10.6 Torque  (Page 2/6)

Now let’s consider how to define torques in the general three-dimensional case.

From the definition of the cross product, the torque $\overrightarrow{\tau }$ is perpendicular to the plane containing $\overrightarrow{r}\text{and}\overrightarrow{F}$ and has magnitude

where $\theta$ is the angle between the vectors $\overrightarrow{r}$ and $\overrightarrow{F}$ . The SI unit of torque is newtons times meters, usually written as $\text{N}·\text{m}$ . The quantity $r_{\perp }=r\text{sin}\theta$ is the perpendicular distance from O to the line determined by the vector $\overrightarrow{F}$ and is called the lever arm    . Note that the greater the lever arm, the greater the magnitude of the torque. In terms of the lever arm, the magnitude of the torque is

The cross product $\overrightarrow{r}\times \overrightarrow{F}$ also tells us the sign of the torque. In [link] , the cross product $\overrightarrow{r}\times \overrightarrow{F}$ is along the positive z -axis, which by convention is a positive torque. If $\overrightarrow{r}\times \overrightarrow{F}$ is along the negative z -axis, this produces a negative torque.

If we consider a disk that is free to rotate about an axis through the center, as shown in [link] , we can see how the angle between the radius $\overrightarrow{r}$ and the force $\overrightarrow{F}$ affects the magnitude of the torque. If the angle is zero, the torque is zero; if the angle is $90\text{°}$ , the torque is maximum. The torque in [link] is positive because the direction of the torque by the right-hand rule is out of the page along the positive z -axis. The disk rotates counterclockwise due to the torque, in the same direction as a positive angular acceleration.

Any number of torques can be calculated about a given axis. The individual torques add to produce a net torque about the axis. When the appropriate sign (positive or negative) is assigned to the magnitudes of individual torques about a specified axis, the net torque about the axis is the sum of the individual torques:

Calculating net torque for rigid bodies on a fixed axis

In the following examples, we calculate the torque both abstractly and as applied to a rigid body.

We first introduce a problem-solving strategy.

Calculating torque

Four forces are shown in [link] at particular locations and orientations with respect to a given xy -coordinate system. Find the torque due to each force about the origin, then use your results to find the net torque about the origin.

Strategy

This problem requires calculating torque. All known quantities––forces with directions and lever arms––are given in the figure. The goal is to find each individual torque and the net torque by summing the individual torques. Be careful to assign the correct sign to each torque by using the cross product of $\overrightarrow{r}$ and the force vector $\overrightarrow{F}$ .

Solution

Use $\left|\overrightarrow{\tau }\right|=r_{\perp }F=rF\text{sin}\theta$ to find the magnitude and $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$ to determine the sign of the torque.

The torque from force 40 N in the first quadrant is given by $(4)(40)\text{sin}90\text{°}=160\text{N}·\text{m}$ .

The cross product of $\overrightarrow{r}$ and $\overrightarrow{F}$ is out of the page, positive.

The torque from force 20 N in the third quadrant is given by $\text{−}(3)(20)\text{sin}90\text{°}=-60\text{N}·\text{m}$ .

The cross product of $\overrightarrow{r}$ and $\overrightarrow{F}$ is into the page, so it is negative.

The torque from force 30 N in the third quadrant is given by $(5)(30)\text{sin}53\text{°}=120\text{N}·\text{m}$ .

The cross product of $\overrightarrow{r}$ and $\overrightarrow{F}$ is out of the page, positive.

The torque from force 20 N in the second quadrant is given by $(1)(20)\text{sin}30\text{°}=10\text{N}·\text{m}$ .

The cross product of $\overrightarrow{r}$ and $\overrightarrow{F}$ is out of the page.

The net torque is therefore ${\tau }_{\text{net}}={\displaystyle \sum _i\left|{\tau }_i\right|}=160-60+120+10=230\text{N}·\text{m}\text{.}$

Significance

Note that each force that acts in the counterclockwise direction has a positive torque, whereas each force that acts in the clockwise direction has a negative torque. The torque is greater when the distance, force, or perpendicular components are greater.

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