The forces are applied are to a wheel of radius 0.350 m, as shown in the Fig.E10.4. One force is perpendicular to the rim, one is tangent to it, and the other one makes a $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}$angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?

We have the given data:

The radius of the wheel $=r=0.350\mathrm{m}$.

The magnitude of the forces are:

$F_1=11.9\mathrm{N},F_2=14.6{\mathrm{N}}_1{F}_3=8.50\mathrm{N}$

We have to find the net torque on the wheel due to the three forces.

Consider the counter clockwise direction to be the positive torque.

Then the force ${}^{F_1}$causes 0 torque, the force $F_2$causes a negative torque,

and the force $F_3$causes a positive torque.

Therefore, the total torque has the magnitude of,

localid="1667990785881" $\tau ={\tau }_1-{\tau }_2-{\tau }_3$

The magnitude of the torques that the three forces cause are given by,

$$\begin{gathered} {\tau }_1=r_1{F}_{1}sin\left({\theta }_1\right) \\ {\tau }_2=r_2{F}_{2}sin\left({\theta }_2\right) \\ {\tau }_3=r_3{F}_{3}sin\left({\theta }_3\right) \end{gathered}$$

Where, $r_1=0,r_2=r_3=r=0.350$and ${\theta }_1=0°,{\theta }_2=40.0°,{\theta }_3=90.0°$.

Thus, we get the values of torque as:

_{$$\begin{gathered} {\tau }_1=0, \\ {\tau }_2=(14.6)(0.350)\sin (40°)=3.285N.m \\ {\tau }_3=(8.50)(0.350)\sin (90°)=2.975N.m \end{gathered}$$}

Hence, the net torque magnitude is,

localid="1667990934351" $$\begin{gathered} \tau =0-3.285+2.975=-0.31N.m \\ \therefore \tau =-0.31N.m \end{gathered}$$

Since this torque is negative, its direction is clockwise.