A uniform rope of total length l is at rest o | vedclass

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(C) Let the mass per unit length of the rope be $\lambda$. The length of the rope hanging is $f l$,and the length of the rope on the table is $(1-f)l$

Vedclass · Accepted Answer

$f=1/(1+1/\mu)$

Vedclass · Answer

(C) Let the mass per unit length of the rope be $\lambda$. The length of the rope hanging is $f l$,and the length of the rope on the table is $(1-f)l$. The weight of the hanging part of the rope,which acts as the pulling force $F$,is: $F = (f l) \lambda g$ The normal reaction $N$ exerted by the table on the part of the rope lying on the table is: $N = ((1-f)l) \lambda g$ The maximum static friction force $f_s$ acting on the rope is: $f_s = \mu N = \mu (1-f) l \lambda g$ For the rope to be at rest (in equilibrium),the pulling force must be equal to the limiting friction: $F = f_s$ $f l \lambda g = \mu (1-f) l \lambda g$ Dividing both sides by $l \lambda g$: $f = \mu (1-f)$ $f = \mu - \mu f$ $f + \mu f = \mu$ $f(1+\mu) = \mu$ $f = \frac{\mu}{1+\mu}$ To match the given options,we can rewrite this as: $f = \frac{1}{\frac{1+\mu}{\mu}} = \frac{1}{\frac{1}{\mu} + 1} = \frac{1}{1 + \frac{1}{\mu}}$ Thus,the correct option is $(c)$.

$A$ uniform rope of total length $l$ is at rest on a table with a fraction $f$ of its length hanging (see figure). If the coefficient of friction between the table and the rope is $\mu$,then:

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