The original system is shown in (a).
Since the \(F_1\) and \(F_2\) are parallel, the magnitude of the resultant force is just the sum of the two magnitudes and it points down.
\begin{equation*}
R = F_1 + F_2
\end{equation*}
The resultant moment about point \(A\) is
\begin{equation*}
M_A = F_1 d_1 + F_2 (d_1 + d_2)\text{.}
\end{equation*}
To create the equivalent system (b), the resultant force and resultant moment are placed at point
\(A\text{.}\)
The system in (b) can be further simplified to eliminate the moment at
\(M_A\text{,}\) by performing the process in reverse.
In (c) we place the resultant force \(R\) a distance \(d\) away from point \(A\) such that the resultant moment around point \(A\) remains the same. This distance can be found using \(M = Fd\text{.}\)
\begin{equation*}
d = M_A/R
\end{equation*}
The systems in (a), (b), and (c) are all statically equivalent