## 2 Modeling Using Fully Cartesian Coordinates
### Mass Matrix of a Body in the Plane  
When dealing with mechanical systems in the plain, natural coordinates introduce a set of points to define a body, two \"basic points\"; see [25] for a detailed explanation. These points concentrate the mass of all the bodies in the mechanical system in accordance to their relation among each other, expressed though the mass matrix. To calculate or assemble the mass matrix of the system, it is necessary to calculate the mass matrix of the elements first. In planar systems, the standard model of a body is made with a pair of points, for example, \(A\) and \(B\), as seen in Figure 1. In this figure, it can be noted an inertial fixed reference frame **XY** and a local reference frame **xy** attached to the moving body at the basic point \(A\), its origin. It is also noted that the second basic point \(B\) has its position in local coordinates at \((l_{b},0)\) and that the CoM of the body, point g\({}_{b}\), has local coordinates \((x_{b},y_{b})\). Considering that the body \(b\) has its mass concentrated at g\({}_{b}\) equal to \(m_{b}\) and that its polar moment of inertia with respect to the origin of the local reference frame (point \(A\)) is \(l_{b}\), it is possible to distribute the whole mass into the basic points \(A\) and \(B\) to form the constant mass matrix of a body as follows:  
\[\textbf{M}_{b}= \begin{bmatrix}\textbf{M}_{1b}&\textbf{M}_{2b}\\ \textbf{M}_{3b}&\textbf{M}_{4b}\end{bmatrix}  \tag{1}\]where:  
\[\mathbf{M}_{1b}= \begin{bmatrix}m_{b}-\frac{2m_{b}x_{b}}{l_{b}}+\frac{l_{b}}{l_{b} ^{2}}&0\\ 0&m_{b}-\frac{2m_{b}x_{b}}{l_{b}}+\frac{l_{b}}{l_{b}^{2}}\end{bmatrix}  \tag{2}\]  
\[\mathbf{M}_{2b}= \begin{bmatrix}\frac{m_{b}x_{b}}{l_{b}}-\frac{l_{b}}{l_{b}^{2}} &-\frac{m_{b}y_{b}}{l_{b}}\\ \frac{m_{b}y_{b}}{l_{b}}&\frac{m_{b}x_{b}}{l_{b}}-\frac{l_{b}}{l_{b}^{2}} \end{bmatrix}  \tag{3}\]  
\[\mathbf{M}_{3b}= \begin{bmatrix}\frac{m_{b}x_{b}}{l_{b}}-\frac{l_{b}}{l_{b}^{2}} &\frac{m_{b}y_{b}}{l_{b}}\\ -\frac{m_{b}y_{b}}{l_{b}}&\frac{m_{b}x_{b}}{l}-\frac{l_{b}}{l_{b}^{2}}\end{bmatrix}  \tag{4}\]  
\[\mathbf{M}_{4b}= \begin{bmatrix}\frac{l_{b}}{l_{b}^{2}}&0\\ 0&\frac{l_{b}}{l_{b}^{2}}\end{bmatrix}  \tag{5}\]  
Note that \(J_{b}\) can be expressed in terms of the mass \(m_{b}\); thus, the mass matrix can be expressed solely in terms of the mass of the body. This matrix has been obtained by applying the virtual power method, leading to a direct formulation of the inertia forces [25]. Special Case of a Body with Three \"Basic Points\"  
A model of a body with a greater number of points could be also useful. In this case, to complete the model of the 3 RRR PPM, it was required to develop the mass matrix of a body with three \"basic points\"; see Figure 2.  
Considering that the body \(b\) in Figure 2 has its mass concentrated at \(g_{b}\) equal to \(m_{b}\) and that its polar moment of inertia with respect to the origin of the local reference frame (point \(A\)) is \(J_{b}\), its moment of inertia with respect to the \(x\) axis is \(J_{xb}\) and its moment of inertia with respect to the \(y\) axis is \(J_{yb}\). It is possible to distribute the whole mass into the basic points \(A,B\), and \(C\) to form the constant mass matrix of a body as follows:  
\[\mathbf{M}_{b}= \begin{bmatrix}\mathbf{M}_{1b}&\mathbf{M}_{2b}&\mathbf{M}_{3b }\\ \mathbf{M}_{2b}&\mathbf{M}_{4b}&\mathbf{M}_{5b}\\ \mathbf{M}_{3b}&\mathbf{M}_{5b}&\mathbf{M}_{6b}\end{bmatrix}  \tag{6}\]  
Figure 1: A general model of a body in fully Cartesian Coordinates using two basic points, \(A\) and \(B\). where:  
\[\mathbf{M}_{1b} = \begin{bmatrix}m_{1t}&0\\ 0&m_{1t}\end{bmatrix}  \tag{7}\] \[m_{1t} = \left(\frac{1}{h_{b}^{2}}+\frac{k_{b}^{2}}{l_{b}^{2}h_{b}^{2}}-\frac {2k_{b}}{l_{b}h_{b}^{2}}\right)J_{yb}+\left(\frac{2}{bh}-\frac{2k_{b}}{l_{b}^{2} h_{b}}\right)J_{b}+\frac{1}{l_{b}^{2}}J_{xb}+m_{b}\left(1-\frac{2x_{b}}{l_{b}}- \frac{2y_{b}}{h_{b}}+\frac{2y_{g}k_{b}}{l_{b}h_{b}}\right)\] (8) \[\mathbf{M}_{2b} = \begin{bmatrix}m_{2t}&0\\ 0&m_{2t}\end{bmatrix}