91Ó°ÊÓ

Turning a bolt with a wrench produces a torque vector that drives the bolt forward. The magnitude of the torque vector is $||r||||\mathbf{F}||\sin \mathrm{θ}$, 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 $||r×\mathbf{F}||$. We have to find the magnitude of the torque. and expressing answer in foot-pounds.

A force of 40 lb is applied to a wrench with a 9-inch handle at an angle of${90}^{â—¦}$.

$||r||||\mathbf{F}||\sin \mathrm{θ}$

From the question $r=9-\mathrm{inch},\mathbf{F}=40\mathrm{lb},\mathrm{θ}={90}^o$

$$\begin{gathered} ||r||||\mathbf{F}||\sin \mathrm{θ}=||9||||40||\sin {90}^o \\ ||r||||\mathbf{F}||\sin \mathrm{θ}=9×40×\sin {90}^o \\ ||r||||\mathbf{F}||\sin \mathrm{θ}=9×40×1 \\ ||r||||\mathbf{F}||\sin \mathrm{θ}=360\mathrm{in}-\mathrm{lb} \end{gathered}$$

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

$$\begin{gathered} ||r||||\mathbf{F}||\sin \mathrm{θ}=\frac{1}{12}×360\mathrm{foot}-\mathrm{pounds} \\ ||\mathrm{r}||||\mathbf{F}||\mathrm{²õ¾\pm ²Ôθ}=30\mathrm{foot}-\mathrm{pounds} \end{gathered}$$