A rigid rod of length \(L\) and negligible mass connects two identical cylindrical floats, each possessing radius \(R\) and mass \(m_{a}\). The system floats in a fluid of mass density \(\rho_{f}\). A block of mass \(m_{b}\) is suspended from the center of the span by an elastic spring of stiffness \(k\) as shown, and a downward vertical force of magnitude \(F\) is applied to the suspended mass, as indicated. Derive the equations of motion of the system using Lagrange's Equations.

To derive the equations of motion using Lagrange's equations, we need to find the potential energy (U) and kinetic energy (T) of the system. The potential energy of the floats is due to the buoyant force and the gravitational force, while the kinetic energy is due to the motion of the floats in the vertical direction. The buoyant force on each of the floats is given by Archimedes' principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. The volume of the fluid displaced by each float is given by \(V_{float} = \pi R^2 h\), where h is the submerged height in the liquid. So, the buoyant force is given by \(F_{B}=\rho_{f} g V_{float}\). The potential energy due to the buoyant force is given by: $$U_{B} = F_{B}(R-h).$$ The potential energy due to the gravitational force on the mass m_a and the suspended mass m_b is given by: $$U_{G} = m_{a}gh + \frac{1}{2}kh^{2}.$$ Therefore, the total potential energy of the system is: $$U = U_{B} + U_{G}.$$ Now, let's find the kinetic energy of the floats. Since the mass of the rod is negligible, we can ignore it when calculating the kinetic energy. Then the total kinetic energy of the system is the sum of the kinetic energies of the two floats and the suspended mass m_b: $$T = \frac{1}{2}(2m_{a} + m_{b})\dot{h}^{2}.$$

The Lagrangian (L) is the difference between the kinetic and potential energies of the system: $$L = T - U.$$ Substituting the expressions of T and U: $$L = \frac{1}{2}(2m_{a} + m_{b})\dot{h}^{2} - \left(\rho_{f} g \pi R^2 h (R-h) + m_{a}gh + \frac{1}{2}kh^{2}\right).$$

To find the equations of motion, we need to apply the Lagrange's equation, which is written as: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{h}}\right) - \frac{\partial L}{\partial h} = \frac{\partial F_{external}}{\partial h},$$ where \(F_{external} = F\) is the external force acting on the system. First, let's calculate the partial derivatives: $$\frac{\partial L}{\partial \dot{h}} = (2m_{a} + m_{b})\dot{h},$$ $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{h}}\right) = (2m_{a} + m_{b})\ddot{h},$$ $$\frac{\partial L}{\partial h} = -\rho_{f} g \pi R^2 (2R-2h) - m_{a}g - kh,$$ and now we can plug these expressions into Lagrange's equation: $$(2m_{a} + m_{b})\ddot{h} = \rho_{f} g \pi R^2 (2R-2h) + m_{a}g + kh - F.$$

Finally, we can rearrange the equation by isolating \(\ddot{h}\), which is the accelerating motion of the system: $$\ddot{h} = \frac{1}{2m_{a} + m_{b}}\left[\rho_{f} g \pi R^2 (2R-2h) + m_{a}g + kh - F\right].$$ This equation describes the motion of the floating system under the influence of the elastic force of the spring, the gravitational force of the floats, and the external force F.