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Stainless Steel in Sea Water

  • Thread starter TorMcOst
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  • #1
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I have a bolt (13 kg) of stainless steel (1650 cm``). How much will this weight in sea water?

Thank you in advance!
 

Answers and Replies

  • #2
Born2bwire
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You should be able to use Archimedes principle. The buoyant force is equal to the weight of fluid displaced by the object. Calculate the weight of the volume of equivalen seawater and subtract that off I believe.
 
  • #3
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Thanks a lot for your respond Born2bwire!

This means that my calculation will become something like this:

Weight in Air: 13,000 kg
- Weight of fluid displaced (1650 cm3*1,025): 1,691 kg
= Weight in Sea Water: 11,309 kg

Correct?
 
  • #4
Born2bwire
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Yeah, that's how I would work it. We'll see if anyone here ends up disagreeing.
 
  • #5
Astronuc
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Thanks a lot for your respond Born2bwire!

This means that my calculation will become something like this:

Weight in Air: 13,000 kg
- Weight of fluid displaced (1650 cm3*1,025): 1,691 kg
= Weight in Sea Water: 11,309 kg

Correct?
Does one mean 13 (13.000) kg, or perhaps 13,000 g? kg (g) is a unit of mass, N (Newton) or dyne (used in cgs system) is a measure of weight. Mass * acceleration of gravity would give weight (force) due to gravity.

The apparent mass in seawater would be 11.309 kg (11,309 g). The application of Archimedes principle (buoyancy) is correct.

In English units, one will find pounds mass (lbm) and pounds force (lbf), but strictly speaking they are not the same.
 
  • #6
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Thanks Astronuc!

However I got a little confused. By 13,000kg I meant 13 kg (in air). And my following answer was that this bolt would weight 11,3kg in water. Do you agree or disagree with this calculation?
 
  • #7
Born2bwire
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Thanks Astronuc!

However I got a little confused. By 13,000kg I meant 13 kg (in air). And my following answer was that this bolt would weight 11,3kg in water. Do you agree or disagree with this calculation?
He doesn't like your use of kilograms as a unit of weight. Technically, kilogram is a unit of mass, Newtons would be an appropriate unit of weight.
 
  • #8
Astronuc
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Thanks Astronuc!

However I got a little confused. By 13,000kg I meant 13 kg (in air). And my following answer was that this bolt would weight 11,3kg in water. Do you agree or disagree with this calculation?
As Born2bwire indicated, I was referring to the use of mass as weight. Being a physics forum, we wish to be accurate in such a matter.

I agree with the calculation that 13 kg of steel as described would have the weight equivalent of 11.3 kg (as measured in air), when the steel is immersed in seawater of density (1025 kg/m3), i.e. it appears lighter in water than it would in air.
 
  • #9
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Cheers Astronuc and Born2bwire, this have been to great help for me! And for the record, I will try to be more accurate in my physics from now on!
 

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