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

$F_{AD}=2.94KN$ $F_{AB}=F_{AC}=1.96KN$
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: $F_{AB}=F_{AC}=1.96KN$