There are two monkeys and the monkey B is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it? Take g = 10 m/s 2.

Hint: A is holding the rope tied to the ceiling and B is holding the tail of A. The masses of the monkeys are given and they move up with a common acceleration a. We have to use Newton's equation of motion here and Newton’s second law, also, a free body diagram is must for this problem.

Complete step by step answer:
The free body diagram of the [problem is shown below.

maximum tension in the tail of monkey A can be 30N
$
30-2g-2a=0
\implies 30-20-2a=0
\implies 2a=10
\implies$ a=\[5m/{{s}^{2}}\]

Also,
$
T-5g-30-5a=0 \\
\implies T=50+30+25 \\
\implies T=105N \\
$
So, the maximum force applied by A is 105 N. This is the maximum force with which monkey A should pull the rope so that the combined system moves with maximum acceleration.
Now, applying the condition of minimum force, we get, there will be no acceleration in Monkey A and B ie a= 0
\[\begin{align}
  & {{F}_{\min }}=(M+m)g \\
 & =70N \\
\end{align}\]

Thus, monkey A need to apply force between 70 N and 105 N

Note- When more than one force acts on any system then we have to add all the forces vectorially using either triangle law of vector addition or parallelogram law of vector addition. But this was a special case where the forces were either parallel or in the same direction, so we added them easily.