Exact definition of Newtons III Law Watch

Choochoo_baloo
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Hello all,

Just asking to be 100% sure.

Does the definition of Newton's III as:
"For every force in nature, there is an equal and opposite force"

apply to all forces irrespective whether the two objects (whatever size/mass/charge etc.) physically touch/collide?
I ask because traditional AS textbooks loosely apply it to collisions of billiard balls etc.

I assume it applies to the action of ALL forces, even without physical contact. For example two bodies with mass with no other forces acting.

e.g. a planet and small pebble (no atmosphere). From Gravitational force eqn AND Newton's III, we get say F N on each body, but assuming planets mass as infinite comapred to pebble, F/~infinity = ~ 0 ms^-2, thus it appears the planet does not move whilst the pebble clearly does.

Thank you!
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JosephML
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Newton's third law applies to all forces (gravitational, electrostatic etc)
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Joinedup
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(Original post by Choochoo_baloo)
Hello all,

Just asking to be 100% sure.

Does the definition of Newton's III as:
"For every force in nature, there is an equal and opposite force"

apply to all forces irrespective whether the two objects (whatever size/mass/charge etc.) physically touch/collide?
I ask because traditional AS textbooks loosely apply it to collisions of billiard balls etc.

I assume it applies to the action of ALL forces, even without physical contact. For example two bodies with mass with no other forces acting.

e.g. a planet and small pebble (no atmosphere). From Gravitational force eqn AND Newton's III, we get say F N on each body, but assuming planets mass as infinite comapred to pebble, F/~infinity = ~ 0 ms^-2, thus it appears the planet does not move whilst the pebble clearly does.

Thank you!
well yes it holds when there is no contact eg gravity , electromagnetism etc.

The pebble accelerates towards the CoM of the pebble - earth system with a greater acceleration than the earth does because it has a lower mass. In practice you'd usually assume the pebble does all the falling and the earth stays put because the movement of the earth is unmeasurably small.

Start using >> and << (very much greater / lesser than) and definately avoid saying you assume any mass is infinty in exams though. If you assume the mass of earth is the mass of a pebble multiplied by infinity you could end up calculating an infinite gravitational field strength.
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Choochoo_baloo
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Thanks, nice to be confirmed!

Another problem:

How can a particle, on a SMOOTH inclined plane, with the plane on the Earth's surface, be explained by Newton's III Law?

I see it as follows:
By definition a normal force (albeit a component of the reaction force created through Newton's III) exists due to the particle causing a reaction on the Earth/plane.

But, the model I so far 'believed' was that there is weight on the particle vertically down, and normal perpendicular to plan surface. These are the ONLY two forces.

The traditional explanation of acceleration down slope due to weight component parallel to plane makes sense from a mechanics view, but I would think that actually from a strict Newton's III, the weight has no components, but how can the particles down slope acc, be explained from components of the reaction (ie due Newton's III) - of which Normal is one??

Thanks in advance.
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JosephML
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(Original post by Choochoo_baloo)
Thanks, nice to be confirmed!

Another problem:

How can a particle, on a SMOOTH inclined plane, with the plane on the Earth's surface, be explained by Newton's III Law?

I see it as follows:
By definition a normal force (albeit a component of the reaction force created through Newton's III) exists due to the particle causing a reaction on the Earth/plane.

But, the model I so far 'believed' was that there is weight on the particle vertically down, and normal perpendicular to plan surface. These are the ONLY two forces.

The traditional explanation of acceleration down slope due to weight component parallel to plane makes sense from a mechanics view, but I would think that actually from a strict Newton's III, the weight has no components, but how can the particles down slope acc, be explained from components of the reaction (ie due Newton's III) - of which Normal is one??

Thanks in advance.
What do you mean the weight doesn't have components? It does have components, parallel and perpendicular to the plane. The normal reaction is as a result of N3L and the component of weight perpendicular to the plane. The component parallel to the slope causes the acceleration. As the particle moves down the slope it gets closer to the CoM of the system (earth-particle) likewise the earth will accelerate towards the particle.

Edit: I say it does have components, what I should say is we can turn it into components giving an equivalent effect.
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Choochoo_baloo
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(Original post by JosephML)
What do you mean the weight doesn't have components? It does have components, parallel and perpendicular to the plane. The normal reaction is as a result of N3L and the component of weight perpendicular to the plane. The component parallel to the slope causes the acceleration. As the particle moves down the slope it gets closer to the CoM of the system (earth-particle) likewise the earth will accelerate towards the particle.

Edit: I say it does have components, what I should say is we can turn it into components giving an equivalent effect.
I am still unsure I'm afraid.

You have echoed what most A Level book say.

I thought that the reaction force, by definition of Newton's III, must be equal and opposite to the initial force. In this case weight of particle, equal+opposite to reaction from slope/Earth. But in this frictionless slope model, you divide up the weight up into hypothetical components (as I have been doing for years!), surely the weight (force exerted on particle by Earth/slope), and the normal are NOT equal+opposite, no matter if , as you say, divide the weight into imaginary parallel/perpendicular components.

I guess this is a conceptual misunderstanding I have, rather than mathematical!

Thanks in advance.
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JosephML
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(Original post by Choochoo_baloo)
I am still unsure I'm afraid.

You have echoed what most A Level book say.

I thought that the reaction force, by definition of Newton's III, must be equal and opposite to the initial force. In this case weight of particle, equal+opposite to reaction from slope/Earth. But in this frictionless slope model, you divide up the weight up into hypothetical components (as I have been doing for years!), surely the weight (force exerted on particle by Earth/slope), and the normal are NOT equal+opposite, no matter if , as you say, divide the weight into imaginary parallel/perpendicular components.

I guess this is a conceptual misunderstanding I have, rather than mathematical!

Thanks in advance.
Well the weight isn't force exerted on the particle by the slope that is the normal/reaction. The weight is the force exerted by the particle on the slope. The weight and normal are not equal and opposite, but the component of the weight perp. to the slope and the normal are...

Is this what you mean?
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Choochoo_baloo
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The weight is the force exerted by the particle on the slope. The weight and normal are not equal and opposite
This is my point. Firstly the sloped and Earth are treated as one huge mass.
The weight (in strict reference to the force due to gravity on particle vertically downwards) is NOT the force exerted by the particle on the slope/Earth!!
From any simple free body diagram of Newton's III, considering a free falling particle towards Earth's surface, the weight refers to the gravitational attraction between Earth and particle - but strictly in reference to one of the two equal+opposite forces - the one Earth/slope exerts on particle, not as you said I'm afraid.

You then confirm that the force diagram we are all familiar with in particle/smooth inclined plane, shows the pair of forces, assumed to be duee to Newton's III, NOT equal and opposite.

Further to my second point, even in the simplest of cases, such as two masses in free space, they have equal+opposite attractive forces on each other, towards each other. In this simple application of Newton's III, each still has a resultant force and therefore have non-uniform motion (Newton's II).

So I am thoroughly confused!
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JosephML
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(Original post by Choochoo_baloo)
This is my point. Firstly the sloped and Earth are treated as one huge mass.
The weight (in strict reference to the force due to gravity on particle vertically downwards) is NOT the force exerted by the particle on the slope/Earth!!
From any simple free body diagram of Newton's III, considering a free falling particle towards Earth's surface, the weight refers to the gravitational attraction between Earth and particle - but strictly in reference to one of the two equal+opposite forces - the one Earth/slope exerts on particle, not as you said I'm afraid.

You then confirm that the force diagram we are all familiar with in particle/smooth inclined plane, shows the pair of forces, assumed to be duee to Newton's III, NOT equal and opposite.

Further to my second point, even in the simplest of cases, such as two masses in free space, they have equal+opposite attractive forces on each other, towards each other. In this simple application of Newton's III, each still has a resultant force and therefore have non-uniform motion (Newton's II).

So I am thoroughly confused!
I, too, am very confused. I think that this is the point where it may be better to ask a physicist!
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FireGarden
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I'm not terribly sure I understand your issue, so I shall phrase it as I understand it, and answer anyway incase it happens to be useful.

Problem

Given a mass on a smooth inclined plane, we would then draw three arrows: The weight of the mass, the reaction force of the plane on the mass, and the resultant force (also the direction of motion) going down the plane. The issue we have is that by Newtons Law, we are expecting the reaction to be opposite to weight, although it clearly isn't for an inclined plane. Why is this?

Answer

You are not considering the correct pairs of forces! When we look at weight, this is the force due to gravity the particle experiences. The reaction force we know to exist by Newtons third law, is the weight of the Earth as according to the mass! (which is equal in magnitude, but opposite in direction). The way to think of this lies in the symmetry of the situation. The earth is attracting the mass, so the mass has weight according to the earth. Conversely, the mass is attracting the earth, so the earth has a weight according to the mass! In our situation above, we only talk of weight from the earths perspective, and so we are ignoring the (newtons third law) reaction force.

The reaction force on the plane is a result of the weight, but is not the reaction force to the weight of the mass. The mass is unable to move through the plane (obviously.. it is in the way!). It can only do this, though, because the plane is providing a force to stop the mass moving through the plane. In some situations the plane cannot provide enough force to stop the mass going through the plane.. in this case, the plane breaks! The reaction force is just the force the plane has on the mass to stop it penetrating the plane; and this is equal and opposite to the force the mass is trying to penetrate the plane with. Which will be the component of its weight normal to the plane. The weight is not the force the plane opposes, it opposes the force on the plane the mass can provide because it has weight. It's not as direct as simply "the weight".

I hope this helps; but please don't hesitate if you'd like to ask more questions!
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krisshP
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(Original post by Choochoo_baloo)
Thanks, nice to be confirmed!

Another problem:

How can a particle, on a SMOOTH inclined plane, with the plane on the Earth's surface, be explained by Newton's III Law?

I see it as follows:
By definition a normal force (albeit a component of the reaction force created through Newton's III) exists due to the particle causing a reaction on the Earth/plane.
Weight always acts, regardless if an object is on a plane or not. You do need to remember that in N3L the forces do act on different objects. This is an interesting thread if you are a bit confused with N3L:
http://www.thestudentroom.co.uk/show....php?t=2426135

So Earth exerts a gravitational force on the particle (particle's weight) and so, by N3L, the particle exerts an equal and opposite gravitational force on the Earth.

Also the particle exerts a normal contact force on the slope, so by N3L, the slope exerts an equal and opposite normal contact force on the particle which is the normal force you usually see when drawing the free-body diagrams of a particle on a slope.

(Original post by Choochoo_baloo)
But, the model I so far 'believed' was that there is weight on the particle vertically down, and normal perpendicular to plan surface. These are the ONLY two forces.

The traditional explanation of acceleration down slope due to weight component parallel to plane makes sense from a mechanics view, but I would think that actually from a strict Newton's III, the weight has no components, but how can the particles down slope acc, be explained from components of the reaction (ie due Newton's III) - of which Normal is one??

Thanks in advance.
Weight does have components, one parallel to the plane and the other perpendicular to the plane. Look at this picture:
Name:  image.jpg
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We know that weight and the normal contact forces act on the particle on the inclined plane from the above ^ N3L stuff. Can you see in the picture that both forces that are perpendicular to the inclined plane (R and mgcosø) cancel out, giving no resultant force perpendicular to the inclined plane? But why is there even this normal contact force acting on the particle which causes such a cancellation? Well because of N3L as I explained above (N3L link). By N1L there's no acceleration perpendicular to the plane. Now look at the component of weight parallel to the inclined plane known as mgsinø. This doesn't cancel out with anything, so by N1L there's acceleration down the inclined plane.

That's my best shot at it, I hope it all makes sense I've just finished AS Physics, so I thought I should post.
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Choochoo_baloo
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Also the particle exerts a normal contact force on the slope, so by N3L, the slope exerts an equal and opposite normal contact force on the particle which is the normal force you usually see when drawing the free-body diagrams of a particle on a slope.
Weight does have components, one parallel to the plane and the other perpendicular to the plane. Look at this picture:
Name:  image.jpg
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Size:  19.8 KB
We know that weight and the normal contact forces act on the particle on the inclined plane from the above ^ N3L stuff. Can you see in the picture that both forces that are perpendicular to the inclined plane (R and mgcosø) cancel out, giving no resultant force perpendicular to the inclined plane? But why is there even this normal contact force acting on the particle which causes such a cancellation? Well because of N3L as I explained above (N3L link). By N1L there's no acceleration perpendicular to the plane. Now look at the component of weight parallel to the inclined plane known as mgsinø. This doesn't cancel out with anything, so by N1L there's acceleration down the inclined plane.

That's my best shot at it, I hope it all makes sense I've just finished AS Physics, so I thought I should post.

Thank you very much to all posters, I do appreciate everyones' efforts.

Unfortunately I fear I have not been 100% clear on my exact 'issue'. With respect to quoted poster (AS Physics guy), I've done A Level Physics/Further Maths/about start Cambridge NatSci, so I point out politely that I'm well aware of most of your solution (especially second quote - I wasn't referring to the application of Newtons I and II).

Although I am unsure about a few technicalities, I do know that this sentence is not correct I'm afraid:

Also the particle exerts a normal contact force on the slope, so by N3L, the slope exerts an equal and opposite normal contact force on the particle which is the normal force you usually see when drawing the free-body diagrams of a particle on a slope.
Firstly in the model we treat plane and rest of Earth as a single giant mass (ie as if the plane is a smooth mountain etc).

If you take weight of particle, as the action half, then considering WHOLE system (not only the particle) so a true two component free body force diagram, ie the Earth/plane exert this action force as particle's weight, then there is an EQUAL+OPPOSITE reaction induced upon the Earth/plane mass. I am well aware that the ubiquitous normal contact force comprises only a component of the reaction force (of N3L action-reaction pair). Hopefully I have now made my query crystal clear now!!

Actually here is a rough sketch of my dilemma:
Name:  N3L FINAL image.jpg
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Size:  149.4 KB

Hopefully someone can clear this up for me!!!

Many thanks in advance.
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james.h
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(Original post by Choochoo_baloo)
Thank you very much to all posters, I do appreciate everyones' efforts.

<snip>

Actually here is a rough sketch of my dilemma:
Name:  N3L FINAL image.jpg
Views: 130
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I'll have a stab at this.

Gravity is a force that acts between two massive objects: in this case, the particle and the "other particle" (Earth, in this case - any other contributors, like the mass of the slope, aren't going to have much of a noticeable effect). For convenience, let's call these two particles A and B respectively.

So, we have the gravitational force pulling A toward B...and the reaction force of the slope on A. This reaction force is equal and opposite to the force of gravity mentioned, in accordance with Newton's Third Law (and discounting the effect of any other forces that may be taking part in this scenario). We have a distinct, unequivocal pair of forces here. :yep:

Your problem seems to be that you are decomposing the reaction force into its components, then complaining that you have a spare one left over. You could decompose the reaction force into hundreds and thousands of vectors if you wanted, but the idea of Newton's 3rd Law is that all force contributions cancel out. The decomposition of reaction force into normal and parallel components, as you likely know, is only a convenience to simplify the calculations; if you wanted, you could solve this sort of problem entirely in vector form.

Sorry if I've managed to miss your point as well!
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krisshP
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(Original post by Choochoo_baloo)
Actually here is a rough sketch of my dilemma:
Name:  N3L FINAL image.jpg
Views: 130
Size:  149.4 KB

Hopefully someone can clear this up for me!!!

Many thanks in advance.
Oh from that diagram I understand your issue a lot more. This post is great:

(Original post by james.h)
Your problem seems to be that you are decomposing the reaction force into its components, then complaining that you have a spare one left over. You could decompose the reaction force into hundreds and thousands of vectors if you wanted, but the idea of Newton's 3rd Law is that all force contributions cancel out. The decomposition of reaction force into normal and parallel components, as you likely know, is only a convenience to simplify the calculations; if you wanted, you could solve this sort of problem entirely in vector form.
The whole point/aim of breaking a force down into its components is to simplify the situation we have and make our lives easier. If you have an inclined plane and you have normal contact force already perpendicular to the plane, as usual, then what's the need to overcomplicate things by breaking it down into components?
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Choochoo_baloo
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(Original post by krisshP)
Oh from that diagram I understand your issue a lot more. This post is great:



The whole point/aim of breaking a force down into its components is to simplify the situation we have and make our lives easier. If you have an inclined plane and you have normal contact force already perpendicular to the plane, as usual, then what's the need to overcomplicate things by breaking it down into components?
Sigghhh...!! I am afraid you both are missing the point/wrong about a few things you do say. I know it must appear that I am ungrateful, I am not, it is just that everyone is just re-wording A Level stuff. Which my unhappiness with lack of clarity caused this thread.

As an example, everyone thinks that the normal force (which some may not realise is present at any point of contact, perp to tangent of plane of contact.) is THE reaction force I am eluding to.

It's slightly worrying that most seem to think that the normal force in the basic stationary box on flat surface model, which is obviously half of the action-reaction pair in this situation, then magically also applies in my situation, since someone said the normal is the reaction in this case!! I know that to be incorrect - people must be clear about the strict definition of the reaction force regarding N3L.

Simply my question is this: We take the combined Earth/plane as a large mass. The particle is by definition a point mass.

At a distance, the well understood N3L gives attractive W force on each object.

Once the particle lands on plane, obviously rolling down the slope, the particle AND Earth/plane are still interacting (will forever be), thus N3L still applies.

The normal force is present doe to a point contact (particle is a point mass). Considering only the free body diagram of the particle (ie one of the two object system) we know that it accelerates down slope. Everyone seems to think the free body diagram of only particle is what I am asking about. It is not :-/

In this simplified system, the old weight induced by particle on Earth/plane (which was technically inversely proportional to distance^2), at moment of contact is still the reaction force.

Knowing that normal force is a component of this reaction, I would like to know where/what the remaining forces are so that by vector addition the resultant of normal of the phantom other force(s) = reaction force vector, magnitude same as particle's weight, directed to the particle.

Thanks to all of you for your time - this is what scholarly debate is made of! :-)
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FireGarden
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By your description, after reading it like, 3 times I still think the problem is exactly the one I tried to address, where you are expecting a solution to (normal force from plane) = - [ (weight) + (some other third force, which is the one you are looking for)].

My answer, in short, is that this doesn't exist because you're thinking the normal force is caused by weight itself and some mysterious force, when really it is caused by only a component of weight. There is no "triangle", just [normal force] = - [component of weight perpendicular to plane].
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sankar biswss
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weight and reaction will cancel each other even though they are not pair
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sankar biswss
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weight and reaction will cancel each other even though they are not pair
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DFranklin
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(Original post by Choochoo_baloo)
Does the definition of Newton's III as:
"For every force in nature, there is an equal and opposite force"

apply to all forces irrespective whether the two objects (whatever size/mass/charge etc.) physically touch/collide?
Yes.

More formally, it tends to be written (mathematically) as the sum of all forces is always zero.

Moving onto your later problem:

I think there are 2 issues here. Firstly, it's that we often ignore forces "on the earth", (because they have no measurable effect). So in the standard inclined-plane diagram, we don't put in the reaction to gravity - that force acts on the earth and we ignore it.

A second unfortunate nomenclature issue (with the slope problem and other problems where we talk about touching/colliding objects) is that the "reaction force" isn't (to my mind) really a reaction in a Newton III sense, it's a "two objects don't want to interpenetrate" force that by tradition is called the reaction force. But it is completely separate from the gravitational force and isn't a reaction to it. In many ways "contact force" or "normal force" might be better terms, but it is what it is.

So, the full set of forces in the inclined plane are:

(A) Gravity, acting on the ball. equal and opposite to gravity, acting on the earth (typically not shown on the diagram).

(B) "Reaction" force on the ball with the direction/force necessary to stop the ball going through the plane. equal and opposite to "Reaction" force on the plane (note that in terms of calculating the necessary force to stop interpenetration we can ignore this because it won't move the plane (because it's attached to the earth and the earth is very heavy)).

Finally, regarding the "resolving forces" issue:

This is really about finding the appropriate reaction (contact) force. We resolve the gravitational force parallel/perpendicular to the plane, then the force exerted by the plane to prevent penetration will be exactly that required to cause zero acceleration perpendicular to the plane, so it will be equal-and-opposite-to the perpendicular component of gravity.
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DFranklin
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(Original post by sankar biswss)
weight and reaction will cancel each other even though they are not pair
Arghhh! This thread is 4 years old. You bumping it has you made me spend ages writing a correct reply only to realise after the event that the OP has undoubtedly lost interest long ago!

[Stuff happens - but please do check the date of the most recent reply before replying to a post yourself].
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