CH 12 – Equilibrium
In this chapter we will first examine the conditions for equilibrium of a rigid body.
Then we will briefly describe how we characterize the extent to which a body can be
deformed.
An object is in equilibrium when both the forces and torque add to zero. That is,
F  0
  0
Both the force and torque are vectors. As defined in an earlier chapter, torque is
  r F
and the magnitude of the torque is
  rF sin  ,
where  is the angle between r and F.
The direction of  is perpendicular to the plane formed by r and F as given by the ‘righthand rule’. So, if r and F are in the x-y plane, then  is along the + or – z axis. In
component form, the conditions for equilibrium are
 Fx  0 ,  Fy  0,  Fz  0
 x  0,  y  0,  z  0
In this chapter we will only deal with 2-dimensional problems where the forces are in the
x-y plane. Then
 Fx  0,  Fy  0
 z  0
Note: While torque depends on choice of axis (pivot), if the net torque is zero about one
axis, then it is zero about any other axis. This means that we can selectively choose the
pivot when working a problem to best simplify the solution.
Center of mass and center of gravity
Solving equilibrium problems often requires knowing the center of gravity of a body.
Center of mass is the mass-weighted average position of the object.
1
rcm 
 mi ri
 mi
For an object of uniform density, this is just the geometric center of the object (the
volume-weighted average position).
Center of gravity is the weight-weighted average position.
rcg 
 mi gi ri
 mi gi
Since g is typically constant over the structure of interest, then center of gravity and
center of mass are the same and will be used interchangeably.
Example 1:
A 75-kg diver stands on the end of a 3-m diving board which has a mass of 60 kg and has
a uniform rectangular structure. The diving board is clamped at the end opposite the
diver and 1 m from this end. What are the forces exerted by the clamps on the diving
board?
F2
x1
Wb
Wm
F1
ℓ/2
ℓ/2
Solution:
The weight of the board can be treated as a single force acting at the center of the board.
Since we have two unknowns, F1 and F2, we need two independent equations. From the
1st condition for equilibrium we have
 Fy  0
F2  F1  Wb  Wm  0
F2  F1  ( 135kg )( 9.8m / s 2 )  1323 N
2
The 2nd condition for equilibrium is
  0
When calculating the torques, we need to choose a pivot. Any pivot will work; however,
if we choose one of the clamps as a pivot, then the torque formula simplifies, since then
the torque due to F1 or F2 will be zero. Choosing the left end as the pivot and a sign
convention such that counter-clockwise torques are positive (torque direction out of the
page) and clockwise torques are negative (torque direction into the page), we have from
the 2nd condition for equilibrium
F2  x2  Wb   Wm   0
2
F2  1.5Wb  3Wm  0
F2  ( 1.5 )( 60 )( 9.8 )  ( 3 )( 75 )( 9.8 )  3087 N
Substituting this value for F2 into the force formula, we have
F1  F2  1323 N  3087 N  1323 N  1764 N
Note: We guessed that F1 would be down and F2 would be up. Had we guessed wrong,
then we would have gotten a negative number for the force.
An alternative approach to solving this problem would be to apply the 2nd condition twice
using two different pivots. For example, choosing pivot 2 gives a single equation that
can be used to solve for F1.
Example 2:
A 100-N sign hangs from a horizontal rod which has weight 150 N and length 2 m. The
left end of the rod is attached to a building and the right end is suspended by a cable at an
angle of 30o from the horizontal. Find the force exerted by the building on the left end of
the rod and the tension in the cable. The sign hangs 0.5 m from the right end of the rod.
T
Fwall

30o
ℓ/2
30o
Wrod
xsign
ℓ
3
Wsign
Solution:
Summing the torques about a pivot through the left end of the beam gives
o
 T sin 30  Wrod / 2  Wsign xsign  0
T
Wrod / 2  Wsign xsign
sin 30
o

( 150 )( 2 / 2 )  ( 100 )( 1.5 )
 300 N
( 2 )( 0.5 )
Summing the forces gives
o
 Fx  Fwall ,x  T cos 30  0
Fwall ,x  ( 300 )cos 30o  259.8 N
o
 Fy  Fwall , y  T sin 30  Wrod  Wsign  0
Fwall , y  Wrod  Wsign  T sin 30o  150  100  ( 300 )( 0.5 )  100 N
The magnitude of the wall force is
2
2
2
2
Fwall  Fwall
,x  Fwall , y  ( 259.8 )  ( 100 )  278.4 N
And the angle it makes with the beam is
 Fwall , y 
 100 
 21.1o
  tan 1 

 Fwall ,x 
 259.8 


  tan1 
Note that the angle of the force exerted by the wall is not the same as the cable angle (a
common misconception).
Moduli of Elasticity
Given a sufficient stress, an object will be strained. That is, its dimensions will change.
Stress is defined as force per unit area, while strain is defined as a fractional change in
dimension. The ratio of stress to strain is the modulus of elasticity.
Tensile and Compressive Stress
If an object of length L is pulled (or compressed) with a force normal to its end area, then
the tensile stress if F/A, where A is end area. It will undergo a fractional change in length,
a tensile strain, given by L/L.
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L
A
F
F
L+L
The stress-strain relationship is
F
L
E
,
A
L
where E is called Young’s modulus.
Shear Stress
A force applied parallel to a surface can shear an object as shown.
A
x
F
L
F
The shear stress is F/A, where A is the area of the surface subject to the horizontal force.
The shear strain x/L represents an angular deformation. The shear stress-strain
relationship is
F
x
G
,
A
L
where G is the shear modulus.
Hydraulic Stress
The volume of a body can be changed by applying a fluid pressure to all parts of its
surface.
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V
The hydraulic stress-strain relationship is
p  B
V
V
where B is the bulk modulus. The negative sign means that a positive pressure causes a
negative change in volume.
The moduli of elasticity are materials properties that can typically be found in
engineering tables.
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