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.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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 transferred to the floor is,

$M=\left(dm\right)v=\frac{M}{L}\times dx\times \sqrt{2gx}$ [because the element comes to rest]

So, the force exerted on the floor change in momentum is given y,

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

Because, v = $\frac{dx}{dt}=\sqrt{2gx}$ ( for the chain element)

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

Again, the force exerted due to 'x' length of the chain on the fkoor due to its own weight is given
by, $W=\frac{M}{L}\left(x\right)\times g=\frac{Mgx}{L}$

So, the total forced exerted is given by,

$F=F_1+W=\frac{2Mgx}{L}+\frac{Mgx}{L}=\frac{3Mgx}{L}$