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Hydrostatic Fluid- Linearly Accelerating Slope

  1. Mar 15, 2014 #1
    1. The problem statement, all variables and given/known data
    An open rectangular tank contains water up to about half of its depth. This tank accelerates at a=2.20 m/s[itex]^{2}[/itex] up a slope α (alpha), which causes the free water surface to form an angle θ (theta) with the original horizontal plane.
    What is the α angle of the slope?


    2. Relevant equations

    p=ρh(g[itex]\stackrel{+}{-}[/itex]a)
    tanθ=[itex]\stackrel{a}{g}[/itex]


    3. The attempt at a solution

    θ=tan[itex]^{-1}([/itex][itex]\stackrel{199}{1242}[/itex])=9.1°

    angle between water surface and slope = γ

    γ= α+θ

    tanγ=([itex]\stackrel{a}{g}[/itex])

    tanγ=([itex]\stackrel{2.2}{9.81}[/itex])

    γ=12.64°

    α=γ-θ=12.64-9.1=3.54°




    I am not sure if this is the right approach?

    Any help would be greatly appreciated.
    Thank you, Anna
     

    Attached Files:

    Last edited: Mar 16, 2014
  2. jcsd
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