A uniform chain of length L which hanges part | vedclass

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Question

$\mathrm{m}_{1} \mathrm{g}=\mu \mathrm{m}_{2} \mathrm{g} \quad \mathrm{m}_{1} \rightarrow$ mass of hanged part

$\mathrm{m}_{2} \rightarrow$ mass of

Vedclass · Accepted Answer

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

Vedclass · Answer

$\mathrm{m}_{1} \mathrm{g}=\mu \mathrm{m}_{2} \mathrm{g} \quad \mathrm{m}_{1} ightarrow$ mass of hanged part

$\mathrm{m}_{2} ightarrow$ mass of remaining part

$\mu=\frac{m_{1}}{m_{2}}=\frac{\frac{M}{L} 	imes \ell}{\frac{M}{L}(L-\ell)}=\frac{\ell}{L-\ell}$

A uniform chain of length $L$ which hanges partially from a table, 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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