A uniform chain of length L which hangs parti | vedclass

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(C) Let $M$ be the total mass of the chain of length $L$. The mass per unit length is $\lambda = \frac{M}{L}$.When a length $l$ hangs from the table,t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $M$ be the total mass of the chain of length $L$. The mass per unit length is $\lambda = \frac{M}{L}$.When a length $l$ hangs from the table,the mass of the hanging part is $m_1 = \lambda l = \frac{M}{L} l$.The mass of the part remaining on the table is $m_2 = \lambda (L - l) = \frac{M}{L} (L - l)$.For the chain to be in equilibrium,the downward force due to the hanging part must be balanced by the maximum static frictional force acting on the part on the table.The downward force is $F_g = m_1 g = \frac{M}{L} l g$.The maximum frictional force is $f_{max} = \mu N = \mu m_2 g = \mu \frac{M}{L} (L - l) g$.Equating the two: $\frac{M}{L} l g = \mu \frac{M}{L} (L - l) g$.Solving for $\mu$: $\mu = \frac{l}{L - l}$.

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

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