Question

a very flexible uniform chain of mass m and length l is suspended vertically so that its lower end just touches the surface of a table. when the upper end of the chain is released, it falls with each link coming to rest the instant it strikes the table. the force exerted by the chain on the table at the moment when y part of the chain has already rested on table​

Answer 1

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

Answer 2

heya mate here is your answer.......

hope it helps uh.....