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# 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 that position. Any portion that strikes the floor comes to rest.Assuming that the chain doesnot form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor?

Chetan Mandayam Nayakar
312 Points
10 years ago

kinetic energy of falling part = loss in its potential energy

therefore (M/2)(1-(x/L))v2=Mgx(1-(x/L)), thus v=√(2gx)

magnitude of force = rate at which momentum is destroyed upon impact with the floor

=(d/dt)(M(vdt/L)v)=Mv2/L=2Mgx/L

Chetan
19 Points
3 years ago
Okay the logic is correct sir but in my module the answer is 3Mx/L.. and the question is exactly same ... help
Anand
13 Points
one year ago
the net force will be calculated as the force required to destroy the momentum which is due to gain in velocity (2mgx/l) and also the normal reaction exerted by the rest mass of x length (mgx/l) to hold its weight  = 3mgx/l