Question

a very flexible uniform chain of mass m and length l is suspended vertically so that its lower end just touches a smooth inelastic plane inclined at 45 degree. When it is relesed, find the force exerted by the chain on the plane at any instant t.

Answer 1

     F  = (M/L)g (x + 2y)

Explanation:

Please refer to the attached picture for the diagram.

The force exerted by the chain on the table F = F1 + F2

F1 is the weight of the chain on the table

F2 = rate of change of momentum of the chain at the instant it strikes the table.

So F1 = (M/L) x.g

Let's take a small element dy of the chain at the height y above the table, then mass of the element = (M/L)dy

Velocity of the element on striking the table v = √(2gy)

dp = (M/L)dy * √(2gy)

F2 = dp/dt

     = (M/L) (dy/dt) √(2gy)

v = dy / dt = √(2gy)

Therefore F2 = (M/L).(2gy)

So F = (M/L) x.g + (M/L).(2gy)

     F  = (M/L)g (x + 2y)