Torque can be interpreted from two perspectives, as shown in FIGURE 12.20. First, the quantity $F\text{sin}\varphi$ is the tangential force component $F_t$. Consequently, the torque is

In other words, torque is the product of r with the force component $F_t$ that is tangent to the circular path followed by this point on the wrench. This interpretation makes sense because the radial component of $\vec{F}$ points straight at the pivot point and cannot exert a torque.

FIGURE 12.20 Two useful interpretations of the torque.

A second perspective, widely used in applications, is based on the idea of a moment arm. Figure 12.20 shows the line of action, the line along which the force acts. The minimum distance between the pivot point and the line of action—the length of a line drawn perpendicular to the line of action—is called the moment arm (or the lever arm) d. Because $\text{sin}\left(180\text{°}-\varphi \right)=\text{sin }\varphi$, it is easy to see that $d=r\text{sin}\varphi$. Thus the torque $rF\text{sin}\varphi$ can also be written

EXAMPLE 12.8 Applying a torque

Luis uses a 20-cm-long wrench to turn a nut. The wrench handle is tilted $30\text{°}$ above the horizontal, and Luis pulls straight down on the end with a force of 100 N. How much torque does Luis exert on the nut?

VISUALIZE FIGURE 12.21 shows the situation. The angle is a negative $\varphi =-120\text{°}$ because it is clockwise from the radial line.

FIGURE 12.21 A wrench being used to turn a nut.

FIGURE 12.21 Full Alternative Text

SOLVE The tangential component of the force is

$$F_t=F\text{sin}\varphi =-86.6\text{N}$$

According to our sign convention, $F_t$ is negative because it points in a cw direction. The torque, from Equation 12.21, is

$$\tau =r{F}_t=(0.20\text{ m})\left(-86.6\text{ N}\right)=-17\text{ N m}$$

Alternatively, Figure 12.21 has drawn the line of action by extending the force vector forward and backward. The moment arm, the distance between the pivot point and the line of action, is

$$d=r\text{cos(}30\text{°})=0.17\text{m}$$

Inserting the moment arm in Equation 12.22 gives

$$\left|\tau \right|=Fd=(0.17\text{m})\left(100\text{N}\right)=17\text{N}\text{m}$$

The torque acts to give a cw rotation, so we insert a minus sign to end up with

$$\tau =-17\text{N}\text{m}$$

REVIEW The largest possible torque, if Luis pulled perpendicular to the 20-cm-long wrench, would have a magnitude of $20\text{N}\text{m}$. Pulling at an angle reduces this, so $17\text{N}\text{m}$ is a reasonable answer.