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 4.8 - Further Simplification of a Force and Couple System - Problems - Page 186: 113

Answer

$$ \begin{aligned} & F_R=10.75 \mathrm{kip} \downarrow \\ & d=13.7 \mathrm{ft} \end{aligned} $$

Work Step by Step

$$ \begin{gathered} +\uparrow F_R=\Sigma F_y ; \quad F_R=-1750-5500-3500 \\ =-10750 \mathrm{lb}=10.75 \mathrm{kip} \downarrow \quad \text { Ans. } \\ ↺+M_{R_A}=\Sigma M_A ; \quad-10750 d=-3500(3)-5500(17)-1750(25) \\ d=13.7 \mathrm{ft} \quad \text { Ans. } \end{gathered} $$
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