# 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!

• To whoever added the tag: This is not homework! – Bobson Dugnutt Aug 13 '16 at 15:40
• The tag is "homework and exercises." This is an exercise. – sammy gerbil Aug 13 '16 at 19:11

I think you are intending to apply a constant force $F$ of magnitude $F_s(\alpha=0)$. This is countered by the horizontal components of the reaction forces $N_1$ and $N_2$, which add up to a variable force $F_s(\alpha)$. So it is the net force on the rod $F-F_s$ which is variable, not the applied force.

I think there is no "trick" here : all you are doing is stating that $m\ddot x$ is equal to the net horizontal force on the rod, $F-F_s$, which varies with $\alpha$.

The question you need to ask is : Does the variable reaction force have the same static value $F_s$ when the rod is moving?

I think the answer to this question is No. If the rod is accelerating and there is a component of this acceleration perpendicular to the ground at P and to the incline at Q, then the normal reaction force will not have the static value. This is like weighing an object in a lift which is accelerating up or down. The acceleration of the lift affects the "weight" measured as the tension in a spring or the reaction force provided by the floor of the lift.

• @Floris : Thank you for prompting me to re-think my answer, which I have amended. – sammy gerbil Aug 15 '16 at 13:47
• I think your new answer is a great deal better than your old one was. – Floris Aug 15 '16 at 13:49

You can always do one integration of an equation of the form $$\frac{d^2}{dt^2}x(t) = F(x(t)).$$

Write $dx/dt = v$, so $d^2x/dt^2 = dv/dt$, and multiply both sides of the equation by $2v$. \begin{gather} 2v\frac{dv}{dt} = 2vF(x) \\ \frac{d}{dt}(v^2) = 2vF(x) \\ \begin{aligned} v^2 &= 2\int vF(x)\, dt \\ &= 2\int F(x) \frac{dx}{dt}\, dt \\ &= 2\int F(x)\, dx \end{aligned} \end{gather} That equation is stating principle that energy is conserved. $v^2$ is a measure of the kinetic energy of the system and $\int F(x)\,dx$ is a measure of the external work done on it.

You then get an equation of the form $$\frac{dx}{dt} = G(x)$$ You can separate the variables, but you might not be able to integrate it using elementary functions.