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Hydrostatic force on a plane surface

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data

    im required to find the width of the strip of integration across the triangular lamina w(s) and then integrate to find the area

    I am given w(s) = x_right(s) - x_left(s) between 0<s<λh and λh<s<h




    2. Relevant equations



    3. The attempt at a solution

    i have found the equation i think for the first inequality using similar triangles

    w(s) = (d-b)s/λh + ds/h (do i need to find anonther equation for w(s) with for the second inequality?)

    due to the orientation of the triangle im not sure whether i have to split the region up into two sections when integrating to find the area

    i would really appreciate it if someone could point me in the right direction i dont want any solutions

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



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 28, 2012 #2
    You could split it into two sections and integrate over y, then sum the results. You have to find the equations of the lines that form the three sides of the triangle. Your limits of integration should go from -h to -lambda*h and from -lambda*h to 0 for the second integral.
     
  4. Apr 28, 2012 #3
    your function "w(s) = (d-b)s/λh + ds/h" cant be correct on 2 accounts.
    In the first term, why is d-b used and not b-d?. In the second term the dimensions are not correct. Some length is missing?
    "im not sure whether i have to split the region up" If the functions are different, they must be done separately.
     
  5. Apr 28, 2012 #4
    thank you both for your replies, yes your right i copied it up wrong i meant to put (b-d)

    how do you man some legnth is missing? a h canceled when calculating w(s). thats fine i was just unsure about the integration but i think that may help thanks again
     
  6. Apr 29, 2012 #5
    The term ds/h is dimensionless, whereas w(s) has units of length. (EDIT) Sorry, I misread the d as a delta. (second edit) Doesn't ds/h give the length xl(s). But that isn't exactly the length you want to include in w(s).
     
    Last edited: Apr 29, 2012
  7. Apr 29, 2012 #6
    erm for the first inequality w(s) would be given by xr(s) - xl(s)? so it would be (b-d)s/λh + h - ds/h - h = (b-d)s/h - ds/h. this would be a dimension of width as the h and λh would cancel with the s to leave b-d and d?
     
  8. May 2, 2012 #7
    I agree your first term, but the second term ds/h is not right. Can you do similar triangles on the triangles Odh and s xls h. ( I think there is a 1-lambda in there somewhere)
     
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