Will a Sealed Steel Container Float in the Ocean?

[mailmas]

Homework Statement

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?

Homework Equations

FBuoyancy = density*volume*g
weight = mg
m = density*volume
density of salt water = 1030 kg/m^-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?

 

[FactChecker]

Homework Statement

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.
Volume = (6.06)(2.59)(2.43) = 38.139

Wsteel = (8050)(38.139)(.02meters)(9.81) = 60,237.1 N <- I don't understand what to do about the thickness.

You should not be using the volume of the cube here. You should be adding up the area of the sides and multiplying by the side thickness. That would be the volume of the steel of the sides.

 

[mailmas]

You should not be using the volume of the cube here. You should be adding up the area of the sides and multiplying by the side thickness. That would be the volume of the steel of the sides.

Oh alright. The original problem said it was 20 feet high so I'm going to assume I multiply .02m by 2.59m and 2.43m. Then multiply that number by 6.
V = (.02)(2.59)(2.43) = .1258m^3
6V = .755m^3
Then,
Wsteel = (8050)(.755)(9.81) = 59,641N

 

[FactChecker]

I think you should handle each of the 6 sides separately using the correct dimensions for each side. The 6.06 meters is approximately 20 feet, so that is where that came from. The inner volume is not exactly the same as the outer volume because of the metal thickness, but that may not be enough to matter.

 

[mailmas]

I think you should handle each of the 6 sides separately using the correct dimensions for each side. The 6.06 meters is approximately 20 feet, so that is where that came from. The inner volume is not exactly the same as the outer volume because of the metal thickness, but that may not be enough to matter.

I think I understand my mistake now. But I don't know which measurement is the width or length so I just chose one to calculate:

Area of Long Part = (6.06)(2.59) = 15.69m^2
4*15.69 = 62.79

Area of Top/Bot = (2.59)(2.43) = 6.3m^2

6.3+62.79 = 69
V= 69(.02) = 1.38m^3

 

[mailmas]

So weight of steel is: Wsteel = (8050)(1.38)(9.81) = 109000N

 

[haruspex]

Area of Long Part = (6.06)(2.59) = 15.69m^2
4*15.69 = 62.79

How many sides have dimension 6.06 x 2.59?

Area of Top/Bot = (2.59)(2.43) = 6.3m^2

How many sides have those dimensions?
How many remaining sides are there?

 

[mailmas]

How many sides have dimension 6.06 x 2.59?

How many sides have those dimensions?
How many remaining sides are there?

4 Long Parts and 2 for Top and Bottom. I forgot to multiple by 2 for the 6.3.

2(6.3)+62.79 = 75.3
V= 75.3(.02) = 1.5m^3

Wsteel = (8050)(1.5)(9.81) = 119000N

 

[FactChecker]

But I don't know which measurement is the width or length so I just chose one to calculate:

But you do know there are two sides of each, right?

 

[mailmas]

But you do know there are two sides of each, right?

Oh okay yeah I get it.

(6.06)(2.59)(2) = 31.39
(6.06)(2.43)(2) = 29.45
(2.43)(2.59)(2) = 12.59

Atot = 73.42
V = 1.46

Wsteel = (8050)(1.46)(9.81) = 116000N

 

[haruspex]

Oh okay yeah I get it.

(6.06)(2.59)(2) = 31.39
(6.06)(2.43)(2) = 29.45
(2.43)(2.59)(2) = 12.59

Atot = 73.42
V = 1.46

Wsteel = (8050)(1.46)(9.81) = 116000N

Looks ok.

 

[mailmas]

Looks ok.

Thanks. Do you see anything else wrong with the answer?

 

[haruspex]

Thanks. Do you see anything else wrong with the answer?

No.