Solve the following :


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.


$\underset{M}{\text { Consider }} \lambda=$ Mass per unit length of chain


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 $=F_{1}+F_{2}=\frac{3 M g x}{l}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now