Is this trick for finding the eq.s of motion of a rod being pushed up a incline valid?
So I'm trying to find the equations of motion for the point $Q$ (as seen in the figure below) along an incline with angle $\beta$ with the horizontal, when the rod from $P$ to $Q$ is pushed by a constant and horizontal force $F$. Assume no friction.
I've calculated that the horizontal force, $F_s$, that must be applied at $P$ in order to keep the rod static at an angle $\alpha$ is $$F_s=\frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta +1}F_G$$ which has a maximum when $\alpha=0$ in the interval $[0,\beta]$.
The "trick" is now to say that we can calculate the eq.s of motion for $Q$ resulting from the constant $F$ on $P$ by calculating the eq.s of motion for $Q$ when the force (before $F$) is now the variable $F-F_s$.
For example, by choosing $F$ to be the minimum force required to set the rod in motion, $$F=F_s(\alpha=0)=\frac{\sin \beta}{\cos \beta +1}F_G,$$ we would obtain (where $x$ is along the incline) $$m\ddot{x}=F-F_s=F_G\sin \beta \left( \frac{1}{\cos \beta-1} - \frac{\cos \alpha}{\cos \alpha \sin \beta +1}\right)$$ Now we express $\alpha$ by $x$ as $$\cos\alpha =\frac{x}{L}\left(\cos \beta -1 \right)+1$$ to obtain the rather nasty looking D.E. $$\ddot{x}(t)=g\sin \beta \left[\frac{1}{\cos \beta-1} - \frac{1}{\sin \beta+\left(\frac{x(t)}{L}\left(\cos \beta -1 \right) \right)^{-1}} \right]$$
Q1: Is this method of "pretending" we apply a force which is the force "left over" from what is required to keep the rod static valid? If it is, could you please guide me to some resources that expands upon the idea (it is still rather fuzzy in my head)? If it isn't, what is wrong with it?
Q2: Can the D.E. be solved? Can this problem be solved using only elementary mechanics-methods (i.e., no Lagrangians and such)? How would you do it?
Looking forward to your inputs, thanks! 1002