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: 117

Answer

$\begin{aligned} & F=1302 \mathrm{~N} \\ & \theta=84.5^{\circ} \\ & x=7.36 \mathrm{~m}\end{aligned}$

Work Step by Step

$ \begin{aligned} & \pm F_{R x}=\Sigma F_x ; \quad F_{R x}=450 \cos 60^{\circ}-700 \sin 30^{\circ}=-125 \mathrm{~N}\\&=125 \mathrm{~N} \leftarrow \\ & +\uparrow F_{R y}=\Sigma F_y ; \quad F_{R y}=-450 \sin 60^{\circ}-700 \cos 30^{\circ}-300=-1296 \mathrm{~N}=1296 \mathrm{~N} \\ & F=\sqrt{(-125)^2+(-1296)^2}=1302 \mathrm{~N} \\ & \theta=\tan ^{-1}\left(\frac{1296}{125}\right)=84.5^{\circ} \quad P \\ & ↺+M_{R A}=\Sigma M_A ; \quad 1296(x)\\&=450 \sin 60^{\circ}(2)+300(6)+700 \cos 30^{\circ}(9)+1500 \\ & x=7.36 \mathrm{~m} \end{aligned} $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.