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# Rotational Equilibrium Study Guide (page 2)

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Updated on Sep 26, 2011

#### Example 1

Consider a seesaw of length L and a child playing on it. Show the force, the level arm, and the pivot point. Find the torque acting on the end of the seesaw.

#### Solution 1

Figure 9.9 shows the line of action, the pivot point, the weight of the child playing on the seesaw, and the lever arm.

Because the board has a length L, the lever arm is And the torque is:

### Value of Torque

The value of the torque in Figure 9.11 (a) is considered positive, or counterclockwise (CCW), whereas the second picture shows a negative, or clockwise (CW) torque.

#### Example 2

Now consider that the child stepping up from the board and starting to push on it with a force equal to his weight and in the direction shown in the figure. Is the torque the same as before?

#### Solution 2

Figure 9.10 shows the line of action, the pivot point, the force of the child playing on the seesaw, and the lever arm. The lever arm in this case is:

T = F · l = F · (L/2) · cos α

T = m · g · (L/2) . cos α

In this case, the torque is smaller than when the child exerts her weight perpendicular to the plank.

In these two cases, the applied force has the same magnitude but different directions. Hence, the effect on the object will be different.

Another observation that we need to make about torque is that, similar to force, it is a vector: For the same values of the force and lever arm, we might end up with different directions of rotation. Let's look at the following example. In Figure 9.11(a), the rotation is counterclockwise (CCW), whereas in Figure 9.11(b), although the same value of force and lever arm was used, the object rotates clockwise (CW).

### Rotational Equilibrium

A rigid object is in rotational equilibrium when the net external force and the net torque are both zero.

Net F = 0

Net τ = 0

The two conditions have to be satisfied in the same time.

## Equilibrium of a Rigid Object

We have seen that there are cases when one, two, or more forces can give a zero or nonzero torque, depending on the value of the lever arm. In the example with the rotating wheel, once the two forces are applied such that their extension goes through the center of rotation (and the lever arm is zero) and the forces are equal and opposite, there is no rotation and therefore no torque. This situation is a rotational equilibrium, and to establish the state, we need to talk about two conditions: net external force and net torque.

#### Example 1

Two kids playing soccer hit the ball at the same time with equal and opposite forces directed along the same diameter of the ball. Does the ball start to spin around its own axis?

#### Solution 1

There is no data, so no conversion to SI is necessary.

F = –F

Net τ = ?

Because the line of action passes through the center of rotation, any thing perpendicular to it from the center would be zero. Hence, there is zero lever arm and zero torque for each of the forces.

Net F = F + ( –F) = 0 N

Net τ = F · 0 + F · 0 = 0 N · m

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### Related Questions

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