Talk:Apparent weight/Archive 1

Page

I have started the page on apparent weight in order help solve some of the ambiguities I have seen with my students (that is rather common in many textbooks) on what a scale or balance really measures. One think is sure: a spring does not measure the weight of a body itself but another force that also acts on the body that is often referred to as the apparent weight. The difference between these two notions appears explicitely when the weighing scale used does not indicate the value for the weight that would be expected. The apparent weight is therefore not the weight. When a body is at rest in the vicinity of the Earth surface, there are at least 2 forces that acts upon this body. One is the weight, the force exerted on the body due to its own mass and Earth mass, and another one exerted, in most conditions by the surface on which the body is laying. This other force, often called the normal force (as it is perpendicular in most conditions to the surface, itself often horizontal) is sometimes wrongly called the reaction force to the weight. The values of these 2 forces are the same, in certain conditions only, when the acceleration is zero, that is rather common, leading to the confusion that should not appear in the definition of the apparent weight.

Indeed, let us consider a body of mass 'm' that lays on a horizontal surface near the sruface of the Earth. The forces that are exerted on this body are :

its weight, $W_{m}$ that is the force of gravity, which value is given by

$W_{m}=G{\frac {m\cdot M_{E}}{d^{2}}}$ , where

a force exerted by the surface on the body, that exists otherwise the body would fall through the surface.

The Newton law tells us that the total force exerted on the body,

--Nicop 08:04, 28 Oct 2004 (UTC)

The article is totally confusing to me. I do not find apparent weight a useful concept. To me the total "supporting reaction force" is the weight whether this is from the floor, buoyant fluids, or a spring scale. Accelerating frames do indeed change weight not just "apparent weight". --Op. Deo 07:52, 6 Jun 2005 (UTC)

If you define weight as the scale reading it changes, but if you define weight as the force of gravity on a body, then that is not at all changing in each frame. In order to have a consistent terminology then, weight is defined as the latter, and so we need the concept of apparent weight to express how much we feel we weigh/what a scale reads in order to distinguish them.

The fundamental problem is that every physics student knows enough to know that weight is caused by gravity, but the everyday usage is that weight == scale reading, leading to the wrong impression that scales should read the same in in accelerating frames because they "measure mass." The whole thing is just another example of how physics has devoloped a precise definition for something and students of physics necessarily have some difficulties as they readjust their lexicon.

I agree though that the current article is lacking and could benefit from some examples. --Laura Scudder | Talk 15:10, 6 Jun 2005 (UTC)

Thanks for the comment. I happend by this page and the concept of apparent weight surprised me. As you say it is a student problem. It is one which I have not thought about since my early teens.

In Newtonian physics we work with mass, the gravitational field and the resulting force, and the force required to produce acceleration. I dont think there is a precise single definition of weight. It would be of course just a name for one of the forces relevant in a typical problem. I dont think that I have never found giving one of the forces the name weight to be particularly useful, except if I was using the word weight in its traditional everyday sense. That is in a description of the mass of something that was required only to be of low precision rather than for use in a physics equation. ( I dont use weight in my normal repertoire of equations)

Here is an example - I might say one apparatus is heavier than second, and I might describe the processes of determining their mass as weighing them with scales of some type (either force meters - or mass balances). But I would be doing this because I was doing some everyday activity such as determining what postage was due to be paid on the apparatus. If I needed the weight in order to do some engineering calculations concerning a cable to support the apparatus I would convert the result of my weighing experiment to force units, or if I wanted place my apparatus in a rocket to go to space I would use the weighing to determine the mass units.

After looking at this page I wandered around and looked at the weight page. I was disturbed to see that section 2 "Weight as Force" brought in apparent weight. I then looked at their talk page and was quite shocked at the confused approach.

What are these wiki articles intended to be? Articles for the general reader? Tutorial articles for science students? Rigorous scientific articles? Each question requires a different approach, which is hopelessly muddled in the present article on weight. As afar as apparent weight is concerned. I would not touch it for the geneal reader. I would just say that weight as measured by a spring balance changes. For the early student I accept that some teachers might want to use the concept of apparent weight. But for the physicist and particular engineers such as space engineers then we dont really use weight at all in our equations.

I am a novice at Wiki, and came here very recently to contribute some history that was missing. I was disturbed to find that the place seems infested with fellow scientists and IT people (perhaps not surprising of course). I have aready done a few edits on science topics which I felt were very incomplete and I have nore some science articles which are crying out for me to write. However, I am seriously doubting whether I should jump with my opinions into the article about weight and apparent weight. Op. Deo 19:22, 6 Jun 2005 (UTC)

Wikibooks tries to accumulate cohesive textbooks, unlike wikipedia, which is of course an encyclopedia instead of a textbook. The goals of wikipedia science articles vary a lot between for instance apparent weight and interaction picture, but Wikipedia:WikiProject Science tries to set some guidelines for popular articles at least. --Laura Scudder | Talk 01:50, 7 Jun 2005 (UTC)

I cleaned up a little and began to add examples. More coming within the day. StuTheSheep 18:15, Jun 20, 2005 (UTC)

Added example, clarified some terms. I'm having issues with formatting the equations and could use some help, so I'm going to add a cleanup tag to the talk page. StuTheSheep 14:14, Jun 22, 2005 (UTC)

I found this page very useful in understanding "g force" and how the human sensation of increased or decreased "apparent" weight (or weighlessness for that matter) relates to the body undergoing acceleration. Thank you for a very concise explanation, and the example. Well done. A satisfied reader.

Hey shouldn't m/s² be expressed as m/s^-2? or does it mean the same thing?

m/s²=m.s^-2

Let start from definitions and physical nature of the force "weight": weight is electro-magnetic force which shows the strength of interaction between the object (whose weight we are measuring) and the support. In majority of cases that interaction is connected/related to gravitational attraction between the object and other body (may be the Earth) - and therefore many students (even PhD students - I have seen some on their defenses, that is one of my questions to check whether PhD student is ready for defense) think that weight is a gravitational force (which is WRONG). The simple explanation is that we need to consider TWO pairwise interactions: between the object and the support (electro-magnetic force), and between the object and the Earth (gravitational force). There are two forces acting on the object: the first one from the support on the object (usually points UP) -- call it Normal force, the second one is from the Earth on the object (usually points DOWN :-) -depends on definition DOWN-) -- call it Gravitational attraction force (or just simple gravitational force). If we assume acceleration of the object a = 0 m/s^2, the absolute values of two forces acting on the object should be same, therefore Normal force is equal to Gravitational force. But forces are not coming along. For each real force there is counter-force (see the 3-rd Newton's Law). The counter-force for Normal force is also of electro-magnetic type, acts from the object on the support and called WEIGHT. If a = 0, gravitational force = normal = weight. Confusion comes from here. However, if acceleration is not 0, gravitational force is not longer equal to normal force = weight, so weight is not equal to gravitational force. E.g. weightless condition does not imply zero gravitational force. (Weightless condition implies Weight=Normal force=0 -- the object does not act on support -- like walls/floor of spaceship, etc. But there IS gravitational force action on the object from the Earth/any other planet, etc.) As for 'apparent weight' there is no such concept in Theoretical Physics. First of all -- what is the nature of that force (what type of interaction is responsible for that force) -- is it gravitational? is it electro-magnetic? If it is gravitational (as sometimes claimed in some textbooks)-- it should NOT change with the motion of the object and weight should be constant -- which is not the case. If it is electro-magnetic interaction -- how it can change the type of interaction? Just to be convenient for explanation? Couple of my friends joked that the 'concept' was introduced to tell below-2.00-GPA-students to grasp the concept of weight. It is possible to continue about the 'apparent weight' and its internal contradictions, but that will not contribute to the understanding of the notion of 'weight'.

Centrifugal force? On a physics page? SMACK! That needs to be fixed... —Preceding unsigned comment added by 68.62.186.235 (talk) 21:26, 1 October 2007 (UTC)[reply]

I just made a change that explains why it is centrifugal rather than centripetal force. It has to do with the convenience of using a rotating reference frame when considering the forces acting on a object on the Earth's surface. I will admit that I was thinking it should be centrifugal at first, but I came to this final conclusion after reading more about it. undefinedvalue (talk) 05:13, 7 December 2007 (UTC)[reply]

Confusion about purpose

I am confused about why apparent weight is being discussed is such detail when it is just another name for the normal force when a gravitational force is involved (at least to my understanding). The examples provided just seem to include the various forces that you could consider acting on an object that might alter the normal force acting on that object. From this point of view the article seems to simply be a general physics lesson. I think that this article does not need the extensive "lesson" section, where instead a single example that clearly shows what apparent weight is would keep this article from becoming "too long; didn't read." undefinedvalue (talk) 05:49, 7 December 2007 (UTC)[reply]

This article pushes a confusing original-research term, and the whole thing should be scrapped and merged with g-force

This whole article on apparent weight should be scraped, since it involves a distinction (which involves original research and original nomenclature, so far as I can see) the essence of which is silly and unneeded. Somebody has gotten the foolish idea that there is something more "real" about the weight of an object at 1-g and at rest, than its weight in any other circumstances (in a rocket or on the moon, etc), and wants to call the 1-g weight the "weight" or "real weight" (LOL) or "actual weight" or "really-truly-true weight (foot stomp)." This is the reason we have the concepts of rest mass and invariant mass in physics. If you want an object's weight you can measure it on a scale, or you can calculate it as the product of its rest mass and the proper acceleration in its rest frame (as measured by an accelerometer). This will provide a number which is Lorentz invariant (as both rest-mass and proper acceleration are invariant, thus so is their product). Which is not suprising because the weight of an object is the same for all observers. But there is nothing more "real" about the weight of an object on the Moon than on Earth. They are different numbers, to be sure, but one is no more real than the the other. Why confuse the reader? S B H arris 19:31, 7 August 2009 (UTC)[reply]

The term "apparent weight" is used in the physics literature, to distinguish between "weight". The weight of an object in some given gravitational field is a fixed number, while the apparent weight is minus the normal force acting on the object. You can reformulate that in terms of the stress tensor integrated over the surface of the object.

But that's not the definition given in the article, and it's also not useful enough to write an article about. If I'm hanging in a harness that takes 5/6th of my weight off a scale that I'm standing on, my "apparent weight" could be said to be the same as if I was on the moon. WOW! But so what? The straps digging into my crotch tell me that my total weight is the same as ever, no matter what the scale says. I've just forgotten to add spring-scales to the lines holding me up! See my comments in the TALK section for g-force where I point out that various bouyancy forces (which are all induced by differential surface pressures) are just a spread-out version of this parachute harness. They don't make weight go away, but they do redistribute it so that (one of) your scales doesn't see it. But it's still there, somewhere. S B H arris 01:23, 8 August 2009 (UTC)[reply]

I think, "g-force" is a special case of apparant force when an object is accelerating. It doesn't apply when an object is at rest. E.g. consider the case of floating frogs in a strong magnetic field given in this article. Count Iblis (talk) 20:14, 7 August 2009 (UTC)[reply]

G-force most certainly applies at rest, as you would agree if you were on a planet with a 2.5-g gravity field, and you were being squashed by the force needed to resist gravity (say you were on a couch in a balloon floating in the atmosphere of Jupiter). This is force. It is stress. But it's produced by the couch and your motion, not by the gravity of Jupiter (jump from your balloon and it disappears in your first seconds of fall, until you begin to feel aerodynamic forces). The magnetically levitated floating frogs are rather a special case where most g-forces might indeed not exist, since the water is levetated directly, much as happens with gravity, not by outside pressures. In that case, an acclerometer where the test weights were made of little balls of water, would indeed measure close to zero-gee. So the definition of being "accelerometer-measured" acceleration continues to hold. This is one of the few cases were an acclerometer does measure weight/mass, but does NOT measure proper acceleration. Another is where the test masses are charged and the weight from their proper acceleration is removed and adjusted for by an electric field (a design actually used in sensitive gradiometers/gravitimeters). However, absent gross effects of this type from divergent E or B fields acting on charged or diamagnetic objects, the idea of weight/mass = proper acceleration = g-force = acceleration away from free-fall is a pretty good definition. S B H arris 01:23, 8 August 2009 (UTC)[reply]

Apparent weight is not the same as g-force, however this could be merged with weight. It is a term used a lot in physics and the current article would totally confuse my students. Sophia ♫ 22:31, 27 May 2010 (UTC)[reply]

Apparent weight is used a lot in physics?? WHO uses it? And whatever do they mean by it? S B H arris 00:06, 28 May 2010 (UTC)[reply]

Ok more precise! The g-force article defines it as the acceleration that a body undergoes. Apparent weight would be the product of this and the body's mass, measured in Newtons. That is why I don't think those two articles should be merged. As for "who" uses the term "apparent weight", I will find refs if you want but in the UK it is a common term, certainly amongst the physicist I know to refer to any "weight" that is not equivilent to that of the body at rest. Sophia ♫ 12:54, 28 May 2010 (UTC)[reply]

But what you speak of W = mass*g-force, that W is weight (or at least they have the same magnitude, and in that sense are the same). Thus, Weight = -(mass*g-force). If you take m*g-force as a vector and weight as a vector, one has the opposite sign of the other, since one is a counterforce of the other. But my weight on the moon is different from my weight on the Earth, and it would be silly to think of one of them as my "apparent weight" and the other as my "real weight" (or just "weight"). They are both my weight, and there is nothing "apparent" about either of them (or, if you like, nothing more apparent about one than the other). The very word "apparent" makes false suggestions by implication. Out in deep space or in free fall, I weigh nothing! Not apparently nothing, but really, really, truly honestly nothing. Which is what it's called weightlessness, which is where zero-g (meaning zero-g-force) and zero gravity redirect. Thus:

{\vec {W}}=-m{\vec {g}}

Where the vector-g is the g-force.

I see that I'm going to have to go over and fix the weight article, now. Yes, the apparent weight article needs to go. The NIST definition is just the same as the math above. It's minus mass times proper acceleration. S B H arris 22:51, 29 May 2010 (UTC)[reply]

"Weight" and "Apparent weight"

See Talk:Weight#"Weight" and "Apparent weight": Contradiction and Confusion. —Preceding unsigned comment added by 86.136.27.202 (talk) 21:05, 2 February 2010 (UTC)[reply]

Mistakes in recent rewrite?

The most important thing that went wrong in the new version was the ommision of the fact that apparent weight is entirely due to internal stresses in the body and is thus caused by short range forces exerted on the body. This crucial fact was omitted and a whole chain of erroneous reasoning was build on that. Count Iblis (talk) 14:11, 29 May 2010 (UTC)[reply]

The concept of apparent weight is also meaningful when e.g.an object in an elevator is falling to the floor of the elevator.--Patrick (talk) 16:12, 29 May 2010 (UTC)[reply]

I don't see how introducing an effective weight is useful to explain a larger acceleration toward to floor.... Count Iblis (talk) 16:31, 29 May 2010 (UTC)[reply]

No, what went wrong is the failure to realize that "weight" is the counterforce (reaction force) to whatever force induces proper acceleration, which is the acceleration away from the state or path of free fall. That proper acceleration is the same as g-force. It is what an accelerometer measures. These concepts of (proper acceleration and g-force) are somewhat subtle concepts, but once grasped, weight is easy. For any mass, weight is simply the counterforce vector that points in the opposite direction to the proper acceleration, and differs from it by the factor of mass:

{\vec {W}}=-m{\vec {g}}

Where the vector-g is the g-force.

There is no such thing as "apparent weight", or at least no place for it. There is just "weight." All the "short range" stuff is due to the fact that gravitation does not cause proper acceleration, or g-force, and isn't measured by an accelerometer. Thus, it contributes to g-force and proper acceleration only indirectly, via the mechanism that some mechanical force must induce g-force in a gravitational field, if the object is not in free-fall. And since every mechanical force exists as an action-reaction pair, the weight is the reaction force to that mechanical force.

Doing a read through the g-force article, which is finally correct, will help everybody here (also a read through the proper acceleration article, which is about the same thing, but more technical). The g-force for an object sitting on the ground is 1 g (9.81 m/sec^2), upwards. The proper acceleration is 1 g, upwards. The mechanical force that causes this proper acceleration is provided by the floor/ground, and is $m{\vec {g}}$ (a vector force), upwards. The opposing downward force, which "explains" in this frame why the object undergoes no coordinate acceleration, but is motionless (zero net force), is the weight. Which is $-m{\vec {g}}$ , a vector pointing downwards. Remove the floor (or let the elevator cable snap), and the coordinate acceleration now is non-zero (as the object goes into free fall), but the g-force, proper acceleration, and the weight, all now go to zero. We have weightlessness and zero-g (meaning zero g-force). It's actually quite simple once you start from the proper acceleration and "accelerometer-measured" point of view. S B H arris 23:11, 29 May 2010 (UTC)[reply]

In a accelerating lift the weight per unit mass of an object seems to have a certain value. This manifests itself as the negative of the g-force when it rests on the floor, and as the acceleration with respect to the lift when it is falling (with the g-force being zero). It seems odd to say that this value is undefined or irrelevant if the object is not resting on the floor of the lift.--Patrick (talk) 23:36, 29 May 2010 (UTC)[reply]

Sure, but the value of the weight when the item contacts the floor, is determined by the fact that the floor must push on it with enough force to keep it in the accelerated frame of the lift. In a building on earth that acceleration is due to the acceleration needed to counteract gravity (an acceleration upward, which must be produced by the floor) plus the physical (coordinate) upward motion of the lift (which must also be imparted by the floor). They add in the upward direction (that is what an accelerometer affixed to the elevator would measure as total acceleration, and the force they produce on a mass is the normal force), and the object's "weight" is the counterforce in the opposite direction, which is necessary to produce ZERO motion in the lift frame. To help you see this, it's better to remove the messy Earth and gravity, and have your "lift" be a chamber in a rocket in space, with an item (perhaps you in a spacesuit) is floating in the middle of it. Now, does the object (or you) have weight? No! Turn on the rocket at 1 g, and you might think that everything suddenly "acquires" weight. Not so, till you contact something. The floor rushes up toward you, to be sure, but until you hit it, you feel fine, and weightless. Basically (moving our point of view to the inertial frame of the floating object or floating you) you're still hanging in space, and until the rocket comes up and hits you, you're weightless. See the point? It's undefined. If the floor of the rocket had a hole in it, you'd simply pass though and the rocket would pass on its way and nothing ever would hit you, and you never would feel any "weight." So it is indeed an arbitrary idea. "Weight" only appears when something must "push" you out of free-fall, or the free-fall frame (inertial frame).

I might add that the acceleration that an elevator floor comes toward you if you start out in free fall at the ceiling, depends what the elevator is doing. If it's falling toward the Earth with acceleration -a relative to the shaft, the floor comes toward you with slower acceleration than 1 g (it will be g-a, all the way to zero, if a = g and the elevator is in free-fall, too). If the elevator is stuck in the shaft, it's exactly 1 g. If the elevator is doing up the shaft with some acceleration a relative to the shaft, then you see the floor come toward you with acceleration g + a. But in all those cases, you don't care until you hit the floor. If the floor has a hole in it, the elevator keeps going right on by, upward, and you pass through going toward the earth at 1 g (relative to the shaft). In none of these cases do you have any weight. When you contact the floor, your weight is m(g+a) for a rising elevator, m(g-a) for a falling elevator, and mg for a stuck one. This is just your weight, and is what a scale would read. There is no such thing as "apparent weight." This entire article should be deleted. There's nothing to merge WITH. It's plain WEIGHT that is related to g-force, not apparent weight.S B H arris 01:14, 30 May 2010 (UTC)[reply]

What is wrong with defining apparent weight as the integral of sigma(n-hat) over the boundary of the object? To explain this in more detail, note that the local tension in an object in some arbitrary x-hat direction is denoted as as sigma(x-hat). This is defined to be the force per unit area with normal x-hat that the side that x-hat points to exerts on the opposite side accross the surface. Then integrating sigma(n-hat) where n-hat is the outward normal over the boundary of the object gives you exactly what you would want to call "apparent weight". Count Iblis (talk) 01:20, 30 May 2010 (UTC)[reply]

What? How is this different from minus the mass of the object, times its proper acceleration? S B H arris 02:39, 30 May 2010 (UTC)[reply]

As many times that you say it should go - it is a term that is widely used and would benefit from a user friendly explanation. Weight is not acceleration as you obviously know, which is why I thought that merging it with the g-force article would be confusing. I'll go have a look at the work that you have done there to see if it is a better fit, but whatever, it is a term that needs explaining. [1] [2] [3] [4] Sophia ♫ 11:53, 30 May 2010 (UTC)[reply]

The surface integral of the sigma(n-hat) taken just within the body is not mass times acceleration, because this only takes into account short range contact forces. The whole point of apparent weight is that you want to have a measure of how heavy you feel. If you are accelerated at a million g in a gravitational field you would still feel weightless. It is only when you are accelerated via short range contact forces that you'll feel anything. This is because your whole body has to be acclerated and if the external force only acts on the atoms on your skin, then thee atoms in your skin have to exert a force on the atoms that are a bit deeper etc. etc. So, the force will be transferred to your whole body via internal stresses. Therefore, you you consider the surface integral of the sigma(n-hat) taken just within the body, you'll get a measure of the of apparent weight. Count Iblis (talk) 14:09, 30 May 2010 (UTC)[reply]

Sigh. Look, you guys, will you please just READ the articles on g-force and proper acceleration (which is measured in g-force units)?? These are actually well-known and long used terms in physics and engineering (google them), just as "apparent weight" is not. Proper acceleration and g-force, when multiplied by mass, are equal in magnitude to the acceleration that produces "weight". Not only "weight" as defined by NIST, but also the "apparent weight" that you seem concerned about, here.

g-force and proper acceleration are not the same as ordinary acceleration, which is coordinate acceleration, dv/dt. Rather, these accelerations (for g-force is not a force but an acceleration also) refer to acceleration AWAY FROM the free-fall or inertial conditions, also called the inertial frame. Thus, they automatically ARE the acceleration you FEEL, the short-range stress-mechanical forces inside your body, and (when multiplied by mass) are the weight you measure (though in the opposite direction, as a vector). These terms automatically exclude any forces and accelerations due to gravitation: an object falling under the influence of gravity is in free fall, and has no proper acceleration, experiences no internal stresses (absent tides), and has no g-forces. Also, if it's an accelerometer, it will read zero. It is is in zero-g. It has no weight. It has no apparent weight. What could be clearer?

The equivalence principle of Einstein guarantees that an object in accelerated inertial motion due to free fall in a gravitational field, will feel the same as a person in free fall in conditions where there is no gravitation. Thus, it is only acceleration AWAY from free fall which produces weight. Gravitationally induced free fall produces no weight, and gravity per se produces no g-force. Nor does it produce a reading on an accelerometer-- it's the mechanical opposing forces that do that.

Multiplied by MASS, the proper accceleration or g-force gives you a vector which is the normal force on a flat surface (the stress tensor if you like, but that's simply what Count Iblis is talking about). It is also called the specific force. In magnitude it is exactly the weight, but acting in the opposite direction. It's the force that keeps the body from moving in an inertial, or free-fall path. Example: for a person standing on the force, this specific force is provided by the floor, acts upward, and prevents the person from going into inertial motion (which on the Earth, would be a fall toward the center of the Earth). The opposing vector force is the person's weight, which explains, in the accelerated Earth-surface frame, why the person is not moving (there are two opposing forces acting on him-- his weight force pointing down, and the normal force or stress tensor pointing up). There is no "apparent weight" here; there is just weight. On the moon, it would all be 1/6th as much for all forces. Bouyancy, as I explained above, does not affect weight at all, so long as the scale is chosen properly (since bouyancy merely spreads out the AREA acted on by the forces, but does not CHANGE the total forces, or weights, in the slightest).

On a rocket in deep space accelerating at 1-g, it's all the same. Again, as on the surface of the Earth, the occupants would be in free fall if it were not for the floor which provides the contact-force that is transmitted through their shoes and into their body, and gives them a proper acceleration (which in this case, for slow speeds, is the same as their flat-space coordinate acceleration dv/dt). Interestingly, in special relativity, as they go faster with the same rocket push, their dv/dt and coordinate acceleration falls, but their proper acceleration and weight remain constant (it's Lorentz invariant). Again, accelerometers (think of one on a rocket approaching the speed of light) read proper acceleration, which is the acceleration which produces g-force and weight, but NOT coordinate acceleration (not surprisingly). How fast the rocket accelerates, depends on what frame you observe it from, but the weight the passengers feel is what they feel, and has only one answer, and that's what their accelerometer and their bathroom scales tell them. And of course it is what they feel.

Anyway, you guys are laboring to fix up a definition which has no definition and is meaningless, or else (if you do it the other way) you are laboring to introduce a concept which already has good and long-accepted term in physics that refers to it (weight), and no less than three terms for the magnitude of the associated acceleration that acts on mass to produce it (specific force, g-force, proper acceleration).

If you want to continue to argue this point, take any of the specific situations above, and show me (with math) how your definition differs from weight, and how the magnitude of the key stress-producing acceleration you refer to differs from specific force, g-force, and proper acceleration. S B H arris 19:04, 30 May 2010 (UTC)[reply]

Comment I can't see any difference between this article and weight. Apparent weight is just the weight that an object exerts on scales, which is how weight is usually measured anyway. When most people weigh themselves on scales, they're measuring apparent weight, since there's about 100g of buoyancy from the air (depending on altitude).

I also find that the weight article and the apparent weight article in total have less than 30k of actual, readable text anyway (and I'm sure there's large overlaps), so there's a strong case for merging this back to there. Weight is heavily influenced by g-force anyway. - Wolfkeeper 20:16, 30 May 2010 (UTC)[reply]

Actually, weight is ENTIRELY determined by g-force (proper acceleration). It is "local scale weight" that is changed by buoyancy, but local scale weight is also changed by somebody partly supporting the object by any means that the local scale doesn't "see." So what?

Let me try again with this. Suppose you and I are standing next to each other and we're both standing on separate bathroom scales. I reach over and pull up on your belt and take 2 lbs off your weight. Do you now weigh less? Good question. Notice that I now seem to weigh 2 lbs MORE. But weight for our system has been conserved and didn't change. All I did was hide it from your local scale.

Buoyancy effects do exactly this. They merely transfer some of the weight of an object to the surrounding large area of Earth's surface (which supports a slightly heavier atmosphere and experiences more pressure) so it's not seen. But it's still there. So it's rather odd not to count it. There is a difference between an object's weight and an object's "badly measured weight." The last coming from measuring the weight of something that has something else partly supporting it. The NIST definition, which defines weight in terms of forced needed to oppose proper acceleration, avoids all that. My proper acceleration is produced by the ground (and anything else holding me up), all pushing up on my MASS , and my weight is the exactly-equal force that pushes back. That force is the force that would be necessary to put me into free fall in my ground frame. Remove my supports, and that weight-force WOULD put me into free fall! S B H arris 20:34, 30 May 2010 (UTC)[reply]

Sure, that's the physics. But we're talking about concepts in encyclopedias. Conceptually, apparent weight is closer to weight than g-force. I mean you could merge weight, g-force, apparent weight and proper acceleration into one big article. They're all almost exactly the same thing, or incredibly interrelated. But we would have article length issues if we did that. But apparent weight goes into the weight article just fine.- Wolfkeeper 20:53, 30 May 2010 (UTC)[reply]

Yes it does. But if we put it there, can we not let "apparent weight" as a term apply only to bad weights measured by instruments which had no way to correct for bouyancy? That's the biggest use I see for the term on Google (http://physics.info/buoyancy/) IF you subtract out the many "mirror" definitions that stem from our bad article right here on WP in the first place. Otherwise I fear we're going to promote the idea that only what you weigh on the surface of the Earth is your "weight", and that what you weigh on the moon or in a rocket or elevator is just your "apparent weight," and is somehow false therefore. No. Your weight on a scale which misses you buoyancy lift is "wrong" because the scale doesn't cover the whole surface of the Earth, so the scale fails to measure the true weight, which is the force pushing downward to the Earth. And ISO agrees. The problem is that the scale isn't entirely under you, is all, not that your weight is less. You weigh just as much floating in water as you weigh floating in a water bed! (It's just that the support is even better.) The other different weights for you in other differently-accelerated reference-frames, however, are perfectly correct, and all correspond with the NIST definition of weight. S B H arris 21:02, 30 May 2010 (UTC)[reply]

I would think we could do that (but who knows in this place- herding cats is often far simpler, at least you can stick them in cat boxes, whereas unaccountably the policies don't let me do that with other editors.)- Wolfkeeper 21:56, 30 May 2010 (UTC)[reply]

One other thing I noticed, actually the apparent weight article is already subarticled off the weight article anyway.- Wolfkeeper 21:56, 30 May 2010 (UTC)[reply]

Yes it is, and doing so constitutes a POV fork to promote a wrongheaded and bad term (or at least a wrongheaded definition for it). And this kind of thing is starting to creep into the weight article now, where the idea is being promoted that if you have no weight in a free-falling elevator, that this is only your "apparent weight" and not your "real weight." That is utter crap, and is what happens when we allow abominations like the apparent weight article to proliferate to try to cover situations other than buoyancy and perhaps other types of mechanical support that a reference scale (which is supposed to be measuring the weight) doesn't "see." S B H arris 18:55, 25 December 2010 (UTC)[reply]

Setting assumption

This articles assumes weight on Earth or near it, or in a location that is similar to that environment, shouldn't there be coverage of apparent weight in more extreme conditions? 76.66.193.224 (talk) 07:09, 3 June 2010 (UTC)[reply]

Redirect

This article contains WP:OR not supported by sources, therefore I'm redirecting to weight. Gerardw (talk) 10:52, 10 May 2011 (UTC)[reply]

I disagree. Count Iblis (talk) 14:15, 10 May 2011 (UTC)[reply]

Disagree about what? That it should be redirected, or the lack of sources?Gerardw (talk) 20:56, 10 May 2011 (UTC)[reply]

I disagree with redirecting. Redirecting to "Weight" is problematic, because that article doesn't cover this topic well, it actually has a small section on it that refers to this article. About sourcing, I would agree that a few sources for the terminology could be added, but that's a trivial matter. The content is not OR in the usual sense of the word (i.e. editors here doing there own research which yield results that are either inconsistent with current knowledge or which yield valid results that are new).

What is going on here is that editors have been freely giving elementary, high school level examples, which is allowed per common practice here on Wikipedia, (see e.g. Methods of contour integration); obviously, copying examples from textbooks is not allowed, so sourcing according to the most puritan Wiki-standards is not possible here. Thing is that the article is still reporting on the known physics knowledge as it exists in the real world; the examples are necessary to explain the concept. The whole point of an example to clarify things, so demanding verbatim verifiability of an example is besides the point. Count Iblis (talk) 21:20, 10 May 2011 (UTC)[reply]

I'm not familiar with the concept, it seems kind of bogus to me, and there are no sources listed ... NIST site referenced doesn't use the term "apparent weight." Gerardw (talk) 21:28, 10 May 2011 (UTC)[reply]

I think a google search will bring up quite a few webpages, so this at least shows that the concept is used and is notable. Now, it is true that one may criticize the concept, e.g. SBHarris has done so, and argued that the article on G-Force covers basically the same thing. However, what we should not do on Wikipedia is write on physics (terminology) from our preferred point of view, if that would mean not covering existing (and legitimate, so no crank sources allowed) points of views.

A good example is "relativistic mass". I don't like this concept at al, as it is synomymous with the energy. But in the real world, this term is used. I can argue all day long that in my advanced theoretical physics textbooks this is not used, but as long as high school level textbooks and high school teachers use this, I cannot delete "relativistic mass" from wiki-articles. Count Iblis (talk) 21:49, 10 May 2011 (UTC)[reply]

Okay. Must have added this in the past 30 years. Not sure it actually improves understanding of anything but agree NPOV means we have to include it. Gerardw (talk) 21:56, 10 May 2011 (UTC)[reply]

I don't mind having the article, so long as the thing notes that the concept is NOT well-defined, and has about as many meanings as authors. I suppose it would be a good place to split some of these out. Some people use "apparent weight" to be what your scale WOULD see if your scale saw all support (so it's not affected by buoyancy). Other people use the term to mean what your spring scale DOES see, so it IS affected by buoyancy (if the scale is placed wrong, or is small and misses the weight of the bouying fluid). Other people want to correct for inertial effects, so they say your apparent weight in the space shuttle is zero, but your weight is some other number (this basically trades the definitions of weight and apparent weight). And so on. The ISO defines weight as -m (g-force), which neglects bouyancy effects, but does take into account all acceleration/inertial "forces." Definitions which differ from this need to be carefully defined, and split out, and an article on "apparent weight" would do that. It it such a messy idea that I haven't attempted it. Does a man standing in 2 feet of water, or a penny submerged in a beaker of water, have less "apparent weight" than if the water wasn't there? I dunno. By which scale? A scale weighing the whole pool of water or beaker will see the full weight of the added object, but a scale at the bottom of the pool will not. It's not apparent to me what is the "apparent weight", because it's not a clearly-defined and widely-used term that means ONE particular thing. S B H arris 23:51, 10 May 2011 (UTC)[reply]

I agree with these points of criticism. The problem is that the "definitions" used are based on phenomenolgy too much and one needs to have a good theoretical starting point that captures exactly what one wants. My personal definition would involve the integral of the stress tensor over the surface of the body, but taken just inside the body. Then, if you are accelerated by any long range force that acts uniformly per unit mass, then that won't contribute to the apparent weight. Short range forces acting only on the boundary of the body, will cause deformations leading to stresses inside the body.

From the point of view of "apparent weight", you are weightless when floating in the ISS, but also if you manage to use extremely strong magnetic fields to float here on Earth. Count Iblis (talk) 16:51, 11 May 2011 (UTC)[reply]

Seems to me we need RS that have definitions of the various uses of the term. Gerardw (talk) 19:33, 11 May 2011 (UTC)[reply]

[5] here it says your apparent weight is what the scale says in an elevator. Thus your apparent weight in orbit would be zero, but not your weight. [6] here an associate professor of physics says the same: that apparent weight is what you feel and the scale measures, but your REAL weight doesn't change in an elevator! All of this different from the ISO definition, and all of it suggesting that weightlessness in orbit is only an illusion-- it's only apparent, and not "real." Duh. The Boston U physics department in this next bit claims that apparent weight is what you have after you remove bouyancy forces. [7]. It never occurs to them that you might not have the scale right under the bouyed-up object. Do they think that if I pull up on your belt, that you weigh less? That your apparent weight is less, just because some is on your belt and not all is on your feet? S B H arris 20:21, 11 May 2011 (UTC)[reply]

Comedic gold

We know that the weight of an object in its own reference frame (the one where it is at rest) is just the negative of its mass times its proper acceleration. Or, it is minus mass times the g-force. W = -m (g-force).

However, this article now defines the different concept of "apparent weight" as: The mass of an object times the g-force is the sum of all forces except gravity, so the apparent weight with respect to a structure is the sum of external non-gravitational forces, plus the mass of the object times the acceleration with respect to the structure minus the g-force. What?

The emperor has no clothes. This article has no idea at all what it's talking about. Which is not surprising since what it's talking about has no good definition in physics. S B H arris 18:19, 11 November 2011 (UTC)[reply]

As far as I can see, according to Weight your first statement is not necessarily correct. The first definition in that article says that weight is W = mg, so it remains unchanged however a body is accelerated, provided only that g does not vary. The article then goes on to give an alternative and completely different "operational" definition that it says is "the same as what some other sources term the object's 'apparent weight'". Therein lies the problem, and the reason, presumably, why we originally had two different articles. I half remember (but too lazy to go back and check the history now!) that at one time the Weight article did not admit the "apparent weight" definition, but now that it does, it would seem to make sense to merge this article to Weight. What say you? 81.159.104.115 (talk) 03:31, 19 November 2011 (UTC)[reply]

At the time I argued that the definition of W = -mg is somewhat silly. As opposed to the (to me, correct) alternative: W = -m(g-force) = -m (proper acceleration). The definition W = -mg where you must calculate (not measure) g, is bizarre. This implies that your "weight" in a fighter jet at 12 g's, as you're being crushed and black out, actually has not changed from when you're standing on the ground. And also that your weight in a freely-falling elevator, as you float around, has not changed-- that your weightlessness there is also an illusion. You're not weightless in free fall according to this, even though no experiment can tell that you're not weightless (absent tides). You're supposed to calculate "g" and use it to find your weight, which you cannot measure (!). And that in the space shuttle, you are not actually "weightless" but have 90% of the weight you do on the ground (since g is supposedly about 90% of what it is on the ground, although the method of calculating "g" for this situation is another comedy routine, as we must all pretend that it involves distance only, and not motion).

None of these definitions are in line with experience, or the way the word "weight" is commonly used, and in particular with the way the word weightless is used (as in a weightless astronaut). W = -mg also does not agree with the equivalence principle, which treats all type of proper acceleration as exactly the same, since there is no way to physically distinquish between them locally (if Einstein is right). And the operational definition doesn't try to (though I got argument on this also). However, the point is that I was outvoted and the result is what you see: two different contradictory definitions in the main article on weight and a host of other definitions in THIS article, I think as some kind of atonement for that. Or perhaps as a result of the confusion of that. I see no way of fixing it so long as the W = -mg people require us all to accept W = -mg, and require us to define "g" as GM/r^2, and no correction for the Earth's rotation allowed, or for motion, or anything else. THEY get to define "g." Not according to measurement or what they feel, but according to what they please. Feh. S B H arris 04:32, 19 November 2011 (UTC)[reply]

W = mg, as I initially understood it from Weight, also contradicts my everyday understanding of "weight". However, looking further, I'm a bit confused about what g is supposed to represent in that definition. In the lead section of Weight it explains that W = mg implies, for example, that objects in free-fall still have their usual "full" weight, just as in the examples you give. The assumption seems to be that, at or near Earth's surface, g = 9.8 m/s^2 whatever the object is doing. However, later in that article, the spookily similar ISO definition, F_g = mg, seems to allow that g varies depending on the reference frame, so that, for example, objects in free-fall have no weight when weighed in the "natural" reference frame. Is it definitely the intention that there should be this difference? Why doesn't the W = mg definition work the same as the F_g = mg definition? 86.177.104.26 (talk) 12:43, 19 November 2011 (UTC)[reply]

You are correct that the W = mg assumes "g" is 9.81 m/s^2 at the Earth's surface no matter what the object is doing (including falling or orbiting) and of course no matter what the Earth's surface is doing (since it can be rotating). This definition just defines weight as a sort of mass with different units, but one that you can only detect by looking at the force that mass exerts due to gravitation at the Earth's surface, when not in motion. I think this weight definition could be corrected to a different value for the Moon or Mars, so long as no motion was involved. Basically it demands that "weight" is something that can only be imparted by a gravitational field to a motionless object.

In the F_g = mg definition of the ISO, the "g" is not the same g as for the first definition (except in the special case where you ARE standing on non-rotating Earth or other non-rotating astronomical body). Instead, the ISO "g" is the local acceleration of free-fall in the object frame, a more general concept which is -(proper acceleration) or -(g-force) or -(four-acceleration). These are all the same, since (by definition) "free-fall" is the geodesic path (also the inertial motion path) where all these quantities are ZERO. The acceleration you take away from this free-fall path through 4-dimensional space-time, is what your accelerometer measures, and what you feel as g-force. Multiplied by your mass, it's the force that makes you deviate from a geodesic (inertial) path. Think of the pilot-seat of a rocket out in space, pushing you away from what would ordinarily be free fall. The equal-and-opposite reaction force to that of the seat (the push of your butt against the pilot seat) is your ISO weight, which is where the minus sign comes in. The ISO gets away from the minus sign by defining the weight as the force that would be needed TO GET YOU INTO FREE FALL, instead of the same vector with opposite sign, which is the force (see four-force) that is the force that is making you DEVIATE FROM FREE FALL (or making you deviate from an inertial path or a geodesic path-- same thing). Otherwise, it's the same concept as all those links.

I think some people would like to use the ISO definition as "apparent-weight" (this article) but they still have to decide what they want to do about buoyancy, which doesn't enter in to the ISO definition. A balloon has the same ISO weight whether in air or vacuum, since the "weightlessness" of a balloon in air is only due to lack of scales underneath it, since its weight continues to be distributed over the planet's surface. The same is true for all weights that seem to be reduced by suspension or bouyancy-- this is a "weight scale placement" problem, not an actual weight reduction. A passenger jet on the runway has the same weight as one in flight, but instead of being distributed to the runway, it's distributed to all the land under the jet (ultimately the entire Earth surface). The air provides "lift" and something solid provides that reaction force to that air, and so on. S B H arris 20:50, 19 November 2011 (UTC)[reply]

So is it really the consensus view (even if not your view personally) that Weight should lead with a definition that gives orbiting astronauts and people in free-falling lifts the same weight as they have when they are standing on the surface of the Earth (well, give or take the slight reduction in gravity a few hundred miles up)? Is that the way that physics books mostly teach it? I am no expert, but it does not seem very sensible to me. 86.181.169.8 (talk) 21:58, 19 November 2011 (UTC)[reply]

Yep! I kid you not! It's not a slight reduction, as the 200 miles of altitude on top of 3960 gets g down to 90% of ground value, due to the square. Now you see why this is so darkly humorous (the title of this section). Some undergrad physics books teach it that way and others not, but enough teach it the way you note that I was not able to argue for getting the stupid and limited definition out of the damned article. So now we only have "apparently weightless" astronauts, per official Wikipedia.

Of course, doing it this way makes what college intro-physics texts the "last word" on defining "weight" (an approach I think has its own problems) but I was outvoted on that also. Finally, I suspect that even into-undergrad-physics books that loosely define weight as -mg where g is forever gravitational g and nothing else, would re-think if they considered the ramifications (like astronauts that really, really have a "weight" 90% as much as on the ground-- something almost NOBODY really believes, or teaches). But the authors of these things weren't very careful, because they didn't have to be. And no, it's not very sensible, but here we are. S B H arris 00:34, 20 November 2011 (UTC)[reply]

Yes, it shouldn't even be an article -- just a small section in Weight. Gerardw (talk) 09:02, 21 November 2011 (UTC)[reply]

We can't correct the real world mess about this issue. The best thing is to keep this article and explain here and in the Weight article what the issues are as best as we can. Count Iblis (talk) 15:56, 21 November 2011 (UTC)[reply]

Fair enough, but I do feel that this article is overblown. It seems to be making a huge and complicated mountain out of something that I feel could be explained much more simply. I'm really struggling to make sense of some parts of it. 86.176.208.18 (talk) 18:50, 21 November 2011 (UTC)[reply]

Does it have a single reference that actually discusses "apparent weight."??? — Preceding unsigned comment added by Gerardw (talk • contribs) 20:00, 21 November 2011 (UTC)[reply]

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Let start from definitions and physical nature of the force "weight":

weight is electro-magnetic force which shows the strength of interaction between the object (whose weight we are measuring) and the support. In majority of cases that interaction is connected/related to gravitational attraction between the object and other body (may be the Earth) - and therefore many students (even PhD students - I have seen some on their defenses, that is one of my questions to check whether PhD student is ready for defense) think that weight is a gravitational force (which is WRONG). The simple explanation is that we need to consider TWO pairwise interactions: between the object and the support (electro-magnetic force), and between the object and the Earth (gravitational force). There are two forces acting on the object: the first one from the support on the object (usually points UP) -- call it Normal force, the second one is from the Earth on the object (usually points DOWN :-) -depends on definition DOWN-) -- call it Gravitational attraction force (or just simple gravitational force). If we assume acceleration of the object a = 0 m/s^2, the absolute values of two forces acting on the object should be same, therefore Normal force is equal to Gravitational force. But forces are not coming along. For each real force there is counter-force (see the 3-rd Newton's Law). The counter-force for Normal force is also of electro-magnetic type, acts from the object on the support and called WEIGHT. If a = 0, gravitational force = normal = weight. Confusion comes from here. However, if acceleration is not 0, gravitational force is not longer equal to normal force = weight, so weight is not equal to gravitational force. E.g. weightless condition does not imply zero gravitational force. (Weightless condition implies Weight=Normal force=0 -- the object does not act on support -- like walls/floor of spaceship, etc. But there IS gravitational force action on the object from the Earth/any other planet, etc.) As for 'apparent weight' there is no such concept in Theoretical Physics. First of all -- what is the nature of that force (what type of interaction is responsible for that force) -- is it gravitational? is it electro-magnetic? If it is gravitational (as sometimes claimed in some textbooks)-- it should NOT change with the motion of the object and weight should be constant -- which is not the case. If it is electro-magnetic interaction -- how it can change the type of interaction? Just to be convenient for explanation? Couple of my friends joked that the 'concept' was introduced to tell below-2.00-GPA-students to grasp the concept of weight. It is possible to continue about the 'apparent weight' and its internal contradictions, but that will not contribute to the understanding of the notion of 'weight'.

71.65.243.149 08:12, 22 June 2007 (UTC)[reply]

Last edited at 08:12, 22 June 2007 (UTC). Substituted at 14:15, 1 May 2016 (UTC)