Lifting a wet object off a table When there is a film of water in between an object and the tabletop it rests on, it is difficult to pull the object directly upward. However, when you slide the object, it comes off easily. What is the physics behind this?
 A: It's essentially the same as any "vacuum attaching pad" or whatever they are called. The water is incompressible, so if you want to lift the glass, the only way that can happen, is to suck the water inwards, and that's very hard to do if the gap is very thin.
Imagine a simplified geometry, where the glass is rectangular $a\times b$, and the water can only be sucked inwards from one side. Let $b$ be the tranverse direction in which there is no velocity. Then, conservation of volume gives you
$$\Phi(x)=-xb \dot{h}$$
where $x$ is between $0$ and $a$ and the left hand side tells you about the volume flux of water. Observe how the flux is increasing toward the entry into the hap.
Due to viscosity $\eta$, this flux is related to the longitudinal pressure gradient.
$$\Phi(x)=-\frac{bh^3}{6\eta}\frac{dp}{dx}=-xb \dot{h}$$
and integration gives you the pressure profile (the pressure is the most negative at the farthest point from the edge -- for a circular glass that would be at the centre, but we simplified here)
$$p(x)=(x^2-a^2)3\eta\frac{\dot{h}}{h^3}$$
The total force on the bottom of the glass is the integral of this pressure over the surface area:
$$F=b\int p\,d x = -2ba^3\eta\frac{\dot{h}}{h^3}$$
You see this force is proportional to the velocity with which you are raising the glass. There are also additional contributions due to surface tension and in very thin water layers even van der Waals forces. But let's estimate how much this viscosity-driven force could be. Let's say for water $\eta=9\times 10^{-4}{\rm Pa\, s}$, layer thickness $h=0.1{\rm mm}$, a square glass $a=b=100{\rm mm}$. Then, for lifting the glass by $\dot{h}=1{\rm mm/s}$, the force equals $F=180{\rm N}$, which is equivalent to lifting a 18kg weight. Quite a lot of force. Of course, shape of the glass matters, for circular glass, you have a different geometry with water being sucked from all sides, so the prefactor is different, but its probably [something of order of 1]*radius^4 instead of $2ba^3$. The magnitudes will be similar, however. The worst contributor is the gap thickness: once you manage to lift it a bit, the power of 3 means the force gives up very quickly, leading to the "plop" sound and you feeling a sudden jerk when the force suddenly goes away.
For sliding the glass, you need to overcome the viscosity force at your velocity, so $F=\eta a b v/h$, which is $0.09{\rm N}$ for $v=1{\rm m/s}$ - a considerable velocity and negligible force!
