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 that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.

Answer 1

Let us consider a small element at a distance 'x' from the floor of length 'dy'

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

So, the velocity with which the element will strike the floor is,

$v=\sqrt{2gx}$

$\therefore$ So, the momentum transfered to the floor is,

$M=\left(dm\right)v=\frac{M}{L}.dx\sqrt{2gx}$

[because the element comes to rest]

So, the force exerted on the floor changes in momentium is given by,

$F_1=\frac{dM}{dt}=\frac{M}{L}.\frac{dx}{dt}\sqrt{2gx}$

Because, $v=\left(\frac{dx}{dt}\right)=\sqrt{2gx},$

(for the chain element.)

$F_1=\frac{M}{L}{\left(\sqrt{2gx}\right)}^2$

$=\frac{M}{L}.2gx=\frac{2Mgx}{L}$

Again, the force exerted due to 'x' length of the chain on the floor due to its own weight is given by,

$W=\frac{M}{L}\left(x\right)\times g=\frac{Mgx}{L}$

So, the total force exerted is given by

$F=F_1+W$

$=\frac{2Mgx}{L}+\frac{Mgx}{L}$

$=\frac{3Mgx}{L}$