb
y
R=
Using Eq. (1):
642.95 604.09 100sinα=+ or 22.9α=° 

97. %%%%%%%%%Problem 3.114 How to calculate the equivalent force of a proposed
%%%%%%%%%system of forces

%%Considering an equivalent force R has two components Rx & Ry for the
%%proposed system of forces ie. Summation of all forces in X direction is
%%equal to Rx and Summation of all forces in Y direction is equal to Ry
Rx=90*cos(65/57.3) + 400*cos(65/57.3) - 100*cos(30/57.3)
Ry=90*sin(65/57.3) + 400*sin(65/57.3) + 100*sin(30/57.3) + 160

%%Calculating equivalent force angle by using tan(theta)=Ry/Rx
theta = atan(Ry/Rx)*57.3 %%%%theta in Degrees

%%Calculating the application point of the R over the edge AB
%%Summation of the system moments = the equivalent moment by the equivalent
%%force R we can take the moments about any arbitrary point ex. point A

%%Assume D is the distance between the equivalent force and point A

D =(160*46 + 90*sin(65/57.3)*66 + 400*sin(65/57.3)*66 + 90*cos(65/57.3)*36 +
400*cos(65/57.3)*26)/Ry


%%Considering D = 66 in -----> the application angle of the equivalent
%%force and alfa angle could be calculated as following:-

Ry =(160*46 + 90*sin(65/57.3)*66 + 400*sin(65/57.3)*66 + 90*cos(65/57.3)*36 +
400*cos(65/57.3)*26)/66

sinalfa=(90*sin(65/57.3) + 400*sin(65/57.3) + 160 - Ry)/(- 100)

alfa=asin(sinalfa)*57.3

Rx=90*cos(65/57.3) + 400*cos(65/57.3) - 100*cos(alfa/57.3)

R = (Rx^2 + Ry^2)^0.5

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306


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 RΣ− − = =FjkR
and
:( )
R
OO CO
ΣΣ×+Σ=MrFMM

0.6 0 0.1 0.4 0.3 0 ( 30 32 ) N m
0840 0 080
R
O
++ −−⋅=
ijk ijk
jkM

(15.6Nm) (2Nm) (82.4Nm)
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
Mλ==− ⋅− ⋅Mjk
Then pitch
155.379 N m
0.47741 m
116 N
M
p
R
⋅
== =
or 0.477 mp= 

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307
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 mz
z−=
=−
or
0.1857 mz=−
From k:
44.208 84
0.52629 mx
x−=−
=
or
0.526 mx=
The axis of the wrench intersects the xz-plane at

0.526 m 0 0.1857 mxyz===− 

100. %%%%%%%%%Problem 3.137 How to calculate the equivalent wrench
%%(Force - Couple)

%%of a proposed system of forces and moments

%%Calculating the equivalent force at the orgin
Rx = 0
Ry = - 84
Rz = - 80
R = (Rx^2 + Ry^2 + Rz^2)^0.5
%% ie. R = 0i + (Ry)j + (Rz)k

%%%Calculating the unit vector of R
lamdaR=[Rx Ry Rz]./(Rx^2 + Ry^2 + Rz^2)^0.5

%%Calculating the equivalent moment at the orgin in form of two components
%%in Y & Z directions
Mox = 84*0.1 - 80*0.3
Moy = - 30 + 80*0.4
Moz = - 32 - 84*0.6
MoMAG = (Mox^2 + Moy^2 + Moz^2)^0.5
Mo = [Mox Moy Moz]
%% ie. Mo = (Mox)i + (Moy)j + (Moz)k

%%%Calculating the component of Mo in the direction of R
M1MAG = lamdaR(1)*Mo(1) + lamdaR(2)*Mo(2) + lamdaR(3)*Mo(3)
%%%Scalar value of M1 on N.m


M1 = M1MAG*lamdaR %%%%Vector form in N.m

M2 = Mo - M1 %%%% in N.m

Pitch = M1/R %%%% in meters

%%%Calculating the shift of the wrench in xz plane through M2 = rxz X R
%%% | i j k |
%%% M2 = Determinate | rx ry rz| = (M2x)i + (M2y)j + (M2z)k
%%% | Rx Ry Rz|


rx = M2(3)/Ry %%% in meters
rz = - M2(1)/Ry %%% in meters

101. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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308


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
Σ− − + =


FkijR

(8.00 lb) (15.00 lb) (26.4 lb)=− − −Rijk 
and
31.4 lbR=
We have
/
:
R
BABA A B B
Σ×++=Mr FMMM

815
0 10 0 220 238 264 220 14(8 15 )
17 17
0 0 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

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309
PROBLEM 3.138* (Continued)

and
11
(62.818 lb in.) (117.783 lb in.) (207.30 lb in.)
R
Mλ==− ⋅− ⋅− ⋅Mijk
Then pitch
1246.56 lb in.
7.8522 in.
31.4 lb
M
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=== 

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341

PROBLEM 3.158
A concrete foundation mat in the shape of a regular hexagon of side
12 ft supports four column loads as shown. Determine the magnitudes
of the additional loads that must be applied at B and F if the resultant
of all six loads is to pass through the center of the mat.

SOLUTION
From the statement of the problem, it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that

0or 0 0
Oxz
MMΣ= Σ=Σ=M
For the applied loads:

Then 0: (6 3 ft) (6 3 ft)(10 kips) (6 3 ft)(20 kips)
(6 3 ft) 0
xB
F
F
F
Σ= + −
−=M
or
10
BF
FF−= (1)

0: (12 ft)(15 kips) (6 ft) (6 ft)(10 kips)
(12 ft)(30 kips) (6 ft)(20 kips) (6 ft) 0
zB
F
F
F
Σ= + −
−−+=M

or
60
BF
FF+= (2)
Then Eqs.
(1) (2)+  35.0 kips
B
=F

and
25.0 kips
F
=F


104. Chapter 4
Equilibrium of Rigid Bodies


4.1 INTRODUCTION
We saw in the preceding chapter that the external forces acting on a rigid body can
be reduced to a force-couple system at some arbitrary point O. When the force and
the couple are both equal to zero, the external forces form a system equivalent to
zero, and the rigid body is said to be in equilibrium.



Resolving each force and each moment into its rectangular components, we can
express the necessary and sufficient conditions for the equilibrium of a rigid body
with the following six scalar equations:



4.2 REACTIONS AT SUPPORTS AND CONNECTIONS FOR A 2
DIMENSIONAL STRUCTURE
In the first part of this chapter, the equilibrium of a two- dimensional structure is
considered; i.e., it is assumed that the structure being analyzed and the forces
applied to it are contained in the same plane. Clearly, the reactions needed to
maintain the structure in the same position will also be contained in this plane. The
reactions exerted on a two- dimensional structure can be divided into three groups
corresponding to three types of supports, or connections as following:

105. Table 4.1

106. 4.3 STATICALLY INDETERMINATE REACTIONS.
PARTIAL CONSTRAINTS
In the two examples considered in the preceding section (Figs. 4.2 and 4.3) , the
types of supports used were such that the rigid body could not possibly move under
the given loads or under any other loading conditions. In such cases, the rigid body
is said to be completely constrained. We also recall that the reactions
corresponding to these supports involved three unknowns and could be determined
by solving the three equations of equilibrium. When such a situation exists, the
reactions are said to be statically determinate. Consider Fig. 4.4a, in which the
truss shown is held by pins at A and B. These supports provide more constraints
than are necessary to keep the truss from moving under the given loads or under
any other loading conditions. We also note from the free-body diagram of Fig.
4.4b that the corresponding reactions involve four unknowns. Since, only three
independent equilibrium equations are available, there are more unknowns than
equations; thus, all of the unknowns cannot be determined. While the equations
∑????????????
????????????=0 and ∑????????????
????????????=0 yield the vertical components B y and A y, respectively,
the equation ∑????????????
???????????? =0 gives only the sum A x + B x of the horizontal components of
the reactions at A and B. The components A
x and B x

are said to be statically
indeterminate. They could be determined by considering the deformations
produced in the truss by the given loading, but this method is beyond the scope of
statics and belongs to the study of mechanics of materials.

107. The supports used to hold the truss shown in Fig. 4.5a,b consist of rollers at A and
B. Clearly, the constraints provided by these supports are not sufficient to keep the
truss from moving. While any vertical motion is prevented, the truss is free to
move horizontally. The truss is said to be partially constrained.

The examples of Figs. 4.6 and 4.7 lead us to conclude that a rigid body is
improperly constrained whenever the supports, even though they may provide a
sufficient number of reactions, that are arranged in such a way that the reactions
must be either concurrent or parallel.

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345


PROBLEM 4.1
Two crates, each of mass 350 kg, are placed as shown in the
bed of a 1400-kg pickup truck. Determine the reactions at each
of the two (a) rear wheels A, (b) front wheels B.

SOLUTION
Free-Body Diagram:


2
2
(350 kg)(9.81 m/s ) 3.4335 kN
(1400 kg)(9.81 m/s ) 13.7340 kN
t
W
W
==
==

(a) Rear wheels
: 0: (1.7 m 2.05 m) (2.05 m) (1.2 m) 2 (3 m) 0
Bt
MW W W AΣ= + + + − =

(3.4335 kN)(3.75 m) (3.4335 kN)(2.05 m)
(13.7340 kN)(1.2 m) 2 (3 m) 0A
+
+−=


6.0659 kNA=+ 6.07 kN=A

(b) Front wheels: 0: 2 2 0
yt
FWWWABΣ= −−− + + =

3.4335 kN 3.4335 kN 13.7340 kN 2(6.0659 kN) 2 0B−− − + +=

4.2346 kNB=+ 4.23 kN=B


109. %%Problem 4.1 How to calculate the ground reactions on a truck of 1400 kg
%%As long as the case is in equilibrium state it is available to use the
%%relevant equilibrium equations ---> Summation of Forces = 0 & Summation
%%of Moment at any point = 0

%%Calculating the moments about point A
%% Rb*3 - 1400*1.8 - 350*(3.75 - 2.8) + 350*(4.5 - 3.75) = 0

Rb = ((1400*1.8 + 350*(3.75 - 2.8) - 350*(4.5 - 3.75))/3) %%% in kg
Ra = (2*350 + 1400 - Rb) %%% in kg

Rb = (Rb/2)*9.8 %%%As we have 2 wheels in front in N
Ra = (Ra/2)*9.8 %%%As we have 2 wheels in rear in N

%%Calculating the maximum available distance between C & D
%%Rb should be estimated to equal 0 to calculate the maximum available
%%distance CD keeping weight in D location.

CD = (1400*1.8 + 350*(3.75 - 2.8))/350 - 2.8 + 3.75 %%%% in meters

for I=1:1:91

CD1(I) = I*0.1;

Rb1(I) = ((1400*1.8 + 350*(3.75 - 2.8) - 350*(2.8 + CD1(I) - 3.75))/3); %%%
in kg

Ra1(I) = (2*350 + 1400 - Rb1(I)); %%% in kg
Rb1(I) = (Rb1(I)/2)*9.8; %%%As we have 2 wheels in front in N
Ra1(I) = (Ra1(I)/2)*9.8; %%%As we have 2 wheels in rear in N
end

figure(1);
plot(CD1,Rb1,'o',CD1,Ra1,'*');
xlabel('CD Distance in meters '),grid on;ylabel('Reaction on one wheel in
front in N'),grid on;

xlim([0 10])
ylim([0 11000])

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346


PROBLEM 4.2
Solve Problem 4.1, assuming that crate D is removed and that
the position of crate C is unchanged.
PROBLEM 4.1 Two crates, each of mass 350 kg, are placed
as shown in the bed of a 1400-kg pickup truck. Determine the
reactions at each of the two (a) rear wheels A, (b) front wheels B .

SOLUTION
Free-Body Diagram:


2
2
(350 kg)(9.81 m/s ) 3.4335 kN
(1400 kg)(9.81 m/s ) 13.7340 kN
t
W
W
==
==

(a) Rear wheels
: 0: (1.7 m 2.05 m) (1.2 m) 2 (3 m) 0
Bt
MW W AΣ= + + − =

(3.4335 kN)(3.75 m) (13.7340 kN)(1.2 m) 2 (3 m) 0A+−=

4.8927 kNA=+ 4.89 kN=A

(b) Front wheels: 0: 2 2 0
yt
MWWABΣ= −−++=

3.4335 kN 13.7340 kN 2(4.8927 kN) 2 0B−− + +=

3.6911 kNB=+ 3.69 kN=B


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347


PROBLEM 4.3
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
FBΣ= =
0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0
B
Ma a AΣ= − − + + =

(40 160)
12
a
A
−
=
(1)

0: (40 lb)(6 in.) (50 lb)(12 in.) (30 lb)( 12 in.)
(10 lb)( 20 in.) (12 in.) 0
A
y
Ma
aB
Σ= − − − +
−++ =

(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


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348


PROBLEM 4.4
For the bracket and loading of Problem 4.3, determine the smallest
distance a if the bracket is not to move.
PROBLEM 4.3 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
Ma a AΣ= − − + + =

(40 160)
12
a
A
−
=


0: (40 160) 0Aa=−= 4.00 in.a= 

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349


PROBLEM 4.5
A hand truck is used to move two kegs, each of mass 40 kg.
Neglecting the mass of the hand truck, determine (a) the vertical
force P that should be applied to the handle to maintain
equilibrium when
35 ,α=° (b) the corresponding reaction at each
of the two wheels.

SOLUTION
Free-Body Diagram:

2
1
2
(40 kg)(9.81 m/s ) 392.40 N
(300 mm)sin (80 mm)cos
(430 mm)cos (300 mm)sin
(930 mm)cos
Wmg
a
a
b
αα
αα
α
== =
=−
=−
=


From free-body diagram of hand truck,

Dimensions in mm

21
0: ( ) ( ) ( ) 0
B
MPbWaWaΣ= − + = (1)
0: 2 2 0
y
FPWBΣ= − + = (2)
For
35α=°

1
2
300sin35 80cos35 106.541 mm
430cos35 300sin 35 180.162 mm
930cos35 761.81 mm
a
a
b
=°−°=
=°−°=
=°=

(a) From Equation (1):

(761.81 mm) 392.40 N(180.162 mm) 392.40 N(106.54 mm) 0P −+=

37.921 NP= or 37.9 N=P

(b) From Equation (2):

37.921 N 2(392.40 N) 2 0B−+= or 373 N=B


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350


PROBLEM 4.6
Solve Problem 4.5 when α = 40°.
PROBLEM 4.5 A hand truck is used to move two kegs, each of
mass 40 kg. Neglecting the mass of the hand truck, determine
(a) the vertical force P that should be applied to the handle to
maintain equilibrium when
α = 35°, (b) the corresponding reaction
at each of the two wheels.

SOLUTION
Free-Body Diagram:


2
1
2
(40 kg)(9.81 m/s )
392.40 N
(300 mm)sin (80 mm)cos
(430 mm)cos (300 mm)sin
(930 mm)cos
Wmg
W
a
a
b
αα
αα
α
==
=
=−
=−
=



From F.B.D.:

21
0: ( ) ( ) ( ) 0
B
MPbWaWaΣ= − + =

21
()/PWa ab=−
(1)


0: 2 0
y
FWWPBΣ= −−++ =


1
2
BW P=−
(2)
For
40 :α=°

1
2
300sin 40 80cos40 131.553 mm
430cos 40 300sin 40 136.563 mm
930cos 40 712.42 mm
a
a
b
=°−°= =°−°=
=°=

(a) From Equation (1):
392.40 N (0.136563 m 0.131553 m)
0.71242 m
P
−
=


2.7595 NP= 2.76 N=P

(b) From Equation (2):
1
392.40 N (2.7595 N)
2
B=−
391 N=B


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351


PROBLEM 4.7
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
MAΣ= + − =

1761.11lbA=+ 1761lb=A

(b) Rear wheels: 0: 1700 lb 3200 lb 2(1761.11 lb) 2 0
y
FBΣ= − − + + =

688.89 lbB=+ 689 lb=B


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352


PROBLEM 4.8
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
FAΣ= =
0: (15 lb)(28 in.) (20 lb)(22 in.) (35 lb)(14 in.)
(20 lb)(6 in.) (6 in.) 0
B
y
M
A
Σ= + +
+−=

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
Σ= + +
−−=

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
yB C
FA FΣ= − − = − +− =
−+ −=

0 0 (Checks)=

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353


PROBLEM 4.9
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 BΣ= −− − − + =

600 2800 16 0aB−+ +=

(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≤≤ 

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354


PROBLEM 4.10
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
FBΣ= =

y
BB=

0: (50 N) (100 N)(0.45 m ) (150 N)(0.9 m ) (0.9 m ) 0
A
Md d dBdΣ= − −− −+ −=

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
MA dΣ= − −+ =

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 mmd≥ 
Since
180 N,A≤ Eq. (2) yields

(0.9)180 90 72
0.40 m
180 180
d
−
≤==
400 mmd≤ 
Range:
150.0 mm 400 mmd≤≤ 

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355


PROBLEM 4.11
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 P = 0.

SOLUTION

0: (3.00 kN)(0.500 m) (2.25 m) (3.00 m) 0
BD
MT QΣ= + − =

0.500 kN (0.750)
D
QT=+ (1)

0: (3.00 kN)(2.75 m) (2.25 m) (0.750 m) 0
DB
MT QΣ= − − =

11.00 kN (3.00)
B
QT=− (2)
For cable B not to be slack,
0,
B
T≥ and from Eq. (2),

11.00 kNQ≤
For cable D not to be slack,
0,
D
T≥ and from Eq. (1),

0.500 kNQ≥
For neither cable to be slack,

0.500 kN 11.00 kNQ≤≤ 

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356


PROBLEM 4.12
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 4 kN and neglecting the weight of the beam,
determine the range of values of Q for which the loading is safe
when P = 0.

SOLUTION

0: (3.00 kN)(0.500 m) (2.25 m) (3.00 m) 0
BD
MT QΣ= + − =

0.500 kN (0.750)
D
QT=+ (1)

0: (3.00 kN)(2.75 m) (2.25 m) (0.750 m) 0
DB
MT QΣ= − − =

11.00 kN (3.00)
B
QT=− (2)
For
4.00 kN,
B
T≤ Eq. (2) yields

11.00 kN 3.00(4.00 kN)Q≥− 1.000 kNQ≥−
For
4.00 kN,
D
T≤ Eq. (1) yields

0.500 kN 0.750(4.00 kN)Q≤+ 3.50 kNQ≤
For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11,

0.500 kN 3.50 kNQ≤≤ 

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357


PROBLEM 4.13
For the beam of Problem 4.12, determine the range of values of Q
for which the loading is safe when P = 1 kN.
PROBLEM 4.12 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 4 kN and neglecting the weight of
the beam, determine the range of values of Q for which the loading
is safe when P = 0.

SOLUTION

0: (3.00 kN)(0.500 m) (1.000 kN)(0.750 m) (2.25 m) (3.00 m) 0
B D
MT QΣ= − + − =

0.250 kN 0.75
D
QT=+ (1)

0: (3.00 kN)(2.75 m) (1.000 kN)(1.50 m)
(2.25 m) (0.750 m) 0
D
B
M
TQ
Σ= +
−− =

13.00 kN 3.00
B
QT=− (2)
For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN.
Making
04.00 kN
B
T≤≤ in Eq. (2), we have

13.00 kN 3.00(4.00 kN) 13.00 kN 3.00(0)Q−≤≤−


1.000 kN 13.00 kNQ≤≤ (3)
Making
04.00 kN
D
T≤≤ in Eq. (1), we have

0.250 kN 0.750(0) 0.250 kN 0.750(4.00 kN)Q+≤≤ +


0.250 kN 3.25 kNQ≤≤ (4)
Combining Eqs. (3) and (4),
1.000 kN 3.25 kNQ≤≤ 

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358


PROBLEM 4.14
For the beam of Sample Problem 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 30 kips and that the
reaction at A must be directed upward.

SOLUTION
0: 0
xx
FBΣ= =

y
B=B

0: (3 ft) (9 ft) (6 kips)(11 ft) (6 kips)(13 ft) 0
A
MPBΣ= − + − − =

348kipsPB=− (1)

0: (9 ft) (6 ft) (6 kips)(2 ft) (6 kips)(4 ft) 0
B
MAPΣ= − + − − =

1.5 6 kipsPA=+ (2)
Since
30 kips,B≤ Eq. (1) yields

(3)(30 kips) 48 kipsP≤− 42.0 kipsP≤ 
Since
030kips,A≤≤ Eq. (2) yields

0 6 kips (1.5)(30 kips)1.6 kipsP+≤≤

6.00 kips 51.0 kipsP≤≤ 
Range of values of P for which beam will be safe:

6.00 kips 42.0 kipsP≤≤ 

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359


PROBLEM 4.15
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
MTΣ= − − =

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
FCTΣ= −=

1600 N 0 1600 N
xx
CC−==+ 1600 N
x
=C

0: 240 N 240 N 0
yyy
FCTΣ= −− − =

1200 N 480 N 0
y
C−−=

1680 N
y
C=+ 1680 N
y
=C


46.4
2320 NCα=°
=
2.32 kN=C
46.4° 

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360


PROBLEM 4.16
Solve Problem 4.15, assuming that 0.32 m.a=
PROBLEM 4.15 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: (0.32 m) (240 N)(0.4 m) (240 N)(0.8 m) 0
Cx
MTΣ= − − =

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
FCTΣ= −=

900 N 0 900 N
xx
CC−==+ 900 N
x
=C

0: 240 N 240 N 0
yyy
FCTΣ= −− − =

1200 N 480 N 0
y
C−−=

1680 N
y
C=+ 1680 N
y
=C


61.8
1906 NCα=°
=
1.906 kN=C
61.8° 

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361


PROBLEM 4.17
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
MTΣ= − =

150.0 lbT= 
(b)
3
0: 100 lb (150.0 lb) 0
5
xx
FCΣ= + + =

190 lb
x
C=− 190 lb
x
=C


4
0: lb) 0
5
yy
FCΣ = + (150.0 =

120 lb
y
C=− 120 lb
y
=C


32.3α=° 225 lbC= 225 lb=C
32.3° 

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362


PROBLEM 4.18
The lever BCD is hinged at C and attached to a control rod at B.
Determine the maximum force P that can be safely applied at D if the
maximum allowable value of the reaction at C is 250 lb.

SOLUTION
Free-Body Diagram:

0: (5 in.) (7.5 in.) 0
C
MTPΣ= − =

1.5TP=

3
0: (1.5 ) 0
5
xx
FPC PΣ= + + =

1.9
x
CP=− 1.9
x
P=C


4
0: (1.5 ) 0
5
yy
FC PΣ= + =


1.2=−
y
CP 1.2=
y
PC


22
22
(1.9 ) (1.2 )
xy
CCC
PP
=+
=+


2.2472CP=
For

250 lb,C=

250 lb 2.2472P=

111.2 lbP= 111.2 lb=P


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363


PROBLEM 4.19
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
CA B D E
MF FΣ= − =

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
FCΣ= − + =

432 N
x
C=+

4
0: (720 N) 600 N 0
5
1176 N
yy
y
FC
C
Σ= − + − =
=+

1252.84 N
69.829
C
α
=
=°


1253 N=C
69.8° 

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364


PROBLEM 4.20
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
xA BxxA B
FFCCFΣ= − + = =

4
0: 0
5
yA By DE
FFCFΣ= − + − =

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= 

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386


PROBLEM 4.38
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 . Determine the reactions at A, B, and C .

SOLUTION
Free-Body Diagram:

0: cos30 (120 lb)cos60 0
x
FAΣ= °− °=

69.28 lbA= 69.3 lb=A

0: (8 in.) (120 lb)(16 in.)cos30
(69.28 lb)(8 in.)sin30 0
B
MCΣ= − °
+°=


173.2 lbC= 173.2 lb=C
60.0° 
0: (8 in.) (120 lb)(8 in.)cos30
(69.28 lb)(16 in.)sin 30 0
C
MBΣ= − °
+°=

34.6 lbB= 34.6 lb=B
60.0° 
Check: 0: 173.2 34.6 (69.28)sin30 (120)sin 60 0
y
FΣ= − − °− °=

00(check)= 

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389


PROBLEM 4.41
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 .θ=°

SOLUTION
Free-Body Diagram:
0: cos30 20 40 0
y
FEΣ= °− − =

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
Σ= −
−+°=
D
M
CE

80 3 69.282(0.5)(3) 0C−− + =

7.9743 lbC= 7.97 lb=C


0: sin30 0
x
FE CDΣ= °+−=

(69.282 lb)(0.5) 7.9743 lb 0D+−=

42.615 lbD= 42.6 lb=D


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390


PROBLEM 4.42
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
θ 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 θΣ= − − =

60
cos
E
θ
= (1)

0: (20 lb)(4 in.) (40 lb)(4 in.) (3 in.)
60
+sin3in.0
cos
D
MC
θ
θ
Σ= − −

=


1
(180 tan 80)
3
C
θ=−
(a) For
0,C= 180tan 80θ=

4
tan 23.962
9
θθ== ° 24.0θ=° 
From Eq. (1):
60
65.659
cos 23.962
E==
°


0: sin 0
x
FDCE θΣ= −++ =

(65.659) sin 23.962 26.666 lb==D
(b)
0 26.7 lb==CD
65.7 lb=E 66.0° 

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391


PROBLEM 4.43
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 lbW=
From F.B.D. of beam AD :
0: 0
xx
FDΣ= =

0: 40 lb 40 lb 100 lb 0
yy
FDΣ= −−+ =

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
MMΣ= − +
+=

20.0 lb ft
D
M=⋅ or 20.0 lb ft
D
=⋅M


(b)
90 lbW=
From F.B.D. of beam AD :

0: 0
xx
FDΣ= =

0: 90 lb 40 lb 40 lb 0
yy
FDΣ= + − − =

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
MMΣ= − +
+=

30.0 lb ft
D
M=− ⋅ or 30.0 lb ft
D
=⋅M


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392



PROBLEM 4.44
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
MWΣ= −
+−⋅=

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
MWΣ= −
++⋅=

max
104.0 lbW= or 88.0 lb 104.0 lbW≤≤ 

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393

PROBLEM 4.45
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
FAΣ= =

0: 0
yy
FAWΣ= −=

78.480 N
y
=A

0: (1.6 m) 0
AA
MMWΣ= − =


(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
FAWΣ= −=

78.480
x
=A
0: 0
yy
FAWΣ= −=

78.480
y
=A
(78.480 N) 2 110.987 N==A 45°
0: (1.6 m) 0
AA
MMWΣ= − =

(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
FAΣ= =

0: 2 0
yy
FAWΣ= − =

2 2(78.480 N) 156.960 N
y
AW== =

0: 2 (1.6 m) 0
AA
MMWΣ= − =

2(78.480 N)(1.6 m)
A
M=+ 251.14 N m
A
=⋅M


157.0 N=A
251 N m
A
=⋅M 

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394



PROBLEM 4.46
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, determine the reaction at C.

SOLUTION
Free-Body Diagram:


0: (20 N) 0
xx
FCΣ= + =

20 N
x
C=−
0: (20 N) 0
yy
FCΣ= − =

20 N
y
C=+

28.3 N=C
45.0° 

0: (20 N)(0.160 m) (20 N)(0.055 m) 0
CC
MMΣ= + + =



4.30 N m
C
M=− ⋅ 4.30 N m
C
=⋅M


136. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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396


PROBLEM 4.48
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
FEΣ= ∴ =
0: 3600 lb 1200 lb 4000 0
yy
FEΣ= − − − =

8800 lb
y
E=

8.80 kips=E
; 36.0 kip ft
E
=⋅M 




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397
PROBLEM 4.48 (Continued)


(b) 0: (3600 lb)(12 ft) (1200 lb)(6.5 ft) 0
EE
MMΣ= + + =


51,000 lb ft
E
M=− ⋅


00
xx
FEΣ= ∴ =
0: 3600 lb 1200 lb 0
yy
FEΣ= − − =

4800 lb
y
E=

4.80 kips=E
; 51.0 kip ft
E
=⋅M 

138. %%Problem 4.48 how to calculate the reactions at point E for a balanced
%%system as shown in figure
%%The is balanced, therefore the summation of forces in both directions X,
%%Y and the summation of moment at any point = 0.
%%Calculating the moment about point E as following:-
%%Assuming the reaction torque at E is in counterclock wise direction.

for I=1:1:150
T(I) = I*100 %%% T in lbs
Me(I) = T(I)*3.75 - 3600*12 - 1200*6.5;


%%Calculating Ry at point E from the summation of forces in Y direction,
%%considering Rx = 0 from the FBD.

Ry(I) = T(I) + 3600 + 1200;
end

figure(1);
plot(T, Me,'o');
xlabel('Tension in Cable ADCF '),grid on;ylabel('Reaction Couple in
lb.ft'),grid on;

xlim([0 14000])
ylim([- 60000 0])

figure(2);
plot(T, Ry,'*');
xlabel('Tension in Cable ADCF '),grid on;ylabel('Reaction Ry in lb'),grid on;
xlim([0 6000])
ylim([4000 10000])

%%In case the cable is anchored at a point located 1 ft above point E.
%%in this case tension T is eliminated from the equations directly.


Me0 = - 3600*12 - 1200*6.5


%%Calculating Ry at point E from the summation of forces in Y direction,
%%considering Rx = 0 from the FBD.

Ry0 = 3600 + 1200

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398



PROBLEM 4.49
For the rig and crate of Prob. 4.48, 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=⋅ 

141. %%Problem 4.49 how to calculate the reactions at point E for a balanced
%%system as shown in figure
%%The is balanced, therefore the summation of forces in both directions X,
%%Y and the summation of moment at any point = 0.
%%Calculating the moment about point E as following:-
%%Assume that Me at x=1.5ft is equal to - Me at x=17.5 ft
%% i.e. T*3.75 - 3600*1.5 - 1200*6.5 = - (T*3.75 - 3600*17.5 - 1200*6.5)

T = (3600*(17.5 + 1.5) + 1200*(6.5 + 6.5))/(3.75 + 3.75)

Me = T*3.75 - 3600*1.5 - 1200*6.5
Me0 = T*3.75 - 3600*17.5 - 1200*6.5
Memin = T*3.75 - 3600*((17.5+1.5)/2) - 1200*6.5
%%Calculating Ry at point E from the summation of forces in Y direction,
%%considering Rx = 0 from the FBD.

Ry = T + 3600 + 1200

%%In Case that we Assume that Me at x=1.5ft is equal to 1.5*Me at x=17.5 ft

T1 = (3600*(- 17.5*1.5 + 1.5) + 1200*(- 6.5*1.5 + 6.5))/(1*3.75 - 1.5*3.75)

Me1 = T1*3.75 - 3600*1.5 - 1200*6.5
Me10 = T1*3.75 - 3600*17.5 - 1200*6.5

%%In Case that we Assume that Me at x=1.5ft is equal to - 1.5*Me at x=17.5 ft

T2 = (3600*(17.5*1.5 + 1.5) + 1200*(6.5*1.5 + 6.5))/(1*3.75 + 1.5*3.75)

Me2 = T2*3.75 - 3600*1.5 - 1200*6.5
Me20 = T2*3.75 - 3600*17.5 - 1200*6.5

142. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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440


PROBLEM 4.83
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


143. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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441



PROBLEM 4.84
A uniform rod AB of length 2R rests inside a hemispherical bowl of radius
R as shown. Neglecting friction, determine the angle
θ 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
α of triangle DOA is the central angle corresponding to the inscribed angleθof
triangle DCA.

2αθ=
The horizontal projections of
,( ),
AE
AE x and ,( ),
AG
AG x are equal.

AE AG A
xxx==
or
( )cos 2 ( )cosAE AGθθ=
and
(2 )cos2 cosRR θθ=
Now
2
cos 2 2cos 1θθ=−
then
2
4cos 2 cosθθ−=
or
2
4cos cos 2 0θθ−−=
Applying the quadratic equation,

cos 0.84307 and cos 0.59307θθ== −

32.534 and 126.375 (Discard)θθ=° = ° or 32.5θ=° 

144. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
448


PROBLEM 4.91
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
Cx y
DD TΣ= − × + + − ×+ − ×− =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
FΣ= 0
xx
CDT++= 1600 1200 400 N
x
C=−=

0:
y
FΣ= 720 0
yy
CD+− = 480 720 1200 N
y
C=+=

0:
z
FΣ= 0
z
C=
(b)
(400 N) (1200 N) ; (1600 N) (480 N)=+ =− −Ci jD ij 

145. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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465


PROBLEM 4.105
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
MΣ=

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
MΣ= × +× +×− =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= 

146. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
466
PROBLEM 4.105 (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
Σ= − + = =
Σ= + + − = =
Σ= − − = =


(1.800 kN) (4.80 kN)=+Cjk 

147. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
469



PROBLEM 4.107
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
MΣ=
Free-Body Diagram:


( 6 ft) (7 ft) (6 ft) 11ft
( 6 ft) (7 ft) (6 ft) 11ft
(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
A BBDBBEC
MTTΣ= ×+×+×− =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= 

148. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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470
PROBLEM 4.107 (Continued)


66
0: (1100 lb) (1100 lb) 0
11 11
xx
FAΣ= − − =

1200 lb
x
A=

77
0: (1100 lb) (1100 lb) 840 lb 0
11 11
yy
FAΣ=++−=

560 lb
y
A=−

66
0: (1100 lb) (1100 lb) 0
11 11
zz
FAΣ= + − =

0
z
A= (1200 lb) (560 lb)=−Aij 

149. Chapter 5
Distributed Forces
Centroids and Center of Gravity

5.1 INTRODUCTION
We have assumed so far that the attraction exerted by the earth on a rigid body
could be represented by a single force W. This force, called the force of gravity or
the weight of the body, was to be applied at the center of gravity of the body.
Actually, the earth exerts a force on each of the particles forming the body. The
action of the earth on a rigid body should thus be represented by a large number of
small forces distributed over the entire body. You will learn in this chapter,
however, that all of these small forces can be replaced by a single equivalent force
W. You will also learn how to determine the center of gravity, i.e., the point of
application of the resultant W, for bodies of various shapes. In the first part of the
chapter, two-dimensional bodies, such as flat plates and wires contained in a given
plane, are considered. Two concepts closely associated with the determination of
the center of gravity of a plate or a wire are introduced: the concept of the centroid
of an area or a line and the concept of the first moment of an area or a line with
respect to a given axis. You will also learn that the computation of the area of a
surface of revolution or of the volume of a body of revolution is directly related to
the determination of the centroid of the line or area used to generate that surface or
body of revolution (theorems of Pappus-Guldinus). The determination of the
centroid of an area simplifies the analysis of beams subjected to distributed loads
and the computation of the forces exerted on submerged rectangular surfaces, such
as hydraulic gates and portions of dams. In the last part of the chapter, you will
learn how to determine the center of gravity of a three-dimensional body as well as
the centroid of a volume and the first moments of that volume with respect to the
coordinate planes.

150. 5.2 CENTER OF GRAVITY OF A TWO-DIMENSIONAL BODY
Let us first consider a flat horizontal plate (Fig. 5.1). We can divide the plate into n
small elements. The coordinates of the first element are denoted by x
1 and y 1, those
of the second element by x
2 and y 2, etc. The forces exerted by the earth on the
elements of the plate will be denoted, respectively, by Δ W
1, ΔW 2, . . . , Δ W n

�????????????
????????????=????????????=∆????????????
????????????+∆????????????
????????????……..∆????????????
????????????
.
These forces or weights are directed toward the center of the earth; however, for all
practical purposes they can be assumed to be parallel. Their resultant is therefore a
single force in the same direction. The magnitude W of this force is obtained by
adding the magnitudes of the elemental weights.
To obtain the coordinates ????????????� and ????????????� of the point G where the resultant W should be
applied, we write that the moments of W about the y and x axes are equal to the
sum of the corresponding moments of the elemental weights,

�????????????
????????????=????????????�????????????=????????????
????????????∆????????????
????????????+????????????
????????????∆????????????
????????????……..????????????
????????????∆????????????
????????????

�????????????
????????????=????????????�????????????=????????????
????????????∆????????????
????????????+????????????
????????????∆????????????
????????????……..????????????
????????????∆????????????
????????????

Fig. 5.1 Center of Gravity of the Plate

151. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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553


PROBLEM 5.1
Locate the centroid of the plane area shown.

SOLUTION




2
,inA
,inx ,iny
3
,inxA
3
,inyA
1 8 0.5 4 4 32
2 3 2.5 2.5 7.5 7.5
Σ 11 11.5 39.5

XAxAΣ=

23
(11 in ) 11.5 inX = 1.045 in.X= 
YA yAΣ=Σ
(11) 39.5Y= 3.59 in.Y= 


152. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
554


PROBLEM 5.2
Locate the centroid of the plane area shown.

SOLUTION
For the area as a whole, it can be concluded by observation that

2
(72 mm)
3
Y=
or
48.0 mmY= 


Dimensions in mm





2
,mmA
,mmx
3
,mmxA
1
1
30 72 1080
2
××= 20 21,600
2
1
48 72 1728
2
××= 46 79,488
Σ 2808 101,088

Then XA xA=Σ (2808) 101,088X = or 36.0 mmX= 

153. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
555


PROBLEM 5.3
Locate the centroid of the plane area shown.

SOLUTION


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 126 54 6804×= 9 27 61,236 183,708
2
1
126 30 1890
2
××=
30 64 56,700 120,960
3
1
72 48 1728
2
××=
48 16− 82,944 27,648−
Σ 10,422 200,880 277,020

Then XA xAΣ=Σ

22
(10,422 m ) 200,880 mmX = or 19.27 mmX= 
and YA yAΣ=Σ

23
(10,422 m ) 270,020 mmY = or 26.6 mmY= 

154. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
556


PROBLEM 5.4
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
Σ 54 306 279

Then XA xA=Σ
(54) 306X = 5.67 in.X= 
(54) 279
YA yA
Y
=Σ
= 5.17 in.Y= 

155. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
557


PROBLEM 5.5
Locate the centroid of the plane area shown.

SOLUTION

By symmetry, XY=
Component
2
,inA
,in.x
3
,inxA
I Quarter circle
2
(10) 78.54
4
π
= 4.2441 333.33
II Square
2
(5) 25−=− 2.5 62.5−
Σ 53.54 270.83


23
: (53.54 in ) 270.83 inXA xA XΣ=Σ =
5.0585 in.X= 5.06 in.XY== 

156. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
558


PROBLEM 5.6
Locate the centroid of the plane area shown.

SOLUTION


2
,inA
,in.x ,in.y
3
,inxA
3
,inyA
1 14 20 280×= 7 10 1960 2800
2
2
(4) 16ππ−=− 6 12 –301.59 –603.19
Σ 229.73 1658.41 2196.8

Then
1658.41
229.73
xA
X
A
Σ
==
Σ

7.22 in.X= 

2196.8
229.73
yA
Y
A
Σ
==
Σ

9.56 in.Y= 

157. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
559


PROBLEM 5.7
Locate the centroid of the plane area shown.

SOLUTION

By symmetry, 0X=

Component
2
,inA
,in.y
3
,inyA
I Rectangle (3)(6) 18= 1.5 27.0
II Semicircle
2
(2) 6.28
2
π
−=− 2.151 13.51−
Σ 11.72 13.49

23
(11.72 in. ) 13.49 in
1.151in.
YA yA
Y
Y
Σ=Σ
=
=

0X=
1.151in.Y= 

158. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
560


PROBLEM 5.8
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
π
= 25.465 95.435
3
72.000 10×
3
269.83 10×
3
2
(60) 2827.4
4
π
−=− –25.465 25.465
3
72.000 10×
3
72.000 10−×
Σ 7200
3
72.000 10−×
3
629.83 10×

Then
3
(7200) 72.000 10XA x A X=Σ =− × 10.00 mmX=− 

3
(7200) 629.83 10YA y A Y=Σ = × 87.5 mmY= 

159. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
561

PROBLEM 5.9
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
π
−=− 163.169 43.169
3
720.86 10−×
3
190.716 10−×
Σ 5707.1
3
525.08 10×
3
132.724 10×

Then
3
(5707.1) 525.08 10XA xA X=Σ = × 92.0 mmX= 

3
(5707.1) 132.724 10YA yA Y=Σ = × 23.3 mmY= 

160. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
562

PROBLEM 5.10
Locate the centroid of the plane area shown.

SOLUTION
First note that symmetry implies 0X= 


2
,inA
,in.y
3
,inyA
1
2
(8)
100.531
2π
−=− 3.3953 –341.33
2
2
(12)
226.19
2π
= 5.0930 1151.99
Σ 125.659 810.66

Then
3
2
810.66 in
125.66 inyA
Y
AΣ
==
Σ
or
6.45 in.Y= 

161. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
563


PROBLEM 5.11
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
π
−=− 0.76394 −3.8880
Σ 12.9106 17.7120

Then
3
2
17.7120 m
12.9106 myA
Y
AΣ
==
Σ
or
1.372 mY= 

162. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
564


PROBLEM 5.12
Locate the centroid of the plane area shown.

SOLUTION


Area mm
2

,mmx ,mmy
3
,mmxA
3
,mmyA
1
31
(200)(480) 32 10
3
=×
360 60
6
11.52 10×
6
1.92 10×
2
31
(50)(240) 4 10
3
−=×
180 15
6
0.72 10−×
6
0.06 10−×
Σ
3
28 10×
6
10.80 10×
6
1.86 10×

:XA xAΣ=Σ
32 63
(28 10 mm ) 10.80 10 mmX×=×
385.7 mmX= 386 mmX= 
:YA yAΣ=Σ
32 63
(28 10 mm ) 1.86 10 mmY×=×
66.43 mmY= 66.4 mmY= 

163. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
565

PROBLEM 5.13
Locate the centroid of the plane area shown.

SOLUTION


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 (15)(80) 1200= 40 7.5
3
48 10×
3
910×
2
1
(50)(80) 1333.33
3
=
60 30
3
80 10×
3
40 10×
Σ 2533.3
3
128 10×
3
49 10×

Then XA xA=Σ

3
(2533.3) 128 10X =× 50.5 mmX= 

3
(2533.3) 49 10
YA yA
Y
=Σ
=× 19.34 mmY= 

164. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
566


PROBLEM 5.14
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
Σ 5.3333 17.0650 10.6670

Then XA xAΣ=Σ

23
(5.3333 in ) 17.0650 inX = or 3.20 in.X= 
and YA yAΣ=Σ

23
(5.3333 in ) 10.6670 inY = or 2.00 in.Y= 

165. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
567


PROBLEM 5.15
Locate the centroid of the plane area shown.

SOLUTION
Dimensions in mm


2
,mmA
,mmx ,mmy
3
,mmxA
3
,mmyA
1 47 26 1919.51
2
π
××= 0 11.0347 0 21,181
2
1
94 70 3290
2
××=
−15.6667 −23.333 −51,543 −76,766
Σ 5209.5 −51,543 −55,584

Then
51,543
5209.5
xA
X
A
Σ−
==
Σ

9.89 mmX=− 

55,584
5209.5
yA
Y
A
Σ−
==
Σ

10.67 mmY=− 

166. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
568


PROBLEM 5.16
Determine the x coordinate of the centroid of the trapezoid shown in terms of
h
1, h2, and a.

SOLUTION


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
Σ
12
1
()
2
ah h+
2
121
(2)
6
ah h+



21
126
1
122
(2)
()ah hxA
X
Aahh+Σ
==
Σ+

12
12
21
3hh
Xa
hh+
=
+


167. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
631


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


I
II
III
1
(150lb/ft)(9 ft) 675 lb
2
1
(120 lb/ft)(9ft) 540lb
2
675 540 1215 lb
: (1215) (3)(675) (6)(540) 4.3333 ft
R
R
RR R
XR x R X X
==
==
=+ = + =
=Σ = + =

(a)
1215 lb=R
4.33 ftX= 
(b) Reactions: 0: (9 ft) (1215 lb)(4.3333 ft) 0
A
MBΣ= − =

585.00 lbB= 585 lb=B


0: 585.00 lb 1215 lb 0
y
FAΣ= + − =

630.00 lbA= 630 lb=A


168. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
632


PROBLEM 5.68
Determine the reactions at the beam supports for the given loading.

SOLUTION
First replace the given loading by the loadings shown below. Both loading 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
MB CΣ=− + + =

270 N=B 270 N=B


0: 675 N 300 N 270 N 0
y
FAΣ= − + + =

105.0 N=A 105.0 N=A


169. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
633


PROBLEM 5.69
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
MBΣ= − − + + =


740 lbB=

740 lb=B



0: 740 lb 3000 lb 600 lb 0
y
FAΣ= + − − =


2860 lbA=

2860 lb=A



170. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
634


PROBLEM 5.70
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
FAΣ= − + =

575 lb=A


0: (800 lb)(2 ft) (225 lb)(5 ft) 0
AA
MMΣ= − + =

475 lb ft
A
=⋅M



171. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
635


PROBLEM 5.71
Determine the reactions at the beam supports for the given
loading.

SOLUTION

I
II
1
(4 kN/m)(6 m)
2
12 kN
(2 kN/m)(10m)
20 kN
R
R
=
=
=
=


0: 12 kN 20 kN 0
y
FAΣ= − − =

32.0 kN=A

0: (12 kN)(2 m) (20 kN)(5 m) 0
AA
MMΣ= − − =

124.0 kN m
A
=⋅M


172. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
636


PROBLEM 5.72
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
FAΣ= =

0: 1800 N 4800 N 0
yy
FAΣ= + − =
or
3000 N
y
A= 3.00 kN=A



15
0: (3 m)(1800 N) m (4800 N) 0
4
AA
MM

Σ= + − =



or
12.6 kN m
A
M=⋅ 12.60 kN m=⋅
A
M



173. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
637

PROBLEM 5.73
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
FAΣ= =

0: 800 lb 200 lb 0
yy
FAΣ= − − =
or
1000 lb
y
A= 1000 lb=A



0: (3 ft)(800 lb) (16.5 ft)(200 lb) 0
AA
MMΣ= − − =

or

5700 lb ft
A
M=⋅ 5700 lb ft
A
=⋅M


174. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
638


PROBLEM 5.74
Determine the reactions at the beam supports for the given
loading when w
O = 400 lb/ft.

SOLUTION



I
II
11
(12ft) (400lb/ft)(12ft) 2400lb
22
1
(300 lb/ft)(12 ft) 1800 lb
2
O
Rw
R
== =
==

0: (2400 lb)(1 ft) (1800 lb)(3 ft) (7 ft) 0Σ= − + =
B
MC

428.57 lb=C 429 lb=C


0: 428.57 lb 2400 lb 1800 lb 0Σ= + − − =
y
FB

3771lb=B 3770 lb=B


175. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
639


PROBLEM 5.75
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
MwΣ= − = 900 lb/ft
O
w= 
(b) Corresponding value of
I
:
R

I
6(900) 5400 lb==R

0: 5400 lb 1800 lb 0Σ= − − =
y
FB 7200 lb=B


176. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
640


PROBLEM 5.76
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 aBΣ= − − −+ =
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
Σ= − +−




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= 

177. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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641
PROBLEM 5.76 (Continued)

(b) We have
0: 0
xx
FAΣ= =
From Eq. (1):
600 300(0.53590)
yy
AB=
=+

761 N= 761 N==AB


178. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
642


PROBLEM 5.77
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
+  
Σ= − − − + =
  
  

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
FAΣ= =

0: 900(1) N 300(4 1) N 750 N 0
yy
FAΣ= − − − + =
or
1050 N
y
A= 1050 N=A


179. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
643


PROBLEM 5.78
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
GD E
MwΣ= − − + =

or
3150 N/m
DE
w= 
and
0: (0.8 ) N 1800 N 3600 N 3150 N 0
yB C
FwΣ= − − + =
or
2812.5 N/m
BC
w= 2810 N/m
BC
w= 

180. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
644


PROBLEM 5.79
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 (a)
the value of w
A so that w BC = w DE, (b) the corresponding
values of w
BC and w DE.

SOLUTION
(a)

We have
I
II
1
(6 m)( N/m) (3 ) N
2
1
(6 m)(1200 N/m) 3600 N
2
(0.8 m)( N/m) (0.8 ) N
(1 m) ( N/m) ( ) N
AA
BC BC BC
DE DE DE
Rww
R
Rw w
Rw w
=⋅
==
==
==
Then
0: (0.8 )N (3 )N 3600N ( )N 0
yB CA D E
Fww wΣ= − − + =
or
0.8 3600 3
BC DE A
ww w+= +
Now
5
2000
3
BC DE BC DE A
ww ww w=  == + (1)
Also,
0: (1 m)(3 N) (3 m)(3600 N) (4 m)( N) 0
GA DE
Mw wΣ= − − + =

or
3
2700
4
DE A
ww=+ (2)

Equating Eqs. (1) and (2),

53
2000 2700
34
AA
ww+=+
or
8400
N/m
11
A
w=

764 N/m
A
w= 
(b) Eq. (1)

5 8400
2000
311
BC DE
ww

== +



or
3270 N/m
BC DE
ww== 

181. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
645


PROBLEM 5.80
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)(1m)(10 kg/m )(9.81m/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 0Σ= − =
x
FH


44.145 kN=H 44.1 kN=H




0: 141.26 47.09 39.24 0Σ= − − − =
y
FV


227.6 kN=V

228 kN=V



182. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
646
PROBLEM 5.80 (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) 0Σ= −Σ+ =
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° 

183. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
647


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 triangular section of water shown. (Thickness = 1 m.)


33 2
(7.2m)(10 kg/m )(9.81m/s )=p

33 2
1
33 2
2
33 2
3
33 22
(4.8 m)(7.2 m)(1 m)(2.4 10 kg/m )(9.81 m/s )
3
542.5 kN
1
(2.4 m)(7.2 m)(1 m)(2.4 10 kg/m )(9.81 m/s )
2
203.4 kN
1
(2.4 m)(7.2 m)(1 m)(10 kg/m )(9.81m/s )
2
84.8 kN
11
(7.2 m)(1m)(7.2 m)(10 kg/m )(9.81m/s )
22
=×
=
=×
=
=
=
==
=
W
W
W
PAp
254.3 kN


0: 254.3 kN 0
x
FHΣ= − = 254 kN=H




0: 542.5 203.4 84.8 0
y
FVΣ= − − − =


830.7 kNV=

831 kN=V



184. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
648
PROBLEM 5.81 (Continued)

(b)
1
2
3
5
(4.8 m) 3 m
8
1
4.8 (2.4) 5.6 m
3
2
4.8 (2.4) 6.4 m
3
x
x
x
==
=+ =
=+ =


0: (2.4 m) 0
A
MxVxWPΣ= −Σ+ =

(830.7 kN) (3 m)(542.5 kN) (5.6 m)(203.4 kN)
(6.4 m)(84.8 kN) (2.4 m)(254.3 kN) 0
(830.7) 1627.5 1139.0 542.7 610.3 0
(830.7) 2698.9 0
x
x
x
−−
−+ =
−−−+=
−=


3.25 mx=

(to right of A )


(c) Resultant on face BC
:
Direct computation:

33 2
2
22
2
(10 kg/m )(9.81 m/s )(7.2 m)
70.63 kN/m
(2.4) (7.2)
7.589 m
18.43
1
2
1
(70.63 kN/m )(7.589 m)(1 m)
2
Pgh
P
BC
RPAρ
θ==
=
=+
=
=°
=
=
268 kN=R
18.43° 
Alternate computation: Use free body of water section BCD.


268 kN−=R
18.43°

268 kN=R
18.43° 

185. Chapter 6
Analysis of Structures
Simple Truss (Pin Joints & Two-Forces Members)

6.1 INTRODUCTION
The problems considered in the preceding chapters concerned the equilibrium of a
single rigid body, and all forces involved were external to the rigid body. We now
consider problems dealing with the equilibrium of structures made of several
connected parts. These problems call for the determination not only of the external
forces acting on the structure but also of the forces which hold together the various
parts of the structure. From the point of view of the structure as a whole, these
forces are internal forces. Consider, for example, the crane shown in Fig. 6.1a ,
which carries a load W. The crane consists of three beams AD , CF, and BE
connected by frictionless pins; it is supported by a pin at A and by a cable DG. The
free-body diagram of the crane has been drawn in Fig. 6.1b . The external forces,
which are shown in the diagram, include the weight W, the two components A
x
and A
y

of the reaction at A, and the force T exerted by the cable at D . The internal
forces holding the various parts of the crane together do not appear in the diagram.
If, however, the crane is dismembered and if a free-body diagram is drawn for
each of its component parts, the forces holding the three beams together will also
be represented, since these forces are external forces from the point of view of each
component part (Fig. 6.1c).
Fig. 6.1 Simple Truss

186. 6.2 ANALYSIS OF TRUSSES BY THE METHOD OF JOINTS
The simple truss can be considered as a group of pins and two-force members. The
truss shown, whose free- body diagram is shown in Fig. 6. 2a, can thus be
dismembered, and a free-body diagram can be drawn for each pin and each
member (Fig. 6.2b). Each member is acted upon by two forces, one at each end;
these forces have the same magnitude, same line of action, and opposite sense.
Furthermore, Newton’s third law indicates that the forces of action and reaction
between a member and a pin are equal and opposite. Therefore, the forces exerted
by a member on the two pins it connects must be directed along that member and
be equal and opposite. The common magnitude of the forces exerted by a member
on the two pins it connects is commonly referred to as the force in the member
considered, even though this quantity is actually a scalar. Since the lines of action
of all the internal forces in a truss are known, the analysis of a truss reduces to
computing the forces in its various members and to determining whether each of its
members is in tension or in compression.


Fig. 6.2 Free Diagram of a Simple Truss

187. Forces on each joint can be illustrated through the Free -Body Diagram as it is
shown in the following figure Fig. 6.3, consequently the unknown forces can be
obtained from equalizing the summation of all forces on a specified joint in both
direction X, Y to zero.


Fig. 6.3 Free Body Diagram of Truss Joints

188. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
743


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
FBΣ= = = B
0: (3.2 m) (48 kN)(7.2 m) 0
Cx
MBΣ= − − =


108 kN 108 kN
xx
B=− = B


0: 108 kN 48 kN 0
x
FCΣ= − + =


60 kNC=

60 kN=C

Free body: Joint B :

108 kN
54 3
BCAB
FF
==



180.0 kN
AB
FT= 


144.0 kN
BC
FT= 



Free body: Joint C
:



60 kN
13 12 5
AC BC
FF
==


156.0 kN
AC
FC= 

144 kN (checks)
BC
F=

189. %%Problem 6.1 How to calculate all the forces in each member as shown in
%%figure using the technique of FUNCTIONS as the following:-


%%Calculating the reactions acting on the entire truss from FBD
%%As the summation of moments about any point = Zero, then
%%Taking the moment about point B to calculate the reaction acting on
%%joint C as the following:-
Cx = (48*4)/3.2

%%As the summation of forces = Zero in any direction,thus By = 0,
%%calculating Bx as following:-
Bx = Cx + 48



%%Calculating the FBD at joint B
%%Three Balanced Forces
theta1 = atan(4/3)
theta2 = (pi/2)
theta3 = atan(3/4)
%%F the force between theta1 & theta2
F1 = abs(Bx)
[F2,F3] = TBF(theta1, theta2, theta3, F1);
Fbc = F2
Fab = F3


%%Calculating the FBD at joint C
%%Three Balanced Forces
theta1 = atan(7.2/3)
theta2 = (pi/2)
theta3 = atan(3/7.2)
%%F the force between theta1 & theta2
F1 = abs(Cx)
[F2,F3] = TBF(theta1, theta2, theta3, F1);
Fcb = F2
Fca = F3

190. %%Problem 6.1 How to calculate all the forces in each member as shown in
%%figure using the technique of FUNCTIONS as the following:-
%%Three Balanced Forces
function [F2,F3] = TBF(theta1, theta2, theta3, F1)
F2 = (F1/sin(theta3))*sin(theta1);
F3 = (F1/sin(theta3))*sin(theta2);

191. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
744


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
Reactions:

0: 1260 lb
A
MΣ= = C


0: 0
xx
FΣ= = A


0: 960 lb
yy
FΣ= = 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
yA C
FFΣ= − − =

1200 lb
AC
F= 1200 lb
AC
FC= 

3
0: 720 lb (1200 lb) 0 (checks)
5
x
FΣ= − =

192. %%Problem 6.2 How to calculate all the forces in each member as shown in
%%figure using the technique of FUNCTIONS as the following:-


%%Calculating the reactions acting on the entire truss from FBD
%%As the summation of moments about any point = Zero, then
%%Taking the moment about point A to calculate the reaction acting on
%%joint C as the following:-
Cy = (300*63)/15

%%As the summation of forces = Zero in any direction,thus Ax = 0,
%%calculating Ay as following:-
Ay = 300 - Cy



%%Calculating the FBD at joint A
%%Three Balanced Forces
theta1 = (pi/2)
theta2 = atan(15/20)
theta3 = atan(20/15)
%%F the force between theta1 & theta2
F1 = abs(Ay)
[F2,F3] = TBF(theta1, theta2, theta3, F1);
Fac = F2
Fab = F3


%%Calculating the FBD at joint C
%%Three Balanced Forces
theta1 = atan(48/20)
theta2 = atan(15/20)
theta3 = pi - theta1 - theta2
%%F the force between theta1 & theta2
F1 = abs(Cy)
[F2,F3] = TBF(theta1, theta2, theta3, F1);
Fac = F2
Fcb = F3

193. %%Problem 6.1 How to calculate all the forces in each member as shown in
%%figure using the technique of FUNCTIONS as the following:-
%%Three Balanced Forces
function [F2,F3] = TBF(theta1, theta2, theta3, F1)
F2 = (F1/sin(theta3))*sin(theta1);
F3 = (F1/sin(theta3))*sin(theta2);

194. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
745


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

22
22
3 1.25 3.25 m
34 5m
AB
BC
=+ =
=+=

Reactions:
0: (84 kN)(3 m) (5.25 m) 0
A
MCΣ= − =


48 kN=C


0: 0
xx
FACΣ= −=


48 kN
x
=A


0: 84 kN 0
yy
FAΣ= = =


84 kN
y
=A

Joint A:

12
0: 48 kN 0
13
xA B
FFΣ= − =


52 kN
AB
F=+ 52.0 kN
AB
FT= 

5
0: 84 kN (52 kN) 0
13
yA C
FFΣ= − − =

64.0 kN
AC
F=+ 64.0 kN
AC
FT= 
Joint C
:



48 kN
53
BC
F
=
80.0 kN
BC
FC= 

195. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
746


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
Free body: Truss:
From the symmetry of the truss and loading, we find
600 lb==CD


Free body: Joint B :
300 lb
215
BCAB
FF
==


671lb
AB
FT= 600 lb
BC
FC= 
Free body: Joint C
:

3
0: 600 lb 0
5
yA C
FFΣ= + =


1000 lb
AC
F=−

1000 lb
AC
FC=




4
0: ( 1000 lb) 600 lb 0
5
xC D
FFΣ= − + + =

200 lb
CD
FT=



From symmetry:

1000 lb , 671lb ,
AD AC AE AB
FF CFF T== ==

600 lb
DE BC
FF C==



196. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
747


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: (24) (4 2.4)(12) (1)(24) 0
Dy
MFΣ= −+ − =


4.2 kips
y
=F


0: 0
xx
FΣ= =F


0: (1 4 1 2.4) 4.2 0
y
FDΣ= −+++ + =

4.2 kips=D

Joint A:

0: 0
xA B
FFΣ= = 0
AB
F= 

0: 1 0
yA D
FFΣ= −− =

1kip
AD
F=− 1.000 kip
AD
FC= 
Joint D
:
8
0: 1 4.2 0
17
yB D
FFΣ= −+ + =

6.8 kips
BD
F=− 6.80 kips
BD
FC= 

15
0: ( 6.8) 0
17
xD E
FFΣ= − + =


6kips
DE
F=+

6.00 kips
DE
FT=



197. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
748
PROBLEM 6.5 (Continued)

Joint E
:



0: 2.4 0
yB E
FFΣ= − =


2.4 kips
BE
F=+

2.40 kips
BE
FT=


Truss and loading symmetrical about
c.L

198. %%Problem 6.5 How to calculate all the forces in each member as shown in
%%figure using the technique of FUNCTIONS as the following:-


%%Calculating the reactions acting on the entire truss from FBD
%%As the summation of moments about any point = Zero, then
%%Taking the moment about point F to calculate the reaction acting on
%%joint F as the following:-
Dy = (1*24 + 4*12 + 2.4*12)/24

%%As the summation of forces = Zero in any direction,thus Fx = 0,
%%calculating Fy as following:-
Fy = 1 + 4 +1 +2.4 - Dy


%%Calculating the FBD at joint D
%%Three Balanced Forces
theta1 = (pi/2)
theta2 = (pi*3/2)
theta3 = atan(6.4/12)
theta4 = (0)

%%theta1, theta2, theta3, theta4 are related to F1, F2, F3, F4
%%respectively.
F1 = Dy
F2 = 1
[F3,F4] = FBF(theta1, theta2, theta3, theta4, F1, F2);
Fdb = F3
Fde = F4

%%Calculating the FBD at joint B
%%Three Balanced Forces
theta1 = atan(6.4/12)
theta2 = (pi*3/2)
theta3 = 2*pi - atan(6.4/12)
theta4 = pi/2

%%theta1, theta2, theta3, theta4 are related to F1, F2, F3, F4
%%respectively.
F1 = - Fdb
F2 = 4
[F3,F4] = FBF(theta1, theta2, theta3, theta4, F1, F2);
Fbf = F3
Fbe = F4

199. %%Problem 6.5 How to calculate all the forces in each member as shown in
%%figure using the technique of FUNCTIONS as the following:-
%%Four Balanced Forces
%%F1 F2 F3 F4 are four forces acting on a single joint
%%theta1, theta2, theta3, theta4 are related to F1, F2, F3, F4
%%respectively.

function [F3,F4] = FBF(theta1, theta2, theta3, theta4, F1, F2)
A = [cos(theta3) cos(theta4);sin(theta3) sin(theta4)];
C = [- (F1*cos(theta1)+F2*cos(theta2)) -(F1*sin(theta1)+F2*sin(theta2))]'
F = inv(A)*C
F3 = F(1)
F4 = F(2)

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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
749


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


22
22
512 13ft
12 16 20 ft
AD
BCD
=+=
=+=

Reactions
: 0: 0
xx
FDΣ= =
0: (21ft) (693 lb)(5 ft) 0
Ey
MDΣ= − =

165 lb
y
=D


0: 165 lb 693 lb 0
y
FEΣ= − +=

528 lb=E


Joint D:
54
0: 0
13 5
xA DD C
FFFΣ= + =
(1)

12 3
0: 165 lb 0
13 5
yA DD C
FFFΣ= + + =
(2)

Solving Eqs. (1) and (2) simultaneously,

260 lb
AD
F=− 260 lb
AD
FC= 

125 lb
DC
F=+ 125 lb
DC
FT= 

201. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
750
PROBLEM 6.6 (Continued)

Joint E
:

54
0: 0
13 5
xB EC E
FFFΣ= + = (3)

12 3
0: 528 lb 0
13 5
yB EC E
FFFΣ= + + =
(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. 209).
400 lb
AC
FT= 

125.0 lb
BC
FT= 
Joint A
:
54
0: (260 lb) (400 lb) 0
13 5
xA B
FFΣ= + + =

420 lb
AB
F=− 420 lb
AB
FC= 

12 3
0: (260 lb) (400 lb) 0
13 5
0 0 (Checks)
y
FΣ= − =
=

202. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
751


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
Free body: Entire truss:

0: 2(5 kN) 0
xx
FCΣ= + =

10 kN 10 kNx
xC=− = C



0: (2 m) (5 kN)(8 m) (5 kN)(4 m) 0
C
MDΣ= − − =

30 kN 30 kND=+ = D



0: 30 kN 0 30 kN 30 kN y
yy yFC CΣ= + = =− = 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
xB D
FFΣ= + =


55 kN
BD
F=−

11.18 kN
BD
FC=




2
0: 20 kN ( 5 5 kN) 0
5
yB C
FFΣ= − − − =


30 kN
BC
F=+

30.0 kN
BC
FT=




203. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
752
PROBLEM 6.7 (Continued)

Free body: Joint C
:



0: 10 kN 0
xC D
FFΣ= − =


10 kN
CD
F=+

10.00 kN
CD
FT=




0: 30 kN 30 kN 0 (checks)
y
FΣ= − =

204. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
753


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
Reactions:

0: 16 kN
Cx
MΣ= = A


0: 9 kN
yy
FΣ= = A


0: 16 kN
x
FΣ= = C



Joint E:

3 kN
54 3
BE DE
FF
==



5.00 kN
BE
FT= 


4.00 kN
DE
FC= 

Joint B
:

4
0: (5 kN) 0
5
xA B
FFΣ= − =

4 kN
AB
F=+ 4.00 kN
AB
FT= 

3
0: 6 kN (5 kN) 0
5
yB D
FFΣ= − − − =

9 kN
BD
F=− 9.00 kN
BD
FC= 
Joint D
:

3
0: 9 kN 0
5
yA D
FFΣ= − + =

15 kN
AD
F=+ 15.00 kN
AD
FT= 

4
0: 4 kN (15 kN) 0
5
xC D
FFΣ= − − − =

16 kN
CD
F=− 16.00 kN
CD
FC= 

205. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
754


PROBLEM 6.9
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
FΣ= = 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
xB DB E
FFFΣ= + + =

or
24 20 30,000 lb
BD BE
FF+=− (1)

733
0: (1500) 600 0
25 5 5
yB DB E
FFFΣ= − + − =
or
7 15 7,500 lb
BD BE
FF−=− (2)

206. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
755
PROBLEM 6.9 (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
xD F
FFΣ= + =

1200 lb
DF
F=− 1200 lb
DF
FC= 

77
0: (1200 lb) ( 1200 lb) 600 lb 0
25 25
yD E
FFΣ= − − − − =

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= 
Note: Compare results with those of Problem 6.11.

207. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
756


PROBLEM 6.10
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
FΣ= =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 

208. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
757
PROBLEM 6.10 (Continued)

Free body: Joint B
:

444
0: (1500 lb) 0
555
xB DB C
FFFΣ= + + =

or
1500 lb
BD BE
FF+=− (1)

333
0: (1500 lb) 600 lb 0
555
yB DB E
FFFΣ= − + − =
or
500 lb
BD BE
FF−=− (2)
Add Eqs. (1) and (2):
22000lb
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
xD F
FFΣ= + =

1000 lb
DF
F=− 1000 lb
DF
FC= 

33
0: (1000 lb) ( 1000 lb) 600 lb 0
55
yD E
FFΣ= −− − − =

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= 

209. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
758


PROBLEM 6.11
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
FAΣ= =
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= 

210. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
759
PROBLEM 6.11 (Continued)

Free body: Joint C
:
3
0: 10.5 0
5
yC D
FFΣ= − =

17.50 kN
CD
FT=



4
0: (17.50) 44.625 0
5
xC E
FFΣ= + − =


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

211. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
760


PROBLEM 6.12
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
FΣ= =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
:

32.25 2.25
0: (11.08 kN) 0
5 2.462 2.462
xB C B D
FF FΣ= + + =
(1)


4 11.08 kN
0: 3 kN 0
5 2.462 2.462
BD
yBC
F
FFΣ= − + + − =
(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= 

212. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
761
PROBLEM 6.12 (Continued)

Free body: Joint C
:
44
0: (2.8125 kN) 0
55
yC D
FFΣ= − =


2.8125 kN,
CD
F=

2.81 kN
CD
FT=



33
0: 10.125 kN (2.8125 kN) (2.8125 kN) 0
55
xC E
FFΣ= − + + =


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= 

213. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
762


PROBLEM 6.13
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
MHΣ= − − −
−− =


4.50 kN=H


0: 0
0: 9 0
94.50
xx
yy
y
FA
FAH
A
Σ= =
Σ= +−=
=−
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= 

214. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
763
PROBLEM 6.13 (Continued)

Free body: Joint B
:

22 1
0: (7.8262 kN) 0
55 2
xB D BC
FF FΣ= + + =

or
0.79057 7.8262 kN
BD BC
FF+=− (1)

11 1
0: (7.8262 kN) 2 kN 0
55 2
yB D BC
FF FΣ= + − − =

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: (1.8861 kN) 0
52
yC D
FFΣ= − =


1.4911 kN
CD
F=+ 1.491 kN
CD
FT= 

11
0: 7.00 kN (1.8861 kN) (1.4911 kN) 0
25
xC E
FFΣ= − + + =


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
xD FD E
FFFΣ= + + − =

or
0.79057 5.5900 kN
DF DE
FF+=− (1)

111 2
0: (6.3355 kN) (1.4911 kN) 2 kN 0
525 5
yD FD E
FFFΣ= − + − − =
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= 

215. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
764
PROBLEM 6.13 (Continued)

Free body: Joint F
:

12
0: (3.3544 kN) 0
25
xF G
FFΣ= + =


4.243 kN
FG
F=− 4.24 kN
FG
FC= 

11
0: 1.75 kN (3.3544 kN) ( 4.243 kN) 0
52
yE F
FFΣ= − − + − − =


2.750 kN
EF
F=

2.75 kN
EF
FT=




Free body: Joint G
:

111
0: (4.243 kN) 0
222
xG HE G
FFFΣ= − + =

or
4.243 kN
GH EG
FF−=− (1)

111
0: (4.243 kN) 1.5 kN 0
222
yG HE G
FFFΣ= − − − − =
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=

216. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
765


PROBLEM 6.14
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
MAΣ= + − =

2250 NA=+ 2250 N=A


0: 2250 N 0
yy
FFΣ= + =

2250 N 2250 N
yy
F=− = F


0: 800 N 800 N 0
xx
FFΣ= − − + =


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
xC F
FFΣ= − =

1600 N
CF
F=+ 1600 N
CF
FT= 

0: 2250 N 0
yE F
FFΣ= − =

2250 N
EF
F=+ 2250 N
EF
FT= 

217. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
766
PROBLEM 6.14 (Continued)

Joint C
:
8
0: 1200 N 1600 N 0
17
xC E
FFΣ= − + =


850 N
CE
F=−

850 N
CE
FC=



15
0: 0
17
yB CC E
FFFΣ= + =


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
xB E
FFΣ= − − + =


400 N
BE
F=−

400 N
BE
FC=



15
0: 1500 N 2250 N (850 N) 0
17
0 0 (checks)
y
FΣ= − + =
=

218. Chapter 7
Shear and Bending Moment of Beams



7.1 SHEAR AND BENDING MOMENT IN A BEAM
Consider a beam AB subjected to various concentrated and distributed loads (Fig.
7.8a). We propose to determine the shearing force and bending moment at any
point of the beam. In the example considered here, the beam is simply supported,
but the method used could be applied to any type of statically determinate beam.
First we determine the reactions at A and B by choosing the entire beam as a free
body (Fig. 7.8b); writing MA = 0 and MB = 0, we obtain, respectively, R
B and R A

.



Fig. 7.1 Shear Force and Bending Moment in a Beam Fig. 7.2 Positive Directions of Shear Force and Bending Moment

219. 7.2 SHEAR AND BENDING -MOMENT DIAGRAMS
Now that shear and bending moment have been clearly defined in sense as well as
in magnitude, we can easily record their values at any point of a beam by plotting
these values against the distance x measured from one end of the beam. The graphs
obtained in this way are called, respectively, the shear diagram and the bending
moment diagram. As an example, consider a simply supported beam AB of span L
subjected to a single concentrated load P applied at its midpoint D (Fig. 7.3a). We
first determine the reactions at the supports from the free-body diagram of the
entire beam (Fig. 7.3b); we find that the magnitude of each reaction is equal to P/2.
Next we cut the beam at a point C between A and D and draw the free-body
diagrams of AC and CB (Fig. 7.3c). Assuming that shear and bending moment are
positive, we direct the internal forces V and V’ and the internal couples M and M’
as indicated in Fig. 7.2a. Considering the free body AC and writing that the sum of
the vertical components and the sum of the moments about C of the forces acting
on the free body are zero, we find V = +P/2 and M = +Px/2. Both shear and
bending moment are therefore positive; this can be checked by observing that the
reaction at A tends to shear off and to bend the beam at C as indicated in Fig. 7.2b
and c. We can plot V and M between A and D (Fig. 7.3e and f); the shear has a
constant value V = P/2, while the bending moment increases linearly from M = 0 at
x = 0 to M = PL/4 at x = L/2.
Cutting, now, the beam at a point E between D and B and considering the free
body EB (Fig. 7.3d), we write that the sum of the vertical components and the sum
of the moments about E of the forces acting on the free body are zero. We obtain V
= -P/2 and M = P(L - x)/2. The shear is therefore negative and the bending
moment positive; this can be checked by observing that the reaction at B bends the
beam at E as indicated in Fig. 7.2c but tends to shear it off in a manner opposite to
that shown in Fig. 7.2b. We can complete, now, the shear and bending-moment
diagrams of (Fig. 7. 3e and f); the shear has a constant value V =-P/2 between D
and B, while the bending moment decreases linearly from M = PL/4 at x = L/2 to
M = 0 at x = L.
It should be noted that when a beam is subjected to concentrated loads only, the
shear is of constant value between loads and the bending moment varies linearly
between loads, but when a beam is subjected to distributed loads, the shear and
bending moment vary quite differently.

220. Fig. 7.3 Shear Force and Bending Moment Diagrams

221. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1041


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



0: 0
y
FwxVΣ= − −=

Vwx=−

1
0: 0
2
x
MwxM

Σ= +=




21
2
Mwx=−



max
||VwL= 

2
max1
||
2
MwL=


222. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1042


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
Σ= − −=

2
0
1
2
x
Vw
L
=−


2
0
1
0: 0
23
y
xx
FwM
L
Σ= + = 


3
0
1
6
x
Mw
L
=−





max 0
1
||
2
VwL=



2
max 01
||
6
MwL=


223. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1043


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
33
D
LL
MPPAL 
Σ= − −=
 
 


/3P=A

0: 0
xx
FDΣ= =

0: 0
3
yy
P
F PPDΣ= −++ =


/3
y
DP=− /3P=D
(a) Shear and bending moment. Since the loading consists of concentrated loads,
the shear diagram is made of horizontal straight-line segments and the
B. M. diagram is made of oblique straight-line segments.
Just to the right of A:

1
0: 0
3
y
P
FVΣ= −+=

1
/3VP=+ Σ

11
0: (0) 0
3
P
MMΣ= − =
1
0M= Σ
Just to the right of B:

2
0: 0,
3
y
P
FVPΣ= −+−=

2
2/3VP=− Σ

22
0: (0) 0
33
PL
MM P

Σ= − + =



2
/9MPL=+ Σ
Just to the right of C:
3
0: 0
3
y
P
FPPVΣ= −+−=

3
/3VP=+ Σ
33
2
0: (0) 0
33 3
PL L
MM PP
Σ= − + − =
 

3
/9MPL=− Σ

224. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1044
PROBLEM 7.31 (Continued)

Just to the left of D:

4
0: 0
3
y
P
FVΣ= −=

4
3
P
V=+
Σ

44
0: (0) 0
3
P
MMΣ= −− =
4
0M= Σ





(b)
max
|| 2/3;VP=
max
|| /9MPL= 

225. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1045


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) Shear and bending moment.
Just to the right of A:


1
;VP=+
1
0M= Σ

Just to the right of B
:
2
0: 0;
y
FPPVΣ= −−=
2
0V= Σ
22
0: 0;
2
L
MMP

Σ= − =



2
/2MPL=+ Σ

Just to the left of C :
3
0: 0,
y
FPPVΣ= −−=
3
0V= Σ
33
0: 0,
2
L
MMPPL

Σ= + −=



3
/2MPL=+ Σ






(b)
max
||VP= 

max
|| /2MPL= 

226. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1046


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
MLAMΣ= −=

0
y
M
L
=
A


0: 0
yy
FACΣ= − +=

0
M
L
=
C

Along AB:

0
0
0: 0
y
M
FV
L
M
V
L
Σ= − −=
=−


0
0
0: 0
J
M
MxM
L
M
Mx
L
Σ= +=
=−

Straight with
0
at
2
M
MB=−
Along BC:
00
0: 0
y
MM
FVV
LL
Σ= − −= =−


0
00
0: 0 1
K
M x
MMxM MM
LL 
Σ= + −= = −



Straight with
0
at 0 at
2
M
MBMC==
(b) From diagrams:
max 0
|| /VML= 

0
max
|| at
2
M
MB= 

227. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1047


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
FPVΣ= −−=

VP=−




0: 0Σ= +−=
Jx
MMPPL

()MPLx=−




(a) The V and M diagrams are obtained by plotting the functions V and M .

(b)
max
||VP=




max
||MPL=



228. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1048


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) Just to the right of A:

11
015kN0
y
FV MΣ= =+ =
Just to the left of C
:

22
15 kN 15 kN mVM=+ =+ ⋅
Just to the right of C
:

33
15 kN 5 kN mVM=+ =+ ⋅
Just to the right of D
:

44
15 kN 12.5 kN mVM=− =+ ⋅
Just to the right of E
:

55
35 kN 5 kN mVM=− =+ ⋅
At B
: 12.5 kN m
B
M=− ⋅


(b)
max
| | 35.0 kNV =
max
| | 12.50 kN mM =⋅ 

229. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1049


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
MBΣ= − − − =


37.5 kNB=+

37.5 kN=B



0: 0
xx
FAΣ= =

0: 37.5kN 40kN 32kN 16kN 0
yy
FAΣ= + − − − =

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
FVΣ= − −=
2
10.5 kNV=+



22
0: (50.5 kN)(0.6 m) 0MMΣ= − =
2
30.3 kN mM=+ ⋅



Just to the right of D :

3
0: 50.5 40 32 0
y
FVΣ= − −−=
3
21.5 kNV=− 

33
0: (50.5)(1.5) (40)(0.9) 0MMΣ= − + =
3
39.8 kN mM=+ ⋅



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1050
PROBLEM 7.36 (Continued)

Just to the right of E
:

4
0: 37.5 0
y
FVΣ= + =
4
37.5 kNV=−



44
0: (37.5)(0.2) 0MMΣ= − + =
4
7.50 kN mM=+ ⋅


At B : 0
BB
VM== 

(b)
max
|| 50.5kNV= 










max
| | 39.8 kN mM =⋅ 

231. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1051


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
MEΣ= − − − =


16 kipsE=+

16 kips=E


0: 0
xx
FAΣ= =

0: 16 kips 6 kips 12 kips 4.5 kips 0
yy
FAΣ=+−−− =

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
FVΣ= − −=
2
0.50 kipsV=+



22
0: (6.50 kips)(2 ft) 0MMΣ= − =
2
13 kip ftM=+ ⋅



Just to the right of D :

3
0: 6.50 6 12 0
y
FVΣ= −−−=
3
11.5 kipsV=+ 

33
0: (6.50)(4) (6)(2) 0MMΣ= − − =
3
14 kip ftM=+ ⋅



232. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1052
PROBLEM 7.37 (Continued)

Just to the right of E
:


4
0: 4.5 0
y
FVΣ= − =
4
4.5 kipsV=+



44
0: (4.5)2 0MMΣ= − − =
4
9kip ftM=− ⋅


At B : 0
BB
VM== 

(b)
max
|| 11.50kipsV= 













max
|| 14.00kipftM =⋅ 

233. Chapter 8
Friction
Pulleys, Tension Belts and Journal Bearings

8.1 INTRODUCTION
In the preceding chapters, it was assumed that surfaces in contact were either
frictionless or rough. If they were frictionless, the force each surface exerted on the
other was normal to the surfaces and the two surfaces could move freely with
respect to each other. If they were rough, it was assumed that tangential forces
could develop to prevent the motion of one surface with respect to the other. This
view was a simplified one. Actually, no perfectly frictionless surface exists. When
two surfaces are in contact, tangential forces, called friction forces, will always
develop if one attempts to move one surface with respect to the other. On the other
hand, these friction forces are limited in magnitude and will not prevent motion if
sufficiently large forces are applied. The distinction between frictionless and rough
surfaces is thus a matter of degree. This will be seen more clearly in the present
chapter, which is devoted to the study of friction and of its applications to common
engineering situations. There are two types of friction: dry friction, sometimes
called Coulomb friction, and fluid friction. Fluid friction develops between layers
of fluid moving at different velocities. Fluid friction is of great importance in
problems involving the flow of fluids through pipes and orifices or dealing with
bodies immersed in moving fluids. It is also basic in the analysis of the motion of
lubricated mechanisms. Such problems are considered in texts on fluid mechanics.
The present study is limited to dry friction, i.e., to problems involving rigid bodies
which are in contact along nonlubricated surfaces. In the first part of this chapter,
the equilibrium of various rigid bodies and structures, assuming dry friction at the
surfaces of contact, is analyzed. Later a number of specific engineering
applications where dry friction plays an important role are considered: wedges,
square-threaded screws, journal bearings, thrust bearings, rolling resistance, and
belt friction.

234. Experimental evidence shows that the maximum value F m

of the static-friction
force is proportional to the normal component N of the reaction of the surface
(Fig.8.1), w e have:
????????????
????????????=????????????
????????????????????????
Where m
s is a constant called the coefficient of static friction. Similarly, the
magnitude F
k

????????????
????????????=????????????
????????????????????????
of the kinetic-friction force may be put in the form:


Where m
k is a constant called the coefficient of kinetic friction. The coefficients of
friction ms and m
k

do not depend upon the area of the surfaces in contact.


Fig. 8.1 Friction Force with respect to the Acting one

8.2 ANGLES OF FRICTION
It is sometimes convenient to replace the normal force N and the friction force F
by their resultant R . Let us consider again a block of weight W resting on a
horizontal plane surface. If no horizontal force is applied to the block, the resultant
R reduces to the normal force N (Fig. 8.2a). However, if the applied force P has a
horizontal component P
x which tends to move the block, the force R will have a
horizontal component F and, thus, will form an angle f with the normal to the
surface (Fig. 8.2b). If P
x is increased until motion becomes impending, the angle
between R and the vertical grows and reaches a maximum value (Fig. 8.2c). This
value is called the angle of static friction and is denoted by ∅
Rs
. From the geometry
of Fig. 8.2c, we note that:

235. ????????????????????????????????????(∅
????????????)=
????????????
????????????
????????????=
????????????
????????????????????????
????????????=????????????
????????????

If motion actually takes place, the magnitude of the friction force drops to F
k;
similarly, the angle f between R and N drops to a lower value ∅
Rk
????????????????????????????????????(∅
????????????)=
????????????
????????????
????????????
=
????????????
????????????????????????
????????????=????????????
????????????
, called the angle
of kinetic friction (Fig. 8.2d). From the geometry of Fig. 8. 2d, we write:



Fig. 8.2





Fig. 8.3

236. 8.3 BELT FRICTION
Fig. 8.4 shows the relation between the Minimum T
1 and larger T 2 considering the
friction coefficient between the belt and the pulley of m
s

as noted by the following
equations.
Fig. 8.4

????????????????????????�
????????????
????????????
????????????
????????????
�=????????????
????????????????????????

Or

????????????
????????????
????????????
????????????
=????????????
????????????
????????????????????????


8.4 JOURNAL BEARINGS. AXLE FRICTION
Journal bearings are used to provide lateral support to rotating shafts and axles.
Thrust bearings, which will be studied in the next section, are used to provide axial
support to shafts and axles. If the journal bearing is fully lubricated, the frictional
resistance depends upon the speed of rotation, the clearance between axle and
bearing, and the viscosity of the lubricant. As indicated in Sec. 8.1, such problems
are studied in fluid mechanics. The methods of this chapter, however, can be
applied to the study of axle friction when the bearing is not lubricated or only
partially lubricated. It can then be assumed that the axle and the bearing are in
direct contact along a single straight line (Fig. 8.5).

????????????=????????????????????????????????????????????????????????????(∅
????????????)

237. Fig. 8.5

238. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1221


PROBLEM 8.1
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when
θ = 25° and P = 750 N.

SOLUTION
Assume equilibrium:
0: (1200 N)sin 25° (750 N)cos 25° 0
x
FFΣ= + − =

172.6 NF=+ 172.6 N=F


0: (1200 N)cos 25° (750 N)sin 25° 0
y
FNΣ= − − =

1404.5 NN=
Maximum friction force:
0.35(1404.5 N) 491.6 N
ms
FNμ== =
Since
,
m
FF< block is in equilibrium

Friction force:
172.6 N=F
25.0° 

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1222


PROBLEM 8.2
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when
θ = 30° and P = 150 N.

SOLUTION
Assume equilibrium:

0: (1200 N)sin 30° (150 N)cos 30° 0
x
FFΣ= + − =


470.1 NF=− 470.1 N=F


0: (1200 N)cos 30 (150 N)sin 30 0
y
FNΣ= − °− °=

1114.2 NN=

(a) Maximum friction force:
0.35(1114.2 N)
390.0 N
ms
FNμ=
=
=
Since F is
and ,
m
FF> block moves down 
(b) Actual friction force:
0.25(1114.2 N) 279 N
kk
FF Nμ== = = 279 N=F
30.0° 

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1223


PROBLEM 8.3
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when P = 100 lb.

SOLUTION
Assume equilibrium:
0: (45 lb)sin 30 (100 lb)cos 40 0
x
FFΣ= + °− °=

54.0 lbF=+

0: (45 lb)cos 30 (100 lb)sin 40 0
y
FNΣ= − °− °=

103.2 lbN=
(a) Maximum friction force:

0.40(103.2 lb)
41.30 lb
ms
FN
μ=
=
=

We note that
.
m
FF> Thus, block moves up


(b) Actual friction force:

0.30(103.2 lb) 30.97 lb,
kk
FF Nμ== = =  31.0 lb=F
30.0°



241. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1224


PROBLEM 8.4
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when P = 60 lb.

SOLUTION
Assume equilibrium:

0: (45 lb)sin 30 (60 lb)cos 40 0
x
FFΣ= + °− °=

23.46 lbF=+


0: (45 lb)cos 30 (60 lb)sin 40 0
y
FNΣ= − °− °=

77.54 lbN=
(a) Maximum friction force:

0.40(77.54 lb)
31.02 lb
ms
FNμ=
=
=

We check that
m
FF< . Thus, block is in equilibrium

(b) Thus
23.46 lbF=+

23.5 lb=F
30.0°


Note: We have
0.30(77.54) 23.26 lb.
kk
FNμ== = Thus .
k
FF> If block originally in motion, it will keep
moving with
23.26 lb.
k
F=

242. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1225


PROBLEM 8.5
Determine the smallest value of P required to (a) start the block up the
incline, (b) keep it moving up, (c ) prevent it from moving down.

SOLUTION
(a) To start block up the incline:

1
0.40
tan 0.40 21.80
s
s
μ
φ
−
=
==°

From force triangle:

45 lb
sin 51.80 sin 28.20P
=
°° 74.8 lbP= 
(b) To keep block moving up:


1
0.30
tan 0.30 16.70
k
k
μ
φ
−
=
==°


From force triangle:

45 lb
sin 46.70 sin 33.30°P
=
° 59.7 lbP= 

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you are using it without permission.
1226
PROBLEM 8.5 (Continued)

(c) To prevent block from moving down
:

From force triangle:

45 lb
sin 8.20 sin 71.80°P
=
° 6.76 lbP= 

244. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1231


PROBLEM 8.10
Determine the range of values of P for which equilibrium of the block shown
is maintained.

SOLUTION
FBD block:
(Impending motion down):


11
1
tan tan 0.25
(500 N) tan (30° tan 0.25)
143.03 N
ss
P
φμ
−−
−
==
=−
=
(Impending motion up):


1
(500 N) tan (30° tan 0.25)
483.46 N
P
−
=+
=
Equilibrium is maintained for
143.0 N 483 NP≤≤ 

245. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1232


PROBLEM 8.11
The 20-lb block A and the 30-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
θ for which motion is impending.

SOLUTION
Since motion impends,
s
FNμ= between A + B
Free body: Block A

Impending motion:

1
0: 20cos
y
FN θΣ= =

1
0: 20sin 0θμΣ= − − =
xs
FT N

20sin 0.15(20cos )Tθθ=+

20sin 3cosTθθ=+ (1)
Free body: Block B


Impending motion:

1
0: 30sin 0
xs
FT N θμΣ= − + =

1
30sin
s
TN θμ=−

30sin 0.15(20cos )θθ=−

30sin 3cosTθθ=− (2)
Eq. (1)-Eq. (2):
20sin 3cos 30sin 3cos 0θθ θθ+− +=

6
6cos 10sin : tan ;
10
θθθ== 30.96θ=° 31.0θ=° 

246. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1233


PROBLEM 8.12
The 20-lb block A and the 30-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
θ for which motion is impending.

SOLUTION
Since motion impends,
s
FNμ= at all surfaces.
Free body: Block A

Impending motion:

1
0: 20cos
y
FN θΣ= =

1
0: 20sin 0θμΣ= − − =
xs
FT N

20sin 0.15(20cos )Tθθ=+

20sin 3cosTθθ=+ (1)
Free body: Block B


Impending motion:

21
0: 30cos 0
y
FN N θΣ= − − =

2
22
30cos 20cos 50cos
0.15(50cos ) 6cosθθθ
μ θθ
=+=
== =
s
N
FN


12
0: 30sin 0
xs s
FT NN θμ μΣ= − + + =

30sin 0.15(20cos ) 0.15(50cos )Tθθ θ=− −

30sin 3cos 7.5cosTθθ θ=−− (2)
Eq. (1) subtracted by Eq. (2):
20sin 3cos 30sin 3cos 7.5cos 0θθ θθ θ+− ++ =

13.5
13.5cos 10sin , tan
10
θθθ
°
==
53.5θ=° 

247. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1236


PROBLEM 8.15
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, determine (a) the largest allowable value of α, (b) the
corresponding magnitude of the force P.

SOLUTION
(a) Free-body diagram
If the crate is about to tip about C , contact between crate and ground is only at C 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
φμ
−−
==
=°

Since the crate is a 3-force body. P must pass through E where R and W intersect.

0.4 m
1.1429 m
tan 0.35
1.1429 0.5 0.6429 m
0.6429 m
tan
0.4 m
s
CF
EF
EH EF HF
EH
HB
θ
α
===
=−= −=
==


58.11α=° 58.1α=° 
(b) Force Triangle

0.424
sin19.29 sin128.82
PW
PW
==
°°

2
0.424(40 kg)(9.81 m/s ),P= 166.4 NP= 
Note: After the crate starts moving,
s
μ should be replaced by the lower value .
k
μ This will
yield a larger value of
.α

248. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1237


PROBLEM 8.16
A 40-kg packing crate is pulled by a rope as shown. The coefficient of static
friction between the crate and the floor is 0.35. If α = 40°, determine (a) the
magnitude of the force P required to move the crate, (b) whether the crate will
slide or tip.

SOLUTION
Force P for which sliding is impending
(We assume that crate does not tip)

2
(40 kg)(9.81 m/s ) 392.4 NW==

0: sin40 0
y
FNWPΣ= −+ °=

sin 40NWP=− ° (1)

0: 0.35 cos40 0
x
FNPΣ= − °=
Substitute for
N from Eq. (1): 0.35( sin 40 ) cos 40 0WP P−°− °=

0.35
0.35sin 40 cos 40W
P
=
°+ ° 0.3532PW= 
Force P for which crate rotates about C

(We assume that crate does not slide)

0: ( sin 40 )(0.8 m) ( cos 40 )(0.5 m)
C
MP PΣ= ° + °

(0.4 m) 0W−=

0.4
0.4458
0.8sin 40 0.5cos 40W
PW
==
°+ ° 
Crate will first slide


0.3532(392.4 N)P= 138.6 NP= 

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1337


PROBLEM 8.103
A 300-lb block is supported by a rope that is wrapped 1
1
2
times around a horizontal
rod. Knowing that the coefficient of static friction between the rope and the rod is
0.15, determine the range of values of P for which equilibrium is maintained.

SOLUTION


1.5 turns 3 rad
β π==
For impending motion of W up,

(0.15)3
(300 lb)
1233.36 lb
s
PWe e
μβ π
==
=

For impending motion of W down,

(0.15)3
(300 lb)
72.971 lb
s
PWe e
μβ π− −
==
=

For equilibrium,
73.0 lb 1233 lbP≤≤ 

250. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1339


PROBLEM 8.105
A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of
static friction is 0.25, determine (a) the smallest value of the mass m for which
equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION
We apply Eq. (8.14) to pipe B and pipe C .

2
1
μβ
=
s
T
e
T
(8.14)
Pipe B
:
21
,
AB C
TW TT==

2
0.25,
3
s
π
μβ
==

0.25(2 /3) /6A
BC
W
ee
T
ππ
== (1)
Pipe C
:
21
,,0.25,
3
BC D s
TT TW
π
μβ
== = =

0.025( /3) /12BC
D
T
ee
W
ππ
== (2)
(a) Multiplying Eq. (1) by Eq. (2):

/6 /12 /6 /12 / 4
2.193
ππ ππ π +
=⋅ = = =
A
D
W
ee e e
W


50 kg
2.193 2.193 2.193 2.193
A
W
gAD A
D
WW m
Wm
g
= ====


22.8 kgm= 
(b) From Eq. (1):
2
/6
(50 kg)(9.81 m/s )
291 N
1.688
A
BC
W
T
e
π
== = 

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1340


PROBLEM 8.106
A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of
static friction is 0.25, determine (a) the largest value of the mass m for which
equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION
See FB diagrams of Problem 8.105. We apply Eq. (8.14) to pipes B and C .
Pipe B :
12
2
,,0.25,
3
AB Cs
TW T T
π
μβ
== = =

0.25(2 /3) / 62
1
:
μβ ππ
===
s BC
A
TT
eee
TW
(1)
Pipe C
:
12
,,0.25,
3
BC D s
TT TW
π
μβ
====

0.25( /3) /122
1
:
s D
BC
TW
eee
TT
μβ ππ
=== (2)
(a) Multiply Eq. (1) by Eq. (2):

/6 /12 /6 /12 /4
2.193
D
A
W
ee e e
W
ππ ππ π +
=⋅ = ==

0
2.193 2.193 2.193(50 kg)
AA
WWmm=== 109.7 kgm= 
(b) From Eq. (1):
/6 2
(50 kg)(9.81 m/s )(1.688) 828 N
BC A
TWe
π
== = 

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1346


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
ππ
βπ
ππ
βπ
=°+°=+=
=°−°=−=



Since
,
BA
ββ< 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
μβ
π
==
==

12
0: (0.12 m)( ) 0
AA
MM TTΣ= + −=

(0.12 m)(450 N 157.914 N) 35.05 N m
A
M=−=⋅

35.1 N m
A
M=⋅ 

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1347

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
μπ π
==
=


180 radians
β π=°=
(a) Torque: 0: (675.15 N)(0.06 m) (224.84 N)(0.06 m)
A
MMΣ= − +

27.0 N mM=⋅ 
(b)
12
0: 900 N 0
x
FTTΣ= +− =


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= 

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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1349


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
μβ π
β π==°=
== =
=

Free body: Control bar


12
0: (12 in.) (4 in.) (25 lb)(28 in.) 0
C
MT TΣ= − − =


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
MMTTΣ= + − =

(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
μβ π
βπ==
== =
=

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you are using it without permission.
1350
PROBLEM 8.115 (Continued)

Free body: Control rod


21
0: (12 in.) (4 in.) (25 lb)(28 in.) 0
C
MT TΣ= − − =

11
1
2
2
2.1933 (12) (4) 700 0
31.363 lb
2.1933(31.363 lb)
68.788 lbTT
T
T
T −−=
=
=
=

Return to free body of drum


12
0: (8 in.) (8 in.) 0
E
MMTTΣ= + − =

(31.363 lb)(8 in.) (68.788 lb)(8 in.) 0M+−=

299.4 lb in.M=⋅ 299 lb in.M=⋅ 

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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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1351


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
μβ μπ
μπ
β π==°=
==
=

Free body: Control rod



12
0: (28 in.) (12 in.) (4 in.) 0
C
MP T TΣ= − + =

11
28 12 (4) 0PTeT
μπ
−+ =
For self-locking brake:
0P=

11
12 4
3
ln3 1.0986
1.0986
0.3497
μπ
μπ
μπ
μ
π
=
=
==
==
s
s
s
s
TTe
e
0.350
s
μ= 