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 20 lb is applied to a wrench with a 6-inch handle at an angle of ${60}^{â—¦}$.

$||r||||\mathbf{F}||\sin \mathrm{θ}$

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

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

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