Question

A uniform rope of mass \[m\] hangs freely from a ceiling. A monkey of mass \[M\] climbs up the rope with an acceleration \[a\]. The force exerted by the rope on the ceiling is

Answer 1

Hint:We are asked to find the total force exerted by the rope on the ceiling or the net tension on the rope. Find the tension due to the rope itself and tension due to the monkey on the rope. Then combine these two results to find the net tension on the rope.

Complete step by step answer:
Given, mass of the rope, \[m\].Mass of the monkey, \[M\].Acceleration of the monkey when it climbs the rope, \[a\]. Let \[T\] be the net tension of the rope or the total force exerted by the rope on the ceiling and \[g\] be the acceleration due to gravity.We now draw a free body diagram for the given problem.

The force exerted by the rope on the ceiling will be the force exerted by the rope itself and the force exerted by the monkey that is,
\[T = {T_1} + {T_2}\] (i)
where \[{T_1}\] is the force exerted by the rope itself or tension due to the rope and \[{T_2}\] is the force exerted by the monkey on the rope or tension on the rope due to the monkey.
The tension due to the rope, \[{T_1}\] can be written as by observing the free body diagram,
\[{T_1} = mg\] (ii)
For the monkey, we can write by observing the free body diagram,
\[{T_2} - Mg = Ma\]
\[\Rightarrow {T_2} = Ma + Mg\] (iii)
Now substituting equations (ii) and (iii) in (i), we get
\[T = mg + Ma + Mg\]
\[\therefore T = mg + M\left( {a + g} \right)\] (iv)

Therefore, the force exerted by the rope on the ceiling is \[mg + M\left( {a + g} \right)\].

Note: In such types of questions, draw a free body diagram before proceeding for calculations. Free body diagram is a diagram showing all the forces with its direction acting upon the system. Also, if the rope was given to be massless then net tension or force on the rope would depend only on the monkey.