Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems - Page 82: 16

Answer

a) $\vec D=16\;\hat i−22\;\hat j$ b) See the figure below. c) $27.2,\;306^\circ$

Work Step by Step

a) We need to find $\vec F$ in component forms whereas $$\vec F=\vec A-4\vec B$$ Plugging the known; $$\vec F=(4\;\hat i−2\;\hat j)-4(−3\;\hat i+5\;\hat j )=4\;\hat i+12\;\hat i−2\;\hat j-20\;\hat j$$ $$\boxed{\vec D=16\;\hat i−22\;\hat j}$$ b) See the figure below. We drew the 3 vectors there. c) The magnitude of $\vec F$ is given by applying the Pythagorean theorem. $$|\vec F|=\sqrt{F_x^2+F_y^2}=\sqrt{16^2+(-22)^2}=\color{red}{\bf 27.2}$$ and its direction is given by $$\tan\alpha_F=\dfrac{F_y}{F_x}$$ Thus, $$\alpha_F=\tan^{-1}\left[\dfrac{F_y}{F_x}\right]=\tan^{-1}\left[\dfrac{-22}{16}\right]=\bf -54^\circ$$ Thus, $$\alpha_F=\color{red}{\bf 306^\circ}$$
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