Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems: 44

Answer

(a) $F_3 = 2.9~\hat{j}~N$ (b) $F_4 = -1.6~\hat{i}~N$

Work Step by Step

(a) $F_1+F_{2y}+ F_3 = 0$ $F_3 = -F_1-F_{2y}$ $F_3 = [-(-5.0~\hat{j})-(6.0)~sin(20^{\circ})~\hat{j}]~N$ $F_3 = 2.9~\hat{j}~N$ (b) $F_{2x}+F_4= 4.0~\hat{i}~N$ $F_4= 4.0~\hat{i}~N-F_{2x}$ $F_4= 4.0~\hat{i}~N-(6.0)~cos(20^{\circ})~\hat{i}~N$ $F_4 = -1.6~\hat{i}~N$
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