A uniform rope lies on a horizontal table suc | 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 $\lambda = M/L$.The length of the hanging part is $

Vedclass · Accepted Answer

$0.33$

Vedclass · Answer

(C) Let the total length of the rope be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.The length of the hanging part is $l = 0.25L = L/4$. The mass of the hanging part is $m = \lambda l = M/4$.The length of the part on the table is $L - l = 0.75L = 3L/4$. The mass of this part is $M' = \lambda(3L/4) = 3M/4$.The downward force causing the rope to slide is the weight of the hanging part: $F_g = mg = (M/4)g$.The limiting friction force opposing the motion is $f_s = \mu_s N = \mu_s M' g = \mu_s (3M/4)g$.At the point of sliding,the downward force equals the limiting friction force: $(M/4)g = \mu_s (3M/4)g$.Solving for $\mu_s$: $\mu_s = (M/4) / (3M/4) = 1/3 \approx 0.33$.

$A$ uniform rope lies on a horizontal table such that a part of it hangs over the edge. The rope begins to slide down when the length of the hanging part is $25\%$ of its entire length. The coefficient of friction between the rope and the table is

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