Let's say we have a body at rest.
enter image description here

The forces acting on this body are:

  1. gravitational force and
  2. normal force.

The pressure of the air is $1bar = 100kPa $ and lets assume that the surface of the body is $1m^2$.

The force of air acting on this body is : $ P = \frac{F_{AIR}}{S} \implies 100kPa \cdot S = 10^5 N$

At the first glance, this really isn't the force we can neglect so simply.

But the buoyancy of air is acting on the body, right? How can those two forces cancel each other out?


The forces that occur due to pressure always act normal to the surface that they are in contact with. For the guy lying on the ground, there is air pressure on top of him, on the sides of him, and under him. This means that the atmospheric forces pushing him up roughly balance the atmospheric forces pushing him down, leading to practically no net force pushing down on him. However, because atmospheric pressure increases with decreasing altitude, there is a slightly higher pressure at the ground than at the top of the guy's chest, leading to a very slight buoyant force on that guy. The magnitude of that buoyant force for a 70 kg individual is in the range of 0.25 pounds (0.11 kg), so it is normally ignored. Depending on a particular physics problem that you are trying to solve, that small difference may need to be considered.


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