4.3 Density and pressure

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 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.

For a static fluid, the pressure increase caused by a depth h is p=ρgh, from weight per unit area of the fluid column.

Derive it using ρ=m/V and p=F/A; state whether the pressure is gauge pressure or includes atmospheric pressure.

Water 3.0 m below the surface adds about 2.94×10⁴ Pa gauge pressure when ρ=1000 kg m⁻³ and g=9.8 m s⁻².

Depth, not container shape, sets the pressure in a connected static fluid; the equation is not valid when density varies without adjustment.

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.

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.

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.