A chain of mass $M$ and length $l$ is suspended vertically with its lowest end touching a table. The chain is released and falls onto the table. What is the force exerted by falling part of the chain on the table?(Neglect the size of individual links)

Here's what I did. Momentum of a size $dx$ of chain, which was initially at a height $x$ from the bottom of the chain, when it is just touching the table is given by $$p = M\sqrt{2gx}\frac{dx}{l}$$

This momentum is lost in time $$t = \frac{dx}{\sqrt{2gx}}$$

Force exerted by table on chain and chain on table is given by $$F = \frac{p}{t} = 2Mg\frac{x}{l}$$

As it turns out, this is the right answer

But, the author asks to solve this using Work-Energy Theorem. Here's what I did:

Work done by table on chain: $$F.dx$$ Energy of part of chain lost to this work done: $$Mgx\frac{dx}{l}$$

Equating the above two equations, I get $$F = Mg\frac{x}{l}$$

This is off by the right answer by a factor of two. What goes wrong when I use Work-Energy Theorem, although I get the right answer when I use momentum?


when you apply work energy principle you need to apply change in height of the center of mass of the length dx of chain , considering uniform mass distribution center of mass of chain of length dx is (dx/2). Force F act on a displacement of center of mass of chain i.e. $F.(\frac{dx}{2})=Mgx\frac{dx}{L}$. Hence $F=2Mgx/L$

  • $\begingroup$ okay...That's what I thought! Thanks for confirmation! $\endgroup$ – Cheeku Mar 14 '13 at 14:19

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