University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises - Page 30: 1.60


The magnitude of $F$ is 90.2 N. The direction of the force is $75.6^{\circ}$ south of west.

Work Step by Step

We can find the horizontal component $F_x$ of the fourth force. $F_x = 100.0~cos(30.0^{\circ}) - 80.0~sin(30.0^{\circ}) - 40.0~cos(53^{\circ}) = 0$ $F_x = -86.6~N + 40.0~N + 24.1~N = -22.5~N$ We can find the vertical component $F_y$ of the fourth force. $F_y = 100.0~sin(30.0^{\circ}) + 80.0~cos(30.0^{\circ}) - 40.0~sin(53^{\circ}) = 0$ $F_y = -50.0~N -69.3~N + 31.9~N = -87.4~N$ We can use $F_x$ and $F_y$ to find the magnitude of $F$. $F = \sqrt{(-22.5~N)^2+(-87.4~N)^2} = 90.2~N$ We can find the angle $\theta$ south of west. $tan(\theta) = \frac{F_y}{F_x}$ $\theta = tan^{-1}(\frac{87.4}{22.5}) = 75.6^{\circ}$
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