Question

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

• A. Between 50 N and 80 N
• B. Between 30 N and 80 N
• C. Between 70 N and 105 N
• D. Between 60 N and 90 N

Answer 1

The correct option is C Between 70 N and 105 N

The monkey B (the child of monkey A) is stationary with respect to the monkey A, and both moves with some common acceleration,$a$.

Now, the force with which monkey B should be pulled upwards so as to be lifted with acceleration a, must be equal to, $m(g+a)$, where $m$ is the mass of monkey B. Since the maximum tension in the tail of monkey A can be $30N$.

So, $30=m(g+a)=2(10+a)$

$\Rightarrow a=5m/s^2$.

This is the maximum acceleration with which both can move upwards as combined system. If acceleration is greater then this value, the tail of monkey A will break. Now the force which monkey A should apply on rope so that monkey A gets this force acted upon him as the reaction force must be sufficient to lift the monkey A and the child monkey B with the common acceleration of $5m/s^2$.

$F_{max}=(M+m)(g+a)=7\times 15=105N$.

This is the maximum force with which monkey A should pull the rope so that combined system moves with maximum acceleration. The condition of minimum force acting on the combined system of the monkey and child will arise when both move up zero acceleration. In that case, $F_{min}=(M+m)g=70N$.

Thus the monkey A should pull the rope with the force of $70N$ to move up with maximum acceleration, such that the tension in his tail is just equal to the breaking strength of the tail.