When a person is standing still, the action of the Earth on their body is a gravitational force. What is the reaction force? Is it the person's weight acting on the Earth?

  • $\begingroup$ Yes It called normal force. Its an electromagnetic force. Read resnick halliday for details. $\endgroup$ – Prathyush Nov 14 '16 at 15:58
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    $\begingroup$ The normal force of the ground on the person is not the third law reaction to the gravitational force. Reaction forces are always of the same type. In this case: the Earth pulls gravitationally on the person, the person pulls gravitationally on the Earth. $\endgroup$ – Paul T. Nov 14 '16 at 15:59
  • $\begingroup$ @Prathyush Makes a common mistake. PaulT is correct. I emphasize this for any novices who might be reading. $\endgroup$ – garyp Nov 14 '16 at 16:08
  • $\begingroup$ @PaulT. is correct my bad. I got confused. $\endgroup$ – Prathyush Nov 14 '16 at 16:19
  • $\begingroup$ related physics.stackexchange.com/q/45653 $\endgroup$ – Paul T. Nov 15 '16 at 1:14

Newton's third law applied to gravity

Newton's 3rd law is always true in classical physics. For a person standing on the ground the Earth pulls on the person with gravity, so the person pulls on the the Earth with gravity. 3rd law force pairs are always of the same type (in this case gravity) and always act on different objects. For your example one acts on the person, and the other acts on the Earth.

$$ \vec{F}_\mathrm{Earth\rightarrow person} = - \vec{F}_\mathrm{person\rightarrow Earth} \quad (3^\mathrm{rd}\,\mathrm{law})$$

What about the normal force?

The normal force pushing up on the person prevents them from falling. This force contributes to the net force acting on the person. To understand it we must consider Newton's 2nd law, $\vec{F}_\mathrm{net} = m \vec{a}$. If the person is not accelerating, they have zero net force acting on them. All of the forces in a Newton's 2nd law statement act on the same object.

$$ \vec{F}_\mathrm{Earth\rightarrow person} + \vec{N}_\mathrm{ground\rightarrow person} = m_\mathrm{person}\, \vec{a}_\mathrm{person} = 0 $$

$$ \vec{F}_\mathrm{Earth\rightarrow person} = - \vec{N}_\mathrm{ground\rightarrow person} \quad (2^\mathrm{nd}\,\mathrm{law})$$

The equality of the normal force and the gravitational force depends on there being zero acceleration. If the person is accelerating, then the normal force is not equal to the gravitational force. You can experience this at the start or end of an elevator ride, when you speed up and slow down.

By Newton's third law the ground pushes with a normal force on the person, so the person pushes with a normal force on the ground.

$$ \vec{N}_\mathrm{ground\rightarrow person} = -\vec{N}_\mathrm{person\rightarrow ground} \quad (3^\mathrm{rd}\,\mathrm{law})$$

It's worth noting that the name 'normal force' refers to forces that push perpendicularly to a surface. 'Normal' is a mathy word for perpendicular. The ground pushes perpendicularly relative to the surface of contact (the person's feet). Anytime you push on something, you apply a normal force to the surface.

  • $\begingroup$ The third law is not always true. In the conventional statement it is violated in Lorentz force interactions, notably the force between two moving charged particles. However, the more "important", and more general law, conservation of momentum, does hold for electromagnetic interactions once you include the momentum of the field itself. I don't know how one would modify the conventional statement of the third law to cover electromagnetic interactions, as it introduces a third entity, the field. $\endgroup$ – garyp Nov 14 '16 at 16:24
  • $\begingroup$ True... I added a `classical physics' caveat. Although, that seems above the scope of this question. $\endgroup$ – Paul T. Nov 14 '16 at 16:27
  • $\begingroup$ I agree that the details are out of scope, but a novice reading this might glom on to the word always. I don't think "classical" is quite the right restriction, as it can be argued that field energy/momentum is classical. The third law fails for the Lorentz force even at low speed. Maybe "Newtonian (i.e., mechanical, not including electromagnetic interactions)" would be a better restriction? $\endgroup$ – garyp Nov 14 '16 at 16:33
  • $\begingroup$ @garyp - That's right. The electromagnetic field is also classical. What about the 3rd law in the case of a classical field force? $\endgroup$ – freecharly Nov 14 '16 at 17:34
  • $\begingroup$ @freecharly Third law is basically equivalent to momentum conservation. So if you have a system has a translation in-variance you can look the the no-ether charge corresponding to translation invariance. I am not sure about this with out doing the calculation involved, but you maybe able to rewrite this in a force form. $\endgroup$ – Prathyush Nov 14 '16 at 20:47

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