Pick your class, read the steps, then try them right in the lab —
change one slider at a time and watch the rays and graphs respond.
Class 9 — Pressure in a liquid
Concept: pressure is force per unit area. In a liquid it grows with depth: P = P0 + ρgh, and at one depth it is the same in every direction.
- Open Pressure & Pascal's law, water, gauge at 2.0 m: ρgh = 1000 × 9.8 × 2 = 19.6 kPa, and with the atmosphere (101.3 kPa) the total is 120.9 kPa = 1.19 atm.
- Move the gauge to 10 m: ρgh = 98 kPa and the total pressure is 199.3 kPa, about two atmospheres. Every 10 m of water adds one atmosphere.
- Choose mercury at 0.76 m: ρgh = 13 600 × 9.8 × 0.76 = 101.3 kPa, as much as the whole atmosphere. A denser liquid gives more pressure at the same depth.
- The four red arrows are all equal: the pressure pushes equally in all directions at a point.
Takeaway: P = P0 + ρgh; it depends only on the depth, not on the shape of the vessel.
Class 9 — Buoyancy and Archimedes' principle
Concept: a body in a fluid feels an upward force equal to the weight of the fluid it displaces. It floats if its average density is less than the fluid's.
- Open Buoyancy & Archimedes: a 20 cm cube (8 L) of density 600 kg/m3 in water. Its weight is 4.8 × 9.8 = 47.0 N. Fully under, the buoyant force would be 1000 × 0.008 × 9.8 = 78.4 N, more than the weight, so it rises and floats with 60% under the surface (600 ÷ 1000).
- Change the density to 900 (ice): 90% under, 10% above. This is why most of an iceberg is hidden.
- Choose mercury and a density of 7800 (iron): it floats with 57.4% under, because mercury is denser than iron.
- Back in water, set 2700 (aluminium): it sinks, and held under it weighs 211.7 − 78.4 = 133.3 N, so it appears 37% lighter. Ships float because the hollow hull gives a low average density.
Takeaway: B = ρliquidVsubmergedg; floats if ρbody < ρliquid, with fraction submerged = ρbody ÷ ρliquid.
Class 11 — The barometer and Pascal's hydraulic press
Concept: a barometer balances the air pressure against a column of liquid, P = ρgh. In a closed liquid, a pressure change is passed on undiminished (Pascal's law).
- Choose The barometer with mercury: h = 101 325 ÷ (13 600 × 9.8) = 0.760 m = 760 mm. With water it would have to be 10.34 m tall.
- Raise the height above sea level to 8800 m (near Mount Everest): the pressure falls to 35.5 kPa and the mercury column to 267 mm. Air pressure falls to about half at 5.8 km.
- Choose The hydraulic press: A1 = 10 cm2, A2 = 400 cm2, F1 = 50 N. The pressure is 50 kPa everywhere, so F2 = 50 × 400 ÷ 10 = 2000 N: a force multiplied by 40.
- The big piston moves 40 times less: 50 cm pushed in on the small side lifts the load by only 1.25 cm. The work done is 25 J on both sides, so energy is conserved.
Takeaway: F2 = F1A2 ÷ A1, and d2 = d1A1 ÷ A2.
Class 11 — Continuity and Bernoulli's principle
Concept: in the steady flow of an incompressible liquid, A1v1 = A2v2, and P + ½ρv2 + ρgz is constant along a streamline.
- Open Continuity & Bernoulli: A1 = 20 cm2, A2 = 8 cm2, Q = 1.0 L/s. v1 = 0.001 ÷ 0.002 = 0.5 m/s and v2 = 1.25 m/s: the water is 2.5 times faster in the narrow part.
- With P1 = 8 kPa the pressure in the throat is P2 = 8000 + ½ × 1000 × (0.25 − 1.5625) = 7344 Pa. The pressure fell by 656 Pa as the water sped up; see the lower water level in the middle tube.
- Double the flow to 2.0 L/s: v1 = 1 m/s, v2 = 2.5 m/s and P2 = 5.38 kPa. The faster the flow, the lower the pressure in the throat.
- Raise the narrow part by 0.5 m: P2 = 7344 − 4900 = 2444 Pa (it also loses ρgz).
- The Venturi meter reads the flow from the difference in tube heights: v1 = A2√(2gΔh ÷ (A12 − A22)) = 0.5 m/s, the same as Q ÷ A1.
Takeaway: where the speed is high the pressure is low; this explains the lift on a wing and the action of a spray gun.
Class 11 — Torricelli's theorem
Concept: water leaves a hole at depth h with the speed v = √(2gh), the same as a body falling freely through h.
- Open Torricelli's jet with H = 1.00 m and the hole at y = 0.50 m. The depth of the hole is h = 0.5 m, so v = √(2 × 9.8 × 0.5) = 3.13 m/s. The jet falls through 0.5 m in t = √(2 × 0.5 ÷ 9.8) = 0.319 s and lands R = vt = 1.00 m away.
- Move the hole to y = 0.25 m: h = 0.75 m, v = 3.83 m/s, t = 0.226 s and R = 2√(hy) = 0.866 m. Then y = 0.75 m: the range is again 0.866 m.
- The range R = 2√(h y) is greatest when the hole is half-way up (y = H ÷ 2), where Rmax = H = 1.00 m. Holes at heights y and H − y have the same range.
- As the tank drains the depth h falls and the jet gets weaker and shorter. For a tank 300 times the area of the hole the level takes about 96 s to fall from 1.00 m to the hole.
Takeaway: v = √(2gh); R = 2√(h y); the maximum range equals the water depth H.
Class 11 — Viscosity, Stokes' law and Poiseuille's formula
Concept: a ball falling through a liquid reaches a terminal speed vt = 2r2(ρ − σ)g ÷ 9η. The flow through a tube is Q = πΔPr4 ÷ 8ηL.
- Open Viscosity: a steel ball of radius 2 mm in glycerine (η = 1.49 Pa·s). vt = 2 × (0.002)2 × (7800 − 1260) × 9.8 ÷ (9 × 1.49) = 3.82 cm/s, reached within a few milliseconds (τ = 4.7 ms). The Reynolds number is 0.129, so Stokes' law holds.
- Make the ball bigger: at 3 mm vt = 8.60 cm/s (it grows as r2, so 1.5 times the radius gives 2.25 times the speed). At 4 mm the formula gives 15.3 cm/s but Re = 1.04, above 1, and Stokes' law starts to fail.
- Change the liquid to honey (η = 7 Pa·s): vt = 0.79 cm/s for the 2 mm ball. In water the formula gives an absurd 59 m/s (Re = 237 000): the flow is turbulent and the law does not apply.
- Switch to the tube: water, ΔP = 4 kPa, r = 1 mm, L = 50 cm gives Q = π × 4000 × (10−3)4 ÷ (8 × 0.001 × 0.5) = 3.14 mL/s. Double the radius to 2 mm: Q = 50.3 mL/s, sixteen times as much (r4).
Takeaway: F = 6πηrv; vt ∝ r2; Q ∝ r4 and Q ∝ ΔP ÷ (ηL).
Class 11 — Surface tension and capillary rise
Concept: the surface of a liquid is like a stretched skin with a tension T (force per unit length). It makes liquids rise in narrow tubes and makes drops and bubbles round.
- Open Surface tension (capillary): water, T = 0.0728 N/m, tube radius 0.5 mm, contact angle 0°. h = 2T cosθ ÷ ρgr = 2 × 0.0728 ÷ (1000 × 9.8 × 0.0005) = 2.97 cm.
- Halve the radius to 0.25 mm: the water rises twice as high, 5.94 cm. The rise is inversely proportional to the radius.
- Choose mercury (T = 0.465 N/m, θ = 140°): cos 140° is negative, so the mercury is pushed down by 1.07 cm.
- Choose drops and soap bubbles: a water drop of radius 5 mm has an excess pressure 2T ÷ r = 29.1 Pa inside; a soap bubble of the same radius has two surfaces, 4T ÷ r = 4 × 0.025 ÷ 0.005 = 20 Pa. A smaller bubble has a larger excess pressure.
- Choose a soap film: a 5 cm wire is pulled by the film with F = 2Tl = 2 × 0.025 × 0.05 = 2.5 mN, and stretching it by 3 cm stores 0.075 mJ as surface energy.
Takeaway: h = 2T cosθ ÷ ρgr; ΔP = 2T ÷ r (drop), 4T ÷ r (soap bubble); work = T × increase in area.
Class 11 — Stress, strain and Young's modulus
Concept: stress = force ÷ area, strain = ΔL ÷ L, and in the elastic range Young's modulus Y = stress ÷ strain is a constant of the material.
- Open Elasticity: a steel wire (Y = 200 GPa), 2.0 m long and 1.0 mm in diameter (A = 7.85 × 10−7 m2) with a 100 N load. The stress is 127.3 MPa, the strain 6.37 × 10−4, and the wire stretches ΔL = 1.27 mm. The energy stored is ½FΔL = 63.7 mJ.
- The elastic limit of this steel is 250 MPa, which corresponds to a load of 196 N. Beyond it (the red line on the graph) the wire stretches permanently, and at about 353 N (450 MPa) it breaks.
- Change to copper (Y = 120 GPa): the stretch under the same stress is bigger. Steel stretches less than copper and aluminium: it is stiffer.
- Double the length: ΔL doubles. Double the diameter: the area is four times larger and ΔL falls to a quarter.
Takeaway: Y = (F ÷ A) ÷ (ΔL ÷ L), so ΔL = FL ÷ AY; U = ½FΔL.