Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems - Page 252: 1


The magnitude is about 528N. It's 120 degrees clockwise from force A

Work Step by Step

1) Finding x-component of the force $(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $ $395N-\cos(75^{\circ})485N + \overrightarrow{F}_{x}=0$ $\overrightarrow{F}_{x}= -395N+\cos(75^{\circ})485N$ $\overrightarrow{F}_{x}\approx-252N$ 2)Finding the y-component $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $\sin(75^{\circ})475N + \overrightarrow{F}_{y}=0$ $\overrightarrow{F}_{y}=-\sin(75^{\circ})475N$ $\overrightarrow{F}_{y}\approx-459N$ 3) Finding the magnitude of \overrightarrow{F} $\Vert\overrightarrow{F}\Vert=\sqrt {F^2_x+F^2_y}$ $\Vert\overrightarrow{F}\Vert\approx\sqrt {(-252N)^2+(-459N)^2}$ $\Vert\overrightarrow{F}\Vert\approx528N$ 3) Finding the direction $\arctan{F_y/F_x}\approx60^{\circ}$ $180^{\circ}+60^{\circ}=120^{\circ}$ 120 degrees clockwise from force A
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