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\[ M_{cg}=I_{A}\alpha \] Where \(M\) is the moment of external forces around this center of mass and \(I_{A}\) is the mass moment of inertia around \(A\). But since we want the system to be translating, then \(\alpha =0\). Therefore \begin{align} M & =0\nonumber \\ F_{y}z\sin \theta -F_{x}z\cos \theta & =0\tag{1} \end{align}
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} In vertical direction (where these is no acceleration)\begin{align} F_{y}-W_{AB}-W_{C} & =0\nonumber \\ F_{y} & =W_{AB}+W_{C}\tag{3} \end{align}
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).\[ -Fh+N_{B}L_{B}-N_{A}L_{A}=I\alpha \] For \(\alpha =0\)\[ -Fh+N_{B}L_{B}-N_{A}L_{A}=0 \] And when \(N_{A}=0\)\[ F=\frac{N_{B}L_{B}}{h}\] But \(N_{A}+N_{B}=mg\) or since \(N_{A}=0\) then \(N_{B}=mg\) and the above becomes\begin{align*} F_{\min } & =\frac{mgL_{B}}{h}\\ & =\frac{\left ( 37\right ) \left ( 9.81\right ) \left ( 0.35\right ) }{\left ( 0.71\right ) }\\ & =178.929\text{ N} \end{align*}
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\[ T-m_{B}g=m_{B}a_{y}\] But \(a_{y}=R\alpha \) where \(\alpha \) is angular acceleration of spool. Hence\begin{equation} T-m_{B}g=m_{B}R\alpha \tag{1} \end{equation} For the spool, the equation of motion is \(M=I\alpha \) or\begin{equation} -TR=mr_{G}^{2}\alpha \tag{2} \end{equation} Where \(r_{G}\) is radius of gyration. We have two equations and two unknowns \(\alpha ,T\)., solving gives\begin{align*} \alpha & =\frac{-m_{B}gR}{mr_{G}^{2}+m_{B}R^{2}}\\ T & =m_{B}R\alpha +m_{B}g \end{align*}
Hence\[ \alpha =\frac{-\left ( 6\right ) \left ( 9.81\right ) \left ( 0.18\right ) }{\left ( 4\right ) \left ( 0.11\right ) ^{2}+\left ( 6\right ) \left ( 0.18\right ) ^{2}}=-43.636\text{ rad/sec}^{2}\] And\begin{align*} T & =\left ( 6\right ) \left ( 0.18\right ) \left ( -43.636\right ) +\left ( 6\right ) \left ( 9.81\right ) \\ & =11.733\text{ N} \end{align*}
I will use \(L\) for \(w\) so not to confuse it with \(\omega \). Resolving forces in \(x\) direction\[ -O_{x}=ma_{Gx}\] in the \(y\) direction\[ P+O_{y}=ma_{Gy}\] But \(a_{Gx}=\frac{L}{2}\omega ^{2}\) and \(a_{G_{y}}=-\frac{L}{2}\alpha \) where \(\alpha \) is angular acceleration of gate. Hence the above becomes\begin{align} -O_{x} & =m\frac{L}{2}\omega ^{2}\tag{1}\\ P+O_{y} & =-m\frac{L}{2}\alpha \tag{2} \end{align}
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\[ \alpha =-\frac{3}{2}\frac{\left ( 21\right ) }{\left ( 14\right ) \left ( \frac{213}{32.2}\right ) }=-0.340 \] From (3)\begin{align*} O_{y} & =\frac{mL}{6}\alpha \\ & =\frac{\frac{213}{32.2}\left ( 14\right ) }{6}\left ( -0.3408\right ) \\ & =-5.25\text{ N} \end{align*}
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} In \(y\) direction\begin{equation} F_{y}-2mg=m\left ( \frac{3}{2}L\right ) \alpha \tag{2} \end{equation} Taking moments about the hinge \(O\)\begin{equation} \left ( -mg\frac{L}{2}-mgL\right ) =\left ( \left ( m\frac{L^{2}}{3}\right ) +\left ( \frac{1}{12}mL^{2}+mL^{2}\right ) \right ) \alpha \tag{3} \end{equation} Solving (2,3) for \(F_{y},\alpha \) gives \begin{align*} F_{y} & =40.394\text{ N}\\ \alpha & =-5.52503\text{ rad/sec}^{2} \end{align*}
We are given \(\omega =6.9\) rad/sec., hence from (1) \[ F_{x}=-1342.6\text{ N}\]