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?
