A very flexible uniform chain of mass M and length L is suspended vertically so that its lower and just touches the surface of a table. When the upper end of the chain is released it falls with each link coming to rest the instant it strikes the table. Find the force exerted by the chain on the table at the moment when y part of the chain has already rested on the table.

Explanation

In this problem force on the table will be due to the weight of the chain on the table `(say F_(1))` and also due to the momentum of chain being transmitted at each instant when the chain strikes it `(say F_(2))` so that
`F=F_(1)+F_(2)` ….(i)
Now as mass per unit length of the chain is `(M//L)` so when y length of the chain is on the table,
`F_(1)=(M/Lxxy)g` ....(ii)
Now to calculate `F_(2)` consider element dy at a height y above the table so that its mass will be `(M//L)` dy and it will hit the table, with a velocity `v=sqrt(2gy)` so the momentum transmitted by it to the table,
`dp=(dm)v=(M//L)dy sqrt(2gy)`
So that, `F_(2)=(dp)/(dt)=M/L(dy)/(dt)sqrt(2gy)`
or `F_(2)=M/L 2gy[as (dy)/(dt)=v=sqrt(2gy)]` ....(iii)
Substituting `F_(1)` and `F_(2)` from eqns (ii) and (iii) in (i) we get
`F=3 {M/L yg}=3` [weight of chain on the table]
Note: Here as `y=1/2"gt"^(2), F=3(Mg^(2)t^(2)//2L)`, i.e., force is time dependent and increases non linearly with time.