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.
Solution:
$\underset{M}{\text { Consider }} \lambda=$ Mass per unit length of chain
$=\frac{M}{L}$
After chain's release,
$d m=\lambda d x$
$v=\sqrt{2 g x}$
$\Delta p=d m . v=\frac{M}{L} \cdot d x \sqrt{2 g x}$
$F_{1}=\frac{\Delta p}{\Delta t}=\frac{M}{L} \frac{d x}{d t} \sqrt{2 g x}=\frac{M}{L} .2 g x$
Weight of chain due to $x$ length
Total force $=F_{1}+F_{2}=\frac{3 M g x}{l}$