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

$F_{AB}=98.6N$ $F_{AC}=267N$
We can find the required forces as follows: $W=mg$ $\implies W=(20)(9.81)=1962.N$ and $\theta=tan^{-1}(\frac{1}{2})=26.6$ $\phi=tan^{-1}(\frac{2.5}{2})=51.3$ Now, the sum of the forces in the x-direction is given as $\Sigma F_x=0$ $\implies -F_{AB}cos\phi-F_{AC} cos\theta+300=0$.eq(1) and the sum of the forces in the y-direction is given as $\Sigma F_y=0$ $\implies F_{AB}sin\phi+F_{AC}sin\theta-196.2=0$..eq(2) Solving eq(1) and eq(2), we obtain: $F_{AB}=98.6N$ and $F_{AC}=267N$