A uniform chain of length L and mass M overhangs a horizontal table with its two third part n the table. The friction coefficient between the table and the chain is `mu`. Find the work done by the friction during the period the chain slips off the table.

Explanation

We see that a portion of the chain is lying on the table top Let the masss of that portion be `m_(1)` Let the mass of the remaining (hanging) portion of the chain be `m_(2)` since the chain is at the point of slipping the weight of the hanging portion of the chain conterbalance the maximum static friction forde `f_(max)` between `m_(1)` and the surface

`rArr m_(2) g = f_(max) `
` m_(2) g = mu N_(1)`
where `N_(1) - m_(2) g = 0` for the equilibrium of the portion of chain lying as on the table
`rArr m_(2) g - mu m_(2) g = 0`
`rArr mu = (m_(1))/(m_(2)) = ((M)/(L)x)/((M)/(L)(L- x)`
`rArr mu = (x//L)/(1 - x//L) rArr mu =(eta)/(1 - eta)`