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A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1=18.0 N\text{F}_{ 1 } = 18.0 \ \text{N}, F2=26.0 N \text{F}_{ 2 } = 26.0 \text{ N }, and F3=14.0 N\text{F}_3 = 14.0 \ \text{N}. The plate and all forces are in the plane of the page.

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We have a square metal plate with side length of L=0.180L=0.180 m o, it is pivoted about an axis through point OO at its center and perpendicular to the plate as shown in the following figure. The magnitude of the forces shown in the figure are F1=18.0F_{1}=18.0 N, F2=26.0F_{2}=26.0 N, and F3=14.0F_{3}=14.0 N. We need to find the net torque on the plate, first let the counterclockwise be the positive direction, then the force F1F_{1} causes a negative torque, the force F2F_{2} causes a positive torque, and the force F3F_{3} causes a positive torque, so the total torque has a magnitude of,

τ=τ2+τ3τ1\tau =\tau_{2}+\tau_{3}-\tau_{1}

the magnitude of the torques that the forces F1F_{1}, F2F_{2}, and F3F_{3} cause, are,

τ1=r1F1sin(θ1)τ2=r2F2sin(θ2)τ3=r3F3sin(θ3)\begin{align*} \tau_{1}&=r_{1} F_{1} \sin (\theta_{1} )\\ \tau_{2}&=r_{2} F_{2} \sin (\theta_{2} )\\ \tau_{3}&=r_{3} F_{3} \sin (\theta_{3} ) \end{align*}

where r1=r2=r3=rr_{1}=r_{2}=r_{3}=r, θ1=135\theta_{1}=135^{\circ}, θ2=135\theta_{2}=135^{\circ} and θ3=90\theta_{3}=90^{\circ}, so,

τ1=rF1sin(θ1)τ2=rF2sin(θ2)τ3=rF3sin(θ3)\begin{align*} \tau_{1}&=r F_{1} \sin (\theta_{1} )\\ \tau_{2}&=r F_{2} \sin (\theta_{2} )\\ \tau_{3}&=r F_{3} \sin (\theta_{3} ) \end{align*}

from the graph we can find r=(L/2)2+(L/2)2=L/2r=\sqrt{(L/2)^{2}+(L/2)^{2}}=L/\sqrt{2}, so,

τ1=(0.180 m)(18.0 N)2sin(135)=1.62 Nmτ2=(0.180 m)(26.0 N)2sin(135)=2.34 Nmτ3=(0.180 m)(14.0 N)2sin(90)=1.782 Nm\begin{align*} \tau_{1}&=\frac{(0.180 \mathrm{~m}) (18.0 \mathrm{~N})}{\sqrt{2}} \sin (135^{\circ})=1.62 \mathrm{~N\cdot m}\\ \tau_{2}&=\frac{(0.180 \mathrm{~m}) (26.0\mathrm{~N})}{\sqrt{2}} \sin (135^{\circ})=2.34\mathrm{~N\cdot m} \\ \tau_{3}&=\frac{(0.180 \mathrm{~m}) (14.0\mathrm{~N})}{\sqrt{2}}\sin (90^{\circ})=1.782 \mathrm{~N\cdot m} \end{align*}

the net torque magnitude is,

τ=2.34 Nm+1.782 Nm1.62 Nm=2.502 Nm\tau=2.34\mathrm{~N\cdot m}+1.782 \mathrm{~N\cdot m}-1.62 \mathrm{~N\cdot m}=2.502 \mathrm{~N\cdot m}

τ=2.502 Nm\boxed{\tau=2.502 \mathrm{~N\cdot m}}

since this torque is positive then its direction out of the page\textbf{out of the page}.

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