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

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ANSWER::

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

Now , dm = Mdx / L

The velocity with which element will strike the floor , v = √(2gx)

Therefore , momentum transferred to floor is ,

M = (dm)v = [M x dx x √(2gx) ] / L

Now , the force exerted on floor change in momentum is given by ,

F₁ = dM / dt = [ M x dx x √(2gx) ] / (L x dt)

And ,

v = dx / dt = √(2gx) ( for element of chain)

F₁ = [ M x √(2gx) x √(2gx) ] / L

F₁ = M(2gx) / L

F₁ = 2Mgx / L

Now , again the force exerted due to x length of chain on floor due to its own weight is ,

W = Mgx / L

Therefore , total force exerted is ,

D = F₁ + W = 2Mgx / L + Mgx / L = 3Mgx / L

Hope it helps!