# The monkey B shown in figure (5-E20) 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/s2.

Given: 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/s2

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.

T1 = weight of monkey B T1 = 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

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