If \theta = 60 \degree and \textbf{F} = 450 \ \text{N}, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-1
Solution:
The parallelogram law and the triangulation rule are shown in the figures below.
Considering figure (b), we can solve for the magnitude of \textbf{F}_R using the cosine law.
\begin{align*}
\textbf{F}_R & = \sqrt{700^2+450^2-2\left( 700 \right)\left( 450 \right)\cos45^{\circ}}\\
& = 497.01 \ \text{N}\\
& = 497 \ \text{N}
\end{align*}Then we use the sine law to solve for the interior angle \theta.
\begin{align*}
\frac{\sin \theta}{700} & = \frac{\sin 45^{\circ}}{497.01}\\
\sin \theta & =\frac{700\ \sin 45^{\circ }}{497.01}\\
\theta & = \sin^{-1} \left( \frac{700\ \sin 45^{\circ }}{497.01} \right)\\
& \text{This is an ambiguous case }\\
\theta & = 84.81^\circ \ or \ \theta =95.19^\circ \\
\end{align*}In here, the correct angle measurement is \theta = 95.19^{\circ}.
Thus, the direction angle \phi of \textbf{F}_R measured counterclockwise from the positive x-axis, is
\begin{align*}
\phi & = \theta +60^\circ \\
& = 95.19^\circ +60^\circ \\
& = 155^\circ
\end{align*}
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