91Ó°ÊÓ

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. ${\mathrm{θ}}_A=135.0^o$
8. ${\mathrm{θ}}_B=90.0^o$
9. ${\mathrm{θ}}_C=160.0^o$
   

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$\mathit{τ}\mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}$

$$\begin{gathered} \mathit{τ}\mathbf{=}\mathit{r}\mathbf{×}\mathit{F} \\ \mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{θ} \end{gathered}$$

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

$\begin{array}{rcl}\mathrm{τ}& =& r×F\\ & =& rF\sin \mathrm{θ}\\ & =& {\mathrm{τ}}_A+{\mathrm{τ}}_B+{\mathrm{τ}}_C\\ \mathrm{τ}& =& r_A{F}_A\sin {\mathrm{θ}}_A-r_B{F}_B\sin {\mathrm{θ}}_B+r_C{F}_C\sin {\mathrm{θ}}_C\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{τ}& =& 8m×10N×\sin 135-4m×16N×\sin 90+3m×19N×\sin 160\\ & =& 12.06N.m\\ \mathrm{τ}& ≈& 12N.m\end{array}$

Hence the torque is, 12N.m .