A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is ___________ m.

Solution

Physics sir · 4 steps

At the limit, the hanging weight equals the maximum friction on the part still on the table.

1. Set up. Let L = 6 m, let x be the hanging length and λ the mass per metre of the chain. The hanging part pulls with weight λ x g. The part on the table, of length L - x, presses down with normal force λ (L - x) g.
2. Limiting condition. The hanging length is largest when static friction reaches its maximum value μ N. Only the part on the table feels friction, so the pull of the hanging part must equal μ λ (L - x) g.

   λ x g = μ λ (L - x) g ⇒ x = (μ L)/(1 + μ)

3. Put in numbers. x = (0.5 × 6)/(1 + 0.5) = 3/1.5 = 2 m. Both λ and g cancel, so the answer does not depend on the mass of the chain.
4. So. The maximum length of chain that can hang over the edge is 2 m.

The common slip: 3. Using the weight of the whole chain as the normal force gives x = μ L = 3, not 2. Only the length still on the table presses on it.

The idea to keep. For a chain hanging off a rough table, friction acts only on the part on the table. Use x/(L - x) = μ, which gives x = μ L/(1+μ).

Answer from the NTA final answer key. Solution by PrepWiser.