Answer
$F_{AD}=2.94KN$
$F_{AB}=F_{AC}=1.96KN$
Work Step by Step
We can determine the required tensions in the cables as follows:
$F_{AD}=-\frac{2}{3}F_{D}\hat i+\frac{2}{3}F_{AD}\hat j+\frac{1}{3}F_{AD}\hat k$
The sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies -\frac{2}{3}F_{AD}+F_{AB}=0$.eq(1)
The sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies \frac{2}{3}F_{AD}-F_{AC}=0$..eq(2)
The sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies \frac{1}{3}F_{AD}-W=0$
$\implies \frac{1}{3}F_{AD}-(100)(9.81)=0$
This simplifies to:
$F_{AD}=2.94KN$
We plug in the value of $F_{AD}$ in eq(1) and eq(2) to obtain:
$F_{AB}=F_{AC}=1.96KN$
