Preview

Fluid Statics

Q: A wood block (ρ = 600 kg/m³, V = 0.002 m³) is placed in water (ρ = 1000 kg/m³). What fraction is submerged?

At equilibrium: ρ_obj × V_total = ρ_fluid × V_submerged. Fraction = ρ_obj/ρ_fluid.

Q: What is the buoyant force on a 0.001 m³ steel ball (ρ = 7800 kg/m³) fully submerged in water?

F_b = ρ_water × g × V_submerged = 1000 × 9.81 × 0.001 ≈ 9.81 N — the fluid's density prices it, the steel's does not (that only decides sink vs float).

Q: Set the Dense Fluid preset (ρ = 1200 kg/m³, depth 4.0 m, P₀ = 120 kPa). Compute the pressure at the bottom, then check the Pressure at Depth row.

P = P₀ + ρgh = 120000 + 1200 × 9.81 × 4.0 = 167,088 Pa ≈ 167.09 kPa — the panel shows exactly this.

Q: From the default state, drag Surface Pressure from 101 to 180 kPa. By how much does the pressure at the bottom rise, and why?

Pascal's principle: the full ΔP₀ = 79 kPa is transmitted undiminished to every depth — P₀ enters P = P₀ + ρgh additively.

Hydrostatic pressure, Pascal's principle, and the buoyancy ledger

Every meter of water above you is a bill you cannot refuse. Stand at the bottom of a pool and the water column overhead charges you 9,810 pascals per meter — a constant rate, set by nothing but density and gravity, because each layer of fluid must carry the weight of every layer above it. Add the atmosphere's own 101 kilopascals pressing on the surface, and the total reads like a ledger: P = P₀ + ρgh, one surface term plus one depth term, added — never multiplied. That single line runs more of the world than it appears to. It is why dam walls thicken toward their base, why your ears pop a few meters down, and why a ship's hull feels the sea press harder on its belly than on its deck. That pressure difference is also the secret origin of buoyancy: fluid pushes up on an object's bottom harder than it pushes down on its top, and the net push equals exactly the weight of the fluid the object shoved aside — Archimedes' price tag, F_b = ρ_fluid·g·V. It prices the fluid, not the object: that is how a steel ship floats while a steel nail sinks, same metal, different displaced volume. And there is one more law hiding in the plus sign. Raise the pressure on the surface — a piston, a pump, a deeper atmosphere — and the entire increase arrives undiminished at every depth, Pascal's principle, the bookkeeping behind every hydraulic brake and jack on Earth. In this lab you hold all three levers: the fluid's density, the depth, and the surface pressure. Watch the color of the water column deepen as you descend, read the pressure ledger to the hundredth of a kilopascal, and catch the buoyant force repricing itself while the object's weight never blinks.

What you'll be able to do

Close the hydrostatic equation on the panel's own rows at the shipped default state (ρ = 1000 kg/m³, h = 2.5 m, P₀ = 101 kPa): P = P₀ + ρgh = 101000 + 1000 × 9.81 × 2.5 = 125525 Pa → the Pressure row reads '125.53' kPa, decomposing as 101.0 kPa of surface pressure plus 24.53 kPa of gauge pressure, with a constant gradient dP/dh = ρg = 9.81 kPa per meter (OpenStax §14.1 Eq. 14.4/14.8, verified) — the sim's own quiz Q1 prices the same gradient at 3 m (29430 Pa), and Example 14.1's dam prices it at 40 m (392 kPa)
Separate absolute from gauge pressure under the depth lever: dragging Total Depth 2.5 → 5.0 m at defaults doubles ONLY the ρgh term (24.53 → 49.05 kPa) while the surface term is untouched, so the Pressure row climbs 125.53 → 150.05 kPa — not the 251.05 kPa a 'double the depth, double the pressure' reading predicts; each added meter bills exactly ρg = 9.81 kPa ('Each meter of water adds ~9,810 Pa', the sim's quiz Q1 feedback), because the fluid is incompressible and the gradient is constant (§14.1, verified)
Audit Pascal transmission and the buoyancy ledger as two separate books: Surface Pressure 101 → 150 kPa raises the bottom reading by the FULL 49.00 kPa (125.53 → 174.53 kPa — Δp_everywhere = Δp_top, §14.3 verified, the sim's quiz Q3 keyed answer), while the Buoyant Force row prices only the displaced fluid (F_b = ρ_fl g V: 5136.50 N fresh → 5264.92 N sea at ρ = 1025) and the Object Weight row stays pinned by the object's own density (W = 800·g·V = 4109.20 N) — their ratio 800/ρ_fl (< 1 throughout the slider domain) is the float condition, with fraction submerged = ρ_obj/ρ_fl = 78% at the sea preset (§14.4, verified; Example 14.4 digit anchor)

Formulas

P = P_0 + \rho g h

Hydrostatic pressure — increases linearly with depth h

F_b = \rho_{fluid} \cdot g \cdot V_{displaced}

Archimedes' principle — buoyant force equals weight of displaced fluid

F_{net} = F_b - W_{object}

Net force on a submerged object — determines sink or float

\frac{\rho_{obj}}{\rho_{fluid}} \leq 1 \Rightarrow \text{float}

Float condition — object density at or below fluid density

Make a prediction

Default state (Fresh Water): ρ = 1000 kg/m³, depth 2.5 m, P₀ = 101 kPa — the Buoyant Force row reads 5136.50 N and the Object Weight row reads 4109.20 N for the orange sphere. Now press the Sea Water preset (ρ = 1025 kg/m³, depth 3.0 m). What happens to those two force rows?

No grading here — commit to a guess, then scroll down and test it yourself.

This is a Pro experiment

Upgrade to Pro to access this experiment — or keep learning with one of our free labs.

View Pro plans Browse free experiments

Explore

Audit the factory state row by row (it is also the Fresh Water preset): Pressure at Depth 125.53 kPa decomposing as Surface 101.0 kPa plus 24.53 kPa of gauge pressure, Depth 2.5 m, Density 1000 kg/m³, Buoyant Force 5136.50 N, Object Weight 4109.20 N. Match the water column's color — light blue at the surface, deep blue at the bottom — against the Pressure legend, and note the amber probe plane sits fixed mid-column (registered quirk: it tracks nothing; the rows carry the truth).
Price the depth lever: drag Total Depth from 2.5 up to 5.0 m and read the Pressure row — 125.53 → 150.05 kPa. The gauge part doubled (24.53 → 49.05 kPa) but the total did not, because the 101.0 kPa surface term rides along unchanged. Each added meter bills exactly 9.81 kPa (quiz Q1's 'Each meter of water adds ~9,810 Pa'). Then drop to 0.5 m and confirm the meter-rate again: 105.91 kPa.
Run the Pascal audit: drag Surface Pressure from 101 to 150 kPa and watch BOTH pressure rows — the Surface row climbs 49.0 kPa and the Pressure at Depth row climbs the FULL 49.00 kPa with it (125.53 → 174.53 kPa), undiminished at the bottom, while the ρgh part of the book is untouched. This is quiz Q3 staged live — and the bookkeeping behind every hydraulic brake.
Keep the buoyancy ledger across the three presets in turn: press Fresh Water → Sea Water → Dense Fluid and read the two force rows each time — Buoyant Force reprices 5136.50 → 5264.92 → 6163.80 N with the fluid's density while Object Weight never leaves 4109.20 N. Watch the orange sphere bob higher in denser fluid (fraction submerged 800/ρ: 80% → 78% → 67%). Then open the Quick Quiz and price all three questions against what your own sliders just showed you.

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: Buoyant Force rises to 5264.92 N — exactly ×1.025, tracking the fluid's density — while Object Weight stays pinned at 4109.20 N: buoyancy prices the displaced fluid, weight prices the object

The Buoyant Force row computes F_b = ρ_fluid·g·V with the sphere's fixed V = (4/3)π·0.5³ = 0.5236 m³: 1025 × 9.81 × 0.5236 = 5264.92 N, exactly 1.025× the fresh-water 5136.50 N. Only the fluid's density was repriced — the preset's extra half meter of depth never enters the formula, because buoyancy is the pressure DIFFERENCE across the object's own height, and that difference is the same at 3 m as at 30 m (OpenStax §14.4: the net upward force equals the weight of fluid displaced, wherever the object sits in the column). The Object Weight row computes W = ρ_obj·g·V from the object's own density (800 kg/m³ — the same hardcode the HTML ships): it reads 4109.20 N before and after, because nothing about the object changed. Option B's premise is half true — the fluid really does press harder on every square centimeter of the sphere (the Pressure row climbs 125.53 → 131.17 kPa) — but weight is mg, priced by the object's own mass; the fluid can push on the sphere, it cannot make the sphere heavier. Option C names a real birth story of buoyancy (pressure grows with depth — §14.4 Figure 14.20) but misreads the account: what matters is the top-to-bottom difference across the object, which is depth-independent for a fully submerged object. The panel's own ratio certifies the split: W/F_b = 800/1025 = 0.78 — the sphere floats with 78% submerged (quiz Q2's exact arithmetic).

Quiz (0/4)

A wood block (ρ = 600 kg/m³, V = 0.002 m³) is placed in water (ρ = 1000 kg/m³). What fraction is submerged?

What is the buoyant force on a 0.001 m³ steel ball (ρ = 7800 kg/m³) fully submerged in water?

Set the Dense Fluid preset (ρ = 1200 kg/m³, depth 4.0 m, P₀ = 120 kPa). Compute the pressure at the bottom, then check the Pressure at Depth row.

From the default state, drag Surface Pressure from 101 to 180 kPa. By how much does the pressure at the bottom rise, and why?

You can now

Close the hydrostatic equation on the panel's own rows at the shipped default state (ρ = 1000 kg/m³, h = 2.5 m, P₀ = 101 kPa): P = P₀ + ρgh = 101000 + 1000 × 9.81 × 2.5 = 125525 Pa → the Pressure row reads '125.53' kPa, decomposing as 101.0 kPa of surface pressure plus 24.53 kPa of gauge pressure, with a constant gradient dP/dh = ρg = 9.81 kPa per meter (OpenStax §14.1 Eq. 14.4/14.8, verified) — the sim's own quiz Q1 prices the same gradient at 3 m (29430 Pa), and Example 14.1's dam prices it at 40 m (392 kPa)
Separate absolute from gauge pressure under the depth lever: dragging Total Depth 2.5 → 5.0 m at defaults doubles ONLY the ρgh term (24.53 → 49.05 kPa) while the surface term is untouched, so the Pressure row climbs 125.53 → 150.05 kPa — not the 251.05 kPa a 'double the depth, double the pressure' reading predicts; each added meter bills exactly ρg = 9.81 kPa ('Each meter of water adds ~9,810 Pa', the sim's quiz Q1 feedback), because the fluid is incompressible and the gradient is constant (§14.1, verified)
Audit Pascal transmission and the buoyancy ledger as two separate books: Surface Pressure 101 → 150 kPa raises the bottom reading by the FULL 49.00 kPa (125.53 → 174.53 kPa — Δp_everywhere = Δp_top, §14.3 verified, the sim's quiz Q3 keyed answer), while the Buoyant Force row prices only the displaced fluid (F_b = ρ_fl g V: 5136.50 N fresh → 5264.92 N sea at ρ = 1025) and the Object Weight row stays pinned by the object's own density (W = 800·g·V = 4109.20 N) — their ratio 800/ρ_fl (< 1 throughout the slider domain) is the float condition, with fraction submerged = ρ_obj/ρ_fl = 78% at the sea preset (§14.4, verified; Example 14.4 digit anchor)