Home experiment question I want to make a home experiment where I roughly explain the phenomenon of water remaining in a straw if you close one end of the straw.
So I'm thinking that the weight of the water is pulling it down, but the pressure underneath the straw is keeping it up. If there was no finger on the top, the pressure would also be pushing it down so it would fall.
Would this be a valid way to show this?
Mass of water: volume*density: $\pi r^2*H*1000$
Weight of water: $\pi r^2*H*1000*9.8$
Pressure*area = force
force= $101325*\pi r^2$ (weight of water)
And I somehow want to combine this with the upward force of pressure in the room.
Am I on the right track?
 A: This is not altogether correct. In particular, it does not matter what the pressure of the room is (unless you reach extreme values), because actually what is holding the water column is not some "high" value of the pressure at the bottom of it, it is the fact that the pressure at the top of the water column (at the water-finger interface, or in the air pocket between water and finger) is lower than the atmospheric pressure. It is exactly the room pressure minus $\rho g h$, where $\rho$ is the density of water, $g$ gravitational constant and $h$ the height of the column.
This is true whatever the radius of the straw. However, you need two more ingredients to keep the system stable : 
First, the air pocket should remain incompressible : this is true for small enough water columns, but there'll be some height $h$ above which it will expand under the weight of water. And even if there is no air, if you get to pressures lower than 0 at the top, the phenomenon of cavitation will nucleate a bubble of void that will expand. 
Second, if the straw is too wide, the bottom surface will grow unstable and water will flow e.g. on one side, allowing bubbles to rise on the other. What actually maintains the stability there is a force called surface tension. You can lower it by adding a surfactant (dish washing soap e.g.) to the water, you'll see that the biggest straw possible is then smaller!
A: Yes, what you have formulated is fine. The pressure acting on the water from the bottom of the straw will be equal to the weight of the water times the cross section area. So $101325\pi r^2$ is the force acting from below also. That is precisely why there is an equilibrium and the water is not falling.
A: Weight of water is too more to get hold from bottom just by h$\rho$g pressure but actually when you put thumb you cutoff above pressure and then the pressure at bottom then dominates and hold water in straw.You can find the force now using formula above and it will be right to say that the force just due to pressure on bottom surface is $1.01325*10^5*\pi r^2$
