Archimede's Principle problem

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In summary, the problem involves a hollow cubical box floating in a lake with 1/3 of its height submerged, and the walls of the box have negligible thickness. Water is poured into the box and the question asks for the depth of the water at the instant the box begins to sink. The answer from the book is 0.2 meters, calculated by considering the buoyancy force and volume of displaced water. However, there seem to be some logical problems with the question, as the mass of the box is not taken into account. Some suggest that the box has a mass of 88.2N, while others point out that the question implies the volume of water is cubical. Ultimately, the correct answer is that the
  • #1
shortydeb
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Homework Statement


A hollow cubical box is 0.30 meters on an edge. This box is floating in a lake with 1/3 of its height beneath the surface. The walls of the box have a negligible thickness. Water is poured into the box. What is the depth of the water in the box at the instant the box begins to sink?

The answer from the book that i got the problem from is 0.2 meters.

Homework Equations


buoyancy force = (density of the water)(volume of displaced water)(gravity)

The Attempt at a Solution


First I did the buoyancy force, which i think is: (1000 kg/m^3)(1/3)(0.3)^3 m^3)(9.8 m/s^2) = 88.2 N.

But what I don't get is, if the walls of the box have negligible thickness, wouldn't that mean the mass of the box is basically zero? So how can the box even have 1/3 of its volume in the water when it has no weight?

Also if the buoyancy force is 88.2 N just holding the box in equilibirium, wouldn't that mean the box weighs 88.2 N...Any help at all would be really appreciated..!
 
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  • #2
There do seem to be a few logical problems with this question (it seems like they are treating the box to have zero mass), but I think they just want you to calculate the height of water necessary to overcome an 88.2N buoyant force. Doing so gives:

(88.2N)/(9.8m/s^2)(1000kg/m^3) = V = .009 m^3

h = (.009m^3)^1/3 = .208m
 
  • #3
You don't have to calculate anything. If the box is massless, it can't sink if it's not full of water, can it? Even if it is full of water, will it sink if it's massless? I have some problems with this question as well. But at least you can avoid the roundoff error by answering that you simply need to fill the box with water before it can even begin to 'sink'.
 
  • #4
Dick said:
You don't have to calculate anything. If the box is massless, it can't sink if it's not full of water, can it? Even if it is full of water, will it sink if it's massless? I have some problems with this question as well. But at least you can avoid the roundoff error by answering that you simply need to fill the box with water before it can even begin to 'sink'.

my initial reaction was the same as yours, however i do believe that the intent of the question is to consider that the box does have mass.

As the OP stated, this box mass would be 88.2N (without checking your maths)
 
  • #5
Rrrright. That was stupid of me. Think I should take a nap. Thanks! The box doesn't have zero mass, it just doesn't weigh much.
 
  • #6
thanks for the help.

There do seem to be a few logical problems with this question (it seems like they are treating the box to have zero mass), but I think they just want you to calculate the height of water necessary to overcome an 88.2N buoyant force. Doing so gives:

(88.2N)/(9.8m/s^2)(1000kg/m^3) = V = .009 m^3

h = (.009m^3)^1/3 = .208m




doesnt that imply the volume of water is cubical?
 
  • #7
shortydeb said:
thanks for the help.

There do seem to be a few logical problems with this question (it seems like they are treating the box to have zero mass), but I think they just want you to calculate the height of water necessary to overcome an 88.2N buoyant force. Doing so gives:

(88.2N)/(9.8m/s^2)(1000kg/m^3) = V = .009 m^3

h = (.009m^3)^1/3 = .208m

doesnt that imply the volume of water is cubical?

Yes, it does. And no, it's not cubical. You know the box is floating with 0.1m of it's height submerged when empty. If you pour 1cm of water in, how much does the box sink? Try and answer that without calculating anything. Just think about it. Then answer the original question.
 

What is Archimedes' Principle problem?

Archimedes' Principle problem is a concept in physics that explains the buoyant force experienced by an object when it is submerged in a fluid. It states that the buoyant force acting on an object is equal to the weight of the fluid that the object displaces.

What is the formula for Archimedes' Principle problem?

The formula for Archimedes' Principle problem is Fb = ρVg, where Fb is the buoyant force, ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

How does Archimedes' Principle problem relate to real-life scenarios?

Archimedes' Principle problem is applicable in real-life scenarios such as explaining why objects float in water, why ships can stay afloat, and why hot air balloons rise in the air. It is also used in designing and building ships, submarines, and other watercraft.

What factors can affect the buoyant force in Archimedes' Principle problem?

The buoyant force in Archimedes' Principle problem can be affected by the density of the fluid, the volume of the object, and the acceleration due to gravity. It is also influenced by the shape and size of the object, as well as the depth and pressure of the fluid.

How is Archimedes' Principle problem used in scientific experiments?

Archimedes' Principle problem is often used in scientific experiments to determine the density of an object or a fluid. It can also be used to calculate the weight of an object by measuring its buoyant force when submerged in a fluid. Additionally, it is used to explain the behavior of objects in different fluids.

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