1. Topic 2.2 Extended F – Torque, equilibrium, and stability
Torque τ is the rotational equivalent of force.
Consider a door, as viewed from above.
WALL r2 F0 r1 F1 F2 θ2 θ3 θ1
The location of the force will determine the ease with which the door opens.
Easier Harder F2 F1 F0 r2
> r1
> r0
© 2006 By Timothy K. Lund
It is clear that the turning force τ is proportional to the distance r from the pivot point.
If we now choose one r, the angle θ that F makes with r will also determine the ease with which the door opens.
Easier Harder θ1
> θ2
> θ3
The turning force τ is dependent on the angle between the force and the position vector.

2. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: Torque has the same units as a force times a distance (work). But torque is not work and so is never written as joules (J).
TORQUE
0.25 m
WALL
30°
80 n
In fact, if θ is zero, so is the turning force τ.
Finally, the turning force τ is proportional to the size of the force.
Putting it all together we have
© 2006 By Timothy K. Lund
Definition of torque τ
τ = rF sin θ
Suppose we apply a force of 80 n a distance of 25 cm from the hinge, at an angle of 30° with respect to r. Then the torque is
Question: Why do we use sine instead of cosine?
τ = rF sin θ
= (0.25 m)(80 n) sin 30°
= 10 n·m

3. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: We call the line PARALLEL to the force the force line of action.
FYI: We call the line PERPENDICULAR to the force the lever arm or the moment arm.
TORQUE
FYI: Even though all of the forces are identical, each force produces a different rotation. F
Consider a disk that can rotate about its center.
Suppose we apply an identical force F to the disk at four different points.
Which force produces the greatest torque? F
Which force produces the least torque? F F
Which force/s produce/s a ccw torque?
Although F is the same at each point, r is in a different direction: B
Which force/s produce/s a cw torque? C r
The length of the lever arm is r sin  so that if we want, we can rewrite our torque equation like this: F A r r
© 2006 By Timothy K. Lund  r D 
τ = rF sin θ
moment or lever arm
τ = F(r sin θ)
Thus
line of action
"The torque is the force times the lever arm."

4. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: The disk is in rotational equilibrium but the two forces still cause the disk to have a net force up and to the left. It is NOT in translational equilibrium.
EQUILIBRIUM
FYI: If an object is in rotational equilibrium then its angular speed will NOT change. If it is not spinning it will continue to not spin. If it is spinning at a particular angular speed, it will continue to spin at that speed.
Consider a disk that can rotate about its center.
Suppose we apply two forces of equal value F to the disk at the two points shown.
The force at A wants to turn the wheel ccw. The force at B wants to turn the disk cw. Thus
FYI: A body is said to be in MECHANICAL EQUILIBRIUM when the conditions for BOTH rotational and translational equilibrium are satisfied. F
Condition for Rotational Equilibrium
 = 0 A
If we reverse the force at B the conditional for rotational equilibrium is not satisfied since both forces now want to produce a ccw torque:
© 2006 By Timothy K. Lund F
The force at A and the force at B now cancel, though, so that the center of mass does not accelerate. Thus B F
Condition for Translational Equilibrium
F = 0

5. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: We say that an object is in static equilibrium when it is in mechanical equilibrium AND it is not moving. Both of these situations are examples of static equilibrium.
Topic 2.2 Extended F – Torque, equilibrium, and stability
EQUILIBRIUM
Suppose a uniform beam of mass m and length L is placed on two scales, as shown.
It is expected that each scale will read the same, namely half the weight of the beam.
Now we place a block of mass M on the beam, closer to the left-hand scale. M
© 2006 By Timothy K. Lund X
It is expected that the left scale will read higher than the right one, because the block is closer to it.

6. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: We must use an extended diagram, since there are torques present.
Topic 2.2 Extended F – Torque, equilibrium, and stability
EQUILIBRIUM
To analyze an extended system we use what we will call an extended free-body diagram.
L/2 x N1 Mg mg N2
Find N1 and N2 in terms of x, L, m, and M and g.
From our balance of forces we have
© 2006 By Timothy K. Lund
ΣFy = 0
N1 + N2 – Mg – mg = 0
Note: We have ONE equation with TWO unknowns. We must use the balance of torques to obtain the other equation.

7. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: If a system is in static equilibrium you can use ANY point as the pivot.
Topic 2.2 Extended F – Torque, equilibrium, and stability
NOTE: Mg and mg want to rotate the lever arm in the POSITIVE direction, and N2 provides the balancing NEGATIVE torque.
EQUILIBRIUM
Find N1 and N2 in terms of x, L, m, and M and g.
L/2 x + N1 Mg mg N2
In order to use our balance of torques we need to choose a pivot point.
I have arbitrarily chosen the left end of the board as my pivot point.
© 2006 By Timothy K. Lund
I have arbitrarily chosen CLOCKWISE to be a positive torque.
Στ = 0
0·N1 + x·Mg + (L/2)mg – L·N2 = 0
FYI: Choosing the pivot point at the application point of a force eliminates that force from the balance of torques equation.

8. Topic 2.2 Extended F – Torque, equilibrium, and stability
Rotational Motion and Equilibrium 8-2 Torque, Equilibrium and Stability
Topic 2.2 Extended F – Torque, equilibrium, and stability
Note: If x = L/2, N2 reduces to the expected (M + m)g/2 (exactly half the total weight). N1 will reduce to exactly the same thing.
EQUILIBRIUM
Find N1 and N2 in terms of x, L, m, and M and g.
L/2 x + N1 Mg mg N2
We now resolve our system of equations:
0·N1 + x·Mg + (L/2)mg – L·N2 = 0
© 2006 By Timothy K. Lund
N2 = g xM L m 2
N1 + N2 – Mg – mg = 0
N1 = (M + m)g - N2

9. Topic 2.2 Extended F – Torque, equilibrium, and stability
Find N1 and N2 if x = 2 m, L = 20 m, the mass of the plank is 50 kg, and the mass of the block is 100 kg.
We have
N2 = g xM L m 2
= ·10
2·100 20 50 2
= 350 n
N1 = (M + m)g - N2
= ( )
= 1150 n
© 2006 By Timothy K. Lund
Question: Why is N1 > N2?

10. Topic 2.2 Extended F – Torque, equilibrium, and stability
Improper structural analysis may result in DEATH!
EQUILIBRIUM
In the previous example all of the torques were caused by perpendicular forces, so no angles were used (since sin 90° = 1).
Now consider the boom crane shown here:
WARNING
The critical components of the crane need to be designed to withstand the forces (and torques) the crane is subjected to.
© 2006 By Timothy K. Lund
We need to know what tension the support cables must withstand.
We need to know what forces the support pin must withstand.

11. Topic 2.2 Extended F – Torque, equilibrium, and stability
Let’s do some GOOD analysis!
Here are the variables: θ
Our goal is to find the three green forces, and the net force on the pin. mg Mg FH
An extended free-body diagram simplifies our analysis. m FV T
If L is the length of the beam, L/2 is the effective position of the beam’s weight Mg.
© 2006 By Timothy K. Lund θ x
L/2 mg Mg
Let x be the distance mg is from the pin. FH FV

12. Topic 2.2 Extended F – Torque, equilibrium, and stability
Suppose θ = 30°, the length of the beam is 20 m, and the mass of the beam is 400 kg.
30° x 10 mg
60°
Then our diagram reduces to…
60° x
ΣFx = 0 FH FV
T - FH = 0
4000 n
T = FH
ΣFy = 0
© 2006 By Timothy K. Lund
FV – mg = 0
FV = mg
Στ = 0
10·4000 sin 60° + xmg sin 60° - 20·T sin 30° = 0
T = xm

13. Topic 2.2 Extended F – Torque, equilibrium, and stability
Our three unknowns in terms of x and m are thus:
30° x 10
T = FH = xm mg
60°
60°
FV = m x FH FV
4000 n
As a specific example, if m = 2000 kg, and x = 15 m, then…
FV = ·2000
= n
© 2006 By Timothy K. Lund
T = FH = ·15·2000
= n
To find the net force on the pin…
FPIN2 = FV2 + FH2
FPIN2 =
FPIN = n

14. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: The fun begins when we wish to remove the conditions of static equilibrium, as in this example...
Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: Of course, after the collapse the debris reestablish static equilibrium conditions.
EQUILIBRIUM
Consider a typical building shown here
Vertical supports called columns transfer the weight of the building to the ground.
Horizontal supports called spans transfer the weight of the floors to the columns.
© 2006 By Timothy K. Lund
The building is in static equilibrium.
If explosive charges are detonated at the base of the columns, the building is no longer in equilibrium.

15. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: The UNSTABLE configuration has a built-in ANTI-RESTORING FORCE that tries to remove the mass from its equilibrium position.
Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: The STABLE configuration has a built-in RESTORING FORCE that tries to return the mass to its equilibrium position.
STABILITY
FYI: The NEUTRAL configuration has neither built-in force.
Before we talk about stability we have to define three types of equilibrium.
Consider two bowls and a level table top.
If we carefully place a ball on each surface it will be in equilibrium:
Now we gently displace each ball to the right and observe the results...
© 2006 By Timothy K. Lund
STABLE
NEUTRAL
UNSTABLE
EQUILIBRIUM
EQUILIBRIUM
EQUILIBRIUM

16. Topic 2.2 Extended F – Torque, equilibrium, and stability
FYI: As long as the center of gravity is DIRECTLY above the BASE OF CONTACT, the extended object is in equilibrium.
FYI: If we rotate the stable extended body far enough, it will become unstable:
STABILITY
Now we can look at extended solids.
Consider the three solids, all of which are in equilibrium. cg mg cg mg mg cg cg mg mg mg mg mg mg
© 2006 By Timothy K. Lund
FYI: mg provides a restoring torgue so that this body is in STABLE EQUILIBRIUM.
FYI: mg provides NO torque so that this body is in NEUTRAL EQUILIBRIUM.
FYI: mg provides an anti-restoring torque so that this body is in UNSTABLE EQUILIBRIUM.