- #1
agro
- 46
- 0
Let us define a pool. Viewed from top, we see a rectangle of sides 16 ft and 10 ft. When we look from the side (so the 16 ft side is perpendicular to our view), it is a trapezoid with width 16 ft and 4 ft, 16 ft being the height. It is roughly like this:
From top:
From one side (and the other opposite side):
The two other sides are rectangular (4 ft x 10 ft and 8 ft x 10 ft). The floor (OK, it's also a "side") is a rectangle with dimension 10 ft x [itex]\sqrt{16^2+4^2}[/itex] ft.
Let the pool be filled completely with water with weight density [itex]\pi[/itex] = 62.4 lb/ft^3.
The question is: what is the fluid force on the bottom of the pool?
Let us use the second image for this discussion. I define an vertical x-axis with x = 0 located at 4 ft below the water surface and the positive x is below x = 0. I define h(x) = 4 + x as the height of water surface at a particular x location and w(x) = 10 as the width of the pool.
To find the fluid force F, I use the equation
[tex]
F = \int_0^4\pi h(x)w(x)\sec t\;dx
[/tex]
Where t is the angle between the x-axis and the inclined pool floor.
Since sec t = [itex]\sqrt{16^2+4^2}/4[/itex],
[tex]
F = \int_0^4 (62.4) (4 + x) (10) \frac{\sqrt{16^2+4^2}}{4} dx
[/tex]
However the answer given by my book isn't the same as mine (the book's answer is bigger). Since I ruled out the possibility of miscalculation the integral, it must be my formulation of the integral (or the book's answer is wrong).
However I can't see what's wrong with my integral... I tried doing it "low level", formulating the Riemann Sums first, but it results in the integral I use.
What's wrong?
Thanks a lot.
From top:
Code:
16 ft
------------------
| |
| |
| | 10 ft
------------------
From one side (and the other opposite side):
Code:
16 ft
------------------
4ft | |
... |
... | 8 ft
... |
...|
The two other sides are rectangular (4 ft x 10 ft and 8 ft x 10 ft). The floor (OK, it's also a "side") is a rectangle with dimension 10 ft x [itex]\sqrt{16^2+4^2}[/itex] ft.
Let the pool be filled completely with water with weight density [itex]\pi[/itex] = 62.4 lb/ft^3.
The question is: what is the fluid force on the bottom of the pool?
Let us use the second image for this discussion. I define an vertical x-axis with x = 0 located at 4 ft below the water surface and the positive x is below x = 0. I define h(x) = 4 + x as the height of water surface at a particular x location and w(x) = 10 as the width of the pool.
To find the fluid force F, I use the equation
[tex]
F = \int_0^4\pi h(x)w(x)\sec t\;dx
[/tex]
Where t is the angle between the x-axis and the inclined pool floor.
Since sec t = [itex]\sqrt{16^2+4^2}/4[/itex],
[tex]
F = \int_0^4 (62.4) (4 + x) (10) \frac{\sqrt{16^2+4^2}}{4} dx
[/tex]
However the answer given by my book isn't the same as mine (the book's answer is bigger). Since I ruled out the possibility of miscalculation the integral, it must be my formulation of the integral (or the book's answer is wrong).
However I can't see what's wrong with my integral... I tried doing it "low level", formulating the Riemann Sums first, but it results in the integral I use.
What's wrong?
Thanks a lot.