Lesson 2: Pressure at a Point Inside a Liquid

2.1 Observations

• Push a cork under water and release it: it rises back to the surface.
• So the water exerts an upward force on the submerged cork; this force comes from the difference in pressure acting on the bottom and the top of the cork.

• At any point inside a liquid the pressure acts in all directions.
• The force produced by liquid pressure on a surface is always perpendicular to that surface.
• The pressure increases with depth.

2.2 Deriving P = P_a + ρgh

• Imagine a horizontal plate of area A at depth h below the free surface of a liquid of density ρ.
• The plate carries a vertical column of liquid above it, of height h and cross-section A.

• Volume of the column = A h
• Mass of the column = ρ A h
• Weight of the column = F_g = ρ A h g

• Because the liquid is at rest and incompressible, the force on the plate due to the liquid equals the weight of this column.
• So the pressure due to the liquid alone is

P = F/A = ρ A h g / A → P = ρ g h (N/m²)

If the liquid surface is open to the atmosphere (atmospheric pressure P_a), this pressure is added, so the total (absolute) pressure at depth h is

P = P_a + ρ g h

The quantity ρgh is the gauge pressure — the pressure due to the liquid only.

The pressure at a point depends on:

• the depth h (greater depth → greater pressure),
• the density of the liquid ρ (greater density → greater pressure at the same depth),
• g and the pressure at the surface P_a. It does not depend on the shape or width of the container nor on the area of its base, nor on the amount of liquid.

2.3 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.

2.4 Worked examples

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.

2.5 Balance of two immiscible liquids in a U-tube

• Pour some water into a U-shaped tube.
• Then pour oil (which does not mix with water) into one arm until the oil/water separating surface is at A in that arm.
• Let the oil column above the separating level have height h_o and the water column above the same horizontal level in the other arm have height h_w.
• The points at the same level in the same liquid have the same pressure.
• Applying this to the two points at the level of the separating surface:

P_a + ρ_o g h_o = P_a + ρ_w g h_w → ρ_o h_o = ρ_w h_w or ρ_o / ρ_w = h_w / h_o

So by measuring h_o and h_w we can find the relative density of the oil (ρ_o/ρ_w), and its density if the density of water is known.

Example 1. The oil column is 10 cm and the water column is 8 cm: relative density of the oil = 8/10 = 0.8; ρ_o = 800 kg/m³.

Example 2 (unequal arms, book-style). A U-tube has a narrow arm of cross-section 1 cm² and a wide arm of 2 cm². Water is in the tube, and 6 cm of oil of density 900 kg/m³ is poured into the narrow arm. Height of the water above the separating level: h_w = ρ_o h_o/ρ_w = 900 × 6/1000 = 5.4 cm. (The cross-sectional areas of the arms do not matter, only heights.)

Example 3 (water and mercury). Mercury (13 600 kg/m³) forms a column of 5 cm in one arm; the water column above the mercury level in the other arm: h_w = 13 600 × 5/1000 = 68 cm.