## 3 Results
### Analysis of the Actions of Some Mechanisms Articulated at One End on the Vertebral Fragments
#### 3.2.2 Analysis of Arm Action (Lever Loading)  
Since the lever from Figure 10 is the element in the system on which both the external load (\(F\)) and the actuation transmitted through the piston act, the loading of this lever was analyzed in particular.  
For this, a lever is isolated; it will be loaded with the stresses to which it is subjected, and both the forces that will be transmitted to the rod and the one that acts along the lever, i.e., the force that stresses the lever in compression, are determined.  
The arm tilt angle \(\beta\) in the initial position is 0, as depicted in Figure 11.  
Figure 11: Lever loading scheme: (**a**) an isolated lever; (**b**) lever loads.  
Figure 9: The aspect of the arm with the highest load: (**a**) three arms; (**b**) arms arms.  
The upper end of a lever is considered to be acted upon by the force required by the mechanism denoted by \(F\) (the load). We noted the reactions in the two joints with \(R_{1}\) and \(R_{2}\). Let \(F_{\text{as}}\) be the axial force that will act on the rod and that is distributed evenly on the two levers. This results in \(F_{\text{as}}/2\). We use \(l\) to denote the calculated length of the lever and \(\alpha\), the angle made by the lever with the horizontal axis.  
Writing the equilibrium equations for the lever, we have  
\[R_{1}=\frac{F_{\text{as}}}{2} \tag{1}\]  
\[R_{2}=F \tag{2}\]  
\[F\cdot l\cdot\cos(\alpha)=R_{1}\cdot l\cdot\sin(\alpha) \tag{3}\]  
Also, by solving the above system, we obtain  
\[F\cdot l\cdot\cos(\alpha)=\frac{F_{\text{as}}}{2}\cdot l\cdot\sin(\alpha) \quad\Rightarrow\quad 2\cdot F\cdot\cos(\alpha)=F_{\text{as}}\cdot\sin(\alpha) \quad\Rightarrow\quad F_{\text{as}}=\frac{2\cdot F\cdot\cos(\alpha)}{\sin( \alpha)} \tag{4}\]  
or:  
\[F_{\text{as}}=2\times F\times\cot(\alpha) \tag{5}\]  
It is observed that for a value of the angle \(\alpha=90^{\circ}\), \(\cos(\alpha)=0\). That is, \(F_{\text{as}}=0\). This means that the rod is no longer stressed axially.  
If \(\alpha\to 0\), then \(\sin(\alpha)\to 0\) and \(F_{\text{as}}\rightarrow\infty\).  
If \(\alpha=45^{\circ}\), it turns out that \(F_{\text{as}}=2\times F\).  
Interpreting the above, we say that the axial force in the rod varies from zero to infinity, corresponding to the variation in the position of the lever from its vertical position to its horizontal position.  
In the closed position (the position in which the mechanism is implanted), the levers are close to horizontal, and therefore, we have high values from the cotangent function. Thus, the initial value of the angle \(\alpha\) is significant.  
Regarding the axial force in the lever \(F_{\text{ap}}\), it is determined from the equilibrium equations corresponding to Figure 11b that  
\[F_{\text{ap}}=\frac{F_{\text{as}}}{2}\cdot\cos(\alpha)+R_{2}\cdot\sin(\alpha) \tag{6}\]  
Substituting, we obtain  
\[F_{\text{ap}}=\frac{2\cdot F}{2}\cdot\frac{\cos(\alpha)}{\sin(\alpha)}\cdot \cos(\alpha)+\frac{2\cdot F}{2}\cdot\sin(\alpha)\Rightarrow F_{\text{ap}}= \frac{F}{\sin(\alpha)}\cdot(\cos(\alpha)^{2}+\sin(\alpha)^{2})\Rightarrow F _{\text{ap}}=\frac{F}{\sin(\alpha)} \tag{7}\]  
This is the force that will apply to the lever during compression.  
Second Key Finding  
It is observed that both for the pushing force (actuation force) and for the tension in the lever, the highest values are at the beginning of the mechanism actuation when the device is in the closed position. This implies that to have as small values as possible for the axial force acting on the levers (and indirectly for the actuation force of the mechanism), the angle \(\alpha\) must have as large of a value as possible. This will lead to lower demands for other components of the assembly.