The body in Fig. 10-40 is pivoted at O. Three forces act on F_A = 10N it: at point A,8.0m from O; F_B = 16N at B,4.0m from O ; F_C = 19Nandat C,3.0m from O. What is the net torque about O ?

1. Magnitude of position vector 1 is, $r_A=8.0m$.
2. Magnitude of position vector 2 is, $r_B=4.0m$.
3. Magnitude of position vector 3 is, $r_C=3.0m$.
4. $F_A=10N$
5. $F_B=16N$
6. $F_C=19N$
7. ${\theta }_A=135.0^o$
8. ${\theta }_B=90.0^o$
9. ${\theta }_C=160.0^o$
   

Torqueis a turning action on a body about a rotation axis due to a force. If force is applied at a point, then total torque is the cross product of radial vector and force exerted on the body. The magnitude of torque is$\mathit{\tau }\mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

$$\begin{gathered} \mathit{\tau }\mathbf{=}\mathit{r}\mathbf{\times }\mathit{F} \\ \mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta } \end{gathered}$$

If torque causes counterclockwise rotation, the torque is considered as positive and if torque causes clockwise rotation, the torque is considered as negative.

$\begin{array}{rcl}\tau & =& r\times F\\ & =& rF\sin \theta \\ & =& {\tau }_A+{\tau }_B+{\tau }_C\\ \tau & =& r_A{F}_A\sin {\theta }_A-r_B{F}_B\sin {\theta }_B+r_C{F}_C\sin {\theta }_C\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\tau & =& 8m\times 10N\times \sin 135-4m\times 16N\times \sin 90+3m\times 19N\times \sin 160\\ & =& 12.06N.m\\ \tau & \approx & 12N.m\end{array}$

Hence the torque is, 12N.m .