Pascal law and hydrostatics

A fluid at rest cannot support shear stress, so at a given point the pressure is the same in all directions and is defined as $p = F/A$.

Pressure increases with depth:

$$p = p_0 + \rho g h,$$

where $p_0$ is the pressure at the surface, $\rho$ is the fluid density, and $h$ is the depth.

Gauge pressure is $\rho g h$, while absolute pressure includes the atmospheric term $p_0$.

Pascal's law: a change in pressure applied at the boundary of an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

Therefore, in a hydraulic press $F_1/A_1 = F_2/A_2$: the force is multiplied by the same factor as the area.

Incompressibility requires $A_1 d_1 = A_2 d_2$, so the force gain is paid for by a distance loss: $F_1 d_1 = F_2 d_2$.

At what depth in water is the gauge pressure equal to 49 kPa? Take $\rho = 1000$ kg/m³ and $g = 9.8$ m/s².

In a hydraulic press, a small piston of area $A_1 = 1.0 \times 10^{-3}$ m² acts with a force $F_1 = 50$ N. The large piston has area $A_2 = 0.25$ m². Find the force $F_2$ and the displacement of the large piston if the small piston moves down by $d_1 = 0.20$ m.

Water and oil are at rest in a U-tube. In one arm water stands to a height $h_w = 0.20$ m above the interface level, and in the other arm oil stands to a height $h_o = 0.25$ m. Find the density of the oil if the density of water is $\rho_w = 1000$ kg/m³.

Safe model of a hydraulic press. Take two medical syringes of different diameters without needles and connect them with a flexible tube filled with water (no air bubbles). Push the small syringe's plunger and note the force delivered by the large plunger; record the displacements of both plungers with a ruler. Check the relations $F_2/F_1 = A_2/A_1$ and $d_1/d_2 = A_2/A_1$.

Simulation for depth. Because measuring pressure at large depths at home is hazardous, use an interactive fluid-pressure simulation: vary the depth and the fluid density, record $p$, and plot $p$ versus $h$. Do not use mercury, open flames, or sealed vessels under pressure.

Show that pressure at a point is the same in all directions, that it grows linearly with depth, and that the force ratio in a hydraulic press equals the ratio of piston areas.

A three-part diagram: (1) a container of fluid with a marked point from which equal-length arrows point in all directions; (2) a graph of $p$ versus $h$ — a straight line with slope $\rho g$ starting at $p_0$; (3) two pistons of areas $A_1$ and $A_2$ connected by fluid, with arrows $F_1$ and $F_2$ and marked displacements $d_1$ and $d_2$.

At a garage, a student pumps a manual hydraulic jack. They press the small piston with force $F_1=200\ \text{N}$; the small piston area is $S_1=5\ \text{cm}^2=5\cdot10^{-4}\ \text{m}^2$. The large piston has area $S_2=200\ \text{cm}^2=0.020\ \text{m}^2$ and lifts a car of mass $800\ \text{kg}$ (weight about $7.85\cdot10^3\ \text{N}$). The small piston moves down by $h_1=0.20\ \text{m}$.

The scenario uses Pascal's law: pressure $p=F_1/S_1=200/(5\cdot10^{-4})=4\cdot10^5\ \text{Pa}$ is transmitted unchanged through the oil to the large piston. Then $F_2=pS_2=4\cdot10^5\cdot0.020=8.0\cdot10^3\ \text{N}$, greater than the weight of $800\ \text{kg}$, so the car rises. Incompressibility of the oil links displacements: $S_1h_1=S_2h_2$, hence $h_2=0.005\ \text{m}$.

A dam holds water to depth $H=20\ \text{m}$. At depth $h=20\ \text{m}$, the gauge pressure of the water is $p=\rho gh=1000\cdot9.81\cdot20=1.96\cdot10^5\ \text{Pa}$. Absolute pressure adds the atmosphere: $p_{\text{abs}}=1.96\cdot10^5+1.01\cdot10^5=2.97\cdot10^5\ \text{Pa}$. On every square metre at the bottom, the water pushes with force $F=pS=1.96\cdot10^5\ \text{N}$.

Here the hydrostatic law $p=\rho gh$ operates: pressure grows linearly with depth, so the lower part of a dam is thicker and stronger. Pascal's law adds that at the same depth pressure is the same in all directions, so water pushes both perpendicular to the wall and upward on gates.

In a pool, a swimmer dives to depth $h=10\ \text{m}$. The pressure created by water alone is $p_{\text{water}}=\rho gh=1000\cdot9.81\cdot10=9.81\cdot10^4\ \text{Pa}$. Absolute pressure with the atmosphere is about $1.99\cdot10^5\ \text{Pa}\approx2\ \text{atm}$. At depth $5\ \text{m}$, the water pressure is only $4.9\cdot10^4\ \text{Pa}$.

The scenario practices the hydrostatic formula $p=\rho gh$ and Pascal's law: water pressure is transmitted in all directions, so the eardrum feels an extra pressure of about $9.8\cdot10^4\ \text{Pa}$ already at $10\ \text{m}$. Comparing $5\ \text{m}$ and $10\ \text{m}$ shows the linear dependence of pressure on depth.