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

$F_{DA}=10.0lb$ $F_{DB}=1.11lb$ $F_{DC}=15.6lb$
We can find the force in the supporting cables as follows: $F_{DA}=\frac{3}{4.5}F_{DA}\hat i-\frac{1.5}{4.5}F_{DA}\hat j+\frac{3}{4.5}F_{DA}\hat k$ and $F_{DC}=-\frac{1.5}{3.5}F_{DC}\hat i+\frac{1}{3.5}F_{DC}\hat j+\frac{3}{3.5}F_{DC}\hat k$ Now, the sum of the forces in the x-direction is given as $\Sigma F_x=0$ $\implies \frac{3}{4.5}F_{DA}-\frac{1.5}{3.5}F_{DC}=0$..eq(1) The sum of the forces in the y-direction is given as $\Sigma F_y=0$ $\implies -\frac{1.5}{4.5}F_{DA}-F_{DB}+\frac{1}{3.5}F_{DC}=0$..eq(2) The sum of the forces in the z-direction is given as $\Sigma F_z=0$ $\implies \frac{3}{4.5}F_{DA}+\frac{3}{3.5}F_{DC}-20=0$..eq(3) Solving eq(1), eq(2) and eq(3), we obtain: $F_{DA}=10.0lb$ $F_{DB}=1.11lb$ $F_{DC}=15.6lb$