Question

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:
A. $\dfrac{l}{\mu }$
B. $\dfrac{l}{{\mu + 1}}$
C. $\dfrac{{\mu l}}{{\mu + 1}}$
D. $\dfrac{{\mu l}}{{\mu - 1}}$

Answer 1

To solve this question, first we will assume the mass of the rope and the length of the hanging part and then find the forces acting on the rope that is in the form of weight of the hanging part and the portion lying on the table.

Complete step by step solution:
Let the mass of the rope be $M$ .
Let the length of the hanging portion is $x$ .
Then the forces acting on the rope are:
$w = weight\,of\,the\,hanging\,part = (\dfrac{M}{l}x)g$
$w = weight\,of\,the\,portion\,lying\,on\,the\,table \\\ w = (\dfrac{M}{l}g)(l - x) \\\$
$w = {({f_s})_{\max }} = {\mu _s}N$
or,
$(\dfrac{M}{l}g)x = {\mu _x}\dfrac{M}{l}(l - x)g \\\ \Rightarrow \dfrac{x}{l} = {\mu _x}[\dfrac{{l - x}}{l}]\,or\,\dfrac{x}{l} = {\mu _x}[1 - \dfrac{x}{l}] \\\ \Rightarrow \dfrac{x}{l} = m{\mu _s} - {\mu _s}\dfrac{x}{l} \Rightarrow (1 + {\mu _s})\dfrac{x}{l} = {\mu _s} \\\$
$\therefore x = \dfrac{{{\mu _s}l}}{{1 + {\mu _s}}}\,or\,x = \dfrac{{\mu l}}{{1 + \mu }}$
Hence, the correct option is (C.) $\dfrac{{\mu l}}{{\mu + 1}}$

Note:
Then they are just called forces. They are in newtons. Weight, since it is a force, is also in newtons. The mass that goes along with that weight is in kilograms. Typically, when a force is applied to mass, that causes the mass to accelerate.