# a chain is lying on a rough table with a friction 1/n of its length hanging down from the edge of table if it is just of the point of sliding down from the table then the coefficient of friction between the table and the chain is

Rajdeep
231 Points
5 years ago
HELLO THERE!

Let the total length of the chain be L.
Mass of length L = M.

So, mass of unit length of the chain = $\dpi{80} \frac{M}{L}$

It is given that,  $\dpi{80} \frac{1}{n}$ of the chain is hanging down from the table.

So, length of chain that is hanging = $\dpi{80} \frac{L}{n}$

Mass of the hanging part = $\dpi{80} \frac{M}{L}\times \frac{L}{n} = \frac{M}{n}$

Length of the chain which is on the table = $\dpi{80} L - \frac{L}{n} = L(1-\frac{1}{n}) = L(\frac{n-1}{n})$

Mass of the chain which is on the table = $\dpi{80} \frac{M}{L} \times L(\frac{n-1}{n}) = M(\frac{n-1}{n})$

Now, Force acting downwards due to weight of the chain will be equal to the frictional force acting on the part of chain placed on the table.

Force acting downwards due to weight of the chain $\dpi{80} \frac{M}{n}g$

Frictional force acting on the chain which is on the table = $\dpi{80} M(\frac{n-1}{n})\mu$

Now,
$\dpi{80} M(\frac{n-1}{n})\mu = \frac{M}{n}g \\\\or, \mu = \frac{g}{n-1}$