Pressure difference in a bucket sitting in an elevator

[mikezietz]

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]

It is \rho (g-a) \delta h after taking incompressibility into account.
So, needs no change.


spacetime
www.geocities.com/physics_all

 

[Andrew Mason]

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 F_dn3 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