P = F/A · pressure in fluids · atmospheric pressure · upthrust · floating · P = ρgh
🔢 Calculate pressure using P = F/A and state its SI unit (Pascal, Pa)
💧 Explain how pressure varies with depth in a fluid using P = ρgh
🌍 Describe atmospheric pressure and how it changes with altitude
⬆️ Define upthrust and explain it in terms of pressure difference
🚢 Apply the principle of floating: upthrust equals weight
📊 Solve multi-step problems combining P = F/A and P = ρgh
1 · Pressure: P = F/A
Pressure is defined as the force acting perpendicular to a surface per unit area. It tells us how concentrated a force is.
Pressure (Pa) = Force (N) ÷ Area (m²)
P = F / A
Symbol
Quantity
SI Unit
P
Pressure
Pascal (Pa) = N/m²
F
Force (perpendicular)
Newton (N)
A
Cross-sectional area
m²
1 Pa = 1 N/m². A larger contact area means a smaller pressure for the same force. This is why snowshoes stop you sinking — they spread your weight over a large area, reducing pressure on the snow.
Rearrangements:
F = P × A A = F / P
Pressure can also be measured in kilopascals (kPa) where 1 kPa = 1000 Pa, or atmospheres (atm) in everyday contexts.
2 · Pressure in Fluids & P = ρgh
A fluid is any substance that can flow — liquids and gases. In a fluid at rest, pressure acts equally in all directions at any given point. Pressure increases with depth because there is more fluid above pushing down.
The pressure at depth h in a fluid of density ρ:
P = ρgh
Symbol
Quantity
SI Unit
P
Pressure due to fluid column
Pa
ρ (rho)
Density of fluid
kg/m³
g
Gravitational field strength
N/kg (≈ 9.8 or 10)
h
Vertical depth below surface
m
P = ρgh gives the additional pressure due to the fluid column only. To find absolute pressure at depth h, add atmospheric pressure: Ptotal = Patm + ρgh.
Key facts about fluid pressure:
Pressure depends only on vertical depth, not on the shape of the container.
At the same depth in a connected fluid, pressure is equal in all directions.
Denser fluids (larger ρ) exert greater pressure at the same depth.
This principle is used in hydraulic systems — pressure applied at one point is transmitted throughout the fluid.
3 · Atmospheric Pressure
The Earth's atmosphere is a layer of gas extending hundreds of kilometres above the surface. Its weight creates a pressure on everything at Earth's surface. This is atmospheric pressure.
Standard atmospheric pressure at sea level ≈ 101 325 Pa ≈ 101 kPa ≈ 1 atm
Atmospheric pressure decreases with altitude because:
There is less air (fewer gas molecules) above you as you go higher.
The column of air above is shorter and less dense.
At the top of Mount Everest (~8850 m), pressure is roughly 34 kPa — about one third of sea-level pressure.
Atmospheric pressure is not constant — it varies with weather and altitude. A barometer measures atmospheric pressure. Pilots and mountaineers must account for reduced air pressure at altitude.
Gas molecules in the atmosphere are constantly moving and colliding with surfaces. Each collision exerts a tiny force; the total of billions of collisions per second on every surface creates atmospheric pressure. At sea level this is equivalent to a 10 m column of water pressing down on you — roughly 10 N on every cm² of surface.
4 · Upthrust & Archimedes' Principle
When an object is submerged (or partially submerged) in a fluid, the fluid exerts pressure on every surface of the object. Because pressure increases with depth, the upward force on the bottom face is greater than the downward force on the top face. The net upward force is called upthrust.
Upthrust = weight of fluid displaced by the object
U = ρfluid × Vdisplaced × g
This is Archimedes' Principle. The upthrust equals the weight of fluid that occupies the same volume as the submerged part of the object.
Upthrust acts upwards through the centre of buoyancy (the centroid of the displaced volume). It does not depend on the weight of the object itself — only on the volume displaced and the fluid density.
Why does pressure difference create upthrust? Consider a cuboid submerged in water:
Pressure on the bottom face: Pbottom = ρg(h + d) where d is the height of the cuboid
Pressure on the top face: Ptop = ρgh
Net upward force = (Pbottom − Ptop) × A = ρg·d·A = ρg·V = weight of displaced fluid ✓
5 · Floating, Sinking & Density
Whether an object floats or sinks depends on the balance between its weight (downwards) and the upthrust (upwards).
Condition
Result
Weight > Upthrust
Object sinks
Weight = Upthrust
Object floats (or is neutrally buoyant)
Weight < Upthrust
Object rises / floats at surface
An object floats when it displaces fluid equal in weight to its own weight. At the surface, it sinks until the weight of displaced fluid equals its own weight — then equilibrium is reached.
Density comparison shortcut:
If ρobject < ρfluid → object floats
If ρobject > ρfluid → object sinks
If ρobject = ρfluid → neutrally buoyant (hovers at any depth)
Submarines use ballast tanks: filling tanks with water increases overall density → sinks; pumping water out decreases density → rises. Fish use swim bladders to adjust their density. Hot air balloons heat air to reduce density below surrounding air density.
A floating object: Upthrust = Weight of object, so ρfluid × Vdisplaced × g = m × g, giving Vdisplaced = m / ρfluid
Example 1 · Basic Pressure Calculation A box weighing 360 N rests on the floor. Its base measures 0.40 m × 0.30 m. Calculate the pressure it exerts on the floor.
1Identify the formula: P = F / A
2Calculate the area: A = 0.40 × 0.30 = 0.12 m²
3The force is the weight: F = 360 N
4Substitute: P = 360 ÷ 0.12 = 3000 Pa
Pressure = 3000 Pa (3.0 kPa)
Example 2 · Fluid Pressure at Depth A diver descends to a depth of 25 m in seawater (density = 1025 kg/m³). Calculate the additional pressure due to the seawater at this depth. (g = 9.8 N/kg) Then find the total pressure on the diver if atmospheric pressure = 101 000 Pa.
1Use P = ρgh for fluid pressure only
2P = 1025 × 9.8 × 25
3P = 1025 × 245 = 251 125 Pa ≈ 251 000 Pa (251 kPa)
Additional fluid pressure ≈ 251 000 Pa (251 kPa) Total pressure ≈ 352 000 Pa (352 kPa)
Example 3 · Upthrust Calculation A solid metal sphere of volume 2.0 × 10⁻³ m³ is fully submerged in fresh water (density = 1000 kg/m³). Calculate the upthrust acting on the sphere. (g = 9.8 N/kg) The sphere has a mass of 15.6 kg. Determine whether it will float or sink.
1Upthrust = weight of water displaced = ρ × V × g
2U = 1000 × 2.0 × 10⁻³ × 9.8 = 2.0 × 9.8 = 19.6 N
3Weight of sphere W = m × g = 15.6 × 9.8 = 152.88 N
4Compare: W (152.88 N) >> U (19.6 N) → Weight exceeds upthrust
Upthrust = 19.6 N The sphere sinks because its weight (152.88 N) is much greater than the upthrust (19.6 N).
Example 4 · Floating Object A wooden block of mass 4.8 kg floats on fresh water (density = 1000 kg/m³). Calculate the volume of water displaced. (g = 9.8 N/kg) The block has dimensions 0.40 m × 0.30 m × 0.08 m. What fraction of the block is submerged?
1When floating: Upthrust = Weight → ρwater × Vdisplaced × g = m × g
5Check: density of wood = m/V = 4.8 / (9.6 × 10⁻³) = 500 kg/m³ — less than water ✓ floats
Volume displaced = 4.8 × 10⁻³ m³ 50% of the block is submerged (density of wood = 500 kg/m³ = half that of water)
Q1. A force of 500 N acts on an area of 0.025 m². What is the pressure?
Q2. Which statement about pressure in a fluid is correct?
Q3. Calculate the pressure at a depth of 8.0 m in fresh water. (ρ = 1000 kg/m³, g = 9.8 N/kg) Give your answer in Pa.
Q4. An object is fully submerged in water. Its weight is 30 N and the upthrust is 30 N. What will happen to the object?
Q5. A ship of mass 8 000 000 kg floats on seawater (ρ = 1025 kg/m³, g = 9.8 N/kg). Calculate the volume of seawater displaced. Give your answer in m³.
Challenge 1 · Multi-step Pressure Problem A hydraulic press has a small piston of area 5.0 cm² (5.0 × 10⁻⁴ m²) and a large piston of area 800 cm² (0.080 m²). A force of 40 N is applied to the small piston. (a) Calculate the pressure transmitted through the fluid. (b) Calculate the force exerted by the large piston. (c) Explain why this system is useful in engineering.
(a) P = F/A = 40 ÷ (5.0 × 10⁻⁴) = 80 000 Pa
(b) F = P × A = 80 000 × 0.080 = 6400 N
(c) The system acts as a force multiplier — a small input force (40 N) produces a much larger output force (6400 N) because pressure is transmitted equally throughout the fluid (Pascal's principle) and the output area is much larger than the input area. This is useful for car brakes, lifts and presses where large forces are needed from small inputs.
Challenge 2 · Upthrust & Density An aluminium block (density = 2700 kg/m³) has dimensions 0.10 m × 0.10 m × 0.05 m and is fully submerged in oil of density 850 kg/m³. (g = 9.8 N/kg) (a) Calculate the volume of the block. (b) Calculate the weight of the block. (c) Calculate the upthrust on the block. (d) Calculate the resultant force on the block and state its direction.
(a) V = 0.10 × 0.10 × 0.05 = 5.0 × 10⁻⁴ m³
(b) m = ρV = 2700 × 5.0 × 10⁻⁴ = 1.35 kg W = mg = 1.35 × 9.8 = 13.23 N ≈ 13.2 N (downwards)
(c) U = ρoil × V × g = 850 × 5.0 × 10⁻⁴ × 9.8 = 850 × 4.9 × 10⁻³ = 4.165 N ≈ 4.17 N (upwards)
(d) Resultant = W − U = 13.23 − 4.165 = 9.07 N downwards The block sinks because its weight exceeds the upthrust (ρAl = 2700 > ρoil = 850 kg/m³).
Challenge 3 · Depth from Pressure A submarine hull experiences a total water pressure of 4.12 × 10⁵ Pa. Atmospheric pressure is 1.01 × 10⁵ Pa and the density of seawater is 1025 kg/m³. (a) Calculate the gauge pressure (pressure due to seawater alone). (b) Calculate the depth of the submarine. (g = 9.8 N/kg) (c) Explain why submarines have thick hulls made of high-strength steel.
(b) Using P = ρgh → h = P / (ρg) = (3.11 × 10⁵) / (1025 × 9.8) h = 311 000 / 10 045 = 30.96 m ≈ 31 m
(c) At depth, the total pressure on the hull is very large (over 4 × 10⁵ Pa in this case, increasing further at greater depth). The hull must withstand this compressive force without deforming or imploding. Thick high-strength steel provides the structural integrity needed to resist these enormous inward forces, keeping the crew safe inside at near-atmospheric pressure.
Challenge 4 · Floating & Partial Submersion (Extended) An iceberg floats in seawater. The density of ice is 917 kg/m³ and the density of seawater is 1025 kg/m³. (a) Show that approximately 89% of an iceberg is submerged below the waterline. (b) An iceberg has a total mass of 2.5 × 10⁸ kg. Calculate the volume of the iceberg above the waterline in m³. (g = 9.8 N/kg) (c) Explain the danger this poses to ships.
(a) When floating: ρice × Vtotal × g = ρseawater × Vsubmerged × g
→ Vsubmerged/Vtotal = ρice/ρseawater = 917/1025 = 0.8946 ≈ 89.5% submerged ✓
(c) Because ~89% of the iceberg is hidden below the surface, ships can see only a small part of the total ice mass. The much larger submerged portion extends outwards and downwards, potentially further than the visible tip suggests. A ship could steer away from the visible ice but still strike the hidden underwater mass, causing catastrophic hull damage (as occurred with the Titanic).