Question

Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center and a very light board that is 3.2 m long (Fig. 9–57). Two playmates are already on either end. Boy A has a mass of 45 kg, and boy B has a mass of 35 kg. Where should girl C, whose mass is 25 kg, place herself to balance the seesaw?

Short answer

She should place herself at 0.64 m right to the middle of the seesaw.

Step-by-step solution

Concepts

In equilibrium, the net force in the x and y directions should be zero, and the torque about any point is zero.For this problem, you should place her on the side of the boy with the lower mass.

Given data

The mass of boy A is \({m_{\rm{A}}} = 45\;{\rm{kg}}\).

The mass of boy B is \({m_{\rm{B}}} = 35\;{\rm{kg}}\).

The mass of girl C is \({m_{\rm{C}}} = 25\;{\rm{kg}}\).

The length of the seesaw is \(L = 3.2\;{\rm{m}}\).

Calculation

The free-body diagram of the problem is given below.

In the balancing situation, the torque is zero. Then,

\(\begin{array}{c}\left( {{m_{\rm{A}}}g \times \frac{L}{2}} \right) - \left( {{m_{\rm{B}}}g \times \frac{L}{2}} \right) - \left( {{m_{\rm{C}}}g \times x} \right) = 0\\\left( {45\;{\rm{kg}} - 35\;{\rm{kg}}} \right) \times \frac{{3.2\;{\rm{m}}}}{2} = \left( {25\;{\rm{kg}}} \right) \times x\\x = 0.64\;{\rm{m}}\end{array}\).

Hence, she should place herself at 0.64 m right to the middle of the seesaw.