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.

AB = 4 m

OA = 2 m

OB = 2 m

Force at A = 10 N

Force at B = 10 N

As we know ,

Moment of force = F × r

Moment of force about o at point A = 10 × 2

= 20 N m Clockwise

Moment of force about o at point B = 10 × 2

= 20 N m Clockwise

Total moment of forces about the centre O

= 20 + 20 = 40 N m

Total moment of force about the pivot O is 40 Nm (clockwise)