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 4.4 - Principle of Moments - Problems - Page 138: 5

Answer

$$ 4.125 \mathrm{kip} \cdot \mathrm{ft} \text { (Counterclockwise) } $$$$ 2.00 \mathrm{kip} \cdot \mathrm{ft} \text { (Counterclockwise) } $$$$ 40.0 \mathrm{lb} \cdot \mathrm{ft} \text { (Counterclockwise) } $$

Work Step by Step

$$ \begin{aligned} ↺+\left(M_{F_1}\right)_B & =375(11) \\ & =4125 \mathrm{lb} \cdot \mathrm{ft}=4.125 \mathrm{kip} \cdot \mathrm{ft} \text { (Counterclockwise) } \\\\ ↺_{+}\left(M_{F_2}\right)_B & =500\left(\frac{4}{5}\right)(5) \\ & =2000 \mathrm{lb} \cdot \mathrm{ft}=2.00 \mathrm{kip} \cdot \mathrm{ft}(\text { Counterclockwise) } \\\\ ↺_{+}+\left(M_{F_3}\right)_B & =160 \sin 30^{\circ}(0.5)-160 \cos 30^{\circ}(0) \\ & =40.0 \mathrm{lb} \cdot \mathrm{ft}(\text { Counterclockwise) } \end{aligned} $$
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