A uniform chain of mass m and length L is held vertically is 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 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
When a length x falls the part of the chain on the floor applies force gMx/L.
The remaining part of the chain is falling-
The remaining part will move distance x
it will fall with velocity – v^2-u^2=2*g*x
=> v^2 = 2*g*x
=> v=√2gx
momentum of small mass dm = dx*M/L *√2gx
Force = change of momentum / change in time (dt)
force of small mass dm = dx/dt * M/L * √2gx
=> v*M/L*√2gx
=> 2gx *M/L
Total force = Mgx/L + 2Mgx/L
= 3Mgx/L