A square raft of mass \(m\) and side \(L\) sits in water of specific gravity \(\gamma_{w}\). A uniform vertical line force of intensity \(P\) acts downward at a distance \(a\) left of center of the span. (a) Use Lagrange's Equations to derive the 2-D equations of motion of the raft. (b) Check your answers using Newton's Laws of Motion.

Let the raft's center of mass position coordinates be \((x, y)\) and its rotation angle be \(\theta\). Thus, the generalized coordinates are \((x, y, \theta)\). First, we need to find the kinetic energy \(T\) of the raft. Since the raft is a rigid body, its mass moment of inertia \(I\) about its center of mass is given by \(I = \frac{mL^2}{6}\). Therefore, the kinetic energy \(T\) is the sum of the translational and rotational kinetic energies: \(T = \frac{1}{2}(m\dot{x}^2 + m\dot{y}^2)+\frac{1}{2}I\dot{\theta}^2 = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2+\frac{L^2}{6} \dot{\theta}^2)\) Next, we calculate the potential energy \(V\). The potential energy due to the force \(P\) is given by \(V = -P(y-a\sin\theta)\), where \(y\) is the vertical position of the raft's center of mass and \(-a\sin\theta\) is the vertical distance between the line of action of the force and the center of mass.

The Lagrangian \(L\) is given by \(L = T - V = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2+\frac{L^2}{6}\dot{\theta}^2)+P(y-a\sin\theta)\) Using Lagrange's equations: \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_i}}\right) - \frac{\partial L}{\partial q_i} = 0\) We obtain the 2-D equations of motion: 1. For \(q_1 = x\): \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x}=0 \implies m\ddot{x} = 0\) 2. For \(q_2 = y\): \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right) - \frac{\partial L}{\partial y}=0 \implies m\ddot{y} = -P\) 3. For \(q_3 = \theta\): \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta}=0 \implies \frac{mL^2}{6}\ddot{\theta} = Pa\cos\theta\) These are the 2-D equations of motion of the raft.

Now we will use Newton's second law to write the force balance equations for the raft. 1. For \(x\)-direction: \(\sum F_x = m\ddot{x} \implies 0 = m\ddot{x}\) 2. For \(y\)-direction: \(\sum F_y = m\ddot{y} \implies -P = m\ddot{y}\) 3. For rotation: \(\sum M_{cm} = I\ddot{\theta} \implies -Pa\cos\theta = \frac{mL^2}{6}\ddot{\theta}\)

Comparing the force balance equations obtained from Newton's second law with the equations of motion from Lagrange's equations, we see that they are the same: 1. \(m\ddot{x} = 0\) 2. \(m\ddot{y} = -P\) 3. \(\frac{mL^2}{6}\ddot{\theta} = Pa\cos\theta\) Thus, the 2-D equations of motion derived using Lagrange's equations are correct as they match the force balance equations from Newton's laws of motion.