College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 2 - Section 2.3 - Vector Addition Using Components - Practice Problem - Page 33: 2.3

Answer

The magnitude of this force is $157~N$ and it is directed at an angle of $57.2^{\circ}$ below the +x-axis.

Work Step by Step

We can find the magnitude of this force: $F = \sqrt{F_x^2+F_y^2} = \sqrt{(85~N)^2+(-132~N)^2} = 157~N$ We can find the angle $\theta$ below the positive x-axis: $tan~\theta = \frac{132~N}{85~N}$ $\theta = tan^{-1}(\frac{132~N}{85~N})$ $\theta = 57.2^{\circ}$ The magnitude of this force is $157~N$ and it is directed at an angle of $57.2^{\circ}$ below the +x-axis.
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