Variation of pressure with elevation
Consider a hypothetical differential cylindrical element of fluid of cross sectional area A and height (z2 - z1).
Upward force due to pressure P1 on the element = P1A
Downward force due to pressure P2 on the element = P2A
Force due to weight of the element = mg = rA(z2 - z1)g
Equating the upward and downward forces,
P1A = P2A + rA(z2 - z1)g
P2 - P1 = - rg(z2 - z1)
Thus in any fluid under gravitational acceleration, pressure decreases, with increasing height z in the upward direction.
Equality of pressure at the same level in a static fluid:
Equating the horizontal forces, P1A = P2A (i.e. some of the horizontal forces must be zero)
Equality of pressure at the same level in a continuous body of fluid:
Pressures at the same level will be equal in a continuous body of fluid, even though there is no direct horizontal path between P and Q provided that P and Q are in the same continuous body of fluid.
We know that, PR = PS
PR = PP + rgh à 1
PS = PQ + rgh à 2
From equn.1 and 2, PP = PQ
General equation for the variation of pressure due to gravity from point to point in a static fluid:
Resolving the forces along the axis PQ,
pA - (p + dp)A - rgAds cos(q) = 0
dp = - rgds cos(q)
or in differential form,
dp/ds = - rgcos(q)
In the vertical z direction, q = 0.
Therefore,
dp/dz = -rg
This equation predicts a pressure decrease in the vertically upwards direction at a rate proportional to the local density.