Action-reaction pairs, concerning gravitational force &normal reaction

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Discussion Overview

The discussion revolves around the concepts of action-reaction pairs in the context of gravitational force and normal reaction forces, particularly focusing on the scenario of a person standing at rest on the surface of the Earth. Participants explore the implications of these forces on the state of rest and the resultant forces involved.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant questions how they can remain at rest on the Earth's surface despite the gravitational force acting on them, suggesting a resultant force greater than zero due to the interaction with the ground.
  • Another participant asserts that the net force on the individual is zero because the gravitational force downward is balanced by the normal force upward, clarifying that these forces are not action-reaction pairs.
  • Some participants emphasize that the action-reaction pair exists between the person and the ground, where the person pushes down on the ground and the ground pushes up on the person, but these forces do not cancel each other out as they act on different bodies.
  • There is a discussion about the implications of Newton's laws, with one participant expressing confusion about how stationary objects can obey Newton's third law while appearing to contradict the first law.
  • Participants clarify that the normal reaction force is equal to the gravitational force, leading to a net force of zero on the person, but they also explore the concept of resultant forces between the person and the Earth.

Areas of Agreement / Disagreement

Participants express differing views on the nature of action-reaction pairs and their implications for forces acting on a stationary object. There is no consensus on the interpretation of these forces and their roles in maintaining equilibrium.

Contextual Notes

Some participants highlight the need for clarity regarding the definitions of forces involved, particularly distinguishing between gravitational forces and normal forces, as well as the conditions under which action-reaction pairs operate.

FieldvForce
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I would like to know how it is possible that I can be at rest on the surface of the Earth (relative the the Earth of course).

My main cause for confusion is the action reaction pair that occurs when I stand on the surface of any solid material.

The normal reaction nornal reaction a-r pair ensures that I don't pass through the ground and that the ground doesn't pass through me, however would that not mean that there is a resultant force greater 0.

I'll elaborate, say I weigh 80kg and so I am being pulled toward the centre of the Earth with a force of just under 800N, the Earth is being pulled toward me with a force of 800N also, though my acceleration is greater than that of the Earth's; the normal reaction - normal reaction a-c pair cancels out and the Earth and I approach the other's center with a resultant force of 1600N between us, yet I remain at rest, why is this?
 
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You remain at rest--unaccelerated--because the net force on you is zero. The forces on you are the gravitational pull of the Earth pulling you down and the normal force from the floor pushing you up. These forces are not 'action/reaction' pairs.
 
Doc Al said:
You remain at rest--unaccelerated--because the net force on you is zero. The forces on you are the gravitational pull of the Earth pulling you down and the normal force from the floor pushing you up. These forces are not 'action/reaction' pairs.

There is a action reaction pair between myself and the ground, the normal reaction from myself that results from the normal reaction of the ground, this pair itself is brought about indirectly by gravity in this particular situation.
 
FieldvForce said:
There is a action reaction pair between myself and the ground, the normal reaction from myself that results from the normal reaction of the ground, this pair itself is brought about indirectly by gravity in this particular situation.
Yes, there is an action-reaction pair: You push down on the ground and the ground pushes up on you. What about it? (Neither of those is a gravitational force, of course.)
 
Doc Al said:
Yes, there is an action-reaction pair: You push down on the ground and the ground pushes up on you. What about it? (Neither of those is a gravitational force, of course.)

Well I just meant surely that action reaction pair is what is responsible for the resultant force of 0N.

What remains is the resultant force of 1600N bringing my centre and the Earths centre together.
 
FieldvForce said:
Well I just meant surely that action reaction pair is what is responsible for the resultant force of 0N.
No, not at all. Action/reaction forces act on different bodies. Only one of those forces acts on you, so you need another force acting on you--gravity--to produce a net force of zero.

Action/reaction pairs, since they act on different bodies, never 'cancel out'.
 
Doc Al said:
No, not at all. Action/reaction forces act on different bodies. Only one of those forces acts on you, so you need another force acting on you--gravity--to produce a net force of zero.

Action/reaction pairs, since they act on different bodies, never 'cancel out'.

Ah, thanks a lot that really quashes the paradoxical notion of stationary objects obeying Newtons third law but not his first.

Just to delineate.

Gravity is pulling me towards the center of the Earth, as a result I meet with the ground (without sufficient KE to break through) with a force of 800N.

The normal reaction of the ground is 800N and so the net force acting on me is 800 - 800N

The normal reaction normal reaction a-r pair is comprised of my normal reaction against the ground (so that it doesn't pass through me) and it's normal reaction against me (so that I don't pass pass through it).So taking down as the positive direction.We have F (me) =800N (i.e mg) - 800N (i.e normal)
We have F (ground) = not 0 or?
 
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