# Determine the magnitudes of F1

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:

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.

$F_1\,=\,\left\{F_1\cos60^0i+0j+F_1\sin60^0k\right\}\,=\,\left\{0.5F_1i+0j+0.87F_1k\right\}$

$F_2\,=\,\left\{\dfrac{3}{5}F_2i-\dfrac{4}{5}F_2j+0k\right\}\,=\,\left\{0.6F_2i-0.8F_2j+0k\right\}$

$F_3\,=\,\left\{-F_3\cos^30^0i-F_3\sin30^0j+0k\right\}\,=\,\left\{-0.87F_3i-0.5F_3j+0k\right\}$

$F\,=\,\left\{0i+800\sin30^0j-800\cos30^0k\right\}\,=\,\left\{0i+400j-693k\right\}$

Now we can write our equation of equilibrium. All forces added together must equal zero because the system is in equilibrium.

$\sum \text{F}\,=\,0$

$F_1+F_2+F_3+F\,=\,0$

As each force added together must equal zero, each individual component (x, y, z-components) added together must also equal zero.

x-components:

$0.5F_1+0.6F_2-0.87F_3\,=\,0$

y-components:

$-0.8F_2-0.5F_3+400\,=\,0$

z-components:

$0.87F_1-693\,=\,0$

Solving the three equations gives us:

$F_1\,=\,796$ N

$F_2\,=\,147$ N

$F_3\,=\,564$ N