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mikezietz
Nov22-04, 05:56 PM
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.

spacetime
Nov23-04, 03:31 AM
It is \rho (g-a) \delta h after taking incompressibility into account.
So, needs no change.


spacetime
www.geocities.com/physics_all

Andrew Mason
Nov23-04, 05:45 PM
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.
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 m_1g = \rho h_1Ag, the second is m_2g = \rho \triangle hAg and the uppermost is m_3g = \rho h_3 Ag 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:

m_3g - F_{dn3} = m_3a = 0

When the bucket falls with acceleration a,

m_3g - F_{dn3} = m_3a
m_3(g - a) = F_{dn3}

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

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

AM