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Determine the manitude and coordinate direction angles of \(\mathrm{F}_{2}\) so that the resultant of the two forces is zero.

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

\( +z \uparrow \)

\( xy{\longrightarrow} \)

 

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

\( 180 \cos 15 \sin 60+F_{2 x}=0 \)

\( ∴ F_{2 x}=150.6 N \)

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

\( 180 \cos 15 \cos 60+F_{2 y}=0 \)

\( ∴ F_{2y} =-86.9 N \)

\( +\uparrow \ \sum F_{2}=0 \)

\( -180 \sin 15+F_{2 z}=0 \)

\( ∴ F_{2z} =46.6 N \)

\( \overrightarrow {F_{2}}=(-150.6 i-86.9 j+46.6 k) N \)

\( \left|F_{2}\right|=180 \mathrm{N} \)

\( \begin{array}{l}{\alpha=\cos ^{-1}\left(\frac{-150.6}{180}\right)=147^{\circ}} \\ {\beta=\cos ^{-1}\left(\frac{-86.9}{180}\right)=119^{\circ}} \\ {\gamma=\cos ^{-1}\left(\frac{46.6}{180}\right)=75^{\circ}}\end{array} \)

 

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