Hydrostatic Force in swimming pool

In summary, the conversation discusses the calculation of the force on one of the sides of a swimming pool with an inclined bottom and varying depths. The correct set up for the integrals is shown and the error in determining the contact width for the differential area is addressed. The relationship between the contact width and the depth of the pool is determined to be linear.
  • #1
johnhuntsman
76
0
A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9 ft. If the pool is full of water, estimate the force on one of the sides.

They mean one of the trapezoidal sides, not the rectangular ones. Anyway, I've drawn it out and decided it out to be thought of as two plates: one recangular (with dimensions 40ft x 3ft) and one triangular (a right triangle with a height of 6ftand length 40ft).

The positive x-axis is oriented such that it is pointing south.

Using similar triangles I can get the "depth" of the upside down triangular plate to be [itex]3x/20[/itex] in terms of x. Since it is 3ft below the surface (there's still the rectangular plate on top) the "depth" of the triangle to be [itex]3+3x/20[/itex].

I have my integrals set up like so:

[itex]ρg=62.5 lb./ft^2[/itex]

[itex]\int_{0}^{3} 62.5x(40)~dx + \int_{3}^{9} 62.5(3+3x/20)(40)~dx[/itex]

My answer is incorrect. The correct answer is 4.88E4 lbs. Assuming that my integration is correct, how did I mess up the set up to get the wrong answer?

Blurry picture of a poor diagram:
http://s1096.beta.photobucket.com/user/TVRobot/media/DSCN57191.jpg.html?sort=3&o=1
 
Last edited:
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  • #2
johnhuntsman said:
A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9 ft. If the pool is full of water, estimate the force on one of the sides.

They mean one of the trapezoidal sides, not the rectangular ones. Anyway, I've drawn it out and decided it out to be thought of as two plates: one recangular (with dimensions 40ft x 3ft) and one triangular (a right triangle with a height of 6ftand length 40ft).

The positive x-axis is oriented such that it is pointing south.

Using similar triangles I can get the "depth" of the upside down triangular plate to be [itex]3x/20[/itex] in terms of x. Since it is 3ft below the surface (there's still the rectangular plate on top) the "depth" of the triangle to be [itex]3+3x/20[/itex].

I have my integrals set up like so:

[itex]ρg=62.5 lb./ft^2[/itex]

[itex]\int_{0}^{3} 62.5x(40)~dx + \int_{3}^{9} 62.5(3+3x/20)(40)~dx[/itex]

My answer is incorrect. The correct answer is 4.88E4 lbs. Assuming that my integration is correct, how did I mess up the set up to get the wrong answer?

Blurry picture of a poor diagram:
http://s1096.beta.photobucket.com/user/TVRobot/media/DSCN57191.jpg.html?sort=3&o=1

In the second integral, I got a different integrand. Instead of (3+3x/20), I got x(9-x)/6
 
Last edited:
  • #3
Chestermiller said:
In the second integral, I got a different integrand. Instead of (3+3x/20), I got (9-x)/6

How'd you get that?

[Edit] Or perhaps more importantly, does that get you to 4.88E4 lbs.? [Edit]
 
  • #4
johnhuntsman said:
How'd you get that?

[Edit] Or perhaps more importantly, does that get you to 4.88E4 lbs.? [Edit]

I corrected a factor in my previous post. It should read x (9-x )/6. With this, I get a force of 48750 lb.

You must have made an error in determining the contact width for the differential area on the triangular section. The contact width I got was w = 40 (9 - x) / 6, so that when x = 3, w = 40, and when x = 9, w = 0.
 
  • #5
Chestermiller said:
I corrected a factor in my previous post. It should read x (9-x )/6. With this, I get a force of 48750 lb.

You must have made an error in determining the contact width for the differential area on the triangular section. The contact width I got was w = 40 (9 - x) / 6, so that when x = 3, w = 40, and when x = 9, w = 0.

So what you're saying is that w is to 40ft what 9 - x is to 6ft?

Or rather, "what width of small triangle is to width of large triangle, height of small triangle is to height of large triangle."
 
  • #6
johnhuntsman said:
So what you're saying is that w is to 40ft what 9 - x is to 6ft?

Or rather, "what width of small triangle is to width of large triangle, height of small triangle is to height of large triangle."

Yes. That's pretty much how I did it, but, in retrospect, it's probably easier to say that when x = 3, w = 40 and when x = 9, w = 0, and that the relationship between w and x is linear, so fit it with a straight line.
 
  • #7
Chestermiller said:
Yes. That's pretty much how I did it, but, in retrospect, it's probably easier to say that when x = 3, w = 40 and when x = 9, w = 0, and that the relationship between w and x is linear, so fit it with a straight line.

Alright thanks. I appreciate it.
 

1. What is hydrostatic force and how does it affect swimming pools?

Hydrostatic force refers to the pressure exerted by a fluid at rest on a submerged object. In the case of a swimming pool, this force is created by the weight of the water in the pool and can cause significant structural damage if not properly managed.

2. How can hydrostatic force be controlled in a swimming pool?

To control hydrostatic force in a swimming pool, a hydrostatic relief valve is typically installed. This valve allows excess water to escape from under the pool and equalize the pressure inside and outside of the pool. Additionally, regular maintenance and monitoring of the pool's water level can help prevent excessive hydrostatic pressure.

3. What are the potential risks of not managing hydrostatic force in a swimming pool?

If hydrostatic force is not managed in a swimming pool, it can cause significant damage such as cracks, bulges, and even complete failure of the pool's structure. This can lead to costly repairs and potential safety hazards for swimmers.

4. Are there any safety precautions to take regarding hydrostatic force in swimming pools?

Yes, it is important to regularly check and maintain the pool's water level to prevent excessive hydrostatic pressure. Additionally, if you notice any signs of damage or excessive pressure such as cracks or bulges, it is important to address them immediately and consult a professional if needed.

5. Can hydrostatic force affect the cleanliness of a swimming pool?

Yes, if hydrostatic force is not managed, it can cause damage to the pool's structure, which can lead to leaks and other issues that can affect the cleanliness of the pool. It is important to regularly check and maintain the pool's water level to prevent these potential issues.

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