## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Problems - Page 31: 18

#### Answer

$F=19.2N$ $\theta= 2.37^{\circ}$ $\searrow$

#### Work Step by Step

We are asked to determine the magnitude of the resultant force, $F_R$, and its direction, $\theta$ We can apply the law of cosines to find the magnitude of $F'$, which is the result of $F_1$ and $F_2$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{20^2+50^2-2*20*50*\cos(70^{\circ})}$ $c=47.07$ $F'=47.07N$ Using this result, we can apply the law of sines to find $\theta '$ $\sin70^{\circ} / 47.07 = \sin \theta' / 20$ Solving for $\theta'$, we obtain: $\theta'=23.53^{\circ}$ We can apply the law of cosines to find the magnitude of $F_R$, which is the result of $F'$ and $F_3$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{47.07^2+30^2-2*47.07*30*\cos(13.34^{\circ})}$ $c=19.18$ $F=19.2N$ Using this result, we can apply the law of sines to find $\theta$ $\sin13.14^{\circ} / 19.18 = \sin \phi / 30$ Solving for $\phi$, we obtain: $\phi=21.15^{\circ}$ $\theta= \theta' -\phi = 23.53^{\circ}-21.15^{\circ}=2.37^{\circ}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.