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leorik
New Member
USA
2 Posts |
 Posted - 08/23/2007 : 20:46:43
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Talking to a guy, we sort of keep going back and forth on this subject.
He made a statement saying "WEIGHT IS DIRECTLY PROPORTIONAL TO DENSITY"
to which I thought for .5 seconds and corrected it with mass in place of weight. My logic being if in 0G (space) weight is 0. which means if his statement held true, that density can never change in 0G.
Is there a way to prove by formulas that weight is or is not directly proportional to density?
I used the formula m=dv
he took that into a weight formula with the following
quote:
mass = density * volume
weight = mass * Ag (acceleration due to gravity)
then
mass = weight / Ag
substitute into original equation
weight / Ag = density * volume
multiply both sides by Ag
weight = density * Ag * volume
I'm looking for something solid to put this to rest :)
thank you! |
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uberclay
Advanced Member
Canada
159 Posts |
Posted - 08/24/2007 : 00:05:32
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To say that something is directly proportional means that it is a constant multiple. We can say that weight (w) is directly proportional to mass (m) because w = m*g, where g is the local acceleration of gravity. If m were to increase by a factor of 2, so would the weight (w = 2*m*g).
You can test the relationship between weight and density the same way, using the equation you provided. |
Edited by - uberclay on 08/24/2007 00:06:19 |
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leorik
New Member
USA
2 Posts |
Posted - 08/24/2007 : 00:47:01
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OK, so testing with the formula that I provided.
weight = density * Ag * volume
Let's say that gravity is 0.
weight = d*0*v
so weight = 0. No matter what density is, when gravity is 0, weight is 0.
This does not hold true, however, with mass.
see what I'm saying? :) |
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tkhunny
Advanced Member
USA
1001 Posts |
Posted - 08/24/2007 : 08:14:40
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Given non-zero gravity, the weight of identically-sized objects, is directly proportional to density.
1) Weight does not exist without gravity. If you can get it to zero, it isn't perfectly proportional.
2) Take any object with weight. Squeeze it really hard. It will have greater density. The weight has not changed.
This "guy" isn't a boyfriend, is it? You may hurt his feelings! Good luck, in any case. |
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skeeter
Advanced Member
USA
5634 Posts |
Posted - 08/24/2007 : 12:43:54
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If volume is a non-zero constant, then density is directly proportional to mass.
d = k *m ... where k = 1/V
If the acceleration due to gravity is a non-zero constant, then mass is directly proportional to weight.
m = k *W ... where k = 1/g
so ...
d = k*m = k*k*W = CW ... where C = k*k = 1/(gv). |
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Subhotosh Khan
Advanced Member
USA
9114 Posts |
Posted - 08/25/2007 : 10:20:34
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In general though, we should say "Weight is proportional to density assuming mass and acceleration due to gravity are constant".
The assumption should be stated - because those in general are "variable" - and not constant. |
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Subhotosh Khan
Advanced Member
USA
9114 Posts |
Posted - 08/25/2007 : 21:55:06
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That is why I used the term "should" - instead of "must".
In general, while doing any experiment we should be careful about defining "controlled" variable and "uncontrolled" variable (sometimes lumped with constant). That is one of lessons I learnt very hard way while doing Designed Experiments.
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tkhunny
Advanced Member
USA
1001 Posts |
Posted - 08/26/2007 : 23:13:57
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quote:
So, as F = G(m1)(m2).(r^2), F is proportional to m1...
This is a formula for a model. It is a useful model. We can build and understand things if we use this model.
Take a close look at the standard model/formula for luminescence. It has a singularity as zero!!
These formulas are models for physical concepts. These formulas are not the physical concepts themselves. The formula proves only the model, not quite the physical concept.
Formulas and models can be excellent reproductions of physical phenomona within the Domain of the model. Back to luminescence...it isn't intended to work at zero, so who cares if there is a singularity at zero, but we should state this deficiency when we talk about it. In the case at hand, appropriate Domains must be established. Without explicit assumptions, it really is a silly question or argument. What is the Domain? Is it proportional? It is if we pick a Domain where it is proportional, sure. Now pick some other Domain.
You have not refuted either of my points, 1) or 2). |
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deff
Junior Member
USA
5 Posts |
Posted - 08/27/2007 : 03:48:35
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Weight is proportional to mass with the "multiplicative factor" of gravity (weight=mass x acceleration).
mass is proportional to density with the "multiplicative factor" being volume.
So, simple as it sounds, as long as gravity and te volume of the object dont change, yes, they are proportional. Now yu can go into details, and break it down even more and say that (g=GM/r^2) and say that Mass of the earth and the distance form the earth's core cant change either, or you could go even farther if you want, but essentially, yes, for all comon purposes they are proportional.
weight = Mass x acceleration
density = mass/volume
density = weight/acceleration x volume
So density = Weight x 1/(acceleration x volume)
and we are assuming that acceleration die to gravity and volume are constants. |
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tkhunny
Advanced Member
USA
1001 Posts |
Posted - 08/27/2007 : 08:15:55
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| ...which is not the original question. The original question was an unqualified expression. No fair adding qualifications. The original statement simply is wrong, as the original poster indicated. |
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