Question

A uniform chain of length $L$ is lying on the horizontal table. If the coefficient of friction between the chain and the table top is $\mu$, what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on the table?

• A. $\frac{L}{(1+\mu )}$
• B. $\frac{\mu L}{(1+\mu )}$
• C. $\frac{L}{(1-\mu )}$
• D. $\frac{\mu L}{(1-\mu )}$

Answer 1

Answer: (B)

$\frac{\mu L}{(1+\mu )}$

Let $M$ be the mass of the chain and $L$ its length.

If a length $l$ hangs over the edge of the table, the force pulling the chain down is $\frac{Ml}{L}g$.

The force of friction between the rest of the chain of lenth $(L-l)$ and the table is $\frac{\mu M(L-l)}{L}g$.

For equilibrium the two forces must be equal:

$\Rightarrow$$\frac{Ml}{L}g=$$\frac{\mu M(L-l)}{L}g$

$\Rightarrow l=\mu (L-l)$

$\Rightarrow l=\frac{\mu L}{1+\mu }$