College Physics (4th Edition)

Published by McGraw-Hill Education

Chapter 2 - Section 2.3 - Vector Addition Using Components - Practice Problem - Page 36: 2.4

Answer

(a) $F_x = 49.1~N$ $F_y = 2.95~N$ (b) $F = 49.2~N$ (c) The sum of these force is directed at an angle of $3.4^{\circ}$ above the +x-axis.

Work Step by Step

(a) We can find the horizontal component of the sum of the forces: $F_x = (22.0~N)~cos~30.0^{\circ}+ (22.0~N)~cos~30.0^{\circ}+ (22.0~N)~cos~60.0^{\circ}$ $F_x = 49.1~N$ We can find the vertical component of the sum of the forces: $F_y = (22.0~N)~sin~30.0^{\circ}+ (22.0~N)~sin~30.0^{\circ}+ (-22.0~N)~sin~60.0^{\circ}$ $F_y = 2.95~N$ (b) We can find the magnitude of the sum of the forces: $F = \sqrt{F_x^2+F_y^2} = \sqrt{(49.1~N)^2+(2.95~N)^2} = 49.2~N$ (c) We can find the angle $\theta$ above the positive x-axis: $tan~\theta = \frac{2.95~N}{49.1~N}$ $\theta = tan^{-1}(\frac{2.95~N}{49.1~N})$ $\theta = 3.4^{\circ}$ The sum of these force is directed at an angle of $3.4^{\circ}$ above the +x-axis.

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