Question

With this can I get the answer to the following

Answer 1

$\begin{gathered} b)\text{ Equillibrium }equation\text{ for forces} \\ +\uparrow\Sigma F_y=0 \\ Fa+Fb-F1-F2-F3=0 \\ Fa+Fb-50KN-30KN-41KN=0 \\ Fa+Fb-121KN=0 \\ Fa+Fb=121KN\text{ (1)} \\ c)\text{Equillibrium }equation\text{ for moments} \\ Counter\text{clockwise is positive } \\ +M_A=0 \\ (1m)Fb-(50KN)(0.25m)-(30KN)(0.5m)-(41KN)(0.75m)=0 \\ (1m)Fb-12.5\text{KNm-}15\text{KNm-}30.75\text{KNm}=0 \\ (1m)Fb-58.25KNm=0 \\ (1m)Fb=58.25KNm\text{ (2)} \\ d)\text{ System equation} \\ Fa+Fb=121KN\text{ (1)} \\ (1m)Fb=58.25KNm\text{ (2) } \\ \text{From (2) solving Fb} \\ Fb=\frac{58.25KNm}{1m} \\ Fb=58.25KN \\ \text{With the value of Fb in (1)} \\ Fa+Fb=121KN\text{ } \\ Fa+58.25KN=121KN\text{ } \\ Fa=121KN\text{ -58.25KN} \\ Fa=62.75\text{ KN} \\ \text{The force Fa is 62.75Kn and force Fb is 58.25KN} \end{gathered}$