Question

A meter scale is kept on a knife edge at $50cm$ mark. A weight of $20gf$ is hanging at $80cm$ mark and a weight of $50gf$ is hanging at $60cm$ mark . Where should a weight of $55gf$ be placed to make the meter scale balanced?

Answer 1

Step 1: Given data is,

Length of the scale = $100cm$

Balanced at = $50cm$

Weight of $20gf$ hanging at $80cm$

Weight of $50gf$ hanging at $60cm$

We have to find the mark where the $55gf$ of weight should be placed.

For balancing the scale torque should be equal on both sides.

Let x be the mark where $55gf$ is hanging

Let v be the velocity

Step 2 :

By conservation of angular momentum from point A ,

$$\begin{gathered} \Rightarrow 20\times v\times \left(80-50\right)+50\times v\times \left(60-50\right)=55\times v\times \left(50-x\right) \\ \Rightarrow 20\times 30+50\times 10=55\times \left(50-x\right) \\ \Rightarrow 1100=55\times \left(50-x\right) \\ \Rightarrow 1100=2750-55x \\ \Rightarrow 55x=2750-1100 \\ \Rightarrow 55x=1650 \\ \Rightarrow x=\frac{1650}{55} \\ \Rightarrow x=30cm \end{gathered}$$

Therefore, the weight of $55gf$ should be placed at $30cm$ mark for the scale to be balanced.