Answer
$W=265lb$
Work Step by Step
We can determine the required maximum weight as follows:
$F_{AB}=-\frac{3}{7}F_{AB}\hat i-\frac{6}{7}F_{AB} \hat j+\frac{2}{7}F_{AB}\hat k $
$F_{AC}=\frac{2}{7}F_{AC}\hat i-\frac{6}{7}F_{AC}\hat j+\frac{3}{7}F_{AC}\hat k$
and $F_{AD}=\frac{3}{5}F_{AD}\hat j+\frac{4}{5}F_{AD}\hat k$
The sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies -\frac{3}{7}F_{AB}+\frac{2}{7}F_{AC}=0$..eq(1)
and the sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies -\frac{6}{7}F_{AB}-\frac{6}{7}F_{AC}+\frac{3}{5}F_{AD}=0$.eq(2)
The sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies \frac{2}{7}F_{AB}+\frac{3}{7}F_{AC}+\frac{4}{5}F_{AD}-W=0$..eq(3)
From eq(1), we have $F_{AC}=\frac{3}{2}F_{AB}$.eq(4)
We plug in this value and $F_{AD}=250lb$ in eq(2) to obtain:
$-\frac{6}{7}F_{AB}-\frac{6}{7}(\frac{3}{2}F_{AB})+\frac{3}{5}(250)=0$
This simplifies to:
$F_{AB}=70lb$
From eq(4), we have $F_{AC}=\frac{3}{2}(70)=105lb$
Now we plug in the known values in eq(3) to obtain:
$\frac{2}{7}(70)+\frac{3}{7}(105)+\frac{4}{5}(250)-W=0$
This simplifies to:
$W=265lb$
