Physics
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.
(i) Calculate the torque produced about the top support point of the rope ladder due to the student's weight. [Take g = 10 m s-2]
(ii) Explain why the ladder starts swinging when the student leans backward and how the student should climb to minimize the swinging.
Force
24 Likes
Answer
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 x Distance of the student from the rope
= 55 x g x 0.35 m
= 55 x 10 x 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.
Answered By
16 Likes
Related Questions
Two forces each of magnitude 10N act vertically upwards and downwards respectively at the two ends A and B of a uniform rod of length 4m which is pivoted at its mid-point O as shown. Determine the magnitude of resultant moment of forces about the pivot O.
Figure shows two forces each of magnitude 10N acting at the points A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of the two forces about the point (i) A, (ii) B and (iii) O, situated exactly at the middle of the two forces.
An aluminium rod of length 50 cm whose mass per unit length is 0.15 kg/m is joined with a steel rod of length 50 cm whose mass per unit length is 0.45 kg/m as shown in the figure below.
At what point should the rod be supported so that it remains in equilibrium horizontally?
The diagram shows a uniform bar supported at the middle point O. A weight of 40 gf is placed at a distance 40cm to the left of the point O. How can you balance the bar with a weight of 80 gf?
