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My Mechanics textbook claims that the sum of the work by internal forces is not generally zero.

translated to English the paragraph reads:

Notice about the work by internal forces: the work by the internal forces is in general not equal to zero although the sum of the internal forces is always equal to zero. Since the work also depends on the relative displacement of the two pointmasses in the material system, the work provided by these 2 internal forces will generally not be equal to zero.

Other sources claim that the work done by internal forces is always zero. Why does a car engine not do work if the wheels don't slip? (answer by Mark Eichenlaub)

Then I didn't do any work on the system at all since, by definition, only external forces can do work.

http://in.answers.yahoo.com/question/index?qid=20110131130859AAMLJLd

the net work done by internal forces is zero since they do not cause motion of the object.

To me it would seem that these sources contradict eachother. Do they? If so, which one is correct? If not, please explain why they don't contradict eachother.

Community
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Edward Stumperd
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  • You state a claim from your textbook -- what is the textbook? You refer to a claim by your professor, but you don't say what it is that your professor is claiming; you only state a claim by your textbook and a claim made on a web page. –  Aug 11 '13 at 21:32
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    The relevant claim the textbook is making is the added paragraph. The link is just one example of a source that I found to state the opposite. I'll try to make my question more direct and clear. – Edward Stumperd Aug 11 '13 at 21:40
  • The point is that you haven't told us the title and authors of the textbook. In the PDF, I don't find anything to support your characterization. What I do find this this: Since the body is rigid and the internal forces act in equal and opposite directions, only the external forces applied to the rigid body are capable of doing any work. This is a completely different statement from your characterization, which didn't say anything about rigid bodies. –  Aug 11 '13 at 21:42
  • I tried to make the claim clearer and add sources that made what I felt is the opposite claim in a more direct way. I didn't include the name of the textbook as it's a Flemish book, which is not internationally published and can't be found online for reference. – Edward Stumperd Aug 11 '13 at 22:02
  • Your new sources still don't say what you've been claiming. For example, the sentence you took from Mark Eichenlaub physics.SE answer is taken out of context in a misleading way. –  Aug 11 '13 at 22:28
  • Internal forces on a flexible cord, are going to stretch it, and thus do work. – John Alexiou Aug 12 '13 at 01:21
  • When you rub your hands against each other, heat is produced. Internal force friction is doing work. – udiboy1209 Aug 12 '13 at 05:08
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    related: http://physics.stackexchange.com/q/8453/ –  Aug 13 '13 at 19:28

4 Answers4

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Well, I think the issue must be that all those references that claim that internal forces don't do work must refer to RIGID BODIES or PARTICLES. If that's the case, they're right. But in general, for deformable bodies, that's not true and your textbook is right.

  • In a rigid body, all the external work done to the body is transformed into kinetic energy. The work of the internal forces is zero, since any couple of internal forces will act in opposite directions (the former fact is known as the work-energy theorem)

  • In a general body (think of continuum mechanics), part of the external work will be done to compensate for some internal work, and not all of it will become kinetic energy. For simplicity, let's assume that the internal forces come from a potential and let's see an example:

Let our system be a spring that is compressed by an external force F, until it reaches a steady position. There is an external work on the system because the force creates a displacement, but in the end the spring is steady, so there is no increase in the kinetic energy between the initial and final states. Where did all this energy go? It compensated the work of the internal forces of the spring. Because these internal forces derive from a potential, the work of the internal forces amounts exactly the change in the potential.

In general we have, with usual sign conventions (but that changes a lot, it is better to think carefully on the problem than to just substitute) $$ - W_e = dK - W_i = dK + dU $$ Where $W_e$ is the external work, $W_i$ the internal, $dK$ the change in kinetic energy and $dU$ the change in the internal energy, that in the example is just the change in the potential.

JVaz
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Work is a subtle concept that can be approached in various ways. There are ambiguities that can be reduced somewhat through the careful use of language. Ultimately, however, these ambiguities can only be eliminated by specifying a theory of physics and stating the definition within that theory.

Suppose our theory is Newton's laws applied to frictionless gear trains. Then the forces are normal forces exerted by a gear tooth A on another gear tooth B. By Newton's third law, the forces are equal in magnitude and opposite in direction. Since these are normal forces, the displacements of A and B are equal. Work can be defined either as $dW=F\cdot dx$ or as the transfer of energy by a mechanical force; since these are normal forces, the two definitions are equivalent. By Newton's third law, A's work on B cancels B's work on A.

Suppose instead that our theory is Maxwell's theory applied to oppositely charged particles A and B moving in a vacuum. We release A and B at some distance from one another. Their subsequent motion is that they oscillate about their common center of mass, radiating electromagnetic waves (and periodically passing through each other). There is clearly no way we can define work so as to make A's work on B always cancel B's work on A. Energy is going out into the electromagnetic waves, and there is no way to count this energy into the definition of work, since it's nonmechanical.

  • Suggestion: There is a drag on the electrons as they accelerate- the "radiation resistance" arising from the electron's delayed self field - so the way the energy is going into the EM field IS mechanical. However, your point is that there is no compensating mechanical work done BY the field - hence the loss to the field. I know this is what you mean, but the way you've said it might sound a bit weird to the OP. – Selene Routley Aug 11 '13 at 23:24
  • @WetSavannaAnimalakaRodVance: In the example with the two charges, I don't think it's possible to patch things up so as to preserve the notion that the flow of energy to radiation is due to the mechanical work done by a force from the radiation back-reacting on the charges. For example, the Lorentz-Dirac force $(2/3)kq^2x'''/c^3$, which involves the third derivative of the position, vanishes at certain times, and these are not the same as the times when the rate of kinetic energy change of the charges vanishes, since that rate is proportional to $x'x''$. [...] –  Aug 11 '13 at 23:35
  • [...] The fundamental issue is that an electromagnetic wave is spread out over space, so it doesn't have a single, well-define position, and therefore there's no way to define its displacement $dx$. –  Aug 11 '13 at 23:40
  • Hmmm. I'll have to think about this one. You're likely right - I seem to have crises of understanding of the EM field (even though it is most of my day job) about once every ten years and I'm due for one now! I make the comment because you can derive the force from the Larmor formula by momentum balance - but I have not looked at the problem from the more basic Lorentz force equation - probably one runs into self energy problems in so doing. – Selene Routley Aug 11 '13 at 23:49
  • @WetSavannaAnimalakaRodVance: I would have to spend some more time working it out carefully, but I think there may be some ambiguities involved, which are only resolved when you integrate over a full period. –  Aug 12 '13 at 00:10
  • All in all a fascinating problem that launched a few careers - an excellent example for the OP! – Selene Routley Aug 12 '13 at 00:11
  • What about a bomb that explodes in free space? Is this also out of the scope of Newton's laws as kinetic energy is produced seemingly out of nothing? – User Jan 17 '18 at 09:27
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Changing our notions of the universe has always been a challenge..."The sum of all internal forces is not zero. If the sum of the internal forces is zero, the universe not would be dynamic"

Is known of the mechanics that when a body exerts a force of field towards another material body, the same body receives equal force and contrary to that exerted, and this is how it is understood the law of universal gravitation formulated by Newton, this that well is true for bodies that possess inertial values similes in magnitude stops being necessary when used bodies whose inertial magnituds may vary in thousandfold the value of the smaller body, for saying a number. This is the case of the system formed by the Earth and an object, as is a tennis ball, this will receive variation in its quantity of motion to the moment to be put down in free fall, but it is not possible to say the same of the Earth, in this case the variation in its motion is null in practice, since this has not moved thousandth none of distance and less to experienced some type of acceleration. Now analyzing the existence of such repose and displacement of the object, is true to say that the body higher energetic content exerts field influence to the interior of the body that we consider as the system to studying.

Then, a body is considered in state of rest when the sum of all the forces acting in it is zero, and is also applicable to situations in the empty space when the field of predominant acceleration is slightly influential for a given time interval, what it wants to say that in the charted of a continuous curve, rather it approximates to a straight line. However a body free of rotation for a given time interval, is considered in a state of absolute rest when in it no external forces are acting or influences of field, so is this who exercises the predominant field on other material bodies that compose a system, thereby generating the point of reference for obtaining an absolute space, is to say an euclidean space. Furthermore, expanding our field of vision, in any system the sum of the internal forces never reaches to be zero, though the body is in equilibrium, what must not get confused to idealize a isolated system, that in the practice, as this way the experience affirms, is not possible to find a system with such characteristics, due to that not exist some material that may isolate the gravitation. Is for this reason that the interaction between systems is the key for understanding of the Principle of Internal Force, which is capable of solving a series of problems own of the mechanics and the electrodynamics.

G. Diez-Acevedo
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  • There seem to be a few errors in grammar & word choice in your middle paragraph. Also, you seem to contradict yourself, saying both it is and is not possible. Could you clarify your position? – Kyle Kanos Dec 21 '14 at 03:41
  • Ok sorry! I am not English!

    "The life without errors would be undisputed, and the communication would be in vain"

    That the bodies mysteriously accelerate through space, it doesn't mean that an external force is the responsible, if not that owes to that the sum of all the internal forces is never in complete equilibrium.

    More information about equations, write me briefly gonzalo.diezacevedo@gmail.com

     – G. Diez-Acevedo Dec 22 '14 at 03:07
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No. W=o for internal forces is valid for all bodies irrespective of their rigidity. Because a nonrigid body can b deformed only due to external forces. Not due to internal forces. Because internal forces act along the same line. With equal and opposite direction causes a net force to b 0

tryst with freedom
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sami javed
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    Out of curiosity, would it really have been too much extra work for you to say W=0, not W=o and to include the e's after your b's? I know typing an extra letter is difficult, but I think it is worth the effort to make your post look respectable – Jim May 19 '15 at 15:41