A uniform chain of length L is lying on a hor | vedclass

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(B) Let $\lambda$ be the linear mass density of the chain. The total length of the chain is $L$. Let $l^{\prime}$ be the length of the chain hanging o

Vedclass · Accepted Answer

$\frac{\mu L}{(1+\mu)}$

Vedclass · Answer

(B) Let $\lambda$ be the linear mass density of the chain. The total length of the chain is $L$. Let $l^{\prime}$ be the length of the chain hanging over the edge.Then,the length of the chain remaining on the table is $(L - l^{\prime})$.The mass of the hanging part is $m_h = \lambda l^{\prime}$,and the mass of the part on the table is $m_t = \lambda (L - l^{\prime})$.The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda l^{\prime} g$.The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - l^{\prime}) g$.For the chain to be in equilibrium,the pulling force must be equal to the maximum frictional force:$\lambda l^{\prime} g = \mu \lambda (L - l^{\prime}) g$$l^{\prime} = \mu (L - l^{\prime})$$l^{\prime} = \mu L - \mu l^{\prime}$$l^{\prime} (1 + \mu) = \mu L$$l^{\prime} = \frac{\mu L}{(1 + \mu)}$

$A$ uniform chain of length $L$ is lying on a horizontal table. If the coefficient of friction between the chain and the table top is $\mu$,what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on the table?

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