In Figure 11.26, if ${\mathit{r}}_{\mathbf{A}}\mathbf{=}\mathbf{3}\mathit{m}\mathbf{,}$ and $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}\mathbf{,}$ what is the magnitude of the torque about locationAincluding units? If the force in Figure 11.26 were perpendicular to ${\overrightarrow{\mathbf{r}}}_{\mathbf{A}}$ but gave the same torque as before, what would be its magnitude?

The force that can cause an object to twist along an axis is measured as torque. In linear kinematics, force is what causes an object to accelerate. Torque is also responsible for the angular acceleration. As a result, torque can be defined as the linear force's rotational equivalent.

Given:

Force acting on the ranch,$F=4N$

Force is at an angle,$\theta =30°$

Length of the ranch,$r_A=3m$

The figure given below shows the direction of the force on the wrench.

Magnitude of the torque relative to location A is given by formula:

Torque, $\tau =r_{A}F\sin \theta$

Substitute the values in the above equation

$$\begin{gathered} \tau =\left(3m\right)\left(4N\right)\sin 30° \\ =6 \text{N.m} \end{gathered}$$

Hence, the magnitude of torque at location A is $6 \text{N.m}$.

To produce the same torque, the amount of the force acting on the torque relative to position A is given by-

$F=\frac{\tau }{r_A\sin \theta }$

$$\begin{gathered} F=\frac{(6N.m)}{\left(3m\right)\sin 90°} \\ =2N \end{gathered}$$

Hence, the magnitude of force at location A is $2 \text{N}$.