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1. Chapter 4
Engineering Mechanics –ME 1102
Sunil Kumar Singh and Rishi Raj
Courtesy: TMH

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Application
Engineers designing this crane will need to determine the forces
that act on this body under various conditions.

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Introduction
•The necessary and sufficient conditions for the static equilibrium of a
body are that the forces sum to zero, and the moment about any point
sum to zero:()   === 00 FrMF
O
•Equilibrium analysis can be applied to two-dimensional or three-
dimensional bodies, but the first step in any analysis is the creation of
the free body diagram
•For a rigid body, the condition of static equilibrium means that the
body under study does not translate or rotate under the given loads
that act on the body

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Free-Body Diagram
The first step in the static equilibrium analysis of a
rigid body is identification of all forces acting on
the body with a free body diagram.
•Select the body to be analyzed and detach it
from the ground and all other bodies and/or
supports.
•Include the dimensions, which will be needed
to compute the moments of the forces.
•Indicate point of application and assumed
direction of unknown forces from reactions of
the ground and/or other bodies, such as the
supports.
•Indicate point of application, magnitude, and
direction of external forces, including the rigid
body weight.

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Reactions at Supports and Connections for a Two-Dimensional Structure
•Reactions equivalent to a
force with known line of
action.

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Reactions at Supports and Connections for a Two-Dimensional Structure
•Reactions equivalent to a
force of unknown direction
and magnitude.
•Reactions equivalent to a
force of unknown
direction and magnitude
and a couple of unknown
magnitude.

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Practice

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Practice
A B
C D
150 kN
150 kN
150 kN
150 kN
Choose the most
correct FBD for the
original problem.
B is the most correct, though C is also
correct. A & D are incorrect; why?
Discuss why each
choice is correct or
incorrect.

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Equilibrium of a Rigid Body in Two
Dimensions
•For known forces and moments that act on a
two-dimensional structure, the followings are
true:Ozyxz MMMMF ==== 00
•Equations of equilibrium become   === 000
Ayx MFF
where A can be any point in the plane of
the body.
•The 3 equations can be solved for no more
than 3 unknowns.
•The 3 equations cannot be augmented with
additional equations, but they can be replaced  === 000
BAx MMF

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Sample Problem 4.1
A fixed crane has a mass of 1000 kg
and is used to lift a 2400-kg crate. It
is held in place by a pin at A and a
rocker at B. The center of gravity of
the crane is located at G.
Determine the components of the
reactions at A and B.
SOLUTION:
•Create a free-body diagram for the crane.
•Determine the reactions at B by solving
the equation for the sum of the
moments of all forces about A. Note
there will be no contribution from the
unknown reactions at A.
•Determine the reactions at A by
solving the equations for the sum of
all horizontal force components and
all vertical force components.
•Check the values obtained for the
reactions by verifying that the sum of
the moments about B of all forces is
zero.

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Sample Problem 4.1
•Create the free-body diagram.
•Check the values obtained.
•Determine B by solving the equation
for the sum of the moments of all
forces about A. () ()
()0m6kN523
m2kN819m510
=−
−+=
.
..B:M
A kN1107.B+=
•Determine the reactions at A by solving the
equations for the sum of all horizontal forces
and all vertical forces.00 =+= BA:F
xx kN1.107−=
xA 0kN523kN8190 =−−= ..A:F
yy kN333.A
y+=

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Sample Problem 4.4
The frame supports part of the roof of
a small building. The tension in the
cable is 150 kN.
Determine the reactions at the fixed
end E.
SOLUTION:
- Discuss the steps for solving this
problem.
•Apply the equilibrium equations
for the reaction force components
and couple at E.
•Create a free-body diagram for the
frame and cable.

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Sample Problem 4.4
•The free-body diagram was
created in an earlier exercise.
•Apply one of the three
equilibrium equations. Try
using the condition that the
sum of forces in the x-direction
must sum to zero. ( )0kN150
5.7
5.4
:0 =+= xx EF kN 0.90−=
xE ( )0kN150936cos:0
o
=+= .EF
xx
•Which equation is correct?( )0Nk150
57
6
:0 =+=
.
EF
xx
A.
B.
C.
D.( )0kN150936sin:0
o
=+= .EF
xx
E.( )0kN150936sin:0
o
=−= .EF
xx kN 0.90−=
xE
•What does the negative sign signify?
•Discuss why the others are incorrect.

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Sample Problem 4.4
•Now apply the condition
that the sum of forces in
the y-direction must sum
to zero. kN200+=
yE
•Which equation is correct?
A.
B.
C.
D.
E.
•What does the positive sign signify?
•Discuss why the others are incorrect.() ( )0kN150936sinkN204:0
o
=−−= .EF
yy ( ) ( )0kN150
57
6
kN204:0 =+−=
.
EF
yy ( ) ( )0kN150
57
6
kN204:0 =−−=
.
EF
yy ( ) ( )0kN150
57
6
kN204:0 =−+=
.
EF
yy 
E
y
=+200 kN ( ) ( )0kN150936coskN204:0
o
=−−= .EF
yy

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Sample Problem 4.4
•Finally, apply the condition
that the sum of moments about
any point must equal zero.
•Three good points are D, E, and F.
Discuss what advantage each point has
over the others, or perhaps why each is
equally good. =:0
EM ( ) ( )
( ) ()
( ) 0m5.4kN150
5.7
6
m8.1kN20m6.3kN20
m4.5kN20m7.2kN20
=+−
++
++
EM mkN0.180 =
EM
•Assume that you choose point E to
apply the sum-of-moments condition.

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Equilibrium of a Two-Force Body
•Consider a plate subjected to two forces F
1 and F
2.
•For static equilibrium, the sum of moments about A
must be zero. The moment of F
2 must be zero. It
follows that the line of action of F
2 must pass
through A.
•Similarly, the line of action of F
1 must pass through
B for the sum of moments about B to be zero.
•Requiring that the sum of forces in any direction be
zero leads to the conclusion that F
1 and F
2 must
have equal magnitude but opposite sense.

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Equilibrium of a Three-Force Body
•Consider a rigid body subjected to forces acting at
only 3 points.
•Assuming that their lines of action intersect, the
moment of F
1 and F
2 about the point of intersection
represented by D is zero.
•Since the rigid body is in equilibrium, the sum of the
moments of F
1, F
2, and F
3 about any axis must be
zero. It follows that the moment of F
3 about D must
be zero as well and that the line of action of F
3 must
pass through D.
•The lines of action of the three forces must be
concurrent or parallel.

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Sample Problem 4.6
A man raises a 10-kg joist, of
length 4 m, by pulling on a rope.
Find the tension T in the rope
and the reaction at A.
SOLUTION:
•Create a free-body diagram of the joist.
Note that the joist is a 3 force body acted
upon by the rope, its weight, and the
reaction at A.
•The three forces must be concurrent for
static equilibrium. Therefore, the reaction
R must pass through the intersection of the
lines of action of the weight and rope
forces. Determine the direction of the
reaction R.
•Utilize a force triangle to determine the
magnitude of the reaction R.

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Sample Problem 4.6
•Create a free-body diagram of the joist.
•Determine the direction of the reaction R.()
( )
( )
6361
4141
3132
tan
m2.313 m51508282
m515020tan m4141)2545(cot
m4141
m828245cosm445cos
ooo
2
1
oo
.
.
.
AE
CE
..BDBFCE
..CDBD
.AFAECD
.ABBF
===
=−=−=
==+=
===
===
 
6.58=

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Sample Problem 4.6
•Determine the magnitude of the reaction R.
38.6sin
N 1.98
110sin4.31sin
==
RT N 8.147
N9.81
=
=
R
T

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Equilibrium of a Rigid Body in Three
Dimensions
•Six scalar equations are required to express the conditions for the
equilibrium of a rigid body in the general three dimensional case. = = =
===
000
000
zyx
zyx
MMM
FFF
•These equations can be solved for no more than 6 unknowns which
generally represent reactions at supports or connections or unknown
applied forces.
•The scalar equations are conveniently obtained by applying the vector
forms of the conditions for equilibrium,()   === 00 FrMF
O

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Reactions at Supports and Connections for a Three-Dimensional Structure

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Reactions at Supports and Connections for a Three-Dimensional Structure

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Sample Problem 4.8
A sign of uniform density weighs
1200 N and is supported by a ball-
and-socket joint at A and by two
cables.
Determine the tension in each cable
and the reaction at A.
SOLUTION:
•Create a free-body diagram for the sign.
•Apply the conditions for static
equilibrium to develop equations for
the unknown reactions.

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Sample Problem 4.8
•Create a free-body diagram for the sign.( )
( )kji
T
kji
T
7
2
7
3
7
6
3
2
3
1
3
2
++−=
=
−+−=
=
EC
ECEC
BD
BDBD
T
EC
EC
T
T
BD
BD
T
Since there are only 5 unknowns, the sign is
partially constrained. It is free to rotate about
the x axis. It is, however, in equilibrium for
the given loading.

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Sample Problem 4.8( )
( )( )
0 m.N 1440771080:
0514061:
0N 1200 m1.2
0:
0N 1200:
0:
0N 1200
7
2
3
2
7
3
3
1
7
6
3
2
=−+
=−
=−++=
=+−
=−++
=−−
=−++=


ECBD
ECBD
ECEBDBA
ECBDz
ECBDy
ECBDx
ECBD
T.T.
T.T.
TTA
TTA
TTA
k
j
jiTrTrM
k
j
i
jTTAF ( )( )( )kjiA N 100.2N 449.7N 1500.7
N 1400.8N 450
−+=
==
ECBD
TT
Solve the 5 equations for the 5 unknowns,•Apply the conditions for
static equilibrium to
develop equations for the
unknown reactions.

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Equilibrium of a Rigid Body in Two
Dimensions
•For known forces and moments that act on a
two-dimensional structure, the followings are
true:Ozyxz MMMMF ==== 00
•Equations of equilibrium become   === 000
Ayx MFF
where A can be any point in the plane of
the body.
•The 3 equations can be solved for no more
than 3 unknowns.
•The 3 equations cannot be augmented with
additional equations, but they can be replaced  === 000
BAx MMF
❑Rigid body is completely constrained
❑Reactions are statically determinate

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Statically Indeterminate Reactions
•More unknowns than equations
•A
x + B
x = 0
•Can however be determined separately by considering the deformations produced
in the truss.
•We will discuss this in the second part of the course - Mechanics of Materials

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Let us try this
•Write the following equations and solve
•σ??????
??????=0
•σ??????
??????=0
•σ??????
??????=0
P
x= P
y= 5 kN, Q
x= Q
y= 3 kN, S
x= S
y= 1 kN
W = 10 kN
All arms are 1 m long (except for diagonal)
Rigid body is constrained, but the reactions at its support are not statically determinate

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Statically Determinate Reactions
•Fewer unknowns than equations - partially constrained
•Equilibrium cannot be ascertained under the given loading conditions

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Let us try this
•Write the following equations and solve
•σ??????
??????=0
•σ??????
??????=0
•σ??????
??????=0
P
x= P
y= 5 kN, Q
x= Q
y= 3 kN, S
x= S
y= 1 kN
W = 10 kN
All arms are 1 m long (except for diagonal)
Rigid body is not fully constrained (is partially constrained), but the reactions at its
support are determinate

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Statically Indeterminate Reactions
•So, it is clear that if the number of equations of equilibrium are not
equal to the number unknowns:
➢either the rigid body is not completely constrained
➢or the reactions at its support are not statically determinate
➢or both
•While necessary, does it also seem to be the sufficient condition?
•Can you think of a situation where the number of unknowns is equal to
the number of equations but the system either not completely
constrained or the reactions at its supports are statically
indeterminate?

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Statically Indeterminate Reactions
•Equal number unknowns and equations, but improperly constrained
•The reactions are statically indeterminate

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Let us try this
•Write the following equations and solve
•σ??????
??????=0
•σ??????
??????=0
•σ??????
??????=0
P
x= P
y= 5 kN, Q
x= Q
y= 3 kN, S
x= S
y= 1 kN
W = 10 kN
All arms are 1 m long (except for diagonal)
Rigid body improperly constrained, and reactions at support are statically indeterminate

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Sample Problem 4.3
A loading car is at rest on an inclined
track. The gross weight of the car and
its load is 25 kN, and it is applied at at
G. The cart is held in position by the
cable.
Determine the tension in the cable and
the reaction at each pair of wheels.
SOLUTION:
•Create a free-body diagram for the car
with the coordinate system aligned
with the track.
•Determine the reactions at the wheels
by solving equations for the sum of
moments about points above each axle.
•Determine the cable tension by
solving the equation for the sum of
force components parallel to the track.
•Check the values obtained by verifying
that the sum of force components
perpendicular to the track are zero.

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Sample Problem 4.3
•Create a free-body diagram( )
( )
kN 10.5
25sinkN 25
kN 22.65
25coskN 25
−=
−=
+=
+=


y
x
W
W
•Determine the reactions at the wheels.( ) ( )
( )0mm 1250
mm 150kN 22.65mm 625kN 10.5:0
2
=+
−−=
R
M
A kN 8
2
+=R ( ) ( )
( )0mm 1250
mm 150kN 22.65mm 625kN 10.5:0
1
=−
−+=
R
M
B kN 2.5
1
=R
•Determine the cable tension.0TkN 22.65:0 =−+=xF kN 22.7+=T

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Sample Problem 4.4
A 6-m telephone pole of 1600-N used
to support the wires. Wires T
1 = 600 N
and T
2 = 375 N.
Determine the reaction at the fixed
end A.
SOLUTION:
•Create a free-body diagram for the
telephone cable.
•Solve 3 equilibrium equations for the
reaction force components and
couple at A.

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Sample Problem 4.4
•Create a free-body diagram for
the frame and cable.
•Solve 3 equilibrium equations for the
reaction force components and couple.010cos)N600(20cos)N375(:0 =−+= xx AF N 238.50+=
xA () ( ) 020sinN 37510sin600NN1600:0 =−−−= yy AF N 1832.45+=
yA  =:0
AM ( ) ( )
0 (6m)
20cos375N)6(10cos600N
=
−+ mM
A N.m00.1431+=
A
M

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Suggested Problems for Practice
•Problem 4.F2, Problem 4.1, Problem 4.10, Problem 4.15, Problem
4.23, Problem 4.27, Problem 4.31, Problem 4.33, Problem 4.52,
Problem 4.58, Problem 4.73, Problem 4.75, Problem 4.F7, Problem
4.95, Problem 4.105, Problem 4.128, Problem 4.149 Problem 4.152

40. In the study of the equilibrium of rigid bodies, i.e. the situation
when the external forces acting on a rigid body form a system
equivalent to zero, we have
 F = 0  M
O =  (r x F)
Resolving each force and each moment into its rectangular
components, the necessary and sufficient conditions for the
equilibrium of a rigid body are expressed by six scalar equations:
F
x = 0 F
y = 0 F
z = 0
M
x = 0 M
y = 0 M
z = 0
These equations can be used to determine unknown forces
applied to the rigid body or unknown reactions exerted by its
supports.
Chapter 4 EQUILIBRIUM OF RIGID BODIES

41. When solving a problem involving the equilibrium of a rigid
body, it is essential to consider all of the forces acting on
the body. Therefore, the first step in the solution of the
problem should be to draw a free-body diagram showing
the body under consideration and all of the unknown as
well as known forces acting on it.

42. F
x = 0 F
y = 0 M
A = 0
where A is an arbitrary point in the plane of the structure.
In the case of the equilibrium of two-dimensional structures,
each of the reactions exerted on the structure by its supports
could involve one, two, or three unknowns, depending upon the
type of support.
In the case of a two-dimensional structure, three equilibrium
equations are used, namely

43. F
x = 0 F
y = 0 M
A = 0
These equations can be used to solve for three unknowns. While
these three equilibrium equations cannot be augmented with
additional equations, any one of them can be replaced by
another equation. Therefore, we can write alternative sets of
equilibrium equations, such as
F
x = 0 M
A = 0 M
B = 0
where point B is chosen in such a way that the line AB is not
parallel to the y axis, or
M
A = 0 M
B = 0 M
C = 0
where points A, B, and C do not lie in a straight line.

44. Since any set of equilibrium equations can be solved for only
three unknowns, the reactions at the supports of a rigid two-
dimensional structure cannot be completely determined if they
involve more than three unknowns; they are said to be
statically indeterminate. On the other hand, if the reactions
involve fewer than three unknowns, equilibrium will not be
maintained under general loading conditions; the structure is
said to be partially constrained. The fact that the reactions
involve exactly three unknowns is no guarantee that the
equilibrium equations can be solved for all three unknowns. If
the supports are arranged is such a way that the reactions are
either concurrent or parallel, the reactions are statically
indeterminate, and the structure is said to be improperly
constrained.

45. A
B
F
1
F
2
Two particular cases of rigid body equilibrium are given
special attention. A two-force body is a rigid body subjected
to forces at only two points. The resultants F
1 and F
2 of these
two forces must have the same magnitude, the same line
of action, and opposite sense.

46. A
B
F
1
F
2
C
D
F
3
A three-force body is a rigid body subjected to forces at
only three points, and the resultants F
1, F
2 ,and F
3 of
these forces must be either concurrent or parallel. This
property provides an alternative approach to the solution
of problems involving a three-force body.

47.  F = 0  M
O =  (r x F)
When considering the equilibrium of a three-dimensional
body, each of the reactions exerted on the body by its supports
can involve between one and six unknowns, depending upon
the type of support.
In the general case of the equilibrium of a three-dimensional
body, the six scalar equilibrium equations listed at the beginning
of this review should be used and solved for six unknowns. In
most cases these equations are more conveniently obtained if
we first write
and express the forces F and position vectors r in terms of
scalar components and unit vectors. The vector product can
then be computed either directly or by means of determinants,
and the desired scalar equations obtained by equating to zero
the coefficients of the unit vectors.

48. As many as three unknown reaction components
can be eliminated from the computation of  M
O
through a judicious choice of point O. Also, the
reactions at two points A and B can be eliminated
from the solution of some problems by writing
the equation  M
AB = 0, which involves the
computation of the moments of the forces about
an axis AB joining points A and B.

49. If the reactions involve more than six unknowns,
some of the reactions are statically indeterminate; if
they involve fewer than six unknowns, the rigid body
is only partially constrained.
Even with six or more unknowns, the rigid body will
be improperly constrained if the reactions associated
with the given supports either are parallel or intersect
the same line.