Fluid Pressure and Fluid Force in Calculus II

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  • #1
Shay10825
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Hi. I need help on how to set up the integral for these problems.

1. A cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. Find the fluid force on a circular end of the tank if the tank is half full, assuming that the diameter is 3 feet and the gasoline weighs 42 pounds per cubic foot.
Answer: 94.5 lbs

Thanks
 
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Answers and Replies

  • #2
Fermat
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1. You have to find the force on the bottom half of a circular area. In effect, you will be finding the force (local pressure at the depth * element area) on an elemental strip (horizontal) and summing them (i.e. calculus integration). Have you done any of this yet ? i.e making up horizontal strips.
 
  • #3
Shay10825
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Fermat said:
Have you done any of this yet ? i.e making up horizontal strips.

No i have not. Why do you make horizontal strips? How would you set up the integral?
 
  • #4
Fermat
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2. It just looks like you should take the average/effective pressure acting on the porthole as the pressure of seawater at a depth of 15 ft.
And multiply that be the area of the porthole.
 
  • #5
Shay10825
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yeah I got number 2 a couple of seconds ago
 
  • #6
Shay10825
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How do I set up number 1?
 
  • #7
Fermat
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Shay10825 said:


No i have not. Why do you make horizontal strips? ...
That's what you do when doing integral calculus from basics.

Have you not done any of that stuff before? You should have, if you've been given that question.
 
  • #8
Shay10825
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Fermat said:
That's what you do when doing integral calculus from basics.

Have you not done any of that stuff before? You should have, if you've been given that question.

THe last thing we did was Centroids and for all of the problems I just use the formula in the book. But for this problem i keep using the book's formula but I can't get the correct answer. I don't make strips.
 
  • #9
Fermat
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You're not getting the right answer because the pressure varies over the depth of the fluid, and you can't use an average pressure because the area varies non-uniformly over the depth. That's why you have to use integral calculus and strips of area to work it out.

Have you never worked out the area under a curve using cakculus , where you use strips of small areas?
 
  • #10
Shay10825
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I used the formula:

F= w integral[ h(y)*L(y) dy
 
  • #11
Fermat
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Could you tell me what h(y) and L(y) are?

That integral looks like it will solve some problems - but what area is involved? square, circular?
 
  • #12
Shay10825
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h(y) = the depth of the fluid

L(y) = the horizontal length of the region y
 
  • #13
Fermat
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Ok.

What YOU have to do now is work out the correct expressions for h(y) and L(y).

I'll start you off. h(y) is h = y, where y is the depth of fluid below the half-way line.

L(y) is the length of the horizontal line at the depth y. i.e L is a chord of the circle, the mid-point of which is at a distance y from its centre.

The w in your integral expression should be the pressure at the depth y - is that correct ?

Edit: no - now I think w is the density of the fluid.
 
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