Buoyancy of Cube Question

In summary, a large sealed container full of air with internal dimensions of 6.06 x 2.59 x 2.43 m, made of steel with a thickness of 2.00 cm, falls into the ocean. Using the equations for buoyancy and weight, it is determined that the container will float as the weight of the steel (116,000 N) is less than the buoyancy force of the saltwater (385,000 N).
  • #1
mailmas
46
0

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?
 
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  • #2
mailmas said:

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.
 
  • #3
FactChecker said:
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
 
  • #4
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.
 
  • #5
FactChecker said:
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
 
  • #6
So weight of steel is: Wsteel = (8050)(1.38)(9.81) = 109000N
 
  • #7
mailmas said:
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?
mailmas said:
Area of Top/Bot = (2.59)(2.43) = 6.3m^2
How many sides have those dimensions?
How many remaining sides are there?
 
  • #8
haruspex said:
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
 
  • #9
mailmas said:
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?
 
  • #10
FactChecker said:
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
 
Last edited:
  • #11
mailmas said:
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.
 
  • #12
haruspex said:
Looks ok.
Thanks. Do you see anything else wrong with the answer?
 
  • #13
mailmas said:
Thanks. Do you see anything else wrong with the answer?
No.
 

1. What is the definition of buoyancy?

Buoyancy is the upward force that a fluid exerts on an object that is submerged in it. It is caused by the difference in pressure between the top and bottom of the object.

2. How does the shape of an object affect its buoyancy?

The shape of an object can greatly affect its buoyancy. Objects with a larger surface area will experience a greater upward force, making them more buoyant. This is why a flat object, like a raft, can float on water even though it is denser than the water itself.

3. What is the relationship between the density of an object and its buoyancy?

The density of an object plays a critical role in determining its buoyancy. An object that is less dense than the fluid it is submerged in will float, while an object that is more dense will sink. This is because the more dense object will experience a greater downward force than upward force from the fluid.

4. How does the weight of an object affect its buoyancy?

The weight of an object also plays a role in its buoyancy. The greater the weight of an object, the more downward force it will experience, making it less buoyant. This is why heavier objects will sink in water, while lighter objects will float.

5. What factors can impact the buoyancy of a cube?

The buoyancy of a cube can be impacted by several factors, including its density, weight, and shape. The density and weight of the cube will determine whether it will float or sink, while the shape of the cube can affect the magnitude of its buoyancy. Other factors such as the density and viscosity of the surrounding fluid can also impact the buoyancy of the cube.

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