Question

A uniform chain of mass $m$ and length $L$ is held vertically is such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming the chain does not form a heap on the floor. Calculate the force exerted by it on the floor when the whole chain is on the floor.

Answer 1

Lets take a small element at a distance $x$ of length $dx$.

The velocity with which element will strike the floor,

$v=\sqrt{2gx}$

Momentum transferred to floor is,

$dP=dmv=\left[\frac{M}{L}xdx\sqrt{2gx}\right]$

$F_1=\frac{dp}{dt}$

and, $v=\frac{dx}{dt}=\sqrt{2gx}$

$F_1=\frac{M\sqrt{2gx}\times \sqrt{2gx}}{L}$

$F_1=\frac{2mgx}{L}$

Now, the force entered by length $x$ due to its own weight is,

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

Therefore,

Total force=$\frac{mgx}{L}+\frac{2mgx}{L}$

$F=\frac{3mgx}{L}$