Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 4 - Force System Resultants - Section 4.6 - Moment of a Couple - Problems - Page 165: 91

Answer

$$ F=832 \mathrm{~N} $$

Work Step by Step

$$ \begin{aligned} \mathbf{M}_C & =\mathbf{r}_{A B} \times(F \mathbf{k}) \\ & =(0.2 \mathbf{i}+0.3 \mathbf{j}) \times(F \mathbf{k}) \\ & =\{0.2 F \mathbf{i}-0.3 F \mathbf{j}\} \mathrm{N} \cdot \mathrm{m} \\ M_C & =F \sqrt{(0.2 F)^2+(-0.3 F)}=0.3606 F \\ 300 & =0.3606 F \\ F & =832 \mathrm{~N} \end{aligned} $$
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.