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Q5P

Expert-verifiedFound in: Page 355

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**

$\tau =rF$

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${F}_{3}$

The net torque about the point O perpendicular is

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

${\tau}_{O}$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${F}_{3}$

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

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

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

${\tau}_{O}=30.0N\times m$

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

The direction of the net torque is in counter clockwise direction

The net torque about the point C perpendicular is

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

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

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

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

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

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

${\tau}_{C}=36.0N\times m$

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

The direction of the net torque is in counter clockwise direction

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