Action-reaction pairs, concerning gravitational force &normal reaction

[FieldvForce]

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?

 

[Doc Al]

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.

 

[FieldvForce]

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.

 

[Doc Al]

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.)

 

[FieldvForce]

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.

 

[Doc Al]

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'.

 

[FieldvForce]

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?