Lecture #7 Part a
2D Rigid Body Equilibrium
(Reference Chapter 5,
sections 1, 2, 3 and 4 – 2D)
Example of Rigid Body Equilibrium


Can we solve for the Reaction at each wheel on
this truck using Translational Equilibrium?
Is this problem Statically Indeterminate?
Sample Problem – Truck





Given: Mass of the truck is 310.56 Slugs.
Force pulling the truck is 800 Lbs attached at the bumper 1.5 ft off the
ground and acting parallel to the ground.
The distance between the wheels is 15 ft
Center of Gravity is 5 ft in front of the rear wheel
There is an equal force on each tire toward the back of the truck to offset
the pulling force
Free Body Diagrams show all
Unknown External Forces
Unknown external forces are also known
as constraints
 Constraints on Space Diagrams need to
be replaced with the correct reactionary
forces on the Free Body Diagram
 In a real world problem, the type of
constraint at each reaction point needs to
be determined

APPLICATIONS
The truck ramp has a weight of 400 lb.
The ramp is pinned to the body of the truck and held in the
position by the cable. How can we determine the cable tension
and support reactions ?
How are the idealized model and the free body diagram used to
do this?
Which diagram above is the idealized model?
APPLICATIONS
(continued)
Two smooth pipes, each having a mass of 300 kg, are
supported by the tines of the tractor fork attachment.
How can we determine all the reactive forces ?
Again, how can we make use of an idealized model and a
free body diagram to answer this question?
CONDITIONS FOR RIGID-BODY EQUILIBRIUM
(Section 5.1)
In contrast to the forces on a particle, the
forces on a rigid-body are not usually
concurrent and may cause rotation of the
body (due to the moments created by the
forces).
Forces on a particle
For a rigid body to be in equilibrium, the
net force as well as the net moment
about any arbitrary point O must be
equal to zero.
 F = 0 (no translation)
Forces on a rigid body
and  MO = 0 (no rotation)
Write scaler equations on board
THE PROCESS OF SOLVING RIGID BODY
EQUILIBRIUM PROBLEMS
For analyzing an actual physical system, first we need to
create an idealized model (above right).
Then we need to draw a free-body diagram
(FBD) showing all the external (active and
reactive) forces.
Finally, we need to apply the equations of
equilibrium to solve for any unknowns.
FREE-BODY DIAGRAMS
(Section 5.2)
Idealized model
Free-body diagram (FBD)
1. Draw an outlined shape. Imagine the body to be isolated
or cut “free” from its constraints and draw its outlined
shape.
2. Show all the external forces and couple moments. These
typically include: a) applied loads, b) support reactions,
and, c) the weight of the body.
FREE-BODY DIAGRAMS
(continued)
Idealized model
Free-body diagram
3. Label loads and dimensions on the FBD: All known
forces and couple moments should be labeled with
their magnitudes and directions. For the unknown
forces and couple moments, use letters like Ax, Ay,
MA, etc.. Indicate any necessary dimensions.
SUPPORT REACTIONS IN 2-D
A few example sets of diagrams s are shown above. Other
support reactions are given in your textbook (Table 5-1).
As a general rule, if a support prevents translation of a body in a
given direction, then a force is developed on the body in the
opposite direction.
Similarly, if rotation is prevented, a couple moment is exerted on
the body in the opposite direction.
Unknown External Forces – 2D
Unknown External Forces – 2D
EXAMPLE
Given: The operator applies a vertical
force to the pedal so that the
spring is stretched 1.5 in. and the
force in the short link at B is
20 lb.
Draw: A an idealized model and freebody diagram of the foot pedal.
CONCEPT QUIZ
1. The beam and the cable (with a frictionless pulley at D) support an
80 kg load at C. In a FBD of only the beam, there are how many
unknowns?
A) 2 forces and 1 couple moment
B) 3 forces and 1 couple moment
C) 3 forces
D) 4 forces
CONCEPT QUIZ
2. If the directions of the force and the couple moments are
both reversed, what will happen to the beam?
A)
B)
C)
D)
The beam will lift from A.
The beam will lift at B.
The beam will be restrained.
The beam will break.
GROUP PROBLEM SOLVING
Draw a FBD of the crane boom, which is supported by a
pin at A and cable BC. The load of 1250 lb is suspended
at B and the boom weighs 650 lb.
GROUP PROBLEM SOLVING (continued)
FBD
GROUP PROBLEM SOLVING
Draw a FBD of member ABC, which is supported
by a smooth collar at A, roller at B, and link CD.
GROUP PROBLEM SOLVING (continued)
FBD
ATTENTION QUIZ
1. Internal forces are not shown on a free-body diagram because the
internal forces are_____. (Choose the most appropriate answer.)
A) Equal to zero
B) Equal and opposite and they do not
affect the calculations
C) Negligibly small
D) Not important
2. How many unknown support reactions
are there in this problem?
A)
2 forces and 2 couple moments
B)
1 force and 2 couple moments
C)
3 forces
D)
3 forces and 1 couple moment
Remaining slides = Section 5.3/5.4 –
2D Rigid Body Equilibrium!!!
(Finally)
READING QUIZ (Section 5.3/5.4)
1. The three scalar equations  FX =  FY =  MO = 0, are ____
equations of equilibrium in two dimensions.
A) Incorrect
B) The only correct
C) The most commonly used
D) Not sufficient
2. A rigid body is subjected to forces as
shown. This body can be considered
as a ______ member.
A) Single-force
B) Two-force
C) Three-force
D)
Six-force
APPLICATIONS
A
The uniform truck ramp has a weight of 400 lb.
The ramp is pinned at A and held in the position by the cable.
How can we determine the forces acting at the pin A and the
force in the cable ?
APPLICATIONS (continued)
A 850 lb of engine is supported by three chains, which are
attached to the spreader bar of a hoist.
You need to check to see if the breaking strength of any of the
chains is going to be exceeded. How can you determine the
force acting in each of the chains?
EQUATIONS OF EQUILIBRIUM
(Section 5.3)
A body is subjected to a system of forces
that lie in the x-y plane. When in
equilibrium, the net force and net moment
acting on the body are zero (as discussed
earlier in Section 5.1). This 2-D condition
can be represented by the three scalar
equations:
 Fx = 0
 Fy = 0
 MO = 0
where point O is any arbitrary point.
Please note that these equations are the ones most commonly
used for solving 2-D equilibrium problems. There are two
other sets of equilibrium equations that are rarely used. For
your reference, they are described in the textbook.
TWO-FORCE MEMBERS & THREE FORCE-MEMBERS
(Section 5.4)
The solution to some equilibrium problems can be simplified
if we recognize members that are subjected to forces at only
two points (e.g., at points A and B).
If we apply the equations of equilibrium to such a member, we
can quickly determine that the resultant forces at A and B must
be equal in magnitude and act in the opposite directions along
the line joining points A and B.
EXAMPLE OF TWO-FORCE MEMBERS
In the cases above, members AB can be considered as two-force
members, provided that their weight is neglected.
This fact simplifies the equilibrium
analysis of some rigid bodies since the
directions of the resultant forces at A and B
are thus known (along the line joining
points A and B).
STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS
1. If not given, establish a suitable x - y coordinate system.
2. Draw a free body diagram (FBD) of the object under
analysis.
3. Apply the three equations of equilibrium (E-of-E) to
solve for the unknowns.
IMPORTANT NOTES
1. If there are more unknowns than the number of independent
equations, then we have a statically indeterminate situation.
We cannot solve these problems using just statics.
2. The order in which we apply equations may affect the
simplicity of the solution. For example, if we have two
unknown vertical forces and one unknown horizontal force,
then solving  FX = 0 first allows us to find the horizontal
unknown quickly.
3. If the answer for an unknown comes out as negative number,
then the sense (direction) of the unknown force is opposite to
that assumed when starting the problem.
EXAMPLE 2D Equilibrium:
Given: The 4kN load at B of
the beam is supported
by pins at A and C .
Find: The support reactions
at A and C.
Plan:
1. Put the x and y axes in the horizontal and vertical directions,
respectively.
2. Determine if there are any two-force members.
3. Draw a complete FBD of the boom.
4. Apply the E-of-E to solve for the unknowns.
EXAMPLE (Continued)
FBD of the beam:
AY
4 kN
1.5 m
1.5 m
AX
A
C
45°
B
FC
Note: Upon recognizing CD as a two-force member, the number of
unknowns at C are reduced from two to one. Now, using E-o-f E, we get,
+ MA = FC sin 45  1.5 – 4  3 = 0
Fc = 11.31 kN or 11.3 kN
 + FX = AX + 11.31 cos 45 = 0;
AX = – 8.00 kN
 + FY = AY + 11.31 sin 45 – 4 = 0;
AY = – 4.00 kN
Note that the negative signs means that the reactions have the opposite
direction to that shown on FBD.
CONCEPT QUIZ
1. For this beam, how many support
reactions are there and is the
problem statically determinate?
A) (2, Yes)B) (2, No)
C) (3, Yes) D) (3, No)
F
2. The beam AB is loaded and supported as
shown: a) how many support reactions
are there on the beam, b) is this problem
A
statically determinate, and c) is the
structure stable?
A) (4, Yes, No)
B) (4, No, Yes)
C) (5, Yes, No)
D) (5, No, Yes)
F
F
Fixed
support
F
F
B
GROUP PROBLEM SOLVING
2D Equilibrium
Given: The jib crane is supported by
a pin at C and rod AB. The
load has a mass of 2000 kg
with its center of mass
located at G.
Assume x = 5 m.
Plan:
Find: Support reactions at B
and C.
a) Establish the x – y axes.
b) Draw a complete FBD of the jib crane beam.
c) Apply the E-of-E to solve for the unknowns.
GROUP PROBLEM SOLVING (Continued)
FAB
4m
Cx
4
5
3
0.2 m
5m
Cy
2000(9.81)
N
FBD of the beam
First write a moment equation about Point C.
Why point C?
+  MC = (3 / 5) FAB 4 + (4 / 5) FAB 0.2 – 2000(9.81)  5 = 0
FAB = 38320 N = 38.3 kN
GROUP PROBLEM SOLVING (Continued)
FAB
4m
Cx
4
5
3
0.2 m
5m
Cy
2000(9.81) N
FBD of the beam
FAB = 38320 N = 38.3 kN
Now solve the FX and FY equations.
 + FX = Cx –

(4 / 5) 38320 = 0
+ FY = – Cy + (3 / 5) 38320 – 2000(9.81) = 0
Solving these two equations, we get
Cx = 30656 N or 30.7 kN and Cy = 3372 N or 33.7 kN
ATTENTION QUIZ
1. Which equation of equilibrium allows
you to determine FB right away?
100 lb
AX
A)  FX = 0
B)  FY = 0
C)  MA = 0
D) Any one of the above.
2. A beam is supported by a pin joint
and a roller. How many support
reactions are there and is the
structure stable for all types of
loadings?
A) (3, Yes)
B) (3, No)
C) (4, Yes)
D) (4, No)
A
B
AY
FB
Sample Problem – 2D Truss

Solve this for the
Unknown Reactions
at Point A and B
2m
1.5m
1.5m
Sample Problem - Table

Given that the table
weighs 50 lbs, what is
the maximum weight
of the person sitting at
it before it tips over?
F5-1: Find Support
Reactions
Find Support Reactions
24,000 lb
5:20. The train car has a weight of 24,000 lb and a center of gravity at
G. It is suspended from its front and rear track by six tires located at A,
B, and C. Determine the normal reactions on these tires if the track is
assumed to be a smooth surface and an equal portion of the load is
supported at both the front and rear tires.
NC
NA
NB
5.21: Determine the horizontal and vertical
components of reaction at pin A and the tension
developed in cable BC used to support the steel
frame.
5-40: The assembly has a weight of 250 lb and center of
gravity at G1. If it is intended to support a maximum load of
400 lb placed at point G2, determine the smallest
counterweight W that should be placed at B in order to prevent
the platform from tipping over.
5.47: The motor has a weight of 850 lb. Determine
the force that each of the chains exerts on the
supporting hooks at A, B and C. Neglect the size
of the hooks and thickness of the beam.
What’s better? Attaching the rope at A or B? FBD!!!!!
Example: 2D – 2 force and 3 force members:
Understanding the Question
The last part of doing Rigid Body
Equilibrium problems is understanding
what needs solved.
 This has to be done by careful
examination and engineering judgment
(for lack of a better word)

Practical Examples of Statics:
Washing Machine
Isolator
Accelerome
ter location
Left/Rear
Accelerome
ter location
Right/Front
Accelerome
ter location
Left/Front
Washing Machine
Isolator
Acceleration Level at .67 kg Imbalance
(Drum Speed = 1068 rpm)
Accelration (g)
0.6
0.5
0.4
0.3
0.2
0.1
0
30 duro
50 duro
70 duro
Foot Mount Option
Vertical acceleration, left front
no mount
Woofer Isolator
Face Mask Shock Isolator:
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damped soft butyl.
Butyl
elastomer
bonded to
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Case Study –Industrial Caster
Isolation
mount
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Torsion Load= 300 lb-ft
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