20. Calculate the total torque acting on thebodysfigure (10-E2) about

1.	20. Calculate the total torque acting on thebodysfigure (10-E2) about the point O.10 N15 N37°90°4 cm1509 1305 N20 NFigure 10-E2
Answer» Torque about a point = Total force × perpendicular distance from the point to that force. Let anticlockwise torque = + veAnd clockwise acting torque = –veForce 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 NTherefore, torque at O due to this force= 10 × 4 × 10–2= 0.4 N-m (clockwise) Force acting at the point A is 20 NTherefore, 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.

20. Calculate the total torque acting on thebodysfigure (10-E2) about the point O.10 N15 N37°90°4 cm1509 1305 N20 NFigure 10-E2

Answer»

Torque about a point = Total force × perpendicular distance from the point to that force.

Let anticlockwise torque = + veAnd clockwise acting torque = –veForce 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 NTherefore, torque at O due to this force= 10 × 4 × 10–2= 0.4 N-m (clockwise)

Force acting at the point A is 20 NTherefore, 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.

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