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

Since chain is unifrom the mass of 'y' part of the chain will be `((M)/(L)y)`. When this part reaches the table its total force exerted must be equal to the weight of y part resting on table +Force due to the momentum impared
`F =(M)/(L)yg + (((M)/(L)dy)sqrt(2gy))/(dt) = (Mg)/(L) y + (M)/(L) v.sqrt(2gy)`
`(":'(dy)/(dt) =v) = (Mg)/(L) y+(M)/(L) sqrt(2gy).sqrt(2gy) = 3(My)/(L)g` .