# Why can we neglect the force of air (atmosphere) acting on the body in free-body diagram?

Let's say we have a body at rest.

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?