Please note that I used the same approach that
@wrobel at #21 and, to be fair, I checked it before getting the answer by myself.
I modified Chestermiller's nice diagram so that it can be applied to this approach
Where ##O## is at rest relative to the lab frame.
We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.
At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)
$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$
Where ##\vec F_i^S## is the specified force at a given point and ##\vec r_i## is the distance from ##O## to the given point.
Let's go step by step
1) Compute generalized forces
The generalized forces in this problem are:
$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$
$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$
Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:
$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$
$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\theta$$
2) Determine the kinetic energy ##(T)## of the system
Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame ##XYS##. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it ##T_1##) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it ##T_2##).
- Let's determine ##T_1##: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get
$$T_1 = \frac 1 2 (4m) \dot x^2$$
- Let's determine ##T_2##: Let's first get the coordinates of the center of mass of the ##AB## rod
$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$
Then the velocity coordinates are
$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$
Or in the more usual form: ##\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y##.
The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the ##AB## rod, the squared modulus of the velocity (##|\vec v_{T_2}|^2##) is.
$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$
Besides, the rods possesses moment of inertia with respect to their center of mass (##\frac{mb^2}{12}## each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.
Thus ##T_2## is given by
$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$
Thus the kinetic energy ##(T)## of the system is
$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$
3) Compute the equations of motion related to each generalized coordinate
The Lagrange equations for a general standard system are:
$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$
- For ##x## we get
$$4m \ddot x = F$$
- For ##\theta## we get
$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$
4) Get ##a_c##
Based on the diagram we see that
$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$
Thus, using the equations of motion we've got, we end up with
$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$
