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

$m=102Kg$
We can determine the required maximum mass as follows: $F_{AD}=-\frac{2}{3}F_{AD}\hat i+\frac{2}{3}F_{AD}\hat j+\frac{1}{3}F_{AD}\hat k$ 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 $\Sigma F_z=0$ $\implies \frac{1}{3}F_{AD}-m(9.81)=0$.eq(3) It is given that the maximum force is $3KN$, so let $F_{AD}=3KN$. Then we plug in this value in eq(1) and eq(2) to obtain: $F_{AB}=F_{AC}=2KN$ Now we plug in the known values in eq(3) to obtain: $\frac{1}{3}(3000)-m(9.81)=0$ This simplifies to: $m=102Kg$