91Ó°ÊÓ

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

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

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

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

Therefore,

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

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:

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

Hence, the net torque magnitude is,

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

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