The vertical force \textbf{F} acts downward at A on the two-membered frame. Determine the magnitudes of the two components of \textbf{F} directed along the axes of AB and AC. Set \textbf{F} = 500 N.
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-4
Solution:
Draw the components of the force using the parallelogram law. Then the triangulation rule.
Solving for FAC using sine law.
\begin{align*}
\frac{\text{F}_\text{AC}}{\sin \ 45^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\
\text{F}_\text{AC} & = \frac{500 \ \text{N} \ \sin45^\circ }{\sin\ 75^\circ }\\
\text{F}_\text{AC} & =366.0254 \ \text{N}\\
\text{F}_\text{AC} &\approx 366 \ \text{N}
\end{align*}Solve for FAB using sine law.
\begin{align*}
\frac{\text{F}_\text{AB}}{\sin \ 60^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\
\text{F}_\text{AB} & = \frac{500 \ \text{N} \ \sin60^\circ }{\sin\ 75^\circ }\\
\text{F}_\text{AB} & =448.2877 \ \text{N}\\
\text{F}_\text{AB} &\approx 448 \ \text{N}
\end{align*}
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