# Action reaction forces: what is the reaction to tension?

1. Jun 6, 2012

### fisico30

hello forum,

consider an object attached to a rope and rotating in a horizontal loop. The tension T in the rope provides the centripetal force points toward the center of the loop.
But forces always exist in pair (action-reaction).

what is the other force in the pair where one of the force is the tension T?

The object, due to its inertia, naturally tends to move along a line tangent to the circular loop. Because of that, the object pulls on the rope causing a tension in it...But we cannot say that that force is inertia, since inertia is not a force....

It is not even the centrifugal force, which is only an apparent, fictitious force that only exists because the object is inside a rotating system. It is not a physical force either...

thanks,
fisico30

2. Jun 6, 2012

### Studiot

Draw three free body diagrams the - rope anchorage, the rope and the body.

Consider the point of attachment of the rope to the body.

What internal force acts on the end of the rope?

What is it's motion?
In particular is it tavelling along its length (ie tangential to the circular motion)

So what is the net force acting on that end of the string?

Can you now see the reaction from the body on the end of the rope?

3. Jun 6, 2012

### fisico30

Well, the tension T is the centripetal force...
Somehow the object, with its inertia, causes a force that is equal and opposite in direction to T....let's call that force F.

So F and T are the action-reaction pair.....It is interesting how the object, with its inertia to move along the tangent to the trajectory, causes a force F that is perpendicular to its motion....

In an old thread I found here on physics forum, I discovered that people used to call that force F the centrifugal force....
but that is an old way of calling it...
Today the centrifugal force is a fictitious force that shows up in rotating frames of reference...

thanks,
fisico30

4. Jun 6, 2012

### Ken G

The picture you are describing can be used, but it is an approach that is fraught with peril! It is essentially the picture of entering the non-inertial frame of the object, and then we would indeed call F the "centrifugal force," even today (this is not some kind of old-fashioned terminology). You say we can't do that, because that's not a real force, but it is what you are calling for when you want the forces on the object to balance.

The safer approach is to stay in the non-inertial frame in which the object is going in a circle, and just note that objects going in a circle are accelerating-- so the forces on them do not balance. You are asking for a force to balance T, but the forces on the object don't balance, so there is no force that balances T.

It sounds like the reason you want the forces to balance is you are looking for an action/reaction pair, but remember, if action/reaction pairs always balanced on every object involved, then forces could never do anything-- like they could never make objects go in a circle! It's only in the frame of the object that the forces balance, and that's not an action/reaction pair, it's a real force / fictitious force pair (like centripetal/centrifugal).
An action/reaction pair is always the same type of force, so the pair force of a tension is always also a tension. They also never act on the same object, they are a relationship between two different objects, so they only balance when you regard both objects as part of the same ("closed") system. The action/reaction tension force is a pair of forces that act on any two segments of the rope if you mentally divide the rope in half. With an analysis, you can see that this will also mean that effectively, the tension action/reaction is a pair of forces, each at opposite ends of the rope, that pull toward the center of the (massless) rope. This pair of forces is there because the rope is slightly stretched, and is acting like a spring.

Last edited: Jun 6, 2012
5. Jun 6, 2012

### fisico30

Thanks Ken G.

But at the point of contact between the rotating object and the rope, the tension T is the force ON the object caused by the rope, and the object must exert a force on the rope. The two forces form the action-reaction pair.

One end of the rope is constrained (fixed) and the other end is attached to the rotating object...

I am sorry, but I am still not clear on your conclusion: isn't the rope exerting a force on the object, hence the object will exert a force on the rope (considering the system rope/object)?

thanks,
fisico30

6. Jun 6, 2012

### Studiot

Yes of course the object exerts a reaction force on the rope. That is the whole point.

Ken said

The reaction acts on the rope, not the object.

The only force acting on the object is the 'action' (pull) from the rope.

Thus, as Ken said, the object is not in equilibrium under the action of external forces.

However the rope is in equilibrium (remember I asked if it translates) because the two forces acting on the rope are the reactions from the object at one end and the point of attachment at the other.

Since we are not interested in calculations about the rope we usually ignore this.

For calculations about the object we have two choices.

1) Newtons method.

Since the object is not in equilibrium we cannot apply the rules of equilibrium but must apply N2 (as Ken said) and calculate the acceleration due to that force. The calculation is completed in dynamic form.

2) D'Alamberts method.

Create a quasi equilibrium by applying a balancing ficticious force and solve using the rules of force equilibrium.

This leads to the same ultimate solution.

I repeat, draw the three free body diagrams.

Edit A little tip here:

You are making the classic error of mixing up internal and applied forces.

There is no tension acting on the object. Tension is strictly an internal force.
The object is regarded as a point mass and therefore has no internal forces.
External forces may be applied as a push or a pull and it is often difficult to identify which. In this case we can.
The rope has dimensions - it is not a point -so has internal forces transmitting the applied reactions at each end. These are of course the tensions. And obviously the tension is equal to the one of reaction pair at each end ie the pull on the object.
It is very easy to do here because it is conventional to skip this step and just say the tension in the string pulls the object.

Last edited: Jun 7, 2012
7. Jun 6, 2012

### fisico30

"....However the rope is in equilibrium (remember I asked if it translates) because the two forces acting on the rope are the reactions from the object at one end and the point of attaachment at the other...."

I agree on that. You say that there is a reaction from the object...that is the force I am talking about...

the object is not in equilibrium since there is a net, centripetal force, the pull from the rope....which is the tension....

I am confused when you say that "There is no tension acting on the object."...
I can see how there is the same tension along the whole rope, at every point (even if the rope is inhomogenenous) since the rope does not move....But the end of the rope provides the pull on the object....that pull is equal to the tension...

8. Jun 7, 2012

### Studiot

Well at least you read some of what I wrote, but you are still confusing internal and external (applied) forces. That is why I distinguished between the tnesion within the rope and the pull on the object. But yes, of course these have the same magnitude.

Did you draw the free body diagrams, I don't see them posted or described.

Hint internal forces do not appear unless there is cut section somewhere.

Meanwhile consider this.

Two blocks of weight W are stacked on top of each other and rest on a table.

Is the bottom block in compression or tension?

Is the top block in compression or tension?

9. Jun 7, 2012

### Ken G

Yes, and both forces in the pair are considered tension. To get a tension action/reaction pair, you can divide the rope arbitrarily in two parts, including the case you mention-- where one part is the very end of the rope.

10. Jun 7, 2012

### Studiot

Ken, why do you consider the external force to be tension?

In this case the direction is obvious but there are situations in which it is not possible to distinguish.

11. Jun 7, 2012

### Ken G

The forces referred to in the action/reaction pair is the force of the rope on the object, and the force of the object on the rope. This pair is essentially the "end of rope" version of a similar pair everywhere within the rope. The latter is certainly considered tension, and I think so is the former, i.e., the force a rope exerts on an object is generally called tension, so it's natural to also call tension the force of the object on the rope.

The centrifugal force is something different, that was answered above.
For a (massless) rope, the tension pair always points along the local direction of the rope. You are talking about rigid objects that can support lateral stresses, which does make it a lot harder.

12. Jun 8, 2012

### Studiot

OK here is why I said that we normally skip a step because the force on the object equals the tension.

It is impossible to put an object in tension (or compression) by applying a single force.
Indeed we are using the single force on the object here to provide an acceleration. It is illogical to say we are applying only one force to the object and then say it is a tension.

Tension requires the application of two external forces.

Here are the free body diagrams to illustrate what I mean.

Block A is neither in compression or tension. Block B is in compression. Since all the forces are colinear there are no moments involved. Block A has one pull and one push. Block B has one pull and two pushes.

Fig2 represents the object whirling on a string.

The rope is in tension and has two opposing external forces acting on it so it is in equilibrium.
These two forces are the reactions at the attacment points of the rotation centre and the object. Both are pulls, but the object cannot be in tension because this is the only force acting on it. The rotation centre attachment must in turn have a fixing force or it would also move.

Fig3 represents a pinned frame or chain stretched out sideways by forces applied at A and B.
The frame is in tension, but the applied forces may be pulls applied by ropes as in 3.
Or they may be pushes applied by jacking against a foundation as in Fig4

Fig4 therefore represents a case where the internal force is a pull but the applied (jacking) force is a push.

Both figs3 and 4 could represent a fairground or other machinery whirling arm with an object at the end like the rope. Would you consider the force on that object to equal the internal tensions in the frame?

Fig5 since you mentioned transverse forces the same situation can pertain here. Here is another pinned frame on supports with transverse reactions at A and B. The reactions are pushes.

Fig6 shows how to achieve the same supports via hanging chains.

So your statement that action and reaction are of the same type only holds true if you do not directly pair internal forces against external applied ones.

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13. Jun 8, 2012

### Ken G

I'm not sure I could really follow that, but I assure that action/reaction pairs are always forces of the same type. This is certainly true at the atomic level, and it is also true when we group effects together into catch-all terms like friction, tension, the normal force, and so on.

14. Jun 8, 2012

### Studiot

What didn't you follow?
I am sorry if it was not clear.
I have shown how to create tension within a body by applying opposing pushes to its ends.

The point is that tension is not a single force it is always a force pair. It cannot exist as a single force. If I apply a single force to a free body please demonstrate how it could create tension in that body. I maintain it would create acceleration.

Since the original question is about dynamics here is another example.

Consider weights, W1 & W2 hanging from respective ends of a light rope wound over a frictionless pulley.
What is the tension in the rope?
Consider the forces acting on each weight. Are you suggesting that W1 & W2 are tensions ?

Many students have trouble with this type of question and I think that is partly due to the short cut in equating the pull directly to the tension.

edits in italics

Last edited: Jun 8, 2012
15. Jun 9, 2012

### Ken G

Of course, that's just Newton's third law.
I have no idea why I would need to demonstrate that. The tension is not a force in the body, it is a force on the body (and inside the rope, of course).
Certainly not, they are gravity forces. However, a force exerted by a rope is generally regarded as tension. Granted, you could reserve the term for internal forces in the rope, and call the action/reaction pair at the very end of the rope something different, but there's just no point in doing that and it is not generally done that way (since the magnitudes of the tension action/reaction pair are everywhere the same in a massless rope, including at both ends).
The pull of gravity has nothing to do with the pull of tension, although certainly there are situations where the two might be required to balance each other. In the problem you give, the tension is determined by a different requirement, that the accelerations of the two masses have the same magnitude. I still don't see the relevance to anything I said above, the force exerted by a rope is still considered to be the tension in the rope (which is everywhere the same, including at both ends, if the rope is regarded as massless). It's the same for a spring force, which also comes in action/reaction pairs and is found internally in the spring as well as at its ends. A spring force is a whole lot like a tension, all that is different is the connection between the force and the length of the spring vs. the length of the rope (the former is a linear relationship, the latter is very steep).

Last edited: Jun 9, 2012
16. Jun 9, 2012

### Studiot

I'm glad you agree at last.

You did want to specify that the action/reaction pair ar of the same type. If you do this you are forced to allow the special pair at the interface.

17. Jun 9, 2012

### fisico30

Thanks everyone.

Studiot: I think the bottom object is under compression while the top one is under tension...(but not really sure).

As far as the rope/rotating object situation goes: the rope pulls on the object and the object pulls on the rope (Newton's 3rd law).
The rope is massless and the net force on it is zero: all the force are equal and opposite. The internal forces cancel each other while the two forces T1 and T2 at the end of the rope(one at the pivotal point and the other at the connection point with the object) are equal and opposite. T2 is caused by the object and points radially out. The rope, in the meantime, causes a radially inward force on the object (centripetal force) which is equal and opposite to T2..

I am a still really off?

The rope is under "tension" due to the moving object. If the objet was not moving the rope would fall flat...
The rope is not parallel to the floor: one of its components is supporting the weight of the object. I am still amazed by this! It is easy to understand with force diagrams maybe, but it is still not each to understand conceptually...somehow the fact that the object rotates stretches the rope and causes it to support it, make it float....(gyroscopic effect)

thanks,
fisico30

18. Jun 9, 2012

### Ken G

I would have agreed at any time, including right from the start.
I still say that an action/reaction pair is always the same type, and I see no contradictions to this widely held view. I don't even see why you seem to see a contradiction. I see nothing "special" about the pair at the interface that distinguishes it from the pairs throughout the rope, which is exactly why they are all regarded as "tension."

19. Jun 9, 2012

### Ken G

No, I'd say you have it exactly.
Yes, picturing this is actually one of the values of the concept of "centrifugal force" (which is still a useful concept you see invoked all the time). The centrifugal force is what we call a "coordinate force" (which is a better term than "fictitious force", which is somewhat denigrating!), which means it is a term that acts as though it were a force, but in a noninertial reference frame.

It stems from our choice of a rotating coordinate system, and is there because we need to include it to get coordinate velocities that don't change with time in the absence of any real forces. By "coordinate velocity", I mean rate of change of a distance component in our rotating coordinate system, which is not the same as a real velocity (which is a vector and is coordinate independent). For example, the rock on the string has a coordinate velocity of zero in the rotating coordinates-- its coordinate is staying the same, so it is not "moving" at all in those coordinates.

But you well know that to have a rotating coordinate velocity stay zero, it must have a real net force on it, called the centripetal force. The point of the centrifugal force is to be able to imagine we have a force balance, since the coordinate velocity is not changing, so that there's no acceleration in the coordinate velocity components. So if we manually tack on the centrifugal force in our rotating coordinates, and include it in F in F=ma, we can treat a as the coordinate acceleration, and not worry that having zero coordinate acceleration does not mean the object isn't really accelerating. In short, centrifugal force is a way to pretend that F=ma pertains to coordinate acceleration a, whereas if we limit F to real forces, then F=ma only pertains to the coordinate-free vector acceleration (which you might call the real acceleration).

The reason I'm resuscitating the centrifugal force in this way is that it is very useful for understanding the behavior of the spinning rock, which indeed does seem mysterious seen from our inertial perspective. Once the rock reaches its steady-state revolution, we can enter the rotating frame, and draw an effective free-body diagram, tacking on the centrifugal force, and use it to completely understand why the rock does what it does, exactly as if the rock was not revolving at all, and had a real force on it that is the same as the centrifugal force. It's a way to make F=ma work the same in a rotating frame as it does in a non-rotating frame, so if you can understand a stationary rock's behavior in a non-rotating frame with a real force on it, you can understand its behavior in the rotating frame with a centrifugal force on it, as the two situations are equivalent.

20. Jun 10, 2012

### Andrew Mason

The reaction to the centripetal force on the object is the object's force on the earth. Both forces are communicated between the object and the earth via the rope. Since the rope is considered massless its function is only to conduct the forces between the earth and the object.

In any description of the forces involved in the motion of macroscopic bodies you will have forces which result in acceleration of macroscopic bodies and tensions within those bodies and the objects that connect them. To analyse them, you have to separate the tensions from the forces that create accelerations.

A better example would be two masses tethered by a rope of negligible mass and each rotating about the centre of mass of the two-body system. There are two accelerating bodies and equal and opposite forces between them. You do not have to worry about the tension as a separate force but merely as the means by which the rotating masses interact.

AM