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Consider a cuboidal block kept on a table with friction coefficient $\mu$. When calculating the friction, we use $f= \mu n$ where $n=mg$ for a horizontal table. But a pressure due to the atmosphere exists whose net effect is a downward force of magnitude $PA$ where $A$ is the surface area of the upper surface. Why isn't this included in calculating the normal reaction force and hence the friction?

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    $\begingroup$ Sometimes you must. Try sliding a pot over a hot ceramic stove and you see that the heat creates a low pressure area under the pot making it stick (i.e. increasing friction). $\endgroup$ – ja72 May 19 '15 at 17:20
  • $\begingroup$ Hi @user117913. Have I answered your inquiry, or do you have any more confusions. If so, please explicate them so I, or someone more qualified (there are lots) can help. $\endgroup$ – Cicero May 20 '15 at 2:04
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First of all, unless you have sealed the cube, there is technically air under the cube as well, and using the equations of buoyancy you can determine that the effects of air pressure is in fact dependent on the hieght of the cube ($PA = \rho g h A$ is a common approximation). Now, for most cubes in consideration, air pressure is an insignificant effect compared with gravitation, but as you can see, with items like paper with large surface areas and small masses air pressure becomes significant. With a cube, the mass is such that the gravitational force is much larger than the air pressure, so air pressure can be ignored. In reality, when dealing with friction we ignore many small effects because we are using approximate models. Air resistance is one of them.

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