University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.4 - The Cross Product - Exercises - Page 623: 26

Answer

$10\sqrt{2} \ ft\cdot lb$

Work Step by Step

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