91Ó°ÊÓ

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" $\mathrm{Ï„}={\mathrm{Ï„}}_1-{\mathrm{Ï„}}_2-{\mathrm{Ï„}}_3$

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

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

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

Thus, we get the values of torque as:

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

Hence, the net torque magnitude is,

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

Since this torque is negative, its direction is clockwise.