Question

A 0.5-kg uniform meter stick is suspended by a single string at the 30-cm mark. A 0.2-kg mass hangs at the 80 cm mark. What mass hung at the 10-cm mark will produce equilibrium?

Answer 1

Answer:

0.50 kg

Explanation:

From the equilibrium condition, torques must be balanced. Torque is the product of force and distance.

The unknown mass is at a distance of 20 cm from the suspension point (30 cm). The 0.20 kg is at a distance of 0.50 cm from that same point.

Clockwise torques must equal anti clockwise torques.

So (0.20 m)  (M g) = (0.50 m)(0.20 g)

Here M is the unknown mass and g stands for gravity.

⇒ Unknown mass = M = (0.50 × 0.20 ) ÷ (0.20)= 0.50 kg