A uniform chain of length L hangs partly from | vedclass

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(C) Let the total length of the chain be $L$ and the length hanging from the table be $\ell$. The length of the chain lying on the table is $(L - \ell

Vedclass · Accepted Answer

$\frac{\ell}{L-\ell}$

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(C) Let the total length of the chain be $L$ and the length hanging from the table be $\ell$. The length of the chain lying on the table is $(L - \ell)$.For the chain to be in equilibrium,the force of friction must balance the weight of the hanging part.The weight of the hanging part is $W = \lambda \ell g$,where $\lambda$ is the linear mass density of the chain.The normal force exerted by the table on the chain is $N = \lambda (L - \ell) g$.The maximum static friction force is $f_{max} = \mu N = \mu \lambda (L - \ell) g$.Equating the forces for the limiting case: $\lambda \ell g = \mu \lambda (L - \ell) g$.Solving for $\mu$: $\mu = \frac{\ell}{L - \ell}$.

$A$ uniform chain of length $L$ hangs partly from a table and is kept in equilibrium by friction. If the maximum length that can hang without slipping is $\ell$,then the coefficient of friction between the table and the chain is:

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