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(D) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.When $1/3$ of the length hangs ove

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$1/2$

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(D) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.When $1/3$ of the length hangs over the edge,the length of the hanging part is $L_h = L/3$ and the length on the table is $L_t = 2L/3$.The mass of the hanging part is $m_h = \lambda (L/3) = M/3$.The mass of the part on the table is $m_t = \lambda (2L/3) = 2M/3$.The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = (M/3)g$.The normal force on the part of the chain on the table is $N = m_t g = (2M/3)g$.The limiting friction force is $f_s = \mu_s N = \mu_s (2M/3)g$.At the point of sliding,the pulling force equals the limiting friction force: $F_g = f_s$.$(M/3)g = \mu_s (2M/3)g$.Solving for $\mu_s$: $\mu_s = (M/3) / (2M/3) = 1/2$.

$A$ uniform metal chain is placed on a rough table such that one end of the chain hangs down over the edge of the table. When one-third of its length hangs over the edge,the chain starts sliding. Then,the coefficient of static friction is

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