Hi, I am working on modeling a simple rigid body dynamic system through a certain range of motion. Here is the picture of the system.
I am trying to solve the system for the variables $x_1'', y_1'', x_2'',y_2''$, the acceleration variables for points A and B respectively, the angular acceleration which we will call \ddot{\theta} corresponding to the angle seen in the diagram, and the force in the rod which is given as Fab. F is the input force that is held constant.
I could have used a typical approach of basing coordinates and equations around the center of mass but I decided I wanted to try it this way. ALSO, the rod connecting the two masses is considered massless. The masses are given value 1, so is r the rod length for mathematical simplification.
Here are the general equations of motion used. SUM OF FORCES AT $A(x_1,y_1)$
$$x_1'' = -F + F_{AB}\sin(\theta) \tag{eq.1}$$
$$y_1'' = F_{AB}\cos(\theta) - N_{A} = 0 \tag{eq.2}$$
SUM OF FORCES AT B(x_2,y_2)
$$x_2'' = Nb - F_{AB}\sin(\theta) = 0\tag{eq.3}$$
$$y_2'' = -F_{AB}cos(\theta)\tag{eq.4}$$
So far we have 5 variables and 4 equations, we will proceed to the moment equation about B
$$Mb = R_{BA} \times N_{ab} - R_{BA} \times F = I_{A}\ddot{\theta}\tag{eq.5}$$
$$Mb = rN_{A}sin(\theta) - rFcos(\theta) = I_{A}\ddot{\theta} = m_{1}r^2\ddot{\theta}\tag{eq.6}$$
As stated earlier, r and m are equal to 1 for mathematical simplification, so we get
$$N_{A}sin(\theta) - Fcos(\theta) = \ddot{\theta}\tag{eq.7}$$
The last equation is the relative acceleration for point A
$$R_{A} = R_{B} + R_{BA}\tag{eq.8}$$
$$R_{A}'' = R_{B}'' + \ddot{\theta} \times R_{BA} - w^2R_{BA}\tag{eq.9}$$
$$R_{A}'' = [0\vec i - F_{AB}cos(\theta)\vec j] + [-\ddot{\theta}\vec k \times (-rsin(\theta)\vec i - rcos(\theta)\vec j)] - \dot{\theta}^2[-rsin(\theta)\vec i - rcos(\theta)\vec j]\tag{eq.10}$$
$$R_{A}'' = [0\vec i - F_{AB}cos(\theta)\vec j] + [-r\ddot{\theta}cos(\theta)\vec i + r\ddot{\theta}sin(\theta)\vec j] + [r\dot{\theta}^2sin(\theta)\vec i + r\dot{\theta}^2cos(\theta)\vec j]\tag{eq.11}$$
Setting r = 1 as given from before and organizing terms under the correct components we get
$$R_{A}'' = [-\ddot{\theta}cos(\theta) + \dot{\theta}^2sin(\theta)]\vec i + [\dot{\theta}^2cos(\theta) + \ddot{\theta}sin(\theta) - F_{AB}cos(\theta)]\vec j\tag{eq.12}$$
Comparing these components to the components for point A
$$x_1'' = -\ddot{\theta}cos(\theta) + \dot{\theta}^2sin(\theta)\tag{eq.13}$$
$$y_1'' = 0 = \dot{\theta}^2cos(\theta) + \ddot{\theta}sin(\theta) - F_{AB}cos(\theta)\tag{eq.14}$$
Solving for $\dot{\theta}^2$ in the second equation we have
$$\dot{\theta}^2 = -\ddot{\theta}tan(\theta) + F_{AB}\tag{eq.15}$$
Plugging this into the new equation for $x_1''$ we get
$$x_1'' = -\ddot{\theta}cos(\theta) + F_{AB}sin(\theta) - \ddot{\theta}tan(\theta)sin(\theta)\tag{eq.15}$$
$$x_1'' = F_{AB}sin(\theta) + \ddot{\theta}[-tan(\theta)sin(\theta) - cos(\theta)]\tag{eq.16}$$
At this point, I have enough equations to solve for my 6 variables($x_1'',y_1'',x_2'',y_2'',F_{ab},\ddot{\theta}$). I put these into a matrix form and got a nonsense answer, I was hoping to get some help. I understand I can formulate the problem in a different way around the center of mass but tvecs not what I was aiming to do. I want to formulate it with these coordinates. I feel like I got a sign wrong somewhere but I'm stuck trying to find it. Edit: format edited with matrices added
The matrix of coefficients is A $$ \begin{pmatrix} 1 & 0 & 0 & 0 & -sin(\theta) & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & cos(\theta) & 0 \\ 0 & 0 & 0 & 0 & 0.5sin(2\theta) & -1 \\ 1 & 0 & 0 & 0 & -sin(\theta) & tan(\theta)sin(\theta) + cos(\theta) \\ \end{pmatrix} $$
Where X is
$$ \begin{pmatrix} x_1 \\ y_1 \\ x_2 \\ y_2 \\ F_{AB} \\ \ddot{\theta} \\ \end{pmatrix} $$$$ \begin{pmatrix} x_1'' \\ y_1'' \\ x_2'' \\ y_2'' \\ F_{AB} \\ \ddot{\theta} \\ \end{pmatrix} $$
And the output matrix B is given with the input force F = 1
$$ \begin{pmatrix} -1 \\ 0 \\ 0 \\ 0 \\ cos(\theta) \\ 0 \\ \end{pmatrix} $$
