One force or two forces in Newton's third law?

[TurtleMeister]

In Newton's third law, do you consider there to be one force or two forces involved? My thinking is that there is only one force which acts equally in magnitude (conservation of momentum) and opposite in direction for each object. However, many sources that I find on the internet claim that there are two forces. Example:

http://www.physicsclassroom.com/class/newtlaws/u2l4a.cfm

These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is:

For every action, there is an equal and opposite reaction.

Doesn't the action and reaction refer to the objects and not the force(s)?

 

[Meir Achuz]

Newton's 3rd law is about two different forces.
"Doesn't the action and reaction refer to the objects and not the force(s)?" NO, they are the two forces.

You give a quote that starts "These two forces...".
How can you think they don't mean two forces?

 

[TurtleMeister]

You give a quote that starts "These two forces...".
How can you think they don't mean two forces?

Because it is the objects that move, not the force(s). I'm thinking of action and reaction in terms of motion.

I got to thinking about this after reading the following post in another thread:

https://www.physicsforums.com/showthread.php?t=465684&page=4

There is only one way to interpret Newton's third law: the forces involved are exactly equal but opposite, they arise from the same interaction, and they act on different bodies.

Highlight is mine.

I realize that this post does not infer only one force. However, if the two forces arise from the same interaction and the two forces are always equal and opposite, then why can't we think of it as the two faces of the same force. In the same way that a coin must have two sides, a force must have at least two objects that it acts on. It may be just a matter of semantics, but it seems like thinking in terms of two separate forces (which seems to be the correct way) is more confusing than simply thinking in terms of the coin analogy.

 

[Meir Achuz]

You sound completely confused. You have misinterpreted DH.
The two forces can be millions of miles apart.

 

[D H]

You sound completely confused. You have misinterpreted DH.

Yep.

By "one interaction" I meant (for example) that the third law counterpart to the Earth's gravitational force on a person is the person's gravitational force on the Earth. There are two forces here (the Earth's gravitational force acting on the person, the person's gravitational force acting on the Earth), but only one type of interaction (gravity).

Another example: the third law counterpart to the Coulomb force exerted by charged particle A on charged particle B is the Coulomb force exerted by charged particle B on charged particle A. Once again we have two equal but opposite forces acting on two bodies, but only one type of interaction. Yet another example: The third law counterpart to the supermassive black hole at the center of the galaxy exerting a gravitational force on the Sun is that the Sun exerts a gravitational force on the supermassive black hole. As clem said, "The two forces can be millions of miles apart."

On the other hand, the Earth pulling a person downward due to gravity and the Earth pushing a person upward are not a third law pair, for a couple of reasons. One reason is that in both cases the force involve the Earth acting on the person. A true third law pair always involves two forces acting on two different objects. The other is that two very different interactions (gravitation and the normal force) are involved here rather than one.

 

[TurtleMeister]

Thanks for the clarification DH. But now I am confused about the Earth and the person standing on the surface not being a third law pair.

On the other hand, the Earth pulling a person downward due to gravity and the Earth pushing a person upward are not a third law pair, for a couple of reasons. One reason is that in both cases the force involve the Earth acting on the person. A true third law pair always involves two forces acting on two different objects. The other is that two very different interactions (gravitation and the normal force) are involved here rather than one.

Thought experiment:
I am in free fall toward an Earth sized object. The object and myself are the only objects. The object accelerates me toward our mutual center of mass. And I also accelerate the object toward our mutual center of mass. The massive object has the exact same momentum as I do. Is this a third law pair? If not then why? If it is, then why does it stop being a third law pair when I am standing on the surface of the object?

Edit:
Ah, I think I misread you again DH. The Earth pulling a person downward due to gravity and the Earth pushing a person upward (who is standing on the surface) is the same thing. Or in other words, it is the earth's side only of the third law pair. The other side of the pair being the person pulling the Earth upward. Is this the correct interpretation?

 

[cephron]

I don't think it (force due to gravity) ever stops being a third law pair. There is simply an additional third law pair, being the normal force.

Before landing:

Planet ->  force of gravity  <- Person

= 1 third law pair of equal-and-opposite forces.

After landing:

Planet ->  force of gravity  <- Person
Planet <-    normal force    -> Person

= 2 third law pairs.

But to say

force of gravity  <- Person ->  normal force

does not constitute a third law pair, even though the forces in question happen to be equal in magnitude and opposite in direction.

I hope my pseudo-visual explanation helps! :)

 

[TurtleMeister]

Thanks for the explanation cephron. But I'm not sure about the 2 third law pairs. Maybe there is only one third law pair, but we may choose to give it more than one name.

The thing that confuses me about the third law, as worded in the quote of the OP, are the words action and reaction. The choice of those words makes me think of the acceleration of the two objects. If it is referring to a force or forces then why isn't it worded "For every force, there is an equal and opposite force"?

 

[JaredJames]

I take it you know the definition of action and reaction and how it applies to forces?

It is "For every force, there is an equal and opposite force":

Third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction". The action and the reaction are simultaneous.

http://en.wikipedia.org/wiki/Newton's_laws_of_motion

http://www.physicsclassroom.com/class/newtlaws/u2l4a.cfm

You push on an object, the object pushes back. Forces come in pairs.

No amount of word play will change this fact.

 

[D H]

Thought experiment:
I am in free fall toward an Earth sized object. The object and myself are the only objects. The object accelerates me toward our mutual center of mass. And I also accelerate the object toward our mutual center of mass. The massive object has the exact same momentum as I do. Is this a third law pair? If not then why? If it is, then why does it stop being a third law pair when I am standing on the surface of the object?

Gravitation in free fall is a third law pair, and it is also a third law pair when you are on the ground. You are pulling gravitationally on the planet just as much as it is pulling on you. Always.

Edit:
Ah, I think I misread you again DH. The Earth pulling a person downward due to gravity and the Earth pushing a person upward (who is standing on the surface) is the same thing. Or in other words, it is the earth's side only of the third law pair. The other side of the pair being the person pulling the Earth upward. Is this the correct interpretation?

No!

cephron has it right in the post that follows yours. When you are standing on the ground there is another third law pair involved in addition to gravity. Multiple third law pairs, in fact. There are quite a few third law pairs when you are standing still on your bathroom scale.

Besides, unless you are at the North or South Pole, the upward normal force is not exactly equal but opposite to the downward gravitational force. The net force would be zero if the normal force was equal but opposite to gravitation. The net force cannot be zero: You are in uniform circular motion about the Earth's rotation axis. Newton's third law does not say that the third law pair are approximately equal but opposite. It says equal but opposite, period. (In other words, the forces are exactly equal but opposite.)

 

[TurtleMeister]

I take it you know the definition of action and reaction and how it applies to forces?

No. I was only familiar with the traditional definition:

Action: the process or state of acting or of being active: The machine is not in action now. Synonyms: movement, operation.

I did not know that the word had special meaning in Newtons third law.

This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction".

So action in Newtons third law actually means force. Thanks for pointing this out. I think I do have a good understanding of how the third law works. But as I pointed out previously, my misunderstanding may be just one of semantics.

cephron has it right in the post that follows yours. When you are standing on the ground there is another third law pair involved in addition to gravity. Multiple third law pairs, in fact. There are quite a few third law pairs when you are standing still on your bathroom scale.

Another misunderstanding I may have had is that in the past I have always associated the third law pair with objects and not forces. So in my thought experiment there is only one third law pair while I am in free fall (the force pulling me and the force pulling the object). But when I touch the surface another pair appears? How does this other pair, or other pairs if there are more, differ from the single pair while I am in free fall?

 

[D H]

Another misunderstanding I may have had is that in the past I have always associated the third law pair with objects and not forces. So in my thought experiment there is only one third law pair while I am in free fall (the force pulling me and the force pulling the object). But when I touch the surface another pair appears? How does this other pair, or other pairs if there are more, differ from the single pair while I am in free fall?

Newton's third law is about objects and forces.

Backing up a bit, what are Newton's laws? As taught by most, Newton's laws leave concepts such as time, mass, and force as undefined terms. (Similarly, point, line, and plane are undefined terms in Euclidean geometry.) I'll delve into those if you want.

Another concept that isn't quite explicit in Newtonian mechanics is the superposition principle. The net force acting on some object is the vector sum of the individual forces acting on the object. Arnold Sommerfeld, nominated for the Nobel prize a record 81 times (he never got it) and teacher of six Nobelists, saw the superposition principle as so important that he called it Newton's fourth law.

With this in mind, consider the forces acting on / exerted by the Earth's Moon. Newton's law of gravitation says that the Moon is subject to gravitational forces from the Earth and the Sun, and also all of the other planets in the solar system and even everything else in the universe. The superposition principle is what let's us add all of these forces together to yield the net force on the Moon. Here's the rub: You aren't going to be able to find an equal but opposite reaction to this net force. You need to look to the individual forces to find those equal but opposite reactions.

Now let's look at something a bit closer to Earth, a vehicle in low Earth orbit. We have the same gravitational third law pairs as before, but now atmospheric drag also comes into play. Once again the superposition principle let's you compute the net force and once again you won't be able to find an equal but opposite reaction to this net force. You must look to the individual forces to find the third law pairs. Atmospheric drag reduces the vehicle's speed with respect to the Earth, so per Newton's third law the vehicle must speed up the atmosphere (by a tiny, tiny bit).

Now let's get down to the Earth and look at a person at the equator standing still on a scale on the rotating Earth. I'm going to ignore the Sun, the Moon, and all the rest of the stuff in the universe. Even ignoring all that, there are still a whole lot of third law interactions to consider. One way to compute the net force is to once again apply the superposition principle. I'll do that later. An alternative is to use Newton's second law, which is a kinematic theorem. The person is in uniform circular motion, rotating at 1 revolution per sidereal day at a distance of 6378 km from the Earth's rotation axis. This corresponds to a centripetal acceleration of 0.033915 m/s². Assuming the person masses 70 kg, the net force on the person is 2.37405 Newtons. Whatever the myriad of forces that act on the person are, we know from kinematics that the net force is necessarily 2.37405 Newtons directed toward the center of the Earth.

So what are the forces acting on the person? Gravity pulls the person toward the center of the Earth, the normal force pushes the person up, and buoyancy also gives a slight upward force. We can compute the buoyant force and gravity, we know the net force, so now we can calculate the normal force. Finally, the third law tells us what the person is doing to the scale, the Earth, and the atmosphere -- but only if we consider the individual components of the net force.

Note that by putting the person on the equator all of these forces, including the net force, are directed toward or away from the center of the Earth. Things get hairier away from the equator.

 

[cephron]

Well, in the case of your person-in-free-fall-towards-planet, the pairs are actually quite nicely defined and separate from each other.

The force of gravity acts from a distance. Your person and planet are far away from each other, but each exerts a force on each other, F_g and -F_g, respectively. These forces are exactly equal and opposite. You might almost say they are the same force, acting bi-directionally*. This is one third-law pair.

The normal force has nothing to do with gravity. It is in fact a fundamentally different force; it is based not on gravity but on electrostatic repulsion: while the person remains in contact with the planet, the electrons in the soles of his feet are repelling the electrons in the substance of the planet's surface, and vice-versa. These repulsive forces are also exactly equal and opposite, giving us F_n and -F_n. But this repulsion is only significant at very close range.

So, at a distance, attractive forces F_g and -F_g (equal and opposite) are noticeable, but the repulsive forces F_n and -F_n are vanishingly insignificant. It might as well not be there at all, as compared with gravity. Note that F_g and F_n, while probably nearly opposite to each other, are by no means equal. Gravity wins, and the planet and person fall towards each other.

Now when the person "makes contact" with the planet (this is, the nearest part of him to the planet is now less than a nanometer** away from the nearest part of the planet to him), suddenly, the electrons are almost on top of each other and normal force turns out to be very strong. Now F_n is roughly equal to F_g, and in this case they might have the appearance of being exactly equal-and-opposite. Fact is that for any given instant, the chances of F_n and F_g being equal to each other are pretty small. They're simply forces in equilibrium with each other. But, just like when the planet and person were far away from each other, F_n is still exactly equal to -F_n, at any instant, because they are a third-law pair. They are "the same force, acting bi-directionally". Same applies to the two sides of the force of gravity.

Sorry for the wall of text. I hope the thousand words were clearer than the picture! ;)

*This is a question of semantics, and I don't honestly know if it's considered correct/good practise to call F_whatever and -F_whatever "the same force, acting bi-directionally". I hope it conveys the idea, though.

**I don't know the actual distance at which electron repulsion becomes what one could call "significant", or how quickly it falls off.

 

[Allen93]

the upward normal force is not exactly equal but opposite to the downward gravitational force.

I want to clarify what he is saying. Normal forces are perpendicular to wherever the contact is between two objects. The force due to gravity is always pointing towards the center of the Earth, so for example, If a stone sat still on an elevated hill, the force due to gravity and normal forces would not be equal, BUT, the Y axis components (if you imagined all the forces on an x/y plane) of the normal force WOULD equal the force due to gravity. :)

 

[D H]

I want to clarify what he is saying.

I thought I was being very clear already!

Normal forces are perpendicular to wherever the contact is between two objects. The force due to gravity is always pointing towards the center of the Earth, so for example, If a stone sat still on an elevated hill, the force due to gravity and normal forces would not be equal, ...

That is *not* what I was talking about.

BUT, the Y axis components (if you imagined all the forces on an x/y plane) of the normal force WOULD equal the force due to gravity. :)

Obviously I wasn't very clear. In saying this you are implicitly assuming
(a) that the net force is zero, and
(b) that the only forces acting on the rock are gravity and surface interactions.

Neither is true. Look at the rock from the perspective of an inertial observer. The rock is undergoing uniform circular motion about the Earth's rotation axis, ergo the net force cannot be zero except for a rock at one of the poles. For a 1 kg rock, the net force is 0.033915 Newtons at the equator, 0.024039 Newtons at 45 degrees latitude. The corresponding values for my 70 kg person are 2.374 and 1.683 Newtons.

Other forces come into play. There's buoyancy, for example. Buoyancy is small but non-zero force. For a 70 kg person with a density of 1 gram/cc at sea-level buoyancy from the air provides an upward force of about 0.86 Newtons.

 

[JaredJames]

The force due to gravity is always pointing towards the center of the Earth,

I'm going to be deliberately awkward here. Wouldn't gravity always point to the centre of gravity (or would it be centre of mass?) (which may or may not be the centre of the earth)?

I'm only mentioning this because although the Earth has things pretty easy, there are plenty of other scenarios where simply taking the centre of an object isn't an option.

 

[Allen93]

I'm going to be deliberately awkward here. Wouldn't gravity always point to the centre of gravity (or would it be centre of mass?) (which may or may not be the centre of the earth)?

I'm only mentioning this because although the Earth has things pretty easy, there are plenty of other scenarios where simply taking the centre of an object isn't an option.

Sorry, I'll clarify. Every object on EARTH, will have a force due to gravity that always point to the center of the Earth's mass.

 

[D H]

I'm going to be deliberately awkward here. Wouldn't gravity always point to the centre of gravity (or would it be centre of mass?) (which may or may not be the centre of the earth)?

By center of gravity I take it that you mean extending all of the gravitational force vectors over the surface of the Earth toward infinity and finding the point at which all those lines intersect. That point doesn't exist.

The Earth is approximately an oblate spheroid. If it was exactly an oblate spheroid, the gravitational force vector would point toward the center of the Earth at all points along the equator and at the poles. For all latitudes other than 0 and ±90° the gravitational force vector does not point to the center of the Earth. If you extend this force vector so that it intersects the equatorial plane you will find that the intersection is at some point away from the center of the Earth.

The Earth is not exactly an oblate spheroid. Geologists have long used gravimetry as a way of looking for buried resources such as oil fields and mineral deposits. Near the edge of the such deposits the gravity vector will point in some direction other than the nominal direction as calculated by the simple oblate spheroid model.

 

[JaredJames]

Ah, thanks DH.

I was thinking along the lines of the Earth not being perfectly uniform so far as mass goes, so gravity wouldn't point specifically to the centre.

If you imagine for example, the southern hemisphere having a slightly higher mass than the northern then the 'centre point' would be slightly more to the south.

Your explanation seems to cover the issue though.

 

[D H]

Sorry, I'll clarify. Every object on EARTH, will have a force due to gravity that always point to the center of the Earth's mass.

Read post #18.

 

[Allen93]

Read post #18.

Ah, well my assumption was derived from the notion that Earth was a perfect sphere, which I now understand is wrong. Thank you for clarifying. Would you mind enlightening me on my post under the 'Academic Guidance' section? Thanks.

 

[TurtleMeister]

Thanks for the reply DH. Your post was informative and interesting. Especially the superposition principle. I will have to read more about that.

You might almost say they are the same force, acting bi-directionally*.

*This is a question of semantics, and I don't honestly know if it's considered correct/good practise to call F_whatever and -F_whatever "the same force, acting bi-directionally".

Well, that is exactly what I was proposing in my first two posts. But I'm sure it would be considered incorrect by the academic world.