Problem 45.

A 20 kg child and a 30 kg child sit at opposite ends of a 4 m seesaw that is pivoted at its center. Where should another 20 kg child sit in order to balance the seesaw?

Solution:

In this problem we need to use only one equilibrium condition: no rotation of the seesaw (the seesaw is balanced). The second equilibrium condition (no translational motion) needs to be used only if we need to find the normal force at the pivot.

We can introduce any point as the point of possible rotation. We consider point O – the center of the seesaw – as the point of possible rotation. There is no rotation about point O, which means that the net torque about point O is zero.

First, we need to show all forces acting on the seesaw:

1. Gravitational force (more exactly it is normal force, which is equal to the gravitational force) on the 20 kg child. Application point of the force is one end of the seesaw (see figure). The distance to point O is .

2. Gravitational force (more exactly it is normal force, which is equal to the gravitational force) on the 30 kg child. Application point of the force is the other end of the seesaw (see figure). The distance to point O is .

3. We place another 20 kg child on the seesaw. We characterize the position of the child by the distance to point O (this is an unknown distance): . The corresponding force is the gravi tational force on the child.

There is no rotation about point O. Then the net torque about point O is zero:

We substitute the known values and obtain

From this equation we can find the position of the 20 kg child: