Pressure in Solids and Liquids
Senior 2 · Physics
Pressure in Solids and Liquids
Lesson 2 of 5
25 min
Let's begin
Dive to the bottom of a swimming pool and your ears hurt. Punch holes at different heights in a jerrycan full of water and the lowest jet spurts furthest. Dam walls, such as those on hydropower plants, are built much thicker at the bottom than at the top. All of this happens because a liquid presses harder the deeper you go.
Picture a column of liquid of height h and base area A above a point. Its volume is A × h, its mass is ρ × A × h (where ρ is the density), and its weight is ρAhg. Dividing that weight by the area A gives:
P = hρg
Here h is the depth in m, ρ is the density in kg/m³ and g is the gravitational field strength (9.8 N/kg). This is the pressure due to the liquid alone. At an open surface the air presses down too, so the total pressure = atmospheric pressure + hρg.
What is the pressure due to the water alone at a depth of 10 m in Lake Kivu? Take ρ = 1000 kg/m³ and g = 9.8 m/s².
P = hρg = 10 × 1000 × 9.8 = 98 000 Pa (98 kPa). That is more than the whole atmosphere presses at the lake surface: Lake Kivu lies about 1460 m above sea level, where air pressure is only about 85 kPa. A diver at 10 m therefore feels a total pressure of about 85 + 98 = 183 kPa, more than double the pressure at the surface.
A water tank stands so that its water surface is 5.0 m above a tap. What is the water pressure at the tap (above atmospheric)? Take g = 9.8 m/s².
P = 5.0 × 1000 × 9.8 = 49 000 Pa. The higher the tank, the stronger the flow from the tap.
Because points at the same level must have the same pressure, the free surface of a liquid at rest is always horizontal. If one part of the surface were higher, the extra pressure beneath it would push liquid sideways until the levels were equal again.
Open the Under Pressure simulation. Drag the pressure gauge down into the pool and watch the reading rise steadily with depth. Then try the trapezium-shaped pool: at the same depth, the reading is the same whatever the shape.
Tip
"h-rho-g": depth × density × g
Use the vertical depth below the surface, never the length of a sloping pipe. Check whether the question wants the pressure due to the liquid alone or the total pressure (add atmospheric pressure).
Find the pressure due to the water at the bottom of a pool 2.0 m deep (ρ = 1000 kg/m³, g = 10 m/s²).
Solution: P = 2.0 × 1000 × 10 = 20 000 Pa (20 kPa).
At what depth in fresh water is the water pressure 49 000 Pa? (g = 9.8 m/s²)
Solution: h = P ÷ (ρg) = 49 000 ÷ (1000 × 9.8) = 5.0 m.
Holes are made at the top, middle and bottom of a jerrycan full of water. Which jet spurts furthest, and why?
Solution: The bottom jet, because it is deepest below the surface, where the pressure hρg is largest.
Sea water has a density of 1030 kg/m³. Find the pressure due to the water at a depth of 50 m (g = 9.8 m/s²).
Solution: P = 50 × 1030 × 9.8 = 504 700 Pa ≈ 5.0 × 10⁵ Pa.
Summary
Sources and credits
The Rwandan examples, worked examples and practice items are written by Ganzaa.