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Old Dec 10th 2015, 02:06 AM   #1
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buoyancy

The following was the question:

You find your mass to be 42 kg on a weighing machine. Is your
mass more or less than 42 kg?

The answer they have given is my actual mass will be more. I am not able to understand. What upward force is acting on my body?

Kindly enlighten me.

with warm regards,

Aranga
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Old Dec 10th 2015, 07:57 AM   #2
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Although you don't state it, I assume that the scale ("weighing machine") is under water, and that it operates by measuring the downward force of your body under Earth's gravity (this is how standard spring-based bathroom scales work) - is that correct?

If so, think about a free-body diagram of your body on the scale under water. Acting downwards is the force of gravity on your body (W=mg), and acting upward is a buoyancy force equal to the volume of your body under water times the density of water, plus the upward force of the scale that you are standing on. That upward force of the scale is equivalent to 45 Kg under normal Earth gravity, or 45 Kg x 9.8 m/s^2. So the question is - how would that upward force of the scale change if there was no buoyancy force (i.e. if you were not under water)?
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Old Dec 10th 2015, 01:31 PM   #3
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There is also buoyancy in air,
But since the density of air is low (relative to the density of a person) the effect is very small and is usually taken to be negligible.

If you were determining the mass of a balloon, then the buoyancy due to the displaced air would have to be considered.
Or indeed (as ChipP suggests) if you are weighing under water (where the mass displaced is much more comparable to the mass of a person).
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Old Dec 11th 2015, 02:41 AM   #4
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Thank you very much sir. They have not given that the weighing machine is under water. Whether air would have any upward force acting on the body standing on the weighing machine?
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Old Dec 11th 2015, 02:45 AM   #5
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Thank you very much. The replies were very useful. I am able to understand even air can have upward force.
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Old Dec 11th 2015, 08:46 AM   #6
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Floating in Air

Hi arangu1508,

Here's a weighing problem that includes the effect of surrounding air.
It is solved thoroughly, I hope you understand it.

http://www.thermospokenhere.com/wp/0...sculpture.html

Good Luck with your studies. TSH
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Old Dec 16th 2015, 10:45 PM   #7
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thank you very much. It is very useful.
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Old Dec 22nd 2015, 10:19 AM   #8
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Here is another way to look at this problem. When you stand on a scale, the force the scale experiences is equal in magnitude to your weight. What we normally sense as weight is the reaction of the surface on which we stand. The scale is calibrated to read mass and thus actually reads N /g.
Since you re spinning along with the earth, circular motion comes into play, and we need to consider the centripetal force which is the resultant of mg acting towards the centre of the earth and the normal reaction say N acting outwards ( note m here is your actual mass.) Thus ( mv^2 ) / r = mg - N, which gives us N = mg - ( mv^2 ) / r . Dividing throughout by g, we get N/g = m - v^2 / rg. Since N/g is what the scale reads, as your mass, your actual mass m = N/g + v^2 / rg and is thus greater.
And "a buoyancy force equal to the volume of your body under water times the density of water" that should be the weight of the volume of water displaced, so it needs to be multiplied by g
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Old Dec 23rd 2015, 04:40 AM   #9
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Thank you very much. New dimension to the above question. Very useful.
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