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Center of pressure

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    pls refer to photo 2 , (circled part) How to transform equation in 3-20 to become equation in 3-22a ?


    2. Relevant equations


    3. The attempt at a solution
    i have tried in photo 3 , yet can get the same form as in 3-22a . In step 2 , i divided the y_p with Fr
     

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  2. jcsd
  3. Jan 29, 2016 #2

    haruspex

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    I can see an error in your working. In the left hand term, you substituted FR=P0A (wrong), but in the right hand term you correctly used PcA.
    I'll continue to study the text, but it is hard to read in places.

    Edit:
    I eventually figured out that the denominator in the last term of 3-22a reads ##(y_c+P_0/(\rho g \sin(\theta))A##, with which I agree.
    Do you get that with the above correction?
     
    Last edited: Jan 29, 2016
  4. Jan 29, 2016 #3
    no , i still gt stucked here
     

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  5. Jan 29, 2016 #4

    haruspex

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    Put the denominator in the second term back into the form PcA for now.
    Take the second term in the numerator (the one with yc2; you missed out the c subscript) and shift it into the first term after the equals sign. In the numerator there, you can extract a factor yc. The other factor is PcA, so cancels with the denominator.
     
  6. Jan 29, 2016 #5
    how to facrorise yc ? there are 3 terms there , but there are only 2 terms contain yc .... and i cant factorise PoA ...
     

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  7. Jan 29, 2016 #6

    haruspex

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    You have successfully factored out the yc. In the other factor, you have two terms each with a denominator PcA. So you can add together the numerators. I think you'll find those numerators add up to PcA.
     
  8. Jan 29, 2016 #7
    ok , i get my yc now , how to transform the ( (rho)g sin theta ( I xxc) / PcA )into the same form as 3-22a?
     
  9. Jan 29, 2016 #8

    haruspex

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    In that term, expand the Pc in the denominator into P0+ etc. form, then divide through numerator and denominator by (rho)g sin theta.
     
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