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Nedcim

Newton's third law pair

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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."

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Posted (edited)
56 minutes ago, Nedcim said:

"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."

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.

Now try a case where 1 and 2 are not balaning each other:
Accelerate the table upwards (for instance in an accelerating elevator). Then the normal force would change but gravity remains the same. (1) is still action/reaction pair and (2) is still action/reaction pair but (1) and (2) does not balance out.  

Edited by Ghideon
clarified sentence

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2 hours ago, Nedcim said:

The physics book I'm using [...]

OK. But please be aware that sometimes the book that one is using is the problem. Here's my explanation:

If the book from the example is on the table, pushing against it by Earth's gravitational pull, and the table is rigid, so that it transmits the equal and opposite, (but minuscule in relation to Earth's mass) gravitational pull that the book exerts on the Earth, and none of them is accelerated with respect to each other, and all other forces into play are compensated so that the table doesn't break apart, deform, etc.; so that they just act as constraining forces; then I can see no reason why that cannot be considered as a Newton pair. If you picture the "binary system" (book) + (Earth and table) in empty space as a two-particle system, as Ghideon is suggesting, I think you will understand what I mean.

If you don't find my explanation useful, please pay no attention.

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Posted (edited)

Sounds, to me, like the book is wrong

Or not

Edited by Strange
Retraction

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3 hours ago, 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."

Your book is not 'wrong' .

But then again it is not'right' either!

It is just being economical with the truth.

IOW it is oversimplifying.

 

It works like this:

There are two types of (mechanical) force.

Contact forces
Body forces

Newton's third law pairs are both contact forces.

Gravity is a body force and acts collectively on a body through the centre of cravity of that body.

So the internal collective force of gravity acting through the COG down through its line of action pushes on each element of the body below it until it reaches the contact point with the table.
Here it develops a contact action and corresponding reaction at that point.

 

Does this help.

Note it is too early for me to draw a diagram but ask if you need one.

 

Because this is such a mouthful for the same effective result, the process is often shortened (as in your book) to pretend that the force of gravity is the contact action.

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5 minutes ago, studiot said:

It works like this:

There are two types of (mechanical) force.

Contact forces
Body forces

Newton's third law pairs are both contact forces.

Gravity is a body force and acts collectively on a body through the centre of cravity of that body.

So the internal collective force of gravity acting through the COG down through its line of action pushes on each element of the body below it until it reaches the contact point with the table.
Here it develops a contact action and corresponding reaction at that point.

I see your point. So we could say:

"The upward normal force from the table and the downward force that the book exerts on the table (ultimately due to gravity) are an action/reaction pair of Newton's third law."

(And "the downward force due to gravity and the upward force that the book exerts on the Earth due to gravity are an action/reaction pair of Newton's third law.")

So the book may not be factually wrong, but I would say it is didactically wrong for not explaining itself clearly.

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Posted (edited)
3 minutes ago, Strange said:

I see your point. So we could say:

"The upward normal force from the table and the downward force that the book exerts on the table (ultimately due to gravity) are an action/reaction pair of Newton's third law."

(And "the downward force due to gravity and the upward force that the book exerts on the Earth due to gravity are an action/reaction pair of Newton's third law.")

So the book may not be factually wrong, but I would say it is didactically wrong for not explaining itself clearly.

Yes.

The following further interesting facts may be noted.

Consider a brick standing on the tbale.

The contact pairs acting on each small element below the COG are compressive.
The contact pairs acting on each small element above the COG are tensile.

A year or two ago I posted an interesting video made with a high speed camera of a spring being first suspended and then dropped.

There is a measureable time delay, visible when the film is slowed, which can be seen between the action of the internal forces and the release of the support force.

I can't immediately locate it at ATM, so if anyone remmbers it please pipe up.

Edited by studiot

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2 hours ago, studiot said:

Yes.

The following further interesting facts may be noted.

Consider a brick standing on the tbale.

The contact pairs acting on each small element below the COG are compressive.
The contact pairs acting on each small element above the COG are tensile.

A year or two ago I posted an interesting video made with a high speed camera of a spring being first suspended and then dropped.

There is a measureable time delay, visible when the film is slowed, which can be seen between the action of the internal forces and the release of the support force.

I can't immediately locate it at ATM, so if anyone remmbers it please pipe up.

What makes you think that?

2 hours ago, studiot said:

 

A year or two ago I posted an interesting video made with a high speed camera of a spring being first suspended and then dropped.

There is a measureable time delay, visible when the film is slowed, which can be seen between the action of the internal forces and the release of the support force.

I can't immediately locate it at ATM, so if anyone remmbers it please pipe up.

Was it a hanging slinky that was dropped?

 

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35 minutes ago, J.C.MacSwell said:

What makes you think that?

It's basically tidal force, I think

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40 minutes ago, Strange said:

It's basically tidal force, I think

Residual and balanced  lateral stresses aside, it's all net compressive across every horizontal plane, maximum at the bottom of the brick and going to zero at the top.

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31 minutes ago, J.C.MacSwell said:

Residual and balanced  lateral stresses aside, it's all net compressive across every horizontal plane, maximum at the bottom of the brick and going to zero at the top.

Are those infinitely many Newton pairs?

 

Contact forces are fields in disguise.

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4 hours ago, Strange said:

Sounds, to me, like the book is wrong

Why?

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5 minutes ago, swansont said:

Why?

Sounds to me like I was wrong!

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4 hours ago, studiot said:

Your book is not 'wrong' .

But then again it is not'right' either!

It is just being economical with the truth.

IOW it is oversimplifying.

Simplifying as introductory physics tends to do. There’s nothing wrong, AFAICT, with the book’s example. It’s not an action-reaction pair.

 

4 hours ago, studiot said:

It works like this:

There are two types of (mechanical) force.

Contact forces
Body forces

Newton's third law pairs are both contact forces.

Newton’s 3rd law applies to all forces, gravity included.

4 hours ago, studiot said:

Gravity is a body force and acts collectively on a body through the centre of cravity of that body.

So the internal collective force of gravity acting through the COG down through its line of action pushes on each element of the body below it until it reaches the contact point with the table.
Here it develops a contact action and corresponding reaction at that point.

This is overthinking the problem, IMO.

Ghideon’s answer was correct. Everything after that has only confused the issue

 

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13 minutes ago, swansont said:

Ghideon’s answer was correct. Everything after that has only confused the issue

Indeed. I wish I had read that answer first.

(my excuse is that I was on my phone. and I didn't have my glasses. and I hadn't had my first cup of coffee. and...)

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1 hour ago, joigus said:

Are those infinitely many Newton pairs?

 

Contact forces are fields in disguise.

No, but I can't remember what is assumed, classically.

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8 hours ago, J.C.MacSwell said:

What makes you think that?

Good question it was a daft idea.  +1

13 hours ago, Ghideon said:

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

Good point much the best explanation. +1

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Ok. Something I didn't understand in the original proposal. Was the table accelerating upwards? Then, of course, it wouldn't be a Newton pair because the system book-table would be accelerating upwards with a total acceleration. Something that wasn't in the premises. Nevertheless I gave Ghideon +1 as the explanation seemed quite thorough.

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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."

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35 minutes ago, joigus said:

Ok. Something I didn't understand in the original proposal. Was the table accelerating upwards? Then, of course, it wouldn't be a Newton pair because the system book-table would be accelerating upwards with a total acceleration. Something that wasn't in the premises. Nevertheless I gave Ghideon +1 as the explanation seemed quite thorough.

 

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Posted (edited)
26 minutes ago, Nedcim said:

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

Only provided the A-B system is free of external forces acting on it.

14 hours ago, Ghideon said:

Now try a case where 1 and 2 are not balaning each other:
Accelerate the table upwards (for instance in an accelerating elevator). Then the normal force would change but gravity remains the same. (1) is still action/reaction pair and (2) is still action/reaction pair but (1) and (2) does not balance out.  

This is the point.

 

So my advice would be: check your book, the diagram, some sentences, because they may clarify that.

Edited by joigus
typo

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Acceleration doesn't change the principle. If something yields to a force, there still is an equal but opposite force.

Essentially as Ghideon stated as an additional case.

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Posted (edited)
53 minutes ago, J.C.MacSwell said:

Acceleration doesn't change the principle. If something yields to a force, there still is an equal but opposite force.

Yes, you're right, but now the system table-book is acted upon by other external forces, so that the normal does not compensate for the weight alone.

I think that's what the book probably means. Otherwise I don't understand what it means. And that's why I suggested books sometimes are not clear.

Although I have no problem to admit that people with more of an engineering background may find clearer something that, to me, that always tend to break things down in terms of fields and such, become more contorted, actually..

------

</eqs.>

I always think better with eqs. F is the external force that pulls up (the lift.) N is the normal, which equals the reaction -N

\[m_{\textrm{book}}\boldsymbol{a}=\boldsymbol{N}-m_{\textrm{book}}\boldsymbol{g}\]

\[m_{\textrm{table}}\boldsymbol{a}=\boldsymbol{F}-\boldsymbol{N}-m_{\textrm{table}}\boldsymbol{g}\]

The action-reaction between table an book is seen here as compensation between N and -N. If you add both equations you get the motion of total system book+table:

\[\left(m_{\textrm{book}}+m_{\textrm{table}}\right)\boldsymbol{a}=\boldsymbol{F}-\left(m_{\textrm{book}}+m_{\textrm{table}}\right)\boldsymbol{g}\]

But N does not compensate with the weight of anything. It's just the pair N, -N that compensate each other. So (N, weight) is not a Newton pair.

You don't need to write F twice, as it only acts on the table. The book never "knows" of any other force but the normal reaction and its own weight. 

</eqs.>

Edited by joigus
add eqs. / (N,-N) --> (N, weight)

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55 minutes ago, 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."

Gravity is a force the earth exerts on the book. The reaction for this would be the gravitational force the book exerts on the earth.

The reaction force to the normal force exerted by the table on the book would be the normal force exerted by the book on the table.

In both cases, the force pairs are exerted by one object on the other. The forces are the same type. In the example, neither of those is true.

The action-reaction force pairs are in opposite directions and equal in magnitude, which is also true in the example. (and can’t be used to draw a conclusion in this example)

 

 

32 minutes ago, joigus said:

Yes, you're right, but now the system table-book is acted upon by other external forces, so that the normal does not compensate for the weight alone.I

Irrelevant, though, to the issue of action/reaction force pairs.

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Posted (edited)
17 minutes ago, swansont said:

Irrelevant, though, to the issue of action/reaction force pairs

I don't think it is, if what one naively thinks is that (weight-normal reaction) are canceling each other when the system is accelerating upwards. They're not. Forces that cancel are (N, -N) N not equaling "minus a weight" but also involving the pulling force (and so augmenting the apparent weight against the table).

16 hours ago, Nedcim said:

normal force from the table and the downward force due to gravity are not an action/reaction pair of Newton's third law.

And that was the whole point.

As show in the equations of motion above.

Edited by joigus
addition

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