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 mass of the chain is $M = \lambda L$.Let $l'$ be the length of the chain hanging

Vedclass · Accepted Answer

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

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