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.