Hydrostatic friction: why do water droplets stay at rest on an inclined glass surface? 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):

*

*impose no-slip condition,

*minimize total energy with fixed contact area,

*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.
 A: 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. 
A: 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.
