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Let us consider a cup that I'm holding vertically in my hand

I physically don't go through the cup because of the pauli's exclusion principle where electron with opposite wave function(spin)s dont want to be near each other and also due to the electrostatic repulsion of electrons and protons.

Then comes the case where the cup doesn't fall from my hand , it's due to frictional force due to Van der waals force.

How does the Van der waals force act if there is a resistance due to pauli's exclusion principle and both act in opposite directions and do we account for the net force by subtracting the effective "force" from pauli's exclusion principle?

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  • $\begingroup$ the "force" from pauli's exclusion is electromagnetic in nature so you just have 2 electromagnetic forces just acting in opposite direction I suppose $\endgroup$
    – Razz
    Commented Apr 30, 2023 at 16:24

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I'm not sure it's correct to attribute friction to van der Walls forces. VDW cause self-attraction in gases and liquids in non-polar molecules, especially large ones. Friction between solid surfaces is generally due to roughness in the surfaces, tiny hills and valleys resisting each other via perpendicular surface contact. So it is ultimately electromagnetic repulsion, the same as what causes the cup to resist deformation from your hand.

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  • $\begingroup$ Yes , I must have not used Van der Waals force for the electrostatic force of attraction(In this case gravity pulls the cup down and the force of friction acts accordingly and hence the cup doesn't fall , but my main question is do we account for "net force" via Pauli's exclusion and repulsion and attraction so that the remaining force cancels gravity out $\endgroup$
    – Naveen V
    Commented Apr 30, 2023 at 12:41

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