Question

Turning a bolt with a wrench produces a torque vector that drives the bolt forward. The magnitude of the torque vector is $||r||||\mathbf{F}||\sin \theta$, where r is the vector along the handle of the wrench, F is the force vector applied to the handle of the wrench, and θ is the angle between these two vectors. Therefore, the magnitude of the torque is $||r\times \mathbf{F}||$. In Exercises 60 and 61, find the magnitude of the torque. Express each answer in foot-pounds.

A force of 20 lb is applied to a wrench with a 6-inch handle at an angle of ${60}^{◦}$.

Short answer

The magnitude of the torque is $5\sqrt{3}$foot-pounds.

Step-by-step solution

Step 1. Given Information

Turning a bolt with a wrench produces a torque vector that drives the bolt forward. The magnitude of the torque vector is $||r||||\mathbf{F}||\sin \theta$, where r is the vector along the handle of the wrench, F is the force vector applied to the handle of the wrench, and θ is the angle between these two vectors. Therefore, the magnitude of the torque is $||r\times \mathbf{F}||$. We have to find the magnitude of the torque. and expressing answer in foot-pounds.

A force of 20 lb is applied to a wrench with a 6-inch handle at an angle of ${60}^{◦}$.

Step 2. Now finding the magnitude of the torque. 

$||r||||\mathbf{F}||\sin \theta$

From the question $r=6-\mathrm{inch},\mathbf{F}=20\mathrm{lb},\theta ={60}^o$

role="math" localid="1649784087723" $$\begin{gathered} ||r||||\mathbf{F}||\sin \theta =||6||||20||\sin {60}^o \\ ||r||||\mathbf{F}||\sin \theta =6\times 20\times \sin {60}^o \\ ||r||||\mathbf{F}||\sin \theta =6\times 20\times \frac{\sqrt{3}}{2} \\ ||r||||\mathbf{F}||\sin \theta =6\times 10\times \sqrt{3} \\ ||r||||\mathbf{F}||\sin \theta =60\sqrt{3}\mathrm{in}-\mathrm{lb} \end{gathered}$$

Step 3. We have to convert the answer in foot-pounds. 

As we know to convert in foot-pounds take your foot-lb figure and divide by 12.

$$\begin{gathered} ||r||||\mathbf{F}||\sin \theta =\frac{60}{12}\sqrt{3}\mathrm{foot}-\mathrm{pounds} \\ ||\mathrm{r}||||\mathbf{F}||\mathrm{sin\theta }=5\sqrt{3}\mathrm{foot}-\mathrm{pounds} \end{gathered}$$