Talk:Newton's laws of motion

Disputed section: Paragraphs deleted from Third Law section

I deleted these paragraphs: The forces acting between particles A and B lie along parallel lines, but need not lie along the line connecting the particles. One example of this is a force on an electric dipole due to a point charge, when the dipole points in a direction perpendicular to the line connecting the point charge and the dipole. The force on the dipole due to the point charge is perpendicular to the line connecting them, so there is a reaction force on the point charge in the opposite direction, but these two force vectors are parallel and, even when extended to a line, they never cross each other in space. (Delete this whole Paragrph. This is not true)

For long-range forces such as electricity and gravitation, the third law may cease to apply.(Delete this Sentence, even for long distance forces this law applies equally well)

because discussions should be on the discussion page, not in the article. I'm not expressing a view on which contributor is right. --Heron 12:37, 30 January 2006 (UTC)

Newton's third law limitations

The third law is not always true. It fails to hold for electromagnetic forces, for example, when the interacting bodies are far apart or rapidly accelerated and, in fact, it fails for any force which propagate from one body to another with finite velocities.
Source: Mechanics, by Keith R. Symon, University of Wisconsin, Addison-Wesley Publishing Company, Inc., chap. 1-4

But in situations in which bodies influence each other at a distance, as for example through the long-range forces of electricity or gravitation, Newton's third law may cease to apply.
Source: Physics - A new introductory course, Particles and Newtonian Mechanics, by A. P. French and A. M. Hudson, by the Science Teaching Center at the Massachusetts Institute of Technology, p. 13-8.

with cases of extreme acceleration, problems do not fall under newtonian physics; they are analyzied with reletivistic physics. The claim about gravity is outright false. Indeed, earth is gravitationally drawn towards stars hundreds of light-years away, but since gravity drops off as the inverse square of distance, the force is incredibly minute. I have seen at least 4 physics books and none of them ever mention newton's third law not applying. I'm not sure about the electromagnetic forces, but I know gravity is still equal and opposite even at great distances.

College Physics seventh edition, by Raymond A. Serway, Jerry S. Faughn, Chris Vuille, and Charles A. Bennett (published by Thomson Brooks/Cole) says "an isolated force can never occur in nature" (page 90). --Crucible Guardian 08:47, 17 February 2006 (UTC)

If no valuable counter-references is brought in, the Wikipedia:Neutral point of view official policy will have to be applied.
--Aïki 00:29, 31 January 2006 (UTC)

Newton's third law as formulated by Newton is valid only when for electrostatic force and gravitational force instantaneous action at a distance is assumed.

However, it would be very silly to state that newton's third law is wrong because it fails to be a relativistic law of physics.

The demand on a theory of physics is that it forms a self-consistent system, and that it agrees with observation to within the accuracy of available observational data.

On a more abstract level we can see that Newton's third law asserts conservation of momentum. The principle of conservation of momentum is just as important in relativistic dynamics as in newtonian dynamics, so in a more abstract sense Newton's third law is still just as valid. --Cleonis | Talk 14:34, 21 February 2006 (UTC)

Newton's third law does not hold exactly for magnetic forces, even with instantaneous action at a distance. The force between two steady current elements, from the Biot-savart force law, does not obey newton's third law. JohnFlux (talk) 12:49, 23 January 2008 (UTC)

We need to include something about the limitations of the 3rd law. An example of it not holding is two magnetic dipoles pointing in the same direction. If you release them they will begin to rotate in the same direction, i.e. the torques they exert on eachother are not opposite. (Conservation of angular momentum is counter-intuitive to the 3rd law here, it is held because of the angular momentum of the field being opposite that of the dipoles) The way it was explained to me was that the 3rd Law only holds for "central" forces, which are forces that act parallel or anti-parallel to the radial vector between two bodies. 130.64.137.176 (talk) 10:26, 16 December 2008 (UTC)

a question ?

say theres two cars 1 in motion and 1 stationary , the car in motion strikes the stationary car , which car will have the most damage ? would the stationary car absorb the momentum from the car in motion therefore causing more damage to the stationary car? please help me on this one ?

The discussion page is for discussing improvements to the article only.TStein (talk) 20:44, 11 July 2008 (UTC)

Assuming that the cars have the same mass they will sustain similar damage. This regardless of whether they remain welded together or bounce apart. This is most easily seen from the 3rd law 'action and reaction are equal and opposite'. The driver of the moving car sees the other 'approaching' at speed. This is the same in his frame of reference as the driver of the stationary car sees in his. Alacrid 19:32, 15 November 2006 (UTC)

As movement can only every be expressed in relative terms, there is no difference between the stationary car and the moving car. The stationary car is only considered stationary due to the frame of reference. Rolo Tamasi (talk) 18:12, 27 January 2008 (UTC)

Just so there's no misunderstanding, if they remain 'welded' together, then there will be more damage done than if they bounced apart. JohnFlux (talk) 12:42, 23 January 2008 (UTC)

This has the cause and effect backward though. The damage absorbs the energy of impact and keeps it from bouncing. TStein (talk) 20:44, 11 July 2008 (UTC)

Do you have a proof of this theory? Rolo Tamasi (talk) 18:12, 27 January 2008 (UTC)

More Answer: So long as the bodies have the same form and hold the same material properties (say, coefficient of restitution), they will deform identically because, as this is a matter simply of motion in frames of reference, the two bodies are indistinguishable in space. —Preceding unsigned comment added by 130.126.215.2 (talk) 08:45, 27 February 2008 (UTC)

Newton's 3rd law for bodies in motion

Does Newton's third law apply to bodies in motion ? (not in a state of equilibrium).

SPECIFICALLY:

A 1 ton box falling in a column of almost empty air, containing only a few air molecules ?

The box will fall under gravity and exert a force (equal to it's weight) on the air molecule.

The air molecule will exert the SAME force back on the box.

By newton's third law.

Therefore the box should not move at all.

UNLESS the air molecule moves.

But then the system is not in equilibrium and newtons' third law cannot really be held to apply to this system as a whole, correct ?

A body in motion is in equilibrium, it is impossible to distinguish between a body in motion and a body that is stationary as speed can only be determined relative to another body.

The force between the box and the air molecules will only be equal to the weight of the box when the box reaches its terminal velocity, at this speed the forces between the box and the air will equal the weight of the box. The box will then stop accelerating and continue falling at a constant speed. The air molecules will move, the total force on them causing them to accelerate will equal the total force on the box. Rolo Tamasi (talk) 08:36, 22 February 2008 (UTC)

Newton's 3rd Law still holds. Newton's 3rd Law refers to pairs of forces: if object is exerting a force on object 2, object 2 is exerting a force on object 1 that is equal in magnitude and opposite in direction. The presence of acceleration does not alter this. In your example, when the falling box hits an air molecule it exerts a force on the air molecule (that is not equal to the weight of the box) and the air molecule exerts the same force back on the box. This does not result in zero acceleration though since the two forces are not acting on the same object. As for the box, there is the force of gravity acting down and the force due to the air molecule acting up. The mistake in your example was assuming that the force of the box on the air molecule equals its weight. PhySusie (talk) 11:57, 22 February 2008 (UTC)

Except, at terminal velocity the force does equal the weight of the box and 3rd law still applies. 128.91.26.30 was therefore not wrong to say the force equals the weight as long as it is appreciated that this is a specific not a general occasion. I think the error (in addition to the error that a body in motion is not in equilibrium) was to say that when this happens the box should not move at all. That is wrong, when the force equals the weight the box stops accelerating it does not stop moving. Rolo Tamasi (talk) 20:21, 22 February 2008 (UTC)

PhySusie said: "The mistake in your example was assuming that the force of the box on the air molecule equals its weight". My question: Why is the force on the air molecule not equal to the weight of the box ?

To Rolo: I am SPECIFICALLY talking about 1 SINGLE air molecule in my example, so why are you bringing up "air" and terminal velocity ?

The reason I am talking about terminal velocity is that is the only time when the force on the box from impact with the air is equal to the weight of the box, which was the condition stated.

But now I understand your scenario. We are not really talking a bout a box in air at all. We are talking about a single collision between two objects of hugely different mass.

As this is a momentary incident the fact that the box is being accelerated by gravity is also irrelevant. The only relevant issues are that the box and the molecule have a speed differential and a big mass differential and that they collide.

The question is what is the force of that impact? There is little reason why it would be equal to the weight of the box (or, equally, the weight of the molecule).

The calculation of the force of the impact is not easily determined because it depends upon the nature of the collision, specifically the time it takes. If it is zero time the force is infinite but for a zero time – pretty unhelpful stuff!

What we do know is the force will be applied equally to the two objects and thus the effect of the impact on their speeds will be inversely proportional to their relative masses. If we knew the speed differential and the mass relativities and assumed no energy was absorbed by the impact we could calculate the speed changes.

Thus we can see that the molecule will “ping” off the surface of the box experiencing a substantial acceleration while the change of speed of the box will be almost undetectable.

Hope that helps. Rolo Tamasi (talk) 10:49, 23 February 2008 (UTC)

Rolo: Thanks for that very illuminating post. I think I get it now ! —Preceding unsigned comment added by 71.242.39.174 (talk) 23:41, 23 February 2008 (UTC)

Your example appears to confuse force with acceleration. Consider the second law F=ma. If the forces on the air molecule and the box are equal and opposite, then the net change in the acceleration of the box will be negligible. You can just check this with some plausible numbers if you don't believe it. Silly rabbit (talk) 03:47, 23 February 2008 (UTC)

THERE'S A BASIC PROBLEM IN YOUR INTERPRETATION! There are 2 pairs of 3rd law forces (which ARE equal and have opposite directions): 1) Weight exerted by earth on box vs force exerted by box on earth (according to gravitational law) 2) Draw force exerted by molecules on box vs force exerted by box on molecules (which makes them move turbulently) THE DRAG FORCE IS NOT RELATED TO THE WEIGHT AND DON'T HAVE TO BE EQUAL!!!! THE ARE BOTH EXERTED ON THE SAME OBJECT SO THEY CANNOT BE A 3RD LAW PAIR! This confussion is very common. Even our dear friend from NASA (see video on 3rd law) makes this mistake. —Preceding unsigned comment added by Andinosa (talk • contribs) 09:50, 16 April 2009 (UTC)

Better Representation?

Hi I am only a year 7 student, and I could be very wrong, but the picture on the page with the two ice skaters pushing against each other may not be the best representation of the third law. They aren't necessarily of the same strength or weight, therefore, one may be pushed back more. I think that the recoil of a firearm would be a much better representation.Nelsondog (talk) 05:18, 16 March 2008 (UTC)Nelsondog

Thanks for checking- but Newton's 3rd Law applies even if one is more massive or is stronger. The force between them will always be the same. PhySusie (talk) 12:45, 16 March 2008 (UTC)

PhySusie, you might have misread what Nelson is saying. He is saying that a better example than two skaters would be to show the effects of equal and opposite forces on two bodies of different mass, such as a rifle and a bullet, where the smaller body is accelerated by a greater amount. The example of two skaters of roughly equal size might be misinterpreted to give the false impression that the acceleration is always equal and opposite. MarcusMaximus (talk) 04:39, 25 August 2008 (UTC)

I agree with Nelson that there should be a better way to represent the third law. I feel that standing on a frictionless surface one would never be able to exert any force on another object, in the way it is depicted in the figure. The figure implies the use of traction from the surface, which would complicate the description. As Nelson suggests recoil would be a much clearer way to illustrate the 3rd law. K.sateesh (talk) 07:01, 15 February 2009 (UTC)

Abit confused

I'm confused with what d is. From what I remember from school shouldn't say a=dv/dt be a=Δv/Δt ?? and if d==Δ then y use d instead of Δ? —Preceding unsigned comment added by Yellow Onion (talk • contribs) 05:45, 19 March 2008 (UTC)

The statement a = dv/dt refers to the first derivative of the velocity function with respect to time. It gives the slope of the line on a velocity vs time graph at that instant in time. The expression you gave using the delta symbol refers to an average over time (so it is not instantaneous) - which is used as an approximation - particularly in algebra based physics courses. PhySusie (talk) 15:25, 19 March 2008 (UTC)

Question about the Motte translation

I'm a bit confused after reading the Motte translation of Newton's second law: "If a force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively"

If Newtons 'motion' is equal to modern day 'momentum' (as stated elsewhere on the page), isn't his concept of 'force' a bit different from the modern concept? Since Newtons laws state that forces generate *change* in momentum, a double force would create double the *change* in momentum. Also, the last part about applying the force altogether or gradually seems weird - if the force was not applied for the same amount of time, you would not end up with the same momentum. His concept of force here seems to be closer to what we call the 'impulse' (force x time), doesn't it? Mhc (talk) 21:44, 20 March 2008 (UTC)

Descriptive name for second law

I have changed this name from "law of acceleration" to "law of resultant force". My reasons are: (i) Newton never referred to acceleration, (ii) a more historically accurate version would refer to "rate of change of momentum" (see Feynman Lectures, vol. 1) (iii) The law refers explicitly to "resultant force" and Newton focussed on the resultant force for many pages, describing how the resultant could be found by vector addition. Moreover, this law does have the purpose of introducing the term force, which pervades all uses of mechanics. See also Williams to support the notion that force is the defining objective of the second law. It seems silly to mis-translate the law to name it in a form that is is less accurate, both historically, and from the viewpoint of relativity and, in fact, from the viewpoint of this article itself. Brews ohare (talk) 15:52, 15 April 2008 (UTC)

On Google, "Law of resultant force" gives two results (including this Wikipedia article), while "Law of acceleration" gives 58.000 results. Since Wikipedia is an online encyclopedia, and verifiability is essential (see WP:V), I propose to revert to the original "Law of acceleration". Further, regarding your edits, note this is an article made for a broad audience. Specially the lead section and first section are read by many people without an academic background. Concepts like rate of change of momentum are, in my opinion, to difficult to start with in the article lead.

Further please have a look at the discussions on this talk page and it archives. For instance, there have been several discussions on the 3rd law, resulting in the abbreviation "To every action, there is an equal and opposite reaction", which you have changed into "For every action ..." (more strongly implying a causal relationship between the two forces). Crowsnest (talk) 21:34, 16 April 2008 (UTC)

Not guilty - To my knowledge I have not tinkered with the third law.

As for momentum - a link is provided to the article on that subject. Personally, I find momentum an easier topic than acceleration, because it shows up not only in common everyday English, but in many other areas of mechanics, e.g. the law of conservation of momentum.

And, as the citations show, formulation in terms of momentum has taken place beginning with Newton in 1687, and continuing to today. It is by no means a "high brow" or "new" approach, although this formulation has shown to be compatible with relativity (1905, wasn't it?), as pointed out later in the article.

I get 7,370 hits for "law of acceleration" using "Newton" as a qualifier to eliminate bogus references. Of course, my objective in naming the "law of resultant force" was to emphasize the "force" aspect of the law. Using "resultant force" + "Newton" gives 48,100 hits, "resultant force" + "second law" provides 24,200 hits. So resultant force is a biggy. Brews ohare (talk) 15:46, 17 April 2008 (UTC)

You are right about the 3rd law change, my apologies for linking you to that change.

Regarding the 2nd law, as you stated yourself, it is your personal preference to use the momentum approach, but it is uncommon to be learned about Newton's 2nd law that way. Using a constant mass, and acceleration is simpler to start with. Regarding your "Law of resultant force", while Google hits don't say anything about the correctness of a statement, it does say something about what is common. This is not about truth, i.e. whether ultimately the 2nd law is about change of momentum or acceleration, but on how to present something in an encyclopedia, without losing most of your audience already in the lead section. So starting simple, in terms of acceleration, and expand to more general case later on. The phrasing you use is original thought, hardly found elsewhere in this formulation. Crowsnest (talk) 16:04, 17 April 2008 (UTC)

You might look at the template in the article. which shows the equation explicitly in terms of momentum

The "change of momentum" approach is very common, as the citations show. More than common, it's time-honored – it's been used since 1687!  :-)

Whether acceleration or momentum is the "easier" way is open to discussion. For some problems, momentum is easier (and/or more accurate) and for other problems it is acceleration. As for what is more natural or more easily absorbed by the reader (unrelated to advantages for any particular problem), I believe the momentum approach has the edge. Being more fundamental, the reader who goes on to the relativity section, or who wants to look into other topics, like the collisions of billiard balls etc. doesn't have to change gears. The historically interested person doesn't have to ask why Newton's correct initial statement has been changed for a less fundamental version.

As a guess, you have in mind problems where acceleration works best. However, solving such problems is not the universe of all problems, and also may neither have bearing on the ease of access to an encyclopedia article, nor to its value as introduction to other articles, nor to its value as general background of a qualitative sort (e.g. philosophical or historical value).

Have you thought a bit about the possible goals of different readers, possible backgrounds, and what would serve their purposes? Brews ohare (talk) 17:33, 17 April 2008 (UTC)

What I have in mind, when starting with acceleration, is that it directly relates to ones own experience. To get acquaintance with Newton's 2nd law most easily, is by direct experience. Which almost everybody has: in an accelerating or decelerating car, a playground swing, roller coaster, etc. You feel the force, while you see the acceleration. That is what makes it easy to relate to Newton's second law. Just because in most instances our own mass is near constant. Momentum, and momentum change rate other than by acceleration, are not in general a direct physical experience of most people. It needs explanation and verification at a higher level of abstraction. Crowsnest (talk) 18:55, 17 April 2008 (UTC)

One can just start with F=ma, and state under which conditions it applies, and later on extend to the rate of change of momentum. This is not about the fundamentally most correct formulation, but about how to present things. Crowsnest (talk) 18:59, 17 April 2008 (UTC)

You have outlined one approach and argued that it the most transparent. I have argued a different approach and argued that it is (i) at least equally transparent, (ii) more fundamental, (iii) historically predominant, (iv) used in many other articles in Wikipedia including this article, (v) adopted by reputable authors ranging from Newton to Feynman, and (vi) useful to a wider variety of readers, who undoubtedly have a wide range of agendas when consulting this article, ranging from philosophy to physics. Brews ohare (talk) 13:44, 18 April 2008 (UTC)

Since nobody else supports my view in this, I will stop with this discussion. Crowsnest (talk) 14:06, 18 April 2008 (UTC)

Recent reversions

User:Jok2000 reverted my edits without response to my explanations on talk page, and without explanations of his own. With the possible exception of the change in name for the second law, which was discussed above, the reverted changes are totally noncontroversial. I expect comment before, for example, addition of a reference is reverted. I believe these reverts were simply a "knee-jerk" reaction without justification, stated or imagined. I have added citations for the statement of Newton's second law, which is an almost literal translation of the Latin and has been used for centuries. Citations I've added go back to 1911 only. This statement of the law is used in this article itself later on. I've also added reference to Moller for relativity - a definitive reference in this area, for those who may not know these simple relations. Brews ohare (talk) 03:11, 16 April 2008 (UTC)

Tell me more about my "imagined" reasons. The 2nd law is F=ma [1]. So now take out your changes to it. Jok2000 (talk) 21:03, 16 April 2008 (UTC)

The article quotes Newton as "Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.", which I'd guess trumps Wolfram as an authority on the matter of Newton's laws. Moreover, I've provided multiple references to show that this wording persists to this day. The formulation in terms of acceleration is a more limited, less accurate formulation. You might also look at the template in the article. which shows the equation explicitly in terms of momentum Brews ohare (talk) 14:24, 17 April 2008 (UTC)

Strange, I thought it was you who removed that bit. Must have been some vandal. I retract the request. Its a bit funny actually, I repaired this once before in October 2007, if you check the history, to be what's there now. The vandals hit this thing twice a day. I can't be here all of the time, I guess. Jok2000 (talk) 15:18, 17 April 2008 (UTC)

Change the 'Classical mechanics' diagram's sub-legend ?

The ‘Classical mechanics’ diagram here and this article historically misrepresent at least Newton’s second law of motion. Here I copy below a discussion of this issue from the diagram’s Template Talk page [2] for further consideration here as perhaps the most relevant article.

Change the diagram's sub-legend

I propose the diagram's sub-legend 'Newton's second law of motion' be changed to 'The second law of motion of classical mechanics'.

The diagram is mistaken because its sub-legend 'Newton's second law of motion' is historically mistaken and if anything should be rather 'The second law of motion of classical mechanics'.

This is certainly not Newton's second law stated in the Principia, which was that THE change of motion [referred to in the first law] is proportional to the motive force impressed, i.e. Dmv @ F, or F --> Dmv (where 'D' = 'the absolute change', Delta, '@' = 'is proportional to', and '->' is the logical symbol for if... then...).

The misrepresentation of Newton's second law as F = ma or similar has the logical consequence that a = F/m and thus a = 0 when F = 0, whereby Newton's first law would be logically redundant just as Mach claimed it was.

But Newton's second law only deals with changes of motion produced by impressed force such as mentioned in the first law, and does not itself assert there is no change of motion without the action of impressed force as the law F = ma does, where F denotes impressed force rather than inertial force. And in fact both Galileo's 1590 Pisan impetus dynamics and Kepler's 'inertial' dynamics, both of which claimed motion would terminate without the continuing action of what Newton called 'impressed force', denied this principle.

But the logical function and historical purpose of Newton's first law is precisely to assert this principle, that there is no change of motion unless (i.e. If not) compelled by impressed force, and thus whereby Dmv <=> F, rather than just F --> Dmv. (Here <=> is the logical equivalence symbol for 'if and only if', and '-->' the symbol for 'If...then...') Thus Mach’s logical criticism was wrong by virtue of his ahistorical misinterpretation of Newton’s second law as F = ma.

Classical mechanics, whatever that might be, needs to be differentiated from Newton's mechanics.

--Logicus (talk) 18:20, 16 April 2008 (UTC)

Hi Logicus. Well, I don't know. The article on Newton's laws of motion says that the Newton second law is "The Rate of change of momentum is proportional to the resultant force producing it and takes place in the direction of that force". Isnt it the same thing that the formula on the template? (or maybe according to you, both article and template are wrong?) I do not oppose you change the sub-lengend, but honestly i'm not sure i see a true difference. Even if the formula is not as Newton stated it, it's greatly inspired by no :)?? And history remembers it as the Newton second law (improved?). Am I wrong? But your comment is interresting. Is this information on Newton laws of motion article???

[[::User:Frédérick Lacasse|Frédérick Lacasse]] ([[::User talk:Frédérick Lacasse|talk]] · [[::Special:Contributions/Frédérick Lacasse|contribs]]) 13:03, 17 April 2008 (UTC)

First of all, the relevant text is at wikisource, and I'll begin by saying that I disagree with Logicus on her/his proposal. Newton's second law is not stated, as such, mathematically. (Perhaps it is later in the Pricipia, I do not know.) For clarity, its statement under Axioms, or Laws of Motion reads:

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the straight line in which that force is impressed.

(Emphasis mine, in order to make clear that this is certainly an if and only if statement. Note that Logicus has failed to quote that word in presenting his/her argument.) Since Newton used Calculus in conjunction with this law to calculate planetary orbits and such, it is not too crude to use modern calculus notation in the box, even if we choose Leibniz' $\mathrm {d} /\mathrm {d} t$ notation over Newton's dot. For that matter, we use vectorial notation when Newton had none. Therefore, clearly, in modern notation, Newton's second law reads

${\vec {F}}={\frac {\mathrm {d} }{\mathrm {d} t}}m{\vec {v}}$, and I have no problem with identifying this equation (or an equivalent one) as Newton's Second Law or Newton's Second Law of Motion. Or, see Goldstein, Poole, and Safko, Classical Mechanics (3rd ed.) page 1, where $\mathbf {F} ={\frac {d\mathbf {p} }{dt}}\equiv {\dot {\mathbf {p} }}$ is identified as Newton's second law of motion.

Regarding the other stuff you've said about Mach, historical interpretations, and other irrelevant (for the purposes here) things, perhaps this can help. Newton's second law, which can be expressed as $F=ma$ cannot imply that "Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon" (Law 1), without presupposing the existence of inertial frames, which is what the first law, in effect, does. The fundamental disconnect between Newton and Mach, as I understand it, concerns the existence of a preferred inertial frame.

If, unlike I've read in many sources over the years, you have sources that claim Newton never equated force with a time-rate-of-change of momentum, you might be able to begin to find people willing to change their ways. However, the classical mechanics template talk is not the place for that discussion. — gogobera (talk) 20:10, 17 April 2008 (UTC)

Logicus's response to Frederick Lacasse, written before Gogobera's contribution: Thanks Frederick. Yes, BOTH the article on Newton’s laws of motion and the template are wrong, because the article mistranslates the Principia’s second law’s phrase 'mutationem motus' as ‘rate of change of momentum’, whereas it should be ‘The change of motion’, with no reference to any rate of change. It referred to an absolute change of motion as produced by an impulse, as in Cartesian vortical mechanics. May I refer you to Bernard Cohen’s ‘Guide to Newton’s Principia’ in the 1999 Cohen & Whitman Principia new translation for a good discussion of this issue, which was also touched on in the recent BBC Radio 4 ‘In Our Time’ programme on Newton’s Laws of Motion and drew the following comment from a listener published on the BBC website @ http://www.bbc.co.uk/radio4/history/inourtime/inourtime_haveyoursay.shtml

“Andrew, Newton's 3 laws

Simon Schaffer might have done well to see in this archive (dating from the programme on Popper), "if you study the original version of Newton's Second Law - not the modern F=ma - you realise that Newton regarded force as a function of time, equivalent to the modern notion of an impulse. It was change of momentum: mass *or* velocity; thus even if mass increases with increased velocity so does the force required, and Newton holds." The insertion of 'rate' in 'rate of change of motion (momentum)', giving F=ma, isn't a flaw of Newton's - it's a mistranslation of 'mutationem motus'. “

The true difference, as I have already pointed out, is that rather than Newton being illogically foolish in his axiomatisation is respect of stating a logically redundant axiom, namely Law 1, as Mach implied, because it was logically entailed by his Law 2, rather his first law states a logically independent axiom which, for example, ruled out Kepler’s theory of inertia according to which the inherent force of inertia resists and terminates all motion. For in Newton’s theory which revised the keplerian theory of inertia the force of inertia only resists accelerated motion and causes uniform straight motion like impetus did in late scholastic Aristotelian and Galilean impetus dynamics.

This is all important for understanding the logic and history of scientific discovery and such as how and why ‘classical mechanics’ emerged, the project started by Duhem that was the major research project of 20th century history and philosophy of science.. But there seems to be some considerable logical confusion and contradiction in Wikipedia articles about Newton’s dynamics and about classical mechanics and what it is and how it relates to Newtonian mechanics. For example the article on ‘Classical mechanics’ says on the one hand the two are equivalent, but on the other hand they are not equivalent because classical mechanics was created later and goes well beyond Newton’s mechanics, as in the following statements:

“There are two important alternative formulations of classical mechanics: Lagrangian mechanics and Hamiltonian mechanics. They are equivalent to Newtonian mechanics, but are often more useful for solving problems.”

Versus

“While the terms classical mechanics and Newtonian mechanics are usually considered equivalent (if relativity is excluded), much of the content of classical mechanics was created in the 18th and 19th centuries and extends considerably beyond (particularly in its use of analytical mathematics) the work of Newton.”

This confusion needs sorting out, but a bigger job than I have time for. I was just trying to reduce this confusion a little, as it cropped up on the Galileo article, but I now see this diagram is pretty ubiquitous in relevant articles. Sorry just to pick on your otherwise no doubt helpful diagram.

I fear the article on Newton’s laws of motion is currently virtually wall to wall ahistorical nonsense, apparently being devoted to teaching some version of 19th century mechanics or A-level Physics rather than the history of physics.

Re your following comment

“Even if the formula is not as Newton stated it, it's greatly inspired by no :)?? And history remembers it as the Newton second law (improved?). Am I wrong?”

One problem with the first claim that Newton greatly inspired the law F = ma is that the Wikipedia classical mechanics article is currently claiming

“The proportionality between force and acceleration, am important principle in classical mechanics, was first stated by Hibat Allah Abu'l-Barakat al-Baghdaadi,[7] Ibn al-Haytham,[8] and al-Khazini.[9] “

- although I have no idea whether this is true or not.

All in all I think I should implement the proposed edit if you have no further comments or objections. But it is still unsatisfactory given it is unclear from Wikipedia what exactly classical mechanics is, whereby such as Lagrangian and Hamiltonian mechanics and even Newtonian mechanics are said to be alternative formulations of it.

The outstanding pedagogical question for your view is surely that if you claim that Newton’s second law was essentially

F = ma, then why did he think he needed to state the first law as his first axiom ? The simple answer is because it gives the equivalence between change of motion and the action of impressed force that the second law does not, because the second law is only at most Dmv α F and not Dmv = F. Logicus (talk) 20:34, 17 April 2008 (UTC)

-- Further to my observation above that this article seems to be in serious conflict between the aim of teaching some form of ‘classical mechanics’ (or A-level Physics) under its various categories on the one hand of ‘Classical mechanics’, ‘Introductory physics’ and ‘Experimental physics’ etc for example, and on the other hand that of explaining what Newton’s laws of motion were under the categories of ‘History of physics’, ‘Isaac Newton’, ‘Latin texts’ etc., I further suggest that what this article and that on ‘classical mechanics’ misleadingly call ‘Newtonian mechanics’ may in fact rather be Laplacean mechanics as distinct from ‘Newtonian mechanics’ insofar as the latter term refers to the mechanics of one Isaac Newton as expounded in his Principia, and Laplacean mechanics may possibly be what is common to both Lagrangean and Hamiltonian mechanics whereby they are inexplicably both said to be forms of some ‘classical mechanics’, but which is itself never properly identified in any corresponding axiomatic form. For example the article says

“The term classical mechanics was coined in the early 20th century to describe the system of mathematical physics begun by Isaac Newton and many contemporary 17th century workers”.

But one major difficulty of many here is that Newton himself said in his Principia Axioms Scholium that its three laws of motion were already accepted by mathematicians and such contemporary figures as Huygens, Wallace and Wren, but also by his non-contemporary Galileo and also even (essentially correctly) pointed out its first law was even current in Greek antiquity, including in Aristotle’s principle of interminable locomotion in a void (Physics 4.8.215a19-22). --Logicus (talk) 16:01, 20 April 2008 (UTC)

Excuse me if I am being a mite blunt but this appears to be a little self centred. It is not up to Wiki editors to decide what Newton intended. Rather it is to publish the balanced view i.e. on the one hand some think Newton meant F=ma on the other had they don’t. I hope ego is not getting in the way here, it is not nearly as complex as the rhetoric implies. Rolo Tamasi (talk) 20:20, 20 April 2008 (UTC)

Unfortunately, I still do not agree with your position, Logicus. Perhaps you were planning on addressing my comments before you changed the template? I believe we ought to use the term Newton's second law for the template for the reasons stated above in my previous comment. I have given a source claiming the equation to be identified as Newton's second law. Here is a list of sources that identify $F={\dot {p}}$, $F=ma$, $F=m(dv/dt)$, or some other equivalent formulation as "Newton's Second Law":

• Danby, John M.A. Fundamentals of Celestial Mechanics. Richmond, VA: Willmann-Bell, 1992. page 44.
• Feynman, Richard P., Robert B. Leighton, and Matthew Sands. The Feynman Lectures on Physics: Mainly Mechanics, Radiation, and Heat. Reading, MA: Addison-Wesley, 1977. pages 9-1,2.
• Goldstein, Herbert, Charles Poole, and John Safko Classical Mechanics. 3rd ed. New York, NY: Addison Wesley, 2002. page 1.
• Hirose, Akira and Karl E. Lonngren. Introduction to Wave Phenomena. Malabar, FL: Krieger Pub. Co, 2003. page 4.
• Marsden, Jerrold E. and Anthony J. Tromba. Vector Calculus. 5th ed. New York, NY: W.H. Freeman and Co., 2003. page 264.
• Penrose, Roger. The Road to Reality. New York, NY: Alfred A. Knopf, 2005. page 389.
• Thornton, Stephen T. and Jerry B. Marrion. Classical Dynamics of Particles and Systems. 5th ed. Belmont, CA: Brooks/Cole, 2004. page 50.

Wikipedia is not about original research. The previous sources are from advanced and introductory textbooks, as well as books for the mathematical layperson. This list is only limited by my desire to stop pulling books off my shelf. Suffice it to say that the equation presented on this template is commonly known as Newton's Second Law. I don't think there is any disagreement over that fact. The idea that "some think Newton meant F=ma" and some don't is an issue for the history of science. The fact is that today, $F=ma$ is identified as "Newton's second law" universally.

Logicus, your arguments have a tendency to sway away from the issue at hand. For instance, your comment about the inconsistency in Wikipedia's treatment of the equivalence of Newtonian mechanics and classical mechanics. However, to clarify, it should be stated that there is an equivalence between the Newtonian, Lagrangian, and Hamiltonian formulations of classical mechanics. So that statement that says "[Lagrangian and Hamiltonian mechanics] are equivalent to Newtonian mechanics" is true (if sloppily worded). Also, saying that classical mechanics has "extend[ed] considerably beyond … the work of Newton." [emphasis mine] is also true, since Newton's work was limited in scope. However, in principle, mechanical results can be derived using any of the formulations. The amount of work to do so can vary tremendously.

If Logicus wants to add cited material discussing Newton's intentions and their historical interpretation, I have no qualms with that. This discussion began, and should be focused on the use of the term "Newton's second law" to describe the equation on the template. Though my understanding is that Newton did mean his second law to be understood as $F={\dot {p}}$, the point can be made with less ambiguity by discarding historical interpretation: the equation, as written, is commonly identified as Newton's second law. Therefore, I will change the template back. — gogobera (talk) 19:26, 25 April 2008 (UTC)

Logicus to gogobera: You said "Perhaps you were planning on addressing my comments before you changed the template?", but I presumed you would appreciate your criticisms were dealt with by my reply to Lacasse. Apparently not. I shall explain later. —Preceding unsigned comment added by Logicus (talk • contribs) 18:18, 29 April 2008 (UTC)

Second Law and Impulse

I believe that it is historically incorrect to state Newton's Second Law as F=dp/dt. Motte's 1729 translation (as stated in the article) reads The alteration of motion is ever proportional to the motive force impressed.. At first sight this appears to be saying that force is proportional to motion (velocity??) which is the very concept that Newton is supposed to be discarding. However, a reading of Newton's definitions in the Principia makes it clear that by motion he means momentum and by force he means impulse in modern terminology. So what Newton actually said, using modern symbols, is;

$I\propto \Delta p$

Obviously, using SI units and differentiating yields the usual form;

${\frac {dI}{dt}}=F={\frac {dp}{dt}}$

However, it is unhelpful that the article states the law in this form right after quoting Newton's original words. The impulse version of the law should be stated first together with Newton's definition of his terms.

It has been noted in comments on this talk page already by others that what is written in the article about force does not make sense but with no satisfactory explanation ever been given. Anyone want to tell me I'm wrong? SpinningSpark 08:31, 2 May 2008 (UTC)

The set-up does suggest that the mathematical equation is a direct implementation of Newton's words. That should be fixed if your history is accurate. Can you substantiate your beliefs about the historical background?

Whether historical statements of the law should come first is debatable, as history is an interest of only a subset of readers. Also the introduction of impulse is pretty opaque to the modern reader, and a bit of a digression in applications. Its use at the beginning of the article would make it all rather indigestible. An historical section may become necessary to give room to explain about the historical role of impulse and its connection to rate of change of momentum. Brews ohare (talk) 20:37, 3 May 2008 (UTC)

I did not introduce the historical diversion into this article, it was there already. I have no view on whether it should be in this article or another. However, given that it is here, it is necessary qualify Motte's translation with the modern terms and show that the modern form of the law can be derived from Newtons form (it is only one simple step). Several comments furhter up the talk page verify that this is indeed causing confusion. As for my source, it is Newton's Principia, the definition of motion (=momentum) is given at definition 2 here [3], I am hoping you can get through the paywall (just hit cancel at login) or else try this one [4]. As for force, the text given in the article makes the point, If a force generates a motion, a double force will generate double the motion . . . whether that force be impressed altogether and at once, or gradually and successively. That only makes any sense if impulse is meant rather than force. If it really meant force a step function would produce the same motion as a ramp function - clearly nonsense. SpinningSpark 01:02, 4 May 2008 (UTC)

You will find an earnest attempt to implement what I understand from your remarks at Impulse. Brews ohare (talk) 02:13, 4 May 2008 (UTC)

You obviously put some effort into that, but I have changed it because Newton from his wording cleary does not restrict the concept to an infinitesimal time (neither does our own article, or NASA). I have changed the reference too, presumably that's what your ref says. My ref is a pretty low level one - the ones on the impulse article might be better but I have not looked them up. SpinningSpark 02:43, 4 May 2008 (UTC)

Logicus: Your discussion might well benefit from studying Cohen's analysis of this issue in his 1999 Guide to Newton's Principia in the 1999 Cohen & Whitman Principia --Logicus (talk) 14:46, 4 May 2008 (UTC)

A reference to I Cohen is added to the Impulse subsection. Brews ohare (talk) 15:55, 4 May 2008 (UTC)

Equivalence. What equivalence ?

Gogobera claimed above 25 April:"...it should be stated that there is an equivalence between the Newtonian, Lagrangian, and Hamiltonian formulations of classical mechanics. So that statement that says "[Lagrangian and Hamiltonian mechanics] are equivalent to Newtonian mechanics" is true (if sloppily worded). Also, saying that classical mechanics has "extend[ed] considerably beyond … the work of Newton." [emphasis mine] is also true, since Newton's work was limited in scope. However, in principle, mechanical results can be derived using any of the formulations. The amount of work to do so can vary tremendously."

From a logical point of view for any two theoretical systems to be equivalent every axiom and theorem of one system must be an axiom or theorem of the other. But where, when and by whom was any such logical equivalence between Lagrangian, Hamiltonian and Newtonian mechanics ever demonstrated ? If Wikipedia makes such logico-historical claims, they need source documentation. Are any of the various mechanical systems of Lagrange, Hamilton, Laplace, Newton (Principia Ed. 3) and the elusive 'classical mechanics' logically equivalent ? And what were the axioms of each of these systems, especially of the latter three (e.g. is Corollary 1 of the Principia a theorem or really an axiom, as in earlier versions i.e is Newton's proof of theoremhood valid, or invalid as Bernoulli presumed in tryit to reprove it?).--Logicus (talk) 14:46, 4 May 2008 (UTC)

To support my claim, I will quote from Section 7.7 of Stephen Thornton and Jerry Marion's Classical Mechanics (p. 257) titled Essence of Lagrangian Dynamics:

We elected to deduce Lagrange's equations by postulating Hamilton's Principle because this is the most straightforward approach and is also the formal method for unifying classical dynamics.

First, we must reiterate that Lagrangian dynamics does not constitute a new theory in any sense of the word. The results of a Lagrangian analysis or a Newtonian analysis must be the same for any given mechanical system. The only difference is the method used to obtain these results.

…

The differential statement of mechanics contained in Newton's equations or the integral statement embodied in Hamilton's Principle (and the resulting Lagrangian equations) have been shown to be entirely equivalent. Hence, no distinction exists between these viewpoints, which are based on the description of physical effects. [emphasis original]

I am not sure if I can be any clearer, except by presenting, in its entirety, section 7.6 of the same book, titled Equivalence of Lagrange's and Newton's Equations which begins: "As we have emphasized from the outset, the Lagrangian and Newtonian formulations of mechanics are equivalent: The viewpoint is different, but the content is the same. We now explicitly demonstrate this equivalence by showing that the two sets of equations of motion are in fact the same." (p. 254) The text continues with a mathematical development of the claim.

Thornton and Marion do go on to clarify that a philosophical difference exists:

In the Newtonian formulation, a certain force on a body produces a definite motion—that is, we always associate a definite effect with a certain cause. According to Hamilton's Principle, however, the motion of a body results from the attempt of nature to achieve a certain purpose, namely, to minimize the time integral of the difference between kinetic and potential energies. [N.B.] The operational solving of problems in mechanics does not depend on adopting one or the other of these views. [emphasis original](p. 258)

Does this make my position clear? I would welcome anyone else's opinions as well. — gogobera (talk) 23:39, 22 May 2008 (UTC)

Inertial frames

Any particle, regardless of what forces act on it, is at rest relative to the reference frame whose origin is defined to coincide with the particle. The non-trivial point about inertial reference frames in Newtonian mechanics is that they are the same for all physical objects. I corrected the formulation.

Taneli HUUSKONEN (talk) 14:12, 5 June 2008 (UTC)

I'm not sure what change you've made, but I hasten to add that while a particle is at rest, by definition, relative to the frame whose origin coincides with the particle, that frame is frequently not an inertial frame. — gogobera (talk) 22:54, 5 June 2008 (UTC)

Second law in intro

I maintain that F=ma is the form of the law that should be listed in the brief statements of the laws in the intro. It is the one most people are familiar with from sixth grade science class, and it does not require previous knowledge of momentum. All the stuff about special relativity and changing mass can be (and is) explained later. I don't think it matters that Newton didn't write F=ma--it is still a traditional, simple statement of the second law.

Apparently a least a couple of people disagree with me. I suggest we talk about it here rather than in revert edit summaries.Rracecarr (talk) 19:57, 1 July 2008 (UTC)

Sorry, I have reverted you again. The article has been like this for a long time after lengthy discussion on this page. I feel you ought to reach consensus here before re-inserting material that has been challenged once. I do not believe that Wikipedia should present an incorrect version of Newton's laws jsut because it is commonly taught that way in high schools. Certainly not in the lede. The article can and should explain the F=ma simplification but should not start off by giving the impression that that is actually Newton's second law. The reason it is taught that way, by the way, is not because momentum is a difficult concept, but because it avoids using differentiation. The acceleration article, for instance, correctly states that a=dv/dt. But at school I was not taught that formula, I was taught v=u+½at². It would be completely wrong to put that formula in the acceleration article lede, just as it is wrong to put the school formula in the Newton's laws lede.

Also, just as a matter of established procedure, it is completely wrong to delete a citation of the eminent physicist Richard Feynmann in favour of an obscure professor who you have only mentioned in edit summaries and with no proper citation anyone can check. At the very least some consensus is needed before anything changes. SpinningSpark 20:51, 1 July 2008 (UTC)

Ummmm... I didn't delete it. I moved it to a more appropriate location. I completely disagree with you. You are mistaken that the article has been this way for a long time--it was changed on April 15, before which it said F=ma for a much longer time. F=ma is not "wrong". Most people coming to this article are not physicists. The first thing they see re the second law should be the familiar (an adjective backed up by references) F=ma. So, sorry, I have reverted you again. Rracecarr (talk) 01:18, 2 July 2008 (UTC)

It was only that way in April 2008 because you changed it before in February 2007. As it was subsequently reverted by other editors I am not the only one who thinks Newton's second law is F=dp/dt. The very first version of this article has a statement even closer to Newton's original. In any case, it is not really relevant what the article used ot say, what is important is what is right. All five of the references state the momentum version of the second law. Those that mention the acceleration version at all, do so afterwards. You really need to find references with more status than the ones given before you make this cahnge again. SpinningSpark 21:13, 2 July 2008 (UTC)

I think F=ma is supported by the existing references. If you insist, I can add more. What I think is important is that the intro be as accessible as possible without being wrong (and again, F=ma is not wrong), and that it jogs the memory of people who haven't thought about Newton's laws for 30 years. There is plenty of space in the article to elaborate on the more universal form of the law. Rracecarr (talk) 21:22, 2 July 2008 (UTC)

I have reverted again, with the addition of a new reference (the first Google hit for "Newton's second law" as it happens), since you requested it. Rracecarr (talk) 21:30, 2 July 2008 (UTC)

You cannot be serious, a high school tutorial web site is not a suitable source to establish this point. At least it does not trump the existing references which are all solid textbooks, plus a lecture by Richard Feynmann. They all say momentum, only some of them go on to mention F=ma and certainly Newton himself does not even mention acceleration, either in the original latin or Mottes translation. SpinningSpark 21:42, 2 July 2008 (UTC)

Oh, I'm perfectly serious. You must know perfectly well that there are plenty of references to support the statement that Newton's second law is F=ma, AND ALSO that Newton's second law is F=dp/dt. Anyone can hand-pick references. That is not the issue. First, you said "the article has been this way a long time". When I pointed out that it had been the other way for much longer before, suddenly "what it used to say in the article is irrelevant". You chastised me for deleting a reference which I didn't delete. You ask for references, when there are already plenty, and when I add one, you say that I "cannot be serious". You can keep pretending to invent new issues. But the real issue is which version of Law 2 is more appropriate in a brief introductory list of traditional statements of the laws. The first law in the intro just says "no force means constant velocity", which, at face value, is redundant--it's a special case of the second law. I don't see you beefing about that. Below, it's explained that formally, the first law defines inertial reference frames. This level of technicality would be inappropriate in the intro. Similarly, the third law just says "equal and opposite reaction"--no blabber about simultaneity and signal propagation between distant particles in special relativity. You keep changing the second law to a more technical, less recognizable form that is inconsistent with the rest of the list.

Here's an idea. Instead of an edit war, let's find an intro that makes both of us happy. It could be a little longer. The second law statement could be something like: A traditional statement of the second law is that the net force on an object is equal to its mass multiplied by its acceleration (F=ma). A more general form, allowing for changing mass, is that force is equal to the rate of change of momentum.

What do you think? Rracecarr (talk) 22:21, 2 July 2008 (UTC)

Here is an even simpler idea: Get rid of the list in the intro altogether. There is another list immediately below it, with more precise statements of the laws, each of which has a "this law is often stated as" sentence for forms taught at introductory levels. Rracecarr (talk) 22:29, 2 July 2008 (UTC)

I'm glad you now want to discuss rather than revert. Finding a mutually agreeable text is the best way forward. However, I cannot accept your proposal. All serious treatments of this subject give F=dp/dt as the second law. To say anything else requires solid references. I am not picking and choosing to suite my own opinions, that's what all the textbooks say. F=dp/dt is not a generalisation of F=ma to allow for changing mass. The second law applies to bodies, any situation in which mass is changing (the classic example is a rocket using up fuel) that has the second law simplistically applied to it will result in the wrong answer. The right answer is only obtained by a consideration of the momentum of all the masses involved, rocket, fuel exhaust and reaction mass. It is rather the other way round, F=dp/dt is Newton's second law; F=ma is an approximation to the second law in non-relativistic situations. It is in exactly the same category as so many other high school formulae, such as Work Done = Force x Distance. We all know that is really $W=\int _{C}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {l}}$, but schools do not want to teach calculus. Hence there is a need to find non-calculus forms for the expression of velocity, acceleration, work and force. Just because such expressions are familiar from school does not make them the accurate definitions. Certainly articles should explain where the simplified forms come from but we should not allow that form to dominate the article. We are writing an encyclopedia here, not a school homework aid. SpinningSpark 23:32, 2 July 2008 (UTC)

I always wanted to discuss. To be honest, I also want to revert, but as I am sure you are aware I have reached 3RR. There are many things I disagree with you about, but for simplicity, let's take one at a time. Consistency: why do you not have similar problems with the statements of the first and third laws. As I've already pointed out, the third law isn't true relativistically either. What are "action" and "reaction" anyway? Why no mention of force, which is what the third law is really talking about? The reason is, of course, that that is the traditional statement of the law. Rracecarr (talk) 23:44, 2 July 2008 (UTC)

I have tried another option. I put a bit of thought into it, and I think it might satisfy both our concerns. Please don't knee-jerk revert. Rracecarr (talk) 01:31, 3 July 2008 (UTC)

Dispute over statement of the second law

Should the second law be stated as Force = mass x acceleration or Force = rate of change of momentum? SpinningSpark 21:48, 2 July 2008 (UTC)

Force = rate of change of momentum. The simplified version force = mass x acceleration is only true if mass is constant. The correct version and the simplified version and the linke between them can be easily explained even in an intro paragraph. Wikipedia should not perpetuate an erroneous over-simplification by only presenting the simplified version. Gandalf61 (talk) 13:48, 3 July 2008 (UTC)

I think both forms should be shown. Most people (at least in the general public) are more familiar with the F=ma form and would be confused if it was not shown. Most introductory calc-based physics texts introduce Newton's 2nd Law as F=ma first, and then a few chapters later generalize it to the time derivative of the momentum when momentum is introduced. I am in favor of giving both the more familiar form of each of the three laws along with the more general (and more correct) forms along with an explanation.PhySusie (talk) 15:52, 3 July 2008 (UTC)

I'm a third party and I agree completely with Susie. Adding both makes the page more usable to more people, such as those looking for supplemental material for their classes. AzureFury (talk) 19:51, 3 July 2008 (UTC)

Both forms need to be given, and there is nothing wrong with giving F=ma first. In classical mechanics, it is an absolutely correct and general statement because mass is always constant in classical mechanics. Even in special relativity (which should not be the primary focus of this article) the formula can be made to be correct by letting m stand for the relativistic mass, instead of the rest mass. Loom91 (talk) 11:23, 5 July 2008 (UTC)

I'm going to agree with everyone else thus far. It's always presented at F=ma even in the beginning of calculus physics texts and classes. Later on, it is explained in more detail, introducing the calculus based component. I'd say explain both. lesthaeghet (talk) 06:29, 7 July 2008 (UTC)

Use both but start with F=ma. In addition to the reasons stated above, the simpler version, F=ma, in my experience is far more commonly used and needed then the more advanced version which is necessary only for limited cases such as rockets. It (F=ma) is more easily understood by everyone even those who have no had calculus. Finally, and most important in my PoV, is fact that introductory articles should always spiral out from easiest and simplest to most complicated and most general. Only experts benefit from going the other direction.

TStein (talk) 20:44, 7 July 2008 (UTC)

I would express it as the rate of change in momentum and as a = F/m. Writing it as F=(m)(a) implies that force is caused by an acceleration and mass interaction, rather than acceleration coming about because of an interaction between a net force and mass. LonelyBeacon (talk) 05:59, 9 July 2008 (UTC)

I agree with LonlyBeacon to an extent. Implies may be too strong of a word, though; maybe F=ma hints that the force is caused by the acceleration is better. Consider V=IR, one might either argue that I=V/R is better since the current is caused by a voltage. Alternatively one might argue that it should be R=V/I since this is a definition of R. In the end, though, because one can algebraically manipulate the equation to many different equivalent forms so the question is probably moot. F=ma has the advantage of working typesetting better than a=F/m and is slightly easier to algebraically manipulate.

TStein (talk) 03:33, 10 July 2008 (UTC)

From a basic standpoint, the layperson is much more likely to recognize or "get" a description of F=ma because it's simplified and readily applicable to most run-of-the-mill calculations. Shifting into explanations of the concept regarding non-constant masses dips into an area more esoteric but certainly more correct. The best course is to give both. The clearest general form would be, in my estimation: 1)translation of law text 2)simplified concept for stable mass is F=ma 3)wider concept involves momentum without the assumption of constant mass. Keeping the simplification and using it first allows one to wade into the waters instead of diving into the deep end. — Scientizzle 15:20, 11 July 2008 (UTC)

Exactly. F=ma is not only much simpler to understand it is by far the most common use case. If we wanted to be absolutely technically correct then we would have to use the relativistic equivalent of the Ehrenfest Theorem, anyway. Why should we settle for only an approximation like dp/dt=F? TStein (talk) 18:14, 11 July 2008 (UTC)

I suggest stating it as rate of change of momentum, but then mentioning the F=ma common form. As it happens, that's pretty much what the article currently does. As an aside, there's no need for 5 references for it though. Modest Genius ^talk 17:21, 12 July 2008 (UTC)

I agree with the majority of posts regarding the topic. Both points are equally valid. Most people researching entry-level Newtonian physics will more readily recognize F=ma, especially as it is readily manipulated using algebra. Also, this is applicable as a "stepping-stone" to those learning about gravity/acceleration equivalence within relativistic physics. However, the more appropriate definition (in my opinion) details the rate of change of momentum since it does not imply constant mass. In short, both are useful and I think both explanations should be presented on the page. That's just my 2kB. Archon Magnus^{(Talk | Home)} 19:42, 16 July 2008 (UTC)

Again, both points are valid, but I agree with the majority of posts, which suggest stating first F=ma; there is also an official guideline about this, i.e. Wikipedia:Make technical articles accessible. I also agree with User:Modest Genius about the excess of references. --Blaisorblade (talk) 00:44, 17 July 2008 (UTC)

Agreed. Start with F=ma, then explain the general form later. This article must be understandable for somebody who does not know physics at all. --Apoc2400 (talk) 11:56, 18 July 2008 (UTC)

I'm not sure if this aspect was overlooked but,

• momentum (p)= m/v;
• a= d(v)/dt (rate of change of velocity),
• and so rate of change of momentum= d(p)/dt=m*d(v)/dt (since mass is constant)

So, I think both the versions are simply corollaries of the same equation. In response to Gandalf61's point, in Newtonian (classical) physics, the mass always stays constant, right? And, for any law, even though when we tend to neglect it, but a law is always to be stated with full conditions. So, the expanded form of Newton's second law would be "With mass remaining constant, force applied on a body equals a product of mass and acceleration produced", or "With mass remaining constant, a constant force equals the rate of change of momentum of the body". —KetanPanchal^taLK 07:07, 19 July 2008 (UTC)

No - mass is not always constant in classical physics - think of rockets for example. The time rate in change of momentum is the more general equation, F=ma is true for the particular (though very common) situation of constant mass. PhySusie (talk) 12:00, 19 July 2008 (UTC)

Actually PhySusie, it's not conceptually sound to treat rockets as variable mass systems. In classical physics, mass is constant without exception. This has been discussed before. Newton's laws are about particles. In order to prove them for extended bodies, the assumption of constant composition is essential, which implies constant mass. The correct way to treat rockets is using conservation of linear momentum. Loom91 (talk) 21:47, 19 July 2008 (UTC)

I'm pretty sure that classical physics includes all of classical mechanics, including the topics of variable mass systems, extended mass systems and linear momentum. Classical physics is more than Newton's Laws. I agree that F=ma is for constant mass and treats the object as a particle. However, my point was that F=dp/dt is more general, and not necessarily (though it often is) the same equation as F=ma. PhySusie (talk) 12:28, 20 July 2008 (UTC)

One of my original reasons for starting this RFC was that Newton himself states the second law in terms of momentum and does not reference acceleration at all. Admittedly he does not use the differential form of the equation, but most likely because calculus would have been treated with suspicion by his contemporaries. Does anyone believe that if a law is named after Newton it should be true to what he actually said?

SpinningSpark 16:02, 20 July 2008 (UTC)

According to that logic, we should not use "momentum" at all, as that would be putting words in Newton's mouth. Instead we should describe it as "quantity of motion". Rracecarr (talk) 21:47, 20 July 2008 (UTC)

Of course that is not what I meant, the article is on the English Wikipedia and should be written in modern english. Actually, Newton did not use the word "motion" either, he used the Latin word motus. He does however give a very clear definition of motus, there can be no doubt that "momentum" is meant. SpinningSpark 06:42, 21 July 2008 (UTC)

That clear definition is nowhere to be found in this article. I did find it in inertia though. The Latin and the translations in this article only serves to confuse the issue, in my opinion. In any case, it is a moot point. As I state below, few if any of the named laws of physics correspond to their original form. How we teach physics should depend on sound pedagogy and not on history. Whatever way that Newton understood his laws is irrelevant to how they are understood today. TStein (talk) 22:16, 25 July 2008 (UTC)

The present form of the article is fine. It puts the momentum formulation first, which is the accurate version both historically and in practice. It then immediately discusses its reduction to F=ma, so no reader can miss that expression of the law. Momentum is arguably easier to grasp than the acceleration approach, being rooted more strongly in everyday experience. (We bump into things every day, and see what happens. We play pool, hockey, croquet. Everyday experience with momentum shaped the context Newton faced in presenting the laws.) The preference for F=ma may be just an outgrowth of learning from textbooks that use F=ma. Brews ohare (talk) 15:29, 21 July 2008 (UTC)

Both forms should be shown. It may be preferable to show F = ma first because it is better known, at least at an introductory level. In my opinion, it doesn't matter exactly how Newton defined the law, except for the parts of the article that are dealing explicitly with history. It is perfectly normal that laws and equations get redefined, new notation introduced, etc., while keeping the original name. So, what matters is, what is the most common definition today? (I'm not sure.) I disagree with using a = F/m, because, although obviously equivalent, it's not the "iconic" version of the equation. --Itub (talk) 15:45, 23 July 2008 (UTC)

I agree strongly that it doesn't matter how Newton defined the Law. There are many examples of laws associated with people that are different than the author originally wrote them. Maxwell's wrote down his equations completely different then those associated with him. Faraday expressed his law completely in terms of words. (It was Maxwell who translated Faraday's ideas into mathematics and he did so in terms of the vector potential.) Looking at all the laws that are named after people, the law that is stated the same as it was first stated is the exception (if one even exists) rather than the rule. What matters is the most common notation that is in use today with the article organized from the most simple to understand to the most general and most complicated to understand. Science is not frozen in time and Newton's laws are not limited to Newton. Both must change to fit the times or die. TStein (talk) 21:20, 25 July 2008 (UTC)

You should show both, but clearly emphasize that the law is that force exerted is equal to the time rate of change of momentum. That force is also mass times acceleration is the more familiar form, but it is only a special case of the law. RayAYang (talk) 03:20, 30 July 2008 (UTC)

Why is the mass inside the differentiation operator: F=d(mv)/dt rather than F=m*dv/dt=ma? It is unnecessary and only results in a less familiar, more convoluted form of the equation.

Can someone give me an example of a variable mass problem in classical mechanics that requires the d(mv)/dt form of Newton's law in order to work properly? The rocket equation is not a valid example, contrary to popular myth, for reasons I can show if asked. I contend that the form F=ma is not a "special case" of Newton's Law. It is valid even for variable mass problems.

On the other hand, the F=d(mv)/dt form shown in this article yields incorrect results if you account for the time-varying mass in the only way I conceive it could be accounted for in this form. F=d(mv)/dt plus the chain rule results in F=m*dv/dt+v*dm/dt. This implies that a melting ice cube under zero net force will undergo acceleration due to its changing mass. An ice cube of m = 1 kg traveling at v = 1 m/s, but melting at a rate of dm/dt = -0.1 kg/s, would be required to accelerate at dv/dt = 0.1 m/s in order to satisfy the equation under zero net force. This is clearly a false result, since a melting ice cube doesn't accelerate under zero force.

So why does it makes sense to state this form of Newton's law in an article about classical mechanics? It seems to me the correct form for classical mechanics is the vector equation F=ma. MarcusMaximus (talk) 01:53, 15 August 2008 (UTC)

The form F=d(mv)/dt, which is equivalent to F=m*dv/dt+v*dm/dt, implies that an object with nonzero velocity, influenced by a nonzero external force, can either accelerate with constant mass (dv/dt is nonzero and dm/dt is zero) or experience a change in mass without accelerating (dm/dt is nonzero and dv/dt is zero), to satisfy the equation. I don't think this is a physically valid result. MarcusMaximus (talk) 02:09, 15 August 2008 (UTC)

What more, out of five references cited as the source of F=d(mv)/dt (refs 3-7), all of them state that F=ma without deriving it from F=d(mv)/dt. Only one of them mentions that F=d(mv)/dt, from which it forms F=ma, and then it proceeds to use F=ma for the rest of the chapter. The Feynmann lectures (ref 3) states: "The product ma ... is the rate of change of momentum, hence the rate of change of the momentum of a body is equal to the resultant force ... acting upon it and is proportional to the resultant force ..."

I restate that I do not see any good reason to state the law in terms of d(mv)/dt in this article. MarcusMaximus (talk) 17:28, 15 August 2008 (UTC)

Some other thoughts about this page

This is not related to the RFC and I hope this is not included there. I have questions about how this page is formatted and I did not want to jump in and edit here because there seems to be enough active editors here and too many cooks can spoil the broth.

I have some concerns that hopefully can help you improve the page.

• The quotes and the translations from Newton disrupt the flow of the content. Even the translation is often hard to understand since it uses a different scientific language. For example what exactly does Newton mean by motion in his second law (acceleration, velocity, position)? It is encyclopedic which is good. (It is not, nor should it be a textbook.) On the other hand, encyclopedias put such content as side boxes to separate it out from the main content.

• Often the history is mixed in with scientific content (in particular with the 1st law). This is another candidate for a box that people can skip or read separately if they so desire. At the least it should have its own section, like what is common on wikipedia.

• The alternate name for Newton's 1st law (law of inertia) is well established, but the alternate names for the other laws are not. It would not surprise me if one or more texts use them but for most people those names will be a distraction. (Try googling "Newton's second Law" vs "Law of resultant force" for instance).

• Have you considered splitting this article up into the main article plus separate articles covering more of the detail? The second and the third law in particular have the potential to be very long.

TStein (talk) 20:35, 11 July 2008 (UTC)

Constant mass? Special case?

In various places I have seen a statement to the effect that F=d(mv)/dt accounts for systems of variable mass, while F=ma is only valid for constant mass systems. On the contrary, the law stated in terms of F=ma is true regardless of whether the mass is constant. If you use the d(mv)/dt form, you can get incorrect results for variable mass systems, such as a melting ice cube will accelerate under zero net force.

In fact, the claim that this form of the equation accounts for variable mass systems implies that the consequence of an unforced system gaining or losing mass is a change in velocity, which is flat out wrong. It is actually a change in momentum.

Therefore F=ma is not a special case of d(mv)/dt for "constant mass systems", because F=d(mv)/dt does not correctly account for varying mass; it is valid only for constant mass systems (see reference 22[5] on the main page of this article). F=ma is the correct form of Newton's Second Law for any particle or rigid body with constant or varying mass at any point in time within the speed domain of classical mechanics. I think we need to scrub the articles on classical mechanics topics from this false claim, and maybe even give an explicit statement to the contrary. MarcusMaximus (talk) 05:19, 25 August 2008 (UTC)

I should clarify that I am not opposed to stating the law as F=d(mv)/dt so long as it is made clear that one of the fundamental assumptions of classical mechanics is constant mass. MarcusMaximus (talk) 05:47, 25 August 2008 (UTC)

The melting ice cube doesn't change its mass. Unless you count the liquid water as being no longer part of the body; but this is not the case, because the water will continue to move at the same velocity of the ice (assuming no external force on either water or ice), so, in this case, the body consists of both ice and water.

A better example would be imagining a shopping cart which is moving at constant speed, and you drop something heavy in it. The cart will slow down, as the total (horizontal) momentum will stay constant. (You can say that the body you dropped pushes the cart back, but then you are considering the body as being just the empty cart, and its mass is constant.)

Or, you can think of a jet plane, which accelerates because its mass decreases. (You can also say that the fluid jet pushes the plane, but it's just the different ways you can divide the system. In the first description, the plane including the fluid still in it is a body, and the fluid already ejected is another, and they have both variable mass. In the second description, the plane excluding the fluid is a body, the whole of the fluid (both in the plane and in the jet) is another, they have both constant mass, and they push each other according to the third law.)

Also, F = ma breaks down in special relativity, whereas F = d(mv)/dt doesn't.

As for Newton's wording, he wrote Mutationem motus proportionalem esse vi motrici impressae, and the following part ("If a force generates a motion [...] whether that force be impressed altogether and at once or gradually and successively.") only makes sense if motus is momentum and vis is impulse. --A r m y 1 9 8 7 ! !  19:53, 2 September 2008 (UTC)

"The melting ice cube doesn't change its mass." That's precisely my point: the melting ice cube cannot be correctly treated as a variable mass system using F=d(mv)/dt, because this equation does not correctly account for varying mass. It only works correctly if you keep all the same mass in the system, which is by definition a constant mass system.

Although I agree with your statement about the melting ice cube, I don't agree with your two examples.

• The cart's final velocity depends on the velocity of the heavy object. If the object has the same forward velocity as the cart when you drop it in, there is no change in the speed of the cart. The object brings its own momentum into the system with it.
• A jet airplane does not accelerate because it is losing mass; you must be talking about a rocket. The thrust of jet airplane is produced by injecting energy into the air passing through the jet engine to accelerate the air backward and thereby gain momentum in the opposite direction. The mass of fuel injected into the combustion chamber is negligible compared to the mass of air flowing through the engine. And, by the way, a rocket doesn't gain speed because it is losing mass, it gains speed because it is ejecting mass forcefully.

My only point is that in the realm of classical mechanics, F=ma is always true, regardless of whether the mass of the system is changing, and F=d(mv)/dt is true only for constant mass systems, which makes it misleading and pointless to put the mass inside the differential. In fact, I can show you how F=ma is derived by using conservation of momentum on a variable mass system, if you want. If you're really worried about special relativity, we can state the law as F=$\gamma$ma, which is true even for relativistic varying-rest-mass systems (while the other form is only true for relativistic systems of constant rest-mass), but I'm pretty sure that is beyond the scope of Newtonian mechanics. MarcusMaximus (talk) 02:55, 3 September 2008 (UTC)

(Yes, replace "jet plane" with "rocket" above.) In classical mechanics, there is no way for a body to change its mass unless it expels some matter, or some matters joins it. In that case, in the reference frame in which the expelled/new matter is stationary, F = d(mv)/dt is still valid. Re-read the examples above, explicitly considering the reference frame in which the horizontal momentum of the body dropped in the cart is zero, or the momentum of the fluid expelled by the rocket is zero. You will see that, in that reference frame, if F is 0, d(mv) will be zero but dv won't.

On the other hand, you can consider a body as being always composed by the same matter, and, if you consider "cart + heavy object" to be a body, you'll find that it (or better, its center of mass) doesn't accelerate without external force; if instead you consider "cart" as a body and "heavy object" as another, there is a force from the cart to the object which speeds up the latter, and an opposite force from the object to the cart slowing down the cart. Under this description, m is always constant in classical mechanics, so the whole issue is meaningless.

(As for SR, F = γma is only valid if F and v are perpendicular, otherwhise it is F = md(γv)/dt = m(γ³a_∥ + γa_⊥), where a_∥ is the part of acceleration which is parallel to velocity, and likewise for a_⊥. For the varying-m case, see http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html.)

--A r m y 1 9 8 7 ! !  09:33, 3 September 2008 (UTC)

It is a major weakness of the F = d(mv)/dt formula if it is true only in a special subset of inertial reference frames. Dynamics are supposed to be identical in any inertial reference frame. I don't understand what exactly you are claiming, but I have a feeling it is a pathological case that I examined a few weeks ago that can be better treated using a different (and more general) formula for conservation of linear momentum. Could you elaborate on what you mean by "In that case, in the reference frame in which the expelled/new matter is stationary, F = d(mv)/dt is still valid."? Thanks. MarcusMaximus (talk) 18:05, 3 September 2008 (UTC)

It means that F = d(mv)/dt is valid only if any accreted/expelled matter has zero momentum in the reference frame used. But that what's commonly intended when one says "F = d(mv)/dt is also valid for non-constant m". I think that in most cases, identifying bodies so that each body always consists of the same matter eliminates these confusions. (In the other cases, I'd just use conservation of momentum. Just curious, what was the problem you were examining?) --A r m y 1 9 8 7 ! ! ! 18:47, 3 September 2008 (UTC)

It sounds like you're saying F = d(mv)/dt is not a general formula for the momentum of variable mass systems, that it is restricted only to be used in a privileged set of reference frames that move with the accreted/expelled mass. This is useless for analyzing anything but the most trivial academic problems, because for any system where the mass entering or leaving the system has a nonconstant velocity, you have to continuously change your reference frame.

"Identifying bodies so that each body always consists of the same matter" is impractical for some systems that expel amorphous mass produce force, like rockets and satellites, where it would be overwhelming to track every little $\delta m$ of mass after it leaves the nozzle. I'm sure there are many more examples.

The problem I was examining was the classic problem of finding the force exerted on a fixed platform by a falling chain. It is a pathological case where the mass "leaving" the system (the "system" is the falling portion of the chain) decelerates to zero velocity at the same moment it "leaves" the system, and thus surrenders all of its linear momentum in the form of a force exerted on the platform. I suppose this is as you specified, where the system must be analyzed from a reference frame where the delta-mass has zero momentum. I believe this is a silly restriction that proves the non-generality of the equation.

Why not just use the more general linear momentum formula that correctly models any variable mass system? I'm not just talking about the algebraic conservation of momentum equation. There is a differential equation that looks similar to the equation in question that accounts for momentum change due to both external forces and mass with nonzero velocity entering or leaving the system, taking momentum with it. It is valid in any inertial reference frame. And best of all (in my opinion), it collapses to F=ma in any reference frame for systems of constant or varying mass, and it is correct.

$\mathbf {F} +{}^{N}\!\mathbf {v} ^{\delta m}{\frac {dm}{dt}}={\frac {d}{dt}}(m{}^{N}\!\mathbf {v} ^{CM})$.

where ${}^{N}\!\mathbf {v} ^{\delta m}$ is the velocity in inertial frame N of the mass center of the particles instantaneously entering or leaving the system, and ${}^{N}\!\mathbf {v} ^{CM}$ is the velocity in N of the mass center of the remaining mass. The first term on the left is the external force, and the second term is the momentum carried by any mass entering or leaving the system. This is the full definition of the rate of change of system momentum in non-relativistic mechanics.

Obviously you could choose a reference frame where ${}^{N}\!\mathbf {v} ^{\delta m}$ is zero to get an equation for the special case you talked about. This equation is much more general. If you carry out the chain rule on the right side, then combine the $\scriptstyle {dm/dt}$ terms on the left side, you get the expression for thrust and the correct equation of motion of a varying mass rocket (or an ice cube).

(I also corrected two small italics formatting errors in your post.) MarcusMaximus (talk) 20:44, 3 September 2008 (UTC)

Yes, that's all correct, but I fail to understand how that "collapses to F=ma in any reference frame for systems of constant or varying mass". (Unless you require ^Nv^δm to always equal ^Nv^CM. But, for example, that isn't the case when a bullet penetrates a wooden block...) --A r m y 1 9 8 7 ! ! ! 22:26, 3 September 2008 (UTC)

I should have said any inertial reference frame, since that is one of the restrictions on Newton's 2nd Law. It collapses to F=ma when you consider that, by definition,

$({}^{N}\!\mathbf {v} ^{\delta m}-{}^{N}\!\mathbf {v} ^{CM}){\frac {dm}{dt}}=\mathbf {v} _{exit}{\frac {dm}{dt}}$,

which is a transfer of momentum, or a "thrust" type force, exerted on the "main" mass by the entering/departing mass. If you group it with the other external forces for convenience, you get F=ma. MarcusMaximus (talk) 22:48, 3 September 2008 (UTC)

Ok, now I understand what you mean. I'm adding a clarification in the article. --A r m y 1 9 8 7 ! ! ! 19:33, 4 September 2008 (UTC)

I'm glad we've understood each other. Do you see my point that putting the (necessarily constant) mass inside the differential is pointless and misleading? MarcusMaximus (talk) 08:39, 5 September 2008 (UTC)

Well, among other things, F = dp/dt is closer to the original formulation by Newton (though ∆p ∝ I is even closer), it happens to hold in special relativity, and it is useful in some circumstances (even if in some of them it means an infinitesimal momentum divided by an infinitesimal time, rather than the derivative of momentum seen as a function of time: see, for example, Kinetic theory#Pressure — though it uses deltas rather than d's —, or the solution of your falling chain problem). So I think it should be kept in the article, even if clarified when necessary. --A r m y 1 9 8 7 ! ! ! 10:13, 5 September 2008 (UTC)

I can accept what you're saying about it being closer to Newton's original formulation. I question the value of it holding true in special relativity, because this is an article about classical mechanics. I believe it is of no (or negative) value to consider that it works for "some" variable mass problems, pathological cases, where it just happens to be the equation you end up with by canceling one of the terms in the general equation because you happened to pick a preferred reference frame (like the falling chain problem). This is misleading, and it violates the principle that the laws of dynamics hold identically in any inertial reference frame.

So I think that if we present it in the current form, it needs to be stressed in articles that F=d(mv)/dt is only valid for constant mass systems, particularly because there seems to be a widespread belief to the contrary. MarcusMaximus (talk) 18:19, 5 September 2008 (UTC)

Boldface or arrows above?

The article mixes the two styles of symbols for vectors, but I think that, for consistency, we should pick one and use it throughout the article. Personally, I would prefer boldface, (as well as because I like it more) because it can be written without TeX, which don't align well with surrounding text when written inline (compare F with ${\vec {F}}$ or $\scriptstyle {\vec {F}}$). A similar issue is for the differential d which is written both in italics and in roman in this same article. What do you think? --A r m y 1 9 8 7 ! ! ! 22:50, 3 September 2008 (UTC)

I vote for boldface vectors for the same reasons. Also, my favorite textbooks use boldface.

As for the differential, I prefer to put d in italics, but scalar variable names in roman. You apparently prefer the inverse? MarcusMaximus (talk) 01:17, 4 September 2008 (UTC)

As for scalars, it is standard to write them in italics (it is the style that all browsers I've seen use to render the <var> tag, and all or almost all the textbooks and scientific papers I've ever seen use it); as for differentials, there are people using italics and people using roman; personally I prefer the latter for clarity, but italics seems to be far more common, at least on Wikipedia. MOS:MATH says "Both forms are correct; what is most important is to consistency within an article, with deference to previous editors." --A r m y 1 9 8 7 ! ! ! 10:06, 4 September 2008 (UTC)

Fg = mg vandalism

The user 125.24.228.54 likes to change force equations to read $F_{g}=mg$. I reverted the edits on this page. The same thing was done to the drag equation article. MarcusMaximus (talk) 18:33, 7 September 2008 (UTC)

Lack of third law in intro

Don't know how long it's been like this but the basic introductory part of the article, although titled three laws lack the third one. Just thought I'd point that out. —Preceding unsigned comment added by 64.231.185.133 (talk) 00:06, 10 September 2008 (UTC)

Thanks for calling this to our attention. I went back and found the edit where this was done, and it was quite a substantial act of vandalism by user 210.15.221.203. It had removed the entire third law, changed markup language that destroyed the references list, and changed random words throughout the article that made it nonsensical. I have done my best to restore it, but have been about about a dozen edits since then, so I can't be sure I got everything. MarcusMaximus (talk) 04:46, 10 September 2008 (UTC)

Requested semi-protection

This page has an obnoxiously high level of vandalism, so I requested semi-protection from unregistered IP addresses and new users. MarcusMaximus (talk) 04:59, 10 September 2008 (UTC)

Requested full protection

I put in a request for full protection for this page for an indefinite period of time. Since the semi-protection expired two days ago it has already been vandalized and reverted over a dozen times. It is vandalized constantly and unrelentingly, and there is a very good risk of losing valuable content. MarcusMaximus (talk) 07:30, 27 September 2008 (UTC)

Newton's Second Law

I was recently viewing this entry:

http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion

and I noticed a common flaw in the definition of impulse which it critical in understanding Newton's Second Law. Impulse is not just the product of mass times the change in velocity, it is the product of both the change of velocity and the change in mass. If it were just m delta v it would be much more difficult to get a space vehicle into orbit. To prove this to yourself, push a cart with 100 pounds of potatos on it at a constant 5 MPH and then have your buddies add another 100 pound bag of potatos to it every second. You will soon see that while your delta v is 0 your delta m is what's going to stop you in your tracks! When I was in school the profs used to ask, "What is NSL?" The common answer was, "F=ma!" That is a simplification. —Preceding unsigned comment added by 75.70.62.142 (talk) 15:55, 30 December 2008 (UTC)

More to the point, Newton himself, (who you would have thought would have been the one to define the law) stated it (in latin) in terms of the rate of change of mass times velocity (i.e. rate of change of momentum).- (User) Wolfkeeper (Talk) 13:58, 5 June 2009 (UTC)

Page protection failure

This page was recently horribly vandalised by an anon editor because the Template:Classical mechanics template had been trashed. Seems like we might need to request protection for that also. SteveBaker (talk) 03:56, 7 January 2009 (UTC)

Question about Relativistic mass in Relativity section

The relativity section of this article is using relativistic mass in the derivation. This can be confusing for students in the US, at least. Here, the use of relativistic mass has been out of vogue for a while and has been replaced by relativistic Energy. In this understanding m always refers to the 'rest mass' and the relativistic momentum is p = γmv. I debated changing this section but I want to make sure that this notation is not prevalent in other fields and countries first. TStein (talk) 21:06, 23 January 2009 (UTC)

I come from the US, with a background in physics, so I offer no new perspective, but I say go ahead and change it. Rracecarr (talk) 22:36, 23 January 2009 (UTC)

Newton's Second Law video

I watched the video just now and I'm a bit uncertain as to its usefulness in describing Newton's second law. It says a lot of numbers and equations but doesn't explain them much. Whanhee 17:28, 25 January 2009 (UTC)

Non-constant mass in the second law.

I noticed a mistaken claim in the main section about the second law: that it only holds for systems of constant mass.

In fact, F = d(m v)/dt is always true, even if the mass is changing.

It's the simplified version F =m a that is only true for systems of constant mass.

Refer to

H. Goldstein (1980). Classical Mechanics, p.2, eqn.(1-4), (1-5).

The section on open and closed systems appears to be somewhat in error. Tom Lougheed (talk) 02:55, 1 February 2009 (UTC)

This has been discussed a lot before, both on this page and on Talk:Isaac Newton. If you think of a rocket going at constant speed, and add two thrusters canceling each other exactly. Then the net force is 0, but the trusters are expelling mass and thus dp/dt is not 0! That is:

$\sum {\vec {F}}\neq {\frac {d{(m{\vec {v}})}}{d{t}}}$

So it's not valid for changing mass systems.
—Apis (talk) 20:39, 1 February 2009 (UTC)

Tom, refer to the footnotes in that section. They have solid citations backing up the current state of the article. MarcusMaximus (talk) 05:10, 26 February 2009 (UTC)

Just to be clear, and avoid sending readers looking for references from February 2009, the mistake in the two-thruster rocket example above is that it does not account for all the momentum changed by the applied forces. Instead, the equation should be written:

$\sum {\vec {F}}=m_{G}{\vec {a_{G}}}+{\dot {m_{of}}}{\vec {v_{of}}}+{\dot {m_{ob}}}{\vec {v_{ob}}}$

where ${\dot {m_{of}}}{\vec {v_{of}}}$ is the momentum of the mass outflowing forward and ${\dot {m_{ob}}}{\vec {v_{ob}}}$ is the momentum of the mass outflowing backwards. In this case, these last two terms cancel so $\sum {\vec {F}}=0$ and ${\vec {a_{G}}}=0$, as expected. -AndrewDressel (talk) 15:09, 5 June 2009 (UTC)

Andrew, it looks like you are not correctly using F=d(mv)/dt. Your equation seems to claim that the rocket's momentum is constant. We know its mass is decreasing, so it better have increasing velocity. Obviously under zero net force it can't be increasing its velocity, so there's a problem here. Most people would say that in order to prevent this error we just have to include all the mass, including the expelled propellant. But now we have a constant-mass system, just as the article and footnotes say you must use. MarcusMaximus (talk) 05:57, 8 June 2009 (UTC)

I didn't explain the source clearly enough. The author explains that the equation is valid only for an instant in time, hence ${\dot {m_{of}}}$ and ${\dot {m_{ob}}}$ but $m_{G}$, without the dot. -AndrewDressel (talk) 16:22, 8 June 2009 (UTC)

What does it mean for a dynamical equation to be valid at only an instant of time, but not for all time? MarcusMaximus (talk) 17:33, 8 June 2009 (UTC)

It means that "all the terms [must be] evaluated at time t". The equation does not say anything explicit about the rocket's momentum, let alone claim that the rocket's momentum is constant. Instead, the momentum at time t must be calculated as $(m_{G}(t_{0})-t({\dot {m_{ob}}}+{\dot {m_{of}}}))v_{G}(t_{0})$ if ${\dot {m_{of}}}$ and ${\dot {m_{ob}}}$ are constant. -AndrewDressel (talk) 18:22, 8 June 2009 (UTC)

Andrew, I see what you are referring to now. You're basically taking the speed of the rocket v_G times its current mass, assuming that both m(dot)'s and v_G are constant. I think these assumptions don't allow us to say anything about d(mv)/dt generally, because you almost have to assume your conclusion to justify your assumptions. The only question this equation can answer is "What is the momentum of a rocket with constant velocity and constant mass flowing out of opposing thrusters?" It doesn't tell us what we really want to know, which is, "What is the motion of this rocket, given that I know the mass flow rate?" MarcusMaximus (talk) 00:38, 9 June 2009 (UTC)

My intuition doesn't agree with this. If the second law holds good only for systems of constant mass, then the law will be F=ma only. Not rate of change of momentum. And it does explain the propulsion of single thruster rockets-the normal ones i mean. If newton's laws don't hold there, how are you able to say that the rocket will proceed with constant velocity? Sganesh 88 (talk) 06:49, 8 June 2009 (UTC)

Sganesh, the law does define the rate of change of momentum for systems of constant mass. However, the form F=d(mv)/dt does not provide the correct equation of motion for a rocket unless you include all the expelled mass in the system for the entire flight of the rocket--an impractical and unnecessary burden that actually turns the problem into a constant-mass system. This provides a counterexample to those who claim that F=d(mv)/dt applies to varying-mass systems. Anyway, a rocket's motion is correctly described using the law of conservation of momentum, or by using F=ma, and naturally putting the thrust in F and treating m as a variable. This is precisely how I simulate the motion of rockets at work, and it gives an answers that are verified by actual flight test data.

The main thing most people don't seem to consider is that when the mass of a system is allowed to vary, the mass that enters or leaves the system carries momentum with it equal to its mass times its velocity. By this mechanism, the momentum of the system increases or decreases independent of any external force. Therefore, to define the rate of change of momentum of a system only as F excludes the possibility of varying mass. A more general law would set d(mv)/dt equal to the sum of F and the rate of momentum transfer via incoming/outgoing mass. MarcusMaximus (talk) 07:48, 8 June 2009 (UTC)

That is true, but it doesn't really justify the claim that F=ma works for varying mass and F=dp/dt doesn't. Either can be applied to, say, a rocket, as long as you do it right. Either one will give bad results if applied incorrectly. It is basically semantic. Strictly, Newton's second law applies to an individual particle, and the mass of a particle never changes, classically. It is misleading to say that F = d(mv)/dt is general, and F = ma is a valid simplification when mass is constant. But F = dp/dt does have the advantage that it remains valid relativistically. Rracecarr (talk) 18:27, 8 June 2009 (UTC)

Rracecarr,

First, I think it's ok to forget about special relativity for the sake of clarity and pedantic effectiveness in the articles about classical mechanics. Talking about F=d(mv)/dt and varying mass tends to induce people to bring up rockets, not relativity, and rockets (in my experience) are handled wrong more often than not.

Second, you are correct that either equation can be misapplied to get the wrong answer. However, the d(mv)/dt form can only be applied to constant mass objects--it requires keeping track of all the entering/departing matter and calculating the velocity of the centers of mass of all the mass particles or bodies, including those that are no longer of interest (like expended fuel or jettisoned equipment). On the other hand, F=ma can be used correctly on a varying mass object without the bother of tracking all the expelled mass flow and separated bodies--one only needs to know the external forces on the object of interest and the forces exerted by the departing mass, and simply treat m as a variable. MarcusMaximus (talk) 00:38, 9 June 2009 (UTC)

Which method is more "bother" is up for debate. Conservation of momentum gives you v1-v0=ln(m0/m1) without the bother of integrating a variable acceleration over time. The reason I think variable mass should be left out of the article is that it is philosophically messy. Defining clearly what the total force on a particle is when what the "particle" is continually changes can be confusing. What you are actually doing, implicitly, when you apply F=ma to a rocket is applying it separately to each particle that makes up the rocket (none of which change in mass). You simply stop carrying out the calculation for a given fuel particle once it is separate from the rocket. Better to keep it simple and not cloud understanding of the basic physics. Rracecarr (talk) 01:13, 9 June 2009 (UTC)

I apologize if I seem argumentative, but I think it is very important to state clearly that F=d(mv)/dt does not apply to anything but systems of constant mass. There are all kinds of otherwise smart people who think their varying masses are taken care of by using the chain rule to get a term that contains m(dot). I've had debates about this with several physicists and I've even seen physics teachers on these talk pages argue that side. This glaring error even shows up on the NASA website! MarcusMaximus (talk) 02:57, 9 June 2009 (UTC)

I guess I don't have any objection to pointing that out, if you think it's a common error. Honestly though, I don't know who could actually think the the velocity of matter entering or leaving the system could safely be ignored. I think my preference would be to phrase Newton's 2nd law as applying to a particle (where a particle comprises by definition a definite, unchanging amount of matter). Then in some section on "systems of particles" or something, rockets and fluid dynamics could be discussed, and it could be pointed out that you don't actually have to painstakingly apply Newton's 2nd law to every individual particle, but you do have to account for any momentum flux across the system boundary. Rracecarr (talk) 14:15, 10 June 2009 (UTC)

Wow, I just looked at the NASA link. That is amazingly bad. They go through all this rigamarole, with multiple errors (including moving to the rocket frame without accounting for pseudoforces), to come up with m dV = -v dm, which you can write down directly from conservation of momentum. Rracecarr (talk) 14:42, 10 June 2009 (UTC)

Well, I took a stab at fixing up the constant mass stuff and the open systems section. I found and added a good explicit reference about varying mass. I also removed that video. MarcusMaximus (talk) 07:24, 11 June 2009 (UTC)

Getting better, in my opinion. However, I think the first mention of "systems" is abrupt, awkward, and possibly misleading. Saying F=d(mv)/dt applies to systems of constant mass, when up to that point the discussion has focussed only on the effect of force on a "particle" or "object" or "body" is confusing. How do you even define "v" for a system of unconnected particles? Really, as soon as you start talking about systems, the equation needs to include summation symbols, and I think there should be at least a sentence smoothing the transition from talking about a single particle to a system of particles. Rracecarr (talk) 15:14, 11 June 2009 (UTC)

Laws of motion

1st law

Mass always changes its values unless it is in certain conditions but we ignore such changes and what about the radioactive elements which decay at every interval of time and thus reduces its mass while there is no external force acted upon it. Free neutron is the another example which is unstable and decay under 15 min. Similarly no external force was involved in creation of universe (bigbang).

2nd law

Let a body of mass of 1 kg on the surface of ground. We say F=W=mg. Is it possible for a body at rest to have acceleration “g” unit = m/s/s, Is there any rate of change of velocity”?

Nothing can travel at speed greater than speed of light c but if we put certain values of m and F in equation F=ma then we can get a speed viz greater than c provided there is no other attraction. e.g. F=150,000 N and m=0.5 kg. (a= rate of change of speed) but mass doesn’t remains constant if it travels close to the speed of light.

Similarly, if a body is moving in space with constant acceleration 1 m/s/s . It means that its speed increases with 1m/s at each second till it reaches to the speed of light. is it possible ?Further, gravitational time dilation is the effect of time passing at different rates in regions of different gravitational potential which was tested/ confirmed with difference of nanosecond recorded by atomic clocks at different altitudes. Thus if time does not remain constant how can we say that an object of mass is moving with v or a (in constant t).

3rd law

To every action there is equal but opposite reaction but in newton’s law of gravitation the resisting force of falling mass or reaction of gravitational pull is missing. Similarly in creation of the universe (bigbang) there is action but reaction is missing. 96.52.178.55 (talk) 04:59, 15 February 2009 (UTC) zarmewa khattak

This is newton's laws of motion, they don't hold at high speed or the special cases you describe. This was described by Einstein, see special relativity and general relativity.

Newton's laws refer to constant mass systems, if you want to model systems of changing mass, you have to break up the system into smaller "particles" so that you account properly for all mass.

In the case of a body resting on the ground there is a force F=mg acting on the body, but the ground is also exerting a force canceling out the one caused by gravity from earth. So the net force is 0, and thus the acceleration is 0.

In the case of a falling mass (e.g. an apple) the earth causes a force on the falling body, but the falling body causes an equal and opposite force on the earth. The body's mass (m) is small in this case, so the acceleration of it (F/m) is relatively large. However, earth's mass (M) is enormous in comparison, and thus the acceleration of the earth (F/M) is negligible when the force is of the same magnitude. If the other body has a mass comparable to that of the earth, the acceleration of the earth would be noticeable.

In the case of big bang, everything is pushing on each other and expanding, there is no problem there either. If you wonder what caused big bang, no one knows.
—Apis (talk) 11:24, 17 February 2009 (UTC)

You said net force = 0 and thus acceleration is 0. Then how we calculate weight of an object i.e. w=mg

The steel bar ruptures when all internally developed resisting forces fail to resist the applied force. Thus if we apply the same analogy to the paradox, the falling mass (Newton’s apple) should rupture first before falling on earth (bigger mass).

Similarly, you said that falling body causes an equal and opposite force on the earth. You also said g=a=F/m and g=a= F/M .This means that weight of 1 kg of sphere is decreases when it’s mass increases on the surface of earth.

Similarly, you also said acceleration of the earth would be noticeable if an object mass comparable to earth. So do we notice such accelerations in case of celestial bodies?

“in case of big bang everything is pushing each other and expanding”. This means someone is wrong either Newton (law of gravitation) or Sir Einstein. 96.52.178.55 (talk) 19:25, 22 February 2009 (UTC) K

You have some misunderstandings. I'll try to help with some of this: First, you are not quite right that "g=a". It is true that the acceleration due to gravity is equal to g, but that doesn't mean I'm not experiencing gravity unless I am accelerating. Almost to the contrary: if I were in an elevator accelerating down at 9.8 m/s², I would feel weightless within the elevator. I would still have the same mass but my weight with respect to the elevator would be zero.

We measure the weight of an object by measuring the reaction force on a scale. Gravity pulls me down with a force of mg; the scale pushes back (the reaction force); we measure the reaction force.

If nothing were under me, I would accelerate with an acceleration of g. The equal-and-opposite reaction would be the earth accelerating toward me at a rate of F/m_earth where the force on me and on the earth is equal (same amount of force) but opposite (we are pulled toward one another) but the accelerations are drastically different because the earth weighs so much.

As for "in the case of [the] big bang...", this is also not a problem. If you have two people on ice skates pushing on one another, they will accelerate away from each other; this is simple Newtonian mechanics. —Ben FrantzDale (talk) 15:05, 23 February 2009 (UTC)

For every action...

As an amateur student of quantum physics (my PhD is is in cognitive nuerobilogogy) and also one who researches parapsychology, I have always had trouble getting my mind around this particular law. The quotation is indeed simplified but on the surface, in day to day life and in the field, would seem to be self defeating. It is already established that there are n absolutes in physics, that time is not linear and non local and that the actions we take now can effect the past, present and future. How does Newtons law fit in with quantum theory? Dr. K. M. —Preceding unsigned comment added by 74.13.83.108 (talk) 18:30, 17 April 2009 (UTC)

For every action...A Question?

I have an even more pressing point... isn't the third law incorrectly stated here? "If Body A exerts a force on body B, body B will exert a force equal and opposite in the same plane". But this isnt true... they work on different bodies.

Yes they do work on different bodies, as the passage you quote states: one force works on body B and the resulting reaction force works on body A. -AndrewDressel (talk) 21:50, 3 May 2009 (UTC)

Take the "weekly shopping" example: When you push a trolley forwards, you are exerting a forwards force on the ground, but it is the TROLLEY that exerts a backwards force on you... Woodyjojo (talk) 21:09, 3 May 2009 (UTC)

You are mixing two pairs of action/reaction forces. First, when you push on the ground, the ground pushes back with an equal and opposite force. Second, when you push on the cart, the cart pushes back with an equal and opposite force. -AndrewDressel (talk) 21:50, 3 May 2009 (UTC)

Proportionality changes to equality only by choice of units

The text correctly states the second law as "... the net force on a particle of constant mass is proportional to the time rate of change of its linear momentum" but incorrectly summarises it as F = d(mv)/dt.

Newton's proportionality only changes to the usual equality when units are particularly chosen to render the proportionality constant as unity, such as kilograms for mass and Newtons for force. Nh5h (talk) 04:37, 20 May 2009 (UTC) Charles

Simpler introduction

I had a semester of quantum mechanics and a semester of relativity in college, but I've forgotten most of it. When I looked up this entry to refresh my memory, I couldn't get anything from this introduction.

Under WP:MTAA, the entire article should be understandable by a non-specialist. That isn't always possible, but certainly the introduction should be understandable.

It should explain what the three laws are in language that a non-specialist can understand. An intelligent layman, or a high school student taking science for the first time, is a non-specialist.

Does anybody have a suggestion for an easy-to-understand introduction? Does anybody object if I try? Nbauman (talk) 14:08, 31 May 2009 (UTC)

The proper formulation of 2d Newton law will be force to acceleration ratio stays the same if body is stiil the same (empiric law). It is definition of inertial mass. Another comment - this system of differential equations has no analytical solution in 99% of cases. —Preceding unsigned comment added by 140.168.71.18 (talk) 03:37, 1 June 2009 (UTC)

I attempted to make an improvement, and Xxanthippe deemed it unhelpful, but he didn't say why. I can perhaps see the appeal of stating the second law as simply F = ma and avoiding big words such as proportional, but I hope to gain consensus for replacing "action and reaction" with "forces always come in equal and opposite pairs" or the equivalent. The fact that "action and reaction" are archaic and misleading is demonstrated by the need always to clarify that Newton meant forces, as the current article does at least twice, and readily confirmed by external sources. - AndrewDressel (talk) 01:08, 2 June 2009 (UTC)

So, we currently have:

Newton's laws of motion are three physical laws that form the basis for classical mechanics. They are:

1. A body at rest stays at rest, and a body in motion stays in motion, unless it is acted on by an external force.

2. Force equals mass times acceleration (F = ma) (or alternately, force equals change in momentum), and

3. To every action there is an equal and opposite reaction.

I think 1 is fine. I would rather see a little more detail in 2: Force on what, mass of what, acceleration of what? My comments about 3 are above. I believe in the current form it is oversimplified to the point of being meaningless. The fact that force A on body 1 from body 2 has an equal and opposite counterpart, force B on body 2 from body 1 is crucial. -AndrewDressel (talk) 17:38, 7 June 2009 (UTC)

If you had too much detail it wouldn't be an introduction. I don't mind changing it, but under WP:MTAA, it has to be simple enough for a non-specialist to understand.

How do the introductory physics textbooks describe it? I suspect that this isn't the kind of thing a Wikipedia editor can just write out of his head. You have to see how the experts describe it. Nbauman (talk) 04:02, 8 June 2009 (UTC)

While that might prove fruitful, I wouldn't be surprised to find most introductory textbooks missing the mark. They have a captive audience and the rest of the chapter or text to straighten things out. I had to look through a few to find one that agrees with my point that expressing the 3rd law as "action and reaction" is not helpful. You are holding this article to a higher standard, which I think is a good thing. -AndrewDressel (talk) 13:02, 8 June 2009 (UTC)

One thing I've noticed after checking several texts is that I haven't seen an author try to explain them all at once. For example Halliday and Resnick introduce one law at a time and spend a page or two explaining it before moving on to the next. -AndrewDressel (talk) 16:34, 8 June 2009 (UTC)

Let's try to piece it together word by word:

1. A body at rest stays at rest, and a body in motion stays in motion, unless it is acted on by an external force.

a. mentions body to which law applies

b. mentions staying at rest

c. mentions staying in motion (maybe should also say "constant motion")

d. mentions external force (maybe should also say "net external force")

2. Force equals mass times acceleration (F = ma) (or alternately, force equals change in momentum), and

a. perhaps should use momentum first, since that's how Newton said it

b. probably has to include famous (infamous?) F = ma, but maybe only in parentheses along with corresponding text.

c. should mention body, which has mass, which experiences change in momentum or acceleration, and on which force acts.

d. should clarify that "force" means "external net force". "External" can't be too advanced if it gets mentioned in first law.

e. What about "constant mass"? I'm not sure.

f. perhaps should use the word proportional, as Newton did.

3. To every action there is an equal and opposite reaction.

a. mentions "equal and opposite", as it should

b. Should use "force" instead of "action", as laws 1 and 2 do, and as I try to explain above.

c. Should mention bodies that create the forces and that the forces act upon. Even Newton himself followed up in the same sentence with "corporum duorum". As Feynman explains in his lectures, Volume 1, Chapter 12, Section 1, What is a force?, "one of the most important characteristics of force is that it has a material origin."

d. The familiar "action reaction" expression can be provided in parentheses just as I suggest "F = ma" be for the 2nd law.

That leaves me with:

1. A body at rest stays at rest, and a body in motion stays in constant motion, unless it is acted on by a net external force.

2. The net external force on a body is proportional to its change in momentum, or more commonly, force equals mass times acceleration: F = ma.

3. The forces between two bodies are always equal and opposite, or more commonly, for every action there is an equal and opposite reaction.

Let the critiquing begin! -AndrewDressel (talk) 13:53, 8 June 2009 (UTC)

The first filter I use for a Wikipedia introduction is (on the authority of WP:MTAA) whether it is simple enough for a reader who is looking for an introduction; for example, a high school student studying physics for the first time.

If it's confusing to me, then it doesn't pass that filter.

If I have to stop and ask myself, "What exactly does that mean," then it doesn't pass that filter.

At Francois Monad's Nobel prize presentation, the speaker quoted Monad as saying, "In describing genetic mechanisms, there is a choice between being inexact and incomprehensible." and said, "In making this presentation, I shall try to be as inexact as conscience permits." [6]

An introduction has to be simple enough for the intended reader to understand it. If you try to be too exact, you can fail that test.

Sometimes you have to oversimplify in the introduction, and clarify it later. Nbauman (talk) 14:27, 8 June 2009 (UTC)

That might be all well and good, and I think I've read it before somewhere, but it doesn't say anything about the current proposal. -AndrewDressel (talk) 14:57, 8 June 2009 (UTC)

Overall I like your simplified version, Andrew. I think at this high level, though, it is even more effective to leave out words like "net" in front of "force", and possibly even "external", because I think the casual reader doesn't form a distinction between unmodified "force" and the highly qualified "net external force".

I think the third law is hard to understand because it deals with about 12 things in very short order: two bodies, two forces, exterted by which body, on which body, in which direction is each force, and of what magnitude? The bodies, directions, and magnitudes are fairly straightforward and are stated in the law. The biggest points of confusion seem to be the "by which" and "on which", because they are not clearly stated in the law. In practice this means students are unable to comprehend the difference between a 3rd law force that impedes all motion by giving a net force of zero (like pressing against a wall), and a 3rd law force that is commonly called an "inertial force" that is the natural companion of acceleration (like pushing a cart). Students know there are two forces that are equal and opposite somewhere in their free body diagram, but they don't know which body is exerting which force on what. MarcusMaximus (talk) 01:23, 9 June 2009 (UTC)

This sounds like WP:OR. Sources will be needed for such material in the main article. Xxanthippe (talk) 03:10, 9 June 2009 (UTC).

I'm just explaining here on the talk page what I think the challenges are to create a good front page article. MarcusMaximus (talk) 03:16, 9 June 2009 (UTC)

I agree with MarcusMaximus that adding terms like "net" and "external" make it confusing to the ordinary reader. What's the difference between "force" and "net force"? I don't know myself.

When I read an introductory paragraph for the first time, I ask myself what each word means. If I have to ask myself what "net" means, and what "external" means, it's much more difficult for me to understand, if I can understand it at all. Nbauman (talk) 17:23, 9 June 2009 (UTC)

Newton's second law

(Copying this from my personal talk page)

Here's your WP:RS:

"According to Newton's second law, force is the time rate of change of the momentum"

[7]

Rracecarr (talk) 15:02, 19 June 2009 (UTC)

I saw that when I did a Google search. That's a blog. Blogs are not a WP:RS, according to WP:SPS. Who is Leonardo Motta? He's just some guy without any stated credentials who posts for free.

Besides, Motta himself also refers to it as "rate of change of momentum" in that link. So it's just as accurate to say, "rate of change of momentum" as "time rate of change of momentum", according to your own source, and it's simpler.

If you could find a well-known physicist or physics teacher using that phrase while addressing introductory students, I would accept it. Nbauman (talk) 15:26, 19 June 2009 (UTC)

NBauman, if you do a google search with "time rate of change" in quotes, you get hundreds of results from class lecture notes, homework problems, and textbooks in calculus, physics, and engineering, illustrating its widespread use to mean "the rate of change with respect to time". I didn't even think this would be a point of contention; it's in common usage.

[8][9][10][11][12][13][14]

And in particular, Richard Feynman uses it: [15]. Look in the Acceleration section on page 8 of that document to see the statement "Acceleration is defined as the time rate of change of velocity."

At the very least, it should be obvious that "rate of change" is ambiguous because it doesn't specify with respect to what. If you accept both as correct, and the rest of us only accept one, then we should go with the one that all of us accept. MarcusMaximus (talk) 22:30, 20 June 2009 (UTC)

First of all, blogs don't count, and neither do class notes WP:ELNO WP:RS. I am glad to find the chapters of Feynman online, since my copy is buried in storage right now.

I agree that it is accurate to say that force is the time rate of change of momentum in Newton's second law.

The issue is whether it is the best way to introduce the concept to an ordinary reader -- for example, a student studying physics for the first time.

There are lots of concepts in physics that are true but too complicated to explain to a beginner. A good teacher knows what to include and what to omit.

The idea that the change in momentum is the time change of momentum is true but it's not essential to understanding the relationship of force to momentum.

I think anyone reading this for the first time would understand it's about time. After all, velocity is the change in distance per unit time.

Feynman specifies the time rate of change because he's giving a precise definition -- and he's not explaining force to a student who's learning about it for the first time. He's emphasizing time. And apparently he got the point across to you.

I'm not disputing the facts. I'm arguing that there are only so many ideas you can put into an introductory sentence before a reader who is new to the topic gets confused. The question is, do you introduce the concept of time in the introduction, or do you point it out further down after the reader has understood the introductory concepts. It's a question of pedagogy.

I'm not going to change it. I'll wait to see if somebody else comes along and decides to to change it for the sake of clearer reading. Nbauman (talk) 00:14, 24 June 2009 (UTC)

Why do you think time rate of change is unclear? MarcusMaximus (talk) 01:21, 24 June 2009 (UTC)

Where can we apply first law?

It is practically impossible for a single mass to exist in universe and all other masses are always under the influence of gravitation (external force) unequivocally 96.52.178.55 (talk) 06:13, 10 July 2009 (UTC)Quoth Khattak #1

That is a good question. It is primarily used as a qualitative statement of principle, which allows us to say that "the only things causing this object's motion to change are these forces". There are also many situations where gravity is negligible compared to other forces. MarcusMaximus (talk) 01:07, 11 July 2009 (UTC)

Reordered discussion of 2nd Law

I changed the order of the sentences in the discussion of Newton's 2nd law in so it is introduced in a more logical way, at least in my opinion. I'd like to hear feedback on what other editors think. MarcusMaximus (talk) 03:30, 13 July 2009 (UTC)

1- Change in motion is proportional to force impressed and impulse impressed can only produce motion if greater than the resisting force of an object. So is this initial amount of impressed force which was used to overcome the resisting force, excluded or included in the mentioened impressed force viz proportional to change in motion? What is true net picture of this impulse impressed, should it be subtracted if included or added in case excluded or otherwise should be justified clearly if remains the same?

2- Here is the sentence in the article to which I want to draw your attention “The product of the mass and velocity is momentum (which Newton himself called "quantity of motion"). Therefore, this equation expresses the physical relationship between force and momentum.”

Momentum (mxv) and motion are two different things. Momentum is the phenomenon which occurs after the cessation of impulse impressed or it is a potential control exists in a body while in the article the change in motion requires an impulse impressed. Thus "momentum" which is mentioned again and again doesn’t jibe with "motion" lucidly in the article. Since this is caption of Newton’s second commandment of Motion not momentum therefore either momentum should be culled or it needs more careful revising.

I hope I have explained things clearly enough that you can understand what I mean.96.52.178.55 (talk) 05:57, 16 July 2009 (UTC) Khattak #1