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

Let y be the maximum length of the chain, which can be held outside the table without sliding
Length of the chain on the table `= (L-y)`
Weight of this part of chain `W = (M)/(L) (L-y) g`
Weight of hanging part of chain `= W= (M)/(L) yg`
For equilibrium
force of friction `(f) = wt` of hanging part of chain
`mu R = W`
`mu W =W `
`mu (M)/(L) (L-y) g = (M)/(L) yg`
`mu Mg - mu (M yg)/(L) = (M)/(L) gy , mu Mg = (M)/(L) yg (1 + mu)`
`y = (mu L)/((1 + mu ))`
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