Physics 2
Equilibrium and Elasticity
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Static Equilibrium
Sometimes an object is subject to several forces, but it does not accelerate.
This is when the object is in equilibrium. We have done problems like this
before, but we neglected the rotational motion. To incorporate this, we
simply need to add a torque formula to our typical force formulas.
𝐹=0
Separate formula for
each vector component
𝜏=0
Choose any convenient pivot point
Here’s an example:
A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
60°
1.5m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
Fx  0
Fy  0
  0
60°
T
Hy
Hx
1.5m
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force that the hinge exerts on the beam.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
60°
T
Fx  0
Hy
Hx  Tx  0
Hx  T  cos(30 )  0
30°

Hx
1.5m
This could also be sin(60)
We will save this equation and come back to it later.
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force the hinge exerts on the beam.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
60°
T
Fy  0
Hy  Ty  2500N  3500N  0
Hy  T  sin(30 )  2500N  3500N  0
Hy
30°

Hx
1.5m
This could also be cos(60)
We will save this equation and come back to it later.
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force the hinge exerts on the beam.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m
from the far end, as shown. It is supported by a hinge at the wall, and a metal
wire running from the wall to the far end.
Find the tension in the wire, and find the horizontal and vertical components of
the force that the hinge exerts on the beam.
We need to draw a diagram of all the forces, then
write down force and torque equations:
60°
T
  0
Hy
Before we can fill in the torque equation we need to
choose a pivot point. A convenient choice is where
the hinge attaches to the beam. This simplifies the
torque equation because the 2 unknown hinge forces
will not create any torque about that point.
Also, remember the sign convention – clockwise
torques are negative and counterclockwise positive.
30°
Hx
Pivot point here
  0
 (2500N)  (2m)  (3500N)  (2.5m)  (T)(4m)  sin(30 )  0

1.5m
2500N
3500N
T=Tension in wire
Hx and Hy are the components of the
force the hinge exerts on the beam.
T  6875N
Now we can go back and substitute this value into the
other equations to find the hinge forces.
Hx  T  cos(30 )  0
Hy  T  sin(30 )  2500N  3500N  0
Hx  5954N
Hy  2564N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
When a force is applied to an object, it will deform. If it snaps back to its original shape
when the force is removed, then the deformation was ELASTIC.
We already know about springs - remember Hooke’s Law : Fspring = -k•Δx
Hooke’s Law is a special case of a more general rule involving stress and strain.
Stress
 (const.)
Strain
The constant will depend on the material that the object is made from, and it is called an
ELASTIC MODULUS. In the case of tension (stretching) or compression we will call it
Young’s Modulus*. So our basic formula will be:
Stress
Y
Strain
Prepared by Vince Zaccone
*Bonus Question – who is this formula named for?
Click here for the answer
For Campus Learning
Assistance Services at UCSB
To use our formula we need to define what we mean by Stress and Strain.
STRESS is the same idea as PRESSURE. In fact it is the same formula:
Stress 
Force
Area
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
To use our formula we need to define what we mean by Stress and Strain.
STRESS is the same idea as PRESSURE. In fact it is the same formula:
Stress 
Force
Area
STRAIN is a measure of how much the object deforms. We divide the change in the
length by the original length to get strain:
L
Strain 
L0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
To use our formula we need to define what we mean by Stress and Strain.
STRESS is the same idea as PRESSURE. In fact it is the same formula:
Stress 
Force
Area
STRAIN is a measure of how much the object deforms. We divide the change in the
length by the original length to get strain:
L
Strain 
L0
Now we can put these together to get our formula for the Young’s Modulus:
Y
F
L
A
L0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
L0=45m
ΔL=1.1m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
A couple of quick calculations and we can just plug in to our formula:
Y
F
L
A
L0=45m
L0
ΔL=1.1m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
A couple of quick calculations and we can just plug in to our formula:
Y
F
L
A
L0=45m
L0
ΔL=1.1m
F  mg  65kg  9.8 m2   637N
s 

7mm
A  r 2  (3.5  103 m)2  3.85  105 m2
Don’t forget to cut
the diameter in half.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
A couple of quick calculations and we can just plug in to our formula:
Y
637N
Y
5
2
F
L
3.8510 m 
1.1m
45m
L0=45m
A
L0
1.65  107 N2
m
0.024
ΔL=1.1m
 6.88  108 N2
m
F  mg  65kg  9.8 m2   637N
s 

7mm
A  r 2  (3.5  103 m)2  3.85  105 m2
Don’t forget to cut
the diameter in half.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
diam=?
L0=2m
ΔL=0.25cm
400N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
diam=?
L0=2m
ΔL=0.25cm
400N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
A
L0
The only piece missing
is the area – we can
rearrange the formula
diam=?
L0=2m
ΔL=0.25cm
400N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
A
F
L
A
L0
The only piece missing
is the area – we can
rearrange the formula
diam=?
L0=2m
F  L0
Y  L
ΔL=0.25cm
400N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

ΔL=0.25cm
400N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

One last step – we need the diameter, and we have the area:
ΔL=0.25cm
400N
r 2  Acircle
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

One last step – we need the diameter, and we have the area:
r 2  A circle  r 
ΔL=0.25cm
400N
1.6  10 6 m2
 7.14  10 4 m

Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

One last step – we need the diameter, and we have the area:
r 2  A circle  r 
ΔL=0.25cm
400N
1.6  10 6 m2
 7.14  10 4 m

double the radius to get the diameter:
d  1.4  103 m  1.4mm
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2x
e)4 x
a)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2x
e)4 x
a)
We can do this one just by staring at the formula for stress:
Force
Stress 
Area
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2x
e)4 x
a)
We can do this one just by staring at the formula for stress:
Force
Stress 
Area
The force is the same in both cases because it says they use the same weight.
The area is related to the square of the radius (or diameter), so when the
diameter doubles the area goes up by a factor of 4.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2x
e)4 x
a)
We can do this one just by staring at the formula for stress:
Force
Stress 
Area
The force is the same in both cases because it says they use the same weight.
The area is related to the square of the radius (or diameter), so when the
diameter doubles the area goes up by a factor of 4.
Thus the stress should go down by a factor of 4 (area is in the denominator)
Answer c)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Bulk Modulus and Volume Changes
When pressure is applied to an object from all directions, its
volume will change accordingly. Think of squishing a foam ball, or
inflating a balloon.
In this case we use a 3-dimensional version of Young’s modulus.
We call it BULK MODULUS, and it is defined in a similar way:
Bulk Modulus
Pressure change
V
p  B
V0
Volume change
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Bulk Modulus and Volume Changes
Example: When water freezes into ice it expands in volume by 9.05 percent. Suppose a
volume of water is in a household water pipe or a cavity in a rock. If the water freezes, what
pressure must be exerted on it to keep its volume from expanding? (If the pipe or rock
cannot supply this pressure, the pipe will burst or the rock will split.)
The bulk modulus for ice is 8x109 N/m2.
Answser: 6.6x108 N/m2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB