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Pressure difference in a bucket sitting in an elevator

  1. Nov 22, 2004 #1
    if a bucket is on an elevator floor in a inconpressible fluid of density p. When the elevator is accelerating downards, what is the pressure difference between two points in a fluid, separated by a vertical distance of delta h.

    I was thinking that it is p (g - a) delta h, since it should be taking away from g, but could it be g + a, or could the whole equation be pa delta h, p g delta h, or p g a delta h, because of the inconpressiblity and/or the possiblity of the pressure being the same, can i have some hints on this please, thanks.
     
  2. jcsd
  3. Nov 23, 2004 #2
    It is [tex]\rho (g-a) \delta h[/tex] after taking incompressibility into account.
    So, needs no change.


    spacetime
    www.geocities.com/physics_all
     
  4. Nov 23, 2004 #3

    Andrew Mason

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    Think of the bucket of water in terms of three layers of water - just masses sitting on top of each other. The weight of the lowest mass is [itex]m_1g = \rho h_1Ag[/itex], the second is [itex]m_2g = \rho \triangle hAg[/itex] and the uppermost is [itex]m_3g = \rho h_3 Ag[/itex] where A is the area of the bucket (assume the bucket has vertical sides).

    When the bucket is at rest, the top layer (m3) is exerting an downward force Fdn3 of m3g on m2.

    At rest:

    [tex]m_3g - F_{dn3} = m_3a = 0[/tex]

    When the bucket falls with acceleration a,

    [tex]m_3g - F_{dn3} = m_3a[/tex]
    [tex]m_3(g - a) = F_{dn3}[/tex]

    Since [itex]m_3 = \rho h_3 A[/itex], the pressure (divide force by A) is: [itex]P = \rho h_3(g - a)[/itex]

    Do a similar analysis of the pressure on the bottom layer and you will see that the pressure difference is
    [itex]\triangle P = \rho \triangle h (g-a)[/itex]

    AM
     
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