The monkey B, shown in the figure (5−E20), 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/s².

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Solution

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

(i)

(ii)

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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Q. 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 / s^{2}

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 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 $m / s^{2}$ .