Determine the magnitudes of F_1, F_2, and F_3 for equilibrium of the particle.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
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We will first write each force in Cartesian vector form.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
F_2\,=\,\left\{\dfrac{3}{5}F_2i-\dfrac{4}{5}F_2j+0k\right\}\,=\,\left\{0.6F_2i-0.8F_2j+0k\right\}
F_3\,=\,\left\{-F_3\cos^30^0i-F_3\sin30^0j+0k\right\}\,=\,\left\{-0.87F_3i-0.5F_3j+0k\right\}
F\,=\,\left\{0i+800\sin30^0j-800\cos30^0k\right\}\,=\,\left\{0i+400j-693k\right\}
Now we can write our equation of equilibrium. All forces added together must equal zero because the system is in equilibrium.
F_1+F_2+F_3+F\,=\,0
As each force added together must equal zero, each individual component (x, y, z-components) added together must also equal zero.
0.5F_1+0.6F_2-0.87F_3\,=\,0
y-components:
-0.8F_2-0.5F_3+400\,=\,0
z-components:
0.87F_1-693\,=\,0
Solving the three equations gives us:
F_2\,=\,147 N
F_3\,=\,564 N