A homogeneous chain of length `L` lie on a table. If the coefficent of friction is `mu` , find the maximum length `L_(1)` of the chain which can hang from the edge of the table without the chain sliding down.
(b) Consider the situation as shown in the figure. Find teh minimum value of `m` so that the blocks do not move.

Explanation

(a) Mass of chain on the table
`= (M)/(L) (L - L_(1))`
`N = (M)/(L) (L - L_(1)) g`
`f = mu N = (mu M (L - L_(1))g)/(L)`
Since the chain is in equilibrium
`(M)/(L) L_(1) g le f rArr (M)/(L) L_(1) g le (mu M (L - L_(1)g))/(L)`
`L_(1) le mu(L - L_(1)) rArr ( 1 + mu)L_(1) le muL`
`L_(1) le (mu L)/(mu + 1) rArr (L_(1))_(max) = (mu L)/( mu + 1)`
(b) `N = (10 + m)g`
`f = mu N = (1)/(3) ( 10 + m)g`
For the blocks to remain at rest
`6g le (1)/(3) (10 + m)g`
`rArr le(10 + m)`
`m ge 8 , m_(min) = 8 kg`.