Physics Question - ma = mg - Page 1
Forum section: electronics / general

author: bostonman
author_rank: Super Contributor
timestamp: May 13, 2021, 03:03:47 am
content: I'm trying to understand exactly why ma = mg.I took physics, and, the concept of weighing myself on the scale and learning it's really our mass, I'm confused about ma = mg (this began after watching the Big Bang Theory).If my body mass is 75kg, and I get on a scale, my 'weight' is (75 * 9.: 735NIf something with a mass of 75kg accelerates at 9.8, then the 'force' is also 735NObviously ma = mg in this case. After reading, it's equal because we are on Earth. If I weighed myself on the moon, then ma = mg, however, if everything is relative to gravity on the moon, then ma = mg on the moon. For the most part, ma would equal mg anywhere in the universe providing gravity and acceleration are the same.Am I missing something in the translation of why ma = mg?

author: Whales
author_rank: Super Contributor
timestamp: May 13, 2021, 03:14:55 am
content: Algebra:m = massa = accelerationg = acceleration due to gravity ~= 9.8ms-2 on earth's surfaceWe start with your eq:ma = mgDivide both sides by m:a = gYep. Don't think of g as a variable, but instead an exact value. Just like pi isn't a variable, it's a value, even though we write it as π instead of 3.14159265. I can write y = π and I can write a = g; these are both choices I make of subbing in particular values for particular purposes.g on the moon != g on earth, they're completely different values and hence once subbed in the meaning of the equations gets limited to "only valid on X planet's surface". Just like how if I choose m=70kg then the above equation becomes "only valid for human beings" instead of "valid for everything with a mass in a gravitational field".Does this help, or am I going down completely the wrong path?

author: andy3055
author_rank: Super Contributor
timestamp: May 13, 2021, 03:22:01 am
content: Here is a good page to read (ThoughtCo.com):https://www.thoughtco.com/mass-and-weight-differences-606116

author: bostonman
author_rank: Super Contributor
timestamp: May 13, 2021, 03:38:12 am
content: QuoteThat was my point. My 'mass' (using 75kg) is the same throughout the universe and can't change (unless I eat lots of food while venturing around space). a = g doesn't only exist on Earth, but can exist anywhere in the universe too.So saying because we are on Earth, and acceleration due to gravity is 9.8, then ma = mg, isn't really correct.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: May 13, 2021, 03:44:27 am
content: Quote from: bostonman on May 13, 2021, 03:38:12 amYou are right. The equality is not general, it only holds for the special case where a=g. That special case does apply almost universally on this planet and as far as we know no one lives anywhere else (space station visits not counting as living), so you can see where someone got a little ahead of themselves with this one.

author: Whales
author_rank: Super Contributor
timestamp: May 13, 2021, 03:55:57 am
content: Quote from: bostonman on May 13, 2021, 03:38:12 amIt is correct, but only in certain situations. Basically no equation is truly universal, all have limitations.f = ma <-- only directly usable as an approximation between a very large mass (earth) and a very small mass (eg a person) OR between two very distant bodies where their radii are much less than their seperatory distance. Otherwise the results are quite wrong (you have to break each mass down into smaller pieces & integrate the results instead).f = m * 9.8ms-2 <--- all of the above restrictions, but now also only valid on the earth's surface (and technically in some other places in the universe, but this explanation is good enough)f = 75kg * a <--- only valid for things of that weight, eg approx a humanf = (width * length * height * density) * a <-- only valid for rectangular prisms of constant density

author: Anthocyanina
author_rank: Frequent Contributor
timestamp: May 13, 2021, 04:01:29 am
content: It's important to understand the distinction between mass and weight. Your weight on the moon is different because gravity on the moon is a different value than gravity on earth. weight is the force exerted by a mass because of gravity. gravity on earth is called "normal gravity" and the mass is calculated according to earth's gravity, so weight on earth is the same as mass on earth. mass on the moon is the same as mass on earth because the matter that gives it that mass would be the same matter on earth (and now under the influence of the "normal gravity") and its mass is calculated with our planet's gravity

author: IanB
author_rank: Super Contributor
timestamp: May 13, 2021, 04:10:40 am
content: Quote from: Whales on May 13, 2021, 03:55:57 amExcuse me? F = ma is one of the fundamental laws of Newtonian mechanics. It is always exactly true, without exception, in a Newtonian framework. What makes you think it is an approximation?QuoteAgain, what the heck? F = mg is universally true for any object of any mass in a gravitational field. The thing to remember is that g varies with the magnitude of the gravitational field, so it is almost never exactly 9.81 m/s2, although it is approximately that on the Earth's surface.QuoteThis is just a total non-sequitur. It is technically true in some sense, but has no useful meaning.

author: IanB
author_rank: Super Contributor
timestamp: May 13, 2021, 04:17:35 am
content: Quote from: bostonman on May 13, 2021, 03:03:47 amWell, it's not, in general, true. So maybe that's why you are confused?What is true is that mg = mg (obviously). The symbol "g" is just used for the special value of acceleration that applies to a free falling body in a given gravitational field (usually Earth's). In this case the acceleration experienced by that body is given by the symbol "g". In general, "a" can have any value at all, but in the specific case of a free falling body in Earth's gravitational field, "a" = "g". So F = mg is just a special and unique case of the more general F = ma. That's all there is to it.

author: AntiProtonBoy
author_rank: Frequent Contributor
timestamp: May 13, 2021, 04:18:47 am
content: Quote from: bostonman on May 13, 2021, 03:03:47 amWeightandinertial massare different things.Weightandforceare also different things, but closely related.The relationshipF = masimply tells you what is the resulting forceFexerted by massmwhen accelerated at a given ratea.In the context of Earth, gravitygearthis a special case of accelerationa = 9.81. In this case, your weight is related to the force exerted by your body's mass when accelerated by gravity. We standardise this weight such that 1 kg = 9.81 newtons.In other words,weightearth= m * a / 9.81. Sincea = gearthon earth, weight and mass is equivalent in magnitude, that isweightearth= m.On the moon, your mass would still remain the same, but the weight would be less due to different acceleration by gravity,a = gmoon= 1.63.In this case,weightmoon= m * 1.63 / 9.81 = m * 0.17.TL;DR:Generic Newton's force formula:F = m * aWeight force on any planet (substituteawith gravitational acceleration):Fplanet= m * gplanetWeight on any planet in kg:weightplanet= m * gplanet/ 9.81

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 13, 2021, 04:21:53 am
content: IIRC, it's surprisingly not well understood why inertial mass equals gravitational mass, at least to high observational precision (which isn't all that high in the grand scheme of physicsthings, because gravity is difficult to measure).The reason given by General Relativity, is that space itself is moving, accelerating towards gravitational wells. There was an excellent animation of this in a recent YT video, which of course I didn't bookmark, so if someone can remember it please put it here -- in any case, the fact that spacetime itself is curving in towards a mass, is equivalent and indistinguishable to inertial mass being accelerated.What'sweirdabout the picture is not that space is accelerating, but why wearen'twhen we stand on the Earth! Well, the answer is relative, of course -- since space is moving past us here on the surface at ~9.8 m/s^2, something must be accelerating us up through all of it. The force for that eternal acceleration is -- drumroll -- the force of your weight pushing you along, and so on down to the core of the Earth where all the force, from the entire mass of the planet, is bearing against itself; fortunately, planetary matter is not very compressible at these pressures (but it isn't resistant to shear, and so the Earth gets pulled into a sphere, more or less -- it reaches hydrostatic equilibrium).So, obviously that weight exactly equals your inertial weight, because otherwise you'd be dragged along in space at 9.8 m/s^2 relative to the Earth's surface.There are some weird ideas, like Mach effects, which I don't think have been disproven? Or perhaps they're equivalent after all, but we don't quite yet understand how. The deeper mystery is how to integrate General Relativity into the Standard Model, so that we have a complete understanding from the smallest quantum level to the scale of the entire universe. Perhaps then we will have a more complete explanation. That, however, will take some time it seems.Also:Purely from Newtonian mechanics, there is no answer, of course. Newton's laws can be derived from Relativity given suitable approximations (c --> infty), just as Relativity is an approximation of some as-yet-unknown better model, and so on (it's turtles all the way down, at least until we have a more convincing reason to believe otherwise).Tim

author: thermistor-guy
author_rank: Frequent Contributor
timestamp: May 13, 2021, 04:38:07 am
content: Quote from: bostonman on May 13, 2021, 03:03:47 amI read your "ma=mg" as indicating inertial mass and gravitational mass are the same:https://www.einstein-online.info/en/spotlight/inertial-and-gravitational-mass/"Gravity"describes how bodies move through space-time, which is curved by large masses. If you treat gravity as a force, then it turns out that the acceleration of small bodies are the same regardless of mass. This is peculiar when you think about it.Go to the gym. Lift a 10 kg mass, then a 20 kg mass with the same acceleration. The 20 kg weight takes more effort (force) on your part.Drop the 10 kg and 20 kg weight from shoulder height to the gym floor. The two weights move through space-time, with the same acceleration. If you regard gravity as a force, then "gravity" must be exerting more force on the 20 kg weight - twice as much - as on the 10 kg, the same way you did during your lifts. This is odd.Now imagine jumping out a window with the 10 kg weight and releasing it mid-air. You and the weight are travelling with the same acceleration through space-time. But you don't feel any force on you, you don't feel any pressure on your body, the way you do when you lift a weight, or go up in a fast elevator. Is gravity really a force, then? If it is, it's a strange kind of force where you don't feel any :-)Interestingly, near the Earth, space-time is curved mostly by the Earth, but there is also a tiny effect due to the sun and moon. An extremely sensitive pendulum clock will detect it:https://en.wikipedia.org/wiki/Shortt%E2%80%93Synchronome_clockhttp://leapsecond.com/pend/pdf/1986-Mar-AH-Boucheron-Shortt.pdfPhysicists aside, we treat gravity as a force because it matches our everyday language, and makes sense of our everyday experience. But it is a misleading line of thought, according to General Relativity.https://www.universetoday.com/108740/how-we-know-gravity-is-not-just-a-force/

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 13, 2021, 04:52:09 am
content: Quote from: thermistor-guy on May 13, 2021, 04:38:07 amAnd (not to be redundant as I haven't read the link to see if it happens to make this connection already..!) we have a similar case with "centrifugal force": the apparent force to the outside, is actually the rotating object pulling inward. The difference between centrifugal and centripetal is merely a coordinate transformation....And no Mr. Bond, I expect you to die.Tim

author: Brumby
author_rank: Unknown
timestamp: May 13, 2021, 07:21:03 am
content: I am consistently amazed at the tangents and minutiae that appear in response to such a simple question. Some have made good effort at trying to clarify while others have been anything but.This, to me, is the best offering so far:Quote from: AntiProtonBoy on May 13, 2021, 04:18:47 amThis is where a general formula benefits from subscripts to distinguish different scenarios.For example:gearthgmoongmarsgplutogoumuamuaAll these values for g will be different - but they plug into equations the same way - giving results that are correct for their respective situations.But then it is spoiled by this:QuoteFor starters, kg is a unit of mass - not a weight. Secondly, what is that divide by 9.81 all about?

author: thermistor-guy
author_rank: Frequent Contributor
timestamp: May 13, 2021, 12:47:33 pm
content: Quote from: Brumby on May 13, 2021, 07:21:03 amI don't regard it as a simple question. To me, the equivalence principle (inertial and gravitational masses are equivalent) is profound and mysterious.Yeah we can do simple calculations, but lurking behind the "why" of those calculations lies something deep about the nature of reality. Since the OP has studied physics, I assumed the question was about more than simple calculation, and more about the "why". Looks like I've misunderstood.

author: TimFox
author_rank: Super Contributor
timestamp: May 13, 2021, 01:33:49 pm
content: One of the great 19th-century physics experiments was carried out by the Hungarian Baron Eötvös from 1885 to 1909. Seehttps://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_experimentHis group showed that gravitational and inertial mass of different materials was equivalent to a sensitivity of about 10^-8.Note thatg, the gravitational acceleration due to the earth, does change with altitude, and slightly due to variations in the density underfoot, but for most purposes (except for inertial-guidance navigation) it is a constant vector pointing down.Kater's pendulum is a classic way to measureghttps://en.wikipedia.org/wiki/Kater%27s_pendulumIn old laboratory textbooks, the period of the pendulum was measured by having it oscillate in front of a grandfather clock and comparing the frequency to that of the clock pendulum (illustration in above Wikipedia reference).

author: mathsquid
author_rank: Frequent Contributor
timestamp: May 13, 2021, 04:12:26 pm
content: Rather than going about defining different subscripted g's, I'd use the law of universal gravitation and G = 6.67408 x 10-11m3kg-1s-2$$ F= G \frac{m_1m_2}{r^2}$$Plugging in mass and mean radius of the earth gives you$$ F = 6.67408 \cdot 10^{-11} \, \frac{5.971 \cdot 10^{24} m}{(6.371\cdot 10^6)^2} \approx 9.8m$$meters/s2

author: radiolistener
author_rank: Super Contributor
timestamp: May 13, 2021, 06:50:28 pm
content: Quote from: bostonman on May 13, 2021, 03:03:47 amThis is not obvious. This is known as equivalence principle, which says that the gravity mass equals to the inertial mass.From experiments we know that the gravity mass is very close to the inertial mass, the difference is less than 10^-14 or something like that. But we don't know if they are equals exactly or there is a very small difference which we cannot catch due to insufficient measurement resolution. And there is no explanation why they are very close.Nobody knows that. So, this is very not obvious.Probably you will be interested to read the Chapter 28 "Electromagnetic mass" in The Feynman Lectures on Physics Volume 2.It is about the mass of electron. A little quote:Quote

author: TimFox
author_rank: Super Contributor
timestamp: May 13, 2021, 07:54:54 pm
content: Obviously, the gravitational and inertial mass can be defined as equal for one specific material, such as pure gold. Baron Eötvös showed that this was true (within very small experimental error) for a wide range of materials. My favorite statement (attributed to) Albert Einstein is "Raffiniert ist der Herrgott, aber boshaft ist er nicht,”. Surely only a demon would violate the equivalence principle by the teeny margin of 10^-14.

author: AntiProtonBoy
author_rank: Frequent Contributor
timestamp: May 14, 2021, 01:32:52 am
content: Quote from: Brumby on May 13, 2021, 07:21:03 amkg and unit of mass equivalence isonly true on Earth. In other words 1 kg of weight equals to 1 unit of mass exerting exactly 9.81 newtons of force on earth; or alternatively we can also say that 1 kg of weight equals to 1 unit of mass accelerated by 9.81 m/s2. If you want to measure weight on another planet, then you will need to scale by the ratiogplanet/ 9.81.Example: if you weight 100 kg on earth, then you will weigh 16.53 kg on the moon. Here is why:weightearth= m = 100 kggmoon= 1.62weightmoon= m * gmoon/ 9.81weightmoon= 100 * 1.62 / 9.81weightmoon= 16.53 kgWiki has a pretty good explanation on this:In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons on the surface of the Earth, which is its mass multiplied by the gravitational field strength. The object's weight is less on Mars, where gravity is weaker, and more on Saturn, and very small in space when far from any significant source of gravity, but it always has the same mass.https://en.wikipedia.org/wiki/Mass_versus_weightAlso, how do I insert LaTeX formulas in this forum?

author: JohnnyMalaria
author_rank: Super Contributor
timestamp: May 14, 2021, 01:58:59 am
content: Quote from: IanB on May 13, 2021, 04:10:40 amYour own assertion answers your questionIt's an approximation because it's only "exactly true" "in a Newtonian framework". Ergo, it's an approximation because it fails outside of a Newtonian framework.

author: mathsquid
author_rank: Frequent Contributor
timestamp: May 14, 2021, 02:02:43 am
content: Quote from: AntiProtonBoy on May 14, 2021, 01:32:52 amTo do displayed math, just put it between double dollar signs. I don't know if there's a way to do it inline.

author: bostonman
author_rank: Super Contributor
timestamp: May 14, 2021, 03:13:53 am
content: I'm somewhat glad my question wasn't so easily answered because I thought it was too simple. As someone pointed out above, some of this has deviated and dove deep into physics.Although the basic plug and chug formulas make things easy, sometimes they don't make sense after trying to analyze them.One thing we can all agree on is mass remains constant, it can't change throughout the universe. If I want to know my weight on Earth, using 100kg for Mass to make the math easier, then (W= mg) it's 9.8 * 100kg = 980N. If a 100kg object accelerates at 9.8m/s^2 for 1s, then the Force (F = ma) would also be 980N.Therefore ma = mg.Now take the same situation above and place me on the moon. My weight is 162N (moon gravity = 1.62m/s^2) and the 100kg object accelerating at 1.62m/s^2 for 1s is also 162N.Therefore ma = mg.My point is: from my understanding, no special case exists where ma = mg because we are on Earth. From my initial assumption, it seemed a special case existed where ma = mg because we are on Earth and/or a connection existed with 9.8 to make ma = mg.Basically, they can equal each other anywhere in the universe providing the accelerating object accelerates at the same rate as the gravity as to where you are.I don't want to deviate, but I'm a bit uncertain how F=ma and W=mg were discovered. We can calculate Mass because we know gravity, but centuries ago, nobody knew the acceleration of gravity, so they couldn't calculate the Mass being dropped. It's kind of the chicken or the egg I guess.Also, if I place a 100kg block on a table, is the block a Weight with the table acting as a Force in the upward direction, or is the block also a Force? During this discussion, I began realizing, I'm confused as to when to use a Weight and when to use a Force (unless we are talking about say a car driving into an object, then that's a Force).For reference, at 5:20 is where the original question (from The Big Bang Theory) came from:

author: AntiProtonBoy
author_rank: Frequent Contributor
timestamp: May 14, 2021, 03:19:16 am
content: Quote from: mathsquid on May 14, 2021, 02:02:43 amInline test: \$ foo = x \$$$ foo = x^6 $$Edit: looks like inline uses the slash dollar sign syntax.Also LaTeX doesn't work when previewing the post. Dave, is there way to fix this?

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: May 14, 2021, 03:40:49 am
content: Quote from: bostonman on May 14, 2021, 03:13:53 amNot sure that's correct. There's rest mass and relativistic mass.https://en.wikipedia.org/wiki/Mass_in_special_relativity

author: bostonman
author_rank: Super Contributor
timestamp: May 14, 2021, 03:47:02 am
content: Okay.... is it safe to just say Mass remains the same for purposes of discussion to answer the question in hand?

author: IanB
author_rank: Super Contributor
timestamp: May 14, 2021, 04:25:59 am
content: Quote from: bostonman on May 14, 2021, 03:47:02 amLet's say you confirm experimentally that F = ma by applying a known force to an object of known mass and measuring the acceleration. And for all combinations of forces and masses this holds true.Now let's say you drop a known mass from a height and measure its acceleration in free fall. And let's suppose that you measure that acceleration in repeated experiments and it is always 9.81 m/s2. You may suppose from the previous formula that the force acting on it due to gravity should be F = ma, with known mass and measured acceleration.Furthermore, suppose you place the same mass on a spring scale, and confirm that the weight of your given mass on the scale is exactly the same as the force of gravity you calculated from the free fall experiment.By these various experiments you now have all the measurements matching up and agreeing with each other. As a result, you may conclude that F = ma = mg until an experiment shows otherwise.

author: thermistor-guy
author_rank: Frequent Contributor
timestamp: May 14, 2021, 05:21:43 am
content: Quote from: bostonman on May 14, 2021, 03:13:53 amHere is how I think of it.You, a 100 kg mass (rest mass), stand still on your bathroom scales. There is net zero acceleration.The mass of the Earth bends the space near it, and accelerates objects towards it (its centre of gravity) at, say, 9.8 m/s/s.Your scales push you back. You can feel the force, that push-back, under your feet.That force produces an acceleration that nets out (cancels) the acceleration due to the Earth's bending of space.The force required to do that is 980 N (100 kg * 9.8 m/s/s). In everyday language, you say "I weigh 100 kg today".You are in a NASA training aircraft, for astronauts, to experience "weightlessness". The plane goes into a parabolic trajectory.Everything inside the plane seems to float, including you. You are accelerating towards the Earth, at 9.8 m/s/s because of the bend in space, but you feel no force.In everyday language you say "I feel weightless." There is no force to stop you from accelerating towards to Earth.You're at home, outside, and you stand on a strong table. You and the table tend to accelerate towards the Earth, because the Earth bends space towards it.But the net acceleration of both you and the table is zero. The table pushes against your feet with enough force, 980 N, so you have net zero acceleration.The Earth pushes back against the table, a little bit harder, so that the table plus your 100 kg mass have net zero acceleration.The way I think of it, is that you can have acceleration due to a large mass bending space, and you can have acceleration due to a force acting on a body.In school, I was taught that the first cause of acceleration was also due to a force, called gravity. This line of thinking works in the everyday world. It'saccurate enough, much of the time, but conceptually it is misleading.Thank you for your though-provoking question.

author: radiolistener
author_rank: Super Contributor
timestamp: May 16, 2021, 01:13:56 pm
content: Quote from: bostonman on May 14, 2021, 03:13:53 amI'm not sure about that. The quantity of matter in the body will be the same at any point of universe, but if we're talking about mass in context of interaction force with other matter in the universe, it may depends.

author: Nominal Animal
author_rank: Super Contributor
timestamp: May 17, 2021, 09:19:57 am
content: Quote from: bostonman on May 14, 2021, 03:13:53 amAnother way to realize the same thing, is to consider how one can measureweightas opposed tomass.To measureweight, we use a spring, a counterweight (for comparative analysis with known masses), or a strain gauge. They all measure the combined forces (gravity, buoyancy due to surrounding fluid – atmosphere –, and so on). So, "weight" is just an indication of the force pulling the object and the planet closer together.To measuremass, we typically use rotating contraptions, since angular momentum is not dependent on gravity, assuming good enough bearings. By measuring the angular momentum caused by a known force at a known distance from the center of rotation, say a spring, we can determine the actual mass.Which also means that while in microgravity, say on a space station, you are weightless, you are most definitely not massless. If you push a little too hard when you careen down an access tube, and catch yourself awkwardly with one hand, it is exactly as easy to break your hand there as it would be here on Earth, given the same velocity of your body at the point of impact.In particular, trying to move or slow down a large mass, is exactly as hard in microgravity as it is on Earth on frictionless bearings – and air bearings are close to frictionless, which is why those air tracks are often used in freshman physics classes.Also, going up to a low Earth orbit, does not mean you get much "further out" of Earth's gravity well. In fact, at the International Space Station, at about 420 km above sea level, the gravity exerted by Earth is still about 8.6 m/s2; or about 87% of that at the sea level. It's just that due to being in orbit, ISS is in free fall; it basically simply misses ever hitting the Earth.Quote from: bostonman on May 14, 2021, 03:13:53 amI'd say it would be worth your time and effort to understand the concept offorce,potential, andwork.Let's consider gravity a field, a field ofgravitational potential energy, but ignore all spacetime-bending and relativistic effects, keeping things very straightforward. We cannot measure it directly, but its units would be the units of energy. (It would also be a simple scalar field, and not say a vector field with a direction at each point, like magnetic fields have.)Gravitational potential energy is not something we can tap into, either. Gravitational potential energy is just a way to express the kinetic energy loss/gain between points at different gravitational potentials. In a real sense, it does not "exist"; but, unless you stay at the exact same potential, you either gain or spend, kinetic energy.Total energyis the sum of kinetic and potential energy; in the same way one might consider the total amount of funds they have the sum of cash at hand, and the amount of funds available in bank accounts (although there is really no physical guarantee you can access those).(Zero-point energyis a completely different concept, and involves quantum mechanics.)In a pendulum, the kinetic energy is maximum at the equilibrium position (where potential energy is zero); the kinetic energy is zero and potential energy maximum at the extremum positions. Ignoring losses due to friction etc., the total energy stays constant, and describes the system (not at any specific point in time, but how it behaves). So, potential energy is a very useful notion when dealing with anything moving. In a gravitational potential field, the "zero" is at infinite distance.The force exerted by a potential field – be it gravity or anything else – is exactly \$F = -\nabla U\$, where \$U\$ is the field at that particular point, \$F\$ is the force exerted, and \$\nabla\$ is thegradient; in Cartesian coordinates,$$\left\lbrace \begin{aligned}F_x &= \frac{\partial \, U(x, y, z)}{\partial \, x} \\F_y &= \frac{\partial \, U(x, y, z)}{\partial \, y} \\F_z &= \frac{\partial \, U(x, y, z)}{\partial \, z} \\\end{aligned} \right.$$i.e., the 3D analog of tangent of the field. The minus sign is there because the force is always "downhill". Without the minus sign, the force would repel.This is, by the way, exactly how molecular dynamics simulators simulate atom motion. Classical models define potentials (force fields in chemistry terms); quantum mechanical or ab initio methods model electron charges (and atomic nuclei as point-like positive charges) as fields, but using quantum mechanics to model the way those fields interact. Because of the heavy calculation involved, we're limited to thousands or at most tens of thousands electrons, and the boundaries of the simulation must also repeat (cannot be in "empty space").For two point-like objects, we have \$U = -G\frac{M \, m}{r}\$, where \$G\$ is the gravitational constant, \$M\$ and \$m\$ are their masses, and \$r\$ is the distance between the two. Note that the divisor really is the distance; this the gravitational potential fields reachfarindeed. If you calculate its derivative, it is the familiar \$F = -G\frac{M \, m}{r^2}\$ (as a vector quantity, in the direction to the center of mass of the system; this applies to each point separately). So, this is just another way to describe Newtonian physics.Work is the amount of force exerted over a specific displacement, say from \$a\$ to \$b\$. If the force is constant, \$W = (b - a)F\$; if the force varies, \$W = \int_a^b F(p) \, d \, p\$. With respect to the gravitational potential field, it is exactly the difference in gravitational potential energy between the final and initial points. If you move in a gravity field, but stay at the same potential, you do no work; only the change in the potential matters.Orbits – and even driving a car in hilly countryside – work very much like pendulums, since the Earths gravitational field is not really uniform, and besides is affected by the Moon anyway: because you don't stay in the same potential, you trade potential energy with kinetic energy. When you go "downhill", you gain speed (trade gravitational potential energy to kinetic energy); when you go uphill, you slow down (trade kinetic energy to gravitational potential energy). The International Space Station occasionally needs additional speed boosts, but this is because even at 400 km up, there is still a bit left of Earth's atmosphere, slowing the ISS down. (Note that ISS orbital speed is also about 7.66 km/s ≃ 27600 km/h ≃ 17100 mph. Currently, atmospheric drag causes the ISS to slow down by about 0.00000000066 km/s per second.)If we go back to the object on a table, there is a force pulling the object and the center of gravity closer together; a small force due to fluid displacement which is either in the same direction (if denser than air) or opposite direction (if lighter than air). These are countered by the "static" force of the chair and ground and surrounding material involved. (Such systems are examined in the Mechanics subfield of physics – classical mechanics in this particular context. The forces that exist due to materials requiring energy to change shape, and the overall forces being less than that, are often called "static" in this context, because they are opposingdynamicforces – forces that try to change things. In other words, "static" and "dynamic" in this context are just labels that help conceptualize the situation, just like "acceleration" and "deceleration" are just two words describing change in speed or velocity.)It is often easier/shorter to just say that it's the table causing the equal but opposite force, but it's actually all material supporting the objectnotchanging, that contribute to it. At the atomic level, it's due to Pauli exclusion principle; because electrons repel each other, you need to spend a lot of energy to smush atoms closer together. (We call those kinds of forcespressure, by the way.)Since it's me writing this, there may have crept inaccuracies or poor analogs into this post. If you notice them, let me know; definitely do point them out so we can fix them. The last thing I want is to lead others astray in their basic grasp of physics!

author: bostonman
author_rank: Super Contributor
timestamp: June 15, 2021, 03:54:30 am
content: I guess the more one reads about physics, the more questions arise.The episode of The Big Bang Theory where Sheldon teaches "a little physics" to Penny was on last week. Since the initial start of this message, I somewhat learned little 'g' is the gravity on Earth. Now if it's the gravity on Earth, what happens to W=mg if we moved to Mars and blew up Earth?Since little 'g' is gravity on Earth, and Earth ceased to exist, then how can that formula exist?This all goes back to not understanding why ma = mg.As you discussed, an object on the table, but in more detail. Speaking purely of ideal numbers and not the many other forces, if a 100kg box is on the floor, and I care to lift it, the force on it is F=ma. This means I need to exert a force of 980N.One thing that occurred to me the other day is that a = 9.8m/s^2, but to lift it (ignoring initial force to overcome gravity), aren't we technically applying greater than 9.8m/s^2, else, if we didn't, the box would remain still because we are only applying an equal upwards force that is being applied in the downward force by Earth's gravity?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 15, 2021, 06:05:52 am
content: Well g is just a matter of convention. Without Earth, but with g still being an accepted mean value around 9.8m/s^2, it would probably just drop out of favor. Or we'd adopt the analogous g♂for Mars mean surface acceleration. Maybe over time the subscript would drop off and we'd be back to g anyway.The transient force to move a box vertically, is the sum of its weight and inertial force, i.e. depending on how fast you move it. Nothing could be simpler!Tim

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 15, 2021, 06:17:59 am
content: Quote from: bostonman on June 15, 2021, 03:54:30 amNo. The box exerts a force of 980N to the floor. If you exert a force of 980N on the box upwards, that just means the box doesn't move, you are just holding it in place – as if someone else were supposed to slip say a sheet of paper under the box.To lift the box, you need to dowork. If the force against which we do the work is constant, then the workW=Fh.Because acceleration due to gravity drops inversely to the distance squared, and sea level is about 6,371,000 m from the center of gravity, the difference between the force due to gravity at sea level and 1,000 m above sea level is less than 0.0314%. Even at 420 km altitude (International Space Station in low Earth orbit), the force due to gravity is 88% of that at sea level. So, for any kind of lifting by hand or by crane, or even aeroplane, we can approximate the force due to gravity being constant.Work has units of energy; this is the minimum amount of energy needed to displace something against the force, as well as the maximum amount of energy one can extract by letting something be displaced due to the force. So, for lifting on any planet or whatever with gravityg, we do workW=mgh, where them gpart is due to the forceFexerted by gravity on something having massm.Note, this massmis NOT WEIGHT. Mass is not dependent on gravity; weight is. If you have a frictionless surface with an object with massm, you need forceFto accelerate it bya(velocity per time unit, i.e. distance unit per time unit per time unit),F=m a, no matter whether you were on Earth, free fall, Moon, or anywhere else.Weight is what we call the force due to gravity. Typical measurement devices nowadays are springs and strain gauges. Essentially, we take something that deforms preferably linearly depending on the force exerted on it, and then measure the deformation somehow.Now, let's say you have a box on the ground that exerts a downward force of 980 N due to gravity (and we'll exclude buoyancy due to air etc.).If you exert a force of 981 N upwards on it, the forces almost balance out, leaving a sum total 1 N upwards force.This force causes an accelerationa,a=F/m, upwards. I we assumeg= 9.80 m/s2, then the massm= 100 kg, because 1 N = 1 kg m s-2. Therefore, the acceleration would be about 0.01 m/s2. Let's say you kept that up for ten seconds, so the velocity would be 0.01 m/s2× 10 s = 0.1 m/s. If you then reduce your upwards force to 980 N, the acceleration drops to zero, but the velocity remains. If you reduce your upwards force to 979 N, the acceleration reverses, becoming 0.01 m/s2downwards.It does not matter how you apply the force; it does not need to be a wasteful rocket engine. Stuff like a solid four-by-four plank of wood can easily handle a kiloNewton of compressive force; lots more in the direction of the grain. Lots of lifting mechanisms use levers, hoists, pulleys etc. to maximize the energy efficiency, so that to lift a 100 kg by one meter upwards in standard Earth gravity (which takes a theoretical minimum of 100 kg × 9.80 m/s2× 1 m = 980 kg m2s-2= 980 N m = 980 J = 980 W s) you need less than say 2 kJ of energy, typically electrical energy – say, 500 W for four seconds.Note that this also means that if we ignore electrical motor efficiency differences (including internal friction), a 1 kW winch that lifts that 100 kg in two seconds, uses exactly the same amount of energy than a 250 W winch that takes eight seconds to lift the same weight.However, the difference inforce exertedis huge. Because of the mechanical leverage involved, the instant acceleration at startup can be pretty huge – in physics, the change of acceleration is calledjerk– so examining the forces to understand how things happen isn't as important as examining the work done to achieve a specific outcome.(You've watched Dave's videos. The energy needed to do specific work is what we examine in physics, to do the back-of-the-envelope calculations; we don't examine the forces involved.)The forces come into play when we need to worry if the materials we use for the mechanism can take the "load"; whether thestressexerted by the forces exceed the material properties.We can, however, estimate theminimumforces involved in events, if we are interested in.If you have a friend who can lift a 100 kg box one meter high in one second, we can immediately estimate the minimum force they exerted on the box upwards to achieve that. Because the total acceleration is linearly dependent on the force,a=F/m, we first need to estimate the minimum acceleration needed to start from standstill and reach one meter in one second; this happens to be constant acceleration ofa= 2 m/s2; and because the friend had to overcome the force from gravity as well, they exerted a force ofF=mg+ma= 100 kg (9.80 m/s2+ 2 m/s2) = 1180 kg m/s2= 1180 N.Due to human physiology, the initial jerk was probably much higher, and the force exerted at the end somewhat less, but the above is the minimum instantaneous force the friend lifting the box did exert. To find out exactly how much, we'd need to track for example the height of the center of mass of the box as a function of time.(When humans do height jump, they exploit that, by the way: by arching their body in a suitable pattern, they can go over a boundary, while keeping the center of mass below the top of the boundary at all times. Nifty trick, that.)

author: bostonman
author_rank: Super Contributor
timestamp: June 17, 2021, 03:46:30 am
content: QuoteIf my upward force would be 980 N, equal to the downward force (100kg mass and Earth being 9.8m/s^2, why would velocity remain the same?Unless you're talking about no air resistance.

author: Circlotron
author_rank: Super Contributor
timestamp: June 17, 2021, 06:14:38 am
content: Quote from: bostonman on May 13, 2021, 03:38:12 amTo be pedantic, your mass increases proportionally to your velocity. If you at 75kg are moving at 100 km/h or 27.777 m/s you will have 28,935 Joules of kinetic energy. That amount of energy is equivalent to 0.0000000000003215 kg or 0.3215 nanograms of matter. That is how much mass you will gain, so I am led to believe.Edit -> because your mass is now slightly more than 75kg then presumably your kinetic energy is also slightly more than 28,935 Joules.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 17, 2021, 07:21:39 am
content: Quote from: bostonman on June 17, 2021, 03:46:30 amThat.It is important to realize that having some force exist, does not mean energy is being spent/transferred.In the case of a box sitting on the ground, the forces arestatic, and no energy is transferred.Similarly, when the total forces acting on an object – including air resistance, gravity, everything – are equal, the object is in free fall, and retains its velocity.For energy to be transferred via some force, the force has to dowork.

author: thinkfat
author_rank: Unknown
timestamp: June 17, 2021, 07:46:49 am
content: I don't think you need to dig deeply into Relativity to understand this equation. It's just a statement, an important one, though, expressed as a mathematical equation.ma = mg just says that mass behaves equally in a gravitational 'field' as when being accelerated by any force. The equation states that gravity can be understood as a force, causing a mass to accelerate.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 17, 2021, 08:31:06 am
content: Quote from: Circlotron on June 17, 2021, 06:14:38 amKinda, but also no (because you end up in a cyclical forever increasing mass and velocity argument if you follow that).Kinetic energy isn't \$E_K = \frac{1}{2} m v^2\$, it is actually \$E_K = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\$ for a rigid body.The two agree for relatively small velocities \$v\$, but start to differ when velocity \$v\$ gets closer to the speed of light in vacuum \$c\$.Here, \$m\$ is the rest mass, that is invariant of velocity. This is based on the linear momentum as described by Einstein's theory of special relativity.In practice, it means that given a specific kinetic energy, your velocity is smaller than the classical kinetic energy (\$E_K = \frac{1}{2} m v^2\$) would indicate; and given a specific velocity, your kinetic energy is higher than the classical formula suggests.Special relativity describes two related concepts:invariant mass(the mass at rest, described by \$m\$ above), andrelativistic masswhich bends space-time. It is easier to think of relativistic mass as relativistic energy, though, because us humans so closely associate "mass = weight"; and it really is the total energy,E, that bends space-time. As an example, even though photons have no rest mass (\$m = 0\$), they do have linear momentum (as described in special relativity) and kinetic energy; and they do bend space-time.

author: bostonman
author_rank: Super Contributor
timestamp: June 17, 2021, 02:24:12 pm
content: QuoteI think I dug too deeply initially. When I saw this on TBBT, I initially thought it meant the Weight of (as an example) a box at rest on a table, and the table has an upward Force exerting a F=ma equal to the weight of the box W=mg.After digging into it more, I began reading that they are equal because we are on Earth. This didn't make sense to me because a can be any object with acceleration that just so happens to be what the gravity is of that planet (or a point in space).What I make of ma=mg is that an object sitting on the ground is experiencing a force F=ma where 'a' is Earth's gravity, and its weight is W=mg where 'g' is gravity on Earth; therefore ma = mg.If we were on the moon, ma would also equal mg in the example above.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: June 17, 2021, 07:10:21 pm
content: Or, for the case originally cited, a = g. Something that given enough time, even Penny would have realized.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 18, 2021, 08:06:13 am
content: Quote from: bostonman on June 17, 2021, 02:24:12 pmThis is the correct intuitive understanding, I believe.Mass is the property of matter that does not depend on gravity.We call the force due to gravity 'weight', and that does depend on gravity; it is that force that weighing scales etc. measure.On other planets and moons, 'g' differs, so objects have different weights there, despite having the same mass.On Earth, it takes 0.45 seconds to drop 1 meter from standstill, because g = 9.80 m/s².On the Moon, it takes 1.11 seconds to drop 1 meter from standstill, because there g = 1.62 m/s².A one liter bottle of water weighs 1.00 kg on Earth, but only 0.165kg on the moon.Its massm=1.00 kg everywhere, because to accelerate it at 1 m per second per second, 1 m/s², you need a force of 1 N, if we ignore possible drag and friction, on Earth as well as the Moon.It is only the downwards force, 'weight', that varies due to different gravities.Physicists use an air cushion, like an air hockey rail, to create a very nearly frictionless surface to slide tiny carriages on.This is a very illustrative device, giving the correct intuition in this matter in a practical way.No matter in what kind of gravity we are in (as long as the air cushion rail still works), the same mass needs the same force to accelerate at the same rate.It is only the 'weight' and not 'mass' that is affected by gravity.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 18, 2021, 08:24:45 am
content: In case someone is wondering whether 'mass' itself could actually be split into two different quantities based on what I wrote above, wonder no longer: Even Einstein contemplated that.The two "different" aspects of mass are calledgravitational mass("slow mass") andinertial mass("fast mass"). One is the quantity related to gravity (the 'm' when dealing with weight); and the other is the one related to movement and inertia (the 'm' when dealing with actual acceleration).Experiments on Earth and in Earth orbit (microgravity) have shown that the two must be equal to at least 12 significant decimal digits: that even if they were different, their relative difference must be less than 1:1,000,000,000,000. So, for all intents and purposes in the human scale, there is just one 'mass' for each object.A core starting concept in Einstein's theory of general relativity is theequivalence principle: that gravitational mass and inertial mass are equal and indistinguishable.It actually goes even further, as thestrong equivalence principlestates that locally, acceleration and gravity are indistinguishable. (That in a small closed box, given any scientific instruments and experiments you want, it is impossible to distinguish whether the box is standing still on a planet, or in constant acceleration.)All experiments we have managed to devise thus far indicates these are exactly, precisely true.

author: bostonman
author_rank: Super Contributor
timestamp: June 18, 2021, 02:16:04 pm
content: Do we ever really care about Weight in equations unless we want to know based on what an object is on planets?Another words, if a box is resting on a table, it's just all based on Forces, correct? It would be the force in the upward direction the table exerts, and the force in the downward direction the box exerts. Tilt the table 45 degrees, and now there is a X and Y axis force (the force the box has in the X direction and friction in the X direction opposite the box), and a Y, with a Fnet.Do we care what the Weight is in at all?

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 18, 2021, 06:20:51 pm
content: Quote from: bostonman on June 18, 2021, 02:16:04 pmWe derive friction from weight, but that's about it.Quote from: bostonman on June 18, 2021, 02:16:04 pmCorrect.Quote from: bostonman on June 18, 2021, 02:16:04 pmAn illustration should be in order here. The three forces shown are vector quantities, with direction and magnitude.RedFwis the force due to gravity the box exerts on the tilted surface.BlueFtis the static force the surface exerts on the box, because it resists deformation.GreenFfis the force due to friction.If the velocity of the box does not change, thenFw+Ft+Ff= 0.If the velocity of the box does change, thenFw+Ft+Ff=ma, wheremis the mass of the box, andais the acceleration vector. Its direction is opposite to friction,Ff.(This assumes perfectly smooth but not frictionless surface, and assumes the friction and tilt angle are such that the box does not start tumbling. If we'd add angular momentum, we'd see that the friction causes torque, which causes the leading edge of the box to exert larger force than the trailing edge of the box, and even a minor imperfection in the surface can catch the leading edge, increasing the forces such that the box starts rolling. Also, if the center of mass is not supported by the surface the box is on, it will tumble.)(If we add momentum and angular momentum, we have a pretty accurate simple physics engine at hand.)So, if we are trying to model the movement of an object, we care about weight, because it relates to friction and the stress on materials holding it in place (like the table surface above). To calculate acceleration, we use mass.Static friction is basically always larger than dynamic friction. In other words, if the object is not moving with respect to the surface it is on, the friction is much greater than when the object is moving with respect to the surface. This is why objects on tilted surfaces often only need a small nudge, but then keep moving, often accelerating. The initial nudge overcomes the static friction, and the dynamic friction is not large enough to stop the movement.Aside from this, it is important to realize that forces cause acceleration; without forces, velocity stays constant.If we push the abovementioned box to some speed on an inclined surface, and the friction is larger than the other forces combined, then the box slows down: it decelerates (decelerates == acceleration vector opposite to its velocity vector). When the box stops, the friction is exactly the other two forces combined. Because static friction is greater than dynamic friction (consider it a curve that drops quickly to a near constant, as velocity increases), there is often a bit of a jerk at the end.If the box "bounces" backwards a bit, that is due to temporary deformations relaxing; essentially spring-back.

author: RJSV
author_rank: Super Contributor
timestamp: June 20, 2021, 03:12:12 am
content: Ok, uh BRUMBY: (thank you)Is it alright, if my HEAD HURTS now ?? Jeez, but you make the best summary...I agree but interesting to read.Question about 'g' is that it's weird to call it acceleration directly. Perhaps it just has exact same units, being in meters per second, per second. So an object, on a scale, is not moving, per SE, but rather it just calculates out in a very similar math process.How did Newton prove (or did he?) a derivation ?I did college level 'classic' physics. Did Newton use integration, using another formula first ?I find I can't clearly address that...

author: bob91343
author_rank: Super Contributor
timestamp: June 20, 2021, 03:39:37 am
content: Yes, mass changes with velocity. So the proper equation is f = d(mv). Or f = m(dv) x v(dm). Or something like that.

author: RJSV
author_rank: Super Contributor
timestamp: June 20, 2021, 07:59:23 am
content: What I mean is, you have to start with integration, of the conventional little bits of matter, within a sphere (Earth). That's calculus, and I believe you can do this with a 'point mass', like a basketball for example.So you integrate that: The 3-D sphere with the point mass, to get the total force. The actual force equation has m1 x m2 and divided by r cubed (to 3rd pwr). And some sort of constant.After integrating that mess, you've got a force but no motion. Of course, any free mass will begin to 'fall' or accelerate downward. AND, any mass supported by the springs of a scale is going to exert that force as static.At that point you have, simply, Velocity = Accel X TimeTo me, it always looks like either stationary, with a static force, or it's going to be moving, in which case you've got Position / Velocity / and Acceleration.Non-relativistic masses, or slightly relativistic situations are, well, EXACTLY THAT. Although likely sensible to view mass as warping space - time, you need a lot of effect, to get, well, a lot of effect.No swirls around a 'hyper-big' mass as we blast thru hyperspace.....

author: bostonman
author_rank: Super Contributor
timestamp: June 23, 2021, 02:03:13 pm
content: I've been away and haven't had a chance to reply.QuoteTo divert a bit, I know (or believe to be true) that if I lift a box fast or slow, the same work is still done. If W = F * h, and F = ma, why would picking up the box faster not use more Work?Assuming no other forces except Earth gravity and the Force of the box, if I pick up the box 1m in 1s, my acceleration is faster than if I pick up the box 1m in 10s; thus making the Work larger.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 23, 2021, 03:00:00 pm
content: The power is proportionally larger, the work is the same.Mind that biological effort is irrelevant, for example you must spend effort just holding the box out in front of you (while doing no work on it). This doesn't need to be the case, as you could rest it on your knee or shoulders or whatever and not expend any effort beyond keeping it balanced. Or, y'know, rest it on the floor where it's obviously not doing any work.Electrical analogs include a lot of actuators, which incur idle power to hold position. Stepper motors, servos, solenoids, etc. The only reason these dissipate power is wire resistance. If they were wound with lossless wire (superconductor), the voltage drop would be zero while the holding force is proportional to applied current (which can circulate without decay).Tim

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 23, 2021, 03:39:03 pm
content: Quote from: bostonman on June 23, 2021, 02:03:13 pmThe key to understanding the conundrum is recognizing that you accelerate and decelerate. Doing both faster requires more power but the opposite signs mean that extra power doesn't translate into work, Only the net change in potential energy of the box is work.

author: IanB
author_rank: Super Contributor
timestamp: June 23, 2021, 03:40:42 pm
content: Quote from: bostonman on June 23, 2021, 02:03:13 pmPicking up the box faster does use more work. You are (1) doing work to overcome gravity, and (2) doing work to accelerate the box.Usually the work to accelerate the box cannot be recovered and goes to waste (like, for example, heat in the brakes of your car).The ideal minimum work to overcome gravity usually only occurs if you pick up the box infinitely slowly. In thermodynamics this is often called a "reversible" process. The box is never moving at any measurable speed, so you never do any excess work to accelerate it.Conceptually, you could get the work of acceleration back again (like regenerative braking), but this would never be 100% efficient and you always lose some energy in such a process.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 23, 2021, 04:03:21 pm
content: Quote from: bostonman on June 23, 2021, 02:03:13 pmBecause regardless of how long it takes, lifting a kilogram of weight in standard gravity ofg=9.80m/s² upwards one meter requires one Joule of energy. (1 J = 1 kg m s⁻².)Remember, theainW=mahis not the acceleration of movement, but the acceleration due to gravity against which you are doing the work.In physics, you cannot do the old scam of paving a driveway without agreeing about it with the owner, and then charging them for it.Wdoes not describe *how* you did the work, only the *amount* of work. The forces you choose to apply to do work against gravity is your choice; they only affect whether a specific end result can be achieved, and how long that will take, not the work needed to achieve that specific end result.Consider, for example, that instead of lifting the 100 kg box we discussed in previous posts by hand, you use a mechanical hoist. What is, exactly such a hoist? Why, it is nothing but a mechanical device that acts as aforce multiplier. There are many other such devices; hydraulic cylinders being among my favourites. Just by dint of their geometry, a force applied to a control surface, yields a much larger force on the output surface! And other than friction and mechanical wear, they spend no energy to do so! What magic is this?Well, no magic. A force in itself is not something we can exploit in any way. We can let the force do work, and convert or collect almost all of the energy released; or we can spend energy to cause a force to do useful work for us. A force in itself is not useful; it isn't evenreal, tangible, in the human sense.A force has the same utterly annoying feature potential energy has: tousethem, a physical change of some sort must occur. It is not the force or the potential energy itself that we can exploit; only thechangecan be exploited. (This is much more fundamental than one might think, and is at the core of why the misguided dreamers dreaming of zero point energy or free energy will never see their dreams fulfilled: they dream of exploiting something thatis, rather than something thathappens.)Quote from: IanB on June 23, 2021, 03:40:42 pmBullshit. That makes the asinine assumption that the energy spent to accelerate the box isnotrecovered. There is no reason for such an assumption; that energy is not "lost", nor does its amount affect the height lifted.Basically, you just made the claim that because you can do other work at the same time, more work is spent in the original task.Recovering just about 100% from acceleration is not hard, either. Converting it into useful energy in aspecificform (say, electricity instead of say heat) can be hard; but there is nothing physically preventing such conversions from reaching essentially 100%. ("Essentially 100%" in this context means that in the human scale, we can approach infinitely close to the limit, even if certain (but not all) such energy conversions can never reach 100% mathematically.)If you, IanB, amend the claim toa human will spend more energy to lift a box faster, then I obviously agree. For a human, because of the antagonistic makeup of our musculature –– we have separate muscles for gross movements, and smaller, weaker muscles for fine movements –– the optimum "speed" (with respect to energy expenditure) is not zero, and depends on the physiology of the individual. We also have basically no facilities for recovering any of the work our musculature has done for later use.It is rather surprising, given that human is the one land animal species that given enough time, can run down every single other known land animal. There is no other land animal in existence that does distance running better than humans do. (Yes, a healthy trained human can run down a horse. There just aren't that many humans around that can still run the 200 miles or 300 kilometers necessary.) There have been several publicity stunts, "competitions" between a man and a horse, in the last century or so; but the human runners in these competitions are laughably bad compared to e.g. message runners in ancient Greece or Rome.(Consider the Mongol Derby, the longest horse (riding) race in the world. In ten days, you need to ride 1000 kilometers; you do get to switch horses very often, though, as we don't want to kill the horses. However, the record for a human running 1000 km is under six days for men, and under eight days for women. No horse can do that. Look up "ultramarathon"; you'll be surprised at what humans are actually capable of.)Humans cannot really recover any of the energy their musculature spends, other than a little bit of the waste heat, and with mechanical implements. Passive exoskeletons – collections of braces and springs – are already in use, that can double the range a fully laden infantryman can travel under their own power in any given day without compromising their fighting ability. They basically recover the work done by human musculature into springs et cetera, and use that energy to augment future movement. However, a bicycle is even more efficient (as it reduces the work needed in the first place), if there is a road one can use.

author: David Hess
author_rank: Super Contributor
timestamp: June 23, 2021, 04:40:36 pm
content: Go back to the beginning and figure out what is actually being measured. Scales measure *force* and not *mass*. Weight is force.Units of pounds are force. The Imperial unit for mass is the slug, although in the industry it is more useful to talk about pounds force and pounds mass.

author: TimFox
author_rank: Super Contributor
timestamp: June 23, 2021, 07:25:30 pm
content: If you apply more vertical forceFto the massMthan the weightWof that mass, then when you stop applying that force, after moving throughHvertical distance,the mass is still moving upwards, now being slowed by the acceleration of gravityg. The net force on the mass while you are applying it is (F-W), which accelerates it vertically so it may rise.IfF=W, the mass does not accelerate, but may rise at a constant velocity (assuming you apply enough extra to overcome air resistance and other factors not included in this discussion).The kinetic energy of the mass when the force stops will beK= (F-W) x H, and the mass will continue to rise for a further vertical distanceD=K/W, whereupon it stops moving up and begins to fall, accelerated by the acceleration of gravityg.

author: SiliconWizard
author_rank: Super Contributor
timestamp: June 23, 2021, 09:54:57 pm
content: Sorry if it has been said, I didn't read quite all.Your "weight", per se, doesn't just depend on gravity, but also on the fact there is a reaction force (equal but opposed to gravity when you are at equilibrium.)This is this reaction force (to illustrate: that just prevents you from "falling" down to the center of the Earth) that effectively gives you a "weight". Your weight is not an inherent characteristic that just depends on mass and gravity. It "expresses" itself through a reaction force. Without a reaction force that opposes gravity, you effectively have no "weight".The most (normally) well known case is freefall. When in freefall, any massive object has virtually no weight.

author: TimFox
author_rank: Super Contributor
timestamp: June 23, 2021, 09:59:36 pm
content: You have merely re-stated Newton's Third Law.Weight is defined as the force of gravity on a mass, which in a uniform gravitational field (i.e., over a region much smaller than the distance to the heavy source of local gravitation, such as within a meter of the Earth's surface) is equal to the productW = m x g, wheregis the acceleration of gravity that is essentially uniform over this small region. On the moon, a similar region will have a different value for g, and therefore the mass will have a different weight.In freshman physics labs, in order to demonstrate the acceleration of gravity, a useful teaching method is the "Atwood Machine"https://en.wikipedia.org/wiki/Atwood_machine, where two masses with a known relatively small difference are connected by strings through pulleys, so that the net force is the difference in their weights.Yes, if the mass is sitting on a table, the table resists the force of gravity and the mass does not accelerate or move (if the surface is spongy, the mass will sink into the surface until the spring force of the foam rubber balances the weight).However, if you drop the mass, and it falls freely with nothing resisting it until it hits your foot, it accelerates according to Newton's Second Law, where the accelerating force equals the weight (defined above) and the mass is the mass. The velocity when it hits your foot is left as an exercise for the reader.

author: bostonman
author_rank: Super Contributor
timestamp: June 24, 2021, 01:13:39 am
content: Many anomalies exist while picking up a box fast versus slow. I'm trying to leave out the human factor in all this, air resistance (which I think has been ignored in the last replies), etc...Also, my physics background is only two or three courses (physics 1, 2, and 3), so I'm limited in knowledge, but obviously these discussions (and reading) have helped over the years.Leaving out recouping energy or methods of picking up a box as in holding the box out in front of me. If say a 100kg box is on the ground, I hover over it, pick it up exactly vertical 0.5m, and drop it. The Force on the box is 980N (100kg * 9.8m/s^2), and the Work is 490J (0.5m * 980N).Now if I picked up the box the same height in 1s versus 10s, the acceleration is still only 9.8m/s^2 regardless? So the fact I've accelerated the box faster or slower has no bearing on how many Joules my body used (using a conversion from Google: 117 calories).It makes sense to a degree from the chasing the horse analogy. The horse may outrun a human, but it burned the candle at both ends and runs out of energy 100 miles down the road after only 1hr (let's assume the animal and human is unable to take a "breather" and start running again). Whereas the human took 50hrs to run 100miles. In the long run, both have burned the same calories based on whatever factors would go into calculating Joules of a human/animal running - something I'm not asking to be calculated.I know from years of going to the gym that some people think running sprints periodically for short duration burns more calories than those who run a constant regular speed. But I imagine it would be the same thing. The person next to me on the treadmill will have run 10miles in a short time, but they didn't (in theoretical physics definitions) haven't burned anymore calories than me who took five times longer to run the same 10miles.Getting back to the box analogy, as someone said, if I now hold the box (maybe say hugging it against my chest), no Work is done. How could this not be true? My arms would get tired, and I'm assuming (again, leaving out the human factor) that my body is burning calories in order to overcome gravity. By the formula Work = Fd, the Work is 0J, and, if it were a motor holding something, the heat lost would be due to wire resistance but no Work done by the motor, but it seems my body would need energy to hold that box still.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 24, 2021, 05:07:07 am
content: Quote from: bostonman on June 24, 2021, 01:13:39 amYou are confusing work and the energy spent by your body; the same mistake IanB made above.To quote fromWikipedia:Work is the energy transferred to or from an object via the application of force along a displacement.Perhaps the error is my fault, though. I should have been careful to always use the wordonly, as in "the work inonlylifting a box upwards one meter in constant gravity"; but I just didn't think of properly excluding adding (unrelated/unnecessary) kinetic energy to the object. (Kinetic energy is not just linear motion, it includes rotation also.)If we have a box massing ten kilograms, standing still on the floor, with a table top nearby with its top one meter off the floor, and you move the box from the floor to the table, from standstill to standstill, the work done – the energy transferred to the object – is always 10 joules (10 J = 10 kg m² s⁻²), regardless of whether you do so in a fraction of a second, or take a year to do it. Or even if you do it twice, moving the box back to the floor in between.This amount, 10 joules, is also the minimum amount of energy you will need to spend to achieve this result. There is no upper limit, but the difference in energy will not be transferred to the object; it goes somewhere else. Often waste heat.If the box starts from standstill, but does not end at standstill, we can increase the work without limit by transferring extra energy as kinetic energy (classically, 0.5mv², wheremis the mass, andvis the change in velocity, for a non-rotating object).If the box does not start at standstill, but does end at standstill, then the work depends on the direction and speed (and rotation, if rotating) of the box. We could even end up with "negative" work, ending up gaining more energy from the box by stopping it than we need to transfer to it to move it a meter higher. A perfect example is using a trebuchet or similar device, and consider the situation where the box has just been "lobbed" as the starting point. Given a suitable trajectory, the box will end up on the table without any energy transfer, any work at all; the box just converts some of its kinetic energy to potential energy (by moving upwards in the local gravity potential) without any further external assistance.Basically, if we do not assume standstill at both starting and ending points, we can stuff as much kinetic energy in the non-standstill case as we want, and increase the work (energy transferred to or from the object) without limit.I so wish I could explain this stuff better!For simplicity, let's assume that we apply a constant lifting force for as long as possible, then drop the lifting force to zero at the precise moment, so that the box ends up with zero speed at the one meter elevation point. We ignore friction and buoyancy due to air, and both initial and final positions have zero velocity. No rotation either; the box has the same kinetic energy at the end that it had at the start.During the lifting phase, the acceleration (positive upwards) of the box isa-g> 0, wheregis the (downwards) acceleration due to gravity, 9.80 m/s², anda=F/m,Fbeing the lifting force applied andmbeing the mass of the box (10 kg). During the coasting part, the acceleration is -g.When a constant accelerationais applied for durationt, the change in velocity isat, and change in position isvt+0.5at², wherevis the initial velocity when the acceleration started.When a force acts on an object, the integral of the force (vector) over the interval, describes the linear momentum imparted to the object by that force during that interval. This is called theimpulse. Its units are Newton seconds (1 N s = 1 kg m s⁻1).(Remember that the box standing still on the floor has two major forces acting on it: one is gravity, and the other is the static force of the floor resisting deformation. Their impulses cancel out, whenever the box stays in the same location. So, if one does not considerallforces acting on an object, the impulses due to only some of the forces acting on it aren't really relevant in the real world: it is like measuring the velocity of a car with respect to an aeroplane flying overhead. By picking a suitable subset of forces, you can make the impulses whatever you want, just like you can make said car velocity whatever you want by picking a suitable aeroplane as the reference.)I suspect, but have no proof, that the chemical energy spent in human musculature is, as a first approximation, roughly linearly dependent on the impulse the forces generated by those muscles, for a relatively wide (but not full) range of human muscle power. In the absense of a better model, I'll use this.So, to minimise the human energy spent, we don't try to minimise the force, but the integral of the lifting force over time; i.e, we minimize the impulse due to the forces the muscles manage to generate.If you work out the math, and define the upwards lifting forceFasF=kmg, i.e. asktimes the force due to gravity (andk>1), then the lifting impulse is sqrt(2gm²k/(k-1)); the forcekmgis applied for duration sqrt(2/(gk(k-1)). Fork>1, this impulse is a monotonically decreasing function, so in this model, the harder the constant force you can apply, the less chemical energy your muscles will burn.Fork=10, andg=9.80m/s², the duration is just 50 milliseconds (0.048 seconds); fork=5, 100 milliseconds (0.101 seconds); fork=2, 320 milliseconds (0.319 seconds). (Thekscale factor is nice, because then the massmdrops out from the duration. The units do not seem to match, but that's just because the formulae has the fixed height to be lifted upwards, one meter, "hidden" in the constants. If you write it out using proper dimensions, it does work out correctly.)It does make sort-of intuitive sense – at least, if you've ever compared the effort needed to do say pull-ups slowly versus fast. I can do a dozen fast pullups (not the swingy ones – those are cheating via angular momentum –, nor using your legs to kick in the air to get some extra help; just fast but proper ones), but ask me to stay stationary midway, and I'll drop in a few seconds, much faster than it would take to do those pullups. I think. Need to verify at the gym (on separate days).However, this result is not because ofworkdone to the box, i.e. energy transferred to the box, but because we assume the energy spent in the muscles is directly proportional to the impulse applied (by the forces generated by those muscles, as measured where those limbs grasp something).

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 24, 2021, 06:55:06 pm
content: This longish thread has shown how much meat there is in even simple problems. And how important it is to frame the question properly. Things which seem obvious to those with long exposure can be either quite confusing or devilishly simple depending on presentation and/or point of view.Pertinent topics which have been touched on are the relation of weight and mass, the inability to distinguish (at non relativistic speeds) the difference between a fixed acceleration and being in a static gravitational field and the differences between work and power and the difference between power stored in a system through changes in it's gravitational potential and work elsewhere (muscles and the like).Just think of all the complexity a hard problem would bring to the table.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 24, 2021, 08:21:14 pm
content: Classical physics is easy and straightforward. Until you try to truly understand it, or to apply your understanding to solve a problem. It's only easy if you've already worked it all out before.Quantum physics is where you need to discard whatever vestiges of intuition you have left, because intuition and its odd evolved rules it operates by becomes a severe hindrance for proper understanding here.String theory is just mathematicians having a very long running joke on physicists.You can construct similar sequences for any topic you choose. If you disagree with my three choices above, just replace them with something more appropriate in your experience; the difference is just due to the differences in our personal experiences, and sequences that apply to me probably do not apply to you and vice versa.Learning is so darned subjective.

author: RJSV
author_rank: Super Contributor
timestamp: June 24, 2021, 08:38:58 pm
content: THANK YOU, DAVID HESS, for mentioning the British units 'SLUG', as mass, (with the corresponding British unit of force I.E. 'POUNDS').Now my concerns, today, are to merely point out how this thread is bit wild, maybe, for my taste. I mean, you have one person mixing up the two main standard systems, (English and Metric standards).You've got someone stating that speed is same a mass, I think he got mixed up, thinking inertia increase is same as mass increase, plus some other (yahoo) stating or mis-stating that relativistic effects are major in EVERY situation. So, yeah, mass increases with speed, but...arrghph not practically speaking.Like I've tried to teach, I think historically, that 'g' came into the picture, in f=mg as 'no suprise' rather than a 'direct' acceleration! The units match, and heck, it's another 'gravity related' deal.On that subject, probably Newton got to f=ma through some observations / measurements, but also by way of the universal force equation, that is mass x mass divided by separation distance (squared).So...anyone with a history context would help clarify these questions, as I am merely speculating, on Newton's exploration.But what I DON'T see is little 'tendrils' of acceleration IE 'g' shooting up past that stationary weight. 'g' is just in the force-related EQUATION, dude.Word salads. AND then there is the fellow that stated "words don't matter anyway, they are just there to describe stuff."Uh, ok, so if I owe you money, $100, let's just be loose, and call it '100 cents'.

author: JohnnyMalaria
author_rank: Super Contributor
timestamp: June 25, 2021, 03:04:28 pm
content: Quote from: David Hess on June 23, 2021, 04:40:36 pmThis is not true. It depends on which system of measurements you are using. e.g., the British Gravitational system does define pound as a force but the FPS system defines it as a mass. The UK legally defines a pound as 0.454 kilogram and, therefore, is defined as mass.I was educated in the UK in the 70s and 80s. Pounds were always units of mass and pounds-force (lbf) specifically used to indicate force.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 25, 2021, 03:34:58 pm
content: I use either SI units, Planck units, Hartree atomic units, or bananas.The last one covers all those slugs and pounds and pints and stones and wheelbarrows and such quite nicely, thank you.I even use decimal degrees instead of arcminutes and arcseconds, to the exasperation of some of my astronomist friends. (I also like to "accidentally" use the word "logy" instead of "nomy", just to see if they're still listening.)If you use inch ounce-force to measure torque, does the scale depend on whether it is a fluid (fl. oz) or not (notfl. oz)? And what does australia (Oz) have to do with this, anyway?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 25, 2021, 03:36:40 pm
content: What, you don't use bananas for angle too!? 0/10 disappointed.Tim

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 04:25:23 pm
content: To avoid ambiguity, use “oz av” for small forces and “fl oz” (a pronounceable acronym) for volume. 1.5 fl oz is roughly “two fingers”, except here in Chicago where we measure with vertical digits.

author: SiliconWizard
author_rank: Super Contributor
timestamp: June 25, 2021, 05:39:32 pm
content: Quote from: TimFox on June 23, 2021, 09:59:36 pmYes, it all comes from Newton's laws. Nothing special. It's just that the notion of weight is something that is actually not all that intuitive.Quote from: TimFox on June 23, 2021, 09:59:36 pmI'm not sure what an "accelerating force" is. There is force, and there is acceleration.And actually, in freefall, the weight has absolutely no effect. Acceleration is the gravity (if you simplify things of course and suppose there are no friction forces due to air - hence why true experiments must be done in vacuum), and it doesn't depend on mass whatsoever. In turn, the speed doesn't depend on mass either. That's something not really intuitive and that many people, even educated ones, have a hard time with.

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 05:57:09 pm
content: An accelerating force is that in Newton's First Law, usually stated now as "an object at rest will stay at rest, and an object in motion will stay in motion unless acted on by a net external force". In the original, "Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare". See "Newton's Principia for the Common Reader" by S Chandrasekhar, available athttps://oiipdf.com/newtons-principia-for-the-common-readerThe Second Law gives the result of such an external force, usually stated now as "F = m a," whereais the resulting acceleration from that force. Actually, Newton's original statement translates as "F = dP/dt", wherePis now called the "momentum" (original "quantity of motion"), and this version works better in Special Relativity. A discussion of this inhttps://blogs.bu.edu/ggarber/archive/bua-py-25/newtons-second-axiom/clarifies this:"Lex II: Mutationem motus proportionalem esse vi motrici impressae,et fieri secundum lineam rectam qua vis illa imprimitur.In translation (by Andrew Motte) this becomes:LAW II.The alteration of motion is ever proportional to the motive force impressed;and is made in the direction of the right line in which that force is impressed.Newton then goes on to clarify:If any force generates a motion, a double force will generate double themotion, a triple force triple the motion, whether that force be impressedaltogether and at once, or gradually and successively. And this motion(being always directed the same way with the generating force), if the bodymoved before, is added to or subducted from the former motion, accordingas they directly conspire with or are directly contrary to each other ; orobliquely joined, when they are oblique, so as to produce a new motioncompounded from the determination of both."I believe that Newton assumed that the inertial mass and the gravitational mass are equal, and this became a cornerstone of Einstein's theory.Once again, I suggest considering the Atwood Machine, where gravitational acceleration and inertial masses form an interesting demonstration.

author: RJSV
author_rank: Super Contributor
timestamp: June 25, 2021, 06:28:09 pm
content: Sorry TIM, I disagree with the use of the term 'MOTION': That's too loose, why not try use the direct terms, Changes in position= velocity (or called speed or rate) and if accelerating/decelerating then you have another 'rate', of change, that of velocity change.To drive a point home I would ask: "...So then, what are the units of this 'MOTION' thingy that you mentioned ?"Are you going to reply: "...Why.. it's feet per second..of course...".But that's called VELOCITY.!!! Don't misunderstand, this is a friendly, and voluntary activity,JUST BUSINESSand so I gotta speak out. The sarcasm is only all in fun. (Refer to Nominal animal previous post)I'm going to re-name VELOCITY: How about 'Wiggles'? ... Or BANNANAs per sector!Wordsaladwordsalad. (thanks for your patience)

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 06:37:59 pm
content: In terms of Newton's work: "quantity of motion" is what is now called "momentum". Terminology changes through the centuries. In non-relativistic mechanics,P=mv, where the vectorsPandvare, respectively, the momentum and the velocity, while the scalarmis the mass. InPrincipia, if I remember correctly, Newton defined the mass as the product of the volume and density, but modern usage defines the density as the mass divided by the volume. There are a lot of circular definitions in the history of science.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 25, 2021, 07:00:25 pm
content: Quote from: T3sl4co1l on June 25, 2021, 03:36:40 pmNext time it comes up, I swear I'll say"it's two bananas short of ninety degrees". Let the mayhem commence.

author: RJSV
author_rank: Super Contributor
timestamp: June 25, 2021, 07:32:27 pm
content: 'MAYHEM'?, 'BANANNAs'? I'm stealing all my jokes, from NOMINAL ANIMAL!O.K. : BANANNAs it is:(You) are pretty sure that 'Banannas per second' is a ridiculous term, but substituting 'loose' descriptors for actual precisely defined terms of PHYSICS is maybe OK. No, no,no:'MOTION' is just a general umbrella term that could mean either, (or both) SPEED or ACCELERATION.I think what got my attention, was a guy, back a ways, who posted that a parameter of ACCELERATION, well then that had to be a 'velocity' simply because he spotted the equation term: "PER SECOND". Must be a speed, right? ... No one will ever know!Heck, even the term 'Specific Meaning' has no weight here.SOOOO, For this reasons, I voteNOMINAL ANIMAL for PRESIDENT

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 08:09:01 pm
content: Try reading what I actually wrote. I did quote Newton’s original statements for historical purposes, but I carefully used the terms acceleration, momentum, and velocity when writing in my own voice. Modern versions of Newton’s theories are easier to understand and, where they are relevant (non-quantum and non-relativistic) are internally consistent and an accurate description of motion with its various attributes. For further discussion, please see any textbook on freshman physics.

author: RJSV
author_rank: Super Contributor
timestamp: June 25, 2021, 09:03:32 pm
content: Yes, Tim, I was a bit out of line by posting in generalities (like some people seem to 'smoosh' together a bunch of tech jargon, without accurate definitions.) But your latest posts, I like, just maybe a bit of time lag, as I comment addressed back a bit ( posts from couple weeks ago.)But, still, not you but one recent post was directly confusing MOMENTUM (increasing) and saying that was 'WEIGHT' increasing (that being, again, momentum as mass X speed.) That's a pretty simplistic and not only wrong, but mis-leading (due to the fact that 'mass' is in the mix.That (other) person's posts appear to me like an ignorant summary, saying 'No one will ever know' IE. the real use of equations and concrete/specific definitions.Heck, I still see that (helpless) expressed, in the discussion of 'fluid ounces' and how that somehow complicates things, impossibly. Just: "no one will ever know". Yeah and 'Einstein'?!!! Gotcha there.I think I will include 'Einstein'+ in everything I post.Gives it an air of legitimacy.Einstein ! Ha ha, gotcha again.AND Thanks, I mean no insults here.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 25, 2021, 09:50:32 pm
content: The disparagement of traditional systems leaves out poundals and a few other variants. These mass and force quantities are fine if kept straight, but the one real advantage of the metric system for anyone of average or above intelligence is that the metric community has done a better (not perfect) job of keeping them straight. Dimensional analysis is always useful. It was nearly mandatory in the traditional units world and still when referencing older texts. Even metric folk can miss out when using an older text that nonchalantly uses physical units expressed in cgs or slightly obscure units like wavenumber.

author: David Hess
author_rank: Super Contributor
timestamp: June 25, 2021, 10:23:49 pm
content: Quote from: JohnnyMalaria on June 25, 2021, 03:04:28 pmMy point is that it is very easy to get into trouble with the unit pound because it can mean force or mass depending on the system, and literature will not always be explicit.

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 10:43:41 pm
content: I agree about the confusion in English-language contexts between pound as unit of force or weight, and as unit of mass. The usual statement on a US food package label is “net wt 1 lb 2 oz” (on a Cheerios box), which in strict physics language would be “mass that weighs 1 lb 2 oz at mean sea level”. In metric lands, however, I have seen pressure gauges calibrated in “kgf/cm2” instead of N/m2or Pa or dyn/cm2.The infamous NASA mission failure a while back involved changing vendors on small rockets and misreading the force units.By the way, my mnemonic for conversion between N and lb is to remember that 1 kg weighs 9.8 N and 2.2 lb (approximately).

author: RJSV
author_rank: Super Contributor
timestamp: June 25, 2021, 11:17:44 pm
content: Ok, Tim I think you've supplied some KEY info, to clarify portions of this thread.So, if I read right, a definite precise definition and use of 'MOTION' was given by Newton. Fantastic, even tho proves me wrong, somewhat.So then, as MOTION was given the precise definition, as what we today might call 'momentum'.Then, will the units be kg-meters/ seconds (squared) ?Plus, maybe the English system direct equiv would beSlug-feet / second (squared) ?Although I realize, maybe the English system avoids that form, and using 'pounds' as an equation substitute is ok as long as a standard is consistent, and even though the language structure isn't strictly correct??? Did I get that right ?As to relativistic effects, after 2 years college courses on CLASSIC Physics, I enrolled in a nuclear physics intro where the instructor described the so- called 'NEUTRINO PARADOX'.In this effect, Neutrinos streaming out from the sun would last too long: They were measuring too many reaching sea level, before the very short decay time kicked in. (They decay to mu-mesons).The answer came in time dialation where time ticks slower, in the neutrino moving at >90 % light speed.But, again, I'm saying some problems in this thread are due to injecting relativistic effects too early, into a sort-of 'word salad' including getting mixed up because MASS does increase (or equiv perceived mass) at relativistic speeds (90% light speed). So, increase in momentum, and now I realize no worries about calling it ' motion of a body', that increase gets all smushed into word salad that reverts to including relativistic mass increase (due to high speed).WHEW, Thanks Tim, for clarifying your contribution!But still,. Nominal Animal for President

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 11:20:02 pm
content: A couple of side comments about gravitational accelerationgand weight.1. Scales. It was mentioned above that "scales" measure force or weight, not mass. This is certainly true for spring scales (or modern units with strain gauges) that measure force as an extension of a compliant spring. However, "balances" that compare the device under test to a standard mass/weight, using leverage as appropriate in a steelyard balance, compare the masses of the DUT and the standard, since the two will balance regardless of the local value ofg, but the weight of the standard will change when you take the experiment to the Moon.2. Measurement ofg. It is not necessary to know the mass of the object in an experiment to measure the uniform gravitational acceleration in a limited region. Galileo worked out the period of a simple pendulum around 1602, where the period is independent of the bob mass and the amplitude of the swing (so long as the amplitude is small enough to use the approximation sin(x) =x, the error is small), but depends on the length of the string. The periodT = 2pi x sqrt (L/g)in that limit of small amplitude. A more precise experimental method is the Kater Pendulum, where the length in the experiment can be better defined. Seehttps://en.wikipedia.org/wiki/Kater%27s_pendulum

author: TimFox
author_rank: Super Contributor
timestamp: June 25, 2021, 11:23:53 pm
content: Quote from: RJHayward on June 25, 2021, 11:17:44 pmInPrincipia, Newton defined his terms before going into his laws. In modern nomenclature, momentum (vector) is the product of mass (scalar) and velocity (vector),P=mv, and has units of kg m/s, not s2.Force is the product of mass and acceleration, and therefore has the units kg m/s2. Kinetic energyE = (1/2) m v2, where all terms are scalars, has the units of kg m2/s2.When you get relativistic, look up the meaning of "rest mass", which is invariant.

author: bostonman
author_rank: Super Contributor
timestamp: June 27, 2021, 02:38:27 pm
content: QuoteAs someone pointed out, this thread has some confusion over people messing terms and units. Hopefully I'm not one of those people as I'm just trying to grasp (what I feel are) basic concepts that confuse me.I try avoiding the human factor in all this because of the complexity, however, gym topics do help because it's where (at least) I do most "lifting".QuoteI think the "longish thread" is due to me asking the same question, but doing different activities that involve a different part of physics each time. As an example, I started off simply asking why ma=mg based off TBBT. I researched it before asking the question and kept seeing discussions that they are equal because of Earth. This didn't make sense to me because of what I've already stated and now I believe I understand why they are "equal". Then I deviated asking about lifting a box off the floor, and that got moved towards "Work".As for speed of lifting something, this actually also came from TBBT. Sheldon and the group were moving a time machine up the stairs and (I believe) Howard said why don't we push faster. Sheldon replied stating Work done doesn't matter how fast you push the object.That got me thinking because if I pick up a box 1m really fast, it feels like I've done far more work than if I lifted it slowly. Now if I lifted the same box 1m really fast ten times, I'd be out of breath whereas ten times slowly and I'd have far more energy at the end.From what I understand due to this thread, one thing to consider is that I've gave the box kinetic energy picking it up faster because now it needs to transfer that kinetic energy when it stops at 1m than it would be if I picked it up slower. If I understand correctly, it would attribute to me being "out of breath" because I've transferred my kinetic energy to the box and then used energy to stop it at 1m.It does seem acceleration should factor into some equation because if Ke = 1/2mv^2, then without acceleration, how did the object achieve velocity?As for doing pullups, I think when discussing the "human factor", too many factors come into play as for getting out of breath and fatigue muscles. If I'm correct, say I use a wench and a motor to lift a box (i.e. no human interaction at all). Excluding friction on the pulleys, wire resistance, etc.. I imagine the Work done by the motor to lift the box would be equal to the Work done by a human to lift the same box the same height.In other words, the calculation is identical, and I can safely say I burned X calories lifting the box ten times based off converting Joules (i.e. Work) into calories.Getting back to lifting a box very fast ten times, at a distance of 1m, it's the same Work, however, now I've wasted energy because the box needs to stop at 1m thus the energy being thrown out the window; therefore I'd get tired sooner. On the other hand, if I lifted the box very slow, my arms would begin getting tired.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 27, 2021, 05:50:05 pm
content: Quote from: RJHayward on June 25, 2021, 07:32:27 pmI was advised by the reptilians to tell you I politely decline, or else.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 27, 2021, 06:26:14 pm
content: Quote from: bostonman on June 27, 2021, 02:38:27 pmSheldon is correct that the time they take does not affect how much energy they need to transfer to the object to push it upwards in Earth's gravity field.Sheldon is wrong in that the chemical energy required by human bodies to perform such energy transfers, definitely does depend on the rate at which the humans do it.Idon't know what that rate is at all, but provided an example model earlier. It can be derived from basic principles, how muscle cells contract when given chemical energy, and so on.So, it is correct – and the core of the joke is – if that one uses "work" strictly according to the physics definition, and completely ignores the human aspect of it.Which is kinda at the root of all engineer/physicist jokes, I guess.(I once had to walk out on a Physics 101 lecture, when the professor went "and as you can see, physics isn't as dry as one might think it to be, since 'bar' can be considered both an unit of pressure, as well as the place to relax after a hard days work", and the students dutifully emitted a respectful "laugh". I love me dad jokes, and the layered nature of that joke (think about it), and loved the prof, but just couldn't handle the herd mentality. I have hard time with canned laugh tracks, too.)Quote from: bostonman on June 27, 2021, 02:38:27 pmI'm the opposite: I have ample power, but not much stamina. (I think those are the terms, not sure.)Anyway, it is a perfect example of what it means to beefficientat something.If the object is at rest both in the beginning and in the end, it really does not care how long the lifting takes. Its own total energy is changed by the exact same amount anyway (the physical work, i.e.m g h, given massm, acceleration due to gravityg, and the vertical elevationh).From the human or device causing the energy to be transferred to the object, "work" gives the minimum. The less efficient we are at it, the more energy we'll waste.Not using tools to optimize the energy expenditure is like asking friends to help move the object, but tell them to keep one hand behind their back. Inefficient.Giving the object more kinetic energy than necessary, or converting it repeatedly into waste heat (say, you lift the object one stair step at a time, lifting the object higher than the next step, then dropping it down), are just ways of doing it inefficiently.Quote from: bostonman on June 27, 2021, 02:38:27 pmYes. It is easier, though, to think of Work from the perspective of the box: when being lifted upwards, measuring from standstill to standstill so ignoring kinetic energy transfers, the height determines exactly how much total energy the box gains.From the perspective of humans doing the sweaty work, we can either be smart and make sure we spend the minimal energy – and we can do that without any external supplies of energy by using mechanical devices like pulleys –, or we can power through, ignoring the "waste" in chemical energy expenditure.It is unfortunate that we don't have a separate word to describe the energy difference, since it is confusing. (Then again, you wouldn't believe some of the physics misconceptions I've observed in graphic artists who have been taught with "interesting" definitions of "gravity", "volume", "depth", et cetera; perfectly valid and very useful in their own domain (in visual/graphic arts), but not at all valid in others, like physics. Same problem, resulting in utterly hilarious misconceptions.)Quote from: bostonman on June 27, 2021, 02:38:27 pmYes; exactly: because you transferred the energy ten times to the box, did not try to recover any of that energy, you can be sure that by doing it, your body burnedat leastthe amount of chemical energy corresponding to the Work while doing it.

author: RJSV
author_rank: Super Contributor
timestamp: June 27, 2021, 08:43:47 pm
content: ====. Since ACCELERATION term is in an equation, therefore there must be movement present... ====Erroneous because:. Newton started by saying "Let's examine a body in motion ( if I understood the help from Timfox).This thread, in 1740, could still encompass all the non-relativistic aspects of our discussion / friendly argument.Perhaps one my faults is to focus too much on some blatant mis-statements:. But I needed clear examples to quote / make my point.Here is my main concern:There are equations involving an acceleration constant, where movement changes are described.BUT, there are other equations where (an acceleration constant) is simply an expression of gravity-related physics, that is the strength of gravity (planet bulk) affects the outcome of a calculation. You cannot expect an explicit 'change in movement' every time.Perhaps it was stupid of me to expect every post to be perfect, (and putting focus on the occasional mis-statements). The example that bugged me was the post stating that 'MV' increased, therefore MASS increases with velocity increase. Uh, no, it is velocity that got bigger, to cause the parameter 'MV' to increase.AND injecting relativistic effects into this discussion is just extra confusion, especially when the terms are similar, and often identical. (That being the use of the term 'mass increase'.So let me restate that:I read that 'MV' increased with speed, therefore the 'mass increased'.So sloppy, obviously SLOPPY MATH.The I read that at relativistic speeds the apparent mass DOES increase, probably having practical implications.BUT, and this is the even more crucial part, one reader takes that and runs back to (my) first complaint, that it's 'V' that caused 'MV' to increase, in a classical physics introspective. (He) takes the relativistic mass increase (valid), and runs back to the classical mode equation, implying that from a practical, everyday (non-relativistic) frame of view, the mass in 'MV' is wildly increased.... O.K. I added 'wildly' etc. to exaggerate my point.=========================================I recall someone I knew years back who used the term 'Lowest common denominator':. A phrase involving personal interactions among military folks.What he meant (I think) is that there is often a choice:BUT: Do you force highest standards on everyone, or do you 'cave' to the ignorant, misbehaved (and maybe they are incapable of complying anyway.) ?(Thanks Phil M.)===========================================Nom Animal:As for presidential bids, I've seen some great minds come out of FINLAND, there. You maybe too smart for that job.Keep up the good work (helpfully debate).P.S. I've seen other venues / political meetings where speakers actually had to PAY money, for their time at the microphone. Dave J. perhaps you could 'monitize' a longer thread, charging by the word...(Just kidding)

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 27, 2021, 11:47:04 pm
content: Quote from: RJHayward on June 27, 2021, 08:43:47 pmIt is complicated, because whenever we talk about "bending spacetime", we must talk in relativistic terms; but at the same time, Newtonian physics and relativistic physics do and must agree at the simpler limit (zero velocity, uniform gravitational field, and so on), because system behaviour at simpler limit is measurable and physically verifiable.No matter how fine the theory, it is only useful if it predicts behaviour physically observed. That's why many think string theory is "just" math and not physics: it cannot really be used to model something physically measurable, so that its predictions could be compared against physical measurements. That makes it "not physics" in the sense that physics describes how the universe behaves; maths folks (including string theorists) can work out the details and the tools, and philosophy folks can deal with the "why" part.For example:Mass does not bend space-time. Energy does.In relativistic physics, "mass" is rarely used. Instead a quantity called "invariant mass" or "rest mass" is used; this is the mass of the object at rest, as if observed in a frame where the object is not moving. The invariant mass is invariant, because it is the same quantity no matter what observational frame it is used in."Relativistic mass" is the term that best corresponds to what we call "mass" in non-relativistic, or Newtonian, physics. It includes the space-time bending effects.Because special relativity says that energy ≡ mass, (mass-energy equivalence,E=mc²), the above can sound like quibbling: I should not say mass does not bend space-time and energy does, because the two are equivalent.But the above is not quibbling: it is worded so because 'mass' as a term is ambiguous. In making that statement,I was thinking of the wrong kind of mass, you see. See?Simply put, we should avoid claiming mass bends space-time, because term 'mass' is ambiguous; we don't know whether it is used to refer to relativistic mass (which would be correct and match the classical physics definition of mass, but odd because physicists use 'mass' to refer to invariant or rest mass instead), or to invariant or rest mass (which usually 'mass' alone refers to when used in physics). One interpretation is right but odd, the other is wrong but the more common use for the term.A case in point: photons.All electromagnetic radiation from gamma rays to infrared light consists of photons. Photons are massless bosons: theirinvariant massor rest mass is exactly zero (none of that "so tiny we can think of it as zero" stuff; plain and clear zero here), and any number of them can occupy the same state and space (until the energy density is high enough to cause certain interesting stuff to happen). Bosons are the ones who like company, and fermions –– for example electrons being fermions –– are the ones who occupy their own quantum state, and exclude others from it.In special relativity, the energy \$E\$ of an elementary particle is defined as$$E = \sqrt{m_0^2 c^4 + p^2 c^2}$$where \$m_0\$ is the invariant or rest mass, \$c\$ is the speed of light in vacuum, and \$p\$ is the linear momentum of the particle. For particles with nonzero invariant mass, the linear momentum \$p\$ is$$p = \gamma m_0 v = \frac{m_0 v c}{\sqrt{v^2 - c^2}}$$where \$v\$ is the velocity of the particle, and \$\gamma\$ is the velocity-dependent Lorenz factor; I expanded that to get the nice simple expression on the right side, but note that usually it is kept contracted to \$\gamma\$ for brevity, so the right side expression may look unfamiliar to many.Note that for small enough velocities, \$p = m_0 v\$. The correction factor, Lorenz factor \$\gamma = 1 / \sqrt{1 - v^2 / c^2}\$, tells you the error you have for any given velocity if you use Newtonian physics instead of special relativity; see how only the \$p\$ term in total energy has anything to do with particle velocity?The speed of sound in air is about 330 m/s, so about 0.0000011 of the speed of light. At that speed, \$\gamma = 1.000000000000605\$, or less than one part in 1,000,000,000,000 over one. Double-precision floating point numbers – those used for the vast majority of numerical computation on this planet outside financial stuff which uses decimal formats – have barely enough precision to describe that!Also see how momentum is related to total energy (for elementary particles). For collections of particles, we'd need to add their internal energy (stored in their structure, their interactions; including angular momentum). In classical physics, we say kinetic energy \$K = m v^2 / 2\$, but it is important to understand how well that matches the relativistic energy. If we use \$E(p) = \sqrt{p^2 c^2 + m_0^2 c^4}\$, then \$K = E(p) - E(0)\$. Expanding the expression for \$K\$ using a Taylor series for small \$v\$ (around \$v = 0\$), we get \$K = m v^2 / 2 + 3 m v^4 / (8 c^2) + \dots\$. Meaning, even the kinetic energy agrees *exactly* at zero velocity, and for larger velocities, just has relatively small additional terms. Classical and relativistic physics agree at small velocities even here, as expected.Photons have zero invariant or rest mass, and for them, linear momentum is$$p = \frac{h}{\lambda}$$where \$\lambda\$ is the wavelength of that photon. Each photon has a single wavelength that only changes when it interacts with other stuff, transferring energy. It does have other properties like polarization, but those do not affect or contribute to the total energy. Thus, their total energy is$$E = \frac{h c}{\lambda}$$A mind-bending detail is that because the particle velocity depends on the observational frame used for the measurement, so does the total energy of the particle.To un-bend ones mind, think of the Doppler effect: the wavelength of the light we observe does depend on our own velocity with respect to the light. We still see it arriving at the speed of light in vacuum, but its energy is shifted. Because photons, having no rest mass, zip everywhere at the speed of light. (The velocity only drops when they interact with other stuff; exactly how that happens is quantum mechanics. That is why we have a constant for it in vacuum, but it drops when zipping through matter.)In sufficiently small regions, acceleration is indistinguishable from gravity. Because of this, and the fact that Earths gravity well bends space around it, if we were to observe a single photon at a single moment of time (as measured by the position of that photon zipping along at the speed of light) from the surface of the Earth, and from an orbital space station, they would see the photon having different wavelengths. Those two are different frames of reference, so there is nothing wrong in their observations differing (by exactly the relative difference of those two reference frames). That is kinda-sorta one of the basic ideas of relativity.Also, we have already experimentally proven that even photons themselves do bend space-time. That tells usinvariant massor rest mass does not bend spacetime, because photons have none (again, not just "close enough to zero for all intents and purposes", but a clear and pure plain real zero), but we have observed them bending space-time. And relativistic mass is even in principle indistinguishable from the energy, because of Einstein's mass-energy equivalence. Thus, to avoid misconceptions, it is – I claim! – a good idea to think of *energy* as bending space-time, and relativistic mass being equivalent to energy per Einstein. As a bonus, we don't have any issues with photons having zero invariant mass, because they are always zipping along at the speed of light (well, duh), and thus always have energy and thus relativistic mass; the zero invariant mass is just not practically relevant other than when working out the details using math. And as a cherry on top, we easily slide into the habit of assuming 'mass' means 'invariant mass' (or rest mass), and instead of talking about 'relativistic mass' can talk about energy; energy being equivalent to mass, but as a term, not at risk of being confused with something else.Compare to the thought pretzels one has to twist oneself into, if one insists mass is what bends space-time. What 'mass', anyway? No, I think that although you can argue that, that argument is not useful; you need to qualify it with 'invariant' or something else to be precise and not be accidentally misunderstood.It's the same as when graphics artists talk about 'volume' or 'gravity' as an expression of an emotion evokable by visuals. If everyone agrees on their definition, and nobody accidentally tries to infer anything about related stuff outside that very small domain, there is no problem. So there is notechnicalproblem in using those terms. But, hilarity ensues when a graphics artist, genuinely puzzled, asks a physicist why they don't just use cut pastel tones to reduce gravity.It is also okay to now realize that the Higgs boson so much talked about a few years ago, has really nothing to do with the bending of space-time, and everything to do with invariant or rest mass. (Specifically, how "gauge bosons", those that carry the four fundamental forces we know of, have an invariant or rest mass of around 80 GeV/c², while it would be much easier to describe them if they had zero invariant or rest mass.)See? As in so many other things, if you just understand the terms correctly in their proper context, things just start falling into place.They matter much more than just whether or not they are technically correct: they are crucial in building correct understanding in the first place, much beyond technical correctness or minute detail.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 28, 2021, 12:46:13 am
content: As a visceral example: shine enough light into a mirror box, and it becomes heavier.Hmm, I forget if this can be understood in more classical terms -- would have to be something about Doppler effects on the radiation pressure, thus manifesting a difference as apparent mass? -- but all that General Relativity requires is that some energy is contained within an object (including rest-mass energy), and that's its effect on spacetime.The idea of a kugelblitz, is to shine enough light, in from all sides, to a single common point, transmitting enough simultaneous energy to create a black hole. The equations don't care what form of energy enters, only that enough does to create a singularity in spacetime. What's inside that singularity? Mass, energy? Does it transform? Moot question -- there's nothing at all at the singularity, it's a, well, singular point.(I mean, as far as we know -- there has to be something quantum mechanical about that. But it alsocan't matterbecause it's beyond the event horizon.)Tim

author: RJSV
author_rank: Super Contributor
timestamp: June 28, 2021, 02:08:11 am
content: But STILL:I'm saying to the OP 'Boston' question thatmg in f=mgis NOT movement, the 'g' is called 'acceleration constant' for our case on Earth sized object.That's a static force equation, a FIELD having potentialto accelerate IF that's a free body to be moved, a la 'Newton'.The other way; ' f=ma', is for the case when you are pushing something: Your force, again, on a body free to move can cause an accelerated movement. And I believe the physics folks had stated, that the nature of the motion, such as including E= 1/2 mv(squared), cannot be distinguished, as either case has same equations and outcomes.That's why I would tend to suggest that 'g' be called an acceleration constant, for specifying a static situation, like weighing your gold bar.AND, 'a' is a more general term, signifying how (your) hand pushing a free body out in space in any single isolated case, is going to, YES, actually start moving that sucker, accordingly. You can change that by pushing harder or less hard. And modulate the resulting acceleration alongside.? See any mention, of photons, above?? Rest Mass ?? Year 2021 ?No, I'm presenting 'Year 1740' arguments, and that approach allows for all the advanced physics that we have today, as well.

author: bostonman
author_rank: Super Contributor
timestamp: June 28, 2021, 02:20:18 am
content: QuoteThis is what I initially questioned. I hate to return to the same question, and, in a way, I'm asking rhetorically.ma can't equal mg all the time. If F=ma, and, to compliment your hand pushing a free body example, the a can be anything. Only in the case of objects (or free bodies) where gravity is acting on it can ma = mg.Little 'g' should just be called a constant, however, it can't be a constant since I can be standing on a mountain and gravity is less (by a small amount) than someone else standing at sea level.From what I understood before initiating this thread is that them being equal has little meaning and isn't anything to "brag" about because we live on Earth. Anywhere in the universe ma can equal mg.Now as for the in depth, informative, and long replies, what I find is that very smart people dislike providing simple answers because much more exists. If I'm pushing a box, I may just consider the pushing factor, whereas, friction should also be included.Obviously my physics knowledge is low, so I look for the general answers based on simple problems. If I jump right into free bodies, friction, the actual gravity at my height, etc... that will just confuse things causing less focus on the actual question.I tend to provide long, in depth answers too when my friends ask me simple questions. It's never easy for me to say the sky is blue without explaining that it's a shade of blue because.... blah blah

author: RJSV
author_rank: Super Contributor
timestamp: June 28, 2021, 02:34:26 am
content: Original Post person: BostonMan:That last looked good, but I want to ask, was the original 'g' a variable, such as in saying "The 'g' in the Mars example will be smaller.OR, is 'g' just a solid constant, for Earth sized cases ?Gives even more context.Going forward from here, I got no problem, accumulating ever more science 'depth', such as the photon related stuff in posts above. Like the guy said, any innovation in theories must take account what we already know from classical-only.

author: bostonman
author_rank: Super Contributor
timestamp: June 28, 2021, 03:01:14 am
content: QuoteI assume you meant 'that last post looked good'.Little 'g' being a constant in the sense that it's understood we are talking about gravity on Earth. Now 9.8m/s^2 is a general rule for physics 101, but, if I dug a five-mile deep hole, gravity would be less because it becomes negligible () at the center of Earth. Also, if I stood on a five-mile high mountain, it would be less because I'm further from the surface of Earth.This means that to argue gravity on Earth is 9.8m/s^2 would be wrong because it can change depending on your height on Earth. Unfortunately, little 'g' can also mean gravity on Mars (or anywhere in the universe) because if I want to know my Weight on mars, I need to use W=mg. To call little 'g' a constant would be wrong, but, I hate to critique TBBT, but I think saying ma = mg is taken out of context.It's like saying your age is equal to my age, but if your birthday is in March and mine is December, we are the same age for a short period of time. So ma = mg only in cases involving the discussion of Weight on Earth and a free falling object on Earth. Otherwise, a free body may have different acceleration, or, maybe the free body takes place on Mars and now ma = mg thus throwing out 9.8 as little 'g'.From what I learned about Mass, it's not always the same throughout the universe, however, physics 101 tells me it is. From what I learned from this thread is: mass can change due to small tiny factors; none of which I was aware of.If anything, I knew gravity is different at different heights and 9.8 is the general number to use for general calculations, but I didn't know mass changed, but, in any case, the fact this was mentioned reminds me to watch how I speak (and type).My pet peeve is people who argue with statements like 'it's a fact', or 'best ever'. Nothing is a fact unless it can be proven and best ever is just silly.It's a fact I'm sitting at the keyboard typing, it's not a fact that anyone reading this can argue it's a fact I typed this. I could have had someone type it for me, used speech to text, etc...Anyway, my point is that whether this thread went much, much deeper, or I (and others) were completely wrong to state things in such a fashion, just reminded me to speak correctly, and, really, it helped me understand how much more exists to Weight, Mass, etc... Now at least I can say 9.8 is good, but know if I were to become a rocket scientists, more exists than just 9.8.

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2021, 03:26:00 am
content: gis a real acceleration. Since the gravitational force on a dropped mass is proportional to that mass, the acceleration of a falling body will be independent of its mass.Simple algebra: the acceleration of massm1isa=F/m1= (1/m1) xG m1m2/R2=G m2/R2, which we define asg.Here,Gis Newton's universal gravitational constant (notg),m1is the mass of the dropped object,m2is the other mass (Earth or moon or whatever), andRis the distance between the two objects, specifically between their centers of gravity. In our case, that is the radius of the Earth or moon.In a given region where the gravity is uniform, such as a cubic meter at the earth's surface, any object whatsoever, of whatever mass (neglecting air resistance) when dropped freely will be accelerated by an acceleration equal tog.In another region, such as a cubic meter at the moon's surface, the value ofgis different than the value on earth, due to the different mass and radius of the large attractor (moon vs. earth).I gave the equation for the period of a pendulum, which is independent of the mass, and hasgexplicitly in the formula: the period of a given pendulum will therefore be different on the moon or on earth.There are further details, such as the fact that during our drop experiment the motion of the Earth can be neglected, since it is much heavier.

author: RJSV
author_rank: Super Contributor
timestamp: June 28, 2021, 05:28:50 am
content: Yup..., BostonMan;I think we are in pretty good shape, today. I had wondered if 'g' expressed more explicitly would maybe be; g(subscript Mars) for Martian surface, g(subscript Moon) for moon surface measurements, etc. Thanks for clarification.This is all voluntary, remember the eevblog hosts might say "Participate, if you please, but please keep it positive."And timfox providing some 'sane' relief.And as to veryveryveryvery small extra elements to any parameters under discussion: They are just that, and mostly 'negligable' in everyday live. Now, with modern growth, we know about, and we somewhat know which effects take effect strongly, and when such effects (while still there) can be ignored.That guidance would maybe fill out the scope, of OP's general question.Does ma = mg ? Yes, under some certain circumstances, that being adjusting your applied force, and then additional math, if relativistic conditions dictate, (strong, or very weak).It's a volunteer task, blogging, and so makes any frustrations a whole order of magnitude more tolerable!!

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 28, 2021, 08:01:44 am
content: Quote from: RJHayward on June 28, 2021, 02:08:11 amSure; I called it acceleration due to gravity, whichI intendedto be understood in exact same sense as 'acceleration constant'.Perhaps this helps:What is the speed of light, if you are in a dark chamber without a single photon? Does the speed of light still exist? Yes, of course.Does the speed of light in vacuum,c, exist outside photons? Yes, of course. We nowdefinecas exactly 299,792,458 m/s, using that to derive the exact length of one meter. (One second is defined as being equal to the time duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the fundamental unperturbed ground-state of the caesium-133 atom.)Can acceleration exist with zero velocity? Yes, of course. Jump up into the air. At the peak of your trajectory, your velocity is exactly zero, but have about 9.80m/s² of downward acceleration (and that stays pretty much constant from the moment your legs are in fully extended position and no longer touch the ground, until they touch the ground again; as air resistance affects the acceleration a bit).In calculating forces using formulae likeF=m a, one must remember that this does not necessarily involve any movement, because only the sum total of such forces acting on an object are really physically measurable. Indeed, the box sitting still on the floor is subject to forceF1=m gdue to gravity, and to forceF2due to the surface it is sitting on resisting deformation; and from the box being still, we know these forces must cancel out, and are therefore equal but opposite.If we pick a subset of forces, we can sometimes get very unrealistic ideas about the situation. For example, one could estimate the gravitational force exerted on the box by the northern hemisphere of the universe, and then claim that the result somehow must be balanced by the static force due to the ground not deforming under the box. Uh, no, if you pick a silly set of forces, the model itself is then silly, and does not describe reality.Another way to define force is via an energy potential and the vector gradient operator:$$\vec{F} = -\nabla V$$The scalar field \$V\$ is not any kind of energy emanation or all-pervading field that binds us together; it is simply a measure of energy. On isosurfaces where \$V\$ is constant, you do not need to transfer any energy. (If we have a frictionless surface here on Earth, in a vacuum chamber so there is no air resistance, an object on that surface would keep their velocity, until they bump into a wall or something.)For gravitational potential energy between two point-like masses, it is \$V = -G M m / r\$, where \$G\$ is the gravitational constant, \$M\$ and \$m\$ are their masses, and \$r\$ is their distance. Along the line between those point-like masses, gradient is just partial derivative with respect to \$r\$, and \$F = \partial (G M m / r) / \partial r = - G M m / r^2\$.Now, if we combine a few of those constants, mainly via \$g = G M / r^2\$, we end up with \$F = m g\$.What is this magic constant \$g\$? Why, nothing but the gravitational constant [m³ kg⁻¹ s⁻2] times mass  divided by the squared mean radius of Earth [ m⁻² ]. Dimensional analysis shows that the resulting units are [ m s⁻2 ] = m/s². Acceleration. What about the numerical value? 6.674×10⁻¹¹×5.972×10²⁴/(6.371×106)² ≃ 9.82.Ah-ha! If we treat gravity as just one form of potential energy, described as \$V = -G M m / r\$ between point-like particles having masses \$M\$ and \$m\$ separated by distance \$r\$, we end up with each of them seeing a force \$F = m g\$ towards their shared center of mass.So, in this case, \$g\$ has nothing to do with acceleration per se; it is just some constant we derived from Earth's mass and radius, and the gravitational constant \$G = 6.674\cdot10^{-11} \, m^3 \, kg^{-1} \, s^{-2}\$.The reason it turns out to 9.82 m/s² and not 9.80 m/s², is that because of inertia, and Earth spinning around on its axis, objects near the equator have an inertial force (or fictitious force, as it is caused by inertia and not by any potential energy) away from the axis of rotation of about 0.03 m/s² or so. This force (again, which is due to inertia, so not caused by any potential energy) is \$F = m \omega^2 r\$, where \$m\$ is the mass of the object rotating, \$r\$ is the radius of rotation, and \$\omega\$ is the angular velocity. (You can convert \$x\$ RPM to radians per second using \$\omega = 2 \pi x [\text{rad}/\text{s}] / 3600 [\text{RPM}]. \$\omega = 2\pi [\text{rad}/\text{s}]\$ is one turn per second.)Because the gravity field of the Earth is not consistent with sea level, it is just much simpler to fuzz it all and include that inertial effect in the mess for simplicity (because the other stuff like gravitational "anomalies" – just a word meaning the effects due to Earth being lumpy and fluid inside, not smooth and evenly distributed – mean this approximation is already off by so much that including the inertial effects won't make it any worse).But, as shown above, just the fact that we have a term that both has the units of acceleration, and in some cases can be physically measured (by observing a weight falling in vacuum), does not mean that actualmotionhas to be involved. Just like speed of light in vacuum is a constant, that has uses related to photons and stuff, but is not dependent on there being something zipping along at that velocity. Or any velocity: remember \$E = m c^2\$.(I cannot help myself:"But what about electrons orbiting atomic nuclei? Don't they experience centripetal forces?"No, because electrons are delocalized in that fuzzy quantum mechanical way, and not sharp points in orbits around the nuclei like you often see pictured. The shape, or probability density function, varies depending on theorbital. Each electron thus delocalized does have a property we call angular momentum, which is deeply involved in how those orbitals get their shapes; and although this angular momentum is split into different parts (spin angular momentum which kinda works out as if the particle was spinning on its axis, but in key aspects isn't; and orbital angular momentum which kinda works as if the delocalized electron had the properties of rotating around the nucleus but doesn't) some confuse the real-world-sounding terms (which even sometimes behave like the properties with those same names in classical, non-quantum mechanics), and end up being the graphics artists I mentioned, asking the physicist why they don't just use pastel tones to reduce gravity.)It pains me to see intelligent people struggle to comprehend ideas, only because the terms we use are so damn misleading/unintuitive/confusing.Make no mistake, they are that to me too. I have a hard time associating any specific "law" or property or constant with just the name of the person who discovered it, because although of historical interest, those names do not help at all in grasping the concepts they represent; you just have to learn them by rote. And I don't. At least "invariant mass"/"rest mass" versus "relativistic mass" is descriptive; if they were called "Lorenzian mass" and "Einsteinian mass", I think I'd have to quit physics altogether.

author: RJSV
author_rank: Super Contributor
timestamp: June 28, 2021, 08:30:15 am
content: That last post a good one. A Nominal animal PHYSICIST beats an EE Everytime.Especially the parts explaining about moving or non-moving, in a world having forces and acceleration; framing for the reader the distinctions.AND the side-bits on electrons operating in non-intuitive ways, I mean, I had to turn off TV set so I could focus on reading clearly. Nice, and way over my head, but a learning experience.

author: bostonman
author_rank: Super Contributor
timestamp: June 28, 2021, 01:15:21 pm
content: Without getting too deep with math, or an explanation, how do we know an object has X kg of mass, and gravity (rounding off) is 9.8m/s^2 based on the beginning of time?We know a 100kg mass is 100kg, but that's based off 9.8m/s^2, but gravity is derived from a Force, and the Force is derived from a known mass and gravity.If we went back hundreds of years (or a twenty-six-hundred year journey as with TBBT - and I'm not trying to be funny), scientists needed a known gravity to calculate mass; or a known mass to calculate gravity.I'm sure the math has been worked out several times over hundreds of years, but it seems because each factor is derived from the other, any one can be wrong throwing off the other.Same with calculating the mass of planets. I believe it's done by looking at how gravity pulls on surrounding stars (?) and thus mass is calculated, however, we needed to start off with the mass of one planet which would be Earth (I should say I assume we started with Earth). If we got the mass of Earth wrong based off using a mass we thought was X kg, then everything derived since is wrong.

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2021, 01:51:10 pm
content: The mass of an object is completely independent of the local acceleration of gravityg.The weight of an object is directly proportional to the local acceleration of gravityg.It is an interesting philosophical question, which requires serious astrophysical investigation, if the universal gravitational constantGhas changed over time.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 28, 2021, 01:55:38 pm
content: Quote from: bostonman on June 28, 2021, 01:15:21 pmWhile you are sort of conceptually correct, there are literally hundreds of different observations that have to be reconciled and they all converge to the answer we use today. Including even direct measurements of the gravitational attraction between spheres in a laboratory. Some truly elegant experimental work to sort out all of the forces and get the precision required. The problem you are describing is one of the things physicists since Kepler have worked on, even in the last few decades. Polishing the apple so to speak and adding decimal places to the g constant.

author: bostonman
author_rank: Super Contributor
timestamp: June 28, 2021, 02:04:27 pm
content: I assumed (to use your term) we've been polishing the apple for years.At my level, and the level of many on here, I imagine it doesn't make much difference if we use 9.81 or 9.80999, or if a 100kg mass is really 100.00001kg.It still amazes me that hundreds/thousands of years ago, all these "constants" were discovered without computers, calculators, and just a pencil and paper. If I'm not wrong, many constants were quite close to what we have now.

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2021, 02:07:05 pm
content: With respect to the mass of planets, Cavendish is credited with "weighing the earth" in 1797 to 1798.The modern interpretation of his work was to calculate the gravitational constantGby directly measuring the force between two objects in the laboratory, using a torsion balance. Seehttps://en.wikipedia.org/wiki/Cavendish_experimentOnceGis known, the mass of the earth follows from insertingg(measured at the surface of the earth) andR(radius of the earth, originally calculated by the ancients) into the gravitational force equation.Orbital mechanics is another interesting field. When calculating other masses in the Solar System, one needs the dimensions of the orbits. Early astronomers were able to calculate orbital dimensions in terms of the "astronomical unit" (a.u.), the mean radius of the Earth's orbit, but a precise value for the a.u. waited until radar ranging of the actual distance between the Earth and Venus around 1964. Since the Sun is continuously losing mass, the a.u. changes slowly with time.

author: David Hess
author_rank: Super Contributor
timestamp: June 28, 2021, 02:18:52 pm
content: Quote from: bostonman on June 28, 2021, 03:01:14 amCommercial load cells are good enough that the variation of surface gravity on Earth is a measurable error term in calibration.

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2021, 03:30:39 pm
content: Also, inertial navigation systems for guided missiles need to correct for local variations in the gravitational field due to geological effects.Notes:1. Please stop callingga physical constant, in the sense thatG(universal gravitational constant) is a constant. Within a limited region where the change in distance to the center of gravity is negligible,ghas a constant value (uniform gravitational field), but it changes by a large amount when you go to the moon, and by a small amount when you ascent to the top of Everest.2. You do not need to know the mass to determineg, use a pendulum.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 28, 2021, 04:26:20 pm
content: You can measure mass (as opposed to weight) using angular momentum and torsion springs.By 1785, those were already so sensitive that Charles-Augustin de Coulomb could use one to measure electrostatic forces, and came up withCoulomb's law.By 1798, Henry Cavendish used one to directly measure the gravitational attraction between two masses in a laboratory experiment, the famousCavendish experiment, although geologist John Michell designed the experiment (Coulomb following his plans), sometime before 1783, but alas, kicked the bucket too early. (Michell was 69 when he died, though.)With commercially available load cells, a turbo pump for pumping a good vacuum, some glass jars and such, you can definitely measure for example the gravitational attraction between say two marbles, in a freshman physics experiment. The most expensive thing is getting precise mechanics nowadays. A couple of hundred years ago there were watchmakers and others happy to help produce interesting scientific apparatuses, but now, you're better off with old mechanisms and new sensors!Quote from: TimFox on June 28, 2021, 03:30:39 pmHey, I blame the language! We need a word describing a variable that is not being varied in this particular context.For differential calculus, we use "derivative" and "partial derivative"; so would "partial constant" work?Yesterday, I spent a couple of hours admiring the concept and wordgeodesic.Let me waffle on a bit once again, but this time, as a hopefully entertaining but informative musing on how precise, exact terms make life interesting, and mushy overloaded ones a misery. (I am thinking of making a T-shirt with"Why don't you just use pastel tones to reduce gravity?, too.)As a background,geodesicis when you extend the concept of "straight line distance between two points" to surfaces; and from its name, one of the most used one is the "straight line distance" between two points on the surface of a sphere or geoid, like the Earth. On a perfect sphere, that distance is the shorter arc of the great circle – a circle passing through those two points, with its center at the center of the sphere.What happens, if you ask your Californian friend what the distance between LA and Las Vegas is?My bet is that the most common answer is something like "Oh, about four hours, at this time of day."Bitch, I asked you distance, and you gave a time interval. What sort of mindfuck is this?(If you happen to have the cultural context, you'd know that because primary transport in that region is via personal automobiles, and the physical distance is less relevant than the actual travel speeds achievable, it makes more sense to describe distances using the typical time taken to drive that distance, than anything else. Just like we include the completely unrelated inertial term in our "average acceleration on Earth due to gravity", because we don't have any better place to put it, and keeping it there takes care of it very nicely, thank you.)In a different cultural context, that friend might ask for clarification, say "As the crow flies, or?".Bitch, I ain't an ornithologist or an ornithopter; how the hell would I know how corvids fly anyway? They may do so ass first for all I care.Now, let's get to "geodesic".Let's say you're driving from LA to Las Vegas with your friends' kid on the back seat, thinking about normal kid stuff, like how many hydrogen bombs in the 50 to 100 kiloton range would they need to carve a canal from Los Angeles to Las Vegas.The kid asks, "Do you know what the geodesic between Los Angeles and Las Vegas is?"Why, you just looked that up a few days ago for this very thread, so you answer "About 368 km, but there is also an elevation difference of about 525 meters."SEE? No confusion. Straight, unambiguous answer unrelated to ornithology, free of oddities like trying to measure distances using time units, and so on.The kid happily takes the added cratering depth into account, you both smile, and have a nice car drive. Everybody wins.If you try to be physically or geometrically correct but do not know the term, the mistakes and confusion one gets mired in are endless.For example, let's say you ask "What is the distance between Los Angeles and Las Vegas?", I ask for clarification, for which you say "in a direct line, but ignoring the curvature of Earth and the difference in elevation". My answer, not to be snarky but maximum helpfulness, could likely be "Either 368 km or 39707 km, depending on which way you measure it."I know, I'd get bitch-slapped for that (if I wasn't driving, and I don't). But I lack the context necessary to give the "straight answer"; I either have to make assumptions (which are often wrong), keep asking for clarifications, or give an answer that outlines the possible set so the asker can determine what the hell they were trying to find out in the first place.And no, I don't really like "partial constant". But I'm used to it, because I useconstevery day, and it does not mean that something is constant, just that in that particular contextI(or the C code I write that uses that) promises to the compiler to not try and modify that value. Andvolatiledoes not mean "flames or vapours are imminent, get your fire extinguisher ready!", it means "hey compiler, that value may change at any point, so make no assumptions about it okay?". And I have to talk about stuff like "immutable string literals" to get new programmers to understand you really don't get to modify them; that if you do, you broke a promise, and the kernel gives you a segmentation fault in confusion.But Iperfectlyunderstand the frustration with mushy and misused concepts and terminology. If we'd have a better set, one with say posters or web pages intended for different levels of complexity/understanding, that I could point others to and say "usingthese" ..., I would do exactly that. And I'd be darn happy; I'd buy it as a dozen posters and lovingly put them on my walls and give out as gifts; they'd make that much a real-world measurable difference.But scientists and their insistence of attaching peoples surnames to things yielding undescriptive (or in the case of original discoverer being someone else, sheer misleading history-twisting self-aggrandizement) words one has to learn by rote by the dozen and use "correctly" (read differently) in every single little domain, and science becomes a Black Magic only those versed in the Old Powerful Names of Power and History and Power have the sheer single-mindedness to delve into.I mean, I can say something like"My mother in law was being a sheer Cheney this week visiting us. Thank Dog her Trumpy sister could not come! Dealing with those two is like trying to watch The Kardashians and Dr. Phil at the same time. I don't know... I think I might be getting a bit Biden in my old age or something; I've been acting like a complete Dubya."and people would just get it.But use "mass" and "constant", and people start getting more confused and even angry... No, this is not sane or right. Yet, it is the world we have.So, what is one to do? Even writing about this like I've decided to do in this thread gets me flak about wasting others time, and gets me added to new ignore lists.I don't know.I just pick a topic at a time, think hard about it, and especially all the ways I know and think of it can be misunderstood, and if I decide it seems possible/worthwhile/useful, construct something that hopefully helps others wade through the word salad but with new mental tools and understanding at their disposal.

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2021, 04:44:48 pm
content: In technical English, a "parameter" is either a constant that can be varied, or a variable whose value is held constant during the calculation.The popular media keep confusing "parameter" with "perimeter", the boundary around an area."Geodesic" is a very interesting concept. In geometry, it is the distance between two points on themanifold in questionwith the shortest length. In three-dimensional Euclidean geometry, it is merely the straight line between the two points. On the surface of a sphere, it is the great circle (arc with center at the center of the sphere) that connects the two points. Traditional ocean navigation plots the great circle on a globe or appropriate projection flat map, and then transcribes the points onto a piecewise-linear route on a Mercator projection, where any line segment is a line of constant compass heading, appropriate to the helmsman. In a classic textbook on general relativity, one example is an ant crawling on the surface of an apple (not really spherical), where the claim is made that sensing only the spatial derivatives along its short body, the ant will follow a geodesic path. That is a good parable, but I don't know if anyone has done the experiment with a real ant, or if it is literally applicable, since I don't know if the ant has a definite endpoint in its tiny mind (analogous to LA and LV).The traditional measurement of mass is to use a gravitational balance (such as held up by Lady Justice at the courthouse) and compare the mass in question to a standard mass, defined by the King. As a practical matter, a "steelyard" balance (q.v.) uses leverage to compare the mass in question to one or more standards. Technically, this equates the gravitational masses on both sides of the balance, but standard theory and Roland, Baron von Eötvös tell us that gravitational mass and inertial mass are equivalent, hence can be set equal.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 28, 2021, 08:58:53 pm
content: Quote from: TimFox on June 28, 2021, 04:44:48 pmThat works.Before I encountered the termeccentric angle, I consistently called \$\theta\$ "the angular parameter" in$$\left\lbrace\begin{aligned} x &= a \cos \theta \\ y &= b \sin\theta \end{aligned}\right. \quad \iff \quad \left\lbrace\begin{aligned}\varphi &= \arctan\left(\frac{b}{a}\tan\theta\right) \\r &= \frac{ab}{\sqrt{(b \cos\varphi)^2 + (a\sin\varphi)^2}} \\\end{aligned}\right. \quad \iff \quad \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$where the leftmost set gives the Cartesian coordinates for the points on the axis-aligned ellipse as a function of the angular parameter, or eccentric angle \$\theta\$; the middle one is the same in polar coordinates where \$\varphi\$ is the polar angle and \$r\$ the distance to the ellipse from origin in that direction; and the right side being the familiar implicit form in Cartesian coordinates.(Oh wait, it isn't an implicit form, because I left the one on the right side. Whatchamacall this form then? Pre-implicit form? Explicit form without any nekkid stuff?)It is so easy to accidentally forget that just because you have a \$\theta\$ or \$\sin\theta\$ or \$\cos\theta\$ term, does not mean that \$\theta\$ necessarily refers to a particular angle. Eccentric angle is illustrative, because you immediately intuitively know that"Ha! I bet it means it is not exactly the polar angle!".I like "angular parameter" too, because for e.g. Lissajous curves (\$x = A \cos(a \theta + a_0)\$, \$y = B \sin(b \theta)\$) \$\theta\$ really is just a parameter you need to vary to sweep the entire curve.Quote from: TimFox on June 28, 2021, 04:44:48 pmYes; I didn't want to write that myself, because the concept ofmanifoldwould have to be explored first.Case in point: A couple of years ago, I helped a CS or SE major with a raytracer visualizing non-Euclidean space. It was piecewise Euclidean; essentially modeled using simplices (tetrahedra) whose faces are discontinuities yielding an overall non-Euclidean geometry.If you've seen the Cube movies, then you probably know the idea: each simplex is a "room", with arbitrary other "rooms" sharing walls in arbitrary orientations. Aside from the discontinuity (a coordinate transform we used barycentric coordinates for, yielding a pretty useful geodesic) at the faces of each simplex, the "rooms" themselves were Euclidean. It modeled pretty much perfectly the case where you're looking at the back of your own head, except with just simple geometric primitives.I still don't know whether you can consider that a manifold, because of the discontinuities at the tetrahedral mesh faces. "Piecewise Euclidean" suited me better; I don't know what they used in their report/paper/whatever. (In Cube, the walls are always perpendicular, so there are no discontinuities of the sort the tetrahedral mesh has; that I do believe is a proper manifold.)I also tried to avoid dealing with terms like "embedding", because defined the way it was, it is darned near impossible to give a nontrivial simulated model any specific dimensionality. If you take a cube, and "glue" its faces together permuting their orientation, what is its dimensionality? I think there might even be pathological cases (choices of mating face pairs and their relative orientations) where you'd need infinite number of dimensions to embed that into a higher-dimensional Euclidean space.If you consider how the tetrahedral meshes can be constructed, it is easier to define the meshed topology in terms of its internal consistency and structure, and completely ignore any efforts to see say where the tetrahedral lattice nodes would be if you wanted to assign them Euclidean higher-dimensional exact coordinates.I preferred to just show the tools, and leave it to mathematicians to define and help with the correct terminology. But I know for a fact that the darned thing worked, because I've seen the rendered images and fully understand how the math works and why. But put me among a gaggle of mathematicians, and they'll laugh me out of the room as I fumble trying to find the correcttermsWords of Wisdom and Power and I Even Have This Hat You See to describe what math it is based on.If you think about that, it is kinda sad.Reminds me of the joke about a man who wanted to see the Circus Director. The director was busy, so snapped at the man to tell what he could do, and to be quick about it. The man proudly exclaimed he could emulate any bird! The director laughed at the man, saying whistlers and ventriloquists are a dime a dozen, and to get out; he was just wasting the Director's valuable time. Utterly dejected, the man sighed and flew away.(The reason I helped with that, is that I once had a dream test-playing a game I purportedly designed, somewhere between Marathon and Doom, but with exactly the sort of non-Euclidean geometry as described above. Basically, any enclosed tubular region of the game world, the world isn't described in absolute coordinates at all, but as piecewise Euclidean spaces glued together using minimal shear discontinuities. A particular example is of a slightly sloping tunnel/stairwell upwards, where you walk "around the corner" (in a circular arc) some 700-800 degrees, without returning to your original location at all. Not just "bigger on the inside", but no turn or bend ever goes the way you think it does. This is trivial to implement with suitably chosen data structures in almost any 3D graphics/physics engine, but at least in my dream the "uneasiness" of going one way, even coming back and going again the same way, but arriving at a completely different place you think, and not being able to find a discontinuity where the "error" or "trick" occurs (there really is none, they're within rounding error; it's just not really Euclidean, only piecewise Euclidean), was fun-kyy. I don't even like that sort of games, although I've played a little at parties for the social aspect; I have even less interest in designing or creating such games. I see Odd dreams, I guess.)

author: RJSV
author_rank: Super Contributor
timestamp: June 28, 2021, 09:17:57 pm
content: Yes, I agree, having the context of using 'g' in a varied environment, not just at Earth surface, and simplifying to be a smooth ball, uniform density, etc.So then "...local acceleration conditions..." for lack of a better word.Sounding better and better.

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2021, 10:08:48 pm
content: I think the better words are "gis the acceleration of gravity in a region of uniform gravitational field".

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 28, 2021, 11:34:17 pm
content: After following many pages of this I think I have finally realized another of the problems. It is akin to the pound mass pound force problem. Two different things with the same name. g has even more definitions for the same symbol.Some practitioners use g to refer to gravitational acceleration at the surface of the earth. It has a formal value, traceable to the defined value of G, and can be used for many calculations that don't require corrections for centripetal acceleration, mascons and the like.Other practitioners uses the same symbol to refer to the local corrected acceleration (somewhere on or very near the Earth) including all of the correction factors.And still another group uses it it refer to the nominal acceleration due to the dominant local mass, the earth, moon or mars for example.I personally assume that the first usage is intended unless context suggests otherwise. That works in the vast majority of cases, especially if you use the sub rule to check the context more carefully the more decimal places matter.The third usage is uncomfortable to me, but made more palatable by flagging and identifying with a subscript identifying the reference mass. Something which I haven't noticed happening in this thread.

author: RJSV
author_rank: Super Contributor
timestamp: June 29, 2021, 06:27:23 am
content: I vote for a lable of:'REGIONALLY STABLE VOLITILE'to use 'g' in a C code and defined as Volitile to C compiler.but that's a bit snarky / infantile. Hey, some classic jokes provide a sort-of structure, for framing difficult / frustrating problems. (I'd rather that irk, over a direct, energetic encounter with gravity...Oops, I meant 'rapid deceleration (w ground).

author: bostonman
author_rank: Super Contributor
timestamp: June 29, 2021, 01:03:02 pm
content: While I agree that 'g' should not be called a constant, a few errors have taken place throughout this thread.The whole topic is based around ma = mg. If this were a topic of C code and someone used 'g' as a variable, then 'g' would be known as a variable, however, it began (and continued) around ma = mg.Somewhere along the way someone (or at least how I interpreted) needed to state that 'g' is the acceleration of gravity on Earth and 'G' is the Gravitational constant. I was a bit baffled why someone assumed the contributors of this thread were confused over g and G when we clearly separated the lower and upper case.In any case, I feel some began reading the thread halfway and therefore felt 'g' was improperly used. Much like if I began reading halfway through about a programming thread involving 'g' as a variable, I may ask: why are you using the gravity on Earth as a variable.I'll ask the philosophical question though: is anything really a "constant"?If science is always polishing the apple per se, then every 'constant' we know will change. Pi is a constant, but it's a never ending value that hasn't been fully carried out (nor may never). G is the gravitational constant, however, it must have some error margin.If all these numbers are constantly being worked on to be more precise, then are they truly constants? By saying (little) 'g' is a constant may not be completely incorrect.

author: TimFox
author_rank: Super Contributor
timestamp: June 29, 2021, 01:53:26 pm
content: If you definegas the acceleration of gravity at the surface of the earth, where exactly on that surface is that to be measured? It is slightly different at the peak of Everest, in the city of Denver, and at whatever they are calling mean sea level now. I prefergas a parameter measured within a region of interest where the gravitational field is uniform, whether that be on Earth or on the moon, which should be stated or made obvious in context.The importance ofgis that all objects within this region of interest experience thesame valueof gravitational acceleration,independent of mass. This is different from Aristotle's theory that predates Newton.(Of course, all of this neglects air resistance, as in freshman physics discussions. When there is serious resistance, the force is proportional to velocity, and dependent on the object's shape. If you drop an object far enough through a medium, the fall approaches "terminal velocity", where the resistance force equals the gravitational force and the object stops accelerating.)

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 29, 2021, 02:03:56 pm
content: In a practical sense g and G are constants. By agreement of international standards bodies. While the values may change over time, in these two cases, and over human time scales these changes will be within the error bars of the existing definition, somewhat akin to computing more digits of pi.I'll leave the philosophical questions of whether these matter on some cosmic sense, and programming style questions to others.As always, when doing something that has real consequences it is important to thoroughly understand the problem at hand. A programmer who uses a standard constant library which includes g better not use that for a Mars lander. A student taking a test needs to understand what the teacher is asking. Don't give a spherical geometry answer in a course in Euclidean geometry unless you are willing to do a great deal of possibly unsuccessful explaining

author: RJSV
author_rank: Super Contributor
timestamp: June 29, 2021, 02:19:22 pm
content: I did some looking for 'g' subscripts ( although my Android smart phone makes subscripts difficult text to enter).Looking at some May 13 posts, there are subscripts; (moon), (sun), etc. usually byantiprotonboy and also Brumby, so that's helpful.Plus the process suggested kind of relating back to Earth, as a benchmark.

author: bostonman
author_rank: Super Contributor
timestamp: June 29, 2021, 02:20:10 pm
content: I agree with both these last two posts.I feel some contributors began reading halfway through and felt that they needed to explain the difference between 'g' and 'G'.To say 'g' is a constant or not I feel can be open for discussion, however, it is based on where you are on Earth, therefore I feel it's more of a calculation and not a constant.Pi, as an example, doesn't need to be calculated, in fact, it's a button on many calculators. Therefore, it's obviously a constant, but with some long term polishing.

author: TimFox
author_rank: Super Contributor
timestamp: June 29, 2021, 03:43:09 pm
content: Ignoring for a while the controversy about little-g being a legitimate constant, I have some observations about weight and mass, and an historical question at the end of this post.My first formal course in physics, at age 16, was the first time I became aware of the important difference between mass and weight. Prior to that, everything of interest to me was in terms of weight: my own weight, the net weight on a box of cereal, etc. In the US, this also introduced me to the problem of the word "pound", which in physics class was always a unit of force or weight, and the ugly word "slug" for the unit of mass in the same system, which never appeared outside of physics textbooks. Since the space race was underway, we were all interested in weightlessness, how much one would weigh on the Moon, and other topics in gravitation away from the Earth.Now to earlier history and laws: The United States Constitution, Article I, section 8 gives Congress the power to “fix the standard of weights and measurement", where the framers saw the need for uniform measurements across the several States of the Union. At about the same time, the founders of the Metric System had a similar need to unify French measurements across the new Republic. Their work lead to the "BIPM", which I believe translates to "International Bureau of Weights and Measures". Note that in both cases, the language (and further statutes in the US and elsewhere) refers to "weight", rather than "mass". As one should not go to a physicist for legal advice, one might not trust a lawyer for scientific clarity.InPrincipia, Newton defined mass as the product of volume and density. (Translated from the Latin: "The quantity of matter is the measure of the same, arising from its density and bulk conjointly".) The original definition of the gram was 1 cm3of water (at the temperature of its maximum density), although that was later changed to the prototype kilogram.My question: historically, was this original definition of the gram considered "weight" or "mass"? Logically, it would be mass, since the corresponding unit of force was already the "dyne".

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 29, 2021, 05:09:50 pm
content: I can't over emphasize the need to know what problem you are working on. If you are designing a spring scale for the bath room, or calculating how high your model rocket will fly or most of the other problems you will encounter the variation of g over the Earth's surface doesn't matter. Other parts of the problem will have unknowns greater than the differences. You terminal velocity when you jump out of an airplane and how fast you reach it will be far more dependent on air density than on g for example.If you are doing something where the variatíon of g with location matters I hope that your general understanding of the problem is such that you aren't concerned with the existence of a standardized version of the answer.In much the same way several governing bodies define pi as a rational number (like 3.14 or 3.14159). This isn't an attack on mathematics, but a way to avoid legal arguments in trade and commerce how many decimal points to use when computing circumferences and areas

author: TimFox
author_rank: Super Contributor
timestamp: June 29, 2021, 05:49:34 pm
content: "Standardized version of the answer": Is there a legal definition ofg?The classic textbook "Physics", D Halliday and R Resnick, John Wiley & Sons 1966 (one of a series of revisions of this calculus-based freshman physics textbook) in section 5-6 on systems of units (pp 90 ff), discusses the problem of mass, weight, force, and gravity. They state that "in the British engineering system....Legally, the pound is a unit of mass but in engineering practice the pound is treated as a unit of force or weight....The pound-force is the force that gives a standard pound an acceleration equal to the standard acceleration of gravity, 32.1740 ft/sec2." The book later notes that "the acceleration of gravity varies with distance from the center of the earth, and this 'standard acceleration' is, therefore, the value at a particular distance from the center of the earth. Any point at sea level and 45 deg N latitude is a good approximation". In this section, the term "g" is not explicitly mentioned.Moving forward to section 5-8 (pp 93 ff), discussing the vectorsWandg, with the scalar massm, we read "The quantitative relation between weight and mass is given byW=mg. Becausegvaries from point to point on the earth,W, the weight of a body of massm, is different in different localities." In later examples, the authors are careful to state "at a point whereg= 32.0 ft/sec2" or "in a locality whereg= 9.80 m/sec2...in a locality whereg= 9.78 m/sec2" for quantitative discussions of a particular weight and mass.This is the standard discussion from the point of view of physics. This may not matter to backyard rocketry, but is obviously important to interplanetary rocketry.The point of what many here think of as my nitpicking is that in the traditional British units, where objects are measured by weight and that is converted to mass (pound-force vs. pound-mass), a legal definition of the acceleration of gravity is required, but modern physicists do not call that "g", since the actual value ofgvaries from place to place. In metric countries that use SI units, this problem does not occur, since material is sold by mass (in kg), not weight (in N).Again, I emphasize that the importance ofgis that its value applies to all objects in the region of interest, as opposed to older theories where a heavy object drops faster than a lighter object.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 29, 2021, 06:41:10 pm
content: From Wikipedia which has the references.I apologize I did forget to subscript the value properly. The article mentions that this is commonThe standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2 (about 32.17405 ft/s2). This value was established by the 3rd CGPM (1901, CR 70) and used to define the standard weight of an object as the product of its mass and this nominal acceleration. The acceleration of a body near the surface of the Earth is due to the combined effects of gravity and centrifugal acceleration from the rotation of the Earth (but the latter is small enough to be negligible for most purposes); the total (the apparent gravity) is about 0.5% greater at the poles than at the Equator.Although the symbol ɡ is sometimes used for standard gravity, ɡ (without a suffix) can also mean the local acceleration due to local gravity and centrifugal acceleration, which varies depending on one's position on Earth (see Earth's gravity). The symbol ɡ should not be confused with G, the gravitational constant, or g, the symbol for gram. The ɡ is also used as a unit for any form of acceleration, with the value defined as above; see g-force.The value of ɡ0 defined above is a nominal midrange value on Earth, originally based on the acceleration of a body in free fall at sea level at a geodetic latitude of 45°. Although the actual acceleration of free fall on Earth varies according to location, the above standard figure is always used for metrological purposes. In particular, it gives the conversion factor between newton and kilogram-force, two units of force.

author: TimFox
author_rank: Super Contributor
timestamp: June 29, 2021, 06:46:40 pm
content: The information in this Wikipedia article is consistent with that I quoted from Halliday and Resnick.Specifically, adding a subscript togindicates that is measured in a specified location, while I insist that "g" is a general term for the parameter in different locations.Luckily, we need not worry about this in metrology when we use SI units.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 29, 2021, 06:47:31 pm
content: I will also give the example describing aircraft and spacecraft accelerations in g's. In this usage the g is always a standard gravity. And this usage is rarely, if ever, used where precision is required. But the g is a relatable unit, requiring less mental gymnastics than saying that most pilots lose consciousness somewhere in the neighborhood of 59 to 69 meters per second squared.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 29, 2021, 07:04:38 pm
content: Quote from: TimFox on June 29, 2021, 06:46:40 pmAs I read the documents from BIPM, in metrology you don't use g in any of its connotations in metrology. Only units traceable to standard distance and time. The only traceability chains authorized are direct tracing to distance and time, comparison to a reference gravitometer which has used the first method, or comparison to an acceleration value at a specific location measured by a gravitometer that has been calibrated by one of the first two methods.The use of g (in any of its forms) is a convenience for practitioners in any number of fields.Again, horses for courses.

author: TimFox
author_rank: Super Contributor
timestamp: June 29, 2021, 07:05:45 pm
content: In that case, as you properly stated it, "g" is not italic, since it is a unit of measurement, whilegis a parameter, equal to the magnitude of the vectorg.When subjected to an acceleration of 10 g along the vertical direction, the victim will weigh 11 times his normal weight.In college, I did measureg(in southern Minnesota), using a Kater pendulum. In the ancient textbooks, before electronic timers, the period was measured by placing the Kater pendulum in front of a grandfather clock's pendulum and observing the beat frequency. In 1969, I used an -hp- 520 vacuum-tube counter, with Nixie tube readout, that was roughly 19 inches cubed, and a simple photocell.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 29, 2021, 08:17:25 pm
content: I forget to use the correct (and useful) subscripts too, sorry.On this forum, you can write say abp, using..and..; these will show the same way in preview as they will in final posts. abpis written asabbin the editor. To get the teletype look, I use... To stop a bracketed term to be interpreted as markup, you can insert aor a(which literally means nothing in teletype, or nothing in italics, respectively) in a suitable place (after each[the forum software considers the beginning of markup, changing each[foointo[foo), to break up a markup-looking expression without visual side effects. It's like adding zero-width spaces: the sequence the server sees is different, but the end result stays the same visually.This forum also has MathJax support. For inline MathJax, for example \$E = m c^2\$, you bracket it with\$on both sides. (That snippet being written as\$ E = m c^2 \$in the editor.) For block expressions, use$$, for example$$ E = m c^2 \tag{1}\label{NA1} $$, to get $$ E = m c^2 \tag{1}\label{NA1} $$where the added stuff lets you create links to it using\$\eqref{NA1}\$(rendering to \$\eqref{NA1}\$). The equation label namespace is usually the entire HTML page, so I recommend using labels with a prefix unlikely to be used in other posts. That way the visible tags can be the same, but the labels keep them separated.Unfortunately, the forum code lacks the trigger to apply the MathJax translation on preview, so what you see is the bracketed MathJax expressions instead of the rendered results. At minimum, it'd only need a simple JavaScript call to a library function tacked on to the AJAX completion of the content preview load; for normal posts and thread view, a related call (that does not render source markup, but the intermediate form the forum software munges the source markup into) is automatically triggered by page load completion. (Implemented this way, it would not exactly match the way the MathJax module in the forum software does it, but the visual result would be the same for the vast majority of users. The difference would be that preview would lack the MathML translation step done on the server end. For full support, the MathJax server-side munging would need to be added to the preview server-side processing,plusthe rendering trigger added to the AJAX completion. Here, MathJax is so rare I guess Dave and cohorts don't see it worth the effort. I'm not familiar enough with SMF innards to have an opinion, but I do want to point out that any such modification is at risk for adding new bugs, so it isn't just laziness; there is always a risk in modifying forum software.)

author: bostonman
author_rank: Super Contributor
timestamp: June 30, 2021, 01:45:25 am
content: These last few posts about how to write 'g' are interesting.First it shows that smart people need to be precise about how they write, speak, and dislike others who are wrong. It also connects with "discussions" I have with others.I've noticed keeping my mouth shut is very difficult when I hear in correct statements, or feel the need to critique others.No matter what value we call 'g', ma can still equal mg. If I accelerated my car from 0 to 9.8m in one-second, then the Force is the mass times 9.8m/s^2 thus making ma = mg.

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 30, 2021, 06:37:14 pm
content: On the magnitude and intuition of forces.ConsiderPrince Rupert's drops. Toughened teardrop-shaped glass beads, created by dropping molten glass into water.Because the glass shrinks as it cools, energy gets stored in internal stresses – internal forces pushing each atom of the glass (mostly silicon and oxygen; silica being SiO2and glass is silica plus impurities) – areimmense. On the bulbous end, the pressure due to this is on the order of hundreds of megapascals; 700 MPa according to a 2016 paper published in Applied Physics Letters. Again, that force is not exerted by anything external. There is potential energy stored in the structure, yes, but no dynamic forces involved; only the same static force that say the ground exerts on the soles of your foot when you're standing put and not sinking into the ground, just pushing the atoms in the drop against each other and nothing external.That is, if you put a Prince Rupert's drop in a padded box, it won't slowly release its structural tension, no more than standing on a granite floor makes it slowly mellow into marshmallow. Nor will it explode if you put it in vacuum. It'll just stay as it is, barring stuff like annealing due to temperature changes, structural changes due to ion or cosmic ray bombardment, and so on.How much is 700 MPa? Well, 0.1 MPa = 100 kPa = 100,000 Pa = 1 bar by definition, and 1 bar is just slightly under the standard pressure of air at sea level (1.013 bar); or just about exactly the air pressure at 111m elevation at a 15°C temperature. So, we're talking about seven thousand atmospheres of pressure, give or take.Such a glass bead certainly does not look like it has anything of the sort locked in its structure in static forces now, does it? That's because the internal forces, the internal stresses, are perfectly balanced. They have to be, or the bead would change shape. The reason cutting the tail makes the drop explode, is that the fracture thus generated (which propagates in the glass bead at five or six times the speed of sound in air, or 5200 - 6800 km/h), unbalances those forces; and no longer in balance the atoms start moving –– in the human scale, in explosive disintegration.Glass, or even its easier version,silicaor silicon dioxide, SiO2, is not hard to simulate on the atom level, but being amorphic, the structure is difficult to get "right" (the way it is in real-world glass or silica). I don't know if anyone has modeled Prince Rupert's Drops, but their macroscopic size makes it a bit daunting; you'd need serious (distributed) computing power to do it properly. Smaller simulations can obviously show details (say, a model of a small region of the surface, extending towards the center of the drop), and those have surely been done.Anyway, on-the-envelope rough estimates (say, 160pm interatomic distance) means that force per atom based on the pressure is on the order of dozen or two picoNewtons (10-12N). Doesn't sound much, until you realize a silicon atom only weighs 28 u ≃ 28 × 1.6605 × 10-27kg ≃ 4.6×10-26kg. If that force was applied to a single atom without any opposing forces, then the instant acceleration of that atom would be some forty million million standard gravities, or about 4×1014m/s2."Stupendous" comes to mind, even if there is a typo of a few orders of magnitude in there; but it "feels about right" to me, so I won't re-check – it's just the crudest possible napkin math. (Oh, and 1 standard gravity ≝ 9.80 m/s². It's used in some science fiction to describe spaceship accelerations (usually high, as in hundreds of standard gravities of acceleration, since they need inertial dampeners to stop humans from pancaking if you go beyond single-digit standard gravity accelerations, and such high accelerations to get any place or be able to match velocities in any time frame dramatic enough for us humans), and I like how it sounds.)The amount of energy stored as potential energy in the structure of matter –– and note that this is just the static, structural forces that don't do work since matter generally stays the same; we're not talking about the utterly ridiculous amounts of energy involved in the matter itself due to \$E = m c^2\$ –– and the static forces in it, are really completely outside our intuitive scales.

author: TimFox
author_rank: Super Contributor
timestamp: June 30, 2021, 07:21:27 pm
content: A brief comment about the mass-energyE=mc2inherent in ordinary matter:In nuclear reactions and similar problems, only a small part of this energy is available to the consumer, basically due to the small difference between neutron and proton masses, since "baryon number" is conserved. For nuclei, the protons and neutrons are baryons, so there can be transformations between them. The proton is the lightest baryon and things stop there, pending discovery of very long decay times that have been theorized but never observed for proton decay.However, I have a vague recollection from an interesting lecture on black holes I attended in the mid-70's, that if one were in a safe orbit around a black hole, outside the event horizon, and very carefully (adiabatically) and slowly lowered a mass into the black hole on a very strong fishing line, the total energy transferred to the reel would, in fact, bemc2.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 01, 2021, 12:53:44 am
content: The one field where we do already get \$E = m c^2\$ out, is matter-antimatter reactions; specifically,electron-positron annihilation, as used in e.g.positron emission tomographyandpositron annihilation spectroscopy.The positrons are generated by nuclear fission (in a radioactive material, calledtracerorradiotracerin PET and injected into the patient at very small quantities, and just placed next to the material or structure to be analyzed for PAS).It just happens that when a positron and an electron annihilate each other at low energies (meaning have low velocities if measured in the frame where their linear momentums cancel out), the end result is two gamma photons with very easily identified energies, and we canusethose gamma photons for useful stuff.At higher energies, by controlling the amount of kinetic energy, you can generate interesting exotic particles like mesons and W and Z bosons, which all tend to decay very quickly (as in within 10-18s or sooner, much much faster than we for example do time steps in atomic simulations) into leptons and hadrons and neutrinos and such. Interesting stuff for particle physicists, but it's those gamma photons that are very useful in practical applications.The CERNAntiproton Decelerator, among other thinds, produces antihydrogen: atoms consisting of an antiproton and a positron. The numbers are small; in 2011, the ALPHA experiment captured 309 antihydrogen atoms for over fifteen minutes (1000 seconds).In certain Science Fiction settings, antihydrogen is stored as a fuel, and generated in huge accelerator rings, sometimes surrounding planets or stars. It is likely unfeasible (and something like just sieving positrons and antiprotons from highly energetic environments would work better), but not out of the realm of possibility; not realistic, but close enough to be tantalizing.

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 03, 2021, 12:12:07 am
content: Quote from: TimFox on June 25, 2021, 05:57:09 pmI just do not like this term "accelerating force" much, hence my remark.But as to what I said about weight and freefall, and that was also mentioned later on in the thread, I think it's an important, yet not always fully understood point. And not so trivial after all.It's actually the thought of weight having virtually no effet in freefall that led Einstein to formulate the equivalence principle (that, true, was something already more or less "known" before, but in a more restricted way), and eventually led to the theory of general relativity. So, even though it's in itself nothing extremely special when you know Newton's laws, the idea is still key in "recent" physics history.What Einstein noticed thinking about weight in freefall is that, in short, the effect of gravity, for an object in freefall, would sort of cancel gravity. That was seminal enough to have inspired the theory that followed...

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2021, 03:33:02 am
content: Should a force effect an acceleration, pursuant to Newton's Second Law, what should one call it?If a strong wind damages your property, it is called a "damaging wind", etc.

author: bostonman
author_rank: Super Contributor
timestamp: July 03, 2021, 02:27:16 pm
content: QuoteCertain terminology is confusing, especially when you've spent your entire life having a specific naming convention. As an example, Weight has always been known to me as weight, but it's technically Mass. Then you say you're picking up X kg of weight, but you're really using Newtons. Personally it's hard for me to comprehend that a car hitting a wall is X Newtons, but it's easier to think of some logical comparison like it's the equivalent of bench pressing 1000lbs.Anyway, just some input about how spending a lifetime using wrong terminology can get confusing when you need to learn the correct way.Acceleration can be confusing. This entire time we've talked about picking up stuff, so "acceleration" is 9.8 due to gravity. Now what if I'm pushing an object (leaving out anomalies like friction)? I need to accelerate the object in order for it to move, but in this case, am I applying Kinetic energy to perform Work instead of using an acceleration formula?Another Work related question, if I walk on a treadmill at a normal pace, I believe it will read approximately 200 calories are burned (not that important) - obviously this is the computer doing a theoretical calculation based off distance. If I decide to burn calories by picking up a 25kg object several times at a distance of 0.5m (let's forget about Work exerted by putting down the object), that is 245N, which is 122J of Work.Due to laziness, I used an online conversion calculator and got 29 calories burned just by picking up that object once (although it seems high to me).Does this mean I can save myself an hour on the treadmill by picking up that item seven times?

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2021, 02:41:09 pm
content: I must refer you to a good freshman physics textbook to keep the technical terms straight.For example, you don't "apply kinetic energy" to accelerate an object, you apply a force to accelerate the object, which thereafter has kinetic energy.Similarly,weight is not mass. There is confusion in the way that weight is used in commerce (at the deli counter, for example) for what is really mass.If I take 1 kg of sausage from the deli and carry it to the moon, it no longer weighs 9.8 N.Questions about calories (technically, kilocalories) burned during exercise are not Newtonian physics, but physiology: complicated biological processes.In your example of lifting a weight, the (Newtonian physical) energy for one lift is the force (245 N) times the distance of the lift (0.5 m), and your body does 122 J of work.If you lift the weightvery slowly, so the acceleration is negligible, then after one lift, the object has 122 J ofpotential energy, and zero kinetic energy. If you then drop it, it will accelerate downwards and will have 122 J of kinetic energy after dropping 0.5 m, illustrating the conservation of energy.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 03, 2021, 08:06:24 pm
content: I agree with TomFox that you need time with Resnick and Halliday or other good introductory physics course.While improper nomenclature is hurting you, improper concepts are your bigger problem. It will be worthwhile for you to do the experiments described in the text and work the end of chapter problems. In general if you find the problems hard it is because you didn't understand the presented concepts. Go back and reread and redo the experiments.Think of your body as a very inefficient car. It burns something like 1000-1500 calories a day sitting in a chair doing nothing. The work perform lifting a weight or walking a treadmill is almost negligible.

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2021, 08:23:35 pm
content: From my younger days, I remember Resnick and Halliday as having very well written explanations on the basic topics for freshmen who could do elementary calculus.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 03, 2021, 10:16:56 pm
content: Quote from: TimFox on July 03, 2021, 08:23:35 pmThe first several chapters don't even require calculus. The text is written assuming that the course is concurrent with first year calculus.

author: bostonman
author_rank: Super Contributor
timestamp: July 04, 2021, 02:59:50 pm
content: QuoteIgnoring for a moment that apparently Newtonian work is different than biological, I'm uncertain I agree with this statement. If someone has a desk job, and the best they can burn is 1500 calories, ninety-minutes on the treadmill burning 300 plus calories is 20% or more additional burnt calories. That could mean the difference of burning off your lunch, or not.As for thinking I could use lifting an object to calculate calories burned, guess I need to research this.

author: TimFox
author_rank: Super Contributor
timestamp: July 04, 2021, 03:41:41 pm
content: Warm-blooded mammals need to consume energy in order to maintain their body temperature and vital processes, hence the energy consumption at rest and that one can starve to death while totally sedentary.Again, another misnomer in units: when dietitians say "calories", they mean "kilocalories". Neither are SI units, since heat energy and other energy forms are equivalent and are measured in Joules.The "calorie" is defined as the amount of heat energy that will raise the temperature of one gram of water by one Kelvin. Similarly, the "kilocalorie" or "kilogram-calorie" or "large calorie" is the energy required to raise the temperature of one kilogram of water by one Kelvin. These units are convenient when doing wet chemistry. Confusion between small-c and large-C calories persists amongst non-physicists and is enough reason to use Joules for serious work. At least the BTU is well-defined.

author: ucanel
author_rank: Regular Contributor
timestamp: July 04, 2021, 03:44:24 pm
content: Great video that discusses the subject:(in a funny way also)Science AshlyumWhy did i take a scale on an elevator?https://youtu.be/c6KliNGReTQ

author: TimFox
author_rank: Super Contributor
timestamp: July 04, 2021, 03:48:14 pm
content: My favorite descriptions of Einstein's equivalence between a uniform gravitational field and a uniformly-accelerated reference frame take place in elevators. This is a good way to show the deflection of a light beam by gravitation, since the entrance and exit of the beam on the two walls of the elevator depends on the acceleration.

author: RJSV
author_rank: Super Contributor
timestamp: July 04, 2021, 05:34:56 pm
content: ...A good thread going on, so please don't misunderstand, my mild distress, faced with recent BULK (simply stated), I tend to skim past. My reasons being medically based. Various ailments having associated 'Chronic Fatique' limits on time / energy.So, my take on movements and expending 'WORK' in formal defined terms includes the non-linear real world of friction. In casual terms, I learned in basic physics, about how friction expresses stronger incremental increases, as you go up the scale of speed. Maybe not increasing as the SQUARE of applied speed; certainly not linearly, usually somewhere in between. It's not a clean business, for example the parachute jump, with all the attendant air resistance and limiting velocity of the jumper (some 120 mph I believe).But I don't dedicate all needed time, on friction or WORK debates, and there is a reason:The blog posts, on eevblog generally, I've been using as I speculate on improving personal JOB positioning.And so it makes reasonable sense, to skip over some details (here),...especially while having time / energy limits.Heck, all creatures operate within limits, huh ?Thank you. - Rick

author: TimFox
author_rank: Super Contributor
timestamp: July 04, 2021, 05:46:05 pm
content: Friction is complicated. In freshman physics, it is normally approximated as a constant value of force, independent of speed, opposite in direction to the velocity, proportional to the "normal" component of force between the object and base. (Here, "normal" means perpendicular to the surface.) A block sliding on an inclined plane is a good homework exercise, where you need to find the normal force as a component of the gravitational force, and then subtract the frictional force from the gravitational force component in the direction of motion. One can find tables of friction co-efficients, including "waxed hickory on snow". To be careful, one distinguishes static friction (before the sliding starts) and dynamic friction (while the object moves).Later, one learns of viscous friction, also opposite in direction to the velocity, whose value is roughly proportional to the velocity. The "dashpot" (piston moving through a fluid, like an automobile's shock absorber) is used to provide damping in the basic model of forced, damped harmonic oscillator (mass, spring, dashpot, and possibly gravity), which is the model for the R-L-C electrical resonant circuit.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 04, 2021, 07:22:04 pm
content: Quote from: TimFox on July 04, 2021, 05:46:05 pmFriction IS complicated. Some folks refer to pressure drag as a form of viscous friction, with some justification. But this type of friction is "sort of" proportional to the square of velocity. With "sort of" meaning over a limited range of speed, pressure and a few other variables.When examining the force required to move and keep a car in motion you can observe at least four classes of friction. A breakaway force, a constant force at low speeds, a force linearly increasing with speed and one proportional to the square of speed. If the dynamic friction is large and the drag is large the viscous term may not be observed without very careful measurements as the pressure drag may become large before viscous drag causes significant growth in the total drag.The simple concepts and models of physics are extremely useful. But many real world problems require a great deal of observation and thought to determine which simple models apply and how they are used.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 04, 2021, 08:03:29 pm
content: (I'll try to be brief!)Quote from: RJHayward on July 04, 2021, 05:34:56 pmYes.Friction itself can be described as a force opposite to velocity.Usually, we choose a coordinate system where our object is stationary, and use the velocity of the surrounding fluid (fordrag equations) or the velocity on the surface the object is rubbing against.Velocity is not the only thing involved in determining its magnitude. Pressure of the fluid affects its drag coefficient, and the force pushing the object to the surface it slides on affects the coefficient of friction. As mentioned above by Tim Fox, CatalinaWOW and others, its magnitude can be proportional to the square of the velocity, to the magnitude of the velocity, or something in between or only slightly different, depending on the situation. The proportionality is useful for us humans to estimate the effect, but arises from shapes and flows so there is no clear formulae, except for idealised shapes and fluids and surfaces.The 'work' done by those forces describes exactly the amount of energy transferred due to drag and friction.That energy is usually lost in heat and deformation.If you have ever done lathe work or milling, "chip breaking" is important because the tools rely on the removed particles (chips) to carry off the extra heat generated by friction. With good tools and proper setup – surface speed of the cutting bit and the depth of each cut –, your work piece can remain relatively cool, while the metal shavings flying off are so hot they rapidly oxidize and change color. (I love machining videos. Knowing the funky physics that occur there just makes them even more enjoyable.)In his 1980 Science Fiction bookSundiver, David Brin describes a craft skimming in the chromosphere of our Sun, that uses a high-tech version of that to remain cool: a very, very powerful refrigeration laser. The "chips" are then photons instead of matter, with energies corresponding to crazy high temperatures, but the core principle is the same. As with lathe and mill chip breaking, to work, those chips have to be hotter than the ambient and hotter than the object you are trying to keep cool(er).

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 04, 2021, 09:37:11 pm
content: Quote from: bostonman on July 04, 2021, 02:59:50 pmI was flippant when I said almost negligible. If 90 minutes on a treadmill had no impact there would be no market for gyms. But the dominant use of the fuel you dump into your body is keeping the fires burning, powering your brain (the single largest consumer of energy in most humans) and doing other basic functions. Running, jumping and lifting and doing the kind of work described in physics makes relatively minor variations in the fuel consumption. Even for top athletes running ultra marathons or world strongest man champions the work output is far smaller than the housekeeping consumption.In principal you can relate lifting weights to calories but the relationship will be very imprecise. Your body speeds up the background consumption during and after exercise so calories will be burned that aren't directed toward the lifting of the weight. Similarly you lift and drop parts of your body while lifting your weight set, complicating the calculation. That is why you will find the kind of empirical numbers you are referring to on your treadmill. Scientists have basically put people into a calorimeter while performing various types of exercise and work and measured the output. Your mileage will vary. Different individuals will get different results so while one person may burn 300 calories on that treadmill others might burn 200 to 600 (made up numbers, I know there is a range, but haven't studied it). I am quite sure that the marketing people for gyms and exercise equipment bias their quotes to the higher results.

author: bostonman
author_rank: Super Contributor
timestamp: July 09, 2021, 02:52:11 am
content: It's unfortunate calories isn't more precise.Some of these questions do revolve around the gym. For years I wondered how many say bench press reps would equal X minutes on a treadmill.

author: TimFox
author_rank: Super Contributor
timestamp: July 09, 2021, 01:25:59 pm
content: In physics, these terms are precise, but due to historical usage, the popular language confuses them.Precisely: a calorie is the amount of heat that increases the temperature of 1 gram of water by 1 K.A kilocalorie is the heat that increases the temperature of 1 kg of water by 1 K.A BTU is the heat that increases the temperature of 1 lb av of water by 1 Fo, which brings us back to the definition of the pound: mass or weight.

author: bostonman
author_rank: Super Contributor
timestamp: July 09, 2021, 01:30:51 pm
content: QuoteIs this the reason when I first mentioned "calorie" burning at the gym, someone said it's really a kilocarlorie?In other words, people who say they burned "calories" at the gym are really using it incorrectly?

author: TimFox
author_rank: Super Contributor
timestamp: July 09, 2021, 02:18:56 pm
content: That was me.Of course, if you burn calories you certainly burn kilocalories, but a different rate.The SI uses only Joules for mechanical, electrical, and thermal energy. 1 kcal = 4184 J.At the gym, if you lift 1 kg up by 1 m, than is 9.8 J of mechanical work and potential energy. Therefore, lifting a 50 kg dumbbell by 1 m is 490 J, or only 0.117 kcal.Once you leave the quiet rational space of the SI, you venture into a dense thicket of other units at your own risk.

author: bostonman
author_rank: Super Contributor
timestamp: July 09, 2021, 02:35:42 pm
content: QuoteJust tell me if I'm going over the deep end or just reverting to "calories" being incorrectly used. So if I eat a Snicker's Bar at whatever.... 100 calories, does this theoretically mean I have to lift that dumbbell 855 times (100 divided by 0.117) to burn off the snack?Or are calories on nutrition labels using the wrong units?

author: TimFox
author_rank: Super Contributor
timestamp: July 09, 2021, 03:28:59 pm
content: The question about the conversion of food energy to mechanical energy through biological processes is far more complicated than the Newtonian physics behind lifting a mass in a gravitational field.Since your body is very inefficient, it takes more food energy to do a given amount of physical work, therefore it takes less physical work to burn off a given amount of food energy.Nicolas Léonard Sadi Carnot​ in 1824 originated the careful consideration of "heat engines", that convert thermal energy into mechanical energy, just as steam engines (invented in the previous century) were becoming important.(A distinguished family. His father Lazare Nicolas Marguerite, Count Carnot was a French revolutionary and did fundamental work on engineering mechanics, and his nephew Marie François Sadi Carnot was President of France at the end of the 19th century.)Compared to a Diesel engine, the human body is not an energy-efficient engine (but it suffices for our normal purposes).As I stated, what dietitians and candy-bar manufacturers call a "calorie" is technically a "kilocalorie" in physics, and sometimes called a "large calorie" to distinguish it from the "small calorie" as defined properly.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 09, 2021, 03:35:17 pm
content: Just to pound the point in calories are not imprecise, but calculations of their consumption in the human body are. There are a great number of processes going on, many variables associated with each process all of which are difficult to measure and many of which are either not simple enough or well enough understood to write an input-otput equation.

author: TimFox
author_rank: Super Contributor
timestamp: July 09, 2021, 03:41:23 pm
content: Of course, nutritional scientists work hard to develop legitimate quantitative models and calculations of the human body's usage of and needs for food energy. It is a far more complex problem than the Newtonian mechanics of apples dropping from trees. Since calories are popular units for chemists, the nutritional guys have tended to use them.Not my field, man.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 09, 2021, 06:21:19 pm
content: Not only do they work hard, but they do good work. But one size definitely does not fit all and there is a huge population demanding simple answers. Bostonman is on his way to understanding the complexity. Most aren't interested or don't have the capacity.

author: JohnnyMalaria
author_rank: Super Contributor
timestamp: July 13, 2021, 01:16:03 pm
content: Quote from: bostonman on July 09, 2021, 02:35:42 pmSeems about right. An innocent Snickers (or half a pint of chocolate milk, some fruit etc) can ruin the benefits of your short trip to the gym.This shows a table of typical calorie expenditures for various types of exercise. Damn those high energy density peanuts!https://www.brianmac.co.uk/energyexp.htm

author: bostonman
author_rank: Super Contributor
timestamp: July 19, 2021, 02:22:49 am
content: QuoteI appreciate the compliment. I find that many don't care to dive into understanding many items we use in life or general concepts.With me, sometimes I try digging too deep for no reason.Often times I have an appreciation for when the math matches real life situations. Obviously anomalies get thrown in the loop, however, being an "engineer", tossing in approximations can help avoid a lengthily and unnecessary list of calculations.As for kilo-calories, a slight reason exists why I somewhat switched to this area. The first is because were discussing Work. The other reason is due to going to the gym. It would be nice to have an idea that if I pickup something and place it down, that it will equal (on a theoretical level) X minutes on the treadmill.I'd like to think that if I lift weights for an hour and someone brags they walked an hour on the treadmill, that I can scientifically state that taking a Neanderthal approach burned an equal (or more) amount of calories.

author: RJSV
author_rank: Super Contributor
timestamp: July 26, 2021, 02:51:46 am
content: Yes, BostonMan, your direct work has potential (no pun) health benefits. BUT, don't ignore the warm-up stuff; doing light stretches, and movements. Postures such as standing on one leg, alternating, that's great as (we) age.Heck, sincerely, look at some YOGA or TaiChi practice, ,(maybe can skip some of the strength work).The simple stretches eliminate some resistance in body movement.In older days, folks didn't look to burn calories, except maybe their horses'.

author: bostonman
author_rank: Super Contributor
timestamp: July 26, 2021, 02:56:20 am
content: As for this thread, and my questions deviating towards burning calories, this would be totally different.I'm still focused on the math/physics of being able to argue differences when someone brags they did an hour on the treadmill and feel they burned more calories than someone lifting weights.

author: JohnnyMalaria
author_rank: Super Contributor
timestamp: July 26, 2021, 02:31:35 pm
content: Quote from: bostonman on July 26, 2021, 02:56:20 amSimple answer for such folk: I'm not here just to burn calories.

author: TimFox
author_rank: Super Contributor
timestamp: July 26, 2021, 03:38:40 pm
content: The answer for calories burned by different physical exercise routines is not to be found in physics, but in physiology.

author: RJSV
author_rank: Super Contributor
timestamp: July 26, 2021, 05:51:53 pm
content: As to the question, about finding an equivalence between calories for an hour lifting weights VS an hour on treadmill, I recall watching a researcher monitoring a face mask, breathing hose. I think the scientists use the oxygen consumption to get parameters related to calories consumed. My guess, also, is there might be some other processes...ie Pathelogical that allow short bursts (of physical performance) in absence of correlation with oxygen consumption.Oxygen consumption, my speculation, would be able to be directly correlated, with the whole cellular energy equation, almost literally molecule by molecule.I think there must be ways of answering your question, assuming you get all the circumstances for comparing the two excercise activities.For a similar reasoning, I probably would not look at a person's body weight (changes), as that could get complex. My guess is that body water loss, sweat, would make exact calorie studies hard to control, just solely on gross weight.Perhaps, there are military conducted studies, that would supply those performance comparisons.- Rick B

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 26, 2021, 07:40:56 pm
content: Bostonman, don't take this negatively, but you are on a fool's errand. I am not challenging your intelligence or your knowledge of physics or your ability to learn. I am sure that after enough work and study you could separate out the actual work performed in the two activities (note that in the pure physics sense neither involves any work since the runner on the treadmill ends in the same state he began, and the weights also return to their original position). Work being defined as the integral of the scalar force*distance for the two activities. You can sort out all of the inefficiencies in the biological production of that force and distance. Determine the base consumption and the changes in base consumption due to the activity. Figure out how to adjust for the differences in size and physiology between you and the guy you are comparing with. And in the end it will mean nothing. The other guy won't be convinced, and you probably won't either unless the answer turns out in your favor.The only real answer when dealing with a question of the type you are discussing is what works for you. Which has little or nothing to do with physics. And which answer trumps all the gym instructors, exercise books and other advice on the subject.

author: bostonman
author_rank: Super Contributor
timestamp: July 27, 2021, 02:57:44 am
content: I wouldn't and didn't get offended at all.This thread seemed to turn into more general conversation about everyday physics and maybe I got too comfortable tossing out questions. You're correct, I'm sure I can open plenty of physics and body training books, but sometimes just talking to people helps weed out the confusion. Books go only so far which can sometimes cause limitations in learning.In some cases this thread went way beyond general mass and acceleration, and other times it seemed to be just general discussion that helped clear confusion.It's possible I just got too comfortable continuing to talk thinking it was all general discussion while others were providing answers in hopes of ending the thread with a final answer.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 27, 2021, 03:42:48 am
content: If you want an interesting example of everyday physics think of driving a car. Moving along at constant speed you approach a stop sign. What is the optimum braking profile?First step, define optimum. Shortest arrival time? Constant deceleration? Minimum jerk?The, assuming brake friction is constant over time and wheel rotation speed, what pressure profile is required to achieve it.Simple problem with much meat.

author: bostonman
author_rank: Super Contributor
timestamp: July 28, 2021, 03:25:00 am
content: If anyone cares, I had a real world discussion at the gym today about lifting weights.A few guys were doing lat pull downs and someone asked the other guy why he can do a pull up, but, he can't do equal weight on a lat pull down machine.I chimed in stating I've had similar thoughts because many people can do a pull up which involves lifting their own body weight, however, they can't pull down an equal amount.We concluded that many factors come into play such as slightly swinging the body, and, more muscles are being used for a pull up.Anyway, thought everyone would get a chuckle out of this story.

author: bostonman
author_rank: Super Contributor
timestamp: July 31, 2021, 01:45:28 pm
content: BTW, the other day I was thinking about this thread.While I've asked certain basic physics questions, or maybe I assume they are basic, and I can certainly find formulas in a book, I find that discussing certain aspects offers much more than books some time.At least for me, sometimes putting it all together doesn't come natural. While simply picking up an object is physics 101, getting down to the deep details that some responses were, or simply analogies connecting everyday real life stuff, really makes (at least myself) see common tasks in a different light.