Question

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.

Short answer

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

Step-by-step solution

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$F_3$

To find the net torque about perpendicular point

The net torque about the point O perpendicular is

${\tau }_O=OC{F}_{1}cos{\theta }_1+OD(-F_{2}sin{\theta }_2)+OO(-F_{3}sin{\theta }_3)$

${\tau }_O$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 pointC,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