Physics Help Forum Buoyancy Cube Question
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 May 9th 2018, 08:47 PM #1 Junior Member   Join Date: May 2018 Posts: 3 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?
 May 10th 2018, 12:35 AM #2 Senior Member   Join Date: Apr 2017 Posts: 401 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
 May 10th 2018, 12:36 AM #3 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India 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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