University Physics — Year 1 · Bachelor Year 1

21Fluid Statics

A submarine at three hundred meters carries, on each square meter of its hull, the weight of a loaded truck. A dam holds back a lake with a wall that is thick at the bottom and thin at the top, for a reason that has nothing to do with how long the lake is. A ship of a hundred thousand tonnes floats because it pushes aside a hundred thousand tonnes of water, and sinks a hand’s breadth deeper when it sails from the sea into a river. Every one of these is a consequence of a single relation — pressure increases downward in a fluid at rest at the rate $\rho g$ — and this chapter derives it, applies it to liquids and to the atmosphere, and deduces from it the force on a wall and the law of Archimedes.

A submarine in an emergency surfacing drill (US Navy): its ballast tanks blown, it is lighter than the water it displaces and rises — Archimedes’ theorem at 7000\, t. — A submarine in an emergency surfacing drill (US Navy): its ballast tanks blown, it is lighter than the water it displaces and rises — Archimedes’ theorem at $7000\,\mathrm{t}$ .

21.1 Pressure in a fluid at rest

Proposition 21.1 (Isotropy of pressure)

In a fluid at rest, the force exerted by the fluid on a small surface $\dd S$ placed at a point $M$ is $\dd\vect F = P(M)\,\dd S\,\vect n$ , normal to the surface and directed toward it, with a pressure $P(M)$ that depends on the point but not on the orientation of the surface.

Partial proof. A fluid at rest transmits no tangential force (otherwise it would flow): the force is normal. Consider a tiny triangular prism of fluid at $M$ with faces of different orientations: it is in equilibrium under the pressure forces on its faces and its weight; the weight scales as the volume (third power of the size), the pressure forces as areas (second power), so for a small enough prism only the pressure forces count, and their balance in every direction requires the same $P$ on every face. ∎

Theorem 21.2 (Fundamental relation of fluid statics)

In a fluid at rest in the uniform gravity $\vect g = -g\vect e_z$ ( $z$ upward), the pressure depends only on $z$ and

\frac{\dd P}{\dd z} = -\rho g ,

$\rho$ being the fluid’s density at that level. Pressure increases downward; surfaces of equal pressure (isobars) are horizontal, and so is the free surface of a liquid.

Proof. Take a horizontal slab of fluid of area $S$ between $z$ and $z + \dd z$ : it is pushed up by $P(z)S$ from below, down by $P(z + \dd z)S$ from above, and pulled down by its weight $\rho gS\,\dd z$ . At rest: $P(z)S - P(z + \dd z)S - \rho gS\,\dd z = 0$ . Horizontal slabs of fluid have no horizontal force from gravity, so horizontally $P$ is uniform. ∎

Left: equilibrium of a horizontal slab of fluid — the pressure below must exceed the pressure above by the slab’s weight per unit area. Right: communicating vessels — a horizontal line meets the same pressure everywhere in a connected liquid, so the free surfaces stand at one level whatever the shapes.

Corollary 21.3 (Incompressible fluid)

In a liquid of uniform density $\rho$ , the pressure at depth $h$ below a point where it is $P_0$ is

P = P_0 + \rho gh .

Water: $1\,\mathrm{bar}$ more every $10\,\mathrm{m}$ . Two points of a connected liquid at the same level are at the same pressure; the free surfaces of communicating vessels stand at one level; a pressure applied anywhere to an enclosed liquid is transmitted undiminished to every point (Pascal’s principle).

Proof. Integrate $\dd P/\dd z = -\rho g$ with $\rho$ constant; the rest is the horizontality of isobars and the uniqueness of $P$ at a level. ∎

Example 21.4 (Manometers and presses)

A U-tube of mercury ( $\rho = 13\,600\,\mathrm{kg}/\mathrm{m}^{3}$ ) whose columns differ by $h = 760\,\mathrm{mm}$ measures a pressure difference $\rho gh = 1.013 \times 10^{5}\,\mathrm{Pa}$ — the atmosphere, which holds a $10.3\,\mathrm{m}$ column of water. A hydraulic press: a force $F_1$ on a piston of area $S_1$ raises the pressure by $F_1/S_1$ throughout; on the large piston $S_2$ it yields $F_2 = F_1S_2/S_1$ — fifty times more for fifty times the area, at the price of fifty times less travel (energy is conserved: $F_1d_1 = F_2d_2$ ).

21.2 The atmosphere

Proposition 21.5 (Isothermal atmosphere)

For a perfect gas of molar mass $M$ at uniform temperature $T$ in uniform gravity,

P(z) = P_0\,\eu^{-z/H}, \qquad H = \frac{RT}{Mg} ,

the scale height: about $8\,\mathrm{km}$ for air; the pressure halves every $H\ln2 \approx 5.5\,\mathrm{km}$ .

Proof. $\rho = PM/RT$ (equation of state), so $\dd P/\dd z = -(Mg/RT)P$ : a first-order linear equation, $P = P_0\eu^{-Mgz/RT}$ . ∎

Example 21.6 (Thin air)

With $T = 260\,\mathrm{K}$ (an average over the lower atmosphere), $H = 7.6\,\mathrm{km}$ : $P = 0.52P_0$ at $5000\,\mathrm{m}$ , $0.31P_0$ at the summit of Everest — close to the measured $0.54$ and $0.33$ ; the real atmosphere cools with altitude, which the Year 2 volume takes into account. The model also tells why the sea is different: water is eight hundred times denser and nearly incompressible, so its pressure rises linearly, a bar per ten meters, with no scale height.

The isothermal atmosphere: pressure (and density) fall exponentially with height, halving every 5.5\, km. Nine tenths of the air lies below 17\, km. — The isothermal atmosphere: pressure (and density) fall exponentially with height, halving every $5.5\,\mathrm{km}$ . Nine tenths of the air lies below $17\,\mathrm{km}$ .

21.3 Forces on walls

Proposition 21.7 (Force on a vertical wall)

A liquid of depth $h$ against a vertical rectangular wall of width $L$ (atmospheric pressure acting on both faces and cancelling) exerts the horizontal force

F = \tfrac12\rho gh^2L ,

equivalent to a single force applied at the center of pressure, at depth $2h/3$ — the resultant acts low, where the pressure is greatest.

Proof. At depth $x$ the gauge pressure is $\rho gx$ ; on the strip of height $\dd x$ the force is $\rho gxL\,\dd x$ ; integrate from $0$ to $h$ . Its moment about the surface line is $\int_0^h\rho gx^2L\,\dd x = \tfrac13\rho gh^3L$ ; dividing by $F$ gives the lever arm $2h/3$ . ∎

The pressure on a dam grows linearly with depth (arrows), so the resultant acts at two thirds of the depth: the wall must be thickest at the bottom. The force depends on h and L, not on how much water lies behind. — The pressure on a dam grows linearly with depth (arrows), so the resultant acts at two thirds of the depth: the wall must be thickest at the bottom. The force depends on $h$ and $L$ , not on how much water lies behind.

Example 21.8 (A dam)

$h = 20\,\mathrm{m}$ , $L = 100\,\mathrm{m}$ : $F = \tfrac12 \times 1000 \times 9.81 \times 400 \times 100 = 2.0 \times 10^{8}\,\mathrm{N}$ — twenty thousand tonnes, applied $13\,\mathrm{m}$ below the surface, the same whether the lake behind is a pond or a fjord: the hydrostatic paradox.

21.4 Archimedes’ theorem

Theorem 21.9 (Archimedes)

A body immersed in a fluid at rest receives from the pressure forces a resultant — the buoyancy — equal and opposite to the weight of the fluid it displaces,

\vect\Pi = -\rho_{\text{fluid}}V_{\text{immersed}}\,\vect g ,

applied at the center of mass of the displaced fluid (the center of buoyancy).

Proof. The pressure forces on the body’s surface depend only on the shape of that surface and on the fluid outside. Replace the body, in thought, by fluid at rest filling the same volume: that fluid is in equilibrium under its weight and the same pressure forces, so the pressure forces balance its weight — they sum to $-\rho V\vect g$ , applied through its center of mass. Remove the thought experiment; the pressure forces are unchanged. ∎

Corollary 21.10 (Floating)

A body of density $\rho_b$ floats in a fluid of density $\rho_f > \rho_b$ with the fraction $\rho_b/\rho_f$ of its volume immersed; if $\rho_b > \rho_f$ it sinks, with an apparent weight $(\rho_b - \rho_f)Vg$ .

Proof. Weight $\rho_bVg$ equals buoyancy $\rho_fV_{\text{imm}}g$ . ∎

A floating body: its weight, applied at its center of mass G, balances the buoyancy, applied at the center of buoyancy C (the center of the immersed volume). Ice (= 917\, kg/ m3) in sea water (1025\, kg/ m3): 89\% below the surface. — A floating body: its weight, applied at its center of mass $G$ , balances the buoyancy, applied at the center of buoyancy $C$ (the center of the immersed volume). Ice ( $\rho = 917\,\mathrm{kg}/\mathrm{m}^{3}$ ) in sea water ( $1025\,\mathrm{kg}/\mathrm{m}^{3}$ ): $89\%$ below the surface.

Example 21.11 (Three buoyancies)

A hydrometer is a weighted tube that sinks until it displaces its own weight: the denser the liquid, the less it sinks — a density gauge read on its stem. A $2800\,\mathrm{m}^{3}$ balloon of hot air at $373\,\mathrm{K}$ displaces $3.4\,\mathrm{t}$ of cold air and weighs $2.6\,\mathrm{t}$ of hot air: $0.7\,\mathrm{t}$ of lift (Problem 20.1). A ship of $100\,000\,\mathrm{t}$ displaces $97\,600\,\mathrm{m}^{3}$ of sea water; in fresh water ( $1000\,\mathrm{kg}/\mathrm{m}^{3}$ ) it must displace $2.5\%$ more and sinks deeper — the Plimsoll marks on its hull.

Remark 21.12 (Stability)

A floating body tilted by a small angle is restored if the buoyancy, now applied at the shifted center of the new immersed volume, produces a righting moment — which happens when the metacenter (the point where the buoyancy’s line of action meets the body’s axis) lies above $G$ . Ballast low in the hull lowers $G$ and steadies the ship; a top-heavy boat capsizes.

21.5 Exercises

Exercise 21.1 ★

Absolute pressure at $10\,\mathrm{m}$ , $100\,\mathrm{m}$ and $1\,\mathrm{km}$ under the sea ( $\rho = 1025\,\mathrm{kg}/\mathrm{m}^{3}$ ); force on a $1.0\,\mathrm{m}^{2}$ porthole at $300\,\mathrm{m}$ .

Solution

Solution of Exercise 21.1.

$P = P_0 + \rho gh$ : $2.0 \times 10^{5}\,\mathrm{Pa}$ ( $2.0\,\mathrm{bar}$ ), $1.1 \times 10^{6}\,\mathrm{Pa}$ ( $11\,\mathrm{bar}$ ), $1.0 \times 10^{7}\,\mathrm{Pa}$ ( $101\,\mathrm{bar}$ ). At $300\,\mathrm{m}$ : $P = 3.1 \times 10^{6}\,\mathrm{Pa}$ , force $3.1\,\mathrm{MN}$ on $1\,\mathrm{m}^{2}$ (net, against $1\,\mathrm{atm}$ inside: $3.0\,\mathrm{MN}$ ).

Exercise 21.2 ★

Height of a mercury column balancing $1.013 \times 10^{5}\,\mathrm{Pa}$ ; of a water column. Why can a suction pump not lift water more than about $10\,\mathrm{m}$ ?

Solution

Solution of Exercise 21.2.

$h = P/\rho g$ : mercury $0.760\,\mathrm{m}$ ; water $10.3\,\mathrm{m}$ . A pump can at best create a vacuum above the water: the atmosphere then pushes the column up by $10.3\,\mathrm{m}$ , no more.

Exercise 21.3 ★

A hydraulic jack: $100\,\mathrm{N}$ on a $2.0\,\mathrm{cm}^{2}$ piston, load on a $200\,\mathrm{cm}^{2}$ piston; distances moved by each if the small piston travels $20\,\mathrm{cm}$ .

Solution

Solution of Exercise 21.3.

$F_2 = F_1S_2/S_1 = 10\,\mathrm{kN}$ (one tonne); $d_2 = d_1S_1/S_2 = 2.0\,\mathrm{mm}$ (same volume displaced, same work).

Exercise 21.4 ★

A U-tube contains water; oil ( $\rho = 850\,\mathrm{kg}/\mathrm{m}^{3}$ ) is poured into one arm and forms a $12\,\mathrm{cm}$ column. Height difference between the two free surfaces.

Solution

Solution of Exercise 21.4.

Same pressure at the oil–water interface level in both arms: $\rho_og h_o = \rho_wgh_w$ , $h_w = 0.85 \times 12 = 10.2\,\mathrm{cm}$ of water above that level in the other arm; the oil surface stands $12 - 10.2 = 1.8\,\mathrm{cm}$ higher.

Exercise 21.5 ★★

Scale height of the atmosphere at $273\,\mathrm{K}$ ; pressure at $5000\,\mathrm{m}$ and at $8848\,\mathrm{m}$ ; height at which $P = P_0/2$ . Compare with the values quoted in Chapter 20.

Solution

Solution of Exercise 21.5.

$H = RT/Mg = 8.314 \times 273/(0.029 \times 9.81) = 8.0\,\mathrm{km}$ ; $P/P_0 = \eu^{-z/H}$ : $0.54$ at $5\,\mathrm{km}$ , $0.33$ at $8848\,\mathrm{m}$ ; half at $H\ln2 = 5.5\,\mathrm{km}$ — the numbers used in Chapter 20.

Exercise 21.6 ★★

A rectangular dam $100\,\mathrm{m}$ wide holds $20\,\mathrm{m}$ of water. Total force, depth of the center of pressure, and the moment of the water about the base of the dam. Why is a dam curved toward the water?

Solution

Solution of Exercise 21.6.

$F = \tfrac12\rho gh^2L = 1.96 \times 10^{8}\,\mathrm{N}$ ; center of pressure at $13.3\,\mathrm{m}$ depth, i.e. $6.7\,\mathrm{m}$ above the base; moment about the base $F \times 6.7 = 1.3 \times 10^{9}\,\mathrm{N}\,\mathrm{m}$ . Curving the dam toward the water turns the thrust into compression of the arch, carried to the valley sides.

Exercise 21.7 ★★

Fraction of an iceberg below the surface (ice $917\,\mathrm{kg}/\mathrm{m}^{3}$ , sea water $1025\,\mathrm{kg}/\mathrm{m}^{3}$ ); of a log of density $600\,\mathrm{kg}/\mathrm{m}^{3}$ in fresh water; a raft of $2.0\,\mathrm{m}^{3}$ of that wood can carry how many $75\,\mathrm{kg}$ people before its deck is awash?

Solution

Solution of Exercise 21.7.

$917/1025 = 89\%$ ; log $60\%$ . Raft: buoyancy at full immersion $1000 \times 2.0 \times 9.81 = 19.6\,\mathrm{kN}$ , own weight $11.8\,\mathrm{kN}$ : $7.8\,\mathrm{kN}$ spare, ten people… the tenth with wet feet: nine to stay dry ( $800\,\mathrm{kg}$ ).

Exercise 21.8 ★★

A hydrometer of mass $20\,\mathrm{g}$ with a stem of cross-section $0.50\,\mathrm{cm}^{2}$ floats in water with $4.0\,\mathrm{cm}$ of stem emerging. How much stem emerges in brine of density $1100\,\mathrm{kg}/\mathrm{m}^{3}$ ? Sensitivity (cm per unit of relative density)?

Solution

Solution of Exercise 21.8.

Immersed volume $m/\rho$ : water $20\,\mathrm{cm}^{3}$ ; brine $20/1.1 = 18.2\,\mathrm{cm}^{3}$ , $1.8\,\mathrm{cm}^{3}$ less, i.e. $1.8/0.50 = 3.6\,\mathrm{cm}$ more stem: $7.6\,\mathrm{cm}$ emerge. Sensitivity $\approx m/(\rho^2S) \approx 40\,\mathrm{cm}$ per unit of relative density ( $3.6\,\mathrm{cm}$ per $0.1$ ).

Exercise 21.9 ★★

A submarine of volume $3000\,\mathrm{m}^{3}$ has a mass of $2800\,\mathrm{t}$ with empty ballast tanks. Fraction emerging at the surface (sea water); mass of water to take in for neutral buoyancy; what if it then enters a fresh-water estuary?

Solution

Solution of Exercise 21.9.

Displaced volume $2800/1.025 = 2732\,\mathrm{m}^{3}$ : $268\,\mathrm{m}^{3}$ ( $9\%$ ) emerge. Neutral: mass $1025 \times 3000 = 3075\,\mathrm{t}$ : take in $275\,\mathrm{t}$ . Fresh water: buoyancy drops to $3000\,\mathrm{t}$ : $75\,\mathrm{t}$ heavy — it sinks unless $75\,\mathrm{t}$ are pumped out.

Exercise 21.10 ★★★

A child’s balloon holds $10\,\mathrm{L}$ of helium ( $M = 4\,\mathrm{g}/\mathrm{mol}$ ) at $1\,\mathrm{bar}$ and $293\,\mathrm{K}$ ; the rubber weighs $3.0\,\mathrm{g}$ . Net lift; mass of string it can carry.

Solution

Solution of Exercise 21.10.

Air displaced: $10^{-2} \times 1.20 = 12.0\,\mathrm{g}$ ; helium inside $12.0 \times 4/29 = 1.7\,\mathrm{g}$ ; rubber $3.0\,\mathrm{g}$ : net lift $7.3\,\mathrm{g}$ — seven meters of light string.

Exercise 21.11 ★★★

Prove that the center of pressure on a vertical rectangular wall is at $2h/3$ by computing the moment of the pressure forces about the surface line. Where is it for a wall that is only partly submerged, say from depth $h_1$ to $h_2$ ?

Solution

Solution of Exercise 21.11.

Moment $\int_0^h\rho gx\cdot xL\,\dd x = \rho gLh^3/3$ , force $\rho gLh^2/2$ : arm $2h/3$ . From $h_1$ to $h_2$ : moment $\rho gL(h_2^3 - h_1^3)/3$ , force $\rho gL(h_2^2 - h_1^2)/2$ : depth $\frac23(h_2^3 - h_1^3)/(h_2^2 - h_1^2)$ .

Exercise 21.12 ★★★

A cylindrical hydrometer (mass $m$ , cross-section $S$ ) floating in a liquid of density $\rho$ is pushed down by $x$ and released. Show that it oscillates harmonically and give the period; compute it for $m = 20\,\mathrm{g}$ , $S = 0.50\,\mathrm{cm}^{2}$ , water. (Example 14.3 revisited.)

Solution

Solution of Exercise 21.12.

Extra immersion $x$ adds the buoyancy $\rho gSx$ upward: $m\ddot x = -\rho gSx$ , $\omega_0^2 = \rho gS/m$ , $T = 2\pi\sqrt{m/\rho gS} = 2\pi\sqrt{0.020/(1000 \times 9.81 \times 5\times10^{-5})} = 1.3\,\mathrm{s}$ .

Inflating a hot-air balloon: heated air is less dense than the cold air around it, and Archimedes’ thrust on the envelope exceeds its weight.

21.6 Problem: The submarine

Problem 21.1

Weekend problem — a steel hull three hundred meters down: the pressure on its plates, the water it must swallow to sink and spit out to rise, the diver who leaves it, and the tanker that passes overhead

Sea water: $\rho = 1025\,\mathrm{kg}/\mathrm{m}^{3}$ ; fresh water $1000\,\mathrm{kg}/\mathrm{m}^{3}$ ; $P_0 = 1.013 \times 10^{5}\,\mathrm{Pa}$ ; $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ . The submarine’s hull encloses $V = 3000\,\mathrm{m}^{3}$ ; with empty ballast tanks its mass is $m_0 = 2800\,\mathrm{t}$ .

Part I — Pressure on the hull.

Absolute pressure at $300\,\mathrm{m}$ , in pascals and in bars.
Force on a circular hatch of diameter $0.80\,\mathrm{m}$ at that depth; compare with the weight of a truck.
Force on one square meter of hull at the keel if the hull is $10\,\mathrm{m}$ tall and the top is at $300\,\mathrm{m}$ : how different from the top?
Water is slightly compressible ( $\chi_T = 4.6 \times 10^{-10}\,\mathrm{Pa}^{-1}$ ): by what fraction is sea water denser at $300\,\mathrm{m}$ than at the surface? Does it matter for the pressure computed above?
The crew inside breathes air at $1\,\mathrm{atm}$ : why does the hull need to be a thick cylinder, and why does a cylinder resist better than a flat box?
A thin cylindrical shell of radius $R$ and wall thickness $e$ under an external overpressure $\Delta P$ carries a compressive stress $\Delta P\,R/e$ in its wall. For $R = 5.0\,\mathrm{m}$ and $e = 3.0\,\mathrm{cm}$ at $300\,\mathrm{m}$ , compute it and compare with the yield stress of the steel, about $600\,\mathrm{MPa}$ .

Part II — Sinking and floating.

At the surface with empty tanks, what volume of the hull emerges?
What mass of sea water must the ballast tanks take in for the submarine to hover (neutral buoyancy)?
Once neutrally buoyant at $300\,\mathrm{m}$ , the hull is compressed by $0.1\%$ by the pressure. Compute the resulting net force and say whether the equilibrium is stable with respect to depth.
The boat sails neutrally buoyant from the sea into a fresh-water estuary. Net force? Which way does it go, and what must the crew do?
A $20\,\mathrm{t}$ torpedo is fired. What must the ballast system do immediately, and by how much?
Why does a submarine use compressed air to empty its tanks, and why is the air’s pressure the limit on how deep it can blow them?
Estimate the volume of air at $1\,\mathrm{atm}$ needed to empty $275\,\mathrm{m}^{3}$ of tanks at $300\,\mathrm{m}$ (Boyle’s law, $PV$ constant).

Part III — The diver. A diver leaves the surface with $6.0\,\mathrm{L}$ of air in her lungs.

Absolute pressure at $10\,\mathrm{m}$ and $30\,\mathrm{m}$ .
If she descends holding her breath, what is her lung volume at $30\,\mathrm{m}$ ?
A scuba diver breathes air at the ambient pressure at $30\,\mathrm{m}$ and ascends while holding her breath: volume her lungs would need at the surface, and the lesson.
A snorkel $1.0\,\mathrm{m}$ long: pressure difference between the water on the diver’s chest and the air in her lungs; force on a chest of $0.05\,\mathrm{m}^{2}$ ; can she inhale?
The air tank holds $12\,\mathrm{L}$ at $200\,\mathrm{bar}$ : how many liters of air at $30\,\mathrm{m}$ does that make, and for how long at $20\,\mathrm{L}$ per minute?

Part IV — The tanker overhead.

A tanker of $100\,000\,\mathrm{t}$ floats in sea water: volume of water displaced.
Its waterline area is $8000\,\mathrm{m}^{2}$ : by how much does it rise or sink when it enters fresh water?
Loading $10\,000\,\mathrm{t}$ of oil: change of draught.
The empty tanker has a high center of mass; it carries sea water as ballast when empty. Why, in terms of the metacenter?
The tanker’s hull at its keel ( $20\,\mathrm{m}$ draught): pressure and force per square meter, compared with the submarine at $300\,\mathrm{m}$ .
The tanker passes directly over the submarine. Does the submarine feel its $100\,000\,\mathrm{t}$ ? Explain with the pressure field.
Summarize: the one relation that fixed every number in this problem, and the theorem that followed from it.

Solution

Solution of Problem 21.1.

1. $P = 1.013\times10^5 + 1025 \times 9.81 \times 300 = 3.12 \times 10^{6}\,\mathrm{Pa} = 31\,\mathrm{bar}$ .

2. $S = \pi(0.40)^2 = 0.503\,\mathrm{m}^{2}$ : net force $(P - P_0)S = 1.52 \times 10^{6}\,\mathrm{N}$ — 155 tonnes, a loaded truck on a hatch.

3. Keel $10\,\mathrm{m}$ deeper: $+\rho g \times 10 = 1.0 \times 10^{5}\,\mathrm{Pa}$ more, $3\%$ : the hull is loaded almost uniformly.

4. $\Delta\rho/\rho = \chi_T\Delta P = 4.6\times10^{-10} \times 3.0\times10^6 = 0.14\%$ : negligible for the pressure (a few kilopascals out of three million).

5. Thirty bars outside, one inside: the hull is a pressure vessel loaded inward; a cylinder (or sphere) turns the pressure into compression along its wall, which steel resists well, whereas a flat plate would bend.

6. $\Delta P\,R/e = 3.0\times10^6 \times 5.0/0.030 = 5.0 \times 10^{8}\,\mathrm{Pa} = 500\,\mathrm{MPa}$ : close to the yield stress — $300\,\mathrm{m}$ is near the limit for this hull, and deeper boats need thicker or stronger steel (or titanium).

7. Displaced volume $m_0/\rho = 2800/1.025 = 2732\,\mathrm{m}^{3}$ : $268\,\mathrm{m}^{3}$ emerge ( $9\%$ ).

8. Neutral: $m = \rho V = 3075\,\mathrm{t}$ : take in $275\,\mathrm{t}$ .

9. Buoyancy falls by $0.1\%$ of $\rho Vg$ : $0.001 \times 3075 \times 9.81 \times10^3 = 30\,\mathrm{kN}$ downward (three tonnes). Deeper $\Rightarrow$ more compression $\Rightarrow$ heavier: unstable — a submarine must trim continuously (and a steel hull compresses less than water, which helps; a too-flexible hull would sink without return).

10. Buoyancy $1000 \times 3000 \times 9.81 = 29.4\,\mathrm{MN}$ against weight $3075 \times 9.81 \times 10^3 = 30.2\,\mathrm{MN}$ : $0.74\,\mathrm{MN}$ down ( $75\,\mathrm{t}$ ); pump out $75\,\mathrm{t}$ .

11. Immediately $20\,\mathrm{t}$ light: take in $20\,\mathrm{t}$ of water ( $19.5\,\mathrm{m}^{3}$ ) into compensating tanks.

12. Water can only be expelled by something at higher pressure than the sea outside; compressed air in bottles pushes it out — only as long as the bottle pressure exceeds the ambient pressure at that depth.

13. $PV$ constant: $275\,\mathrm{m}^{3}$ at $31\,\mathrm{bar}$ needs $275 \times 31 = 8500\,\mathrm{m}^{3}$ of air at $1\,\mathrm{atm}$ — stored at $200\,\mathrm{bar}$ in $43\,\mathrm{m}^{3}$ of bottles.

14. $2.0\,\mathrm{bar}$ at $10\,\mathrm{m}$ ; $4.0\,\mathrm{bar}$ at $30\,\mathrm{m}$ .

15. $PV$ constant: $6.0/4.0 = 1.5\,\mathrm{L}$ — the ribcage is crushed in; free divers feel it.

16. Air taken at $4\,\mathrm{bar}$ expands fourfold: $24\,\mathrm{L}$ in lungs that hold $6\,\mathrm{L}$ : rupture. Never hold your breath on ascent; exhale continuously.

17. $\Delta P = \rho g \times 1.0 = 1.0 \times 10^{4}\,\mathrm{Pa}$ ; force $1.0 \times 10^{4}\,\mathrm{Pa} \times 0.05 = 500\,\mathrm{N}$ on the chest: the breathing muscles cannot lift fifty kilograms — a snorkel works only in the first decimeters.

18. $12 \times 200 = 2400\,\mathrm{L}$ at $1\,\mathrm{bar}$ , i.e. $600\,\mathrm{L}$ at $4\,\mathrm{bar}$ : $30\,\mathrm{min}$ .

19. $V = 10^8/1025 = 97\,600\,\mathrm{m}^{3}$ .

20. In fresh water it must displace $10^5\ \mathrm{m}^{3}$ : $2400\,\mathrm{m}^{3}$ more, i.e. $2400/8000 = 0.30\,\mathrm{m}$ deeper.

21. $10^4/1.025 = 9756\,\mathrm{m}^{3}$ more: $1.2\,\mathrm{m}$ deeper.

22. Empty, the ship’s $G$ sits high and little hull is immersed: the metacenter may fall below $G$ and the ship capsize in a swell. Sea water low in the hull lowers $G$ and deepens the draught.

23. $\rho gh = 2.0 \times 10^{5}\,\mathrm{Pa}$ , $0.2\,\mathrm{MN}/\mathrm{m}^{2}$ : fifteen times less than the submarine’s $3\,\mathrm{MN}/\mathrm{m}^{2}$ .

24. No: the pressure at the submarine’s depth is $P_0 + \rho gh$ , fixed by the depth alone. The tanker’s weight is borne by the water it displaces, which raises the sea level everywhere by an immeasurable amount; no extra load reaches the submarine.

25. $\dd P/\dd z = -\rho g$ — hence $P = P_0 + \rho gh$ for every depth, force and lung volume here — and Archimedes’ theorem, which is that relation integrated over a closed surface.