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Hydrostatic Force set-up

  • Thread starter SPhy
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  • #1
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Even problem in text book.


Homework Statement



"Set up the integral to find the hydrostatic force on the face of the aquarium water tank, whose cross sectional area can be described by, y = e^-x^2 on 0.5≤x≤4.5 meters, resting at the bottom of the water 4 meters deep. Assume the bound y=0".



The Attempt at a Solution



Using the formula

F = bounds[0,a] ∫ρg(a-y)(w(y))dy , essentially, depth x width x gravity x water density

To find width, write y= e^-x^2 in terms of x, yields, x=√ln(1/y) consider only positive values.

integral so far,

F= bounds[0,a] ∫pg(a-y)√ln(1/y) dy

My issue is finding a. Since the book gives the condition y=0, I assume the height of the water = height of tank, so can I say a=4?

Hopefully that made sense. In many problems done so far that are similar to this one, im finding the force on a window, so my A value on the integral bound is different from my A value in the depth expression. I might be thinking about this problem completely backwards, so any help or suggestions would be welcomed!
 

Answers and Replies

  • #2
haruspex
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It is not clear to me whether y is a vertical or horizontal coordinate. "Resting at the bottom" suggests x and y are both horizontal, but then the problem reduces to finding an area.
 
  • #3
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The textbook gives no diagram. Although on a problem where the Y axis was horizontal, a diagram was given. So I'm assuming the X axis is horizontal here. My guess is the tank is resting at the bottom of the body of water, 4 meters deep, so height of water=height of tank?
 
  • #4
haruspex
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I think I get it. The face is a vertical plate. It is straight along the bottom edge and the vertical edges, but the height of the top edge is given by the exponential formula for y. The bottom edge is at a depth of 4m, and y is always less than 1, so the whole face is submerged. You want the total horizontal force on the plate.
You will need a double integral.
 

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