Two forces equal in magnitude and opposite in direction, acting on an object at two different points, form what is called a couple. Two antiparallel forces with equal magnitudes ${\mathit{F}}_{\mathbf{1}}\mathbf{=}{\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{N}$are applied to a rod as shown in Fig. E11.21. (a) What should the distance I between the forces be if they are to provide a net torque of 6.40 N.m about the left end of the rod? (b) Is the sense of this torque clockwise or counterclockwise? (c) Repeat parts (a) and (b) for a pivot at the point on the rod where $\underset{\mathbf{2}}{\overset{\mathbf{∆}}{\mathbf{F}\mathbf{}}}$is applied.

The two antiparallel forces with equal magnitudes are:

$$\begin{gathered} F_1=F_2 \\ =8.00N \end{gathered}$$

The net torque provided about the left end of the rod is $T_{net}=6.40N.m$.

The distance of the force $F_1$from the left end of the rod is x = 3 m .

The distance between the forces $F_1$and $F_2$is I.

When forces are applied on a longitudinal rod at a distance from the pivot point in such a way that the combined effect is the rotation of the rod about the pivot.

The value of torque can be calculated by multiplying the force value with the distance between the force and the pivot point.

The free-body diagram of the rod is given by:

According to the sign convention, the clockwise torque is taken positive while the counterclockwise torque is taken negative.

Calculating the net torque about the left end of the rod,

$$\begin{gathered} {\tau }_{net}=F_2(x+I)-F_{1}x \\ {\tau }_{net}=F_{2}I(F_2-F_1)x \\ {F}_{2}I={\tau }_{net}-(F_2-F_1)x \\ I=\frac{{\tau }_{net}-(F_2-F_1)x}{F_2} \end{gathered}$$

Putting the values,

$$\begin{gathered} I=\frac{6.40N.m-(8N-8N)\times 3m}{8N} \\ =\frac{6.40N.m-0}{8N} \\ =0.8m \end{gathered}$$

Hence, the distance between the forces is 0.8 m.

The magnitude of the net torque $\left({\tau }_{net}\right)$is positive.

Hence, the direction of the torque is clockwise.

If the pivot is located at the point on the rod where the force $F_2$is applied, then calculating torque about the pivot,

$$\begin{gathered} {\tau }_{net}=F_{1}I \\ I=\frac{{\tau }_{net}}{F_1} \end{gathered}$$

Putting the values,

$$\begin{gathered} I=\frac{6.40N.m}{8N} \\ =0.8m \end{gathered}$$

And, the direction of the net torque is unchanged.

Hence, the distance between the forces is 0.8 m and the direction of the net torque is clockwise.