Determine the magnitude of the resultant force \textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2 and its direction, measured counterclockwise from the positive x axis.
Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3
SOLUTION:
The parallelogram law of the force system is shown.
Consider the triangle AOB.
Using cosine law to solve for the resultant force \textbf{F}_{\text{R}}
\begin{align*}
\textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\
& =393.2 \ \text{lb}\\
& =393\:\text{lb}\\
\end{align*}The value of angle θ can be solved using sine law.
\begin{align*}
\frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\
\sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\
\theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\
\theta & = 37.89^{\circ}\\
\end{align*}Solve for the unknown angle \phi .
\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ} The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.
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