I Acceleration of Hinges

Is at least the result you get from the long solution something simple? Then there is a chance for a simpler way to solve it.

 

I would try to use the center of mass frame, where the kinematics is nicely symmetrical. The center of mass frame is non-inertial, but has a known acceleration, and thus a known inertial force field. Since the inertial force field is uniform, it doesn't create any torques at the bars, around their centers of mass.

 

First you must obtain differential equations of motion not at the single moment but for all ##t##. For ##t>0## the square will deform to a rhombus. So this is a system with two degrees of freedom



Let ##X## be the axis of inertial frame and ##x## be the coordinate of the center of the rhombus. The generalized coordinates are ##\alpha,x##.

We shall assume that the force ##\boldsymbol F## does not depend on time. It does not matter since we will consider the differential equations at initial moment only. Then find the kinetic energy to this system ##T=T(\dot\alpha,\dot x,\alpha,x)## You will see that actually ##T## does not depend on ##x,\alpha##. And write the Lagrange equations
$$\frac{d}{dt}\frac{\partial T}{\partial \dot x}-\frac{\partial T}{\partial x}=Q_x,\quad \frac{d}{dt}\frac{\partial T}{\partial \dot \alpha}-\frac{\partial T}{\partial \alpha}=Q_\alpha.$$
The first equation will be ##4m\ddot x=F## (##m## is the mass of the rod), but the second one is not so evident, you have to calculate for it.

 

I usually give this problem to students, once I proposed it here for methodology discussion https://www.physicsforums.com/threads/a-problem-about-the-frame.871175/