Question

A meter stick is balanced at the 50 cm mark. You tie a 10 kg weight at the 10 cm mark, while 20 kg weight is placed at the 80 cm mark. Where should a 20 kg weight be placed so the meter stick will again be balanced?

Answer 1

The free-body diagram of the meter stick is shown below:

There are three force acting on the meter stick:

1. m₁g is the normal force applied by 10 kg weight placed at 10cm mark or 40cm from the pivot. This force tends to rotate the meter stick in counterclockwise direction about the pivot.

2. m₂g is the normal force applied by 20kg weight placed at 80cm mark or 30cm from the pivot. This force tends to rotate the meter stick in clockwise direction about the pivot.

3. m₃g is the normal force applied by the additional 20kg mass which is placed at distance x from the pivot. This force tends to rotate the meter stick in counterclockwise direction about the pivot.

When balanced, the net torque on the meter stick is zero.

Note that the torque due to the m₂=20kg weight placed at 80cm mark tends to rotate the stick in direction opposite to the torque due to m₁ and m₃, so it is taken as negative.

Additional 20kg weight should be placed 10cm from the pivot, it should be placed at

50cm-x=50cm-10cm=40cm mark.

Note 60cm mark is also 10cm away, but then the meter stick will not be balanced because if we placed the additional 20kg weight at 60cm mark, the torque in the clockwise direction will become greater than the torque in the counterclockwise direction. To balance the meter stick we need to put the 20kg weight on the left of the pivot.