Q.A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Concept: Static Equilibrium — for a balanced lever, the sum of clockwise moments about the fulcrum equals the sum of anticlockwise moments.

Step 1: Identify the forces and distances.

The metre stick’s weight $M g$ acts at its centre (50.0 cm). The two coins (total mass $10\ \text{g}$) are placed at the 12.0 cm mark. The new balance point is at 45.0 cm — this is the fulcrum.

Step 2: Measure lever arms from the fulcrum (45.0 cm).

• Coins: distance from fulcrum = $45.0 - 12.0 = 33.0\ \text{cm}$ (left side, anticlockwise moment).
• Stick’s centre of mass: distance from fulcrum = $50.0 - 45.0 = 5.0\ \text{cm}$ (right side, clockwise moment). …

Why static equilibrium works here

A metre stick balanced at its centre means its own weight acts at the 50 cm mark (the centre of gravity of a uniform stick). When you place coins at the 12.0 cm mark, the stick no longer balances at the centre — you have to shift the knife edge to a new point (45.0 cm) to restore balance. At that new balance point, the total torque about the knife edge is zero. That’s the key: the clockwise torque from the stick’s weight (acting at its centre) must exactly equal the anticlockwise torque from the coins.

Step-by-step solution

1. Define the system and the pivot

The metre stick is uniform, so its centre of gravity is at the 50.0 cm mark. Let the mass of the stick be $M$ grams (we’ll work in grams and centimetres — torque units will cancel). Two coins, each 5 g, are stacked at the 12.0 cm mark, so total coin mass $m = 10$ g.

The stick balances when the knife edge is at the 45.0 cm mark. That’s our pivot.

2. Find the lever arms (distances from the pivot)

• Distance from pivot to the stick’s centre of gravity:

  $50.0 \text{ cm} - 45.0 \text{ cm} = 5.0 \text{ cm}$

  This is on the right side of the pivot.

• Distance from pivot to the coins:

  $45.0 \text{ cm} - 12.0 \text{ cm} = 33.0 \text{ cm}$

  This is on the left side of the pivot.

3. Apply the torque balance condition …

Concept: Torque Balance (Static Equilibrium of a Lever)

Step 1: Identify the new pivot

The stick balances at the 45.0 cm mark, so this is the pivot. The stick's own centre of gravity (uniform stick) is at 50.0 cm; the coins (total mass $10$ g) sit at 12.0 cm.

Step 2: Lever arms from the pivot

Stick's weight: $50.0-45.0=5.0$ cm (to the right). Coins: $45.0-12.0=33.0$ cm (to the left). …