Force exerted by a table on a falling chain?

Question

A chain of length L and mass M is held up above a table. The lowest part of the chain slightly touches the top of the table. When we let go of the chain, it starts falling on the table and we assume that each link comnes to rest the instant it reaches the table.
Find the force exerted by the table on the chain after this has fallen through a distance x.

The distance fallen = x so the rest of the chain that is still in 'upright' position is L-x.
At first when we hold the chain up, the force F=mg is 0 since no mass is touching the table. I understand that the force depends on the length that lies on the table at that moment but how do I calculate this?? Can it be something like Mgx/2L ?

Answer 1

They specified a chain here for a reason. Chains cannot sustain a compressive force. The force on the table is produced only by the latest link to strike the table. The part of the chain that has not yet hit the table is in freefall.



Since they are asking for force over time, this really sounds like a momentum transfer problem. Momentum has units like newton-seconds (force x time). The force is then the momentum per second tansferred to the table. Of course, the momentum is also mass x velocity.



When the distance fallen is x, the link striking the table has fallen a distance of x. It has a velocity based on the usual freefal relation:



v^2 = 2gx



Call the total chain mass M, the length L and there are n links. The mass and length of a link are M/n and L/n.



The momentum transferred by a link is P = (M/n) * v



The time it takes from one link to hit until the next one is the link length divided by velocity:



t = (L/n)/v



The momentum is the product of force and time:



P = F*t so F= P/t



Substituting from above:



F = ((M/n) * v) / (L/n)/v) = (M*v^2)/L



Substuting in the realtion between fall distance and velocity:



F = 2Mgx/L



After reading subsequent answers, I notice that I completely forgot about the static potion of the problem, that is the dead weight of the chain on the table. Since x length is on the table, that amounts to Mgx/L, which is exactly one half the static load. Adding the two gives:



F = 3Mgx/L



You can also work this via energy methods where you assume an elastic but stiff table. The result is the same.

Answer 2

Falling Chain Physics

Answer 3

The chain must be assumed to be uniform since we are not told otherwise. The fraction of the chain's mass that is touching the table is thus equal to the fraction of the length of the chain that is touching the table. What is this fraction? It is x/L. So this mass touching the table is (x/L)*M = Mx/L. Then, of course, you would include g to get the force.

Answer 4

Yes you are on the right track. The amount of chain on the table is x. The portion of the whole chain on the table is x/L; this is the portion of the whole mass of chain on the table, so the mass on the table is m = (x/L)*M. The force exerted by that mass is m*g = g*(x/L)*M. The part of the chain that is still falling does not exert any force on the table (it is in free-fall, or weightless).

Answer 5

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