Part 2 · Chapter 14

Fluids

From the pressure on a diver to the lift on a wing — how substances that flow push, float, and speed up

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • The two quantities that matter for fluids — density \(\rho = m/V\) and pressure \(p = F/A\) — and why pressure is a scalar.
  • How hydrostatic pressure grows with depth: \(p = p_{0} + \rho g h\), and the difference between absolute and gauge pressure.
  • Pascal's principle and the hydraulic lever (\(F_{o} = F_{i}\,A_{o}/A_{i}\)) that magnifies force without magnifying work.
  • Archimedes' principle — a buoyant force equal to the weight of displaced fluid (\(F_{b} = m_{f}g\)) — and the rule for floating.
  • Ideal-fluid flow: the equation of continuity (\(Av = \text{const}\)) and Bernoulli's equation (\(p + \tfrac{1}{2}\rho v^{2} + \rho g y = \text{const}\)).
Section 14-1

What Is Physics?

The physics of fluids underpins hydraulic engineering, which turns up almost everywhere: the cooling system of a nuclear reactor, the blood flow in an aging patient's arteries, the irrigation of farmland, a diver's safety, the wing flaps that let a jet land — even the hydraulics that raise and lower huge Broadway and Las Vegas sets. Before any of that, though, we need to answer a deceptively simple question: what is a fluid?

Section 14-2

What Is a Fluid?

A fluid is a substance that can flow and so conforms to the boundaries of any container. The defining feature, in the language of the previous chapter, is that a fluid cannot sustain a shearing stress — a force tangent to its surface. It can push perpendicular to a surface, but it cannot resist being sheared, so it simply flows. Even slow-moving pitch counts as a fluid; it conforms eventually. We lump liquids and gases together because neither has the rigid long-range atomic order of a crystalline solid like ice — liquid water and steam are, at the molecular level, more alike than either is to ice.

Section 14-3

Density and Pressure

With a rigid body we track mass and force. With a fluid — an extended substance whose properties vary from point to point — it is far more useful to track density and pressure. Density is mass per unit volume, treated as smooth on scales much larger than atoms:

Density (uniform sample)
\[ \rho = \frac{m}{V} \]
A scalar, SI unit kg/m³. Liquids are nearly incompressible (water's density barely changes from 1 to 50 atm), but gases are readily compressible (air's density rises ~50-fold over the same range).
Table 14-1 · Some densities (kg/m³)
Material or objectDensityMaterial or objectDensity
Best laboratory vacuum10⁻¹⁷Iron7.9 × 10³
Air (20 °C, 1 atm)1.21Mercury13.6 × 10³
Ice0.917 × 10³Earth (average)5.5 × 10³
Water (20 °C, 1 atm)0.998 × 10³Sun (average)1.4 × 10³
Seawater (20 °C, 1 atm)1.024 × 10³Sun (core)1.6 × 10⁵
Whole blood1.060 × 10³Neutron star (core)10¹⁸

Pressure is the magnitude of the normal force per unit area that a fluid exerts on a surface:

Pressure of a uniform force on a flat area
\[ p = \frac{F}{A} \]
SI unit: the pascal (Pa = N/m²). At a given point in a fluid at rest, pressure is the same in every direction — it is a scalar. Common conversions: 1 atm = 1.01 × 10⁵ Pa = 760 torr = 14.7 lb/in².
Table 14-2 · Some pressures (Pa)
LocationPressureLocationPressure
Center of the Sun2 × 10¹⁶Automobile tire (gauge)2 × 10⁵
Center of Earth4 × 10¹¹Atmosphere at sea level1.0 × 10⁵
Deepest ocean trench1.1 × 10⁸Normal blood systolic (gauge)1.6 × 10⁴
Spike heels on a dance floor10⁶Best laboratory vacuum10⁻¹²
Section 14-4

Fluids at Rest

Every diver knows pressure rises with depth; every mountaineer knows it falls with altitude. To find the rule, isolate an imaginary cylinder of fluid in static equilibrium and balance the three vertical forces on it — the downward push on its top, the upward push on its bottom, and its own weight. The pressures at the upper and lower faces (at heights \(y_{1}\) and \(y_{2}\), measured positive upward) must differ by exactly the weight of fluid between them:

Pressure change with height, and pressure at depth h
\[ p_{2} = p_{1} + \rho g (y_{1} - y_{2}) \qquad\Longrightarrow\qquad p = p_{0} + \rho g h \]
With level 1 at the surface (pressure p₀) and level 2 a depth h below it. The pressure depends only on depth — not on the shape of the container or any horizontal dimension.
🌊
Absolute vs gauge pressure
p = p0 + ρgh  ·  gauge pressure = ρgh

The total (absolute) pressure at depth \(h\) is the atmospheric pressure \(p_{0}\) pressing on the surface plus the column term \(\rho g h\). The gauge pressure is the excess over atmospheric, \(\rho g h\) — what a tire gauge or blood-pressure cuff reads. It is positive when \(p > p_{0}\) (a tire) and negative when \(p < p_{0}\) (sucking on a straw).

Same level, same pressure. Because pressure depends only on depth, any two connected points at the same level in a fluid at rest are at the same pressure, even if they are far apart horizontally. This single fact is the engine behind U-tubes, barometers, manometers, and the hydraulic lever — set the pressures equal at a shared level and solve.
Section 14-5

Measuring Pressure

Two classic instruments follow directly from \(p = p_{0} + \rho g h\). A mercury barometer is a tube of mercury inverted over a dish; the near-vacuum above the column means atmospheric pressure alone supports the column height \(h\):

Barometer and open-tube manometer
\[ p_{0} = \rho g h \qquad\qquad p_{g} = p - p_{0} = \rho g h \]
Barometer (left): the column height measures atmospheric pressure and does not depend on the tube's cross-section. Manometer (right): the height difference h between the two arms gives the gauge pressure of a gas directly.

The barometer's reading in millimetres of mercury equals the pressure in torr only at the standard \(g = 9.80665\,\mathrm{m/s^{2}}\) and 0 °C; elsewhere small corrections apply, because the column height depends on both \(g\) and mercury's temperature-dependent density.

Section 14-6

Pascal's Principle

Squeeze one end of a toothpaste tube and paste comes out the other end — that is Pascal's principle, stated in 1652: a change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and to the walls of its container. Adding lead shot to a piston raises the pressure by \(\Delta p_{\text{ext}}\) everywhere at once, independent of depth.

🛠
The hydraulic lever
Fo = Fi (Ao / Ai)

A small input force on a small piston produces a large output force on a large piston — the same pressure acting over a bigger area. But the output piston moves a proportionally smaller distance (\(d_{o} = d_{i}\,A_{i}/A_{o}\)), so the work is unchanged: \(W = F_{o}d_{o} = F_{i}d_{i}\). A hydraulic jack lifts a car you could never lift directly — at the cost of many small pumps of the handle.

Section 14-7

Archimedes' Principle

Picture a thin sack of water held submerged in a pool: it neither rises nor sinks, so the surrounding water must push up on it with a force equal to the weight of the water inside. That upward push — present because pressure is greater at the bottom of the sack than the top — is the buoyant force. Replace the sack's contents with a stone or a block of wood of the same shape and the surrounding water cannot tell the difference; the buoyant force is unchanged.

Archimedes' principle
Fb = mf g  (weight of displaced fluid)

A fully or partially submerged body feels an upward buoyant force equal to the weight of the fluid it displaces. If \(F_{b}\) exceeds the body's weight it rises (wood); if less, it sinks (stone).

A body floats when the buoyant force grows to match its weight, \(F_{b} = F_{g}\) — so a floating body displaces its own weight of fluid. When a buoyant force acts, a scale reads less than the true weight; this apparent weight is

Floating condition and apparent weight
\[ F_{g} = m_{f}g \;\text{(floating)} \qquad\qquad \text{weight}_{\text{app}} = \text{weight} - F_{b} \]
A floating body's apparent weight is zero — which is why astronauts rehearse complex tasks underwater, where their apparent weight matches the weightlessness of orbit.
Section 14-8

Ideal Fluids in Motion

Real flow is fearsomely complex, so we model an ideal fluid with four simplifying assumptions: flow is steady (velocity at a fixed point does not change in time, i.e. laminar not turbulent), incompressible (constant density), nonviscous (no internal friction draining kinetic energy to heat), and irrotational (a speck of dust carried along does not spin about its own center).

Streamlines never cross. A tracer of dye or smoke traces out a streamline — the path of a fluid element, with velocity always tangent to it. Two streamlines can never intersect, since an element at the crossing would need two velocities at once. Where streamlines crowd together, the fluid moves faster; where they spread apart, it slows. That visual will pay off twice over in the next two sections.
Section 14-9

The Equation of Continuity

Pinch a garden hose and the water speeds up — the flow speed depends on the cross-sectional area. For a steady, incompressible flow, the same volume that enters a tube segment per second must leave it per second. Equating the volumes gives the equation of continuity:

Equation of continuity (volume and mass flow rate)
\[ A_{1}v_{1} = A_{2}v_{2} \qquad R_{V} = Av = \text{const} \qquad R_{m} = \rho A v = \text{const} \]
The volume flow rate R_V (m³/s) is constant along a tube of flow; multiplying by density gives the constant mass flow rate R_m (kg/s). Narrow the channel and the speed rises to compensate.
Section 14-10

Bernoulli's Equation

Apply conservation of energy to a tube of flowing ideal fluid — accounting for the work done by pressure pushing fluid in and out, plus the change in gravitational and kinetic energy — and you arrive at Bernoulli's equation, the fluid-flow restatement of energy conservation:

Bernoulli's equation (along a tube of flow)
p + ½ρv² + ρgy = constant

Three competing terms: static pressure, kinetic-energy density \(\tfrac{1}{2}\rho v^{2}\), and gravitational term \(\rho g y\). Set \(v = 0\) and it reduces to the hydrostatic law; set \(y\) constant and it predicts the headline result below.

Faster flow, lower pressure. Along a horizontal streamline, if a fluid element speeds up, its pressure must drop — and vice versa. Where streamlines crowd and the flow is fast, the pressure is low. This is why a shower curtain pulls inward, why a curveball curves, and (loosely) why a wing lifts: the higher-pressure slow region pushes toward the lower-pressure fast region. A draining tank obeys the same logic — water exits a hole a depth \(h\) below the surface at \(v = \sqrt{2gh}\), exactly the speed of a free fall through \(h\) (Torricelli's law).
Worked Examples

Putting It to Work

1 Atmospheric pressure — force and weight of air

Problem. (a) What does the air in a living room of dimensions 3.5 m × 4.2 m × 2.4 m weigh at 1.0 atm? (b) What downward force does the atmosphere exert on the top of your head (area ≈ 0.040 m²)?

Solution. Use \(\rho = m/V\) with the density of air (1.21 kg/m³), and \(F = pA\) with \(p = 1.0\,\mathrm{atm} = 1.01 \times 10^{5}\,\mathrm{Pa}\).

Weight = ρVg; force = pA
\[\begin{gathered} mg = \rho V g = (1.21)(3.5 \times 4.2 \times 2.4)(9.8) \approx 420\,\mathrm{N} \\ F = pA = (1.01 \times 10^{5})(0.040) \approx 4.0 \times 10^{3}\,\mathrm{N} \end{gathered}\]

The 4 kN on your head is the weight of the entire air column above it, balanced by the air pressure pushing up from below — which is why you never feel it.

2 Gauge pressure on an ascending scuba diver

Problem. A diver fills his lungs at depth \(L\) and (against instructions) holds his breath while surfacing. At the surface the difference between his lung pressure and the outside pressure is 9.3 kPa. From what depth did he start?

Solution. The trapped lung air keeps the pressure it had at depth, while the outside drops to \(p_{0}\). The leftover difference is the gauge pressure \(\rho g L\).

Solve Δp = ρgL for L (water, ρ = 998 kg/m³)
\[ L = \frac{\Delta p}{\rho g} = \frac{9300}{(998)(9.8)} \approx 0.95\,\mathrm{m} \]

Less than a metre — yet that ~9% of atmospheric pressure is enough to rupture lung tissue and force air into the blood. Exhaling steadily on the way up keeps the pressures equalized and removes the danger.

3 Floating block — submerged depth and acceleration

Problem. A block of density \(\rho = 800\,\mathrm{kg/m^{3}}\) and height \(H = 6.0\,\mathrm{cm}\) floats face-down in a fluid of density \(\rho_{f} = 1200\,\mathrm{kg/m^{3}}\). (a) To what depth \(h\) is it submerged? (b) If pushed fully under and released, what is its acceleration?

Solution. Floating means buoyancy equals weight: \(\rho_{f}LWh\,g = \rho LWH\,g\). Fully submerged, the buoyant force grows (full height displaces fluid), and Newton's second law gives the upward acceleration.

Float balance, then F_b − F_g = ma
\[\begin{gathered} h = \frac{\rho}{\rho_{f}}H = \frac{800}{1200}(6.0) = 4.0\,\mathrm{cm} \\ a = \left( \frac{\rho_{f}}{\rho} - 1 \right)g = \left( \frac{1200}{800} - 1 \right)(9.8) \approx 4.9\,\mathrm{m/s^{2}} \end{gathered}\]

Two-thirds of the block sits below the surface (the density ratio), and when forced under it springs upward at half a \(g\).

4 A falling stream necks down — continuity

Problem. Water leaves a tap and narrows as it falls. The cross-section shrinks from \(A_{0} = 1.2\,\mathrm{cm^{2}}\) to \(A = 0.35\,\mathrm{cm^{2}}\) over a drop of \(h = 45\,\mathrm{mm}\). Find the volume flow rate.

Solution. Continuity (\(A_{0}v_{0} = Av\)) plus free-fall kinematics (\(v^{2} = v_{0}^{2} + 2gh\)) give \(v_{0}\); then \(R_{V} = A_{0}v_{0}\).

Eliminate v, solve for v₀, then R_V
\[\begin{gathered} v_{0} = \sqrt{\frac{2ghA^{2}}{A_{0}^{2} - A^{2}}} = \sqrt{\frac{2(9.8)(0.045)(0.35)^{2}}{(1.2)^{2} - (0.35)^{2}}} \approx 28.6\,\mathrm{cm/s} \\ R_{V} = A_{0}v_{0} = (1.2)(28.6) \approx 34\,\mathrm{cm^{3}/s} \end{gathered}\]

The stream necks because gravity speeds it up, and continuity then demands a smaller area to keep the volume flow rate constant.

5 Flow through a narrowing pipe — Bernoulli

Problem. Ethanol (\(\rho = 791\,\mathrm{kg/m^{3}}\)) flows through a horizontal pipe that tapers from \(A_{1} = 1.20 \times 10^{-3}\,\mathrm{m^{2}}\) to \(A_{2} = A_{1}/2\). The pressure difference between the wide and narrow sections is 4120 Pa. Find the volume flow rate.

Solution. Combine continuity (\(v = R_{V}/A\), with \(A_{2} = A_{1}/2\)) and Bernoulli on the horizontal pipe. The narrow section is faster, so its pressure is lower — meaning \(p_{1} - p_{2} = +4120\,\mathrm{Pa}\).

Bernoulli + continuity, solved for R_V
\[ R_{V} = A_{1}\sqrt{\frac{2(p_{1} - p_{2})}{3\rho}} = (1.20 \times 10^{-3})\sqrt{\frac{2(4120)}{3(791)}} \approx 2.24 \times 10^{-3}\,\mathrm{m^{3}/s} \]

The "faster means lower pressure" rule settled the sign: had we guessed the other way, the square root would have gone imaginary.

Review

Chapter Summary

Density & pressure

\(\rho = m/V\) and \(p = F/A\). Pressure is a scalar; gauge pressure is the difference from atmospheric.

Hydrostatics

\(p = p_{0} + \rho g h\) — pressure depends on depth, not container shape. Same level ⟹ same pressure.

Measuring pressure

Barometer: \(p_{0} = \rho g h\). Open-tube manometer: \(p_{g} = \rho g h\).

Pascal's principle

An applied pressure transmits undiminished throughout the fluid. Hydraulic lever: \(F_{o} = F_{i}A_{o}/A_{i}\), same work.

Archimedes' principle

Buoyant force \(F_{b} = m_{f}g\) (weight of displaced fluid). Floating: \(F_{b} = F_{g}\); apparent weight = weight − \(F_{b}\).

Continuity

\(R_{V} = Av = \text{const}\) and \(R_{m} = \rho A v = \text{const}\) — narrow the channel, speed rises.

Bernoulli's equation

\(p + \tfrac{1}{2}\rho v^{2} + \rho g y = \text{const}\) — energy conservation for ideal flow; faster flow means lower pressure.

Ideal fluid

Steady, incompressible, nonviscous, irrotational. Streamlines never cross; velocity is tangent to them.

Practice

Problems

Statics problems lean on \(p = p_{0} + \rho g h\) and "same level, same pressure"; buoyancy problems on \(F_{b} = m_{f}g\) and the floating condition; flow problems on continuity and Bernoulli together. Useful values: \(1\,\mathrm{atm} = 1.01 \times 10^{5}\,\mathrm{Pa}\), \(\rho_{\text{water}} \approx 1000\,\mathrm{kg/m^{3}}\), \(g = 9.8\,\mathrm{m/s^{2}}\).

  1. Find the pressure increase in the fluid in a syringe when a nurse pushes its circular piston (radius 1.1 cm) with a force of 42 N.
  2. An office window measures 3.4 m by 2.1 m. During a storm the outside pressure drops to 0.96 atm while inside stays at 1.0 atm. What net outward force pushes on the window?
  3. At a depth of 10.9 km in the Marianas Trench, what hydrostatic pressure (in atmospheres) must a vessel withstand? Take seawater density as 1024 kg/m³.
  4. Calculate the hydrostatic difference in blood pressure between the brain and the foot of a person 1.83 m tall. Blood density is \(1.06 \times 10^{3}\,\mathrm{kg/m^{3}}\).
  5. A hydraulic press has a small piston of diameter 3.80 cm and a large one of diameter 53.0 cm. What force on the small piston balances a 20.0 kN force on the large one?
  6. An iron anchor (density 7870 kg/m³) appears 200 N lighter in water than in air. Find (a) its volume and (b) its weight in air.
  7. A block of wood floats in fresh water with two-thirds of its volume submerged, and in oil with 0.90 of its volume submerged. Find the density of (a) the wood and (b) the oil.
  8. What fraction of an iceberg (density 917 kg/m³) is visible when it floats in (a) seawater (1024 kg/m³) and (b) fresh water (1000 kg/m³)?
  9. A garden hose of internal diameter 1.9 cm feeds a sprinkler with 24 holes, each 0.13 cm in diameter. If the water moves at 0.91 m/s in the hose, how fast does it leave the holes?
  10. Water enters a basement at 0.90 m/s and 170 kPa through a 2.5 cm pipe, which then tapers to 1.2 cm and rises 7.6 m to the second floor. Find the (a) speed and (b) pressure upstairs.
  11. A large tank is filled with water to depth \(D = 0.30\,\mathrm{m}\). A hole of area 6.5 cm² in the bottom lets water drain. At what rate (m³/s) does water flow out?
  12. A cylindrical tank holds water to height \(H = 40\,\mathrm{cm}\); a hole at depth \(h = 10\,\mathrm{cm}\) lets water out. (a) How far from the base does the stream strike the floor? (b) At what depth should a hole give the same range?
Tip: for a fluid at rest, the only thing that sets pressure is depth — pick a level, set the pressures on both sides equal, and the unknowns usually fall out (this is the whole trick behind U-tubes and manometers). For a fluid in motion, reach for continuity and Bernoulli together: continuity links the speeds through the areas, and Bernoulli links speed to pressure. And remember the recurring rule of thumb — where the fluid runs fast, the pressure runs low.