Using the reactions obtained in Problem $3,$ determine the forces in the members AC$.$ AD$,$ CD and CE$.$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=$Olongrightarrow$+$

$-$FAx$+5kN=0$

$F_{Ax}=5kN$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_y=0+$

$F_{Ay}-6kN-8kN+Fby$

FAy$-6kN-8kN+9kN=0$

FAY$=5kN$

${\displaystyle \underset{?}{\overset{?}{}}}{M}_A=00+$

$(5kN)(2m)+(6kN)(2m)+(8kN)(4m)-Fby(6m)=0$

$F{b}_y=9kN$