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There are two monkeys A and B of masses 6 kg and 3 kg respectively. Monkey B is holding the tail of monkey A. Monkey A is trying to climb the rope as shown in the figure. The tail of monkey A can tolerate the maximum tension of 45 N. Find the minimum and the maximum force that monkey A should apply on the rope in order to carry the monkey B with it. (Take g = 10 m/sec2)

There are two monkeys A and B of masses 6 kg and 3 kg respectively. Monkey B is holding the tail of monkey A. Monkey A is trying to climb the rope as shown in the figure. The tail of monkey A can tolerate the maximum tension of 45 N. Find the minimum and the maximum force that monkey A should apply on the rope in order to carry the monkey B with it. (Take g = 10 m/sec2) Correct Answer 90 N and 135 N

CONCEPT:

Newton's second law of motion

  • According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the applied unbalanced force.
  • The magnitude of the force is given as,

⇒ F = ma

Where F = resultant force, m = mass and a = acceleration

CALCULATION:

Given mA = 6 kg, mB = 3 kg, Tf = 45 N, and g = 10 m/sec2

Let the maximum possible acceleration that monkey A can apply to be "a".

  • The free body diagram of monkey B is drawn as,

[ alt="F1 Shraddha Prabhu 07.09.2021 D2" src="//storage.googleapis.com/tb-img/production/21/09/F1_Shraddha_Prabhu_07.09.2021_D2.png" style="width: 66px; height: 187px;">     -----Figure (1)

From figure 1,

⇒ Tf = mBa + mBg

⇒ 45 = 3a + (3 ×10)

⇒ a = 5 m/sec2     -----(1)

  • The free body diagram of monkey A is drawn as,

Let the tension in the rope be T due to acceleration "a".

[ alt="F1 Shraddha Prabhu 07.09.2021 D3" src="//storage.googleapis.com/tb-img/production/21/09/F1_Shraddha_Prabhu_07.09.2021_D3.png" style="width: 66px; height: 204px;">     -----Figure (2)

From figure 2,

⇒ T = mAa + mAg + Tf

⇒ T = (6 × 5) + (6 × 10) + 45

⇒ T = 30 + 60 + 45

⇒ T = 135 N

  • This tension is the maximum force that monkey A can apply safely in order to carry monkey B with it.

For the condition of minimum force, there will be no acceleration in Monkey A and B,

∵ a = 0 m/sec2

⇒ Fmin = mAg + mBg

⇒ Fmin = (6 × 10) + (3 × 10)

⇒ Fmin = 90 N

  • So the minimum and the maximum force that monkey A should apply on the rope in order to carry the monkey B with it are 90 N and 135 N respectively. Hence, option 3 is correct.

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