What are the physical principles at play when a glass is stuck to a wet table? I've decided to write a relatively detailed paper on the following situation, but I'm finding the topic quite hard-to-google. Imagine a glass table with water spilled onto it. Once a drinking glass is placed on top, it becomes quite difficult to remove. It is more than the weight of the drinking glass that you have to overcome in order to lift it from the table.
Some more specifics and details:

*

*The contact surface of the drinking glass is a circle.

*Both the table and the glass surfaces are completely flat.

*There is no air trapped between the table and the drinking glass when the latter is placed onto the puddle of water resting on the former.

So, main question: how can we model this situation, taking all relevant things into consideration, in terms of the force required to lift the drinking glass?
Secondary question bombardment:

*

*Does the thickness of the layer of water matter? (If so, in what way?)

*How thick is a layer of water anyway?

*Is atmospheric pressure at play here?

*I assume the surface tension/viscosity of the water too?

 A: Many glasses have a concave bottom, and there is likely to be some air trapped underneath. In any case, when you try to lift the glass, the pressure under the glass drops and you are working against the air pressure from above. Adhesion and surface tension may also play a roll. As you lift, water flows in from the edges to equalize the pressures, so there may be a lag which depends on the rate of flow (which will depend on the pressure difference).
Any tilt of the glass may also make a difference.  For experimentation, you will want to try glasses with different shapes on the bottom.  You will need a gripping device that lets you put the center of support above the center of gravity (or not), and a method of measuring the force as a function of time (and or motion).  To look at the effect of surface tension, use a surface with a hole under the glass.
A: Let's model the glass as a cylinder with radius 4cm and height 10cm.
Air pressure is about 101,000Pa, there is a downward force due to this pressure on the top of the glass, usually balanced by a similar upward force due to the air inside the glass.
The water makes an airtight seal if the surface is flat enough.  When you try and lift the glass, the water stays in contact with it, due to surface tension, let's say the water can 'stretch' 1mm without breaking, when you try and lift the glass straight up - but any higher it breaks.
then from the formula $$PV=nRT$$
the volume is increases by a factor 10.1/10 = 1.01 and the pressure inside the glass decreases by the same factor, to 100,000Pa, giving a pressure difference, between the inside and outside of the glass, of 1000Pa
from $$P=\frac{F}{A}$$
with an area of $\pi\times 0.04^2$, about 0.005 square meters, we find the force needed is $$F = 5N$$ plus the weight of the glass.  About the weight of 5 apples plus the weight of the glass.
A: The two surfaces will be pressed together by atmospheric pressure. For a consumption glass and table unflatness will lead to a small direct glass to glass contact area, leaving the space around these sparse points filled with water. The average thickness of the water layer depends on the flatness of both surfaces. At the sparse contact points the van der Waals interaction is at play but negligible. If the surface are very flat as in half concave or convex mirrors then the direct contact area increases so that the van der Waals interaction binds the surfaces strongly.
A: The effect you have been looking for (or comes closest to it) is the suction cup effect. While it is true, as mentioned in the other answers, that it is the air pressure acting at the flat-disc-shaped cavity, which prevents the drinking glass from being removed from the glass table, this is only half of the story.
The other half deals with the question, why the air pressure at the circumference of the disc-shaped cavity does not push air into the cavity and thereby widening the cavity and allowing the drinking glass to be removed. For that, the effect of capillary action comes into play (especially see Wikipedia's section on "Capillary rise of liquid between two glass plates", which is somehow related to your problem, or at least illustrates it).
In a small gap between two glass surfaces (like in a narrow tube, or between a drinking glass and a glass table...), water tends to be sucked in, i.e. it pushes residual air out of the cavity. This becomes stronger, the smaller the gap between the two glass surfaces is. Therefore, if you decrease the gap, there will come a point, where the outward directed capillary adhesion force balances or exceeds the inward directed air pressure.
Without capillary action, only viscosity of the fluid would prevent easy inflow into the cavity, when a force pulls at it, and it would only slow it down (although maybe considerably), but not stop it. With capillary action, this unstable configuration becomes actually stable, in that the cavity will not open without additional external forces (that eventually exceed the limit that the capillary adhesion forces are able to withstand). Stability is the explanation for the stunning effect that a suction cup can stay in place literally for years (well, that may also be due to the water finally evaporating, leaving only the evacuated, sticky rubber cavity). The suction is stable in the same sense as the meniscus of a water column in a narrow glass tube is stable against gravity. If you blow air into the tube (additional force), you are able to lower the meniscus to its original level, but it does not happen if you do nothing.
With different material pairing, adhesion might recede behind cohesion. For example, a drinking glass standing on a glass table covered with mercury (don't do this at home ;-) ) will not show the suction effect, because mercury will get pushed into the cavity, and thereby adding some more to the already inward directed air pressure.
If there is adhesion, however, and the object that is sucked to the glass surface is elastic, the effect is even more pronounced because the object functions as a valve then, i.e. its elasticity allows the cavity to become blob-like when you press it to the glass (which allows the water in the cavity to escape easily), and to become cusp-like when you pull at it (making it more difficult for the water to escape because the gap that defines the adhesion force becomes even smaller than for a rigid arrangement).
A: While other people have mentioned suction cup/air pressure, a big contribution to this effect that needs to be mentioned is the molecular attraction between the 2 glass surfaces. I am not entirely sure what is the relative magnitude of these 2 effects.
The molecular attraction happens because, the water makes it so that at the glass to glass contact suface, becomes free from any other surface impurities, so the 2 glass surfaces bond with each other at the contact points. This makes them stick to each other.
I quote from Feynman

The same phenomenon can be observed in a simple home-made experiment
with a flat glass plate and a glass tumbler. If the tumbler is placed
on the plate and pulled along with a loop of string, it slides fairly
well and one can feel the coefficient of friction; it is a little
irregular, but it is a coefficient.
If we now wet the glass plate and the bottom of the tumbler and pull again, we find that it binds, and
if we look closely we shall find scratches, because the water is able
to lift the grease and the other contaminants off the surface, and
then we really have a glass-to-glass contact; this contact is so good
that it holds tight and resists separation so much that the glass is
torn apart; that is, it makes scratches.

