A uniform chain of mass `m` and length `l` hanges on a thread and touches the surface of a table by its lower end. Find the force exerted by the table on the chain when half of its length has fallen on the table. The fallen part does not form heap.

Answer

`(3)/(2)`

Explanation

`1.` Weight of the portion `BC` of the chain lying on the table, `W=(mg)/(2)` (downwards) Using `v=sqrt(2gh)`
`2.` Thrust force `F_(t)=v_(r)((dm)/(dt))`
`v_(f)=v`
`(dm)/(dt)=lambdav`
`F_(t)=lambda v^(2)`
(where `lambda =(m)/(l)`, is mass per unit length of chaing)
`v^(2)=(sqrt(gl))^(2)=gl`
`:. F_(t)=((m)/(l))(gl)=mg` (downward)
`:.` Net force exerted by the chain on the table is
`F=W+F_(t)=(mg)/(2)+mg=(3)/(2)mg`
So, from Newton's third law the force exerted by the table on the chain will be `(3)/(2) mg` (vertically upwards).