Question

solveeee itttttttt..........

Answer 1

$\textbf{ Hey..mate here is yur ans...}$
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⏩ Let y be the maximum length of the chain can be hold outside on the table without sliding.
Length of chain on the table = ( L - y )g

⏩ weight of part of the chain on table
=> $\textbf{W' = M / L ( L- y )g}$

⏩ weight of hanging part of the chain
=> $\textbf{W = M / L yg}$

For equilibrium :
⏩ Limiting force of friction = weight of hanging part of the chain

$\nu$R = W => $\nu$W' = W

=> $\nu$M / L ( L - y)g = M / L yg

=> $\nu$L - $\nu$y = y

=> y = $\nu$L / 1 + $\nu$

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$\textbf{HOPE IT HELPS U}$

Answer 2

Hello user

Refer to this attachment for the answer