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Hydrostatic force problem

  1. Sep 19, 2014 #1
    1. A vertical plate consists of rectangular and semicircular portion and has dimensions as shown. It is submerged in a liquid such that the upper edge coincides with the free surface of the liquid. What is the ratio of L/R such that force on the rectangular portion is the same as that on the circular portion?

    UT7mHCE.jpg



    2. Relevant equations
    F=γhA


    3. The attempt at a solution

    For rectangular part, since height of it is L and width of it is 2R
    A=2R*L
    h=L/2
    so force on retangular part is
    Fr=γ*2RL*L/2=γRL2

    for semi circular portion
    h=L+4R/3∏
    A=∏R2/2
    ∴Fsc=γ(L+4R/3∏)*∏R2/2

    Since Fr=Fsc

    2RL2=(L+4R/3∏)∏R2(γ cancelled out)
    .
    .
    L2=∏RL/2+2/3R2

    now I'm stuck here. Am I doing this right? or did I misunderstand the problem?

    English isn't my first language, so bear me.
     
  2. jcsd
  3. Sep 19, 2014 #2

    BvU

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    You're doing just fine. Remember you want an answer for L/R, so if you call that x, all you have to do is rewrite the last equation to get something with x and a few numbers...
     
  4. Sep 19, 2014 #3

    haruspex

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    Looks good so far. How can you turn this into an equation only involving the ratio L/R? How do you solve a quadratic?
     
  5. Sep 19, 2014 #4
    so..

    er.. so I divide both side with R^2

    (L/R)^2=pi*L/(2R)+2/3

    since L/R=x

    x^2=pi/2x+2/3
    x^2-pi/2x-2/3=0

    and this gives me two roots

    x1=1/12(3pi-sqrt(96+9pi^2)=-0.3475...
    x2=1/12(3pi+sqrt(96+9pi^2)=1.91832..

    so I pick the one bigger than 0; which is x2.

    so L/R should be 1.91832, right?
     
  6. Sep 19, 2014 #5

    haruspex

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    Looks right, but I would tend to give the answer in surd form, not reduce it to a numerical approximation.
     
  7. Sep 22, 2014 #6
    whoa.. it is hard to get back to old thread. Thanks!
     
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