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Learn9702Paper 1 · Multiple ChoiceForces, density and pressureDensity, pressure and upthrust
Forces, density and pressure · Subtopic 2 of 2

Density, Pressure and Upthrust

Pressure in a fluid increases with depth: Δp=ρgΔh\Delta p=\rho g\Delta h. Because the pressure on the bottom of a submerged object is greater than on its top, there's a net upward force, the upthrust, equal to the weight of fluid displaced: F=ρgVF=\rho gV. An object floats when the upthrust equals its weight.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 35 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use ρ=mV\rho=\dfrac mV and p=FAp=\dfrac FA.
  • Derive and use Δp=ρgΔh\Delta p=\rho g\Delta h.
  • Explain upthrust as due to a pressure difference.
  • Use F=ρgVF=\rho gV for submerged and floating objects.
Pause and think

A steel ball sinks in water but floats on mercury. What decides which happens?

Section 2 of 14

Core idea

Start hereLifting a friend in a pool

In a swimming pool you can lift a friend easily. The water pushes up on them. That push, upthrust, comes from the water pressure being greater at the bottom of their body than at the top.

The big idea

p=ρghp=\rho gh. Pressure difference across an immersed object gives upthrust =ρfluidgV=\rho_{\text{fluid}}gV: the weight of fluid displaced.

1. Pressure with depth

Upthrust: more pressure on the bottom face than the top
smaller pressure (shallower)bigger pressure (deeper)
Sideways pushes cancel. The bottom is deeper by hh, so the net upward force is ρgh×A=ρgV\rho g h\times A=\rho gV.
p=ρghp=\rho gh
  1. m=ρAhm=\rho Ah
    Column of fluid above area AA.
  2. W=ρAhgW=\rho Ahg
    Weight of the column.
  3. p=WA=ρghp=\frac WA=\rho gh
    Pressure from the column.

2. Upthrust

Pressure in water against depth
5101520501001502002503000depth (m)total pressure (kPa)p = p₀ + ρghatmosphere ≈ 101 kPa
The intercept is atmospheric pressure; every 10 m of water adds about one more atmosphere.
Archimedes from pressure
  1. Fup=ρg(h+L)A,Fdown=ρghAF_{\text{up}}=\rho g(h+L)A,\quad F_{\text{down}}=\rho ghA
    Block of height LL, top at depth hh.
  2. U=ρgLA=ρgVU=\rho gLA=\rho gV
    Difference; hh cancels.
Upthrust = weight of fluid displaced, independent of depth.
Worked exampleAluminium in water

V=0.002V=0.002 m³, ρ=2700\rho=2700 kg m−3^{-3}.

Goal
Apparent weight.
Strategy
Weight − upthrust.
  1. W=0.002×2700×9.81=53.0 NW=0.002\times2700\times9.81=53.0\text{ N}
    Weight.
  2. U=0.002×1000×9.81=19.6 NU=0.002\times1000\times9.81=19.6\text{ N}
    Water displaced.
  3. 53.0−19.6=33.4 N53.0-19.6=33.4\text{ N}
    Apparent weight.
Answer33.433.4 N.
✓ CheckRatio U/W=1000/2700U/W=1000/2700 ✓.
Worked exampleFloating ice

Ice 920 kg m−3^{-3} in sea water 1030 kg m−3^{-3}.

Goal
Fraction submerged.
Strategy
Floating: upthrust = weight.
  1. 1030Vsubg=920Vg1030V_{\text{sub}}g=920Vg
    Balance.
  2. VsubV=0.89\frac{V_{\text{sub}}}V=0.89
    Ratio of densities.
Answer89% under water.
✓ Check"Tip of the iceberg" ✓.
Common misconceptionObject density Upthrust uses the fluid density.

Summary

  • p=ρghp=\rho gh.
  • Upthrust =ρgV=\rho gV.
  • Floating: upthrust = weight.
Key words
Upthrust
Upward force from a fluid.
HistoryTorricelli's barometer. In 16431643 Evangelista Torricelli filled a glass tube with mercury and upended it in a dish. The mercury fell to about 7676 cm, leaving a vacuum above: atmospheric pressure was holding up the column. It was the first barometer, and the first man-made vacuum.
Fun factOnly a tenth of an iceberg shows. Ice is about 917917 kg m⁻³ and seawater about 10251025 kg m⁻³, so a floating iceberg has 9171025≈89.5%\tfrac{917}{1025}\approx89.5\% of its volume underwater. "The tip of the iceberg" is literal physics.
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