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 163: 80

Answer

$M_C=126 \mathrm{lb} \cdot \mathrm{ft}$

Work Step by Step

(a) $ \begin{aligned} \mathbf{M}_C & =\Sigma(\mathbf{r} \times \mathbf{F}) \\ & =\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 0 \\ -50 \sin 30^{\circ} & -50 \cos 30^{\circ} & 0 \end{array}\right|+\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 4 & 0 \\ -\frac{4}{5}(80) & -\frac{3}{5}(80) & 0 \end{array}\right| \\ \mathbf{M}_C & =\{126 \mathbf{k}\} \mathrm{lb} \cdot \mathrm{ft} \end{aligned} $ (b) $ \begin{aligned} ↺+M_C & =-\frac{4}{5}(80)(3)+\frac{4}{5}(80)(7)+50 \cos 30^{\circ}(2)-50 \cos 30^{\circ}(5) \\ M_C & =126 \mathrm{lb} \cdot \mathrm{ft} \end{aligned} $
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