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: 26


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.