I am aware of the actual equations .
But I've always wondered about the constants . Why must constants appear in every Physics equation ? And why can the constant in Newton's 2nd Law be simplified to 1 ?

Thank you for your reply .

What are you doing , currently ? Are you a Physics major ?

Why must constants appear in every Physics equation ?
Idk, probably that's how the world works.
Let say you conduct an experiment to find the relationship between force and the extension of a spring.
After you have get the data, you plot a graph, you'll get something like this:
(Dependant variable on y axis, independent variable on x axis)

So you'll get the equation, F = kx (y = mx) where k is the gradient. The c-intercept is 0, so thats why we do not get F = kx + c.

Some teachers tend to teach proportionality like this:
F α x
To remove α we must add a constant:
F = kx

And why can the constant in Newton's 2nd Law be simplified to 1 ?
The equation F=ma is not derived from F=kma where k = 1.
If you want to write it in proportionality form you'll write the equation F=ma as:
F α a (By assing that mass is constant)
And you get F = ma.
This means that "m" is actually the constant for the equation since m do not change. (For secondary physics level).

Another way to view proportionality is by intuition.
Lets take the Newton's gravitational equation F = -GMm/r^2

From this we can see that the attraction force will of course increase when the masses of the two body increases, therefore we can conclude that F α Mm. And intuitively, we know that when the distance, r increases, the force will decrease. so F α 1/r^2 (Using inverse-square law that newton found, since the body is sphere we cant use 1/r).
By combining these two equations, we get:
F α Mm/r^2
To remove the α, we must add a constant.
F = GMm/r^2.

Now you might be wondering how the heck they figured out G. Try to look up on a brilliant experiment called Cavendish experiment.



And no, I'm not a physics major

I hope you get a light on how classical physics equations are derived.

As fallacy said to err is human, so please fix if there is anything wrong with my explanation.

can you write the deriavation of gravitational potential energy at here

As a unsuccessful wannabe math olympian who never had any formal training in mathematical olympiads, I participated in this year's OMK Sulong 2013. Here are a few of my crude attempts at solving those questions.
The answers that I gave during the actual competition are:

1) 125. Looks quite obvious. So didn't give much thought to it.

3) 8125. Through last 4 digit analysis. the last 4 digits 0625 , 3125, 5625 and 8125 recur for every increment of 4 powers of 5. Since 5555= 4(1388) + 3, 8125 is the answer.

4) Is it 108? Wait, I think I made a mistake here. Not sure what I was thinking back then.[/URL][/IMG]

5) 279? Forgive me for my crude workings. 4! implies the number of permutation of the 4 digits. Minus 3! at combinations including the digit zero because one does not simply start a number with digit 0.
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6) 83. It took me forever to do it by trial and error but I got there.

As for section B,

1) b) Heron's Formula!

3) I tried doing it by induction but I am not sure if I have done it correctly. The working is so long that I feel it is almost certainly wrong.

This post has been edited by Intermission: Aug 11 2013, 02:18 PM

You should be able to find a lots of detailed explanations from many physics textbooks and websites.

The derivation of Einstein's most famous equation E = mc² is purely theoretical, which arises as a direct consequence of his Special Theory of Relativity that involves some algebraic manipulations using the Pythagorean theorem (SPM level). To understand how he derived the equation, you must understand a portion of his Special Relativity; Time Dilation (see the embedded Spaceship figure). The following derivation is highly simplified for the mathematically untrained. To further enhance your understanding, you should watch the YouTube video by minutephysics.

Hey Critical_Fallacy! I am so curious of your education background! Are you currently a researcher in a university? You took up a daunting task of explaining time dilation to the masses.

Bravo. Me too haven't had any formal training but surely you're more success than me.

For Section A, I think I only have one question correct, yup that prime factor.
Actually, 8051 = 8100 - 49 = (90^2) - (7^2) = (90-7)(90+7) = 83*97

Q3 should be easy though but I wrote 625 and the answer is 8125 TvT
Q1, Q4, Q5... just randomly put some answers...

Section B, I hate those prove, prove and prove...

Physicists know some things never change and they call them the fundamental physical constants. Such frequently used constants as the speed of light in vacuum, c, magnetic constant, μ0, electric constant, ε0, Newtonian constant of gravitation, G, Planck constant, h, elementary charge, e, and the Proton-to-electron mass ratio, μ, are assumed to be the same at all places and times in the universe. They form the scaffolding around which the theories of physics are built on, and they define the fabric of the cosmos.

Despite that, one of the most fundamental properties of Newton’s Mechanics, Maxwell’s Macroscopic Electrodynamics, and the Laws of Thermodynamics, is the absence of any physical constants in their basic equations. In fact, all necessary fundamental physical constants appear only at the stage of applications of these theories to specific phenomena. Nevertheless, these constants play an important role in physics and metrology because scientists need reference values for measurements in experimental physics and making theoretical prediction on papers.

Other than Newton’s and Maxwell’s equations, centripetal acceleration (a = v²/r), density (Q = m/V), pressure (P = F/A), and electric power (P = V*I) seem to be self-evident counterexamples to your above question. You can also see other frequently used physics equations for yourself in this link.


Although physical constants can appear to be dimensional or dimensionless, the imaginary constant factor k in your F = kma does not exist. That’s because the equation F = ma is self-contained and Newton did NOT require adding a dimensionless constant factor k = 1 to his mechanics to reproduce that “observation of force”. In fact, as guided by VengenZ in Post #20, Newton’s 2nd Law originally states that the net force on an object is equal to the rate of change of its linear momentum p in an inertial reference frame, which can be manipulated algebraically to be stated in terms of an object's acceleration, i.e. F = ma.

--->

Since the dimensionless constant factor k does not exist in F = dp/dt in the first place, then it follows that k is not needed in F = ma either. If your friends or physics teacher insist on the existence of k = 1, be sure to ask them what they think of a = dv/dt, p = mv, v = dx/dt, and other counterexamples as shown, so that other similar fundamental physics equations cannot be ignored, and they will thus be psychologically compelled to consider their merit.

~ HAVE A NICE DAY! ~

This set of frequently used Fundamental Physical Constants is recommended for international use by CODATA (The Committee on Data for Science and Technology). The full 2010 CODATA set of constants may be found at its website (click the image) at the time of this writing.



For your convenience, a PDF document (Extensive Listing) is available which can be read online or printed.
 Fundamental_Physical_Constants_____Extensive_Listing__2010_.pdf ( 124.83k ) Number of downloads: 1

Thank you for your detailed explanation ! I really appreciate it .




Thanks !


This is what I need ! I've been thinking about this since Form Four . You've cleared everything up . Thank you !

As for why I proposed that there is a constant in F = ma , here's what's written in our textbooks , and reference books . One of the errors in our Physics syllabus . My aim is to major in Physics , then one day hopefully , revamp the Physics syllabus in Malaysia .

Here's the explanation provided by textbooks :

From Experiment 1 , a α F
From Experiment 2 , a α 1/m

The two results are combined .

a α F/m

OR

F α ma

Therefore : F = kma

Unit of force is newton , N .

In order to make the formula as simple as possible , we make k = 1 by defining a force of 1 N as

1 N is the force which gives a mass of 1 kg an acceleration of 1 m s^-2 ,

1 N = (k) (1 kg) (1 m s^-2)
k = 1

Therefore , F = ma

This post has been edited by maximR: Aug 12 2013, 04:22 PM

What you see in the following figure is an excerpt from The Principia: Mathematical Principles of Natural Philosophy (2010), the reprinted text which based on Motte’s 1729 translation of the 1726, 3rd edition of Philosophiæ Naturalis Principia Mathematica, that was the final version corrected by Sir Isaac Newton.




As you can see, Newton never explicitly stated the formula F = ma in his Principia. The “motion” and “impressed” in which Newton used his terminology, and how he understood the second law and intended it to be understood, have been extensively discussed by historians of science. It was until James Clerk Maxwell published a treatise in Matter and Motion in 1876 (see excerpt below) that the modern ideas of how Newton was using his terminology is understood.



According to Maxwell, Newton meant by motion “the quantity of matter moved as well as the rate at which it travels” and by impressed force he meant “the time during which the force acts as well as the intensity of the force”. And so in modern terms, motion is Newton’s name for momentum, which allows Newton’s 2nd Law to be formulated as follows:

→

Now, allow me to make a few additional points after my Post #29. Based on what the mathematicians and physicists understood from the experimental observations of Newton’s 2nd Law: the change of momentum of a body is proportional to the impulse impressed on the body, a proportionality constant k, shall be assigned to the formula F = kma. In order to standardize the unit force and to eliminate the proportionality constant k, Conférence Générale des Poids et Mesures (CGPM) in 1948, adopted the name "newton" with symbol N for the SI unit force to define the amount needed to accelerate 1 kg of mass at the rate of 1 m/s² by choosing a proportionality constant of 1.

In a sense, the proportionality constant k can be seen as a conversion factor if we apply F = kma in other system of measurement, such as the Imperial units. For example, 1 pound-force (lbf) is defined as the amount needed to accelerate 1 pound (lb) of mass at the standard gravity of 32.174 ft/s². Consequently, we can deduced from this observation that k = 1/32.174. Thus, to eliminate the proportionality constant k, the preferred unit of mass is the slug because a slug is defined to have a mass of 32.174 lb, so that 1 lbf = 1 slug × 1 ft/s².

To summarize this post and Post #29, naturally, the proportionality constant k = 1 doesn’t appear in momentum (p = m*v), centripetal acceleration (a = v²/r), density (Q = m/V), pressure (P = F/A), electric power (P = V*I), and a bundle of other primary equations of physics, because the physicists relate the basic physical quantities by defining the formulation of physical laws in a way to achieve universality and generality. All in all, to determine whether a significant proportionality constant is needed or not, at the first place, we should at least understand how strong the correlation of physical quantities is by performing the dimensional analysis, and ensure that the experimental data are consistent.

Oh! It is not as daunting as the quantum mechanics. Coincidentally, 12 August marks the birthday of Erwin Schrödinger, the Nobel prize-winning quantum physicist whose eponymous equation lies at the heart of quantum mechanics.

In a shooting competition, the probability that ali hits the target is 0.4. Find the minimum number of trials that ali needs to make such that the probability that he hits the target at least once is 0.8. << add maths question. How do I do it? Which method should I use? Is it part of binomial distribution?

Another great post !
This is the first time I've read that Newton originally stated that Impulse is directly proportional to the Change in momentum . So he actually considered the time in which a force acts on a body . Interesting .

So if I get this right , the reason why G appears in Newton's Law of Universal Gravitation is because the unit of force is defined as amount needed to accelerate 1 kg of mass at the rate of 1 m/s² , therefore G should be added to ensure that the units on both sides of the equation are the same ? And that after careful experiments , experimental physicists discovered that the force on gravity between two bodies does not vary simply as F = Mm/r^2 , so there must be a constant to account for the interaction between the two bodies ?

And that means Newton never really had accurate calculations since he couldn't determine G , and did not include them in his equations ? ( or did he ? if yes , then how did he come to the conclusion that there is a constant just by empirical observation ? or did he derive it mathematically , or did he think that since the force of gravity is usually very small between two bodies , there must be something that accounts for this , therefore , a constant must exist ? )



This post has been edited by maximR: Aug 13 2013, 01:19 PM

Are you planning to go for it again next year? It will be the last time we are allowed to participate in OMK.

This surprised me when I re-learn physics at AS level too. Actually it is not the formula F=ma that leads to F=(mv-mu)/t but rather the other round when newton first formulated his 3 laws of motion.

I was well aware that Newton did not explicitly state that F = ma ( It's derived from his 2nd law ) , and I've read that force is defined as the rate of change of momentum early on when I first started learning dynamics . Did you really not read anything other than your SPM Physics reference book in form four and form five ?

But I didn't know that he said Ft α mv - mu

This post has been edited by maximR: Aug 13 2013, 10:49 PM

I hope I can. I'm upper six this year, should be last year for me to participate =(

In Newtonian Kinetic Energy, as long as the object is not at microscopic level and the speed involved is very much lower than the speed of light, the mass is considered an absolute, or in a state of constant in an inertial frame of reference.



P.S. Thanks sharp-eyed Krevaki for pointing out the typo in Line 4 of the previous image. Correction has been made.

This post has been edited by Critical_Fallacy: Aug 14 2013, 07:28 PM

Though Newton’s Principia (1687) theorized the presence of the gravitational constant G, it was not until 1798 that the constant was determined through observing the density of the Earth in the Cavendish experiment. Because Newton’s calculations could not use the mathematical value of G, he could only calculate a force relative to another force as a ratio using Kepler’s Third Law. The following propositions for Newton’s Law of Universal Gravitation are the excerpts from The Book III of Principia. If you are interested, you probably find this link useful (Read Proposition VIII. Theorem VIII. Corollary 2).





As you probably already know, most modern physics textbooks states Newton’s law of universal gravitation according Maxwell’s Matter and Motion (1876); “Every portion of matter attracts every other portion of matter, and the stress between them is proportional to the product of their masses divided by the square of their distance.” Maxwell also acknowledged that although the idea of inverse square law had been the prior work of others, but in the hands of Newton, the doctrine of gravitation assumed its final form.



This post has been edited by Critical_Fallacy: Oct 24 2014, 12:05 AM