The monkey B, shown in the following figure, is holding on to the tail of monkey A that is climbing up a rope. The masses of 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 monkey B with it? Take g = 10 m/s^{2}.

#### Solution

Let the acceleration of monkey A upwards be a, so that a maximum tension of 30 N is produced in its tail.

T − 5g − 30 − 5a = 0 ...(i)

30 − 2g − 2a = 0 ...(ii)

From equations (i) and (ii), we have:

T = 105 N (max.)

and a = 5 m/s^{2}

So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.

For minimum force, there is no acceleration of A and B.

T_{1} = weight of monkey B

⇒ T_{1} = 20 N

Rewriting equation (i) for monkey A, we get:

T − 5g − 20 = 0

⇒ T = 70 N

∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N.