A uniform chain of mass M and length L is held verticallyi in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.

Explanation

Let us consider a small element at distance x from the floor of length dy
So, `dm=M/Ldx`
so the velocity with which the element will strike the floor is
`v=sqrt(2gx)` ltbr.`:.` so the momentum transferred to the floor is
`M=(dm)v=M/L.dxsqrt(2gx)`
[because the element comes to rest]
So, the forec exerted on the floor changes in momentum ils givne by
`F_1(dM)/(dt)=M/L.(dx)/(dt)sqrt(2g)`
Because `v=((dx)/(dt))=sqrt(2gx)`
for the chain element
`F_=M/L.(sqrt(2gx))^`
`=M/L.2gx=(2Mgx)/L`
Againforce the force exerted due to x lengh of the chian on the floor due to its own weight is given by
`W=M/L(x)xxg=(Mgx)/L`
so the total force exerted is given by
`F=F_1+W`
`=(2Mgx)/L+(Mgx)/L`
`=(3Mgx)/L`