## Related questions with answers

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 $\text{F}_{ 1 } = 18.0 \ \text{N}$, $\text{F}_{ 2 } = 26.0 \text{ N }$, and $\text{F}_3 = 14.0 \ \text{N}$. The plate and all forces are in the plane of the page.

Solution

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

$\tau =\tau_{2}+\tau_{3}-\tau_{1}$

the magnitude of the torques that the forces $F_{1}$, $F_{2}$, and $F_{3}$ cause, are,

$\begin{align*} \tau_{1}&=r_{1} F_{1} \sin (\theta_{1} )\\ \tau_{2}&=r_{2} F_{2} \sin (\theta_{2} )\\ \tau_{3}&=r_{3} F_{3} \sin (\theta_{3} ) \end{align*}$

where $r_{1}=r_{2}=r_{3}=r$, $\theta_{1}=135^{\circ}$, $\theta_{2}=135^{\circ}$ and $\theta_{3}=90^{\circ}$, so,

$\begin{align*} \tau_{1}&=r F_{1} \sin (\theta_{1} )\\ \tau_{2}&=r F_{2} \sin (\theta_{2} )\\ \tau_{3}&=r F_{3} \sin (\theta_{3} ) \end{align*}$

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

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

the net torque magnitude is,

$\tau=2.34\mathrm{~N\cdot m}+1.782 \mathrm{~N\cdot m}-1.62 \mathrm{~N\cdot m}=2.502 \mathrm{~N\cdot m}$

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

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

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