Answer
$$
F=830 \mathrm{~N}
$$
Work Step by Step
$$
\begin{aligned}
C+M_R=\Sigma M ; \quad-400 & =600\left(\frac{0.5}{\cos 40^{\circ}}\right)-F\left(\frac{0.5}{\cos 40^{\circ}}\right)-250(1) \\
F & =830 \mathrm{~N}
\end{aligned}
$$
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
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 162: 73
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