Four forces are applied to a wheel of radius 20 cm as shown in Fig. The net torque produced by the forces about O is ___Nm 

Magnitude of torque = Fr sin $\mathrm{\theta }$ where $\mathrm{\theta }$ is the angle between the force vector and the radius vector. The torques are (see Fig)

$\begin{array}{l}{\mathrm{\tau }}_1=8\mathrm{N}\times 0.2\mathrm{m}\times \sin {30}^{\circ }=0.8\mathrm{Nm}\text{ (clockwise) }\\ {\mathrm{\tau }}_2=4\mathrm{N}\times 0.2\mathrm{m}\times \sin {90}^{\circ }=0.8\mathrm{Nm}\text{ (anticlockwise) }\\ {\mathrm{\tau }}_3=9\mathrm{N}\times 0.2\mathrm{m}\times \sin {90}^{\circ }=1.8\mathrm{Nm}\text{ (clockwise) }\\ {\mathrm{\tau }}_4=6\mathrm{N}\times 0.2\mathrm{m}\times \sin 0^{\circ }=0\end{array}$

$\therefore$Net torque= + 0.8 Nm - 0.8 Nm + 1.8 Nm + 0 = 1.8 Nm clockwise.