The shape of the lesson
By the end of the lesson, learners can
- Float or sink: Lower a 1 litre block into the liquid and change the two densities.
- Three relations to own: A body floats when its average density is less than the fluid's.
- Why depth, not shape.
- Density: ρ = m / V; pressure: p = F / A.
- Depth: Δp = ρgΔh; add atmospheric for total pressure.
- Upthrust: F = ρgV, the weight of fluid displaced.
One picture every idea returns to
Density is mass per unit volume, ρ = m / V. Pressure is force per unit area, p = F / A, in pascals. In a fluid, pressure increases with depth: Δp = ρgΔh. Because the deeper face of a submerged body is pushed harder than the upper face, there is a net upward force, the upthrust, equal to the weight of the displaced fluid: F = ρgV (Archimedes' principle).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Why things float.. Pressure in a fluid grows with depth, pushing harder on the deeper face of any submerged object. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Float or sink | Lower a 1 litre block into the liquid and change the two densities. Project tab 3 of the lab (Density, upthrust and floating) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Three relations to own | A body floats when its average density is less than the fluid's. | Pairs |
| 21 to 30 min | Teach and check: Why depth, not shape | Why depth, not shape. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Why depth, not shape, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Liquid pressure depends on depth and density only, never on the shape of the container or the area of its base; a narrow tube and a wide tank filled to the same depth have the same pressure at the bottom. | When asked for the total pressure at a depth, add the atmospheric pressure: p = p_atmosphere + ρgΔh. And the upthrust uses the submerged volume and the fluid's density, not the object's. |
Density and pressure: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
A metal cube of side 0.050 m has a mass of 1.0 kg. Find its density.
Find the pressure exerted on the floor by a box of weight 600 N standing on a base of area 0.30 m².
Show that the extra pressure at a depth h in a liquid of density ρ is ρgh, by considering the weight of a column of liquid of cross-sectional area A.
A swimming pool is 2.5 m deep and filled with water of density 1000 kg m⁻³. Taking g = 9.8 m s⁻² and atmospheric pressure as 1.0 × 10⁵ Pa, find the total pressure at the bottom.
A block of volume 2.0 × 10⁻³ m³ is held fully submerged in water of density 1000 kg m⁻³ (g = 9.8 m s⁻²). Find the upthrust on it. If the block weighs 12 N, state whether it would float or sink when released.
An iceberg of density 920 kg m⁻³ floats in seawater of density 1025 kg m⁻³. Find the fraction of the iceberg that is below the surface, and comment on the saying about an iceberg.
Answer key and marking notes
- Volume = 0.050³ = 1.25 × 10⁻⁴ m³. ✓
- ρ = m / V = 1.0 / (1.25 × 10⁻⁴) = 8000 kg m⁻³. ✓
p = F / A = 600 / 0.30 = 2000 Pa. ✓
- A column of height h and area A has volume Ah, mass ρAh, and weight ρAhg. ✓
- This weight presses on the area A at the base. ✓
- Pressure = force / area = ρAhg / A = ρgh. ✓
- Hydrostatic pressure = ρgh = 1000 × 9.8 × 2.5 = 24 500 Pa. ✓
- Total pressure = atmospheric + hydrostatic = 1.0 × 10⁵ + 0.245 × 10⁵. ✓
- Total = 1.2 × 10⁵ Pa (to 2 significant figures). ✓
- Upthrust = ρgV = 1000 × 9.8 × 2.0 × 10⁻³ = 19.6 N. ✓
- The maximum upthrust (19.6 N) is greater than the weight (12 N). ✓
- So when released it rises and floats, settling with part of its volume above the surface. ✓
- Submerged fraction = ρ_ice / ρ_sea = 920 / 1025 = 0.90. ✓
- So about 90% is below the surface, and only about 10% shows, matching the idea of the hidden bulk of an iceberg. ✓
17 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Density is defined as:
- Aweight per unit area
- Bmass per unit volume
- Cforce per unit area
- Dvolume per unit mass
Pressure is defined as the:
- Aforce times area
- Bmass per unit volume
- Cweight per unit length
- Dforce acting per unit area
A 2.0 kg block has a volume of 2.5 x 10^-3 m^3. Its density is:
- A800 kg m^-3
- B5000 kg m^-3
- C1250 kg m^-3
- D0.8 kg m^-3
A force of 200 N acts over an area of 0.50 m^2. The pressure is:
- A100 Pa
- B0.0025 Pa
- C400 Pa
- D250 Pa
The extra pressure at a depth of 5.0 m in water (density 1000 kg m^-3, g = 9.8 m s^-2) is about:
- A5.0 x 10^3 Pa
- B4.9 x 10^4 Pa
- C9.8 x 10^4 Pa
- D2.0 x 10^4 Pa
The pressure at a point in a still liquid depends on the depth and the density, but not on the:
- Avalue of g
- Bliquid used
- Cdepth below the surface
- Dshape of the container
Archimedes' principle states that the upthrust on a body equals the:
- Aweight of the fluid it displaces
- Bweight of the body
- Cvolume of the body
- Ddensity of the body
A body of volume 3.0 x 10^-4 m^3 is fully submerged in water (1000 kg m^-3, g = 9.8 m s^-2). The upthrust is:
- A29 N
- B0.29 N
- C2.9 N
- D294 N
An object floats in a liquid when its density is:
- Agreater than the density of the liquid
- Bless than the density of the liquid
- Cequal to its own weight
- Dequal to the pressure
A block floats with 0.80 of its volume below the surface of water (1000 kg m^-3). The density of the block is:
- A1250 kg m^-3
- B80 kg m^-3
- C1000 kg m^-3
- D800 kg m^-3
The total pressure at a depth of 10 m in water (1000 kg m^-3, g = 9.8) with atmospheric pressure 1.0 x 10^5 Pa is closest to:
- A2.0 x 10^5 Pa
- B1.0 x 10^5 Pa
- C9.8 x 10^4 Pa
- D3.0 x 10^5 Pa
The pascal is equivalent to:
- AN m
- Bkg m^-3
- CN m^-2
- DJ m^-3
In a connected liquid that is still, the pressure is the same at:
- Athe surface and the bottom
- Ball points on the same horizontal level
- Cpoints with the same volume above
- Dnarrow parts only
A submarine dives deeper. The pressure of the surrounding water on its hull:
- Adecreases
- Bstays the same
- Cbecomes zero
- Dincreases
A solid object denser than water is released under water. It sinks because:
- Aits weight exceeds the maximum upthrust
- Bthere is no upthrust on it
- Cthe water exerts no pressure
- Dits volume is too small
A rectangular tank base of area 0.20 m^2 holds water to a depth of 0.30 m (1000 kg m^-3, g = 9.8). The force on the base due to the water is about:
- A290 N
- B2900 N
- C590 N
- D59 N
Checking units, the expression rho g h has the SI base units of:
- Akg m s^-2
- Bkg m^-1 s^-2
- Ckg m^-3
- Dkg m^2 s^-2
Quiz key
Density = mass / volume, with unit kg m^-3.
Pressure = force / area, measured in pascals (Pa = N m^-2).
rho = 2.0 / (2.5 x 10^-3) = 800 kg m^-3.
p = 200 / 0.50 = 400 Pa.
Delta p = rho g h = 1000 x 9.8 x 5.0 = 49000 = 4.9 x 10^4 Pa.
Hydrostatic pressure is rho g h. It is independent of the shape of the container and the total volume.
Upthrust = weight of fluid displaced = rho g V, with V the submerged volume.
Upthrust = 1000 x 9.8 x 3.0 x 10^-4 = 2.94 N, about 2.9 N.
A body floats when it is less dense than the fluid, so the displaced fluid can equal its weight without full submersion.
Submerged fraction = rho_block / rho_liquid, so rho_block = 0.80 x 1000 = 800 kg m^-3.
Hydrostatic = 1000 x 9.8 x 10 = 9.8 x 10^4 Pa. Total = 1.0 x 10^5 + 0.98 x 10^5 = about 2.0 x 10^5 Pa.
1 Pa = 1 N m^-2, force per unit area.
At a given depth (same horizontal level) in a connected still liquid the pressure is equal, which is why a liquid finds its own level.
Pressure increases with depth as rho g h, so it grows as the submarine dives.
Even fully submerged, the upthrust (rho_water g V) is less than the object's weight, so the resultant is downward and it sinks.
Pressure = 1000 x 9.8 x 0.30 = 2940 Pa; force = pA = 2940 x 0.20 = 588 N, about 590 N.
rho g h = (kg m^-3)(m s^-2)(m) = kg m^-1 s^-2, which is the pascal, confirming it is a pressure.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.