Question

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

Answer 1

Moment is given by $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
The force $F_2=12gf$ causes clockwise moment and force $F_1=20gf$ also causes clockwise moment.
Net moment about point O ${\tau }_o=F_2{r}_2+F_1{r}_2$
Or ${\tau }_o=(12\times {10}^{-3}\times 9.8)\times 10\times {10}^{-2}+(20\times {10}^{-3}\times 9.8)\times 6=0.02352Nm$