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 4.7 - Simplification of a Force and Couple System - Problems - Page 176: 110

Answer

$$ \begin{aligned} & \mathbf{F}_R=\{44.5 \mathbf{i}+53.1 \mathbf{j}+40 \mathbf{k}\} \mathbf{N} \\ & \mathbf{M}_{R A}=\{-5.39 \mathbf{i}+13.1 \mathbf{j}+11.4 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m} \end{aligned} $$

Work Step by Step

$$ \begin{aligned} & \mathbf{F}_R=\Sigma \mathbf{F} ; \\ & \mathbf{F}_R=80 \cos 30^{\circ} \sin 40^{\circ} \mathbf{i}+80 \cos 30^{\circ} \cos 40^{\circ} \mathbf{j}-80 \sin 30^{\circ} \mathbf{k} \\ & =44.53 \mathbf{i}+53.07 \mathbf{j}-40 \mathbf{k} \\ & =\{44.5 \mathbf{i}+53.1 \mathbf{j}-40 \mathbf{k}\} \mathbf{N} \\ & \mathbf{M}_{R A}=\Sigma \mathbf{M}_A ; \quad \mathbf{M}_{R A}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0.55 & 0.4 & -0.2 \\ 44.53 & 53.07 & -40 \end{array}\right| \\ & =\{-5.39 \mathbf{i}+13.1 \mathbf{j}+11.4 \mathbf{k}\} \mathbf{N} \cdot \mathrm{m} \\ & \end{aligned} $$
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