The body in Fig. 10-39is pivoted at O, and two forces act on it as shown. If r₁= 1.30m ,r₂= 2.15m , F₁=4.20N , F₂ = 4.90N , ${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{175}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$ , and ${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$, what is the net torque about the pivot?

1. The magnitude of position vector 1 is, r₁ = 1.30m .
2. The magnitude of position vector 2 is, r₂ = 2.15m .
3. F₁ = 4.20N
4. F₂ = 4.90N
5. ${\theta }_1=75.0^o$
6. ${\theta }_2=60.0^o$

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

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

If torque causes counterclockwise rotation, then the torque is considered as positive, and if torque causes clockwise rotation, then the torque is considered as negative.By using the concept of moment of inertia, you find rotational moment of inertia of four mass systems about an axis passing through it at various points.

$\begin{array}{rcl}\tau & =& r\times F\\ & =& rF\sin \theta \\ \tau & =& r_1{F}_1\sin {\theta }_1-r_2{F}_2\sin {\theta }_2\end{array}$

Substitute all the value in the above equation.

$$\begin{gathered} \tau =1.30m\times 4.20N\times \times \sin 75-2.15m\times 4.90N\times \sin 60 \\ \tau =-3.85N.m \end{gathered}$$

Hence the torque is, -3.85N.m .