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Old May 9th 2018, 08:47 PM   #1
kid
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Join Date: May 2018
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Buoyancy Cube Question

1. The problem statement, all variables and given/known data
A large sealed container full of air falls into the ocean. It has internal dimensions of 6.06 2.59 2.43 m, with walls made of steel (8050 kg m3 ) which is 2.00 cm thick. Does it float?

2. Relevant equations
FBuoyancy = density*volume*g
weight = mg
m = density*volume
density of salt water = 1030 kg/m^-3

3. The attempt at a solution

Volume = (6.06)(2.59)(2.43) = 38.139
FBuoyancy = (1030)(38.139)(9.81)
= 385,367.9 N
= 385,000 N

Weight:
Wtot = Wair + Wsteel

Wair = m*g = density*volume*g = (1.29)(38.139)(9.81) = 482.65 N
Wsteel = (8050)(38.139)(.02meters)(9.81) = 60,237.1 N <- I don't understand what to do about the thickness.

Weight = 60,719. 8N which is less than FBuoyancy which means it floats.

Is this right?
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Old May 10th 2018, 12:35 AM   #2
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no ...it ain't right ... you have calculated the internal volume of this box

first calculate the upthrust from the displacement of the water , you need the volume of the external box for this ...clue ... first side of box is 606 +2 +2 cm = 610cm ...

then calculate internal volume , subtract internal from eternal volume to get the volume of steel
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Old May 10th 2018, 12:36 AM   #3
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Posts: 292
You have calculated the inner volume of the cube. To calculate buoyancy you need the external volume of the cube, that is the volume of the displaced liquid.
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