Worked examples — Buoyancy — Archimedes' principle, derivation from pressure difference

This is the D3 companion to the Buoyancy topic note. The parent note built the idea and derived $F_B={\rho }_f{V}_{\text{disp}}g$. Here we hunt down every kind of situation buoyancy can throw at you and solve one of each — so no exam problem ever surprises you.

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The scenario matrix

Every buoyancy question lands in one of these cells. Each worked example below is tagged with the cell it covers.

Cell	Case class	What makes it distinct	Example
A	Fully submerged, ${\rho }_o>{\rho }_f$ (sinks)	$F_B<W$; find net force / acceleration	Ex 1
B	Floating, partly submerged	$F_B=W$; find fraction below surface	Ex 2
C	Apparent weight (weigh in fluid)	Scale reads $W-F_B$; back out ${\rho }_o$	Ex 3
D	Two stacked fluids (layered)	Object straddles a boundary; two displaced volumes	Ex 4
E	Buoyancy in a gas (balloon)	${\rho }_f$ tiny; lift = payload it can carry	Ex 5
F	Degenerate: ${\rho }_o={\rho }_f$ (neutral)	Object hovers anywhere; $f=1$ exactly	Ex 6
G	Limiting case: ${\rho }_o\to 0$ or object all air	$f\to 0$; ship-hull / "why steel floats"	Ex 7
H	Exam twist: floating object holding a load / ice melting	Conservation of displaced volume	Ex 8

The table above is one long list, but all eight cells actually branch from a single yes/no question. The decision tree below is that branch drawn out — trace it before every problem, and it tells you which example to imitate.

Recall The one decision tree behind every cell

Compare $F_B^{\text{full}}={\rho }_{f}Vg$ (buoyancy if the WHOLE object were under) with $W={\rho }_{o}Vg$.

• If $F_B^{\text{full}}\ge W$ it floats → use the float branch: Cell B (Ex 2), Cell F (Ex 6), Cell G (Ex 7), Cell H (Ex 8); submerged fraction $={\rho }_o/{\rho }_f$.
• If $F_B^{\text{full}}<W$ it sinks → use the sink branch: Cell A (Ex 1), Cell C (Ex 3); whole volume displaced.
• Layered fluids (Cell D, Ex 4) and gases (Cell E, Ex 5) run the same test but with the right ${\rho }_f$ per layer.

That maps every table cell to exactly one worked example.

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Ex 1 — Sinking cube: net force and acceleration (Cell A)

Step 1 — Volume of the cube. $V=(0.20{)}^3=8.0\times 10^{-3}{\text{m}}^3$ Why this step? Fully submerged means displaced volume = full volume; we need $V$ for both forces.

Step 2 — Buoyant force (uses FLUID density). $F_B={\rho }_{f}Vg=1000\times 8.0\times 10^{-3}\times 9.8=78.4\text{N}$ Why this step? The water is the pusher, so ${\rho }_f$ — not brass — goes here.

Step 3 — Weight (uses OBJECT density). $W={\rho }_{o}Vg=8500\times 8.0\times 10^{-3}\times 9.8=666.4\text{N}$ Why this step? Weight is what gravity pulls on the actual brass, so its own density.

Step 4 — Mass of the cube. $m={\rho }_{o}V=8500\times 8.0\times 10^{-3}=68\text{kg}$ Why this step? Newton's law needs mass, not weight, in the denominator of $a=F_{\text{net}}/m$; mass is density times volume (see the $m$ definition above).

Step 5 — Net force and acceleration. By Newton's second law, net down = $W-F_B$: $F_{\text{net}}=666.4-78.4=588\text{N (down)},a=\frac{F_{\text{net}}}{m}=\frac{588}{68}=8.65{\text{m/s}}^2$ Why this step? The object accelerates because forces don't balance ($F_B<W$); dividing net force by the mass from Step 4 gives the acceleration.

Verify: Buoyancy is $78.4/666.4=11.8\%$ of weight — small but not zero, so the cube sinks slower than in vacuum ($8.65<9.8$). Units: $\text{N}/\text{kg}={\text{m/s}}^2$. ✓

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Ex 2 — Floating block: fraction submerged (Cell B)

Figure s01 — a floating wooden block cut by the water line. The horizontal teal line at $y=4$ is the water surface. The block below that line is drawn in a lighter orange — this is the drowned part that displaces water. The block above the line is drawn in solid orange — the part poking into air. On the LEFT, a plum double-headed arrow spans the above-water height labelled "40% above (4.8 cm)"; on the RIGHT, a teal double-headed arrow spans the below-water height labelled "60% below (displaces water)" — that right-hand teal length is exactly the $f$ our formula computes. Two single arrows through the block's centre show the balanced forces: teal $F_B$ pointing up, orange $W$ pointing down, equal in length because the block floats.

Step 1 — Floating equilibrium condition. Only the submerged part displaces water: ${\rho }_f(fV)g={\rho }_{o}Vg$ Why this step? Floating means $F_B=W$; buoyancy only counts the drowned volume $fV$ (the teal-arrow region in Figure s01).

Step 2 — Cancel $Vg$, solve for $f$. $f=\frac{{\rho }_o}{{\rho }_f}=\frac{600}{1000}=0.60$ Why this step? $V$ and $g$ appear on both sides — see Density and relative density: it's purely a density ratio. This $f=0.60$ is the teal-arrow length in Figure s01.

Step 3 — Height sticking up. Fraction above $=1-0.60=0.40$, so $0.40\times 12\text{cm}=4.8\text{cm above water}.$ Why this step? For a prism of uniform cross-section, volume fraction = height fraction — that's the plum arrow's $4.8\text{cm}$.

Verify: $60\%$ underwater matches intuition (wood is a bit lighter than water, so it rides fairly low). If ${\rho }_o$ were $999$, $f\to 0.999$ (barely floats). ✓

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Ex 3 — Apparent weight, find density (Cell C)

Step 1 — Buoyant force from the weight drop. $F_B=W_{\text{air}}-W_{\text{apparent}}=147-107=40\text{N}$ Why this step? The scale reads $W-F_B$; the missing $40\text{N}$ is exactly the water's upthrust.

Step 2 — Volume from $F_B={\rho }_{f}Vg$. $V=\frac{F_B}{{\rho }_{f}g}=\frac{40}{1000\times 9.8}=4.08\times 10^{-3}{\text{m}}^3$ Why this step? Invert the buoyancy formula — the statue's own density is unknown, but water's isn't.

Step 3 — Mass and density. $m=\frac{W_{\text{air}}}{g}=\frac{147}{9.8}=15.0\text{kg},{\rho }_o=\frac{m}{V}=\frac{15.0}{4.08\times 10^{-3}}=3676{\text{kg/m}}^3$ Why this step? Density is mass per volume — both now known.

Verify: ${\rho }_o=3676>1000$, so it should sink — consistent with needing to hold it under. Relative density $\approx 3.7$ (a stone/ceramic, not solid metal). ✓

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Ex 4 — Object straddling two fluids (Cell D)

Figure s02 — the two-fluid jar and the pressure-slice hint. The jar has two coloured bands: a lighter orange OIL band on top (${\rho }_1=800$) and a darker teal WATER band below (${\rho }_2=1000$), separated by the horizontal "oil–water boundary" line, with the "oil surface" line above. On the RIGHT, two small arrows show why buoyancy from two layers ADDS: a short orange "oil push" arrow and a longer teal "water push (bigger)" arrow, tagged in plum "pushes ADD" — this is the physics of Step 1. The plum cube ($\rho =1500$) is drawn resting on the jar FLOOR, with a plum annotation "sinks: rho_o > rho_water, no boundary float" — that is the answer we are about to prove, shown up front so the figure and narrative agree: the assumed boundary-float does not happen.

Step 1 — Set up total buoyancy from BOTH fluids (assuming the boundary float). Let $A=(0.10{)}^2=0.01{\text{m}}^2$. Suppose height $x$ is in water and $(0.10-x)$ in oil: $F_B={\rho }_{1}A(0.10-x)g+{\rho }_{2}Axg$ Why this step? Each fluid contributes its own ${\rho }_f{V}_{\text{disp}}g$; the total is their sum because hydrostatic pressure builds up layer by layer — the oil slice adds its push, the water slice adds a bigger one. That's the two-arrow "pushes ADD" hint in Figure s02.

Step 2 — Floating equilibrium: $F_B=W$. $[{\rho }_1(0.10-x)+{\rho }_{2}x]Ag={\rho }_{o}A(0.10)g$ Why this step? If it floats at the boundary, net force is zero.

Step 3 — Cancel $Ag$ and solve for $x$. $800(0.10-x)+1000x=1500\times 0.10\Rightarrow 80+200x=150\Rightarrow x=\frac{70}{200}=0.35\text{m}$ Why this step? Algebra — but $x=0.35\text{m}>0.10\text{m}$ is impossible for a $0.10\text{m}$ cube!

Step 4 — Read the impossible answer. $x>$ cube height means even fully in water it can't be supported: ${\rho }_o=1500>{\rho }_2=1000$. The cube sinks to the bottom of the jar — exactly as Figure s02 already shows. The assumed boundary-float never happens. Why this step? Always sanity-check that $0\le x\le \text{height}$. A degenerate answer is telling you the physics: this cell reduces to Cell A (sinking).

Verify: Max buoyancy $9.8\text{N}<$ weight $14.7\text{N}$ ⇒ sinks. Answer: no equilibrium float; cube rests on the jar floor (consistent with the figure). ✓

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Ex 5 — Buoyancy in a gas: a helium balloon (Cell E)

Step 1 — Buoyant force from displaced AIR. $F_B={\rho }_{\text{air}}Vg=1.29\times 5.0\times 9.8=63.21\text{N}$ Why this step? The balloon shoves aside air; the same $F_B={\rho }_{f}Vg$ applies, with ${\rho }_f={\rho }_{\text{air}}$.

Step 2 — Weight of the gas inside plus skin. $W_{\text{He}}={\rho }_{\text{He}}Vg=0.18\times 5.0\times 9.8=8.82\text{N},W_{\text{skin}}=0.20\times 9.8=1.96\text{N}$ Why this step? Everything the balloon carries pulls down: the helium has real (small) weight too.

Step 3 — Net lift for payload. $F_{\text{lift}}=F_B-W_{\text{He}}-W_{\text{skin}}=63.21-8.82-1.96=52.43\text{N}$ $m_{\text{payload}}=\frac{52.43}{9.8}=5.35\text{kg}$ Why this step? Whatever's left over after floating the balloon+gas can hoist a load.

Verify: Positive lift ⇒ balloon rises, as expected since ${\rho }_{\text{He}}\ll {\rho }_{\text{air}}$. If we'd forgotten the helium's weight we'd overstate lift by $8.82\text{N}$. Units: N/(m/s²)=kg. ✓

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Ex 6 — Neutral buoyancy, the degenerate case (Cell F)

Step 1 — Buoyant force if fully submerged. $F_B={\rho }_{f}Vg=1025\times 2.0\times 9.8=20090\text{N}$ Why this step? Start by asking how hard the fluid pushes when the whole capsule is under — that's the biggest upthrust available, and it uses the fluid density ${\rho }_f$.

Step 2 — Weight. $W={\rho }_{o}Vg=1025\times 2.0\times 9.8=20090\text{N}$ Why this step? We need the downward pull to compare against $F_B$; weight uses the object's own density ${\rho }_o$.

Step 3 — Net force = zero, everywhere. $F_{\text{net}}=F_B-W=0$. The submerged fraction is $f=\frac{{\rho }_o}{{\rho }_f}=\frac{1025}{1025}=1.$ Why this step? $f=1$ means the object is exactly, entirely underwater with zero net force — it hovers at any depth (neutral buoyancy, how submarines and fish tune their depth).

Verify: $F_B=W$ exactly ⇒ $a=0$ ⇒ stays put wherever released. This is the boundary between float (${\rho }_o<{\rho }_f$) and sink (${\rho }_o>{\rho }_f$). ✓

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Ex 7 — Limiting case: why a steel ship floats (Cell G)

Step 1 — Weight of the whole boat. $W=mg=2000\times 9.8=19600\text{N}$ Why this step? Only total mass matters, not what it's made of, for the downward pull; here mass $m=2000\text{kg}$ is given directly.

Step 2 — Maximum available buoyancy (whole hull under). $F_B^{\max }={\rho }_f{V}_{\text{hull}}g=1000\times 3.0\times 9.8=29400\text{N}$ Why this step? This is the biggest upthrust the hull shape can generate. Compare to $W$.

Step 3 — Compare: $F_B^{\max }=29400>W=19600$ ⇒ floats. It only needs to displace enough to match its weight: $V_{\text{disp}}=\frac{W}{{\rho }_{f}g}=\frac{19600}{1000\times 9.8}=2.0{\text{m}}^3,f=\frac{2.0}{3.0}=\mathrm{0.667.}$ Why this step? At equilibrium the boat sinks just deep enough that displaced-water weight equals its own — here $2/3$ of the hull.

Verify: Average density $667<1000$ ⇒ floats. $f={\rho }_o/{\rho }_f=667/1000=0.667$ matches Step 3. ✓

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Ex 8 — Exam twist: ice melting in a glass (Cell H)

Step 1 — Water displaced by the floating ice. Floating ⇒ buoyancy = ice weight: ${\rho }_w{V}_{\text{disp}}g=m_{\text{ice}}g\Rightarrow V_{\text{disp}}=\frac{m_{\text{ice}}}{{\rho }_w}.$ Why this step? The submerged part of the ice displaces exactly the weight of water equal to the whole ice's weight.

Step 2 — Water produced when ice melts. The ice becomes $m_{\text{ice}}$ of liquid water, occupying $V_{\text{melt}}=\frac{m_{\text{ice}}}{{\rho }_w}.$ Why this step? Mass is conserved; melted water has density ${\rho }_w$.

Step 3 — Compare. $V_{\text{disp}}=V_{\text{melt}}$ exactly ⇒ the melt-water precisely fills the hole the ice's submerged part occupied. Level stays the same — no overflow. Why this step? The two volumes are algebraically identical; the "displaced" hole and the "produced" water match.

Step 4 — The iron twist. Iron (${\rho }_{\text{iron}}=7800>{\rho }_w$) doesn't float on its own. While frozen in ice it is held up by the surrounding ice, so the whole lump floats and displaces water equal to the iron's weight: $V_{\text{float}}=m_{\text{iron}}/{\rho }_w$. After melting, the iron drops loose to the bottom and now displaces only its own volume: $V_{\text{sunk}}=m_{\text{iron}}/{\rho }_{\text{iron}}$, which is smaller (because ${\rho }_{\text{iron}}>{\rho }_w$). Less displacement ⇒ the water level falls. Why this step? A dense sinker displaces by weight while floating (held up), but only by its own volume once it sinks — floating always displaces more. So the melt of an iron-laden ice cube lowers the level, unlike plain ice which keeps it constant.

Verify: For pure ice, $V_{\text{disp}}-V_{\text{melt}}=0$ ⇒ unchanged. For iron: floating displacement $\propto 1/{\rho }_w=1/1000$ vs sunk displacement $\propto 1/{\rho }_{\text{iron}}=1/7800$; since $1/1000>1/7800$, floating displaced more, so level drops after melting. ✓

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Active Recall

Connections

• Parent topic (Hinglish)
• Pressure in fluids — hydrostatic pressure
• Density and relative density
• Apparent weight and weighing methods
• Floating bodies and stability — metacentre
• Newton's laws — equilibrium of forces
• Pascal's principle