Answer
$F_{DA}=10.0lb$
$F_{DB}=1.11lb$
$F_{DC}=15.6lb$
Work Step by Step
We can find the force in the supporting cables as follows:
$F_{DA}=\frac{3}{4.5}F_{DA}\hat i-\frac{1.5}{4.5}F_{DA}\hat j+\frac{3}{4.5}F_{DA}\hat k$
and $F_{DC}=-\frac{1.5}{3.5}F_{DC}\hat i+\frac{1}{3.5}F_{DC}\hat j+\frac{3}{3.5}F_{DC}\hat k$
Now, the sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies \frac{3}{4.5}F_{DA}-\frac{1.5}{3.5}F_{DC}=0$..eq(1)
The sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies -\frac{1.5}{4.5}F_{DA}-F_{DB}+\frac{1}{3.5}F_{DC}=0$..eq(2)
The sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies \frac{3}{4.5}F_{DA}+\frac{3}{3.5}F_{DC}-20=0$..eq(3)
Solving eq(1), eq(2) and eq(3), we obtain:
$F_{DA}=10.0lb$
$F_{DB}=1.11lb$
$F_{DC}=15.6lb$
