Question

A uniform chain of length ${}^{\prime }{l}^{\prime }$ is placed on a rough table, with its length $\frac{l}{n}(n>1)$ hanging over the edge of the table. If the chain just begins to slide off the table by itself, the coefficient of friction between the chain and the table is:

• A. $\frac{1}{n+1}$
• B. $\frac{1}{n-1}$
• C. $\frac{n-1}{n+1}$
• D. $\frac{1}{n}$

Answer 1

The correct option is B $\frac{1}{n-1}$

In this case, the weight of the hanging part must be just equal to the frictional force acting on the part on the table.
We know that, $F=\mu N$

Let linear mass density be $\gamma$.

Mass of the chain on the table is $m^{\prime }=l(1-\frac{1}{n})\gamma$

Frictional force, $\mu N=\mu \gamma (l-\frac{l}{n})g$, where, $\gamma$ is the linear mass density of the chain.

Weight of the hanging part, $W=\gamma \frac{l}{n}g$

Equating the above two equations yield,

$\mu =\frac{1}{n-1}$