Question

The upper end of a wire of length 5 m is fixed to the ceiling and a 20 kg mass is attached at its lower end. If the wire makes an angle $60^\circ$ with the horizontal, then the moment of force due to gravity to the upper end of the wire is: (Acceleration due to gravity = 10 m s$^{-2}$)

• A. 200 N m
• B. 500/$\sqrt{3}$ N m
• C. 250/$\sqrt{3}$ N m
• D. 500 N m

Hint:

Torque is $ F \times d $, where $ d $ is the perpendicular distance from the pivot to the line of action of the force.

Answer 1

The Correct Option is D

The wire makes a $60^\circ$ angle with the horizontal, so it is $30^\circ$ to the vertical. Gravitational force on the mass: $ F = 20 \times 10 = 200 \, {N} $. 
Torque about the upper end: 
the perpendicular distance from the ceiling to the line of action of gravity (vertical) is $ 5 \sin 30^\circ = 5 \times 0.5 = 2.5 \, {m} $. Torque: \[ \tau = 200 \times 2.5 = 500 \, {N m} \] This matches option (4).