Torque about a point = Total force × perpendicular distance from the point to that force.
Let anticlockwise torque = + ve
And clockwise acting torque = –ve
Force acting at the point B is 15 N
Therefore torque at O due to this force
= 15 × 6 × 10–2 × sin 37°
= 15 × 6 × 10–2 × 3/5 = 0.54 N-m (anticlock wise)
Force acting at the point C is 10 N
Therefore, torque at O due to this force
= 10 × 4 × 10–2 = 0.4 N-m (clockwise)
Force acting at the point A is 20 N
Therefore, Torque at O due to this force = 20 × 4 × 10–2 × sin30°
= 20 × 4 × 10–2 × 1/2 = 0.4 N-m (anticlockwise)
Therefore resultant torque acting at ‘O’ = 0.54 – 0.4 + 0.4 = 0.54 N-m.
