# Mechanics: how to find the generalized force (specific case)?

I know this may seem like a homework problem at first, but please bear with me...

In this problem, we have two masses sliding without friction in a horizontal and vertical track, connected by a rigid massless link. Choosing $$\theta$$ as the single generalized coordinate, we can derive a single equation of motion, e.g. via the Lagrangian method.

The generalized force in context of Lagrangian dynamics is (often) defined as $$$$\tag{1}Q_i = \sum_{j=1}^m \frac{\partial \mathbf{r}_j}{\partial q_i}\cdot \mathbf{F}_j$$$$ where $$i$$ is the index of the generalized coordinate, $$m$$ is the number of applied forces and $$\mathbf{r}_j$$ is the position vector to the $$j$$th force.

Question-1: Using equation (1), how is the applied force $$F$$ (or it's moment) included as a generalized force when our single coordinate is $$q_1=\theta$$, and what expression will the generalized force take ?

From a course at my university, the generalized force (for 2D systems) is defined $$$$\tag{2}Q_i = \sum_{j=1}^m \frac{\partial \mathbf{r}_j}{\partial q_i}\cdot\mathbf{F}_j+\sum_{j=1}^p\frac{\partial \theta_j}{\partial q_i}M_j$$$$ where $$m,p$$ are nr of applied forces/torques

For this system, choosing $$q_1=\theta$$, we get $$Q_1=M$$.   The answer should be $$Q_1 = FL\sin(\theta)$$, which is kind of intuitive (but not quite)

Question-2: Please clarify the $$Q=FL\sin(\theta)$$  part. Would that also be the case if e.g. $$m_2$$ had no mass ?

Edit: I have derived the equations of motion to be $$(m_1L^2 s_{\theta}^2 + m_2 L^2 c_{\theta}^2)\ddot{\theta} - L^2s_{\theta}c_{\theta}(m_1-m_2)\dot{\theta}^2 + m_2gLc_{\theta} = Q$$ where $$c_\theta=\cos(\theta)$$, $$s_\theta=\sin(\theta)$$, and with kinetic and potential energies $$E_k=\frac{1}{2}m_1L^2s_{\theta}^2\dot{\theta}^2 + \frac{1}{2}m_2L^2c_{\theta}^2\dot{\theta}^2$$ $$E_p = m_2gLs_{\theta}$$

Perhaps this is flawed, and the force $$F$$ should be part of the potential ?

• You have to start with the position vector r1 r2 – Eli Jul 25 at 4:52
• $Q_1 = FL\sin(\theta)$ is a torque arm. Where is your Lagrangian? You can use any coordinates you want. $M$ is undefined. Since $q_{1}$ is an angle, then $Q_{1}q_{1}$ has to be torque. If the $q_{1}$ was a position, then $Q_{1}q_{1}$ would be a force. But your $Q_{1}$ does have dimensions of work. – Cinaed Simson Jul 25 at 6:35
• $Q_1=FL\sin(\theta)$ is the moment from $F$. If the system was rigid with mass center at $m_2$, then this quantity makes sense to me. Yes, $M$ is undefined ! – Ronny Landsverk Jul 25 at 9:27
• I don't want to answer my own question, but I guess you can do $$\mathbf{r}_1 = \left(x_0+L(1-\cos(\theta))\right)\hat{\mathbf{i}}$$ and then we get $Q_1=FL\sin(\theta)$ – Ronny Landsverk Aug 14 at 13:07