Homework Help: Finding the total hydrostatic force that can be withstood

1. Dec 4, 2014

leo255

1. The problem statement, all variables and given/known data

We are given a cubic tank, with a parabolic object y = x^2 (one end is -1, 1 and the other is 1, 1 --> it only goes up to 1). The total pressure it can withstand is 160 lbs. Also, the fluid density is 50 lbs.

Find:

a) Total pressure if water is 2-feet deep.
b) Max height the tank can be filled.

2. Relevant equations

Fluid density is 50 lbs.
Total pressure that can be withstood by the object: 160 lbs

3. The attempt at a solution

I think I got most of this:

For part "a," I'm getting a pressure of 50 (2-y), and for the area of the object, I'm getting 2 root(y) dy.
Thus, I got the integral, from 0 to 1, of 100 (2-y)(root(y)), and then got:
100 [4y^(3/2) / 3 - 2y^(5/2) / 5], from 0 to 1, and the answer I got was about 93.33 lbs of total pressure.

For part "b," I'm a little confused as to how to solve it. I set 160 equal to the line above, and then just divide both sides by 100...
Then, since the powers are negligible with limits of integration of 0 and 1, I get 1.6 = 4y/3 - 2y/5 --> 1.6 = 14y/15 ---> y = 1.7143, and that's definitely not the answer I'm looking for.

Last edited: Dec 4, 2014
2. Dec 4, 2014

Staff: Mentor

How does the "parabolic object" tie into the problem? The situation isn't at all clear from your problem description. A picture would be helpful, but you at least need to give a better description of what the problem is about.

3. Dec 4, 2014

leo255

Fair enough - Here is a very basic picture:

http://postimg.org/image/kcxr9rgwp/

EDIT: I think I had a brain-fart - The picture on the RIGHT is correct. On the left, it should look like the graph picture (y = x^2).

4. Dec 4, 2014

Ray Vickson

Fluid density cannot be 50 lb; it can be 50 lb "per cubic something or other", but you need to specify what that would be. Also: pressure cannot be 160 lb, but it can be 160 lb "per square something or other". So, you need to decide if when you say "pressure" you actually pressure, or do you really mean "total force"? It makes a real difference.

5. Dec 4, 2014

Staff: Mentor

Let's be more precise. The above is the pressure on a thin horizontal strip. When you integrate, you're getting essentially the sum of the pressures on a bunch of thin horizontal strips.

Your integral looks like this:
$$\int_0^1 50(2 - y) 2\sqrt{y}dy$$
The LaTeX for the above is this (but the spaces between the pairs of $symbols has to be removed). Code (Text):$ $\int_0^1 50(2 - y)2\sqrt{y}dy$ \$
Seems OK to me.

Last edited: Dec 4, 2014
6. Dec 4, 2014

Staff: Mentor

No, it's a bit more involved than that. When you were calculating the pressure on a thin horizontal strip, the "head" was (2 - y), the distance from the top of the fluid to the horizontal strip. You need to treat the top level of the water as a variable (don't use y). Write that integral and set it equal to 160, and solve for your variable.

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