Question

A uniform chain of mass $M$ and length $L$ is held vertically in 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 a length $x$ has already reached and rest of the length l - x is in the air.

Answer 1

Let us consider a small element $dx$ at a distance $x$ from the floor

now

$dm=\frac{M}{L}dx$

Velocity with element will strike the floor $v=\sqrt{2gx}$

Therefore momentum transferred to floor is

$p=\left(dm\right)v=\frac{mdx\sqrt{2gx}}{L}$

Now force exerted on the floor due to change in momentum is given by

$F_1=\frac{dm}{dt}=\frac{m_{x}dx\sqrt{2gx}}{Ldt}=\frac{m_x}{L}\sqrt{2gx}\frac{dx}{dt}=\frac{m}{L}\sqrt{2gx}v=\frac{m}{L}\sqrt{2gx}\sqrt{2gx}=\frac{2mgx}{L}$

Now again total force is $=F_1+$ gravitation force due to $x$ length

$=\frac{2mgx}{L}+\frac{mgx}{L}=\frac{3mgx}{L}$