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 this position. Any portion that strikes the floor comes to rest . Assuming that 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.

Question

sumi29 · Answer

Seems fairly easy. Are you having trouble in any particular part of the question? Show us your work. Seems like a questions for an entrance examination...

anonymous · Answer

i am nt sure where i m doing it wrong...as the answer is 3mgx/l

sumi29 · Answer

OK seems easy. Here is how I did it: 
$$v=\sqrt{2gx}$$

sumi29 · Answer

Length of chain lying on ground after time t is:
$$T=0.5.g.t^2$$

sumi29 · Answer

Momentum to the table = mv = $$(m/l)dxv$$

sumi29 · Answer

And the force due to the impact = 2mgx/l

sumi29 · Answer

Weight of chain of length x is mgx/l

sumi29 · Answer

So the net force is 3mgx/l

sumi29 · Answer

Out of curiosity, is this question from HCV or something?

anonymous · Answer

yeah its from hcv  @sumi29
 
btw I didn't get ...y r u adding the wt of the chain...
I mean the force exerted by the chain is the contact force between the chain and the floor...
can u plz elaborate....

sumi29 · Answer

Well that is because there are two forces exerted here. The weight of the chain which has fallen on the ground/table, and the force of the impact.

sumi29 · Answer

Damn I almost forgot all about HCV. You appearing for JEE or what? I did too, but I got EML...lol. But now am at MIT, so the work paid off!!!

anonymous · Answer

ohh yeah!!
now I get it....thanks a ton :)

actly I ws just revising the topic...
gt stuck to this ques...

yeah hard work always pays :)