Why dont we consider force exerted by atmospheric pressure in mechanics? My school has started teaching fluid mechanics, and it really bugs me why we don't consider force exerted by atmospheric pressure in mechanics. I couldn't understand a word my teacher said.
There is another particular question that is based on this doubt. Why don't we consider the force exerted by air when we consider something simple like a block lying on a plane surface? Generally we'd say that the forces acting on it are the normal reaction (by the plane surface) and the weight of the block(by earth). Where is the atmospheric pressure?
In this question:


  
*A piece of wood is floating in water kept in a bottle. The bottle is connected to an air pump. Neglect the compressibility of water. When more air is pushed into the bottle from the pump, the piece of wood will float with
(a) larger part in the water
  (b) lesser part in the water
  (c) same part in the water
  (d) it will sink

The answer provided is option (C). I thought it would be (A), because won't the increased pressure press the block more?
 A: As you know, Archimedes' principle states that the buoyant force experienced is equal to the weight of the displaced fluid.
In the case of your incompressible water, the buoyant force from the water experienced by the block of wood is independent of the air pressure as long as the block is immersed to the same extent: any excess pressure to the entire system would act equally on the top and bottom of the block and cancel out.
But wait - there's more.
The question did NOT say to ignore the compressibility of the AIR. If you increase the density of the air (by raising the pressure), the weight of the displaced air increases. This in turn means that the block experiences a greater buoyant force due to the air, and that it will therefore rise (a little bit) in the water in order to find a new "neutral buoyancy" position.
This is counterintuitive: you raise the pressure and the block rises up out of the water. But I'm pretty sure that's correct. The answer should have been (B).
If we ignore the compressibility for a moment and look just at the pressure of the air, we find that it is "all around us"; not only that, but (barring the effect of gravity) it is the same everywhere. As David Hammen pointed out in his answer, sometimes, in the real world, you have to know what to ignore. If you pick up a ball, you don't have to think of it as a bunch of atoms with electrons forming bonds, obeying Schroedinger's equation, ... you can just think of it as a "ball".
In the same way, for most practical purposes we can think of air as something that
- fills every space in our experiment (unless we stop it)
- has very low density
- has a pressure of about 1 kg/cm3
- will provide a small amount of drag to objects moving through it
Most of the time that is all you need to know about air. In the case of your block, if you have a pressure of 1 bar (normal atmospheric pressure) above the block, that same pressure is exerted on the surface of the water - and so at the very top of the water level, the pressure is also 1 bar. As you go deeper in the water, the pressure is even greater, because of the weight of the water "above" the point where you are measuring.
If you increase the atmospheric pressure, the pressure inside the water increases by the same amount. This means that the difference in pressure between top and bottom of the object is independent of the pressure of the air. And the difference is what gives rise to the buoyancy.
Let's look at this picture:

If the block has surface area (top and bottom) $A$ and the atmospheric pressure is $P$, then the force pushing down on the top is
$$F_1 = P\cdot A$$
The pressure at the bottom of the block $P_2 = P + \rho \cdot g \cdot h$ and the force on the bottom is
$$F_2 = P_2 \cdot A = \left(P + \rho \cdot g \cdot h\right) \; A$$
The difference between these forces is what is experiences as buoyancy, and has to equal the weight of the block:
$$m\cdot g = F_2 - F_1 = \rho \cdot g \cdot h \cdot A$$
As you can see, the term $P$ canceled out.
This is true regardless of the shape of the block: it is sufficient to think of the block as made up of many smaller blocks, each of a regular shape, and add up all the forces due to each "blocklet". For each, the $P$ term will cancel out.
Now with some practice you will "know" when you can ignore pressure - until you do, you can (and should) do the more rigorous analysis to convince yourself that you can ignore it.
Knowing what not to do takes a lifetime of learning: but it can make life so much easier...
A: Physics is not just mathematics. An important part of physics is knowing what to ignore, what not to ignore.
If you pumped so much air into the bottle that conditions became downright Venusian, then yes, you would see a small change in the level at which the piece of wood floats. The block would float a tiny bit higher.  You'd see an even greater deviation if the wood was honeycombed through and through, giving the small pressure gradient from the top of the block to the plane of the waterline a better chance to take hold.  However, the bottle is presumably made out of somewhat cheap glass and it will not fare well when pressurized to that extent.
Presumably the block of wood isn't honeycombed through and through, the quantity of air pumped in is small (compared to Venus surface levels), and you have a limited ability to measure the level at which the block of wood floats. In this case, the answer that is closest to being correct is C.
A: Let the block be in original position.Now when a pressure p is given extra it acts on top of block and also at the base in upward direction,since water exerts pressure in all direction by pascals law.The net is zero so option C may be correct.
