Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$\mu l$

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

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

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

Step-by-Step Solution

System Definition: Let the total length of the rope be $l$ and its linear mass density (mass per unit length) be $\lambda$ . Let $l_1$ be the length hanging over the edge and $l_2 = l - l_1$ be the length lying on the table.
Forces:

Driving Force: The weight of the hanging portion tries to pull the rope down. $F_{driving} = m_{hanging}g = (\lambda l_1)g$ .
Resisting Force: The static friction acting on the portion of the rope on the table opposes the motion. The normal force $N$ is equal to the weight of the portion on the table: $N = m_{table}g = \lambda (l - l_1)g$ . The maximum static friction is $f_{max} = \mu N = \mu \lambda (l - l_1)g$ [Source 64].

Equilibrium Condition: For the rope to be on the verge of sliding (maximum overhang), the driving force must equal the maximum static friction: $(\lambda l_1)g = \mu \lambda (l - l_1)g$
Solving for $l_1$ : Cancel $\lambda g$ from both sides: $l_1 = \mu (l - l_1)$ $l_1 = \mu l - \mu l_1$ $l_1(1 + \mu ) = \mu l$ $l_1 = \frac{\mu l}{1 + \mu }$

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