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 mass per unit length be $\lambda$. The length of the chain hanging from the table is $l$. The len

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the total length of the chain be $L$ and the mass per unit length be $\lambda$. The length of the chain hanging from the table is $l$. The length of the chain lying on the table is $(L - l)$. The weight of the hanging part,which acts as the pulling force,is $F_g = (\lambda l)g$. The normal force acting on the part of the chain on the table is $N = (\lambda(L - l))g$. For the chain to be in equilibrium,the limiting friction must balance the weight of the hanging part: $f_{max} = \mu N = F_g$. Substituting the values: $\mu (\lambda(L - l))g = (\lambda l)g$. Solving for $\mu$: $\mu = \frac{l}{L - l}$.

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

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