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Archimedes Principle & Specific Gravity

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HelgaMan
#1
Nov19-08, 05:06 AM
P: 5
I was reading a lab online and I came across one that dealt with Specific Gravity and Buoyany and stuff.. and then they had these equations:

*Note: p = density, M = mass, V = Volume, g = gravitational acceleration, W = Weight
Subscripts: S = substance, W = water

SG = pS / sW = (MS / VS) / (MW / VW) = (MSg / VS) / (MWg / VW) = (WS / VS) / (WW / VW)

and

SG = (WS / V) / (WW / V) = WS / WW = WS / (buoyant force) = WS / (loss of weight in water)
= WS / (WS - weight of substance in water)

which all makes sense to me, mathmatically.. but then i thought about this part "WS / (buoyant force)" and that kinda throws me off because isnt Archimede's Principle that weight of an object is equal to the buoyant force.. and that makes SG always 1? but im pretty sure thats not the case, though..


well, idk, im sure all you smart people will see something that i dont, lol.
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Doc Al
#2
Nov19-08, 05:12 AM
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Quote Quote by HelgaMan View Post
which all makes sense to me, mathmatically.. but then i thought about this part "WS / (buoyant force)" and that kinda throws me off because isnt Archimede's Principle that weight of an object is equal to the buoyant force.. and that makes SG always 1? but im pretty sure thats not the case, though..
No, it's not. Archimedes' principle states that the buoyant force equals the weight of the displaced fluid, not the weight of the object. Read all about it: Archimedes' Principle
HelgaMan
#3
Nov19-08, 05:19 AM
P: 5
oh, okay, lol.

thanks :D


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