Question

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25 s, then the maximum fraction of the length of the chain that can hang over one edge of the table is

• A. 0.2
• B. 0.25
• C. 0.35
• D. 0.15

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relation between the force of friction and the weight of the chain that prevents it from sliding off the table.

1. The chain is lying on a horizontal table, and the friction acts to resist any sliding motion. The part of the chain that hangs off the table is subject to its weight, which tries to pull more of the chain off the table.
2. The force of friction is given by: F_{\text{friction}} = \mu \, N, where \mu is the coefficient of friction and N is the normal force.
3. For the chain on the table, the normal force N is equal to the weight of the chain portion still on the table, which can be expressed as: N = \left(1 - x\right)mg, where x is the fraction of the chain hanging over the edge and mg is the total weight of the chain.
4. The weight of the overhanging portion of the chain is xmg.
5. For the chain to be on the verge of slipping, the force of the weight of the overhanging part must equal the frictional force: xmg = \mu \left(1-x\right)mg.
6. Simplifying this equation, we get: x = \frac{\mu}{1 + \mu}.
7. Substituting the given coefficient of friction \(\mu = 0.25\): x = \frac{0.25}{1 + 0.25} = \frac{0.25}{1.25} = 0.2.

Therefore, the maximum fraction of the length of the chain that can hang over one edge of the table without sliding is 0.2. This matches the correct answer option given.