Question

A uniform rope of length ℓ lies on a table if the coefficient of friction is μ, 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

Hint: There exists friction between the rope and the table, so we have to consider frictional force into the account. The situation is that a part of the length of the rope says \[{{l}_{1}}\]is hanging and we need to find the maximum length that could exist in equilibrium without falling down. We have to use the concept of equilibrium of forces here.

Complete step by step answer:
Let the rope mass be uniformly distributed and say the total mass of the rope be M, and the total length of the rope is l. Then let us assume the length x is hanging down.
Mass per unit length= \[\dfrac{M}{l}\]
Mass of the hanging part= \[\dfrac{Mx}{l}\]
When the rope exists in equilibrium then the frictional force must be equal to the weight of the hanging rope.
Weight of the hanging rope is \[\dfrac{Mx}{l}g\].
Weight of the rope which is on the table is \[\dfrac{M(l-x)}{l}g\].
Force of friction is given by:
${{f}_{k}}=\mu N \\
\Rightarrow {{f}_{k}}=\mu \dfrac{M(l-x)}{l}g \\ $
Now balancing the force, we get,
$\mu \dfrac{M(l-x)}{l}g=\dfrac{Mx}{l}g \\
\Rightarrow \mu (l-x)=x \\
\Rightarrow \mu l-\mu x=x \\
\Rightarrow \mu l=\mu x+x \\
\Rightarrow \mu l=x(\mu +1) \\
\therefore x=\dfrac{\mu l}{\mu +1} $
So, the maximum length is \[\dfrac{\mu l}{\mu +1}\]

Hence, the correct option is C.

Note:Here there is no sliding and thus, we do not have to consider sliding friction into our account. friction is sometimes of great use like it helps in walking and in application of brakes, but at other times it is evil as a lot of wear and tear takes place due to it. The weight of the body always acts vertically downwards. We know that friction is a force and it always opposes the relative motion between the two objects. Its SI unit is Newtons, N.