Force to lift a pyramid that is sitting in a water tank

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Discussion Overview

The discussion revolves around calculating the vertical force required to lift a pyramid weighing 4000 lb from the floor of a water tank. The pyramid has a base of 6 ft square and an altitude of 4 ft, with water 4 ft deep in the tank. Participants explore the implications of buoyancy, atmospheric pressure, and the geometry of the situation.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states the force by the fluid against the pyramid is denoted as Fv, but expresses uncertainty in their analysis.
  • Another participant questions the size of the opening in the tank floor beneath the pyramid, noting it was not provided in the problem statement.
  • Some participants suggest that the hole should cover most of the pyramid's base, with a gasket to create a seal, which alters the typical buoyancy situation.
  • One participant argues that the hole size is irrelevant, asserting that the problem can be solved without considering the hole or the pyramid's shape, as it is a hydrostatics problem.
  • A participant confirms a buoyant force of 10,000 lb, while another expresses confusion about the atmospheric pressure's role in the calculations.
  • Several participants discuss the interpretation of the pyramid's base dimensions, with one noting a potential miscalculation in the volume of the liquid due to misunderstanding the dimensions.
  • Another participant emphasizes the need for detailed reasoning and calculations to assist in resolving the problem.
  • One participant mentions that hydrostatic balance applies to atmospheric pressure as well, hinting at additional forces acting on the pyramid.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the hole size and the role of atmospheric pressure in the calculations. There is no consensus on the correct approach to solving the problem, and multiple competing interpretations remain.

Contextual Notes

Participants highlight limitations in the problem statement, such as the unspecified size of the opening in the tank and the potential misinterpretation of the pyramid's dimensions. The discussion also reflects uncertainty regarding the application of atmospheric pressure in the context of buoyancy.

paulie
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Homework Statement


A pyramid weighing 4000 lb has a base 6ft square and an altitude of 4ft. The base covers an opening in the floor of a tank in which there is water 4 ft deep. Underneath the floor of the tank and on the water surface there is air at atmospheric pressure. What vertical force is required to lift the pyramid off the floor?

Homework Equations


F=δV

Where F is the force; δ=specific weight of the liquid; V is volume

The Attempt at a Solution


I'm not really sure if I analyze what it should look like, but here is my attempt.

Fv is the force by the fluid against the pyramid
 

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Last edited:
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Please post the pictures directly rather than sending us to imgur

Use the UPLOAD button on the post editor.
slask.png
 

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How big is the opening in the floor of the tank below the pyramid?
 
Chestermiller said:
How big is the opening in the floor of the tank below the pyramid?
It was not stated, there's no figure too provided by the book. :( The answer is provided around 10000
 
I think that you have to assume that the hole covers essentially the entire bottom of the pyramid except for the very edge, with a gasket at the very edge of the pyramid to provide the seal to the bottom. So this is not the usual buoyancy situation because the pressure at the very bottom of the pyramid is atmospheric, rather than hydrostatic. What would the buoyancy force of the pyramid be if, rather than having atmospheric air on the bottom, it were totally surrounded by water?
 
Yeah, I don't think "hole size" matters at all. You are assuming the water isn't free to flow under and lift the pyramid (it's an idealized situation, i.e., there's nowhere for the water to go, at least not until after the pyramid has been lifted off).

Given that this is a hydrostatics problem — there's an easy way to solve this problem without worrying about things such as hole size or even some of the details of the pyramid's shape.
 
I confirm the 10,000 lb.
 
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I think, I'm just missing another 6' in the computation of volume in the liquid.

That "6ft square" means (6)^2 and not 6ft^2. (I assumed that 6ft^2 is the area).

I got around 9990 lbs at the end. Although still not sure how can I use the atmospheric pressure on the problem...
 
paulie said:
I think, I'm just missing another 6' in the computation of volume in the liquid.

That "6ft square" means (6)^2 and not 6ft^2. (I assumed that 6ft^2 is the area).

I got around 9990 lbs at the end. Although still not sure how can I use the atmospheric pressure on the problem...
In order for us to help you, we need to see detail on exactly what you did, and your reasoning.
 
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  • #10
paulie said:
Although still not sure how can I use the atmospheric pressure on the problem...
Hydrostatic balance applies for the atmosphere, too. You may be forgetting a pair of forces on the pyramid ...
 

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