## Engineering Mechanics: Statics & Dynamics (14th Edition)

$F_{AD}=900N$ $F_{AB}=693N$ $F_{AC}=646N$
We can determine the required forces as follows: $F_{DA}=\frac{1}{3}F_{AD}\hat i-\frac{2}{3}F_{AD}\hat j+\frac{2}{3}F_{AD}\hat k$ $\implies \Sigma F_z=0$ $\implies \frac{2}{3}F_{AD}-600=0$ This simplifies to: $F_{AD}=900N$ Now, the sum of the forces in the y-direction is given as $\Sigma F_y=0$ $\implies F_{AB} cos 30-\frac{2}{3}F_{AD}=0$ $\implies F_{AB}cos 30-\frac{2}{3}(900)=0$ This simplifies to: $F_{AB}=693N$ Now, the sum of the forces in the x-direction is given as $\Sigma F_x=0$ $\implies \frac{1}{3}F_{AD}+F_{AB} sin30-F_{AC}=0$ We plug in the known values to obtain: $\frac{1}{3}(900)+693 sin30-F_{AC}=0$ This simplifies to: $F_{AC}=646N$