Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.4 Vectors - 8.4 Assess Your Understanding - Page 627: 76


$82.98 \ N$ and $99.57^{\circ}$

Work Step by Step

The resultant vector $F$ is given by: $F=F_1+F_2$ Where: $F_1=F_{1x}+F_{1y}$ and $F_2=F_{2x}+F_{2y}$ The components of the force vectors are: $F_{1x}=30 \cos 45^{\circ}=15 \sqrt 2 \approx 21.21 \\F_{1y}=30 \sin 45^{\circ}=15 \sqrt 2 \approx 21.21 \\ F_{2x}=70 \cos (120)^{\circ}=-35 \\F_{2y}=70 \sin (120)^{\circ}= 35 \sqrt 3 \approx 60.62$ Now, $F_1=F_{1x}+F_{1y} \approx -13.79 $ and $F_2=F_{2x}+F_{2y} \approx 81.83$ So, the resultant force $F$ magnitude is: $|F|=\sqrt {(-13.79)^2+(81.83)^2}=82.98 \ N$ and $F_y=\tan^{-1} \dfrac{F_y}{F_x} \approx 99.57^{\circ}$
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