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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?`
7 years ago

Chetan Mandayam Nayakar
312 Points
```										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
```
7 years ago
Chetan
19 Points
```										Okay the logic is correct sir but in my module the answer is 3Mx/L.. and the question is exactly same ... help
```
3 months ago
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