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Gravity force term

  1. Jul 5, 2012 #1
    This is not a problem statement this is not homework this is not a textbook exercise. This is my own question about a formula in a textbook. I was given an infraction because I did not post this here.
    1. The problem statement, all variables and given/known data

    zxWrF.png
    Source: Fluid Mechanics 2nd ed. - P. Kundu, I. Cohen.pdf

    I am trying to understand the force balance, specifically the force due to gravity. Why is the (1/2) present?

    2. Relevant equations

    (1/2) * ρ * g * dx * dz


    3. The attempt at a solution

    I tried my own force balance and I agree with the book except for the gravity term:

    ƩF in vertical = 0 = +p2dx - Fgravity - p1ds*cos(θ)
    Fgravity = m*a = m*g = ρ*V*g
    m=V*ρ
    ƩF in vertical = 0 = +p2dx - ρ*V*g - p1ds*cos(θ)
     
  2. jcsd
  3. Jul 5, 2012 #2
    Oh I have received an email regarding the thread I started in the general physics section. Another user replied:

    "m=ρ*A*d <=> (since d=1, for its of unit thickness)
    m=ρ*(dxdz/2) = 1/2 ρ dx dz (since A is that of a triangle) and hence
    F= -1/2 ρ dx dz g"

    So is this correct: ?
    V = s*A
    A=1/2*x*y
    V=1/2*x*y*s
    s=1
    V=1/2*x*y

    How are you able to suddenly invoke differential length?
     
  4. Jul 5, 2012 #3

    Doc Al

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    Staff: Mentor

    It's a triangle, so its volume is 1/2 that of a rectangular element. Area = 1/2dxdz.
     
  5. Jul 6, 2012 #4
    How are you able to suddenly invoke differential length?
     
  6. Jul 6, 2012 #5

    Doc Al

    User Avatar

    Staff: Mentor

    What's sudden about it? The infinitesimal fluid element in your diagram has a triangular cross section with sides dx & dz.
     
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