# Newton's third law pair

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21 hours ago, Ghideon said:

The diagram mentions gravity, that implies a third object: earth.

Earth acts on both objects. The result is equal but oppositely directed normal forces. There is no consideration of a third body (Earth) for most application only the force of gravity. For most practical examples, this gives a better picture of the situation which considers the results of gravity:

I think a better example involving a direct consideration of Earth is if you drop an object, Earth accelerates towards that object but the amount of Earth's acceleration is trivial. In that case there are no resultant forces from gravity.

20 hours ago, J.C.MacSwell said:

The downward gravitational force is not a contact force...unlike the force between table and book , and they cannot be an action-reaction pair...each must have one of their own, just like you and your twin.

The opposing normal forces are the same type. How can there be equilibruim with a net external force and torque of zero without a Newton's third law pair between objects?

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Posted (edited)
1 hour ago, Nedcim said:

The opposing normal forces are the same type. How can there be equilibruim with a net external force and torque of zero without a Newton's third law pair between objects?

Who is claiming there is none?

No one has yet described a set up with no action-reaction pairs, and at least 2 are required for equilibrium if one pair exists.

An object, isolated in deep space, can be in equilibrium with no forces acting on it. No one has suggested we contemplate that scenario.

Edited by J.C.MacSwell

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1 hour ago, Nedcim said:

The opposing normal forces are the same type. How can there be equilibruim with a net external force and torque of zero without a Newton's third law pair between objects?

Equilibrium is irrelevant. All equilibrium (something at rest in its own frame) tells you is that there isn’t a net force on an object, or pair of objects, which says nothing about action-reaction. All it tells you, as J. C. has implied, is that the number of forces acting on an object is not one.

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Just to make sure; is your initial question answered, do you now have full understanding of how the upward normal force from the table and the downward force due to gravity are not an action/reaction pair of Newton's third law? Or would our prefer further analysis of the situation in the opening post?

9 hours ago, Nedcim said:

Earth acts on both objects.

True, earth acts on both objects, the table and the book in the opening post, hence four pairs of third law action/reaction forces in my list.

9 hours ago, Nedcim said:

The result is equal but oppositely directed normal forces.

Yes the magnitude of normal forces is the due to the gravity. (But obviously the normal forces & gravity pull are still not third law pairs.)

9 hours ago, Nedcim said:

I think a better example involving a direct consideration of Earth is if you drop an object, Earth accelerates towards that object but the amount of Earth's acceleration is trivial. In that case there are no resultant forces from gravity.

Better example than what? You remove one pair of action/reaction forces by introducing free fall and only keep gravity. Does it help getting better insight into the answers about the initial situation with book on a table? If so then I guess it is a good example.
It also depends on what you wish to illustrate; are we discussing Newtons laws in a more general case or still working on the initial setup with book and table?

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11 hours ago, Nedcim said:

Earth acts on both objects. The result is equal but oppositely directed normal forces. There is no consideration of a third body (Earth) for most application only the force of gravity. For most practical examples, this gives a better picture of the situation which considers the results of gravity:

I think a better example involving a direct consideration of Earth is if you drop an object, Earth accelerates towards that object but the amount of Earth's acceleration is trivial. In that case there are no resultant forces from gravity.

Congratulations, you have found a far better diagram. +1

And yes we usually ignore the effect on the Earth and just consider the effect on the body, be it book, table or ball.

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On 6/1/2020 at 6:16 PM, J.C.MacSwell said:

Who is claiming there is none?

The example is noted as not being an action/reaction pair but it is in equilibrium which is contradicted with:

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No one has yet described a set up with no action-reaction pairs, and at least 2 are required for equilibrium if one pair exists

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An object, isolated in deep space, can be in equilibrium with no forces acting on it. No one has suggested we contemplate that scenario

How? The are debris in deep space that will impact an net external force on the object, and hence no equilibrium.

On 6/1/2020 at 6:32 PM, swansont said:

Equilibrium is irrelevant. All equilibrium (something at rest in its own frame) tells you is that there isn’t a net force on an object, or pair of objects, which says nothing about action-reaction.

How can equilibrium be irrelevant? The objects in the example are assumed to be in equilibrium. If the are objects are not in equilibrium then must be additional forces that were not given.

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All equilibrium (something at rest in its own frame) tells you is that there isn’t a net force on an object, or pair of objects, which says nothing about action-reaction.

I'm not sure how you are defining something at rest in its own frame? If gravity acts on an object then there is a net external force on that object until equal but oppositely set of action/reaction offsets it:

18 hours ago, Ghideon said:

Just to make sure; is your initial question answered, do you now have full understanding of how the upward normal force from the table and the downward force due to gravity are not an action/reaction pair of Newton's third law? Or would our prefer further analysis of the situation in the opening post?

Yes, again you explained it in your initial comment. +1

It's only for the strictest consideration and mostly inconsequential. The second diagram simply omits the consequence of a third body Earth and labels ball and floor interaction as being a pair of third law forces.

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13 hours ago, Nedcim said:

How can equilibrium be irrelevant? The objects in the example are assumed to be in equilibrium. If the are objects are not in equilibrium then must be additional forces that were not given.

Equilibrium doesn’t change whether or not a pair of forces are an action/reaction pair. The moon and the earth each exert a gravitational force on each other. They are a 3rd law pair, and the system is not in equilibrium

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5 hours ago, swansont said:

Equilibrium doesn’t change whether or not a pair of forces are an action/reaction pair.

True, only by the strictest application of Newton's third law. In practice, they are naturally considered to be action/reaction pair as shown by diagram with the volleyball. In fact, the point where equal but oppositely directed forces offset between objects is called a reaction in connection with action/reaction of Newton's third law.

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2 hours ago, Nedcim said:

True, only by the strictest application of Newton's third law. In practice, they are naturally considered to be action/reaction pair as shown by diagram with the volleyball. In fact, the point where equal but oppositely directed forces offset between objects is called a reaction in connection with action/reaction of Newton's third law.

This doesn’t make much sense. “equal but oppositely directed forces offset between objects”?

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19 hours ago, swansont said:

This doesn’t make much sense. “equal but oppositely directed forces offset between objects”?

Where''s the issue? Forces are offset between objects. Let's look back to results of the last diagram:

"The normal force of the rock counters gravity."

There are equal but oppositely directed normal forces between the rock and the reaction which offsets the external force of gravity.

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1 hour ago, Nedcim said:

Where''s the issue? Forces are offset between objects. Let's look back to results of the last diagram:

"The normal force of the rock counters gravity."

There are equal but oppositely directed normal forces between the rock and the reaction which offsets the external force of gravity.

Offset can be a spatial term.

These forces add to zero, but the addition of another force will not change the identity of the action-reaction force pairs we have identified. Equilibrium is a red herring.

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22 hours ago, swansont said:

Offset can be a spatial term.

Obviously not when offset is being applied to the forces between objects that are in direct contact.

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These forces add to zero, but the addition of another force will not change the identity of the action-reaction force pairs we have identified. Equilibrium is a red herring.

Equilibrium depends on the consequences of Newton's third law. You've already seen a few examples. Here Newton's third law is explictly given in the intro to statics:

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27 minutes ago, Nedcim said:

Obviously not when offset is being applied to the forces between objects that are in direct contact.

One can speak of a force offset from the center of mass.

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Equilibrium depends on the consequences of Newton's third law. You've already seen a few examples. Here Newton's third law is explictly given in the intro to statics:

It's a statics chapter, but you can't otherwise tell anything other than each object exerts a force on the other. That would be true if we took the earth's rotation into account, too.

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21 hours ago, swansont said:

It's a statics chapter, but you can't otherwise tell anything other than each object exerts a force on the other.

Fair enough. It's used to show the justification for how the forces are assumed between objects for equilibrium contrasted with different types of forces but it's from a PDF file. Here's a a webpage that gives the same connection with Newton's third law and equilibruim:

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Newton's third law can be applied to examples where bodies are in equilibrium. A body is in equilibrium when the forces acting on it are all balanced, so there is no overall force acting on the body. This means that the body will either be stationary or moving at a constant speed.

If someone hits a ball with a bat the ball is sent flying in the opposite direction. This is a result of Newton's third law. However, at the instant of contact there are two forces acting on the ball. The normal force from the bat and force of gravity. No different than case with the book on the table which had the normal force from the the table and the force of gravity and it failed to be an action/reaction pair.

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Posted (edited)
47 minutes ago, Nedcim said:

Fair enough. It's used to show the justification for how the forces are assumed between objects for equilibrium contrasted with different types of forces

Here is an interesting question for you to consider, Ned.

What happens if the magnitude of the action and reaction force is exactly zero?

Edited by studiot

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1 hour ago, Nedcim said:

Fair enough. It's used to show the justification for how the forces are assumed between objects for equilibrium contrasted with different types of forces but it's from a PDF file. Here's a a webpage that gives the same connection with Newton's third law and equilibruim:

If someone hits a ball with a bat the ball is sent flying in the opposite direction. This is a result of Newton's third law. However, at the instant of contact there are two forces acting on the ball. The normal force from the bat and force of gravity. No different than case with the book on the table which had the normal force from the the table and the force of gravity and it failed to be an action/reaction pair.

Your argument is inconsistent. The example of the bat hitting the ball is not one of equilibrium, and you correctly apply the third law, so I don’t see why you insist that equilibrium must be present.

If something is not in equilibrium, it means that the pseudoforce that one might be tempted to invoke will not have a reaction pairing.

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On 6/6/2020 at 5:04 PM, studiot said:

Here is an interesting question for you to consider, Ned.

What happens if the magnitude of the action and reaction force is exactly zero?

It is irrelevant to consider the magnitudes of forces acting on different objects.

On 6/6/2020 at 5:29 PM, swansont said:

True, because they involve two different examples. I didn't intend for the examples to be directly related.

In any case, you completely ignored the cited material from the website which disagreed with your claim about equilibrium having no connection with action reaction pair. Why?

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The example of the bat hitting the ball is not one of equilibrium, and you correctly apply the third law, so I don’t see why you insist that equilibrium must be present.

I never insisted on equilibrium for the bat and ball. In any case, they are in equilibrium for an instant but that is unimportant. What matters is that is that there are two forces acting on the ball when it collides with the bat. You noted the third law correctly applies. Yet the two forces acting on the book was the reason that the example was is not a third law pair.

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If something is not in equilibrium, it means that the pseudoforce that one might be tempted to invoke will not have a reaction pairing.

Again, not in equilibrium means there is a net external force or torque on an object. A suitable reaction(s) will counter the net external force for equilibrium.

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7 hours ago, Nedcim said:

It is irrelevant to consider the magnitudes of forces acting on different objects.

'irrelevant'  to what ?

I thought it had meaning and would have introduced you to a physical situation you have not yet considered.

Remember also that forces (of any type) are vectors and that vectors have direction.

There is much relevance in my question to N3 and a much deeper question.

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10 hours ago, Nedcim said:

True, because they involve two different examples. I didn't intend for the examples to be directly related.

Both positions can’t be true. Either equilibrium matters, or it doesn’t. You basically did a proof by contradiction here, and showed that equilibrium doesn’t matter.

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In any case, you completely ignored the cited material from the website which disagreed with your claim about equilibrium having no connection with action reaction pair. Why?

Because it doesn’t really say that. “Newton's third law can be applied to examples where bodies are in equilibrium.“ is a true statement. It can be applied in that case.

Note, however, that it does not say it can only be applied to equilibrium.

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I never insisted on equilibrium for the bat and ball. In any case, they are in equilibrium for an instant but that is unimportant. What matters is that is that there are two forces acting on the ball when it collides with the bat. You noted the third law correctly applies.

No, that’s not what matters. “equilibrium for an instant” is not equilibrium

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Yet the two forces acting on the book was the reason that the example was is not a third law pair.

One of the reasons. And it was because both forces act on the book, and 3rd law pairs act on different objects.

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Again, not in equilibrium means there is a net external force or torque on an object. A suitable reaction(s) will counter the net external force for equilibrium.

It won’t “counter” anything. Reaction forces do not “counter” or “cancel” anything, in this context.

The moon exerts a gravitation force on the earth, and vice-versa. This is an action-reaction force pair. Equal and opposite, same kind of force, source and target switched. There is no equilibrium, because equilibrium is not required.

Such a simple example does not show up in textbook quizzes because it’s too easy of a question. Quizzes use multiple forces to make you think about the concepts, and apply your learning.

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8 hours ago, swansont said:

Both positions can’t be true. Either equilibrium matters, or it doesn’t. You basically did a proof by contradiction here, and showed that equilibrium doesn’t matter.

Again, equilibrium doesn't matter with the bat and ball only the two forces acting on the ball.

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Because it doesn’t really say that. “Newton's third law can be applied to examples where bodies are in equilibrium.“ is a true statement. It can be applied in that case.

Note, however, that it does not say it can only be applied to equilibrium.

The book and desk were in equilibrium but forces on the book were not an action/reaction pair. How can Newton's third law be applied to forces that are not an action/reaction pair?

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No, that’s not what matters. “equilibrium for an instant” is not equilibrium

There is no condition of time for equilibrium.

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One of the reasons. And it was because both forces act on the book, and 3rd law pairs act on different objects.

That dilemma is going to be present with innumerable examples where the force of gravity and another force act on an object.

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It won’t “counter” anything. Reaction forces do not “counter” or “cancel” anything, in this context.

Sure, they will. Refer back to the static equilibrium -derrick example. Explain why that example is wrong?

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The moon exerts a gravitation force on the earth, and vice-versa. This is an action-reaction force pair. Equal and opposite, same kind of force, source and target switched. There is no equilibrium, because equilibrium is not required.required.

There is no equilibrium because both Earth and the moon have rotations that are not constant and change with time because there is a net external force acting on both.

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26 minutes ago, Nedcim said:

Again, equilibrium doesn't matter with the bat and ball only the two forces acting on the ball.

It doesn’t matter in any example.

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The book and desk were in equilibrium but forces on the book were not an action/reaction pair. How can Newton's third law be applied to forces that are not an action/reaction pair?

It doesn’t apply to such situations. The question asked which example was not an action/reaction pair.

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There is no condition of time for equilibrium.

Equilibrium implies a static case. Something that does not change unless external conditions change.

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That dilemma is going to be present with innumerable examples where the force of gravity and another force act on an object.

It’s not a dilemma. It’s a matter of properly applying the third law

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There is no equilibrium because both Earth and the moon have rotations that are not constant and change with time because there is a net external force acting on both.

And yet that’s an action-reaction force. Equilibrium is not a requirement.

26 minutes ago, Nedcim said:

Sure, they will. Refer back to the static equilibrium -derrick example. Explain why that example is wrong?

“wrong”?

What was claimed that could be right or wrong? No action-reaction force pairs are given. All of the forces act on the object.

The net force and action-reaction forces are separate concepts.

On 6/2/2020 at 9:28 PM, Nedcim said:

How can equilibrium be irrelevant? The objects in the example are assumed to be in equilibrium.

Yes, in that example the objects happen to be in equilibrium. In other examples, the objects are not.

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If the are objects are not in equilibrium then must be additional forces that were not given.

It’s irrelevant because you don’t have to have equilibrium to apply the third law. You know this, because you’ve given an example where this was the case. Equilibrium is a red herring.

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On 6/8/2020 at 5:31 PM, swansont said:

It doesn’t apply to such situations. The question asked which example was not an action/reaction pair.

No, the example explictly said that the forces were not an action/reaction pair.

The reason given here was two forces acting on the same object. That's the case with a ball hit by a bat so it should fail to be a third law pair by the same convention.

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Equilibrium implies a static case. Something that does not change unless external conditions change.

Exactly, no condition of time. Hence, static equilibrium could be for an instant.

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It’s not a dilemma. It’s a matter of properly applying the third law

Third law can be applied to forces that are not an action/reaction pair?

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“wrong”?

What was claimed that could be right or wrong? No action-reaction force pairs are given. All of the forces act on the object.

How are you going to eliminate an external force or torque without an action/reaction pair? All of the forces act on the object. Notice how all of those forces have have a corresponding reaction?

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It’s irrelevant because you don’t have to have equilibrium to apply the third law.

True, but as website noted Newton's third law can be applied to equilibrium. Again, how can  Newton's third law be applied with no action/reaction pair?

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On 6/7/2020 at 10:57 PM, Nedcim said:

I never insisted on equilibrium for the bat and ball. In any case, they are in equilibrium for an instant but that is unimportant. What matters is that is that there are two forces acting on the ball when it collides with the bat. You noted the third law correctly applies. Yet the two forces acting on the book was the reason that the example was is not a third law pair.

What instant do you believe equilibrium exists for the bat and ball?

(I don't think it's a single instant when or as you think it is)

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9 hours ago, Nedcim said:

No, the example explictly said that the forces were not an action/reaction pair.

Way to miss the point. If they are not an action-reaction pair, Newton’s 3rd law does not apply. They don’t have to be equal in magnitude, opposite in direction, or the same type of force.

In that example, they happen to be equal and opposite, because the point of such an example is to make you apply your knowledge.

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The reason given here was two forces acting on the same object. That's the case with a ball hit by a bat so it should fail to be a third law pair by the same convention.

The ball hits the bat, the bat hits the ball. The forces are not on the same object. They are an action-reaction pair.

The contact force and gravity, the two forces acting on the ball, are not an action-reaction pair.

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Third law can be applied to forces that are not an action/reaction pair?

Still no.

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How are you going to eliminate an external force or torque without an action/reaction pair? All of the forces act on the object. Notice how all of those forces have have a corresponding reaction?

The force an be unknown or not accounted for. You can be in a rotating frame of reference. Those circumstances will not change the action-reaction relationship of other forces that are present.

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True, but as website noted Newton's third law can be applied to equilibrium.

CAN be. Not MUST be, or CAN ONLY be. And perhaps consider that a BBC website might be a less trustworthy source than someone with a PhD in physics.

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Again, how can  Newton's third law be applied with no action/reaction pair?

Again, it can’t.

And again, what Newton’s laws tell you is that if you are in an accelerated frame, then whatever pseudoforce  you identify will not have an action reaction pair.

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2 hours ago, swansont said:

because the point of such an example is to make you apply your knowledge.

Exactly what you are not doing in ignoring my question.

What are the implications of the (real world) situation where the action and reaction forces are exactly zero?

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