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• Notes

$(\alpha, \beta, \gamma)$

$\cos ^{2} \alpha+\cos ^{2} \beta +\cos ^{2} \gamma=1$

$\vec{F}=F \cos \alpha \hat{\imath}+F \cos \beta \hat{\jmath}+F \cos \gamma K$

The two forces $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ acting at  $A$ have a resultant force of $\mathrm{F}_{R}=\{-100 \mathrm{k}\} \mathrm{N} .$ Determine the  magnitude and coordinate direction angles of $\mathrm{F}_{2}.$

$\stackrel{+x}{\longleftarrow}$

$\stackrel{+y}{\longrightarrow}$

$+z \uparrow$

$\stackrel{x}{\longleftarrow} \sum F_{x}=0.$

$-60 \cos 50 \cos 30+F_{2 x}=0$

$∴ F_{2 x}=33.4 N$

$\stackrel{+}{\longrightarrow} \sum F_{y}=0.$

$60 \cos 50 \sin 30+F_{2 y}=0$

$∴ F_{2y} =-19.3 N$

$+\uparrow \ \sum F_{2}=-100$

$-60 \sin 50+F_{2 z}=-100$

$∴ F_{2z} =-54 N$

$\longrightarrow F_{2}=(33 \cdot 4 i-19 \cdot 3 j-54 k) N$

$\left|F_{2}\right|=66.4 \mathrm{N}$

$\begin{array}{l}{\alpha=\cos ^{-1}\left(\frac{33.4}{66 \cdot 4}\right)=59.8^{\circ}} \\ {\beta=\cos ^{-1}\left(\frac{-19.3}{66.4}\right)=107^{\circ}} \\ {\gamma=\cos ^{-1}\left(\frac{-54}{66.4}\right)=144.40^{\circ}}\end{array}$