Newton's 3rd law and force equilibrium

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Discussion Overview

The discussion centers around the concepts of force equilibrium and Newton's 3rd law as they apply to a rock resting on the ground, exploring the interactions between gravitational and normal forces. Participants also touch on related questions involving spring forces and equilibrium configurations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the relationship between gravitational force and normal force acting on a rock at rest, questioning the reaction to the normal force.
  • Another participant clarifies that there are two forces acting on the rock: gravitational attraction by the Earth and the normal force from the ground, each having corresponding reaction forces.
  • Some participants argue that gravity and normal force are not part of a 3rd law action-reaction pair, as they act on the same body, while others assert that they are equal and opposite forces.
  • A later reply introduces the idea that molecular rigidity affects the interaction between the rock and Earth, suggesting that electrostatic repulsion plays a role.
  • Participants discuss a related question about calculating the compression of a spring between two mass points, emphasizing the need to balance forces and consider the spring's relaxed length.
  • There is a clarification that the force in the spring formula pertains to one side, and the existence of forces on both sides of the spring is acknowledged as necessary for equilibrium.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the classification of gravitational and normal forces as 3rd law pairs, with no consensus reached on this point. The discussion on spring compression also reveals varying interpretations of how to approach the problem.

Contextual Notes

Participants note the importance of understanding the forces acting on the same body versus those acting on different bodies in the context of Newton's 3rd law. Additionally, the discussion on spring compression highlights the need for specific parameters to be defined for accurate calculations.

greypilgrim
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Hi.

Assume a rock lying on the ground. Furthermore, let's assume both the rock and Earth to be rigid bodies. I'm a bit confused about force equilibrium and Newton's 3rd law here. There's Earth's gravity acting on the rock, and by Newton's 3rd law (or just by looking at the law of gravity) the rock's gravity acts on earth. But on the other hand, since the rock is at rest, the gravitational force pulling it towards the c.o.m. of Earth needs to be balanced by a normal force of the ground acting on the rock (which is probably due to electrostatic interactions between the two objects).

So what is the reactio to this normal force of the ground on the rock? Isn't it the rock pushing on the ground because of gravity? But then this force would have two opposing forces, first the attraction of the c.o.m. of the Earth and second the normal force of the ground. I guess there's a flaw in my train of thought, but where?

Thanks
 
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Yes. There are two forces on the rock: The gravitational attraction by the Earth + the normal force by the ground. There are reactions to these two, the gravitational attraction by the rock on the Earth + the normal force exerted (downwards) by the rock on the ground. The two gravitational forces are an action-reaction pair, and the two nrmal forces are an action-reaction pair.
 
greypilgrim said:
Hi.

Assume a rock lying on the ground. Furthermore, let's assume both the rock and Earth to be rigid bodies. I'm a bit confused about force equilibrium and Newton's 3rd law here. There's Earth's gravity acting on the rock, and by Newton's 3rd law (or just by looking at the law of gravity) the rock's gravity acts on earth. But on the other hand, since the rock is at rest, the gravitational force pulling it towards the c.o.m. of Earth needs to be balanced by a normal force of the ground acting on the rock (which is probably due to electrostatic interactions between the two objects).

So what is the reactio to this normal force of the ground on the rock? Isn't it the rock pushing on the ground because of gravity? But then this force would have two opposing forces, first the attraction of the c.o.m. of the Earth and second the normal force of the ground. I guess there's a flaw in my train of thought, but where?

Thanks

It's not clear to me why you're getting confused. Gravity and the normal force are the equal and opposites that Newton's 3rd law demands.

If there were no gravity, then the rock would sit on the Earth without exerting a force and the normal force would be zero.

Note that bodies are not rigid at the molecular level, so gravity will have deformed the rock and the Earth to some degree, which is where the electrostatic repulsion comes from.
 
Chandra Prayaga said:
The two gravitational forces are an action-reaction pair, and the two nrmal forces are an action-reaction pair.

PeroK said:
Gravity and the normal force are the equal and opposites that Newton's 3rd law demands.

Those two statements seem to contradict each other, don't they?
 
PeroK said:
It's not clear to me why you're getting confused. Gravity and the normal force are the equal and opposites that Newton's 3rd law demands.

.
No, gravity and normal force are not part of a 3rd law pair.
The fact that gravity on the rock is balanced by the normal force on the rock does not make these two a pair in the sense of 3rd law.
The fact that they act on the same body is a hint. The forces in a 3-rd law pair newer act on the same body.
 
greypilgrim said:
Those two statements seem to contradict each other, don't they?
nasu said:
No, gravity and normal force are not part of a 3rd law pair.
The fact that gravity on the rock is balanced by the normal force on the rock does not make these two a pair in the sense of 3rd law.
The fact that they act on the same body is a hint. The forces in a 3-rd law pair newer act on the same body.

Very true!
 
I think I get it now, I forgot the normal force of the rock.

A related question: Assume two mass points separated by a spring of some spring constant. How do I correctly calculate how much the spring is compressed?
 
Why is the spring compressed?
 
Because the two mass points attract each other.
 
  • #10
You need to know the spring constant. Then you can balance the gravitational attraction with the force exerted by the compressed spring. This gives the equilibrium configuration.
 
  • #11
You also need the length of the spring in the relaxed (uncompressed) state.
 
  • #12
Yes, suppose those quantities are given. What's unclear to me is if I need to consider the forces from both sides on the spring?
 
  • #13
greypilgrim said:
What's unclear to me is if I need to consider the forces from both sides on the spring?
No, the F in the spring formula is the force at one side.
 
  • #14
Correct. In fact, there are always two forces, one on each side of the spring, otherwise, the spring itself would run away.
 

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