A uniform chain of length L changes partly fr | vedclass

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Question

(c) $\mu = \frac{{{\rm{Lenght \,of \,chain\, hanging \,from \,the \,table}}}}{{{\rm{Lenght\, of\, chain \,lying \,on \,the \,table}}}}$ $ = \frac{l}{{

Vedclass · Accepted Answer

$\frac{l}{{L - l}}$

Vedclass · Answer

(c) $\mu = \frac{{{\rm{Lenght \,of \,chain\, hanging \,from \,the \,table}}}}{{{\rm{Lenght\, of\, chain \,lying \,on \,the \,table}}}}$ $ = \frac{l}{{L - l}}$

A uniform chain of length $L$ changes partly from a table which is kept in equilibrium by friction. The maximum length that can withstand without slipping is $l$, then coefficient of friction between the table and the chain is

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