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Next we look at ${\overrightarrow{F}}_{2}$ . The angle between ${\overrightarrow{F}}_{2}$ and $\overrightarrow{r}$ is $90\text{\xb0}$ and the cross product is into the page so the torque is negative. Its value is
When we evaluate the torque due to ${\overrightarrow{F}}_{3}$ , we see that the angle it makes with $\overrightarrow{r}$ is zero so $\overrightarrow{r}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{\overrightarrow{F}}_{3}=0.$ Therefore, ${\overrightarrow{F}}_{3}$ does not produce any torque on the flywheel.
We evaluate the sum of the torques:
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 $5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$ 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 $80\text{\xb0};$ therefore, ${r}_{\perp}=100\text{m(sin80}\text{\xb0})=98.5\phantom{\rule{0.2em}{0ex}}\text{m}$ .
The cross product $\overrightarrow{\tau}=\overrightarrow{r}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\overrightarrow{F}$ gives a negative or clockwise torque.
The torque is then $\tau =\text{\u2212}{r}_{\perp}F=\mathrm{-98.5}\phantom{\rule{0.2em}{0ex}}\text{m}(5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\text{N})=\mathrm{-4.9}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\text{N}\xb7\text{m}$ .
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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