Chapter 5:
Equilibrium of a Rigid Body
Engineering Mechanics: Statics
Chapter Objectives
To develop the equations of equilibrium for a
rigid body.
 To introduce the concept of the free-body
diagram for a rigid body.
 To show how to solve rigid-body equilibrium
problems using the
 equations of equilibrium.

Chapter Outline
Conditions for Rigid Equilibrium
 Free-Body Diagrams
 Equations of Equilibrium
 Two and Three-Force Members
 Equilibrium in Three Dimensions
 Equations of Equilibrium
 Constraints for a Rigid Body

5.1 Conditions for Rigid-Body
Equilibrium



Consider rigid body fixed in the
x, y and z reference and is either
at rest or moves with reference
at constant velocity
Two types of forces that act on
it, the resultant internal force
and the resultant external force
Resultant internal force fi is
caused by interactions with
adjacent particles
5.1 Conditions for Rigid-Body
Equilibrium


Resultant external force Fi
represents the effects of
gravitational, electrical,
magnetic, or contact forces
between the ith particle and
adjacent bodies or particles not
included within the body
Particle in equilibrium, apply
Newton’s first law,
Fi + fi = 0
5.1 Conditions for Rigid-Body
Equilibrium



When equation of equilibrium is
applied to each of the other particles
of the body, similar equations will
result
Adding all these equations
vectorially,
∑Fi + ∑fi = 0
Summation of internal forces = 0
since internal forces between
particles in the body occur in equal
but opposite collinear pairs
(Newton’s third law)
5.1 Conditions for Rigid-Body
Equilibrium




Only sum of external forces will
remain
Let ∑Fi = ∑F, ∑F = 0
Consider moment of the forces
acting on the ith particle about the
arbitrary point O
By the equilibrium equation and
distributive law of vector cross
product,
ri X (Fi + fi) = ri X Fi + ri X fi =
0
5.1 Conditions for Rigid-Body
Equilibrium





Similar equations can be written for
other particles of the body
Adding all these equations
vectorially,
∑ri X Fi + ∑ri X fi = 0
Second term = 0 since internal
forces occur in equal but opposite
collinear pairs
Resultant moment of each pair of
forces about point O is zero
Using notation ∑MO = ∑ri X Fi,
∑MO = 0
5.1 Conditions for Rigid-Body
Equilibrium



Equations of Equilibrium for Rigid Body
∑F = 0
∑MO = 0
A rigid body will remain in equilibrium
provided the sum of all the external forces
acting on the body = 0 and sum of moments
of the external forces about a point = 0
For proof of the equation of equilibrium,
- Assume body in equilibrium
5.1 Conditions for Rigid-Body
Equilibrium
- Force system acting on the body satisfies the
equations ∑F = 0 and ∑MO = 0
- Suppose additional force F’ is applied to the
body
∑F + F’ = 0
∑MO + MO’= 0
where MO’is the moment of F’ about O
- Since ∑F = 0 and ∑MO = 0, we require F’ = 0
and MO’
- Additional force F’ is not required and
equations ∑F = 0 and ∑MO = 0 are sufficient
5.2 Free-Body Diagrams



FBD is the best method to represent all the
known and unknown forces in a system
FBD is a sketch of the outlined shape of the
body, which represents it being isolated from
its surroundings
Necessary to show all the forces and couple
moments that the surroundings exert on the
body so that these effects can be accounted
for when equations of equilibrium are applied
5.2 Free-Body Diagrams
5.2 Free-Body Diagrams
5.2 Free-Body Diagrams
5.2 Free-Body Diagrams
Support Reactions



If the support prevents the translation of a body
in a given direction, then a force is developed on
the body in that direction
If rotation is prevented, a couple moment is
exerted on the body
Consider the three ways a horizontal member,
beam is supported at the end
- roller, cylinder
- pin
- fixed support
5.2 Free-Body Diagrams
Support Reactions
Roller or cylinder
 Prevent the beam from
translating in the vertical
direction
 Roller can only exerts a
force on the beam in the
vertical direction
5.2 Free-Body Diagrams
Support Reactions
Pin
 The pin passes through a hold in the beam
and two leaves that are fixed to the ground
 Prevents translation of the beam in any
direction Φ
 The pin exerts a force F on the beam in this
direction
5.2 Free-Body Diagrams
Support Reactions
Fixed Support
 This support prevents both
translation and rotation of the beam
 A couple and moment must be
developed on the beam at its point of
connection
 Force is usually represented in x and
y components
5.2 Free-Body Diagrams




Cable exerts a force on the
bracket
Type 1 connections
Rocker support for this bridge
girder allows horizontal
movements so that the bridge
is free to expand and contract
due to temperature
Type 5 connections
5.2 Free-Body Diagrams




Concrete Girder rest on the
ledge that is assumed to act
as a smooth contacting
surface
Type 6 connections
Utility building is pin
supported at the top of the
column
Type 8 connections
5.2 Free-Body Diagrams


Floor beams of this building
are welded together and
thus form fixed connections
Type 10 connections
5.2 Free-Body Diagrams
External and Internal Forces




A rigid body is a composition of particles, both
external and internal forces may act on it
For FBD, internal forces act between particles
which are contained within the boundary of the
FBD, are not represented
Particles outside this boundary exert external
forces on the system and must be shown on FBD
FBD for a system of connected bodies may be
used for analysis
5.2 Free-Body Diagrams
Weight and Center of Gravity




When a body is subjected to gravity, each
particle has a specified weight
For entire body, consider gravitational forces as a
system of parallel forces acting on all particles
within the boundary
The system can be represented by a single
resultant force, known as weight W of the body
Location of the force application is known as the
center of gravity
5.2 Free-Body Diagrams
Weight and Center of Gravity
Center of gravity occurs at the geometric
center or centroid for uniform body of
homogenous material
 For non-homogenous bodies and usual
shapes, the center of gravity will be given

5.2 Free-Body Diagrams
Idealized Models
Needed to perform a correct force analysis
of any object
 Careful selection of supports, material,
behavior and dimensions for trusty results
 Complex cases may require developing
several different models for analysis

5.2 Free-Body Diagrams
Idealized Models



Consider a steel beam used to support the
roof joists of a building
For force analysis, reasonable to assume
rigid body since small deflections occur when
beam is loaded
Bolted connection at A will allow for slight
rotation when load is applied => use Pin
5.2 Free-Body Diagrams
Support at B offers no resistance to horizontal
movement => use Roller
 Building code requirements used to specify the
roof loading (calculations of the joist forces)
 Large roof loading forces account for extreme
loading cases and for dynamic or vibration
effects
 Weight is neglected when it is small compared to
the load the beam supports
5.2 Free-Body Diagrams
Idealized Models




Consider lift boom, supported by pin
at A and hydraulic cylinder at BC
(treat as weightless link)
Assume rigid material with density
known
For design loading P, idealized model
is used for force analysis
Average dimensions used to specify
the location of the loads and supports
5.2 Free-Body Diagrams
Example 5.1
Draw the free-body diagram of the uniform
beam. The beam has a mass of 100kg.
5.2 Free-Body Diagrams
Solution
Free-Body Diagram
5.2 Free-Body Diagrams
Solution
 Support at A is a fixed wall
 Three forces acting on the beam at A denoted as Ax,
Ay, Az, drawn in an arbitrary direction
 Unknown magnitudes of these vectors
 Assume sense of these vectors
 For uniform beam,
Weight, W = 100(9.81) = 981N
acting through beam’s center of gravity, 3m from A
5.2 Free-Body Diagrams
Example 5.5
The free-body diagram of each object is
drawn. Carefully study each solution and
identify what each loading represents.
5.2 Free-Body Diagrams
Solution
5.2 Free-Body Diagrams
Solution
5.3 Equations of Equilibrium



For equilibrium of a rigid body in 2D,
∑Fx = 0; ∑Fy = 0; ∑MO = 0
∑Fx and ∑Fy represent the algebraic sums of the
x and y components of all the forces acting on
the body
∑MO represents the algebraic sum of the couple
moments and moments of the force components
about an axis perpendicular to x-y plane and
passing through arbitrary point O, which may lie
on or off the body
5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations


For coplanar equilibrium problems, ∑Fx = 0; ∑Fy
= 0; ∑MO = 0 can be used
Two alternative sets of three independent
equilibrium equations may also be used
∑Fa = 0; ∑MA = 0; ∑MB = 0
When applying these equations, it is required
that a line passing through points A and B is not
perpendicular to the a axis
5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
 Consider FBD of an arbitrarily shaped body
 All the forces on FBD may be
replaced by an equivalent
resultant force
FR = ∑F acting at point A and a
resultant moment MRA = ∑MA
 If ∑MA = 0 is satisfied, MRA = 0
5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
 If FR satisfies ∑Fa = 0, there is no
component along the a axis and
its line of axis is perpendicular
to the a axis
 If ∑MB = 0 where B does not lies
on the line of action of FR, FR = 0
 Since ∑F = 0 and ∑MA = 0, the
body is in equilibrium
5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
 A second set of alternative equations is
∑MA = 0; ∑MB = 0; ∑MC = 0
 Points A, B and C do not lie on the
same line
 Consider FBD, if If ∑MA = 0, MRA = 0
 ∑MA = 0 is satisfied if line of action of FR passes
through point B
 ∑MC = 0 where C does not lie on line AB
 FR = 0 and the
body is in equilibrium
5.3 Equations of Equilibrium
Example 5.7
The cord supports a force of 500N and wraps over
the frictionless pulley. Determine the tension in the
cord at C and the horizontal
and vertical components at
pin A.
5.3 Equations of Equilibrium
Solution
FBD of the cord and pulley
 Principle of action: equal but opposite reaction
observed in the FBD
 Cord exerts an unknown load
distribution p along part of
the pulley’s surface
 Pulley exerts an equal but
opposite effect on the cord
5.3 Equations of Equilibrium
Solution
FBD of the cord and pulley
 Easier to combine the FBD of the pulley and
contracting portion of the cord so that the
distributed load becomes internal to the
system
and is eliminated from the
analysis
5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
 M A  0;
500 N (0.2m)  T (0.2m)  0
T  500 N
Tension remains constant as cord
passes over the pulley (true for
any angle at which the cord is
directed and for any radius of
the pulley
5.3 Equations of Equilibrium
Solution
   Fx  0;
 Ax  500 sin 30 N  0
Ax  250 N
   Fy  0;
Ay  500 N  500 cos 30 N  0
Ay  933 N
5.3 Equations of Equilibrium
Example 5.8
The link is pin-connected at a and rest a
smooth support at B. Compute the horizontal
and vertical components of reactions at pin A
5.3 Equations of Equilibrium
Solution
FBD
 Reaction NB is perpendicular to the link at
B
 Horizontal and vertical
components of reaction
are represented at A
5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
 M A  0;
 90 N .m  60 N (1m)  N B (0.75m)  0
N B  200 N
   Fx  0;
Ax  200 sin 30 N  0

Ax  100 N
5.3 Equations of Equilibrium
Solution
   Fy  0;
Ay  60 N  200 cos 30 N  0
Ay  233 N
5.4 Two- and Three-Force
Members

Simplify some equilibrium problems by
recognizing embers that are subjected top only 2
or 3 forces
Two-Force Members

When a member is subject to
no couple moments and forces
are applied at only two points
on a member, the member is
called a two-force member
5.4 Two- and Three-Force
Members
Two-Force Members
Example
 Forces at A and B are summed to
obtain their respective resultants FA
and FB
 These two forces will maintain
translational and force equilibrium
provided FA is of equal magnitude
and opposite direction to FB
 Line of action of both forces is
known and passes through A and B
5.4 Two- and Three-Force
Members
Two-Force Members


Hence, only the force
magnitude must be determined
or stated
Other examples of the twoforce members held in
equilibrium are shown in the
figures to the right
5.4 Two- and Three-Force
Members
Three-Force Members


If a member is subjected to only three forces, it is
necessary that the forces be either concurrent or
parallel for the member to be in equilibrium
To show the concurrency
requirement, consider a body
with any two of the three forces
acting on it, to have line of
actions that intersect at point O
5.4 Two- and Three-Force
Members
Three-Force Members



To satisfy moment equilibrium about O, the third
force must also pass through O, which then
makes the force concurrent
If two of the three forces parallel,
the point of currency O, is
considered at “infinity”
Third force must parallel to
the other two forces to insect
at this “point”
5.4 Two- and Three-Force
Members


Bucket link AB on the back
hoe is a typical example of a
two-force member since it is
pin connected at its end
provided its weight is
neglected, no other force acts
on this member
The hydraulic cylinder is pin
connected at its ends, being
a two-force member
5.4 Two- and Three-Force
Members


The boom ABD is subjected to the
weight of the suspended motor at D,
the forces of the hydraulic cylinder at
B, and the force of the pin at A. If the
boom’s weight is neglected, it is a
three-force member
The dump bed of the truck operates
by extending the hydraulic cylinder
AB. If the weight of AB is neglected,
it is a two-force member since it is
pin-connected at its end points
5.4 Two- and Three-Force
Members
Example 5.13
The lever ABC is pin-supported
at A and connected to a short
link BD. If the weight of the
members are negligible,
determine the force of the pin
on the lever at A.
5.4 Two- and Three-Force
Members
Solution
FBD
 Short link BD is a two-force
member, so the resultant forces
at pins D and B must be equal,
opposite and collinear
 Magnitude of the force is
unknown but line of action
known as it passes through B
and D
 Lever ABC is a three-force
member
5.4 Two- and Three-Force
Members
Solution
FBD
 For moment equilibrium, three
non-parallel forces acting on it
must be concurrent at O
 Force F on the lever at B is
equal but opposite to the force
F acting at B on the link
 Distance CO must be 0.5m
since lines of action of F and
the 400N force are known
5.4 Two- and Three-Force
Members
Solution
Equations of
Equilibrium
 0.7 

  60.3
 0.4 
   Fx  0;
  tan 1 
FA cos 60.3  F cos 45  400 N  0
   Fy  0;
FA sin 60.3  F sin 45  0
Solving,
FA  1.07kN
F  1.32kN
5.5 Equilibrium in Three
Dimensions (FBD)
5.5 Equilibrium in Three
Dimensions (FBD)
5.5 Equilibrium in Three
Dimensions (FBD)
5.5 Equilibrium in Three
Dimensions (FBD)
5.5 Equilibrium in Three
Dimensions (FBD)

Ball and socket joint
provides a
connection for the
housing of an earth
grader to its frame

Journal bearing
supports the end of
the shaft
5.5 Equilibrium in Three
Dimensions (FBD)

Thrust bearing is
used to support the
drive shaft on the
machine

Pin is used to
support the end of
the strut used on a
tractor
5.5 Equilibrium in Three
Dimensions (FBD)
Example 5.14
Several examples of objects along with their
associated free-body diagrams are shown. In
all cases, the x, y and z axes are established
and the unknown reaction components are
indicated in the positive sense. The weight of
the objects is neglected.
5.5 Equilibrium in Three
Dimensions (FBD)
Solution
5.5 Equilibrium in Three
Dimensions (FBD)
Solution
5.5 Equilibrium in Three
Dimensions (FBD)
Solution
5.5 Equilibrium in Three
Dimensions (FBD)
Solution
5.6 Equations of Equilibrium
Vector Equations of Equilibrium

For two conditions for equilibrium of a rigid body
in vector form,
∑F = 0
∑MO = 0
where ∑F is the vector sum of all the external
forces acting on the body and ∑MO is the sum of
the couple moments and the moments of all the
forces about any point O located either on or off
the body
5.6 Equations of Equilibrium
Scalar Equations of Equilibrium
If all the applied external forces and
couple moments are expressed in
Cartesian vector form
∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0
∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0
 i, j and k components are independent
from one another

5.6 Equations of Equilibrium
Scalar Equations of Equilibrium
∑Fx = 0, ∑Fy = 0, ∑Fz = 0 shows that the
sum of the external force components
acting in the x, y and z directions must be
zero
 ∑Mx = 0, ∑My = 0, ∑Mz = 0 shows that the
sum of the moment components about the
x, y and z axes to be zero

5.7 Constraints for a Rigid Body

To ensure the equilibrium of a rigid body, it is
necessary to satisfy the equations equilibrium
and have the body properly held or
constrained by its supports
Redundant Constraints


More support than needed for equilibrium
Statically indeterminate: more unknown
loadings on the body than equations of
equilibrium available for their solution
5.7 Constraints for a Rigid Body
Redundant Constraints
Example
 For the 2D and 3D problems, both are
statically indeterminate because of additional
supports reactions
 In 2D, there are 5 unknowns but 3
equilibrium equations can be drawn
5.7 Constraints for a Rigid Body
Redundant Constraints
Example
 In 3D, there are 8 unknowns but 6 equilibrium
equations can be drawn
 Additional equations
involving the physical
properties of the body
are needed to solve
indeterminate problems