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: 75


$80.27 \ N$ and $ -16.2^{\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: $F_{1x}=40 \cos 30^{\circ}=20\sqrt 3 \approx 34.64 \\F_{1y}=40 \sin 30^{\circ}=20 \\ F_{2x}=60 \cos (-45)^{\circ}=30\sqrt 2 \approx 42.43 \\F_{2y}=60 \sin (-45)^{\circ}=- 30 \sqrt 2 \approx 42.43$ Now, $F_1=F_{1x}+F_{1y} \approx 77.07 $ and $F_2=F_{2x}+F_{2y} \approx -22.43$ So, the resultant force magnitude is: $|F|=\sqrt {(77.07)^2+(-22.43)^2}=80.27 \ N$ and $F_y=\tan^{-1} \dfrac{F_y}{F_x} \approx -16.2^{\circ}$
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