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A chain hanging from a string just touches the surface of the weighing machine. The string is burnt, and the chain falls on the weighing machine. What would be the weight registered on the weighing machine?

I came across certain standard solutions that claimed that the weight displayed by the weighing machine would be $3mgx/l$ when $x$ lies on the table provided $0<x<l$. However the solutions seemed to assume that the tension in the part of the chain that is falling down is $0$ without any justification, and I couldn't come up with a satisfactory explanation.

Could anybody explain, based on what principles do we state that the tension in the string should be 0?

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  • $\begingroup$ How about: Nobody is pulling on it apart from both sides, so there can't be any tension? $\endgroup$ – Danu Sep 23 '14 at 7:35
  • $\begingroup$ I am not sure if that is principally sound. There's this external impulsive normal force that acts on the chain when it hits the table. However how can the fact that the chain is not being pulled prove that the tension in the chain is 0 for the chain that is falling? That part does have an acceleration. Your argument could be used to state that the part of the chain that falls down on the machine does not have any tension. $\endgroup$ – BigBang07 Sep 23 '14 at 14:44
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I see that you already have the solution with you. So, I will try to help you with the tension in the chain zero part. The tension here will be zero even if the chain is acceelerating because the acceleration is caused by the gravitational force which is pulling it down and this force is acting on every single atom and molecule of the chain. In other words, at any point of time, the force acting upon any part of the chain with respect any other part of the chain is zero. Hence, there won't be any tension in the chain.

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  • $\begingroup$ Thanks for responding. But I have a doubt. The gravitational force is not the only force that's acting on the chain, there is the impulsive force from the table. Should it not affect the conditions and how is the above argument valid in such a case? $\endgroup$ – BigBang07 Dec 22 '14 at 17:11
  • $\begingroup$ As far as I understand and remember my 10th grade physics, a chain can transmit tension but usually cannot transmit pushing forces. So, the impulsive force which is a pushing force is not transmitted to the next link and cannot cause any tension. $\endgroup$ – Gautham Dec 23 '14 at 5:44
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There is an implicit assumption in this problem that the chain is perfectly flexible. In this limit, the part of the chain that just hit the table cannot transmit any force back up the chain. It just bends out of the way when it stops.

If you consider the falling part of the chain as your "system", its surroundings include the Earth (acting via a distance force) and the part of the chain that is just hitting the table (via a contact force). But that part of the chain is neither pulling nor pushing on your system. So the only force acting on the system is from the Earth, and it falls with a constant acceleration.

If you replace the chain with a stiff length of rope, the assumption of flexibility would no longer hold - the bottom of the rope would indeed be pushing up on the falling part, and the answer would be different.

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Strictly speaking, a tiny tension will build up in a chain falling to Earth. This is due to the part of the chain nearest to earth being attracted more strongly than the farthest part. This differential attraction is referred to as "tidal force". However, for chains with a length much smaller than Earth's radius this tidal force is very small. The way the problem is stated, it clearly is the intention to ignore tidal effects.

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