If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction \theta .
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-2
Solution:
The parallelogram law and the triangulation rule are shown in the figures below.
Considering the figure of the triangulation rule, we can solve for the magnitude of \textbf{F} using the cosine law.
\begin{align*}
\textbf{F} & = \sqrt{700^2+500^2-2\left( 700 \right)\left( 500 \right)\cos105^{\circ}}\\
& = 959.78 \ \text{N}\\
& = 960 \ \text{N}\\
\end{align*}Then we use the sine law to solve for the angle \theta.
\begin{align*}
\frac{\sin \left(90^{\circ}-\theta \right)}{700} & = \frac{\sin 105^{\circ}}{959.78}\\
\sin \left(90^{\circ}-\theta \right) & =\frac{700 \sin 105^{\circ }}{959.78}\\
90^{\circ}-\theta & = \sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right)\\
\theta & = 90^\circ-\sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right) \\
\theta & = 90^\circ-44.79^\circ\\
\theta & = 45.2^\circ\\
\end{align*}
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