Talk:Newton's laws of motion

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suggestion to add this sentence

In fluid mechanics Newton second law is called the linear momentum equation. Can anyone add this point to this wiki article. I found this point in Fluid mechanics Frank M White 7th edition pg:140 — Preceding unsigned comment added by B.NIROSHAN (talkcontribs) 09:36, 26 February 2024 (UTC)

Proportionality

Newton's second law states a "proportionality" between impressed force and change in motion. Why does the article replace the term "proportionality" by "equality"? Note that a proportionality of quantities A and B reads A/B = C = constant, which is clearly different from the equality A = B = A. 2003:D2:972D:D312:5467:1AF8:F3D8:6E2E (talk) 19:37, 1 March 2024 (UTC)

The article quotes the 2nd law using "proportionality". Newton's Definition #2 defines momentum. The modern forms in the article are paraphrases as is clearly stated. I added a ref to Feather, Norman. An Introduction to the Physics of Mass, Length, and Time. United Kingdom, University Press, 1959. Johnjbarton (talk) 22:09, 1 March 2024 (UTC)
Any proportionality can be converted to an equality by the use of the constant of proportionality. If A is directly proportional to B, the equation can be written:
${\displaystyle A=B\times k}$ where k is the constant of proportionality.
For example, in the English engineering system of units, the force and the mass are both measured in pounds, the acceleration is measured in feet per second squared, and the constant of proportionality is the reciprocal of gc which is 32.17 ft/s2.
${\displaystyle F={\frac {m}{g_{c}}}\times a}$ where F is a force in pounds (lbf), m is a mass in pounds (lbm), and a is an acceleration in ft/s2. A force of 1 lbf is required to give a mass of 1 lbm an acceleration of 32.17 ft/s2
An alternative is to define a new unit of force so the constant of proportionality is unity and so can be ignored. This has been done in SI units by defining the newton as the unit of force so that a force of 1 newton is required to accelerate a mass of 1 kilogram by 1 m/s2:
${\displaystyle F=m\times a}$ where F is a force in newtons, m is a mass in kilograms, and a is an acceleration in metres per second squared. Dolphin (t) 05:12, 2 March 2024 (UTC)
Sorry, no. It is not true that "any proportionality can be converted to an equality". For example, take A/B = C = constant, with A = 12, B = 3, C = 4 = constant. A and B are proportional. 2A/2B = 24/6 = 4; 3A/3B = 36/9 = 4, etc.(Euclid's law of equal integer multiples). So how do you obtain A = B ?? Note, by the way, that Newton explains in the Scholium after Lemma X (Principia 1713!) that proportionality deals with "indeterminate quantities of a different kind". So A and B (or "force" and "change in motion") are a priori of a different kind in Newton's teaching, and therefore they can never be equal. 2003:D2:972D:D374:5467:1AF8:F3D8:6E2E (talk) 09:09, 2 March 2024 (UTC)
You wrote the equality in your second sentence: A/B = C. It's not that the two proportional variables can be said to be equal, but that you can write an equation (equality) using the proportionality constant (C in your example). CodeTalker (talk) 00:54, 3 March 2024 (UTC)
If A is directly proportional to B we write:
${\displaystyle A\propto B}$
If we wish, we can then write:
${\displaystyle A=k\times B}$ where k is the constant of proportionality.
In your example, k is 4 so your equation is:
${\displaystyle A=4\times B}$
As you can see, this equation never becomes ${\displaystyle A=B}$ Dolphin (t) 00:57, 3 March 2024 (UTC)
That's what I'm saying. You cannot simply skip the constant of proportionality in order to obtain A = B! But that's what they're doing who erroneously assert that "any proportionality can be converted to an equality". Should this be true we would never have discovered the constant c that governs Maxwell's laws and special relativity, nor would Max Planck have discovered the constant h that governs quantum mechanics. Natural constants are always proportionality constants which cannot be dismissed ad hoc. The same with Newton's second law. If you write it according to A/B = C, you have to realize the proportionality constant c. As a matter of fact, Newton's law stems from Galileo (Newton himself ascribes it to his predecessor, in Principia (1713), Book I, Scholium after Corollary VI to the laws of motion). In Galileo's Discorsi of 1638, you can find this law, and there you will find that the required proportionality constant bears dimensions "element of space over element of time", [L/T]. It is the "parameter" of the spacetime reference system of Galileo's (and Newton's) natural reference system of motion "in space and time". I discovered it already in 1985 (Philos. Nat. 22 nr.3 p. 400). It was only banned from mechanics when in the 18th century Euler and others invented "analytical mechanics", which they made the new theory of motion, replacing Galileo's and Newton's geometric mechanics, and basing it on F = ma. Should we return to Galileo and Newton, respecting their geometric method and the said constant altogether, so that the second law would read F = delta (mv)*c, mechanics would again be rooted in the reality of space and time, and everything in mechanics would change. 2003:D2:972D:D312:89AB:3E1B:33AA:525A (talk) 07:53, 3 March 2024 (UTC)
Perhaps my statement would be less likely to confuse if I change it to “any proportionality can be converted to an equation.” Look above in my previous edit to see an example. Dolphin (t) 11:52, 3 March 2024 (UTC)
No, sorry again. We speak of a proportionality between quantities A and B different in kind. This relationship can be symbolized by A~B, where the proportionality constant is implicit. You can make this constant explicit writing an equation according to A/B = C = constant, or A = B*C. But this equation is not an equality A=B!
So F~(ma) as an equation reads F/ma = C = constant, or F = (ma)*C, but never can you arrive at F = ma! Now, since F = ma is certainly the most basic principle of "classical" mechanics, one must see that classical mechanics (Euler, d'Alembert, Lagrange etc.), working with equations and equalities, is not Newtonian mechanics which works with geometric proportions A~B, or A/B = C, the "second law" reading F~delta(mv), or F/delta(mv) = c [L/T]. 2003:D2:972D:D312:89AB:3E1B:33AA:525A (talk) 14:17, 3 March 2024 (UTC)
This is another version of the much discussed variable-mass issue. We should sort it out. Johnjbarton (talk) 16:16, 3 March 2024 (UTC)
I made some small edits to the article to help avoid this confusion.
However to answer the original question
• Why does the article replace the term "proportionality" by "equality"?
The article, written for modern readers, uses modern definitions of "change of motion of an object" and "force" in which case, by these definitions, the proportionality factor is 1.0.
If you have information to contradict my claim (supported by references in the article), please post or add the reference. Johnjbarton (talk) 16:28, 3 March 2024 (UTC)
Just a comment. You're right stating that for modern readers the equality (equivalence) of "force" and "change of motion" is valid. Actually it is the basis of "classical" continuum mechanics. But, as has been shown, it is not Newtonian! Newton's laws is different. It requires a constant of proportionality that is not a dimensionless 1 (Principia 1713, Scholium after Lemma X). The message then is that Newton's (Galileo's!) theory of motion basically differs from that of "classical" mechanics. 2003:D2:972D:D368:F9CE:C42A:76B8:49DB (talk) 07:54, 5 March 2024 (UTC)

Wiki Education assignment: 4A Wikipedia Assignment

This article was the subject of a Wiki Education Foundation-supported course assignment, between 12 February 2024 and 14 June 2024. Further details are available on the course page. Student editor(s): Wkuehl9947 (article contribs). Peer reviewers: Lupe.b007.

— Assignment last updated by Ahlluhn (talk) 00:57, 31 May 2024 (UTC)

Semi-protected edit request on 21 July 2024

In the paragraph:

"Acceleration can likewise be defined as a limit:${\displaystyle a={\frac {dv}{dt}}=\lim _{\Delta t\to 0}{\frac {v(t+\Delta t)-v(t)}{\Delta t}}.}$Consequently, the acceleration is the second derivative of position,[1] often written ${\displaystyle {\frac {d^{2}s}{dt^{2}}}}$."

Change the equation:

${\displaystyle {\frac {d^{2}s}{dt^{2}}}}$

to:

${\displaystyle {\frac {d^{2}p}{dt^{2}}}}$

So that p matches the position variable name Traviskaufman (talk) 22:44, 21 July 2024 (UTC)

References

1. ^ Cite error: The named reference Thompson was invoked but never defined (see the help page).
No, the variable for position is s. Johnjbarton (talk) 14:39, 22 July 2024 (UTC)