2
$\begingroup$

My understanding about constraint force is that it is a force which limits the geometry of particle's motion. For example, situations such as the particle trapped in a track or limited in domain can be assumed constraint force.

But in this point of view, I couldn't understand why friction is constraint force. In lagrangian formulation, we divide forces into two part, $F= F(applied) + F(constraint)$.

If particle moves in one dimension, and assuming there exists sliding friction, that particle can move anywhere. The sliding friction never restrict the domain that particle can move. so I think the sliding force is applied force, rather than constraint force.

Can anyone clarify why friction is constraint force?

$\endgroup$

bumped to the homepage by Community 18 hours ago

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

  • 2
    $\begingroup$ Friction forces are not conservative forces and hence a Lagrangian description of them is ill-suited. Constraint forces originating from holonomic constraints do not produce work and therefore can be suitably described by a Lagrangian formalism (contrary to friction forces). Thus could you please provide a link where it is said that friction forces can be seen as constraint forces? $\endgroup$ – gatsu Mar 11 '14 at 10:50
  • $\begingroup$ GoldStein, calssical mechanics, p17 or ame-www.usc.edu/bio/udwadia/papers/… or google.co.kr/… in that paper, search with friction $\endgroup$ – user42298 Mar 11 '14 at 11:05
  • 1
    $\begingroup$ Comments to the question (v3): Static (as opposed to kinetic or sliding) friction can be viewed as a constraint force. For more on D'Alembert's principle, see also e.g. physics.stackexchange.com/q/8453/2451 , physics.stackexchange.com/q/82884/2451 , and links therein. $\endgroup$ – Qmechanic Mar 11 '14 at 11:06
  • $\begingroup$ sliding friction in 1D and in absence of potential forces can be thought of as a kinetic energy loss $\dot{K}=-f(\dot{x}(t))$ in its simplest form. If you know or imagine you know $x(t)$ and the loss friction function $f$, you can integrate the above equation and get a dynamical constraint of the form $\dot{x}(t)=g(x,\dot{x},t)$. That's the way I interpret the link you gave me. Please tell me if I am interpreting it in a wrong way. I also wonder about the practical usefulness of such a formalism and vocabulary. $\endgroup$ – gatsu Mar 11 '14 at 13:35
0
$\begingroup$

rolling friction can be considered as constraint force. consider a circular ring rolling down a wedge without slipping. static friction at the contact provides a constraint type relation between x(distance of centre of mass from a fixed point) and the rotational coordinate(theta) of the ring. the constraint equation is nonholonomic- x'=r(theta'); where r is the radius of the ring. this so because the static friction made the ring to rotate.the ring might have slipped, but the presence of static friction constrained the body to rotate without slipping.The virtual work principle is also not voilated for rolling friction. but this not the case with sliding friction. its not a constraint force and the system with sliding friction must be excluded when de alembert equation is considered because the virtual work is not zero in this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.