How do I include friction due to normal force in Lagrange Equations?

Question

I am going through the Goldstein book on classical mechanics and the after he derived the Lagrange equations he used Rayleigh dissipation function to include friction as a generalized force. In school we wrote friction as a function of the normal force: $$F_f = \mu N,$$

but I thought that the normal force was considered a constraint force, so is there any way to include the friction equation with the normal force as a generalized force in stead of the ${\bf F}_f = -k{\bf v}$ equation?

Answer 1

It is not possible to insert friction forces in the Lagrangian for a crucial reason: friction forces of the type you consider have absolute value proportional to the normal component of the reactive force due to the constraint. This force does not appear in the Euler-Lagrange formalism just because it has been constructed to that goal. The E-L formalism only deals with (components of) forces and accelerations tangent to the constraints. In other words, there is no way to translate the relation $|F|= \mu |N|$ in the E-L formalism. The case of friction forces proportional to the velocity can be treated in the E-L formulation as pointed out in the other answer by ZeroTheHero.

Answer 2

how to consider friction force and calculate the equations of motion with E.L method.

the Euler- Lagrange equation are:

$$\begin{align*} &L =T-U\\ &\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\vec{q}}}\right)- \frac{\partial L}{\partial \vec{q}}=\left(\frac{\partial \vec{r}}{\partial \vec{q}}\right)^T\vec{f}_a+ \left(\frac{\partial \vec{g}_h}{\partial \vec{q}}\right)^T\vec{\lambda}_h \end{align*}\tag 1$$

• $T$ kinetic energy
• $U$ potential energy
• $\vec{q}$ generalized coordinates
• $\vec{r}$ position vector
• $\vec{f}_a$ external forces
• $\vec{g}_h$ holonomic constrains
• $\vec{\lambda}_h$ holonomic constrain forces

I) the friction force is considered as external force.

II) the friction normal force is calculate with holonomic constraint equation.

Example:

a mass point slides on a circle path, with a friction between the mass and the path.

beginning with the position vector of the mass:

$$\vec{r}=\left[ \begin {array}{c} \rho\, \left( 1-\cos \left( {\frac {s}{\rho} } \right) \right) \\ \rho\,\sin \left( {\frac {s}{ \rho}} \right) \end {array} \right] $$ where $\rho$ is the circle radius and s the arc length.

the friction force $F\mu$ act towards the path tangent $\vec{t}$ with the sign apposite to the tangent velocity $v_t$

with :

$$\vec{t}=\frac{\partial \vec{r}}{\partial s}=\left[ \begin {array}{c} \sin \left( {\frac {s}{\rho}} \right) \\\cos \left( {\frac {s}{\rho}} \right) \end {array} \right] $$

thus:

$$F\mu=\mu\,|N|\,\text{sign}(-v_t)$$ where $v_t=\vec{v}^T\,\vec{t}\quad,\vec{v}=\vec{\dot{r}}$

so the external force is:

$$\vec{f}_a=F\mu\,\vec{t}$$

to obtain the normal force N , we create a gap $q_N$ toward the normal direction $\vec{n}$, thus we have now two generalized coordinates $\vec{q}=[s,q_N]^T$

the mass position vector is now:

$$\vec{r}\mapsto \vec{r}+q_N\vec{n}$$

with:

$$n=\left[ \begin {array}{c} -\cos \left( {\frac {s}{\rho}} \right) \\ \sin \left( {\frac {s}{\rho}} \right) \end {array} \right] $$

you can now obtain the kinetic energy and potential energy:

$$T=\frac{m}{2}\vec{\dot{r}}^T\vec{\dot{r}}$$ $$U=m\,g\,r_y$$

the holonomic constrain equation is:

$$g_h=q_N=0\quad \Rightarrow\quad \dot{q}_N=0\quad,\ddot{q}_N=0$$

with equation (1) ,you can calculate the equation of motion and the normal force . you have two equations for two unknowns $ \ddot{s}\,,N=\lambda_h$

results:

with equation (1) you get:

$$m\,\ddot{s}+m\,g\,\cos\left(\frac{s}{\rho}\right)-F\mu=0\tag 2$$

and

$$m\,\ddot{q}_N+\frac{m\,\dot{s}^2}{\rho}+ m\,g\,\sin\left(\frac{s}{\rho}\right)-N=0\tag 3$$

with $\ddot{q}_N=0$

$ \Rightarrow$

$$N=m\,g\,\sin\left(\frac{s}{\rho}\right)-\frac{m\,\dot{s}^2}{\rho}$$

$$\ddot{s}-\frac{F\mu}{m}+g\,\cos\left(\frac{s}{\rho}\right)=0$$

Simulation

Answer 3

The constraint provided by the normal force is one that will keep the particle on the surface of an object. The friction force, on the other hand, is NOT a constraint even if the normal force is. Because it is not obtained from a potential, it must be included by hand as a generalized force.

Nevertheless, it is possible to reproduce the correct equations of motion with dissipation of the form $F=- \kappa \dot{x}$ by considering the time-dependent Lagrangian $$ L= e^{\alpha t}\left(\frac{m}{2} \dot{x}^2- \frac{k}{2}x^2\right) $$ You can then derive that, since $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{d}{dt} me^{\alpha t}\dot{x}= me^{\alpha t}(\alpha \dot{x} +\ddot{x}) $$ the resulting equation of motion accounts for dissipation for a suitable choice of $\alpha$.

As $L$ is explicitly time-dependent, the Hamiltonian is not conserved (not unexpectedly).

Answer 4

1. Concerning kinetic friction $$F_f~\equiv~|{\bf F}_f|~=~\mu |{\bf F}_n|,\tag{1}$$ let us for simplicity assume that the surfaces stay in contact at all time, and the contact surface is a plane (possibly inclined), i.e that $F_f$ is a constant.

   • Then the friction force is $$ {\bf F}_f~=~-F_f\frac{\bf v}{|{\bf v}|}.\tag{2}$$

   • The corresponding Rayleigh dissipation function is
     $$ {\cal F}~=~ F_f |{\bf v}|, \qquad {\bf F}_f ~=~-\frac{\partial {\cal F}}{\partial{\bf v}}.\tag{3}$$ Recall that Lagrange equations can readily be modified to accommodate the Rayleigh dissipation function, cf. e.g. eq. (1.70) in Goldstein or this Phys.SE post.

   • On the other hand, one may show that there does not exist a velocity-dependent potential $U$ for the friction force (2), cf. e.g. this Phys.SE post, not even in 1D.

2. Lagrangian formulations of linear friction/drag (which OP also mentions) is discussed in this related Phys.SE post.

References:

1. Goldstein, Classical Mechanics; Section 1.5 eq. (1.70).