Equations of motion with mass attached to end effector and external force

Question

I have to find the equations of motion for the given manipulator:

The links of the manipulator are considered weightless and the equations of motion have to be derived by using the Lagrange equations. The coordinated of the point mass are:

$$ x_m = q_1+l_1+l_2c_2 $$ $$ y_m = l_2s_2 $$ And the Lagrangian of the system is:

$$ L = T - V = \frac{1}{2}\cdot \Big[\dot{q_1}^2-2l_2s_2\dot{q_1}\dot{q_2}+l_2^2\dot{q_2}^2\Big]-mgl_2s_2 $$

Now, in order to find the equations of motion, I applied the Lagrange's Method:

$$ \frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{q_i}}\Big) - \frac{\partial L}{\partial q_i} = \tau_i $$

And the final equations of motion written in matrix form are:

$$ \begin{bmatrix} m & -ml_2s_2 \\ -ml_2s_2 & ml_2^2 \end{bmatrix}\cdot \begin{bmatrix} \ddot{q_1} \\ \ddot{q_2} \end{bmatrix} + \begin{bmatrix} 0 & -ml_2c_2\dot{q_2} \\ 0 & 0 \end{bmatrix}\cdot \begin{bmatrix} \dot{q_1} \\ \dot{q_2} \end{bmatrix} + \begin{bmatrix} 0 \\ mgl_2c_2 \end{bmatrix}=\begin{bmatrix} \tau_1 \\ \tau_2 \end{bmatrix} $$

The above equations of motion have been derived by ignoring the force applied $F_x$. Now, my issue is how to include this force in terms of the Lagrange Method. Should I modify the method like the one below:

$$ \frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{q_i}}\Big) - \frac{\partial L}{\partial q_i} = \tau_i + F_x $$

But in which of the two equations of motion should the force be included ? The force $F_x$ is an external constant force applied in the direction of $x_0$.

Answer 1

In Lagrangian mechanics, you're required to deal with the generalized forces and the concept of virtual work $\delta W$.

In our case we have: $$ \delta W_{F_x} = F_x \cdot \delta q_1 + F_x \left( -l_2s_2\right) \cdot \delta q_2, $$

where the coefficients of the terms $\delta q_i$ are given by $F_x \frac{\partial x_m}{\partial q_i}$. By contrast, $y_m$ doesn't provide any contribution to the virtual work as $F_y = 0$.

This sums up in the final equation as: $$ ...=\begin{bmatrix} \tau_1 + F_x \\ \tau_2 -F_xl_2s_2 \end{bmatrix}. $$

In particular, $\tau_1$ is a force actuating the prismatic joint $q_1$, whereas $\tau_2$ is a torque actuating the revolute joint $q_2$. Therefore, units are consistent.