Buoyancy Cube Question

kid

May 2018
3
0
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?
 
Apr 2017
525
130
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
 
Dec 2008
292
98
Udupi, Karnataka, India
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.