Question

A 55 kg student is climbing a rope ladder hanging from a helicopter. While climbing, the student leans 0.35 m backward from the vertical line of the ladder.

1. Calculate the torque produced about the top support point of the rope ladder due to the student’s weight. [Take g = 10 m/s²]
2. Explain why the ladder starts swinging when the student leans backward and how the student should climb to minimize the swinging.

Answer 1

Given,

Mass of the student = 55 kg

Distance of the student from the rope = 0.35 m

Acceleration due to gravity (g) = 10 m s⁻²

i. The required torque is given by,

Torque = Weight of the student × Distance of the student from the rope
= 55 × g × 0.35 m
= 55 × 10 × 0.35
= 192.5 Nm

Hence, the torque produced about the top support point of the rope ladder is 192.5 Nm.

ii. When the student leans backward, the weight of the body acts at a distance from the vertical line of the ladder which produces a torque about the top support of the ladder, causing it to rotate slightly and start swinging.

To minimize this swinging, the student should keep the body straight and close to the ladder, so that the line of action of the weight passes nearly along the ladder and the torque produced is very small.