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

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

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

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 \(\mathrm{F}_{2}.\)

\( \stackrel{+x}{\longleftarrow} \)

\( \stackrel{+y}{\longrightarrow} \)

\( +z \uparrow \)

 

 

\(\stackrel{x}{\longleftarrow}  \sum F_{x}=0. \)

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

\( ∴ F_{2 x}=33.4 N \)

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

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

\( ∴ F_{2y} =-19.3 N \)

\( +\uparrow \ \sum F_{2}=-100 \)

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

\( ∴ F_{2z} =-54 N \)

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

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

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

 

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