Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 719: 25

Answer

$10\sqrt{3} \ ft\cdot lb$

Work Step by Step

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