## Thomas' Calculus 13th Edition

$10\sqrt{2} \ ft\cdot lb$
Magnitude of torque vector = $|{\bf r}|\cdot|{\bf F}|\cdot\sin\theta$, or $|{\bf r}\times{\bf F}|.$ $|\displaystyle \overrightarrow{PQ}|\cdot|{\bf F}|\cdot\sin\theta=(\frac{8}{12} \ ft)(30\ lb)\cdot\sin 135^{\circ}$ $=(\displaystyle \frac{2}{3} \ ft)(30\ lb)\cdot\frac{\sqrt{2}}{2}=10\sqrt{2} \ ft\cdot lb$