A very flexible uniform chain of mass M and length L is suspended from one end s

Question

A very flexible uniform chain of mass M and length L is suspended from one end so that it hangs vertically, the lower end just touching the surface of a table. the upper end is suddenly released so that the chain falls onto the table and coils up in a small heap, each link coming to rest the instant it strikes the table. find the force exerted by the table on the chain at any instant, in terms of the weight of chain already on the table at that moment. Draw free body diagram showing all forces acting.
[Hint: consider the momentum delivered to the table by the falling chain between times t and t+dt. the chain is so flexible that links don't exert forces on one another; they're all just in free fall]

Explanation / Answer

Hi, There are two contributions to the force the chain exerts on the table : (a) the weight of the chain at rest on the table (b) the force due to the impulse of the falling part of the chain.   From given data, the chain has a mass per unit length of M/L. At any instant of time 't' let the chain be dropped by a distance of x. Then there will be a mass of (M/L)x on the table. So Fw = [(M/L)x] g which is the weight of the chain at rest on the table. Impulse delivered by the falling chain can be found by equation Fimp = dp/dt . If we consider an infinitesimal mass, dm, traveling at velocity, v, hitting the table then Fimp = v(dm/dt).   Hence dm = (M/L) dx where dx is the infinitesimal length of dm. So Fimp = (M/L) v (dx/dt) = (M/L) v^2 (since dx/dt = v.) Now the mass dm has fallen a distance x so it's velocity will be          v= (2gx)^(1/2) (from the kinematic equations). So Fimp = (M/L) 2 g x .    Hence the total force exerted by the table on the chain at any instant of time is given by     Ftot = Fw + Fimp = 3 g x (M/L). (since both forces act in same direction, upwards) Hope this helps you.

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