4.3 Density and pressure
- Syllabus
- 9702–2028–2029
- Topic
- 4.3
- Level
- AS
Learning objectives
4.3.1Density• define and use density4.3.2Pressure• define and use pressure4.3.3Derive, from the definitions of pressure and density, the equation• derive, from the definitions of pressure and density, the equation for hydrostatic pressure ∆p = ρg∆h4.3.4The equation ∆p = ρg∆h• use the equation ∆p = ρg∆h4.3.5The upthrust acting on an object in a fluid is due to a difference in• understand that the upthrust acting on an object in a fluid is due to a difference in hydrostatic pressure4.3.6The upthrust acting on an object in a fluid using the equation F = ρgV• calculate the upthrust acting on an object in a fluid using the equation F = ρgV (Archimedes’ principle)
Density is mass per unit volume and links material amount to size
Density is ρ=m/V. It is a property of a material sample at specified conditions, while mass and volume describe the particular object.
Use consistent units and distinguish an object’s external volume from the volume of material if voids or hollow regions are present.
A 0.54 kg metal block occupying 2.0×10⁻⁴ m³ has density 2700 kg m⁻³, consistent with aluminium.
A larger sample of the same uniform material is not automatically denser; changing mass and volume proportionally leaves ρ unchanged.
Pressure is force per unit area acting normally on a surface
Pressure is p=F/A, where F is the normal force distributed over area A. Its SI unit is the pascal, N m⁻².
Use the force perpendicular to the surface and the actual contact area. The same force produces greater pressure over a smaller area.
A 600 N person standing on 0.030 m² exerts an average pressure of 2.0×10⁴ Pa on the floor.
Pressure is not the total force, and forces parallel to a surface do not contribute to normal pressure.
Derive hydrostatic pressure from a fluid column
Consider a stationary fluid column of uniform density ρ, cross-sectional area A and vertical height difference Δh. The pressure difference supports the column's weight.
- Column volume V = AΔh. 2. From ρ = m/V, its mass m = ρAΔh. 3. Its weight F = mg = ρAΔh g. 4. From Δp = F/A, Δp = (ρAΔh g)/A = ρgΔh.
Δp=ρgΔh
This is a pressure difference between two levels in a fluid at rest with uniform density and uniform g. Atmospheric pressure cancels when it acts equally at both levels; container shape does not enter the derivation.
The pressure difference between two levels in a uniform fluid is ∆p=ρg∆h
Between two points separated vertically by ∆h in a fluid of uniform density, the pressure difference is ∆p=ρg∆h.
Only the vertical separation matters. Include a sign or state which point is deeper; pressure increases downward.
A 0.50 m level difference in oil of density 800 kg m⁻³ gives ∆p≈3.92 kPa at g=9.8 m s⁻².
Do not use the total container height when comparing two points, and do not reverse the pressure order with the depth sign.
Upthrust comes from the pressure difference between the lower and upper surfaces of an immersed object
Fluid pressure is greater at the lower surface than at the upper surface, giving a resultant upward force called upthrust.
The net force is caused by the pressure gradient, not by a mysterious property of the object. Compare upthrust with weight to predict rise, sink or equilibrium.
A submerged block has larger pressure on its bottom face than its top face, so the vertical pressure forces do not cancel.
Upthrust is not always equal to weight; equality occurs only when the object has no vertical acceleration.
The upthrust on a displaced volume of fluid is F=ρgV
For a fully or partly immersed object, upthrust equals the weight of displaced fluid: F=ρ_fluid g V_displaced.
Use the fluid density and displaced volume, not the object’s total volume unless it is fully submerged. Compare the result with object weight.
Displacing 2.0×10⁻³ m³ of water gives F≈19.6 N at g=9.8 m s⁻².
Upthrust depends on displaced volume and fluid density, not directly on the object’s material or mass.