Hydrostatic Force in swimming pool

[johnhuntsman]

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 3x/20 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 3+3x/20.

I have my integrals set up like so:

ρg=62.5 lb./ft^2

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

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

 

[Chestermiller]

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 3x/20 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 3+3x/20.

I have my integrals set up like so:

ρg=62.5 lb./ft^2

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

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

 

[johnhuntsman]

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]

 

[Chestermiller]

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.

 

[johnhuntsman]

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."

 

[Chestermiller]

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.

 

[johnhuntsman]

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.