solución:
\(F_B=30lb\)
\(F_A=0\) \(sen\theta=\frac{3}{5}\)
\(\Sigma F=0\) \(\cos\theta=\frac{4}{5}\)
\(\Sigma M=0\) \(\tan\theta=\frac{3}{4}\)
\(F_{Ax}=F_A\cos\theta=\frac{4}{5}F_A\)
\(F_{Ay}=F_Asen\theta=\frac{3}{5}\)
\(F_{Bx}=F_B\cos60\)
\(F_{By}-F_Bsen60\)
\(\Sigma F_x=0\)
\(-F_{Bx}-F_{Ay}=0\)
\(-30lb-\cos60-\frac{4}{5}F_A=0\)
\(\frac{4}{5}F_A=-30lb\)
\(F_A=\frac{5}{4}\left(-30lb\cos60\right)\)
\(F_A=18.75\)
para B
\(r_{Bx}=6ft;\ F_{Bx}=30lb\cos60\)
\(B_y=0;F_{By}=30lb\ sen60\)
\(M_A=r_{Ax}\ F_{Ay}-r_{Ay}=\left(9ft\right)\left(\frac{3}{5}\right)-\left(0\right)\left(\frac{4}{5}FA\right)=\frac{27}{5}lb\cdot ft\)
para A
\(r_{Ax}=9ft;F_{Ax}=\frac{4}{5}F_A\)
\(F_{Ay}=0;F_{Ay}=\frac{3}{5}F_A\)
\(M_B=rBx\ \ Fby-rBy\ FBx=\left(6\right)\left(30sen60\right)-\left(0\right)\left(30\cos60\right)=155.88lb\)
\(\Sigma M=0\)
\(M_B=M_A=0\)
\(\frac{27}{5}F_A=155.88\)
\(F_A=\frac{155.88}{\left(\frac{27}{5}\right)}\)
\(F_A=28.86\)