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

## 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:

anorlunda
Staff Emeritus
Please post the pictures directly rather than sending us to imgur

Use the UPLOAD button on the post editor.

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paulie
Chestermiller
Mentor
How big is the opening in the floor of the tank below the pyramid?

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

Chestermiller
Mentor
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?

olivermsun
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.

Chestermiller
Mentor
I confirm the 10,000 lb.

olivermsun
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...

Chestermiller
Mentor
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.

olivermsun
olivermsun