Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 1 - Introduction and Mathematical Concepts - Problems - Page 21: 23


a) $36.1^{\circ}$ north of west b) $36.1^{\circ}$ south of west

Work Step by Step

a) We find: $F^{2}=A^{2}+B^{2}$ $F=\sqrt {A^{2}+B^{2}}$ $F=\sqrt {(445N)^{2}+(325N)^{2}}=551N$ $tan\theta=\frac{B}{A}$ $\theta=tan^{-1}\frac{325N}{445N}=36.1^{\circ}$ north of west b) We find: $F^{2}=A^{2}+(-B)^{2}$ $F=\sqrt {A^{2}+(-B)^{2}}$ $F=\sqrt {(445N)^{2}+(-325N)^{2}}=551N$ $tan\theta=\frac{B}{A}$ $\theta=tan^{-1}\frac{325N}{445N}=36.1^{\circ}$ south of west
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