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Question: A monkey of mass 2 kg is climbing up a rope with an acceleration of 3 $ms^{-2}$. Find the tension in...

A monkey of mass 2 kg is climbing up a rope with an acceleration of 3 ms−2ms^{-2}. Find the tension in the rope in the portion AB and BC. (take g = 10 ms−2ms^{-2})

A

20 N, 26 N

B

0 N, 26 N

C

20 N, 0 B

D

None of these

Answer

0 N, 26 N

Explanation

Solution

The problem asks us to find the tension in two portions of a rope, AB and BC, as a monkey climbs up.

1. Analyze the forces on the monkey:

  • Mass of the monkey (m) = 2 kg
  • Acceleration of the monkey (a) = 3 m/s² (upwards)
  • Acceleration due to gravity (g) = 10 m/s²

When the monkey climbs up the rope, it exerts a downward force on the rope, and the rope exerts an upward tension force on the monkey. Let T be the tension in the rope segment that is actively supporting and accelerating the monkey. Using Newton's Second Law (F_net = ma) for the monkey: The upward force is T. The downward force is the monkey's weight, mg. Since the monkey is accelerating upwards, the net force is upwards: T−mg=maT - mg = ma T=mg+maT = mg + ma T=m(g+a)T = m(g + a)

Substitute the given values: T=2 kg×(10 m/s2+3 m/s2)T = 2 \text{ kg} \times (10 \text{ m/s}^2 + 3 \text{ m/s}^2) T=2 kg×(13 m/s2)T = 2 \text{ kg} \times (13 \text{ m/s}^2) T=26 NT = 26 \text{ N}

This tension (26 N) is the force exerted by the rope segment above the monkey's grip point.

2. Interpret the rope segments AB and BC: The diagram shows points A, B, and C along the rope, with A being the lowest point (where the monkey is depicted), B above A, and C as the fixed support.

There are two common interpretations for such problems, especially when the options do not match the most literal interpretation of the diagram:

  • Interpretation 1 (Literal Diagram): If the monkey is at point A as shown, then the rope segments AB and BC are both above the monkey. Assuming the rope is massless, the tension throughout the rope segment AC (which includes AB and BC) would be uniform and equal to the tension supporting the monkey. In this case, T_AB = 26 N and T_BC = 26 N. However, (26 N, 26 N) is not among the given options.

  • Interpretation 2 (Standard Problem Convention): In many physics problems, when points A, B, C are marked along a string/rope and an object is at one of these points (e.g., B), then "portion AB" refers to the segment below the object, and "portion BC" refers to the segment above the object. Given the options, it is highly probable that the question intends this interpretation, implying that the monkey is at point B.

Let's proceed with Interpretation 2, where the monkey is considered to be at point B:

  • Tension in portion AB (T_AB): If the monkey is at point B, then portion AB is the rope segment below the monkey. Since there is no mass attached to this segment below the monkey, and the monkey is climbing up (not pulling anything from below), the tension in this segment is zero. TAB=0 NT_{AB} = 0 \text{ N}

  • Tension in portion BC (T_BC): If the monkey is at point B, then portion BC is the rope segment above the monkey. This segment is supporting the monkey and accelerating it upwards. As calculated above, this tension is: TBC=m(g+a)=26 NT_{BC} = m(g + a) = 26 \text{ N}

Therefore, based on the most plausible interpretation that matches one of the options, the tension in portion AB is 0 N and the tension in portion BC is 26 N.