Answer
$F_{AC}=119lb$
$F_{AB}=79.2lb$
$F_{AD}=283lb$
Work Step by Step
We can write the tension developed in each cable as follows:
$F_{AB}=-\frac{3}{7}F_{AB}\hat i-\frac{6}{7}F_{AB}\hat j+\frac{2}{7}F_{AB}\hat k$
and $F_{AC}=\frac{2}{7}F_{AC}\hat i-\frac{6}{7}F_{AC}\hat j+\frac{3}{7}F_{AC}\hat k$
and $F_{AD}=\frac{3}{5}F_{AD}\hat j+\frac{4}{5}F_{AD}\hat k$
The sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies -\frac{3}{7}F_{AB}+\frac{2}{7}F_{AC}=0$.eq(1)
The sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies -\frac{6}{7}F_{AB}-\frac{6}{7}F_{AC}+\frac{3}{5}F_{AD}=0$.eq(2)
The sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies \frac{2}{7}F_{AB}+\frac{3}{7}F_{AC}+\frac{4}{5}F_{AD}-300=0$eq(3)
Solving eq(1), eq(2) and eq(3), we obtain:
$F_{AC}=119lb$
$F_{AB}=79.2lb$
and $F_{AD}=283lb$
