11
$\begingroup$

This is a non-expert question on a (seemingly simple) text-book topic. The question is about "hydrostatic friction", defined as follows.

Consider a drop of water resting on a flat surface. If the surface is slightly inclined, then the drop will not run off but just stay in place.

Does this phenonemon have a simple description?

"Simple" as in "surface tension is simply described by a constant $\gamma$ which gives energy per unit area, $dE = \gamma \ dA$" or "Coulombic friction force is equal to normal reaction force times the coefficient of static friction $\mu_s$."

EDIT 1:

First answer revived my hope for a simple gravity + surface-tension solution. If the glass plate were horizontal, the droplet "chooses" its energetically optimal contact area. Now the same with tilt (gravity):

  1. impose no-slip condition,
  2. minimize total energy with fixed contact area,
  3. compare two optimal shapes with slightly different contact areas.

I hope there will be a critical angle beyond which the gain of gravitational energy overcomes losses to surface tension. Need more effort to write down an solve the variational problem (in cylindrical geometry for simplicity).

EDIT 2:

Found a recent review article the relevant subject: Rev. Mod. Phys. 81, 739 (2009); full text available on author's website. If this helps, will post an answer.

Improve this question
$\endgroup$
3
  • $\begingroup$ Now that's a good question! Compare to a meniscus (energy issues, which you are already thinking about) and to boundary layer (viscus friction). // too lazy to really think about this right now $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Sep 5, 2011 at 17:41
  • $\begingroup$ Viscosity of a Newtonian fluid applies only if there is motion. But water is hardly non-Newtonian (en.wikipedia.org/wiki/Non-Newtonian_fluid ). I believe the question has more to do with the microstructure of glass surface but don't have anything substantial beyond a vague feeling. $\endgroup$
    – Slaviks
    Commented Sep 5, 2011 at 18:11
  • $\begingroup$ I was thinking that the drops shape would be driven in part by the tendency of the outward parts to get ahead of the inner parts if there was any motion (thus not quite minimum energy), but you could be right. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Sep 5, 2011 at 18:19

2 Answers 2

Reset to default
8
$\begingroup$

I know this is an old question, but for the benefit of people visiting here wondering what the answer was, here it goes:

A droplet can stay at rest on an inclined plate because of small heterogeneities on the surface. This can either be a small roughness (of the order of nano/micrometers) or `dirty' spots where the surface chemistry is locally different.

The existence of these heterogeneities allows the droplet to have a different, bigger, contact angle at the downhill side then at the uphill side (this means that a perfectly smooth, clean surface will not be able to hold any droplets, this can most easily be shown by numerical simulation, because a perfectly smooth and clean surface is very hard to find/make experimentally). This difference in contact angle, thus surface energy, translates into a force pointing upwards, which is therefore able to counteract the gravitational force. Because there is a certain maximum to the front and rear contact angles (which depends on the surface roughness/dirtyness, higher roughness/dirtyness will give a larger difference between the two) the droplet will at some point start to move, at which point you get into the paradoxical discussion that Ron Maimon was talking about.

The balance that you get looks like this: $$ \rho V g \sin \alpha = k \gamma w (\cos\theta_u - \cos \theta_d) $$ where, in order of appearance, you have the density of the liquid, the volume of the droplet, the gravitational acceleration, the sin of the tilt angle, a `fudge' factor of O(1) depending on the detailed shape of the droplet, the width of the base of the droplet and the cosines of the uphill contact angle and the downhill angle.

The phenomenon of the difference in downhill and uphill contact angle is called contact angle hysteresis. If you want to know more about it, you can google for it: there are plenty of good sources out there.

To summarize: local heterogeneities on the surface allow a difference in surface energy at the front and rear of the droplet, thus introducing a force that can counteract gravity.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Fantastic! Thanks for a really useful answer. "contact angle hysteresis" is really the key. $\endgroup$
    – Slaviks
    Commented Feb 11, 2013 at 19:34
1
$\begingroup$

This is a well known paradox: a drop in continuum mechanics can never slide. The reason is that there are no-slip boundary conditions between the fluid and the solid, and this no-slip condition plus the contact-angle condition requires the drop cannot move. The surprise is not that it doesn't move for a while, but that it starts to move at all. This means that there is a violation of no slip in the boundary line, and it is a manifestation of the finite atomic scale, the nonzero Boltzmann's constant.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Aren't these two related but different paradoxes? In the completely static case, I wonder if gravity has any sizebale effect on the contact angle? This probablyis related to the violation of non-slip conditon as the thing starts moving, and thus has something to do with the breakdown of macroscopic continuous dynamics approximations at the contact line. But why do you say "nonzero Boltzmann's constant"? Does it imply that thermal fluctuation are vital here? $\endgroup$
    – Slaviks
    Commented Sep 5, 2011 at 19:10
  • $\begingroup$ I am saying this from memory--- I saw a seminar on this a long time ago, I haven't worked it out myself. It is clear that the no-slip has to be violated to get the droplet to move, since no slip holds on the contact region and bending the surface to extend it requires overcoming the surface tension. But the continuum analysis I saw did not take into account Van-der-Waals attraction of the bulk water to the glass, and this might account for the rolling without requiring a violation of the continuum limit (like a no-slip violation) which reveals the mean-free-path. I will think about it. $\endgroup$
    – Ron Maimon
    Commented Sep 6, 2011 at 1:41
  • $\begingroup$ Maybe gravity alone is able to overcome surface tension and bend the liquid over the critical angle if the drop is large enough? Will try to construct a variational energy balance staying with surface (surface tension) and bulk (gravity) terms. $\endgroup$
    – Slaviks
    Commented Sep 6, 2011 at 5:56
  • 2
    $\begingroup$ "It can never slide" does not imply that it will stay at rest. It is intuitive that the drop will roll rather than slide. You seem to argue this is prevented by the contact angle, but how? $\endgroup$
    – leftaroundabout
    Commented Sep 7, 2011 at 1:45
  • $\begingroup$ The argument I saw said that the contact angle never can become flat, because of the balance of surface tension and gravity. I don't know if it is true, because Van-der-Waals might allow it to roll. I am not certain of the answer yet, I am still thinking about it--- sorry. $\endgroup$
    – Ron Maimon
    Commented Sep 7, 2011 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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