# Normal force and details of the Earth holding up an object

• I
• ago01
Earth? Cause if I had two objects of equal mass but one was twice the size of the other then the larger object would experience four times the gravitational force? Thanks for clearing that up!f

#### ago01

It's a simple application of Newton's third law to show that the Earth indeed does accelerate towards an object as it falls towards earth.

M_o is the mass of the object
M_e is the mass of the earth

From the third law (and ignoring air drag):

M_e * a_e - M_o*g = 0 (with a up-positive coordinate system)

Then

a_e = (m_o/m_e)g

Impossibly small but non-zero. Fascinating.

But I am a little confused then on the reality of the normal force and want to make sure I am clear on it. The normal force is force exerted perpendicular to a surface and is equal, but opposite in direction, to the force imposed on the surface. So the object stays put (in that direction) by virtue of the fact that if this wasn't true then the forces would be imbalanced and the object would rocket off the surface or fall into it.

But take the example of a person standing on some ground to make a scenario. Clearly, this is the earth. It's not a table, or chair, or the side of a wall you are pressing on. Why is it the case the normal force would be equal to M_o*g, and not for example, M_e * a_e? Or to use the chair example, why wouldn't it be M_chair * a_chair?

I am probably just trying to confuse myself more but after learning about energies and going back to read over forces again some of things I just took for granted I'd like to at least have some answer to...if not just to make myself feel better.

Why is it the case the normal force would be equal to M_o*g, and not for example, M_e * a_e?
Both objects are in a force balance, so the normal force acting on the object is simply the negative of the gravitational force.
Note that there is no acceleration if the object rests on the surface.

Lnewqban
It's a simple application of Newton's third law to show that the Earth indeed does accelerate towards an object as it falls towards earth.

M_o is the mass of the object
M_e is the mass of the earth

From the third law (and ignoring air drag):

M_e * a_e - M_o*g = 0 (with a up-positive coordinate system)

Then

a_e = (m_o/m_e)g

Impossibly small but non-zero. Fascinating.

But I am a little confused then on the reality of the normal force and want to make sure I am clear on it. The normal force is force exerted perpendicular to a surface and is equal, but opposite in direction, to the force imposed on the surface. So the object stays put (in that direction) by virtue of the fact that if this wasn't true then the forces would be imbalanced and the object would rocket off the surface or fall into it.

But take the example of a person standing on some ground to make a scenario. Clearly, this is the earth. It's not a table, or chair, or the side of a wall you are pressing on. Why is it the case the normal force would be equal to M_o*g, and not for example, M_e * a_e? Or to use the chair example, why wouldn't it be M_chair * a_chair?

I am probably just trying to confuse myself more but after learning about energies and going back to read over forces again some of things I just took for granted I'd like to at least have some answer to...if not just to make myself feel better.
The normal force is ##\frac{GM_em}{R^2}##, where ##M_e## is the mass of the Earth, ##G## the gravitational constant, ##R## the radius of the Earth and ##m## the mass of the object.

The quantity ##g = \frac{GM_e}{R^2}## is the gravitational acceleration at the Earth's surface.

Lnewqban and vanhees71
Both objects are in a force balance, so the normal force acting on the object is simply the negative of the gravitational force.
Note that there is no acceleration if the object rests on the surface.

I see, so it really is just a consequence of the fact that forces balance and so it must be that it's equal to the gravitational force in the opposite direction.

The normal force is ##\frac{GM_em}{R^2}##, where ##M_e## is the mass of the Earth, ##G## the gravitational constant, ##R## the radius of the Earth and ##m## the mass of the object.

The quantity ##g = \frac{GM_e}{R^2}## is the gravitational acceleration at the Earth's surface.

Interesting. I haven't learned about universal gravitation so this might be a dumb follow up but... I assume that this will equal ##mg##?

But I am a little confused then on the reality of the normal force and want to make sure I am clear on it. The normal force is force exerted perpendicular to a surface and is equal, but opposite in direction, to the force imposed on the surface. So the object stays put (in that direction) by virtue of the fact that if this wasn't true then the forces would be imbalanced and the object would rocket off the surface or fall into it.
Just to be clear: The normal force exerted upward by the surface on the object is always equal and opposite to the force that the object exerts on the surface. Those forces are third law pairs and are always equal and opposite. This is true regardless of whether the object is in equilibrium or not.

The object will be in equilibrium if the forces on the object balance. Those forces are gravity and the upward normal force. Those forces are not third law pairs.

jbriggs444, ago01 and Lnewqban
Those forces are gravity and the upward normal force. Those forces are not third law pairs.

Could you clarify how they are not third law pairs? The normal force is the force opposite the surface. So it's the same force as the gravitational force (in this example), but in opposite direction. Is it because the normal force isn't a reaction force? I guess I'm confused (again) sorry.

It's a simple application of Newton's third law to show that the Earth indeed does accelerate towards an object as it falls towards earth.
...
Why is it the case the normal force would be equal to M_o*g, and not for example, M_e * a_e? Or to use the chair example, why wouldn't it be M_chair * a_chair?
...
You can consider the chair as an object that already fell towards the Earth and collided against its surface, remaining there.
Just like meteorites, dust, organic matter, etc. did before, forming the outmost layer of the planet.
All those things, together with the rest of the planet, accelerate at the same "impossibly small" rate towards our falling object.

In that scenario, our falling object is the only one with freedom to move towards the center of the planet, induced by the gravitational force.
The surface against which our object will collide and eventually will repose on, may be, or not, perpendicular to that gravitational force.
If it does, we can call that force normal to the surface, which will be an important concept when you study friction.
If it does not, then the magnitude of the normal force (to the surface) will be less than the magnitude of the gravitational force, and you will have to calculate that value and direction based on trigonometry.

Could you clarify how they are not third law pairs? The normal force is the force opposite the surface. So it's the same force as the gravitational force (in this example), but in opposite direction. Is it because the normal force isn't a reaction force? I guess I'm confused (again) sorry.
The gravitational force of the Earth on the object and the gravitational force of the object on the Earth are third law pairs.

The two normal forces are also a third law pair.

The fact that the normal force and gravitational force in this case have the same magnitude does not make them a third law pair.

The gravitational force of the Earth on the object and the gravitational force of the object on the Earth are third law pairs.

The two normal forces are also a third law pair.

The fact that the normal force and gravitational force in this case have the same magnitude does not make them a third law pair.

I see, so my confusion is that they have the same magnitude. So there's actual 4 forces and 2 pairs.

There is the gravitational force (the same force experienced while falling). This is a 3rd law pair when you pair it with the Earth being pulled towards the falling object (as described in my post). On the surface you are pushing down as hard as the Earth is pushing up so everything is in balance.

There is also the normal force, which is a contact force. The force you are contacting the surface with is equal to ##-mg##, and so (through some method) the surface applies an equal and opposite contact force on you equal to ##mg##. It just so happens the normal force equals your gravitational force in magnitude.

But these two forces themselves are not third law pairs. They are different. One is a contact force, and one is not.

Wow...I'm really glad I asked this question. I think I understand it. Thank you!

As a side note I found this paper:

https://eric.ed.gov/?id=EJ1148872

That says that a lot of new physics students can't identify these forces properly. Wonder why.

Lnewqban and PeroK
Yes, the normal forces - if you think about it - are fundamentally electromagnetic in nature.

nasu
Could you clarify how they are not third law pairs? The normal force is the force opposite the surface. So it's the same force as the gravitational force (in this example), but in opposite direction. Is it because the normal force isn't a reaction force? I guess I'm confused (again) sorry.
I think @PeroK has straightened you out, but I'll chime in. 3rd law pairs must be the same sort of force. Since the upward normal force is a contact force (electromagnetic in its details) so must the downward force on the surface be a contact force. The third law pair to the gravitational force on the object (the Earth exerting a gravitational pull on the object) must also be a gravitational force: it's the gravitational pull of the object on the earth.

ago01, Lnewqban and PeroK
It just so happens the normal force equals your gravitational force in magnitude.
Yes, in this special case this follows from Newtons 2nd Law: If the object is not accelerating and those two are the only forces acting on it, then it follows that they are equal but opposite.

In contrast, the 3rd Law paired forces are always equal but opposite, regardless of acceleration or existence of other forces. And they are of the same type of interaction, and between the same two objects.

As a side note I found this paper:

https://eric.ed.gov/?id=EJ1148872

That says that a lot of new physics students can't identify these forces properly. Wonder why.

Even the NASA website confuses Newton's 2nd and 3rd Laws:
https://www.grc.nasa.gov/WWW/k-12/WindTunnel/Activities/third_law_motion.html

NASA said:
The book lying on the table is exerting a downward force on the table, while the table is exerting an upward reaction force on the book. Because the forces are equal and opposite, the book remains at rest.

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nasu and PeroK
But I am a little confused then on the reality of the normal force and want to make sure I am clear on it. The normal force is force exerted perpendicular to a surface and is equal, but opposite in direction, to the force imposed on the surface.
The surface exerts a force because the surface is deformed. The force is an attempt to return the surface to its former shape. Note that the force has two components. One component is perpendicular to the surface and is called the normal force. The component parallel to the surface is called friction.

Lnewqban