Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Preliminary Problems - Page 27: 1


$F_R=497N$ $\phi=155^{\circ}$

Work Step by Step

We are given that in the figure, $\theta =60^{\circ}$ and $F=450N$. We are asked to determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. We can treat the forces like a triangle and use the law of cosines: $c^2=a^2+b^2-2a*b*cosC$ $c=\sqrt{a^2+b^2-2a*b*cosC}$ $c=\sqrt{(700N)^2+(450N)^2-2*700N*450N*cos(60^{\circ}-15^{\circ}}$ $c=497N$ $F_R=497N$ Now we can use the law of sines to find the angle between $F$ and $F_r$, $\alpha$ $\frac{\sin \alpha}{700N}=\frac{\sin 45^{\circ} }{497N}$ $\alpha=\arcsin (\frac{\sin 45^{\circ} }{497N}*700N)$ $\alpha=95.19^{\circ}$ We will have to add this to $\theta$ to get the angle from the x-axis $\phi = \alpha +\theta= 95.19^{\circ} +60^{\circ}=155^{\circ}$
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