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Homework Help: Amount of Force that Water Exerts on a Dam

  1. May 3, 2009 #1

    1. The problem statement, all variables and given/known data

    Calculate the total force the Oahe Dam, placed on a structure by the reservoir behind it (in pounds and Newtons).

    Oahe Dam: height=75meters, width=2835meters

    2. Relevant equations

    Pressure = Force / Area
    Force = Pressure x Area
    Hydrostatic Pressure: Atmospheric Pressure + (Density x Gravity x Height)

    Density of Water= 1000kg/m^3
    Gravity= 9.81 m/s^2

    3. The attempt at a solution

    - I can use Calculus on this, I know that... but I'm not in calculus yet, so I'm attempting to solve algebraically at least.

    - I understand that pressure increases as the depth increases... so pressure isn't the same in all levels (right?)

    What I tried doing is calculating pressure using (density x gravity x height) in every height...
    For example,
    (1000*9.8*1m) + (1000*9.8*2m) + (1000*9.8*3m)... so forth until I reach 75m.
    Then I found force by using F=P*A...
    the area would be (75m * 2835m)
    & I get a total of 5.94 x 10^12 N as the force
    BUTTTTT. I think the force is too high... =/

    I should also consider atmospheric pressure, but I don't know where to include that (if I even should!)

  2. jcsd
  3. May 3, 2009 #2
    That is certianly one way to go about doing the problem. However...

    Because 75m isn't very big in magnitude, and it's close to the Earths surface, then the force on a 'cube of water' due to gravity. Can be considered;


    Because our relationship for g is linear. Then we can suitably approximate it over the entire height of the dam.

    So half of 75m is 37.5m

    The presure at this depth is Rgh. So we find that the pressure is 3.68x105 and is the average value for the pressure over the entire dams area. Hence with the area in total being 2.13x105m2 the force is approximatly 7.82x1010 note I left out the atmospheric pressure. If we include that 101.3kPa is 1 Atmos.

    Then the pressure at average depth is Rgh + Atmos. or 4.69x105, however remember that this same pressure is on the other side of the dam pushing back as well against the weight of water. Hence we don't really need to include it in the calculation.

    The trouble with your calculation is that you are summing up sections of water that you have already calculated.

    i.e. (1000*9.8*1m) + (1000*9.8*2m) + (1000*9.8*3m)...

    The second term includes the first term, and the third includes the second and first. Which makes it inaccurate.

    If you want to do it that way, it's an expansion of the Trapezium Rule and you should be doing each 'slice' seperatly. However I am quite confident that doing that is overcomplicating the problem. But I could be wrong.

    Hope that helps,
  4. May 3, 2009 #3
    U used d formula P = density x g x h very right n haths has correctly pointed out ur mistake of mixing up sections of previous layers in next terms..now d solution of ur problem s simple.. D approach s to find Pressure acting at the bottom layer n then using F=P.A.. - Now u r adding each LAYERS' pressure n thats only right when u have only one variable or otherwise u need calculus to sum-up d values(known as Integration).. Luckily u can Ignore change in Gravity as h <<<< R...so u get only one variable n that s height..so now simply do P=density x g x h(=75m) n then add to it d atmospheric pressure n then do F=P.A..well done..u did it :-)
  5. May 3, 2009 #4


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    Homework Helper

    Welcome to PF!

    Hi kawaiixsarang! Welcome to PF! :smile:
    Yes, that's basically the correct method …

    and we can tidy it up as (1000*9.8)*(1 + 2 + … + 74) …

    now :rolleyes: … what is the formula for sum in the second bracket? :wink:
    Atmospheric pressure will be the same on both sides of the dam (well, except for 75 feet of air, whose weight is negligible), so you can ignore it. :smile:
  6. May 3, 2009 #5
    Re: Welcome to PF!

    So I used the formula: n(n+1)/2 ; where n is the height of the dam.
    When I sum up all the numbers on the right bracket I get a total of 2850.
    And I multiplied that by 1000*9.8 and I get 2.793 x 10^7...
    & I found F=PxA but it doesn't workkkk because I still get the same answer from the beginning 5.94 x 10^12 :/

    What am I doing wrong??
    Last edited: May 3, 2009
  7. May 3, 2009 #6
    thank you.

    i tried doing your suggestion, so I got:

    1000 * 9.8 * 75 = 7.35 x 10^5 as pressure
    Then I added it on to 101,300 N/m^3 (atmospheric pressure)
    and I did F=PxA like you said...
    which gave me 1.778 x 10^11

    but according to Haths... it should be 7.8 x 10^10

    What am I doing wrong??? =/
  8. May 3, 2009 #7
    Yes, that would give an answer ~75 times too high.

    What you have done is add up the forces exerted on each
    square metre in a vertical strip 1m wide.
    To get the average pressure (force per square metre) you need to divide
    by 75.
    Or simply observe that there are 2835 of these metre wide vertical
    strips in the wall, so multiply your sum by 2835 to get the total force
    on the dam.

    Last edited: May 3, 2009
  9. May 3, 2009 #8
    As tiny Tim said, the contribution that atmospheric pressure makes
    to the water pressure is cancelled out by the atmospheric pressure
    on the outside of the wall.

    1000*9.8*75 gives the pressure at the bottom.
    What you want is the average pressure, which is half of this.

    Now multiply by area and you get 7.8 x 10^10N
  10. May 3, 2009 #9
    Thank you very much for your help!! =)

    I'm assuming all of you tried giving me different ways to solve this problem??
  11. May 4, 2009 #10


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    Homework Helper

    Hi kawaiixsarang! Thanks for the PM! :smile:

    Easier to say (1 + 2 + … + 74) = 1/2 74*75 …

    remember the general formula (1 + … + n) = 1/2 n(n+1)? :wink:

    So the total is very nearly 1/2 752

    and if you subdivided it even more, it would "tend to" exactly 1/2 752

    However, I've just noticed another more intuitive way of doing it (without caclulus) …

    turn the dam on it side! :biggrin:

    then the force on each part of the dam can be represented by a weight of water of height h …

    that gives you a triangle of water of height h and width h, and the total force is the total weight fo the triangle, which is … ? :smile:
  12. May 4, 2009 #11
    Thank youuu! =)

    I understand now. When I solve the entire problem I get out: 7.813e10 =)
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