Assuming the length of a chain to be L and coefficient of static friction `mu`. Compute the maximum length of the chain which can be held outside a table without sliding.

Explanation

If y is the maximum length of the chain which can hang outside the table without sliding , then for equilibrium of the chain , the weight of hanging part must be balanced by force of friction on the portion on the table :

`W = f_(L) " " .......` (i)
But from Fig . 7.37 ,
`W = (M)/(L)`yg and `R = W' = (M)/(L)(L-y)g`
So that , `" " f_(L) = muR = mu(M)/(L)(L-y)g`
Substituting these values of W and `f_(L)` in Eqn. (i) we get
`(M)/(L) yg = mu(M)/(L) (L-y)`g
which on simplification yields `y = mu L (1 + mu)`.