IIT JEE Physics (1978-2018: 41 Years) Topic-wise Complete Solutions

# Balancing the Scale on Fingers

## Objective

To demonstrate the concept of centre of mass.

## Procedure

Support a meter stick horizontally with two fingers (of two hands). Slide your fingers in and they will both meet at the center of mass. To explain this, you need to understand friction and equilibrium. In static equilibrium, net force and net torque on the scale is zero.

Give scale to a volunteer. Hold the scale horizontal by placing finger of one hand below one end of metre scale and finger of another hand below the mid point of scale. Ask him to move the finger at the end of the scale. It is easy do so. Now ask him to move the finger at the mid-point. He is unable to do so. Why?

## Discussion

The stick is in equilibrium. The net force and net torque on the stick is zero. The forces acting on the scale are gravitational pull $$mg$$, and reactions forces from two fingers, say $$N_1$$ and $$N_2$$. Let fingers are at distances $$x_1$$ and $$x_2$$ from the centre of mass. In equilibrium, $$N_1+N_2=mg$$ and $$N_1 x_1=N_2 x_2$$. These conditions give more reaction force on the finger which is close to the centre of mass. Thus, frictional force on this finger is more as compared to the other finger. Thus, when you start moving the fingers, the finger which is far away from the centre of mass starts moving first.