# Why are Normal forces not constant?

1. Jul 20, 2013

### Jherek

If I put a 1 kg mass on the floor, the floor exerts a 10 N force on it, balancing gravity. If I then replace that mass with a 10 kg one of the same area, the floor now exerts 100 N on it. What is it that makes it exert different forces? I know of Newton's third law of opposite and equal reactions, but that doesn't seem to explain it, it just seems to say 'because it does'. Also, I know that the electrons in the floor and the electrons in the object are repelling each other, but why does the upward force vary?

2. Jul 20, 2013

### Claude Bile

Netwon's 3rd Law = conservation of momentum.

Claude.

3. Jul 20, 2013

### robphy

Instead of a rigid floor, consider a trampoline surface whose tension can be tuned up to approach the rigid floor.

4. Jul 20, 2013

### Staff: Mentor

Newton's third law tells me that if I push on a spring hard enough to compress it, the spring is pushing back on me with an equal force.

You can think of the floor, or any other rigid surface for that matter, as a very very very stiff spring that obeys the spring law $F=K\Delta{x}$ where $\Delta{x}$ is the amount that it is compressed. Exert a 10 newton force on it, and it compresses a tiny amount; exert a 100 newton force on it and it compresses roughly ten times as much.

As for how the electrons repelling each other come into the picture? Well, the closer you push the electrons together, the stronger the repulsive force between the electrons will be, so it stands to reason that the force will increase with the amount of compression. (However, I do have to warn you that the analysis of solid materials at the atomic level is a very complex problem, so what I just said is a very hand-wavy oversimplified explanation. For any macroscopic solid object and over a fairly wide range of forces, just go with the "very very very stiff spring" model).

5. Jul 20, 2013

### robphy

It should probably be noted that
the focus should be on Newton's Second Law from the Free-Body Diagram of the object feeling that normal force and that object's acceleration.

In this case, the normal force is a force of constraint...
which is determined to be:
perpendicular to the surface of contract, with size whatever it needs to be to satisfy Newton's Second Law (and the [constraint] geometry of the problem... e.g. the object must stay on the surface).

If I push down or pull up [by an attached string] on the object, I can change the size of the normal force.

Last edited: Jul 20, 2013
6. Jul 20, 2013

### technician

Newton's third law usually written as action and reaction are equal and opposite and act on different bodies.
Action could be the weight of the mass then reaction would be the normal reaction.

7. Jul 20, 2013

### Staff: Mentor

True.

In most textbooks, "weight" is defined as the gravitational force exerted by the earth on the object; the "reaction" (Newton's 3rd law pair) to that force is of course the gravitational attraction exerted by the object on the earth.

8. Jul 20, 2013

### WannabeNewton

The reaction force due to the weight is just the gravitational force exerted on the Earth; it is not the normal force.

I see Doc Al beat me to it. I must quicken my pace! Tra la laaaa

9. Jul 20, 2013

### technician

Totally agree

10. Jul 20, 2013

### dextercioby

Not quite true. The mass exerts 10 N on the floor and the floor exerts 10 N back, as per the 3rd law. Gravity is not balanced.

11. Jul 20, 2013

### Staff: Mentor

I think he had it right. Gravity exerts a downward force of 10 N; the floor exerts an upward force of 10 N, which balances gravity and produces equilibrium per the 2nd law.

12. Jul 21, 2013

### dextercioby

I was referring to the 3rd law and how the forces are paired.

13. Jul 21, 2013

### Curious3141

Just to add something to make things clearer (hopefully):

Action-reaction forces (as per Newton's Third Law) refer to forces acting on *different* but interacting bodies.

When one is talking about forces balancing each other (or in general, summing forces to find the resultant), these are forces acting on the *same* body.

In this case, there are two action-reaction pairs:

1) Gravitational pull of the earth on the mass (also known as the weight of the mass) - action acting on the mass; gravitational pull of the mass on the earth - reaction acting on the earth.

2) Contact (compressive) force exerted by the mass on the surface of the earth - action acting on the earth surface; normal force exerted by the earth surface on the mass - reaction acting on the mass.

Those action-reaction pairs are what Newton's Third Law predict.

The forces on each object balance out, but to pair the balancing forces, we have to switch them around:

1) Gravitational pull of the earth on the mass (also known as the weight of the mass) vs normal force exerted by the earth surface on the mass: both acting on the mass, equal in magnitude, but opposite in direction, so net force on mass = zero.

2) Gravitational pull of the mass on the earth vs contact (compressive) force exerted by the mass on the surface of the earth: both acting on the earth, equal in magnitude, but opposite in direction, so net force on earth = zero.

So there is no net force on either the earth or the mass, but this cannot directly be attributed to Newton's Third Law, which talks about something else.

14. Jul 21, 2013

### technician

If you have a mass simply resting on th Earth's surface then what you have written seems perfectly reasonable.
In effect the mass is just part of the Earth.
The interesting cases are when the mass is accelerating. In free fall the gravitational forces still act but there is no reaction ( contact) force.
A mass in an accelerating rocket will have reaction ( contact) forces but these will have no clear connection to gravitational forces.
I hope this adds to the post, I hope that I have not misunderstood the latest post.

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