Answer
$$F=139 \mathrm{lb}
$$
Work Step by Step
$$
\begin{gathered}
↺+M_R=\Sigma M ; \quad 450=200(1.5)+F \cos 30^{\circ}(1.25) \\
F=139 \mathrm{lb}
\end{gathered}
$$
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 161: 71
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