Answer:
Let y be
the maximum length of the chain, which can be held outside the table without
sliding.
Length of the chain oh the table = (L -
y)
Weight of this part of chain = \[W=\frac{M}{L}yg\]
Weight of hanging part-pf chain
= \[W'=\frac{M}{L}\left( L-y \right)g\]
For equilibrium, Fig. 3(HT).7
force of fraction (j) = wt. of
hanging part of chain
\[\mu R=W\]
\[\mu W'=W\]
\[\mu .\frac{M}{L}\left( L-y \right)g=\frac{M}{L}yg\]
\[\mu Mg-\mu \frac{Myg}{L}=\frac{M}{L}yg\]
\[\mu Mg=\frac{M}{L}yg\left( 1+\mu \right)\]
\[y=\frac{\mu L}{\left( 1+\mu
\right)}\]
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