Calculating torque in figure (a) If $r_A=3m,F=8N,$and $\theta =51°$,what is the magnitude of the torque about location $A$, including units? (b) If the force were perpendicular to ${\overrightarrow{r}}_A$but gave the same torque as in the preceding question. What would be its magnitude?

The force that may cause an object to rotate along an axis is measured by torque. In linear kinematics, force is what propels an object forward. An angular acceleration is also caused by torque. As a result, torque can be thought of as the linear force's rotating equivalent.

Torque acting on an object can be expressed as

$\tau =\overrightarrow{r}\times \overrightarrow{F}$

$=rF\sin \theta ......\left(1\right)$

Here

$r\to$Magnitude of the position vector.

$F\to$Magnitude of the force vector.

$\theta \to$Angle between force and position vector.

The given data is-

Magnitude of the position vector for point$A·r_A=3m$

Magnitude of the force acting on the object,$F=8N$

Angle between force and position vector,$\theta =51°$

$=\left(3m\right)\left(8N\right)\sin 51°$

$=18.65N·m$

Hence, the magnitude of the torque acting on the object is $18.65N·m$

(b) As a result, the force and position vectors are perpendicular. As a result, the angle formed by these vectors, $\theta =90°$

Magnitude of the torque acting on the object obtained from part (a) $\tau =18.65N·m$.

Using equation , the force acting on the object is

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

$=\frac{\left(18.65N·m\right)}{\left(3m\right)\sin 90°}$

$=6.217N$

The magnitude of the torque about location A is-$6.217N$