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Nedcim

Newton's third law pair

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11 hours ago, J.C.MacSwell said:

I don't understand what you are saying here.

Gravitational forces always form an action-reaction pair (as do all forces), at least classically. Either can be resisted by a normal force, in whole or in part, except when the normal force is  at right angles to the line of action-reaction.

 

 

With reference to the sketches in my diagram.

  1. Consider Cavendish's experiment but replace his balls by two heavy laminae (plates).
    It is reasonable to say that the gravitational force between these plates is perpendicular (normal) to the opposing faces.
     
  2. But what happens if we turn the plates round?
    What is the line of mutual attraction now perpedicular to?
     
  3. OK so make the plates circular. The force is now perpenducular to the tangent in any orientation.
     
  4. Now shrink one of the plates to a test plate size and move it to the surface of the other plate, now an ellipse.
    This models a body on the surface of the Earth.
    So is the perpendicular to the tangent the line of action of the gravitation force?
    Yes at point A but not at a general point B as shown.
     
  5. Look an my rocket.
    What does normal to the exhaust mean?

N3_1.thumb.jpg.2e2c7e2f38650145f2147d03f3ac33c7.jpg

 

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On 5/25/2020 at 10:04 PM, Nedcim said:

I still have no clearer understanding why the example is not an an action-reaction pair of Newton's third law. The consensus is that Ghideon gave the best explanation but I don't see any distinction from the book's given definition:

"If object A exerts a force on object B, then object B exerts an oppositely directed force of equal magnitude on A."

 

21 hours ago, Nedcim said:

Here's a similar situation to the example I gave but in this case, the website claims the interaction is a consequence of Newton's third law:

"Figure 1. The force arrow for either direction begins at the surface(the black dot). Due to Newton's third law, the normal forces will be equal in magnitude and in opposite direction.

 

Sometimes the following distinction may help: what do we need to know to find an actual action/reaction pair vs. what do we need to know find the magnitude of the force. Let’s try to stay simple and use example from previous diagram, I’ll do it without math this time*.

image.png.fd741c4ac50f27edca19d0d6aa2c4bcc.png

Using the picture we ask two separate questions:
1: “Can you find an action/reaction pair of forces involving the book and the table?
I could answer something like “There is contact force between the table and the book. The book presses down on the table with some unknown force and the table pushes back with an equal and opposite force directed up. The upwards force is called normal force.”
Note that there is no reference to gravity, it is not needed to find the action/reaction pair and not included since it is not part of the action/reaction pair.
Let’s add the second question:
2: “The mass of the book is m. What is the magnitude of the action and reaction force in question 1?
Answer is “mg”. Now I have to know the mass and assume that gravity is causing the force down. But that does not change my first answer for the action/reaction pair.

Now let’s add an analogy to further illustrate the difference between the action/reaction and gravity in this example. We put a black piece of paper between the book and the table.image.png.0b7e0a37f02d7f133f3a73a3da349b43.png

The paper’s mass is low and considered zero in this discussion. What is the action/reaction force between the paper and the table? The answer is pretty much the same as above “The paper presses down on the table with some unknown force and the table pushes back with an equal and opposite force directed up”. No reference to gravity.
What is the magnitude of the force? Now things change. There is no way of telling the magnitude from looking at the table or the paper. We need to look at the mass of the book to get the answer “mg”. 

Last analogy for now: Let’s assume a very light book is clamped down to the table. The clamp’s pressure is high compared to the mass of the book so mass of book and clamp is neglected.

image.png.852777c7f74a76e3cab06011e4e56c00.png

Again, what is the action/reaction forces for book/table? Answer is same as in the first case: The book presses down on the table with some unknown force and the table pushes back with an equal and opposite force directed up.
What is the magnitude of the force? Now we can’t use gravity at all, we have to get the force the clamp presses down with.

 

Hopefully the above illustrates how one can look at the cause of some force vs the action and reaction pair some force causes. 

 

20 hours ago, joigus said:

"contact forces are an illusion."

That is a valid point, I'll try to get time to do some example/comment that fits in the context of OPs question. 

 

*) Other members have already contributed good math, I’ll add formulas if requested.

 

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Posted (edited)
3 hours ago, Ghideon said:

That is a valid point, I'll try to get time to do some example/comment that fits in the context of OPs question. 

 

I know it is. It also helps break the spell. Mechanics becomes crystal clear in your mind once you understand nothing touches nothing and everything is the same. It's all variables in a Lagrangian. As complicated systems of fields. No difference. Maths give you the key to everything physical.

I always work Lagrange --> Newton, not the other way about.

Great value of more clarifying examples +1.

"The equation knows best"

P. A. M. Dirac, as quoted by James Gleick

Edited by joigus
Added piece of Dirac's wisdom

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On 5/25/2020 at 1:43 AM, Ghideon said:

I'l try: The upward normal force from the table and the downward force due to gravity are two different forces acting on the same object. Action/reaction pair is the same force seen from two points of view. In the above case action/reaction could be seen as two separate action/reaction pairs:
1: Gravitational interaction Book-Earth and Earth-book
2: The books force pushing down on the table and the tables push on the book

If everything is at rest then the forces will balance so numerical values of force due to gravity (1) and normal force (2) are same.

All valid. However, in all practical purposes there no difference in the gravitational interaction of the book-Earth and book-table.  It's simply a matter that the book is not contact with Earth, so by convention Newton's third law fails.  

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5 hours ago, Nedcim said:

All valid. However, in all practical purposes there no difference in the gravitational interaction of the book-Earth and book-table.  It's simply a matter that the book is not contact with Earth, so by convention Newton's third law fails.  

I think I'm not fully following the reasoning, can you elaborate? How is there a failure of Newton's third law? 

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31 minutes ago, Ghideon said:

I think I'm not fully following the reasoning, can you elaborate? How is there a failure of Newton's third law? 

 

 

Yes let us find out what's actually troubling Ned. +1

 

10 hours ago, joigus said:

I know it is. It also helps break the spell. Mechanics becomes crystal clear in your mind once you understand nothing touches nothing and everything is the same. It's all variables in a Lagrangian. As complicated systems of fields. No difference. Maths give you the key to everything physical.

I always work Lagrange --> Newton, not the other way about

 

Virtual work is even simpler, but still not as simple as the standard equations of statics.

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1 hour ago, studiot said:

Virtual work is even simpler, but still not as simple as the standard equations of statics.

Answer to ironic Studiot: Everything is zero.

1 hour ago, studiot said:

Virtual work is even simpler, but still not as simple as the standard equations of statics.

Answer to dead-serious Studiot: And the Poisson brackets formulation of mechanics even simpler, plus the sacred causeway to quantum mechanics.

Sorry, I had to work on a quantum superposition of both your selves. ;)

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10 minutes ago, joigus said:

And the Poisson brackets formulation of mechanics even simpler,

Happy to compare my VW solution with your PB one.

:)

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6 hours ago, studiot said:

Happy to compare my VW solution with your PB one.

:)

Here is Dr Seuss' example

The cat sat still on the mat.

If F is the force of the mat on the cat, M the mass of the cat and W the weight of the cat  then

Imagine a virtual downward displacement δh of the cat's bottom.

Then the work done against F = F(δh)

This equals the loss of (gravitational) potential energy of the cat = Mgδh = Wδh

Fδh  =  Mgδh = Wδh

F = Mg = W.

 

It's a smiley day

All the way

:)

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11 hours ago, Ghideon said:

I think I'm not fully following the reasoning, can you elaborate? How is there a failure of Newton's third law? 

There are no failure of Newton's third law, it's the way that is applied in this particularly circumstance for thoroughness, to ensure when the law is used in other applications there are no contradictions. 

How the forces are generated is not important in terms of Newton's third law. What matters is that there two objects, with equal but oppositely directed forces. The net result is a normal force from the desk that is equal but directly opposite the normal force from the book which is Newton's third law. 

10 hours ago, studiot said:

 

 

Yes let us find out what's actually troubling Ned. +

No, troubles but thanks for your concern. Technically, none of Newton's laws have merit on Earth because they were formulated in the context of uniform motion. 

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4 minutes ago, Nedcim said:

There are no failure of Newton's third law, it's the way that is applied in this particularly circumstance for thoroughness, to ensure when the law is used in other applications there are no contradictions. 

Where is the error in the application? I fail to find it but would like to understand.

7 minutes ago, Nedcim said:

Technically, none of Newton's laws have merit on Earth because they were formulated in the context of uniform motion. 

That is not something I see when reading Newton's laws or applying them. Can you elaborate?

 

 

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Posted (edited)
On 5/28/2020 at 11:24 AM, studiot said:

Happy to compare my VW solution with your PB one.

:)

Errrr...

On 5/28/2020 at 11:12 AM, joigus said:

And the Poisson brackets formulation of mechanics even simpler,

I correct myself: It's not simpler. It's far more powerful for a very reasonable price in complication.

Edited by joigus
mistyped

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1 hour ago, joigus said:

Errrr...

I correct myself: It's not simpler. It's far more powerful for a very reasonable price in complication.

Actually though trivial in this case, and not a simple as the plain statement of static equilibrium F = W, the virtual work method has attraction in a similar situation.

I was once challenged by a PhD specialist in electrodynamics (not here) how to derive the controlling equation for an electromagnet picking up a car in a scrapyard.

When we compared solutions, his was 3 pages of advanced mathematics to avoid magnetic forces doing work, and mine was 3 lines long.

(I haven't forgotten the thermodynamic question)

:)

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23 hours ago, Ghideon said:
 

Where is the error in the application? I fail to find it but would like to understand.

The issue is in the consideration of the two forces acting acting on the book. The force of gravity is irrelevant. What matters is its results of the equal but opposite normal forces acting on the desk and book which is Newton's third law. That reconciliation must be used for the conditions of equilibrium. 

Quote

That is not something I see when reading Newton's laws or applying them. Can you elaborate?

Earth is not in uniform motion, so Newton's laws do not wholly apply on Earth. However, in most cases, the effects are miniscule. That's the situation with the example. Your explanation. was entirely necessary for the situation to be complete. However, those effects are so insignificant they are omitted except in special cases.

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Posted (edited)
23 minutes ago, Nedcim said:

The issue is in the consideration of the two forces acting acting on the book.

 

Yes exactly.

 

23 minutes ago, Nedcim said:

The force of gravity is irrelevant.

 

 

Absolutely not. Not even your textbook says this.

 

23 minutes ago, Nedcim said:

. What matters is its results of the equal but opposite normal forces acting on the desk and book which is Newton's third law. That reconciliation must be used for the conditions of equilibrium. 

 

 

Again this is not the case. Two forces, acting on different objects, cannot be 'in equilibrium. Only forces that act on the same object can be in equilibrium.
Newton's third law is not about equilibrium.

 

I think the key to understanding and getting it right is to start with a good diagram.
and I am sorry to say that the diagram from your book is not a good diagram, for all it prettiness.

 

Edited by studiot

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Posted (edited)
42 minutes ago, Nedcim said:

The force of gravity is irrelevant.

By "irrelevant" do you mean that gravity can be replaced with any other downwards force, such as my example with the clamp I posted? Gravity is most relevant if you wish to calculate the magnitude of the action/reaction force in the example. And without gravity (and with no other force replacing gravity), there would be no action/reaction pair of forces between the book and the table in the example.

42 minutes ago, Nedcim said:

Earth is not in uniform motion, so Newton's laws do not wholly apply on Earth.

The second law state that the rate of change of momentum of a body is directly proportional to the force applied. The law describe the relation mass - acceleration - force*. Simply put, F=ma works, a<>0 means motion is not uniform. And the third law does not require uniform motion. It states equal and opposite force. Not equal and opposite force in uniform motion.

Am I misinterpreting what you mean by "uniform"?

 

 

*) At velocity v<<c etc, where the theory is applicable.

 

Edited by Ghideon
grammar and clarification

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20 hours ago, studiot said:

Absolutely not. Not even your textbook says this.

My book gave no such consideration in its explanation of Newton's third law:

"If object A exerts a force on on object B, then object B exerts a oppositely directed force of equal magnitude on A."

Furthermore, the book adds: "Newton's third law is about forces between objects. It says that such forces always oexample.Truairs that it is not possible for an object A to exert a force on object B without B exerting a force back on A."

Again, there is no considerations on how the forces come to be only that they interact between objects. What's your explanation in those terms why the resultant normal forces from the book and table fail to be a Newton's third law pair? 

Quote

Again this is not the case. Two forces, acting on different objects, cannot be 'in equilibrium. Only forces that act on the same object can be in equilibrium.
Newton's third law is not about equilibrium.

Equilibrium is dependent on the consequences of Newton's third law. Equilibrium requires an net external force and torque to to be zero. Explain how that is possible without the results from Newton's third law?

Quote

 I think the key to understanding and getting it right is to start with a good diagram and I am sorry to say that the diagram from your book is not a good diagram, for all it prettiness.

I agree that the diagram is lacking but it's only used to highlight the unique case that is disregarded in most applications.

20 hours ago, Ghideon said:

By "irrelevant" do you mean that gravity can be replaced with any other downwards force, such as my example with the clamp I posted? Gravity is most relevant if you wish to calculate the magnitude of the action/reaction force in the example. And without gravity (and with no other force replacing gravity), there would be no action/reaction pair of forces between the book and the table in the example.

The result of the forces is what matter for Newton's third law. The type of forces or how they come to be is not important. It's the same idea with motion. It's is of no consequence how motion came to be. What is important is how force changes that motion.

Quote

The second law state that the rate of change of momentum of a body is directly proportional to the force applied. The law describe the relation mass - acceleration - force*. Simply put, F=ma works, a<>0 means motion is not uniform. And the third law does not require uniform motion. It states equal and opposite force. Not equal and opposite force in uniform motion.

Am I misinterpreting what you mean by "uniform"?

law does not require uniform motion. It states equal and opposite force. Not equal and opposite force in uniform motion.

Am I misinterpreting what you mean by "uniform"?

 

 

*) At velocity v<<c etc, where the theory is applicable.

It's not how I mean uniform, but how Newton formulated those laws under the assumption of uniform motion. 

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22 hours ago, Ghideon said:

Am I misinterpreting what you mean by "uniform"?

I'm not sure, but I think he means the Earth is not an inertial system due to rotation --> Coriolis and centrifugal fictitious forces?

Is that what you mean, @Nedcim?

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Posted (edited)
34 minutes ago, Nedcim said:

Furthermore, the book adds: "Newton's third law is about forces between objects.

Yes and how many objects is 'objects'  ?

There must be more than one.

But how many objects are there in a free body diagram ?

Just one.

Understanding this is the key to it all, as Ghideon so nicely told you three pages ago.

34 minutes ago, Nedcim said:

It says that such forces always oexample.Truairs that it is not possible for an object A to exert a force on object B without B exerting a force back on A."

and could you repair this English please /

I do not understand the underlined bit.
 

At the end I can see that it says that if A exerts a force on B then B exerts a force on A and I agree with that (as does everybody else).

Edited by studiot

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5 hours ago, Nedcim said:

My book gave no such consideration in its explanation of Newton's third law:

"If object A exerts a force on on object B, then object B exerts a oppositely directed force of equal magnitude on A."

Furthermore, the book adds: "Newton's third law is about forces between objects. It says that such forces always oexample.Truairs that it is not possible for an object A to exert a force on object B without B exerting a force back on A."

Again, there is no considerations on how the forces come to be only that they interact between objects. What's your explanation in those terms why the resultant normal forces from the book and table fail to be a Newton's third law pair? 

 

A book, object 'A', on a table, object 'B', exert equal and opposite forces on each other. They are an action-reaction pair.

The book stays stationary due to gravitational force of the Earth on the book, which though equal and opposite to the force of the table on the book, isn't an action-reaction pair with it.

5 hours ago, Nedcim said:

The result of the forces is what matter for Newton's third law. The type of forces or how they come to be is not important. It's the same idea with motion. It's is of no consequence how motion came to be. What is important is how force changes that motion.

Let's say you had an identical twin standing on the South Pole, while you stood on the North Pole. Each of you would exert an equal but opposite force on the Earth. This would not be an example of an action-reaction pair. 

 

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5 hours ago, J.C.MacSwell said:

Let's say you had an identical twin standing on the South Pole, while you stood on the North Pole. Each of you would exert an equal but opposite force on the Earth. This would not be an example of an action-reaction pair. 

Like it +1

@Nedcim I hope this helpsyou.

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22 hours ago, joigus said:

I'm not sure, but I think he means the Earth is not an inertial system due to rotation --> Coriolis and centrifugal fictitious forces?

Is that what you mean, @Nedcim?

Yes, that part of it. There's must be additional considerations when using Newton's laws in those conditions:

"Newton's laws take on a more complicated, non-intuitive form. But Newton's laws in the rotating frame can be made to look like the regular Newton's laws if we treat the extra pieces in the equations as inertial forces." 

https://wtamu.edu/~cbaird/sq/mobile/2012/12/15/why-is-the-centrifugal-force-talked-about-so-much-if-its-not-real/

Newton's laws treat objects as points with no consideration in size and shape where air resistance is ignored. Also, Newton's laws do not work for sub atomical particles or objects with very high speeds.

22 hours ago, studiot said:

Yes and how many objects is 'objects'  ?

There must be more than one.

There are two objects: desk and book. 

Quote

But how many objects are there in a free body diagram ?

You mean same diagram you've criticized? In any case that diagram is only focused on the the book. It doesn't show the interaction of the book with the table. 

Quote

and could you repair this English please /

I do not understand the underlined bit.

Sure, it easier to attack me for some careless mistakes I made because I'm typing with a phone rather than admit you that you misrepresented what the book said and you can't give an explanation without making a contradiction to the given rule.

You were are absolutely wrong on Newton's third law and its connection with equilibrium. Again, you didn't attempt to give an explanation for your claim. I wonder why?

16 hours ago, J.C.MacSwell said:

A book, object 'A', on a table, object 'B', exert equal and opposite forces on each other. They are an action-reaction pair.

The book stays stationary due to gravitational force of the Earth on the book, which though equal and opposite to the force of the table on the book, isn't an action-reaction pair with it.

The table and book are both stationary. Where's the objection with Newton's third law?

Quote

Let's say you had an identical twin standing on the South Pole, while you stood on the North Pole. Each of you would exert an equal but opposite force on the Earth. This would not be an example of an action-reaction pair. 

The twins are not in contact, so it's not the same situation as the example. 

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Posted (edited)
1 hour ago, Nedcim said:

There are two objects: desk and book. 

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

Here's a list of third-law-force pairs I see in the diagram where a book is laying on a table standing on earth.

First force -- Second force 

Downward force of gravity on the book exerted by Earth -- Upward force of gravity on Earth exerted by the book

Downward force on the table exerted by the book -- Upward force on the book exerted by the table

Downward force on the ground exerted by the table -- Upward force on the table exerted by the ground (one pair at each leg of the table)

Downward force of gravity on the table exerted by Earth -- Upward force of gravity on Earth exerted by the table

Note: third law pairs do not have to be contact forces.

image.png.a71e2476d5800ba7cc3c41cb9c8e9c0c.png

 

 

 

Edited by Ghideon

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1 hour ago, Nedcim said:

 

The table and book are both stationary. Where's the objection with Newton's third law?

 

I have no objection with Newton's third Law.

1 hour ago, Nedcim said:

 

The twins are not in contact, so it's not the same situation as the example. 

OK. Good. Just giving an example where just because forces are equal but opposite it doesn't constitute an action-reaction pair.

So back to your OP...

On 5/25/2020 at 1:50 AM, Nedcim said:

The physics book I'm using gives two examples where one is a third law pair and the other is not. What is the reasoning that this example is not a third law pair:

"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."

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.

So let's look at the actual pairs which Ghideon has just presented:

53 minutes ago, Ghideon said:

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

Here's a list of third-law-force pairs I see in the diagram where a book is laying on a table standing on earth.

First force -- Second force 

Downward force of gravity on the book exerted by Earth -- Upward force of gravity on Earth exerted by the book

Downward force on the table exerted by the book -- Upward force on the book exerted by the table

Downward force on the ground exerted by the table -- Upward force on the table exerted by the ground (one pair at each leg of the table)

Downward force of gravity on the table exerted by Earth -- Upward force of gravity on Earth exerted by the table

Note: third law pairs do not have to be contact forces.

image.png.a71e2476d5800ba7cc3c41cb9c8e9c0c.png

 

 

 

Does this now make sense?

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2 hours ago, Ghideon said:

Note: third law pairs do not have to be contact forces.

+1

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