Let us assume the center of mass of the overall system is at some distance z from point A somewhere between A and C. It does not matter where it is. Therefore the rotational equation of motion for the hanging system is M_{cg}=I_{A}\alpha
Notice the weights do not come into play, since we are taking moments about center of mass of the overall system.
So we just need to find F_{x},F_{y}. These forces are the reactions on point A where it is connected. These can be found by resolving forces in the horizontal and vertical direction. In horizontal direction\begin{equation} F_{x}=\left ( m_{AB}+m_{C}\right ) a_{0}\tag{2} \end{equation}
Plugging (2,3) into (1) and canceling z (as we see, we really did not need to find where z is), gives\begin{align*} \left ( W_{AB}+W_{C}\right ) \sin \theta -\left ( m_{AB}+m_{C}\right ) a_{0}\cos \theta & =0\\ \tan \theta & =\frac{\left ( m_{AB}+m_{C}\right ) a_{0}}{\left ( W_{AB}+W_{C}\right ) } \end{align*}
Plugging the numerical values\begin{align*} \tan \theta & =\frac{\left ( \frac{128}{32.2}+\frac{462}{32.2}\right ) 9}{\left ( 128+462\right ) }\\ & =0.279 \end{align*}
Hence \begin{align*} \theta & =\arctan \left ( 0.279\right ) \\ & =0.272\\ & =15.616^{0} \end{align*}
Taking moments about G (and assuming no friction from the ground as problems says to neglect rotational inertia of wheels, which seems to imply this). -Fh+N_{B}L_{B}-N_{A}L_{A}=I\alpha
And the acceleration is\begin{align*} F & =ma\\ 178.929 & =37a\\ a & =\frac{178.929}{37}\\ & =4.836\text{ m/s}^{2} \end{align*}
\begin{align*} F & =ma\\ \mu N & =ma\\ a & =\frac{\mu N}{m}\\ & =\frac{\left ( 0.51\right ) \left ( mg\right ) }{m}\\ & =\left ( 0.51\right ) \left ( 32.2\right ) \\ & =16.422\text{ ft/s}^{2} \end{align*}
Hence\begin{align*} v & =at\\ t & =\frac{v}{a}\\ & =\frac{18.2}{16.422}\\ & =1.108\text{ sec} \end{align*}
Resolve forces in vertical direction for hanging mass T-m_{B}g=m_{B}a_{y}
Hence \alpha =\frac{-\left ( 6\right ) \left ( 9.81\right ) \left ( 0.18\right ) }{\left ( 4\right ) \left ( 0.11\right ) ^{2}+\left ( 6\right ) \left ( 0.18\right ) ^{2}}=-43.636\text{ rad/sec}^{2}
I will use L for w so not to confuse it with \omega . Resolving forces in x direction -O_{x}=ma_{Gx}
Now the angular acceleration equation for the gate is, taking moments around center of mass\begin{align} O_{y}\frac{L}{2} & =I_{cg}\alpha \nonumber \\ & =\frac{mL^{2}}{12}\alpha \tag{3} \end{align}
From (2) O_{y}=-m\frac{L}{2}\alpha -P, plug this in (3) gives\begin{align*} \left ( -m\frac{L}{2}\alpha -P\right ) \frac{L}{2} & =\frac{mL^{2}}{12}\alpha \\ -P\frac{L}{2} & =\frac{mL^{2}}{12}\alpha +m\frac{L^{2}}{4}\alpha \\ -P\frac{L}{2} & =\alpha \left ( \frac{mL^{2}}{12}+\frac{mL^{2}}{4}\right ) \\ \alpha & =-\frac{P\frac{L}{2}}{\frac{1}{3}L^{2}m}\\ & =-\frac{3}{2}\frac{P}{Lm} \end{align*}
Plug-in numerical values \alpha =-\frac{3}{2}\frac{\left ( 21\right ) }{\left ( 14\right ) \left ( \frac{213}{32.2}\right ) }=-0.340
To find \omega , from \omega =\alpha t=-0.340\,\left ( 2.2\right ) =-0.748 rad/sec, hence from (1)\begin{align*} -O_{x} & =m\frac{L}{2}\omega ^{2}\\ & =\frac{213}{32.2}\left ( \frac{14}{2}\right ) \left ( -0.748\,\right ) ^{2}\\ O_{x} & =-25.929\text{ N} \end{align*}
Resolving forces in x direction, where F_{x},F_{y} are forces in hinge\begin{equation} F_{x}=-m\left ( \frac{3}{2}L\right ) \omega ^{2}\tag{1} \end{equation}
We are given \omega =6.9 rad/sec., hence from (1) F_{x}=-1342.6\text{ N}