# 10.6 Torque  (Page 3/6)

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## Calculating torque on a rigid body

[link] shows several forces acting at different locations and angles on a flywheel. We have $|{\stackrel{\to }{F}}_{1}|=20\phantom{\rule{0.2em}{0ex}}\text{N},$ $|{\stackrel{\to }{F}}_{2}|=30\phantom{\rule{0.2em}{0ex}}\text{N}$ , $|{\stackrel{\to }{F}}_{3}|=30\phantom{\rule{0.2em}{0ex}}\text{N}$ , and $r=0.5\phantom{\rule{0.2em}{0ex}}\text{m}$ . Find the net torque on the flywheel about an axis through the center.

## Strategy

We calculate each torque individually, using the cross product, and determine the sign of the torque. Then we sum the torques to find the net torque.

## Solution

We start with ${\stackrel{\to }{F}}_{1}$ . If we look at [link] , we see that ${\stackrel{\to }{F}}_{1}$ makes an angle of $90\text{°}+60\text{°}$ with the radius vector $\stackrel{\to }{r}$ . Taking the cross product, we see that it is out of the page and so is positive. We also see this from calculating its magnitude:

$|{\stackrel{\to }{\tau }}_{1}|=r{F}_{1}\text{sin}\phantom{\rule{0.2em}{0ex}}150\text{°}=0.5\phantom{\rule{0.2em}{0ex}}\text{m}\left(20\phantom{\rule{0.2em}{0ex}}\text{N}\right)\left(0.5\right)=5.0\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

Next we look at ${\stackrel{\to }{F}}_{2}$ . The angle between ${\stackrel{\to }{F}}_{2}$ and $\stackrel{\to }{r}$ is $90\text{°}$ and the cross product is into the page so the torque is negative. Its value is

$|{\stackrel{\to }{\tau }}_{2}|=\text{−}r{F}_{2}\text{sin}\phantom{\rule{0.2em}{0ex}}90\text{°}=-0.5\phantom{\rule{0.2em}{0ex}}\text{m}\left(30\phantom{\rule{0.2em}{0ex}}\text{N}\right)=-15.0\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

When we evaluate the torque due to ${\stackrel{\to }{F}}_{3}$ , we see that the angle it makes with $\stackrel{\to }{r}$ is zero so $\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{F}}_{3}=0.$ Therefore, ${\stackrel{\to }{F}}_{3}$ does not produce any torque on the flywheel.

We evaluate the sum of the torques:

${\tau }_{\text{net}}=\sum _{i}|{\tau }_{i}|=5-15=-10\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

## Significance

The axis of rotation is at the center of mass of the flywheel. Since the flywheel is on a fixed axis, it is not free to translate. If it were on a frictionless surface and not fixed in place, ${\stackrel{\to }{F}}_{3}$ would cause the flywheel to translate, as well as ${\stackrel{\to }{F}}_{1}$ . Its motion would be a combination of translation and rotation.

Check Your Understanding A large ocean-going ship runs aground near the coastline, similar to the fate of the Costa Concordia , and lies at an angle as shown below. Salvage crews must apply a torque to right the ship in order to float the vessel for transport. A force of $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$ acting at point A must be applied to right the ship. What is the torque about the point of contact of the ship with the ground ( [link] )?

The angle between the lever arm and the force vector is $80\text{°};$ therefore, ${r}_{\perp }=100\text{m(sin80}\text{°}\right)=98.5\phantom{\rule{0.2em}{0ex}}\text{m}$ .

The cross product $\stackrel{\to }{\tau }=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}$ gives a negative or clockwise torque.

The torque is then $\tau =\text{−}{r}_{\perp }F=-98.5\phantom{\rule{0.2em}{0ex}}\text{m}\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\text{N}\right)=-4.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{N}·\text{m}$ .

## Summary

• The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation $|\stackrel{\to }{\tau }|={r}_{\perp }F$ , where ${r}_{\perp }$ is the perpendicular distance from the axis to the line upon which the force vector lies.
• The sign of the torque is found using the right hand rule. If the page is the plane containing $\stackrel{\to }{r}$ and $\stackrel{\to }{F}$ , then $\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}$ is out of the page for positive torques and into the page for negative torques.
• The net torque can be found from summing the individual torques about a given axis.

## Conceptual questions

What three factors affect the torque created by a force relative to a specific pivot point?

magnitude of the force, length of the lever arm, and angle of the lever arm and force vector

Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque.

When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?

The moment of inertia of the wheels is reduced, so a smaller torque is needed to accelerate them.

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