Question

A uniform chain of mass $m$ and length $l$ hangs from a thread and touches the surface of a table by its lower end. Find the force exerted by the chain on the surface when half of its length has fallen on the table. Assume that the fallen part does not form heap.

Answer 1

Given,

$mass=m$

$length=l$

K.E of the falling part =loss in its potential energy

Therefore,

$\left(\frac{m}{2}\right)(1-\frac{x}{L})v^2$

Thus,

$v=\sqrt{2}\left(2gx\right)$

The magnitude of the force is equal to the rate at which momentum is destroyed upon impact wuith the floor.

$\left(\frac{d}{dt}\right)\left(m\right(v\frac{dt}{L}\left)\right)=\frac{m{v}^2}{L}$

Thus,the force exerted by the chain on the surface when half of its length has fallen on the table.

$=\frac{2mgx}{L}$