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 :
, where
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)
I cleaned up a little and began to add examples. More coming within the day. StuTheSheep 18:15, Jun 20, 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)
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)
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)
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)
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)
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)
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)
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 (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 , 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)
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)
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)
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)
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)
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)
This article contains WP:OR not supported by sources, therefore I'm redirecting to weight. Gerardw ( talk) 10:52, 10 May 2011 (UTC)
[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)
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)
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)
In the Fg = 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)
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)
Yes, it shouldn't even be an article -- just a small section in Weight. Gerardw ( talk) 09:02, 21 November 2011 (UTC)
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)
The comment(s) below were originally left at Talk:Apparent weight/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
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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) |
Last edited at 08:12, 22 June 2007 (UTC). Substituted at 14:15, 1 May 2016 (UTC)
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 :
, where
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)
I cleaned up a little and began to add examples. More coming within the day. StuTheSheep 18:15, Jun 20, 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)
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)
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)
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)
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)
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)
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)
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 (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 , 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)
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)
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)
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)
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)
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)
This article contains WP:OR not supported by sources, therefore I'm redirecting to weight. Gerardw ( talk) 10:52, 10 May 2011 (UTC)
[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)
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)
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)
In the Fg = 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)
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)
Yes, it shouldn't even be an article -- just a small section in Weight. Gerardw ( talk) 09:02, 21 November 2011 (UTC)
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)
The comment(s) below were originally left at Talk:Apparent weight/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
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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) |
Last edited at 08:12, 22 June 2007 (UTC). Substituted at 14:15, 1 May 2016 (UTC)