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 139: 15

Answer

$ \begin{aligned} & \left(M_A\right)_C=768 \mathrm{lb} \cdot \mathrm{ft} \\ & \left(M_A\right)_B=636 \mathrm{lb} \cdot \mathrm{ft} \end{aligned} $ Clockwise

Work Step by Step

$ \begin{aligned} & \left.↺+\left(M_A\right)_C=80\left(\frac{4}{5}\right)(12)=768 \mathrm{lb} \cdot \mathrm{ft}\right) \\ & \left.↺+\left(M_A\right)_B=50\left(\cos 45^{\circ}\right)(18)=636 \mathrm{lb} \cdot \mathrm{ft}\right) \\ & =\left(M_A\right)_B \end{aligned} $ Since $\left(M_A\right)_C>\left(M_A\right)_B$ Clockwise
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