6.7: Determine the constraint equations
From the sketch and the coordinate system the constraint equations can be determined, like discussed in Sec. 5.2. In Figure 6.3 we know that block \(A\) always touches the slope with angle \(\alpha\) at point \(A=C\). Since we have \(\tan \alpha=\frac{y_{A}}{x_{A, 0}-x_{A}}\), where \(x_{A, 0}\) is the coordinate at which the slope crosses the \(x\)-axis, we can determine the constraint equation and its time derivatives as:
\[\begin{align} y_{A} & =\left(x_{A, 0}-x_{A}\right) \tan \alpha \tag{6.6} \label{6.6}\\[4pt] \dot{y}_{A} & =-\dot{x}_{A} \tan \alpha \tag{6.7} \label{6.7}\\[4pt] \ddot{y}_{A} & =-\ddot{x}_{A} \tan \alpha \tag{6.8} \label{6.8}\end{align}\]
We can also use natural coordinates to express the constraint equations. It follows from Equation 5.84 and Equation 5.86 with radius of curvature \(\rho=\infty\) because the slope is a straight line that the constraint equations then become:
\[\begin{align} v_{A, n} & =0 \tag{6.9} \label{6.9}\\[4pt] a_{A, n} & =0 \tag{6.10} \label{6.10}\end{align}\]