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

$F=19.2N$ $\theta=2.37^{\circ}$ $\searrow$
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 $F_2$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{20^2+30^2-2*20*30*\cos(73.13^{\circ})}$ $c=30.85$ $F=30.85N$ Using this result, we can apply the law of sines to find $\theta '$ $\sin73.13^{\circ} / 30.85 = \sin (70^{\circ}-\theta') / 30$ Solving for $\theta'$, we obtain: $\theta'=1.47^{\circ}$ We can apply the law of cosines to find the magnitude of $F_R$, which is the result of $F'$ and $F_3$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{30.85^2+50^2-2*30.85*50*\cos(1.47^{\circ})}$ $c=19.18$ $F=19.2N$ Using this result, we can apply the law of sines to find $\theta$ $\sin1.47^{\circ} / 19.18 = \sin \theta / 30.85$ Solving for $\theta$, we obtain: $\theta=2.37^{\circ}$