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Resultant force from water on dam

  • Thread starter HJKL
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  • #1
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Homework Statement

:
[/B]
The cross-sectional profile of a dam is shaped as an isosceles triangle. The dam is standing on horizontal ground and the watermirror is as high as the top of the dam. The dam has height h, lenght l and the distance from the bottom to the watermirror along the dam's side is s. The water's density is ρ.

a) Find the resultant force from the water on the dam.

b) Find the resultant force's point of attack.


Homework Equations


[/B]
Don't know any relevant equations.

The Attempt at a Solution


[/B]
I do have the solution for this, but I don't understand the equations used and where they come from.

Solution:
a)

dA = l dx

dF = ρgy dA
dF = ρgy l dx

Similar triangles:
y/h = (s-x)/s <=> y = (h/s) (s-x)

dF = ρgl (h/s) (s-x) dx

Resultant force:
F = integral of dF from 0 to s
F = integral of ρgl (h/s) (s-x) dx

Then that is solved and we get (1/2)ρghls.

Where does that first equation come from? dA = l dx?

b)
The torque about an axis through the point A:
xF F = integral of x dF from 0 to s = integral of x ρgl (h/s) (s-x) dx
ρgl (h/s) ((1/2) sx^2-(1/3)x^3) from 0 to s
ρgl (h/s) ((1/2 s^3 - (1/3) s^3) = (1/6) ρghls
=> xF =((1/6) ρghls^2)/ ρghls => 1/3 s

The photo attached is the sketch of the dam.
 

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Answers and Replies

  • #2
haruspex
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Where does that first equation come from? dA = l dx?
Do you understand that l is the length of the dam from end to end, i.e. in your diagram it is into the page?
l.dx is the area of a rectangle measuring dx up the slope of the dam and running the length of it horizontally.
 
  • #3
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No I didn't really understand any of it, but I see now. Thanks:)
I still don't understand why we use that equation though. And what it means.
 
  • #4
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Is A the area of that rectangle then? Why do we part it up like that and use dx and dA?
And is there an original equation or is dA = l dx the original equation?
 
  • #5
haruspex
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Is A the area of that rectangle then?
The slope is being divided into short (infinitesimal) elements of length, dx. From each such we can construct an area element, a rectangle measuring dx by l. We call an element of area dA.
is dA = l dx the original equation?
Yes.

The pressure varies according to depth, but over one thin element of area it is almost constant. If y is the depth of the area element dA then the pressure there is ρgy and the element of force, dF, it exerts on the area element is ρgydA.
Next, we need to express y in terms of x and dA in terms of dx: dF=ρg[h(s-x)/s][l.dx]. Now all the variables are in terms of x and dx we can integrate.
 
  • #6
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Ok, now I understand. Thank you!:)
 
  • #7
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Hey! Can somebody explain problem b)? I don't know if it's called point of attack, and I can't find anything when I search for point of attack. But it's the point where the resultant force acts, right? How do I find that?
 
  • #8
haruspex
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Hey! Can somebody explain problem b)? I don't know if it's called point of attack, and I can't find anything when I search for point of attack. But it's the point where the resultant force acts, right? How do I find that?
Use moments.
 
  • #9
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Ok. So M = F*a and in this case a is x? Then I get M = F*x. I already know F from problem a).
In the solution it says xF * F = integral of x * dF, why is that? And what are the two x's?
 
  • #10
haruspex
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in this case a is x?
If you are defining x as the distance up (or down?) the slope to the point of attack, yes.
integral of x * dF
Consider the pressure of the water on the slope as made up of elements of force, dF(x). The dF(x) element acts at distance x from the bottom, so its moment about the bottom is x.dF(x). The total moment is therefore ∫x.dF, and this equates to F.xpoint of attack.
Equivalently, you can consider the pressure P(x) acting on an element width dx at distance x from the bottom. This exerts a force P(x)L.dx and has moment P(x)xL.dx, so the total moment is ∫P(x)xL.dx.
 

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