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 test. 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

Given is a uniform chain of mass M and length L.

Let us consider a small element of chain at a distance x from the floor having length 'dx'.

Therefore, mass of element, $\mathrm{d}m=\frac{M}{L}\mathrm{d}x$ ...(1)
The velocity with which the element strikes the floor is,
$v=\sqrt{2gx}$ ...(2)

∴ The momentum transferred to the floor is,
$P=\left(\mathrm{d}m\right)v=\frac{M}{L}.\mathrm{d}x\sqrt{2gx}$

According to the given condition, the element comes to rest.

Thus, the force (say F₁) exerted on the floor = Change in momentum
$$\begin{gathered} F_1=\frac{\mathrm{d}P}{\mathrm{d}t}=\frac{M}{L}\times \frac{\mathrm{d}x}{\mathrm{d}t}\sqrt{2gx} \\ (\mathrm{for}\mathrm{the}\mathrm{chain}\mathrm{element}) \\ As\frac{\mathrm{d}x}{\mathrm{d}t}=v, \\ {F}_1=\frac{M}{L}\times {\left(\sqrt{2gx}\right)}^2\left[\mathrm{using}\mathrm{equation}\left(2\right)\right] \\ \Rightarrow F_1=\frac{M}{L}2gx=\frac{2Mgx}{L} \end{gathered}$$

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

Thus, the total force exerted is given by.
F = F₁ + W
where W is the weight of chain below the element dx up to the floor.

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