3 Consequences

1. All points at the same horizontal level inside the same liquid at rest have the same pressure.
2. Connecting vessels: a liquid poured into vessels connected at the bottom rises to the same level in all of them, whatever their shapes (the pressure at their common base level must be the same). So the sea level is the same for all connected seas.
3. Dams are built thicker at the base to withstand the greater pressure at greater depths.
4. Divers wear protective suits. Near the surface, a diver blows air into the sinuses to equalise the pressure. At great depths (hundreds of metres) a pressurised diving suit and helmet protect him from the huge external pressure.

Example 1 (lake). Find the gauge pressure and the total pressure at a depth of 12 m in fresh water. (ρ = 1000 kg/m³, g = 9.8 m/s², P_a = 1.013 × 10⁵ N/m²) ρgh = 1000 × 9.8 × 12 = 1.176 × 10⁵ N/m²; total P = 1.013 × 10⁵ + 1.176 × 10⁵ = 2.189 × 10⁵ N/m² (about 2.2 atm).

Example 2 (force on a tank bottom). An oil tank (ρ = 850 kg/m³) is filled to a depth of 2.5 m. Its base area is 4 m². (g = 10 m/s², P_a = 1.013 × 10⁵ N/m²) ρgh = 850 × 10 × 2.5 = 21 250 N/m². Total pressure P = 1.013 × 10⁵ + 0.2125 × 10⁵ = 1.2255 × 10⁵ N/m². Total force on the base = P × A = 1.2255 × 10⁵ × 4 = 4.9 × 10⁵ N. (Force due to the oil alone: 21 250 × 4 = 85 000 N.)

Example 3 (diver). A diver is at 30 m in sea water (ρ = 1030 kg/m³). ρgh = 1030 × 9.8 × 30 = 3.03 × 10⁵ N/m²; total = 4.04 × 10⁵ N/m² ≈ 4 atm.

Example 4 (depth from pressure). At what depth in water is the total pressure 3 atm if the surface pressure is 1 atm? The extra 2 atm = 2 × 1.013 × 10⁵ = ρgh → h = 2.026 × 10⁵/(1000 × 9.8) = 20.7 m.