Question 6.3.2: Figure 1 shows a vertical column of motionless liquid that i...
PROOF THAT HYDROSTATIC PRESSURE INCREASES LINEARLY WITH DEPTH
Figure 1 shows a vertical column of motionless liquid that is part of a larger body of liquid of constant density \rho_{liq} . The column, of height h, extends from depth h_{1} to depth h_{2}, and has a cross sectional area dA. The pressure is p_{1} at depth h_{1} and p_{2} at depth h_{2} . Use this column to prove that hydrostatic pressure increases linearly as a function of depth h in a homogeneous motionless liquid.
Goal Prove that hydrostatic pressure increases linearly as a function of depth.
Given The dimensions of an arbitrary column of liquid; the liquid is motionless and of constant density.
Assume No assumptions necessary.
Draw We draw a free-body diagram of the column of liquid, keeping in mind that in a liquid at rest, the pressures must act perpendicular to any surface (Figure 2).
Formulate Equations and Solve We apply equilibrium equations to the column of fluid in the vertical direction.
\sum{F_{y} }=0\left(\uparrow + \right)
p_{2} dA − p_{1} dA − W = 0 (1)
where W is the weight of the fluid in the column, which is given by
W=\rho_{liq} gV_{tot}=\rho_{liq} gh dA (2)
Substituting (2) into (1) gives
p_{2} dA – p_{1} dA – \rho_{liq} gh dA = 0
Solving for p_{2} gives
p_{2} = p_{1} + \rho_{liq} gh (3)
This is the equation of a line representing p_{2} as a function of depth, with intercept p_{1} and slope ρliq g (Figure 3). Therefore we can conclude that the pressure varies linearly with depth.
Note: Equation (6.21A) is a special case of (3), when p_{1} is located at the surface of the liquid and is therefore p_{o} , the atmospheric pressure.
p = p_{o} +\rho_{liq} gh (6.21A)
