# Newton's third law pair

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21 minutes ago, swansont said:

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

Except to highlight the difference between unbalanced forces and action/reaction pairs.

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25 minutes ago, joigus said:

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).

Whether they cancel doesn’t matter, since cancellation requires that the forces act on the same object. Action/reaction pairs act on different objects. They don’t cancel.

Bringing up misconceptions in discussion often has the unfortunate effect of reinforcing the misconception. When one later recalls the information, the fact that force cancellation were part of the conversation will be remembered. (when I taught we called this the “not” filter. If you tell someone that X is not true, they will remember you saying that X is true.)

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17 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."

1 hour 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."

And I still have no clearer surety of what was exerting what force on what, since you have not told us everything.

Either there was more text or a diagram.

What was the table supporting to exert and upward normal force on?

(That was why I chose a brick.)

And whatever this object being supported let us try to match to the book's definition.

What was object A ?

What was object B ?

Was a third object C introduced in the book definition because in the example there are three objects.

The Earth, the brick and the table.

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Posted (edited)
53 minutes ago, swansont said:

They don’t cancel.

Ok. They don't cancel. They compensate each other so that, when you consider book+table as a whole, the mutual action reaction pair (N, -N) doesn't even appear in the equations of motion for composite system. I don't think that's a misconception. Maybe it's a misnomer. And I even accept the responsibility that it could be a pedagogical faux pas.

But I've been teaching physics for 40 years now and I know only too well that many students spend hours racking their brains trying to figure out what the constraint forces are or whether they obey some kind of law, when from the mathematical point of view, they're only there as dummy parameters that connect one subsystem's eq. with another, and one'd better eliminate them if one wants to solve the problem.

I do have the feeling that you have a point in what your objections to my pedagogical approach are, and you may be right. I'm too used to solving problems from the mathematical perspective. In analytical mechanics constriction forces sometimes don't even push or pull in the same direction, like tensions in a system of pulleys. It can get very confusing. So I try to formulate the problem mathematically, remove the mathematical nuisances, and then follow my nose. That's what I've tried to do here. And it seems to work for me.

But last time I looked the student was still confused. Maybe it was my fault. If that's the case, I apologise.

PD.: Now that I look back to my replies, it's the word "compensate" that I used the most. Only in a hurried response I said "cancel." Again, my apologies.

Edited by joigus

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2 hours 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."

Because the fact that the forces are equal but opposite are coincidental...(giving us a static case due to the forces balancing)

As already suggested, if the table was accelerating upwards they would not be equal.

If the table was frictionless but not perfectly level the forces would not be opposite.

But as mentioned all these scenarios only involve forces that have equal but opposite counterparts. Every action having a reaction. For the static case the force the table exerts on the object balances the Earth's gravitational pull on the object... but that is not the equal but opposite force of Newton's third law. The equal but opposite gravitational force on the brick is the brick's equal but opposite gravitational force on the Earth...and the equal but opposite force of the table on the brick is the force of the brick on the table.

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52 minutes ago, joigus said:

Ok. They don't cancel. They compensate each other so that, when you consider book+table as a whole, the mutual action reaction pair (N, -N) doesn't even appear in the equations of motion for composite system. I don't think that's a misconception. Maybe it's a misnomer. And I even accept the responsibility that it could be a pedagogical faux pas.

Action-reaction pairs don’t compensate for anything, either.

They don’t appear in equation of motion because one of the forces doesn’t act on the body. There is no composite system under discussion. That’s a distraction to the question that was posed. The question asked gives no reason to look at the book+table system.

The issue of misidentifying action-reaction pairs is a quite common misconception that students have, in my experience. It’s best not to feed it.

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Posted (edited)

Swansont, with the utmost respect, as you deserve no less, I disagree.

1 hour ago, swansont said:

Action-reaction pairs don’t compensate for anything, either.

AAMOF they do: they compensate for momentum. Otherwise momentum wouldn't be conserved. For actions-reactions at a distance governed by fields, the question is more involved: Both subsystems can accelerate towards or away from each other so that total momentum is conserved.
But the normal reaction of a table is a constriction force. That's got some peculiarities to it. Not only the subsystems cannot accelerate towards or away from each other: They can't even displace with respect to each other. Precisely because of the cancellation/compensation or whatever term may be acceptable.
1 hour ago, swansont said:

They don’t appear in equation of motion because one of the forces doesn’t act on the body.

AAMOF they do. I've already shown it with enough detail in the mathematics of the problem:

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

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

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

They do appear in the eq. of motion of sub-systems A and B; they don't appear in the eq. of motion of the system A+B. Now, for eqs. 1st and 2nd that's what I call "to appear in the eqs. of motion." For eq. 3rd, the N-N term is what I call "compensation," or whatever term may be acceptable.

1 hour ago, swansont said:

There is no composite system under discussion. That’s a distraction to the question that was posed. The question asked gives no reason to look at the book+table system.

That depends on the completion to the question that was posed, which is only too obviously incomplete as is. As Ghideon has shown, if the system is at rest (a particular case of what he meant: a=0), action and reaction exactly compensate and the weight equals the normal reaction. So it would be a Newton pair, contrary to what the book states.
On the other hand, were it not a Newton pair, it could be for no other reason than that the system is not in dynamical equilibrium. In that case, it's not a distraction, it's the key to the question.
$\textrm{table}+\textrm{book in equilibrium: (}\boldsymbol{a}\textrm{'s}=0\textrm{)}\Rightarrow\boldsymbol{N}_{\textrm{table}},\boldsymbol{W}_{\textrm{book}}\textrm{are a Newton pair}$
$\Rightarrow$
$\boldsymbol{N}_{\textrm{table}},\boldsymbol{W}_{\textrm{book}}\textrm{not a Newton pair}\Rightarrow\textrm{table}+\textrm{book not in equilibrium: (}\boldsymbol{a}\textrm{'s}\neq0\textrm{)}$
In the second case, you cannot but consider both accelerating, as Ghideon very clearly has argued. The simplest case being both having the same acceleration, which is what Ghideon has proposed and I have explicitly shown in detail.

I will fight this to the bitter end until proven wrong. And nothing would please me more than be proven wrong by you.
Edited by joigus
bad rendering of eq.

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2 hours ago, studiot said:

And I still have no clearer surety of what was exerting what force on what, since you have not told us everything.

Either there was more text or a diagram.

What was the table supporting to exert and upward normal force on?

(That was why I chose a brick.)

And whatever this object being supported let us try to match to the book's definition.

What was object A ?

What was object B ?

Was a third object C introduced in the book definition because in the example there are three objects.

The Earth, the brick and the table.

Here's the diagram:

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9 hours ago, Nedcim said:

Here's the diagram:

In a valid free-body diagram, no action-reaction force pairs will appear. Because in a FBD, all forces are acting on the same object, while for action-reaction, it's a force acting between two different objects.

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31 minutes ago, swansont said:

In a valid free-body diagram, no action-reaction force pairs will appear. Because in a FBD, all forces are acting on the same object, while for action-reaction, it's a force acting between two different objects.

Would "forces on two different objects" be more accurate?

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10 hours ago, joigus said:

Swansont, with the utmost respect, as you deserve no less, I disagree.

AAMOF they do: they compensate for momentum. Otherwise momentum wouldn't be conserved. For actions-reactions at a distance governed by fields, the question is more involved: Both subsystems can accelerate towards or away from each other so that total momentum is conserved.

But the normal reaction of a table is a constriction force. That's got some peculiarities to it. Not only the subsystems cannot accelerate towards or away from each other: They can't even displace with respect to each other. Precisely because of the cancellation/compensation or whatever term may be acceptable.

AAMOF they do. I've already shown it with enough detail in the mathematics of the problem:

mbooka=Nmbookg

mtablea=FNmtableg

(mbook+mtable)a=F+NN(mbook+mtable)g

They do appear in the eq. of motion of sub-systems A and B; they don't appear in the eq. of motion of the system A+B. Now, for eqs. 1st and 2nd that's what I call "to appear in the eqs. of motion." For eq. 3rd, the N-N term is what I call "compensation," or whatever term may be acceptable.

I don't see meartha in your equations (i.e. the reaction force to the weight of the book).

And you have not labeled your normal forces to indicate the source of that force. The normal force of the book on the table, or the normal force of the table on the book. Only one of them appears in an equation. Not both.

What is F in the equation? It's not given in the problem.

Quote

That depends on the completion to the question that was posed, which is only too obviously incomplete as is. As Ghideon has shown, if the system is at rest (a particular case of what he meant: a=0), action and reaction exactly compensate and the weight equals the normal reaction. So it would be a Newton pair, contrary to what the book states.

On the other hand, were it not a Newton pair, it could be for no other reason than that the system is not in dynamical equilibrium. In that case, it's not a distraction, it's the key to the question.

Whether the weight and normal force are equal in magnitude has nothing to do with action-reaction.

FAB = -FBA

There are no caveats on that equation. If the book is in freefall, the earth still feels a pull from the book, and the two forces are equal and opposite.

mearthaearth = -mbookabook

10 hours ago, joigus said:

table+book in equilibrium: (a's=0)Ntable,Wbookare a Newton pair
Ntable,Wbooknot a Newton pairtable+book not in equilibrium: (a's0)
In the second case, you cannot but consider both accelerating, as Ghideon very clearly has argued. The simplest case being both having the same acceleration, which is what Ghideon has proposed and I have explicitly shown in detail.

I will fight this to the bitter end until proven wrong. And nothing would please me more than be proven wrong by you.

There is no way I can reconcile this with Newton's third law. The weight of the book is not something that is acting on the table, as the weight of the book is not a gravitational force. The force the book exerts on the table is a normal force.

16 minutes ago, J.C.MacSwell said:

Would "forces on two different objects" be more accurate?

It's a true statement, but it's not just that it's two different objects. It's the same two objects, and category of force. I would say that your statement is less precise.

Gravity acts on the book, and on the table, so that's forces acting on two objects. But those are not an action-reaction force pair.

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Thank you Ned for the diagram. +1

I applaud Swansont's drive back to basics, and your diagram helps com[plete the picture (pun intended)

I had hesitated to introduce the words "Free Body diagram", but there they are large as life in your last post.

So I assume you have studied these at least a little bit.

I take it that the pair of arrows on the right is meant to be the diagram.

Unfortunately in my opinion that is a very poor example of a free body diagram, partly since the body is not shown.
Obviously the table is not the body in question so should not appear in the FBD.

But all forces should be shown in such a way that their point of application to the body is indicated by the head of the arrow.

Neither of the two diagrams in your picture indicate this.

Worse the left hand diagram seems to indicate that the upward reaction on the brick from the table is not in the same line of action as the downward force.
This implies that there is a couple or moment acting, which I'm sure is not intended.
Note the situation as drawn (with the moment) is entirely possible and crucially important in foundation engineering but I don't want to complicate things unless you want to.

To my mind a proper free body diagram draws a line around the boundary of the body isolated and applies boundary forces exactly where they occur on that boundary and body forces where they occur within the body. In this case it looks more this.

The only forces (W & N), shown are those that maintain the body in exactly the same situation as it was before if it was suddenly cut off from the rest of the universe by the dashed boundary. Note this boundary can slice through objects. Note also it is good practice to state the body that is 'free' so there is not confusion. All forces are applied to that body and no other.

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56 minutes ago, studiot said:

Thank you Ned for the diagram. +1

I applaud Swansont's drive back to basics, and your diagram helps com[plete the picture (pun intended)

I had hesitated to introduce the words "Free Body diagram", but there they are large as life in your last post.

So I assume you have studied these at least a little bit.

I take it that the pair of arrows on the right is meant to be the diagram.

Unfortunately in my opinion that is a very poor example of a free body diagram, partly since the body is not shown.
Obviously the table is not the body in question so should not appear in the FBD.

But all forces should be shown in such a way that their point of application to the body is indicated by the head of the arrow.

Neither of the two diagrams in your picture indicate this.

I wonder if this convention is geographic in nature. FBDs I have seen and made look a lot like the one shown.

56 minutes ago, studiot said:

Worse the left hand diagram seems to indicate that the upward reaction on the brick from the table is not in the same line of action as the downward force.

That is unfortunate, but otherwise the arrows would have to overlap, which might be more confusing.

56 minutes ago, studiot said:

This implies that there is a couple or moment acting, which I'm sure is not intended.

Given that action-reaction is typically introduced very early in the curriculum, I am also sure it's not intended.

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5 minutes ago, swansont said:

I wonder if this convention is geographic in nature. FBDs I have seen and made look a lot like the one shown.

That is unfortunate, but otherwise the arrows would have to overlap, which might be more confusing.

Given that action-reaction is typically introduced very early in the curriculum, I am also sure it's not intended.

Have you seen Wikipedia, which is American?

Just like my diagram, but much prettier.

Since you like the Normal reaction passing through the body, would you also advocate the friction to pass through it, rather than along the line of contact?

Quote

Whilst it is clear you were not confused, others here have indicated confusion.

(I even confused myself at first )

However I have to admit I am ashamed of the BBC for producing this rubbish, if this is now what is officially taught in UK schools.
The line of action of the bodyweight does not even pass through the COG in their diagram.

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24 minutes ago, studiot said:

Have you seen Wikipedia, which is American?

I wasn't aware the Wikipedia prevented editing or participation based on geography.

If you Google free body diagrams, you will find plenty of the other kind

Take a peek at one of the permutations of Halliday & Resnick, a very popular into to physics textbook

Quote

Just like my diagram, but much prettier.

Since you like the Normal reaction passing through the body, would you also advocate the friction to pass through it, rather than along the line of contact?

No, and using the dot renders that moot.

Quote

Whilst it is clear you were not confused, others here have indicated confusion.

By the diagram? It was only recently posted.

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I read all the explanations on this topic, but I still see the example as a contradiction to Newton's third law. Here's my reasoning:

Book pushes on desk with force Fab; Desk pushes on book with a Fba.

Equal but oppositely diected forces, with the forces acting on different objects consistent with Newton's third law.

The same approach is applied in statics with the book being replaced by a truss. The forces applied through truss are transmitted to the reactions(desk). The force reactions are opposite but equal to that of the corresponding member.

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. The normal force always makes a 90 degree angle with the surface (perpendicular)."https://energyeducation.ca/encyclopedia/Normal_force

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9 hours ago, swansont said:

It's a true statement, but it's not just that it's two different objects. It's the same two objects, and category of force. I would say that your statement is less precise.

Gravity acts on the book, and on the table, so that's forces acting on two objects. But those are not an action-reaction force pair.

Probably true for the reason you stated, but...

10 hours ago, swansont said:

In a valid free-body diagram, no action-reaction force pairs will appear. Because in a FBD, all forces are acting on the same object, while for action-reaction, it's a force acting between two different objects.

...Isn't that generally true of the forces seen in an FBD? Only the one of the any action-reaction pair, one acting on the free body being of interest, or needing to be considered in the FBD?

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4 minutes ago, J.C.MacSwell said:

Probably true for the reason you stated, but...

...Isn't that generally true of the forces seen in an FBD? Only the one of the any action-reaction pair, one acting on the free body being of interest, or needing to be considered in the FBD?

That is my understanding.

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18 minutes ago, J.C.MacSwell said:

Probably true for the reason you stated, but...

...Isn't that generally true of the forces seen in an FBD? Only the one of the any action-reaction pair, one acting on the free body being of interest, or needing to be considered in the FBD?

Right. the force exerted on the object is in the FBD. The force exerted by the object is not.

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56 minutes ago, Nedcim said:

I read all the explanations on this topic, but I still see the example as a contradiction to Newton's third law. Here's my reasoning:

Book pushes on desk with force Fab; Desk pushes on book with a Fba.

Equal but oppositely diected forces, with the forces acting on different objects consistent with Newton's third law.

The same approach is applied in statics with the book being replaced by a truss. The forces applied through truss are transmitted to the reactions(desk). The force reactions are opposite but equal to that of the corresponding member.

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. The normal force always makes a 90 degree angle with the surface (perpendicular)."https://energyeducation.ca/encyclopedia/Normal_force

That is the definition of a normal force. Sometimes it is a component of force, say for something sitting on an incline plane, the weight would be counterbalanced by the normal force and the force of friction.

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Posted (edited)
11 hours ago, swansont said:

I don't see meartha in your equations (i.e. the reaction force to the weight of the book).

That would be an inertial term (ficticious force if you will) for the Earth moving at upward-lifting acceleration a. A lift pulling from book and table would not accelerate the Earth, as the lift would be "holding its ground" on it. I don't suppose a lift will accelerate the whole Earth upwards with acceleration a.

11 hours ago, swansont said:

And you have not labeled your normal forces to indicate the source of that force.

Fair enough:

$\boldsymbol{N}=\boldsymbol{F}_{\textrm{table-book}}$

$-\boldsymbol{N}=\boldsymbol{F}_{\textrm{book-table}}$

$\boldsymbol{F}_{\textrm{table-book}}=-\boldsymbol{F}_{\textrm{book-table}}$

As I said, constriction forces (for time-independent holonomic constraints) don't need to be indexed very carefully; they're dummies of the dynamical problem. They're better left out. They disappear from the problem almost as fast as one writes them down. And they always involve action and reaction, but a very specific case of it (no relative motion: holonomic, time-independent) makes them astonishingly simple to deal with mathematically, while a pain in the brain from the conceptual point of view.

11 hours ago, swansont said:

What is F in the equation? It's not given in the problem.

Sorry, I mentioned it before in a post that you may have missed:

On 5/25/2020 at 10:38 PM, joigus said:

F is the external force that pulls up (the lift.)

Where do the pull and the lift come from? If you have accepted Ghideon's analysis:

On 5/25/2020 at 7:43 AM, Ghideon said:

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.

(my emphasis)

which I think you have:

On 5/25/2020 at 3:17 PM, swansont said:

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

(my emphasis)

Then you need a force F to account for that upward lifting acceleration. Or maybe Ghideon's analysis started the confusion, which I don't think is the case and your past self doesn't seem to either.

11 hours ago, swansont said:

Whether the weight and normal force are equal in magnitude has nothing to do with action-reaction.

Agree 100 %, that's why I never said that. Only when they are the only forces between book and table, they are action-reaction pair. So there's nothing intrinsically action-reaction between them.

11 hours ago, swansont said:

The weight of the book is not something that is acting on the table

The weight of the book is not acting at any particular point. From a fundamental point of view, there is no such thing as "the weight of the whole book" or the "center of gravity" or "COM" as a particular point where some force is really acting. Every "particle" in the book is experimenting its own weight; in Newton's scheme of things, it's all just point particles. The consideration of the weight of the book acting on the centre of gravity is just a convenient and justifiable simplification because rigid bodies can be treated as having just 6 degrees of freedom. When they're not subject to torques, the picture can be further simplified to represent the extended body as a point particle with only 3 DOF. But it's not true in any fundamental way. And that is a very fundamental misconception.

The latter is a theorem about the dynamics of systems of particles. And it works like a dream. That's why many extremely clever people like yourself can follow that dream and not make mistakes, but if you want to make any fundamental reasoning, or strongly justify definitions, you must be keenly aware that you're in a theoretical dream. And I really hope my prose doesn't spoil my point, because I only intend to be clear.

There are other techniques, much more powerful, like Lagrangian mechanics, which I particularly prefer, because you forget all that nonsense about forces applied to such and such point.

You've got more points in your post, but I'm skipping them for lack of time and because I perceive some "we agree to disagree" basis in them. In the end, it's likely that we only have different points of view on mechanics and work on different systems of utilitarian definitions.

I still think the OP remains confused though, in spite that we all are taking time from our personal and professional lives to help. I've been (or consider myself to) fortunate enough to spend hours on end studying Feynman in the library --while my classmates mocked me-- and learnt that "contact forces are an illusion." And I took it to heart.

I remain your most devoted follower. I learn loads from you. I just don't think you're always right. And I just don't think you're right about this.

And please forget all that nonsense I said about the bitter end. There is no bitterness here. Most of us do what we love most, which is discuss science.

Joigus

Edited by joigus
mistyped

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1 hour ago, J.C.MacSwell said:
2 hours ago, Nedcim said:

The normal force always makes a 90 degree angle with the surface (perpendicular)."https://energyeducation.ca/encyclopedia/Normal_force

That is the definition of a normal force. Sometimes it is a component of force, say for something sitting on an incline plane, the weight would be counterbalanced by the normal force and the force of friction.

Well yes the any 'normal' is at right angles to or perpendicular to something.

But if you are going to accept swansont's assertion that gravitational forces form an 3rd law action / reaction pair 'normal' makes no sense.

The gravitational pair both act along the line joining the centres of gravity of the two objects.
The actual surfaces of thes bodies may not be in contact and may not be at right angles to this line of action.

Furthermore they are always acting  -  they never stop.
Unlike contact forces which only act during contact and cease immediately contact is discontinued.

So it makes sense to split the contact force into a normal and tangential components in a way that may not be appropriate for other forces.

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Posted (edited)
On 5/25/2020 at 11:30 PM, J.C.MacSwell said:

Except to highlight the difference between unbalanced forces and action/reaction pairs.

Exactly!!!!! Sorry I didn't see this before. Because for unbalanced forces it's not an action/reaction pair, you could say they're not a fundamentally coupled Newton pair, which I think is what the book is trying to say. At least, it's the only sense in which I am capable of understanding it. +1

Edited by joigus
Added 3 exclamation marks

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23 minutes ago, studiot said:

Well yes the any 'normal' is at right angles to or perpendicular to something.

But if you are going to accept swansont's assertion that gravitational forces form an 3rd law action / reaction pair 'normal' makes no sense.

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.

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Thanks for feedback on my post and for adding lots of interesting additional points! I’m trying to put together some clarifications and additional explanations for OP but I’m slightly more busy than usual at this time.

Quick note for @joigus and others: The reason for introducing acceleration in my initial post was to illustrate my answer. I could have explained that better, I blame it on

On 5/25/2020 at 3:32 PM, Strange said:

I hadn't had my first cup of coffee.

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