Mechanical System Modeling

Two-Cart System

Applying Newton’s second law to the two carts gives:

$$M_1\frac{d^2{x}_1(t)}{d{t}^2}=f(t)-K(x_1(t)-x_2(t))-F\left(\frac{d{x}_1(t)}{dt}-\frac{d{x}_2(t)}{dt}\right)$$

and

$$M_2\frac{d^2{x}_2(t)}{d{t}^2}=K(x_1(t)-x_2(t))+F\left(\frac{d{x}_1(t)}{dt}-\frac{d{x}_2(t)}{dt}\right)$$

In matrix form:

$$\left[\begin{array}{cc}{M}_1& 0\\ 0& M_2\end{array}\right]\left[\begin{array}{c}{\ddot{x}}_1\\ {\ddot{x}}_2\end{array}\right]+\left[\begin{array}{cc}F& -F\\ -F& F\end{array}\right]\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]+\left[\begin{array}{cc}K& -K\\ -K& K\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]f$$

In state-space form:

$$\mathbf{x}=\left[\begin{array}{c}{x}_1\\ {\dot{x}}_1\\ x_2\\ {\dot{x}}_2\end{array}\right],\mathbf{u}=f$$

And:

$$\begin{array}{rl}\mathbf{A}& =\left[\begin{array}{cccc}0& 1& 0& 0\\ -\frac{k}{m_1}& -\frac{F}{m_1}& \frac{k}{m_1}& \frac{F}{m_1}\\ 0& 0& 0& 1\\ \frac{k}{M_2}& \frac{F}{M_2}& -\frac{k}{M_2}& -\frac{F}{M_2}\end{array}\right]\\ \mathbf{B}& =\left[\begin{array}{c}0\\ \frac{1}{M_1}\\ 0\\ 0\end{array}\right]\\ \mathbf{C}& =\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\end{array}\right]\\ \mathbf{D}& =[0]\end{array}$$

Which then lets us write our system as

$$\begin{array}{r}\dot{\mathbf{x}}=\mathbf{Ax}+\mathbf{Bu}\\ \mathbf{y}=\mathbf{Cx}+\mathbf{Du}\end{array}$$