#### Answer

$F_{AD}=763N$
$F_{AC}=392N$
$F_{AB}=523N$

#### Work Step by Step

We can find the required forces as follows:
$OA=\sqrt{(2)^2+(1.5)^2}=2.5m$
$OB=\sqrt{(2.5)^2+(1.5)^2}=2.91m$
Now, the sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies F_{AC}-F{AD}(\frac{2.5}{2.91})(\frac{1.5}{2.5})=0$
$\implies F_{AC}=\frac{1.5}{2.91}F_{AD}$.eq(1)
Now, the sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies F_{AB}-F_{AD}(\frac{2.5}{2.91})(\frac{2}{2.5})=0$
$\implies F_{AB}=\frac{2}{2.91}F_{AD}$.eq(2)
Now, the sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies F_{AD}(\frac{1.5}{2.91})-W=0$
$\implies F_{AD}(\frac{1.5}{2.91})-40(9.81)=0$
This simplifies to:
$F_{AD}=763N$
We plug in this value of $F_{AD}$ in eq(2) to obtain:
$F_{AB}=\frac{2}{2.91}(763)$
$\implies F_{AB}=523N$
We plug in $F_{AD}$ in eq(1) to obtain:
$F_{AC}=\frac{1.5}{2.91}(763)$
This simplifies to:
$F_{AC}=392N$