A half metre rod is pivoted at the centre with two weights of 20 gf and 12 gf suspended at a perpendicular distance of 6 cm and 10 cm from the pivot respectively as shown below:

The direction of 20 kgf force is reversed. What is the magnitude of the resultant moment of the forces on the rod ?

Explanation

If the direction of 20 gf force is reversed, it will also create a clockwise moment.
`therefore` Resultant moment
= (120 + 120) gf cm (clockwise)
= 240 gf cm