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: 116

Answer

$$ \begin{aligned} & F=798 \mathrm{lb} \\ & \theta=67.9^{\circ} \\ & x=6.57 \mathrm{ft} \end{aligned} $$

Work Step by Step

$$ \pm \Sigma F_{R_x}=\Sigma F_x ; \quad F_{R_{\mathrm{s}}}=-500\left(\frac{4}{5}\right)+260\left(\frac{5}{13}\right)=-300 \mathrm{lb}=300 \mathrm{lb} \leftarrow $$ $$ +\uparrow F_{R_y}=\Sigma F_y \quad F_{R_y}=-500\left(\frac{3}{5}\right)-200-260\left(\frac{12}{13}\right)=-740 \mathrm{lb}=740 \mathrm{lb} \downarrow $$ $$ F=\sqrt{(-300)^2+(-740)^2}=798 \mathrm{lb} $$ $$ \theta=\tan ^{-1}\left(\frac{740}{300}\right)=67.9^{\circ} $$ $$ \begin{array}{cl} ↺+M_{R B}=\Sigma M_B ; \quad & 740(x)=500\left(\frac{3}{5}\right)(9)+200(6)+260\left(\frac{12}{13}\right)(4) \\ & x=6.57 \mathrm{ft} \end{array} $$
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