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Expert-verified Found in: Page 355 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Calculate the net torque (magnitude and direction) on the beam in Figure P11.5 about (a) an axis through O perpendicular to the page and (b) an axis through C perpendicular to the page. a) Therefore, the net torque about the point O perpendicular is $30.0N.m$

The direction of the net torque is in counter clockwise direction

b) Therefore, the net torque about perpendicular is $36.0N.m$

The direction of the net torque is in counter clockwise direction

See the step by step solution

## Step 1: Formula to calculate the Torque

$\tau =rF$

Here is the Torque, is the perpendicular force and is the applied force.

The force that acts on the beam is shown in figure From this figure $OO$is the perpendicular distance for the force${F}_{1}$ ,$OD$ is the perpendicular distance for the force${F}_{2}$,$OO$ is the perpendicular distance for the force${F}_{3}$ ,${F}_{1}cos{\theta }_{1}$ is the horizontal component of the force${F}_{1}$ ,${F}_{2}sin{\theta }_{2}$ is the vertical component of${F}_{2}$ ,${F}_{3}sin{\theta }_{3}$ is the vertical component of the force${F}_{3}$

## Step 2: To find the net torque about perpendicular point

The net torque about the point O perpendicular is

${\tau }_{O}=OC{F}_{1}cos{\theta }_{1}+OD\left(-{F}_{2}sin{\theta }_{2}\right)+OO\left(-{F}_{3}sin{\theta }_{3}\right)$

${\tau }_{O}$is the net torque about the point O,OO is the perpendicular distance for the force${F}_{1}$ ,OD is the perpendicular distance for the force${F}_{2}$ ,OO is the perpendicular distance from the ${F}_{3}$, ${F}_{1}cos{\theta }_{1}$is the horizontal component for ${F}_{1}$,${F}_{2}sin{\theta }_{2}$ is the horizontal component for ${F}_{2}$,${F}_{3}sin{\theta }_{3}$ is the horizontal component for${F}_{3}$

$\begin{array}{l}Substitute2.0mforOC,25Nfor{F}_{1},{30}^{o}for{\theta }_{1},4.0mforOD,10Nfor{F}_{2},{20}^{o}for{\theta }_{2},0\\ mforOO,30Nfor{F}_{3},and{45}^{o}for{\theta }_{,intheequation}\end{array}$

${\tau }_{O}=OC{F}_{1}cos{\theta }_{1}+OD\left(-{F}_{2}sin{\theta }_{2}\right)+OO\left(-{F}_{3}sin{\theta }_{3}\right)andsolvefor{\tau }_{O}$

${\tau }_{O}=\left(2.0m\right)\left(25N\right)cos{30}^{o}-\left(4.0m\right)\left(10N\right)sin{20}^{o}-\left(0m\right)\left(30N\right)sin{45}^{o}$

${\tau }_{O}=30.0N×m$

Therefore, the net torque about the point O perpendicular is$30.0N×m$

The direction of the net torque is in counter clockwise direction

The net torque about the point C perpendicular is

${\tau }_{C}=\left(CC\right){F}_{1}sin{\theta }_{1}+\left(OC\right)\left(-{F}_{2}sin{\theta }_{2}\right)+\left(-OC\right)\left(-{F}_{3}sin{\theta }_{3}\right)$

${\tau }_{C}$is the net torque about the point C,CC is the perpendicular distance for the force${F}_{1}$

$Substitute0mforCC,25Nfor{F}_{1},{30}^{o}for{\theta }_{1},2.0mforOC,10Nfor{F}_{2},{20}^{o}for{\theta }_{2}$

$30NforF,and{45}^{o}for\theta ,intheequation$

${\tau }_{C}=\left(CC\right){F}_{1}sin{\theta }_{1}+\left(OC\right)\left(-{F}_{2}sin{\theta }_{2}\right)+\left(-OC\right)\left(-{F}_{3}sin{\theta }_{3}\right)andsolvefor{\tau }_{C}$

${\tau }_{C}=\left(0m\right)\left(25N\right)cos{30}^{o}-\left(2.0m\right)\left(10N\right)sin{20}^{o}+\left(2.0m\right)\left(30N\right)sin{45}^{o}$

${\tau }_{C}=36.0N×m$

Therefore, the net torque about perpendicular is$36.0N×m$

The direction of the net torque is in counter clockwise direction ### Want to see more solutions like these? 