Answer
$10\sqrt{3} \ ft\cdot lb$
Work Step by Step
Magnitude of torque vector = $|{\bf r}|\cdot|{\bf F}|\cdot\sin\theta$, or $|{\bf r}\times{\bf F}|.$
$|\displaystyle \overrightarrow{PQ}|\cdot|{\bf F}|\cdot\sin\theta=(\frac{8}{12} \ ft)(30\ lb)\cdot\sin 60^{\circ}$
$=(\displaystyle \frac{2}{3} \ ft)(30\ lb)\cdot\frac{\sqrt{3}}{2}=10\sqrt{3} \ ft\cdot lb$
