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

$F_{DE}=392N$ $F_{CD}=340N$ $F_{CA}=243N$ $F_{CB}=275N$
We can find the required tension developed in each cord as follows: $F_{FD}=mg=(20)(9.81)=196.2N$ Now we use the sum of forces at points D and C to find the required values $\Sigma F_y=0$ $\implies F_{DE} sin 30-F_{FD}=0$ $\implies F_{DE} sin30-196.2$ $\implies F_{DE}=392N$ and $\Sigma F_x=0$ $\implies F_{DE} cos 30-F_{CD}=0$ $\implies 392 \space cos30-F_{CD}=0$ $\implies F_{CD}=340N$ Now we find $F_{CA}$ and $F_{CB}$ as $\Sigma F_x=0$ $\implies -\frac{3}{5} F_{CA}+F_{CD}-F_{CB} cos 45=0$ We plug in the known values to obtain: $F_{CA}=243N$ and $\Sigma F_y=0$ $\implies \frac{4}{5}F_{CA}-F_{CB} sin 45=0$ We plug in the known values to obtain: $F_{CB}=275N$