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Homework Help: Hydrostatics with density and area

  1. Apr 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A dam is made which is rectangular and flat in profile. It is a depth of 25m and a width of 100m and holds back fresh water which has a density of 1000 kg m3 .

    What is the total force that the water exerts on the dam?
    2. Relevant equations
    d/dz p=-ρg

    3. The attempt at a solution
    The hint given was: You will need to express the force as a function of vertical position, and then integrate over the possible values of vertical position.

    So the area would be 25(100)=2500 m2.

    You could also do some algebra and eventually get d/dz F=-ρgA and then you could integrate this....but it doesn't get you the correct answer.

    Any help would be much appreciated!
  2. jcsd
  3. Apr 2, 2015 #2
    What is the pressure at depth z, if z is measured downward from the surface?

  4. Apr 2, 2015 #3

    Andrew Mason

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    It is difficult for us to tell you where you are going wrong if you don't show us what you have done.

    The pressure is a linear function of z, let's call it p(z).(Your expression does not appear to contain the z term).

    So the force on the dam over a horizontal strip dz in width at depth z is p(z)A = p(z)wh = p(z)wdz where w is the width of the dam (100 m). Just integrate that over the whole depth (25 m). It should be a second order function of z.

  5. Apr 2, 2015 #4
    Ok thanks! So pressure would be ρgh=(1000)(9.8)(25)= 2.45*105 Pa

    then you could use the equation p=F/A to find F but this would give you 6.125*108 N which is double than the actual answer of 3.1*108 N
  6. Apr 2, 2015 #5
    I asked for the pressure as a function of z (algebraically). I repeat my question.

  7. Apr 2, 2015 #6
  8. Apr 2, 2015 #7
    Ok thanks!

    I'm still a little confused though...So p(z)A is the force right? And if you were to integrate this: p(z)wh = p(z)wdz you would get p(z)wz from 25 to 0?
  9. Apr 2, 2015 #8
    Good. So you can see that the pressure is varying with depth. So you can't just multiply the pressure at the bottom by the total area to get the force. When you did that, you got twice the correct answer. You need to integrate the pressure over the area. The differential area that the pressure at depth z is acting upon wdz. Now use that to integrate the pressure over the area.

  10. Apr 2, 2015 #9
    So F=ρA
    =ρgwx from 25 to zero

    I'm still doing something wrong though...
  11. Apr 2, 2015 #10
    No. ##F =\int pwdz=\int(ρgz)wdz## from 0 to 25. (Do you see why?)

  12. Apr 2, 2015 #11
    Yes I think so because F=P*A which you need to change P into density which gives ρgz. Also the pressure is always changing with the depth you cannot use the fixed area so you need the change in height times the length.

    Thanks for all your help!
  13. Apr 3, 2015 #12

    Andrew Mason

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    It is not exactly a matter of changing p into density. You are trying to determine force on a slice of the dam at a depth of z.

    F = pA where p is the pressure at a depth of z and A is the area of a strip of the dam that is w by dz. Pressure depends only on the depth. ##p(z) = mg/A = \rho Vg/A = \rho Azg/A = \rho zg##

    So ##F = \int_0^h p(z)dA = \int_0^h g\rho zdA = \int_0^h g\rho zwdz = wg\rho\int_0^h zdz##
  14. Apr 3, 2015 #13
    Hi Andrew.

    Are you sure you want to follow this particular line of reasoning? We have already gotten Patricia to accept the relation that ##p(z) = \rho zg##. Going back to the version p = W/A (where A is the horizontal area and W is the weight of the overlying fluid) is, IMHO, counterproductive, because it only works if the walls of the container are vertical. Also, you are using the symbol A for two different area entities, thus possibly adding to the confusion. Please consider removing or editing this portion of the text.

  15. Apr 3, 2015 #14

    Andrew Mason

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    Hi Chet.
    The pressure at depth z is determined by the weight of the water above z per unit area. How else does one determine the pressure at depth z? Because it is a fluid, the pressure is omnidirectional - so we can determine the horizontal force per unit area acting on the dam. What container are you referring to?

  16. Apr 3, 2015 #15
    If the vessel were a truncated cone with the smaller cross section on the bottom, the pressure on the bottom area would not be equal to the weight of the overlying liquid divided by the bottom area. The slanted side walls would help support the weight, since they would provide a component of force in the vertical direction. However, the pressure would always be ρgz, irrespective of the sidewall geometry.

  17. Apr 3, 2015 #16

    Andrew Mason

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    Yes. But the pressure is force/unit area. To determine that the pressure is ρgz you have to determine the weight of the water bearing on a unit of area. How are you suggesting we show that the pressure is ρgz? You can't just use Bernouilli - he had to work it out from first principles.

  18. Apr 3, 2015 #17

    What I'm saying is that we are dealing with a student here who is a little confused. Apparently in her course, they have gotten her to the point where she is now comfortable with p = ρgz, by whatever derivation. But if we take her back to p = W/A macroscopically, and she accepts that this can always be used, then, when she tries to apply it to something like my cone example, she is going to get the wrong answer, and then will be more confused than ever. You and I know how all this ties together mathematically, but she is new to the subject and probably doesn't. I feel happy that she has gotten to the point where she accepts p = ρgz. I'm just hoping that she is not motivated to take a step backward.

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