A uniform rope of length l lies on a table. I | vedclass

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(C) Let the total length of the rope be $l$ and its total mass be $M$. The mass per unit length is $\lambda = \frac{M}{l}$.Let $l_1$ be the length of

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$\frac{\mu l}{1 + \mu}$

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(C) Let the total length of the rope be $l$ and its total mass be $M$. The mass per unit length is $\lambda = \frac{M}{l}$.Let $l_1$ be the length of the rope overhanging the table. The length of the rope on the table is $(l - l_1)$.The mass of the overhanging part is $m_1 = \lambda l_1 = \frac{M}{l} l_1$. The weight of this part acts as the force pulling the rope down: $F_g = m_1 g = \frac{M g l_1}{l}$.The mass of the part on the table is $m_2 = \lambda (l - l_1) = \frac{M}{l} (l - l_1)$. The normal force on this part is $N = m_2 g = \frac{M g (l - l_1)}{l}$.The maximum static frictional force is $f_{max} = \mu N = \mu \frac{M g (l - l_1)}{l}$.For the rope to be in equilibrium without sliding,the pulling force must be equal to the maximum frictional force: $F_g = f_{max}$.$\frac{M g l_1}{l} = \mu \frac{M g (l - l_1)}{l}$.$l_1 = \mu (l - l_1)$.$l_1 = \mu l - \mu l_1$.$l_1 (1 + \mu) = \mu l$.$l_1 = \frac{\mu l}{1 + \mu}$.

$A$ uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$,then the maximum length $l_1$ of the part of this rope which can overhang from the edge of the table without sliding down is

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