A square metal plate$\mathbf{0}\mathbf{.}\mathbf{180}\mathbf{}\mathbf{m}$on each side is pivoted about an axis through point O at its center and perpendicular to the plate (Fig. E10.3). Calculate the net torque about this axis due to the three forces shown in the figure if the magnitude of the forces are ${\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$, ${\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{26}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$and ${\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. The plate and all forces are in the plane of the page.

We have the given data:

Length of the side of metal plate $=L=0.180\mathrm{m}$.

The magnitude of the forces are:

$F_1=18.0{\mathrm{N}}_2{F}_2=26.0{\mathrm{N}}_1{F}_1=14.0\mathrm{N}$

We have to find the net torque about the axis due to the three forces.

Consider the counter clockwise direction to be the positive torque.

Then the force $F_1$causes a negative torque, the force $F_2$causes a positive

torque, and the force $F_3$causes a positive torque.

Therefore, the total torque has the magnitude of,

localid="1667990529847" $\tau =-{\tau }_1+{\tau }_2+{\tau }_3$

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

localid="1667990532728" $$\begin{gathered} {\tau }_1=r_1{F}_{1}sin\left({\theta }_1\right) \\ {\tau }_2=r_2{F}_{2}sin\left({\theta }_2\right) \\ {\tau }_3=r_3{F}_{3}sin\left({\theta }_3\right) \end{gathered}$$

Where, $r-r_2-r_3-r$and ${\theta }_1=135°,{\theta }_2=135°,{\theta }_3=90°$.

Therefore,

$$\begin{gathered} {\tau }_1=r{F}_1\sin \left(135°\right) \\ {\tau }_2=r{F}_2\sin \left(135°\right) \\ {\tau }_3=r{F}_3\sin \left(90°\right) \end{gathered}$$

Now, the $r$can be calculated as:

localid="1667990536101" $r=\sqrt{(L/2{)}^2+(L/2{)}^2}=L/\sqrt{2}$

Thus, we get the values of torque as:

$$\begin{gathered} {\tau }_1=\frac{\left(0.180\right)\left(18.0\right)}{\sqrt{2}}\sin \left({135}^0\right)=1.62N.M \\ {\tau }_2=\frac{\left(0.180\right)\left(26.0\right)}{\sqrt{2}}\sin \left({135}^0\right)=2.34N.M \\ {\tau }_3=\frac{\left(0.180\right)\left(14.0\right)}{\sqrt{2}}\sin \left({90}^0\right)=1.782N.M \end{gathered}$$

Hence, the net torque magnitude is,

localid="1667990539714" $$\begin{gathered} \tau =-1.62+2.34+1.782=2.502N.M \\ \therefore \tau =2.502N.M \end{gathered}$$

Since this torque is positive, its direction is out of page.