# Analytical Modeling of Particle Scratching Process
## 3 Result and Discussion
### Force Ratio  
The forces applied to the particle are affected by three types of parameters, which may impact on force ratio. One kind is related to dimensions including \(k_{1}\), \(k_{2}\), \(r\), and \(h_{doc}\). Another kind is associated with the material constitutive model (yield strength \(\sigma_{y}\) and coefficient of strain strengthening \(n\)). The last one is the coefficient of friction \(\mu\). In order to study the effect of these parameters on force ratio, a MATLAB R2021b program was made according to the analytical model as shown in the appendix to calculate the forces applied on the particle directly, then the force ratio can be obtained. The coefficient of elastic rebound \(k_{1}\) and the coefficient of plastic upheaval \(k_{2}\) represent the relative height of elastic rebound and plastic upheaval, respectively. The friction coefficient \(\mu\) expresses the linear relationship between the normal stress and shear stress, which is common sense, and may influence the force ratio. In the scratching process, the force ratio depends on \(k_{1}\), \(k_{2}\), and \(\mu\) when other parameters are given. A series of experiments were conducted to research how \(k_{1}\), \(k_{2}\), and \(\mu\) affect the force ratio. The experimental setup is listed in Table 2, and the results are shown in Figure 5.  
As shown in Figure 5, the force ratio increases almost linearly with the increase in the coefficient of elastic rebound \(k_{1}\) and the coefficient of plastic upheaval \(k_{2}\). As shown in Figure 5b, the force ratio increases from 0.1932 to 0.2453 as \(k_{2}\) increases from 0.1 to 0.5 (\(k_{1}=0.3\)), an increase of 0.0521; the force ratio from 0.1769 to 0.2491 when \(k_{1}\) from 0.1 to 0.5 (\(k_{2}=0.3\)), an increase of 0.0722. So, \(k_{2}\) has more influence on the force ratio than \(k_{1}\), which can also be seen from Figure 5 intuitively. The conclusion can also be drawn that the smaller the friction coefficient, the larger the force ratio. For example, assuming both \(k_{1}\) and \(k_{2}\) are 0.3, when \(\mu\) is 0.4, 0.5, and 0.6, the corresponding force ratios are 0.2186, 0.1774, and 0.1403, respectively. However, previous studies have shown that the force ratio (\(F_{t}/F_{n}\)) increases with the increase in friction coefficient [18]. Some research has shown that the friction coefficient affects \(k_{1}\) and \(k_{2}\)[19], but this study assumes that \(\mu\), \(k_{1}\), and \(k_{2}\) were independent of each other because the relationships among them are too complex to reflect in the analytical model. This is the reason for the contradiction. The force ratio depends on \(r\) and \(h_{doc}\) when other parameters are given. The experimental setup for researching the effect of \(r\) and \(h_{doc}\) on force ratio is listed in Table 3, and the result is shown in Figure 6.  
\begin{table}
\begin{tabular}{c c c c c c} \hline  & \multicolumn{3}{c}{\(\sigma_{y}\) = 500 MPa, \(h_{doc}\) = 30 \(\mu\)m, \(n\) = 0.5, \(r\) = 0.1 mm} \\ \cline{2-6}  & \multicolumn{3}{c}{\(\mu\) = 0.3/0.4/0.5} \\ \cline{2-6}  & \(k_{11}\) & \(k_{12}\) & \(k_{13}\) & \(k_{14}\) & \(k_{15}\) \\ \hline \(k_{21}\) & (0.1,0.1) & (0.1,0.2) & (0.1,0.3) & (0.1,0.4) & (0.1,0.5) \\ \(k_{22}\) & (0.2,0.1) & (0.2,0.2) & (0.2,0.3) & (0.2,0.4) & (0.2,