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Homework Help: Dam problem!

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a simple model of a dam. A reservoir of water behind the dam is filled to a height h. Assume that the width of the dam is L. As the reservoir behind a dam is filled with water, the pressure that the water exerts on the dam increases. Eventually, the force on the dam becomes substantial, and it could cause the dam to collapse.

    There are two significant issues to be considered: First, the base of the dam should be able to withstand the pressure. This means that the material of which the dam is made needs to be strong enough so that it doesn't crack (compressive strength).

    (a) What is the total force that the water in the reservoir exerts on the dam?

    The second issue has to do with the strength of the foundation of the dam. The force of
    the water produces a torque on the dam. In a simple model, if the torque due to the water
    were enough to cause the dam to break free from its foundation, the dam would pivot
    about its base (point P). The foundation of the dam should be strong enough so that the
    dam does not topple, which means that the material has to be strong enough that the dam
    does not snap (shear strength).

    (b) What is the magnitude of the torque about the point P due to the water in the reservoir?

    2. Relevant equations

    F= P X A

    3. The attempt at a solution

    For part a)
    break dam into little strips, each strip has a dif force...
    dF = PdA = density(g)(L)(h)dh <-- integrate this?

    = 1/2 (density *g * L)(h^2)???

    for part b)
    to find torque, sum up the little pieces?
    Integration of d(Torque) = integration of dFrmoment arms = integration of dFy

    from here I'm not sure where to go??? any help is appreciated :)
     
  2. jcsd
  3. Mar 13, 2010 #2

    Doc Al

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    Staff: Mentor

    OK. But when you are setting up your little strips, don't use 'h' as your variable, since h is a constant equal to the height of the water. (Don't forget about atmospheric pressure.)

    Yes. Use the same little strips as before, but this time find their torque. Set up an expression for dT similar to what you did for dF. (Heed my advice about changing your variable of integration.)
     
  4. Mar 13, 2010 #3
    Alright so if I used y instead of h, that is okay?
    so For part a)
    break dam into little strips, each strip has a dif force...
    dF = PdA = density(g)(L)(y)dy <-- integrate this?

    = 1/2 (density *g * L)(y^2)???

    and that seems about right? I think I can forget about atmospheric pressure because its the same on either side of the dam correct?

    moving on...

    Integration of d(Torque) = integration of dFrmoment arms = integration of dFy = integration of (density * g * L)y dy * y? = integration of (density)(g)(L)y^2dy
    = 1/3(density*g*L)y^3??

    Thanks for the help, correct me if I'm wrong (sure I did something wrong!)
     
  5. Mar 13, 2010 #4

    Doc Al

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    Perfect.

    Once you put in the limits of integration, you'll get it in terms of 'h'. (Not y--y is just a variable.)

    Depends what they are looking for. For compression, I'd think that you'd need to consider it.

    Almost. Hint: Where is y measured from? (Where is y = 0?)
     
  6. Mar 13, 2010 #5
    Oh so you're saying that h and 0 are the limits of the integration so when its integrated, I should get '= 1/2 (density *g * L)(h^2-0)??? I think that makes sense

    In regards to where y is measured from, I guess y is zero at the bottom? Hmm I don't really know where y is measured from
     
  7. Mar 13, 2010 #6

    Doc Al

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    Yes. (But you don't need to write the '-0' part.)

    Then you'd better figure it out. :wink: Hint: How did you determine the water pressure of each strip?
     
  8. Mar 13, 2010 #7
    Ohh hold on. So y would be zero at the surface of the water (y is the thickness) and it gets larger as you go down. Maybe this has to do with h? so maybe I'd have to convert this to h again like in the last problem?
    so 1/3(density*g*L)h^3??
     
  9. Mar 13, 2010 #8

    Doc Al

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    Right!

    Not exactly. Yes, you'll need to end up with 'h', not 'y', just like before. But that's not the correct answer. Hint: What's the torque from each strip? (Note that you are finding torque with respect to the bottom.)
     
  10. Mar 13, 2010 #9
    I thought I already posted the torque from each strip? 1/3(density*g*L)y^3?? (at least thats what I guess) If I'm finding torque with respect to the bottom, then that is the highest positive y value. I guess I'm confused as to where you are taking me here?? Should I not post that integration anymore? Should I take it back before it was integrated?
     
  11. Mar 13, 2010 #10

    Doc Al

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    Yes. Write the torque (dT) of each small strip as a function of y and h. Once you have that correct, then you can integrate.
     
  12. Mar 13, 2010 #11
    I worked with someone from my class on this problem and they said they ended up with this:

    1/2L*density*g*h^3 - 1/3L*density*h^3 = 1/6L*density*g*h^3

    They're pretty positive its right and I see where they get what is subtracted but how did they find 1/2L*density*g*h^3??
     
  13. Mar 14, 2010 #12

    Doc Al

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    If you do as I suggested in my last post and set up the correct expression for dT, then just by doing the integration you'll get both terms. Get the expression for torque correct--the rest is just turning the crank (integrating).
     
  14. Mar 14, 2010 #13
    Integration of torque = integration of dF * moment arms

    so 1/2 * integration of (density *g * L)(y)dy * y??

    thats the best I got, I hope thats (finally) right.

    end up with 1/6 density * g * L *h^3
     
  15. Mar 14, 2010 #14

    Doc Al

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    No, that's not quite right. Since 'y' is zero at the top of the water, it cannot be the moment arm.
     
  16. Mar 14, 2010 #15
    if y is zero, then should I use h instead?
     
  17. Mar 14, 2010 #16

    Doc Al

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    Express the moment arm in terms of y and h. (Remember that h is a constant.)
     
  18. Mar 14, 2010 #17
    moment arm = h-y

    1/2 * integration of (density *g * L)(y)dy * (h-y)

    so integration of density * g * l * y dy * h - integration of density * g * l * y dy * -y

    so 1/2 density * g * l * h^3 - 1/3 density * g * l * y * h^3

    = 1/6 density g * L * h^3!
     
  19. Mar 14, 2010 #18

    Doc Al

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    Yes!

    Good! (Except for that factor of 1/2.)

    Good!
     
  20. Mar 14, 2010 #19
    Thanks for stickin with me!

    :surprised:bugeye::eek:
     
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