Fluids, pressure on a wall to determine max depth

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Homework Statement



A square pool with 100-m-long sides is created in a concrete parking lot. The walls are concrete 90 cm thick and have a density of 2.5 g/cm3. The coefficient of static friction between the walls and the parking lot is 0.49. What is the maximum possible depth of the pool? (The density of water is 1000 kg/m3.)

Homework Equations



P = F/A, rho=m/v

The Attempt at a Solution



I've been at this one for the past two days. I cannot find a way to approach this as I think I have an idea.

The density of the wall is given, therefore I can determine how much pressure the wall can take with a variable h(max depth).

Wit that I can take that as my net force to determine the height with a given amount of water in the pool.

A couple attempts: I converted 2.5 g/cm^3 to 2500 kg/m^3

I used integration to find the total amount of force the wall can handle
Ftotal = ∫ ρgh (100 dh), limits of integration 0 to y
= 490000y2

I'm just completely stumped here.
 

Answers and Replies

  • #2
ehild
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What is the force that balances the hydrostatic force of the water exerted on the wall?

ehild
 
  • #3
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Pascals principle.
Ptotal = Pair + rho*gh
 
  • #4
ehild
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bd what about the static friction? Does it have no effect? Why did the problem mention it at all?

This pool is a bit strange, it is not fixed or glued to the ground, the wall is just built up...

ehild
 
  • #5
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I am completely confused as well. We completed the chapter on fluids in class and never mentioned friction forces. I assumed this is one of the problems providing an extra element to able to recognize it and defer.

My understanding is to determine the force of the wall to be withstood, considering the density being 2500kg/m3. Which is the reasoning for my integral. At the same time I know that Pascal's Principal says the force of a fluid is the same throughout it's container. So there could be another way?

Anyways I referenced a solution online which has me even more confused.

Horizontal pressure on wall:
Pwall = (ρwatergh)/2

Ffriction on wall = μsρwall * (.90) * (100) g

Set two equal to each other:
P * A = Ffriction

watergh)/2 * h * (100) = μsρwallgh * (.90) * (100)

hmax = [ (2)*(.90) μsρwall ] / ρwater

I am currently trying to make sense of this but again, do not recognize the force of friction in this problem.
 
  • #6
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the equation output was 2.2m which is the correct answer by the way... clueless
 
  • #7
ehild
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You have correctly calculated the force the water exerts on the wall. If the pool is filled up to y height with water, this force is 490000 y2. This force is horizontal, and an other horizontal force can balance it. There is static friction that prevents motion in the horizontal direction. The maximum force of static friction is Volume of the wall*density*g*μs. As the height of the wall is not given, assume that it will be filled up to the rim, so the height of the wall is y, the volume of the wall is V=0.9*100*y. So you need to solve the following equation for y:

Hydrostatic force on the wall= Maximum force of static friction .

ehild
 
  • #8
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Hydrostatic force on the wall= Maximum force of static friction(pretty much the sum of all of the forces along the wall) right?

Is there a way to find the solution to this without the force of friction?
 
  • #9
ehild
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Hydrostatic force on the wall= Maximum force of static friction(pretty much the sum of all of the forces along the wall) right?

Is there a way to find the solution to this without the force of friction?
NO. What keeps the wall on place without the force of friction?
I do not understand your problem.
You can write out the formula for the force of friction. You have the other one for the force of the water. They are equal: this is an equation for the height of water. Solve.

ehild
 
  • #10
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Gottcha. I just wanted to throw in this update so anyone looking this up can see a final reference.

There are two things here that needs to be figured out:
1. Horizontal pressure along the wall, which is the integral part, summing up the forces along the wall at a unknown depth so the limits of integration are 0 to y. This force is 490000 y^2 as computed somewhere above.
2. The second part is to realize there is a force of friction Ff from the slab of one block. Ff = μsFN where FN is the normal force = mg of the block. The density of the block is given, can be written in terms of 'm' that will give Ff a force with a variable y (part left over from the volume). The sum of the forces horizontally found in part 1 has to equal Ff to be at a max depth 'y'.
 

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