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

Published by Pearson

# Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Problems - Page 30: 11

#### Answer

$F=10.8kN$ $\phi=3.16^{\circ}$

#### Work Step by Step

We are asked to determine the magnitude of the resultant force, $F_R$, and its direction, measured counterclockwise from the positive x axis. We can apply the law of cosines to find the magnitude of $F_R$. $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{8^2+6^2-2*8*6*\cos(100^{\circ})}$ $c=10.8$ $F=10.8kN$ Using this result, we can apply the law of sines to find the angle between the top rope and $F_R$, ($\theta$) $\sin\theta / 6 = \sin 100^{\circ} / 10.8$ Solving for $\theta$, we obtain: $\theta=33.16^{\circ}$ We will need to subtract $30^{\circ}$ from this answer to get the angle to $F_R$ as measured from the x-axis. $\phi=33.16^{\circ}-30^{\circ}=3.16^{\circ}$

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