A uniform chain of length 2L is hanging in eq | vedclass

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(A) Let $\lambda$ be the linear mass density of the chain. The total mass of the chain is $M = \lambda(2L)$.When the chain is displaced by $x$,the len

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$\frac{x}{L} g$

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(A) Let $\lambda$ be the linear mass density of the chain. The total mass of the chain is $M = \lambda(2L)$.When the chain is displaced by $x$,the length of the chain on the right side is $(L+x)$ and on the left side is $(L-x)$.The mass of the right side is $m_R = \lambda(L+x)$ and the mass of the left side is $m_L = \lambda(L-x)$.Applying Newton's second law to the whole chain (treating it as a system):The net force is the difference in weights: $F_{net} = m_R g - m_L g = \lambda(L+x)g - \lambda(L-x)g = 2\lambda x g$.The total mass of the system is $M = 2\lambda L$.Using $F_{net} = M a$:$2\lambda x g = (2\lambda L) a$$a = \frac{2\lambda x g}{2\lambda L} = \frac{x}{L} g$.

$A$ uniform chain of length $2L$ is hanging in equilibrium over a small,smooth pulley. If end $B$ is given a slightly downward displacement $x$,the imbalance causes an acceleration. The acceleration of the chain when it has been displaced by distance $x$ is:

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