A uniform chain of length $L$ is lying on the horizontal table. If the coefficient of friction between the…

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?
  1. $\frac{L}{(1+\mu)}$
  2. $\frac{\mu L}{(1+\mu)}$
  3. $\frac{L}{(1-\mu)}$
  4. $\frac{\mu L}{(1-\mu)}$

Solution

Let $l^{\prime}$ part of the chain is hanging over the edge of table without sliding.
$\therefore \quad \mu=\frac{\text { Length hanging over the edge }}{\text { Length lying on the table }}$ (As the chain have uniform linear density) $\begin{array}{ll} \therefore & \mu=\frac{l^{\prime}}{\left(L-l^{\prime}\right)} \\ \Rightarrow & l^{\prime}=\frac{\mu L}{(1+\mu)} \end{array}$

Asked in: AP EAMCET 2011

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