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# Rotational Equilibrium Study Guide (page 3)

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#### Example 2

Consider the seesaw again, now with two children: One is 35 kg and the other 42 kg. Is this system in rotational equilibrium? The two sides of the seesaw are of equal length and the total length is 280 cm. If the seesaw is not in equilibrium, what has to be changed so that it equilibrates (keep masses constant)? Consider the seesaw to be of negligible weight.

#### Solution 2

Convert the quantities to SI. Next, draw a diagram and find the equations.

m1 = 35 kg

m2 = 42 kg

L = 280 cm = 2.8 m

Net F = ?

Net τ = ?

The net external force is the sum of the two weights and the normal force. Because the object does not have a linear motion, the three cancel each other out.

Net F = W1 + W2 – N = 0 N

Weight W1 determines a positive torque (CCW), whereas the second weight determines a negative torque.

The weight is perpendicular on the seesaw in both cases, so the lever arm is equal to half of the length of the seesaw.

τ1 = F · l = W1 · l = m1 · g · l = m1 · g · L/2

τ2 = – F · l = – W2 · l = – m2 · g · l = – m2 · g · L/2

Net τ = τ1 + τ2 = m1 · g · L/2 – m2 · g · L/2 (m1m2) · g · L/2 = –96 N · m

Net τ = –96 N · m

The minus sign tells us that the seesaw rotates clockwise, hence the system is not in rotational equilibrium.

In order for the seesaw to remain in rotational equilibrium, the net torque has to be zero. Although we cannot change the mass of the two children, we can change their position on the seesaw.

τ1 = F · x = W1 · x = m1 · g · x

τ2 = – F · (Lx) = – W2 · (Lx) = – m2 · g · (Lx)

NetT = τ1 + τ2 = m1 · g · xm2 · g · (Lx) = [m1 · xm2 · (Lx)] · g = 0 N · m

And we can solve for the unknown, x:

m1 · xm2 · (Lx) = 0

(m1 + m2) · xm2 · L = 0

(m1 + m2) · x = m2 · L

x = 1.5 m

The smaller weight has to have a longer lever arm to equilibrate the torque exerted by the second, larger weight.

Practice problems of this concept can be found at: Rotational Equilibrium Practice Questions

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