Mechanical Properties of Fluids

This chapter has two halves joined by one idea — fluids offer almost no resistance to shear, so force spreads itself out as pressure rather than staying localised as in a solid. The first half studies fluids at rest: how pressure is defined and measured, Pascal's law and its proof, how pressure varies with depth, the atmosphere as a fluid column, and hydraulic machines that use Pascal's law to multiply force. The second half studies fluids in motion: streamline versus turbulent flow, the equation of continuity as mass conservation, and Bernoulli's principle as energy conservation along a streamline, applied to efflux speed, spinning balls, and aircraft wings. Two closing sections cover where real fluids depart from the ideal picture — viscosity, the internal friction that resists flow and produces a terminal velocity for falling bodies, and surface tension, the extra energy of a liquid's free surface that shapes drops, bubbles, wetting, and capillary rise. Work through it in roughly this order: pressure, Pascal's law, depth and atmosphere, hydraulic machines, streamlines and continuity, Bernoulli and its applications, viscosity and Stokes' law, then surface tension and capillarity.

Pressure: force spread over area

When a fluid at rest touches a surface, the force it exerts is always normal (perpendicular) to that surface. If there were any force component parallel to the surface, the object would, by Newton's third law, push back parallel to the fluid, setting the fluid moving — but a fluid at rest cannot support such a shear. This single observation is why an idealised pressure probe (an evacuated chamber with a calibrated spring pressing on a piston) always reads a force acting straight into its face, regardless of how the probe is oriented inside the fluid.

where: $F$ is the magnitude of the normal force on a piston or surface of area $A$, and $P_{av}$ is the average pressure. Pressure is a scalar — despite the vector $F$ in the definition, it is only the component of force normal to the chosen area that enters, and no direction can be attached to "pressure" itself. Its dimensions are $[\text{ML}^{-1}\text{T}^{-2}]$ and its SI unit is the pascal, $\text{Pa} = \text{N/m}^2$, named for Blaise Pascal; a common practical unit is the atmosphere, $1\ \text{atm} = 1.013\times10^5\ \text{Pa}$, the pressure exerted by Earth's atmosphere at sea level. A companion quantity is density, $\rho = m/V$, with SI unit $\text{kg/m}^3$; liquids are nearly incompressible so their density is almost constant, while gas density changes markedly with pressure.

Pascal's law: pressure is the same in every direction

Blaise Pascal observed that the pressure in a fluid at rest is the same at every point at the same height, and further, that pressure acts equally in every direction at a given point — a fact proved below by balancing forces on a small wedge of fluid. A useful consequence is the hydrostatic paradox: three vessels of completely different shapes, connected at the base and filled with water to the same height, hold very different volumes of water yet show identical pressure — and hence identical water level — at the base, because pressure depends only on depth, never on the shape or cross-section of the container above it.

Variation of pressure with depth, and atmospheric pressure

Take a cylindrical column of fluid of base area $A$ between a point 1 at height $h$ above a point 2. Balancing the weight of the fluid column against the pressure forces on its top and bottom faces gives $P_2 - P_1 = \rho g h$ (proved fully below). Placing point 1 at a free surface open to the atmosphere, where the pressure is atmospheric pressure $P_a$, this becomes the working formula for pressure at depth $h$ below any open liquid surface:

where: $\rho$ is the (assumed uniform) density of the liquid, $g$ is the acceleration due to gravity, and $h$ is depth below the free surface. Neither the cross-sectional area nor the container's shape appears — only depth matters, exactly the hydrostatic paradox above. The quantity $P - P_a = \rho g h$ is the gauge pressure; a tyre gauge or a sphygmomanometer reads gauge pressure, not absolute. A mercury barometer applies the same equation with essentially zero pressure above the column, giving $P_a = \rho g h$ with $h \approx 76\ \text{cm}$ of mercury at sea level; a pressure equivalent to $1\ \text{mm}$ of mercury is a torr ($133\ \text{Pa}$), and $1\ \text{bar} = 10^5\ \text{Pa}$. An open-tube manometer uses the same U-tube balance to read a gas supply's gauge pressure as a column height.

Hydraulic machines: Pascal's law at work

Because pressure applied anywhere to an enclosed fluid is transmitted undiminished to every point and every wall of the vessel, a small force on a narrow piston can be turned into a large force on a wide piston, with the pressure $F_1/A_1$ common to both:

where: $F_1$ is the force applied to the small piston of area $A_1$, and $F_2$ is the resulting force on the large piston of area $A_2$. The ratio $A_2/A_1$ is the device's mechanical advantage — exactly how a hydraulic car lift or the master and slave cylinders of a hydraulic brake work, and why a gentle push on a brake pedal produces a large, evenly distributed clamping force at all four wheels.

Streamline flow and the equation of continuity

The study of fluids in motion is fluid dynamics. A flow is steady if, at any fixed point in space, the velocity of successive fluid particles passing that point stays constant in time (though velocity can differ from point to point). The path a fluid particle traces under steady flow is a streamline, whose tangent at every point gives the local fluid velocity; streamlines never cross, since a crossing point would mean a particle there could have two velocities at once. Flow with parallel streamlines is laminar; beyond a critical speed it loses this order and becomes turbulent, breaking into swirling eddies — the "white water" where a fast stream meets rocks. For an incompressible fluid, mass flowing in must equal mass flowing out across any cross-section of a pipe, however its area changes along the length:

where: $A$ is the pipe's cross-sectional area and $v$ is the fluid speed there, so the volume flow rate $Av$ stays fixed along the pipe. Where streamlines crowd together (narrow cross-section) speed rises; where they spread out, speed falls — a pure statement of mass conservation, with no reference yet to pressure or energy.

Bernoulli's principle and its applications

Bernoulli's principle is energy conservation applied to a streamline of a non-viscous, incompressible fluid in steady flow. As a fluid element accelerates through a pipe of changing area and height, the work done on it by the pressure difference converts into changes of kinetic and gravitational potential energy per unit volume:

where: $P$ is the pressure, $\tfrac{1}{2}\rho v^2$ the kinetic energy per unit volume, and $\rho g h$ the gravitational potential energy per unit volume, at a point of height $h$ and speed $v$ on the streamline. Wherever a streamline speeds up (a narrow constriction, or the crowded streamlines above a wing) its pressure must drop to keep the sum constant, and vice versa. Setting $v = 0$ everywhere recovers the hydrostatic result of Eq. (2), so Bernoulli's equation is the moving-fluid generalisation of pressure-with-depth. It holds only for flow with negligible viscosity and compressibility, along a streamline of steady, non-turbulent flow — it cannot be applied across a rapid or strong turbulence, where pressure and velocity fluctuate rather than settle. A first application is Torricelli's law for the efflux speed from a small hole at depth $h$ below an open tank's free surface: applying Eq. (5) between the hole and the (nearly stationary, if the tank is wide) free surface, both at atmospheric pressure, gives $v = \sqrt{2gh}$ — identical in form to a body that has fallen freely through height $h$. A second application is dynamic lift. Air past a non-spinning ball is symmetric above and below, so there is no net force. A spinning ball drags air with it, speeding the flow on one side and slowing it on the other; by Bernoulli's principle the faster side has lower pressure, and the difference pushes the ball sideways — the Magnus effect, why a spinning cricket or tennis ball curves off a plain parabola. An aerofoil-shaped wing is angled so streamlines crowd together above it, raising the air speed there; the resulting pressure difference, pushing up, is the dynamic lift balancing an aircraft's weight in level flight.

Viscosity and Stokes' law

Real fluids resist relative motion between their own layers — an internal friction called viscosity. Sandwich a fluid between two plates, hold the lower one fixed, and drag the upper one at speed $v$: the fluid layer touching each plate moves with that plate's velocity, and the speed increases smoothly from zero at the bottom to $v$ at the top, this ordered layer-on-layer sliding being called laminar flow. The force needed depends on the rate at which shear strain builds up, not on the strain itself:

where: $\eta$ (eta) is the coefficient of viscosity, $F/A$ the shear stress on the moving plate, and $v/l$ the velocity gradient (strain rate) across a fluid film of thickness $l$. Its SI unit is the poiseuille (Pa·s); thin liquids such as water are far less viscous than thick ones such as glycerine or honey. Viscosity of liquids falls with rising temperature (molecules slip past each other more easily), while viscosity of gases rises with temperature (faster random motion increases momentum transfer between layers) — a contrast worth remembering exactly, since it is a favourite fill-in-the-blank. A body moving through a viscous fluid drags the adjoining layer with it and feels a retarding drag force, Stokes' law, proportional to its speed:

where: $a$ is the radius of a small sphere moving at speed $v$ through a fluid of viscosity $\eta$. As a falling sphere (a raindrop) speeds up, this drag grows until, together with the upward buoyant force, it exactly balances gravity, and the sphere falls at a constant terminal velocity, derived in full below.

Surface tension, surface energy and capillary rise

A molecule deep inside a liquid is pulled equally in every direction by its neighbours, giving it a certain (negative) potential energy; a molecule at the free surface has neighbours only below it, so roughly half that attraction is missing, leaving extra potential energy compared with the bulk. A liquid surface therefore has genuine extra energy proportional to its area, and a liquid left alone settles into the shape of least surface area for its volume — a sphere, for a free drop.

where: $S$ is the surface tension, measured by stretching a liquid film (two surfaces) on a sliding bar of length $l$ with internal force $F$; equivalently, $S$ is the force per unit length the surface exerts across any line on it, and the energy needed to create one square metre of new surface (SI unit $\text{N/m} = \text{J/m}^2$). Surface tension, like viscosity, falls with rising temperature. Where a liquid meets a solid, the surface curves to meet it at a characteristic angle of contact $\theta$, set by the balance of the liquid-air, solid-air, and solid-liquid tensions; $\theta$ is acute (the liquid wets the solid, as water does glass) when the solid-liquid tension is small, and obtuse (the liquid beads up, as mercury does on glass, or water on a lotus leaf) when it is large. A curved liquid surface, having extra energy, always sustains more pressure on its concave (inside) than its convex (outside) side — this excess pressure, derived below, is $2S/r$ for a spherical drop or bubble of liquid in gas (one surface), and $4S/r$ for a soap bubble (two surfaces). The same curvature drives capillary rise: in a fine tube of radius $a$ dipped in a wetting liquid, the meniscus curves concave, the pressure just under it is less than atmospheric by $2S\cos\theta/a$, and liquid is pushed up until the raised column's extra weight restores balance, giving

where: $h$ is the height the liquid rises above the reservoir's free surface, $\rho$ is the liquid's density, and $a$ is the tube's inner radius. Rise is larger for finer tubes ($h \propto 1/a$); if $\cos\theta$ is negative (obtuse contact angle, as for mercury in glass), Eq. (9) gives negative $h$ — the liquid is depressed below the reservoir level instead of rising.

Board level - pressure and gauge pressure at depth in a lake

What is the absolute pressure and the gauge pressure on a swimmer $10\ \text{m}$ below the surface of a lake? Take $\rho_{\text{water}} = 1000\ \text{kg/m}^3$, $g = 10\ \text{m/s}^2$ and atmospheric pressure $P_a = 1.01\times10^5\ \text{Pa}$.

Step 1. This is a direct application of $P = P_a + \rho g h$ with $h = 10\ \text{m}$.

Step 2. Substitute: $\rho g h = 1000 \times 10 \times 10 = 1.0\times10^5\ \text{Pa}$, so absolute pressure $P = 1.01\times10^5 + 1.0\times10^5 = 2.01\times10^5\ \text{Pa}$.

Step 3. The gauge pressure is the excess over atmospheric: $P_g = \rho g h = 1.0\times10^5\ \text{Pa} \approx 1\ \text{atm}$. Answer: absolute pressure $\approx 2.01\times10^5\ \text{Pa}$ ($\approx 2\ \text{atm}$), gauge pressure $\approx 1.0\times10^5\ \text{Pa}$ ($\approx 1\ \text{atm}$) — the pressure has doubled over the surface value after just $10\ \text{m}$, which is why divers equalise ear pressure so often near the surface.

Board level - a hydraulic car lift

In a car lift, compressed air exerts a force $F_1$ on a small piston of radius $5.0\ \text{cm}$. This pressure is transmitted to a second piston of radius $15\ \text{cm}$. If the mass of the car to be lifted is $1350\ \text{kg}$ and $g = 9.8\ \text{m/s}^2$, calculate $F_1$ and the air pressure needed.

Step 1. The large piston must support the car's weight, $F_2 = mg = 1350 \times 9.8 = 13{,}230\ \text{N}$.

Step 2. Apply the hydraulic-lift relation, $F_1 = F_2 (r_1/r_2)^2$ since area is proportional to radius squared: $(5.0/15)^2 = 1/9$, so $F_1 = 13{,}230/9 \approx 1.5\times10^3\ \text{N}$.

Step 3. Find the air pressure from $P = F_1/A_1$, with $A_1 = \pi(0.05)^2 = 7.85\times10^{-3}\ \text{m}^2$: $P \approx 1.87\times10^5\ \text{Pa}$. Answer: $F_1 \approx 1.5\times10^3\ \text{N}$; the air pressure needed is $\approx 1.87\times10^5\ \text{Pa}$, almost double atmospheric pressure, delivered by a compressor rather than a person's hand.

Competitive level - dynamic lift on a Boeing aircraft wing (JEE/NEET pattern)

A fully loaded Boeing aircraft has mass $3.3\times10^5\ \text{kg}$ and total wing area $500\ \text{m}^2$. It flies level at $960\ \text{km/h}$. Taking the density of air as $\rho = 1.2\ \text{kg/m}^3$, estimate (a) the pressure difference between the lower and upper wing surfaces and (b) the fractional increase in air speed over the upper surface relative to the lower surface.

Step 1. The weight of the aircraft is balanced by the net upward force from the pressure difference across the wings: $\Delta P \times A = mg$, so $\Delta P = \dfrac{mg}{A} = \dfrac{3.3\times10^5 \times 9.8}{500} \approx 6.5\times10^3\ \text{Pa}$.

Step 2. Ignoring the small height difference between the top and bottom of the wing in Bernoulli's equation, the pressure difference is $\Delta P = \tfrac{1}{2}\rho (v_2^2 - v_1^2) = \tfrac{1}{2}\rho (v_2 - v_1)(v_2 + v_1)$, where $v_2$ is the upper-surface speed and $v_1$ the lower-surface speed.

Step 3. Take the average speed $v_{\text{av}} = (v_1+v_2)/2 \approx 960\ \text{km/h} = 267\ \text{m/s}$ (the plane's own speed, since the wing speeds bracket it closely), so $(v_2-v_1)(v_2+v_1) \approx (v_2-v_1)(2v_{\text{av}})$, giving $\dfrac{v_2 - v_1}{v_{\text{av}}} = \dfrac{\Delta P}{\rho\, v_{\text{av}}^2} = \dfrac{6.5\times10^3}{1.2 \times 267^2} \approx 0.076 \approx 0.08$. Answer: $\Delta P \approx 6.5\times10^3\ \text{Pa}$, and the upper-surface air speed needs to be only about $8\%$ higher than the lower-surface speed to keep a Boeing aloft. Timing note: about 2 minutes. This is a two-formula chain — first find $\Delta P$ from a plain force balance (weight equals lift), then invert Bernoulli's equation for the speed difference. The trap is forgetting the algebraic shortcut $v_2^2 - v_1^2 = (v_2-v_1)(v_2+v_1)$, which turns an otherwise unsolvable single equation in two unknowns into a one-line answer once you approximate $v_2+v_1 \approx 2v_{\text{av}}$.

Competitive level - viscosity of oil from a terminal-velocity measurement (JEE/NEET pattern)

A copper ball of radius $2.0\ \text{mm}$ falls through a tank of oil at $20^{\circ}\text{C}$ and reaches a terminal velocity of $6.5\ \text{cm/s}$. Given the density of oil is $1.5\times10^3\ \text{kg/m}^3$, the density of copper is $8.9\times10^3\ \text{kg/m}^3$, and $g = 9.8\ \text{m/s}^2$, compute the viscosity of the oil.

Step 1. Start from the terminal-velocity formula derived above, $v_t = \dfrac{2a^2(\rho-\sigma)g}{9\eta}$, and rearrange it to isolate the unknown viscosity: $\eta = \dfrac{2a^2(\rho-\sigma)g}{9 v_t}$.

Step 2. Substitute the given values, converting the radius to metres and the terminal speed to metres per second: $a = 2.0\times10^{-3}\ \text{m}$, $\rho - \sigma = (8.9-1.5)\times10^3 = 7.4\times10^3\ \text{kg/m}^3$, $v_t = 6.5\times10^{-2}\ \text{m/s}$.

Step 3. Compute the numerator and divide. Numerator: $2 \times (2.0\times10^{-3})^2 \times 7.4\times10^3 \times 9.8 = 2 \times 4.0\times10^{-6} \times 7.4\times10^3 \times 9.8 \approx 0.580$. Dividing by $9 \times 6.5\times10^{-2} = 0.585$ gives $\eta \approx 0.99\ \text{Pa}\,\cdot\,\text{s}$. Answer: $\eta \approx 0.99\ \text{Pa}\,\cdot\,\text{s}$, close to a poiseuille, in line with a fairly thick machine oil. Timing note: about 90 seconds if the terminal-velocity formula is already memorised. The two traps are unit conversions — radius in millimetres and speed in centimetres per second both need converting to SI before substitution — and sign hygiene: always subtract the lighter density from the heavier one, $(\rho_{\text{sphere}} - \rho_{\text{fluid}})$, and check the sphere is indeed denser than the fluid (sinking, not rising) before trusting a positive answer.

Students lose marks on this chapter in predictable ways. First, treating pressure as though it had a direction — writing $\vec{P}$, or assuming a fluid only pushes "downward" — when pressure is strictly a scalar acting perpendicular to every surface, in every orientation. Second, mixing up gauge and absolute pressure: pressure "on" a submerged object usually means absolute pressure ($P_a + \rho g h$), while the net outward force on a wall usually needs gauge pressure ($\rho g h$), since atmospheric pressure pushes equally from both sides and cancels. Third, confusing the excess-pressure formulas: $2S/r$ for a drop or a gas bubble inside a liquid (one curved surface), but $4S/r$ for a soap bubble in air (two surfaces). Fourth, applying Bernoulli's equation to turbulent or clearly viscous flows (a river's rapids, oil through a narrow pipe) where its founding assumptions — steady, non-viscous, incompressible flow along one streamline — simply fail. Fifth, inverting the continuity relation and believing a wider pipe gives a faster flow, rather than the correct $v \propto 1/A$. Sixth, forgetting that the angle of contact's sign, not just "wetting versus non-wetting" in words, decides capillary rise versus depression. Finally, swapping how viscosity depends on temperature: it falls for liquids and rises for gases as temperature increases.

• Define pressure and explain, using Pascal's law, why it is a scalar despite being force divided by area.
• Calculate absolute pressure, gauge pressure, and pressure at any depth using $P = P_a + \rho g h$.
• Explain a barometer, a manometer, and a hydraulic lift or brake, including $F_2 = F_1(A_2/A_1)$.
• State the equation of continuity, $Av = \text{constant}$, and use it to relate speeds and areas along a pipe.
• State Bernoulli's equation, its conditions, and apply it to efflux speed, aerofoil lift, and the Magnus effect.
• Distinguish streamline (laminar) flow from turbulent flow, and describe the critical speed.
• Define the coefficient of viscosity, its temperature trends for liquids versus gases, and apply Stokes' law.
• Derive and use the terminal-velocity formula $v_t = 2a^2(\rho-\sigma)g / 9\eta$ for a falling sphere.
• Explain surface tension as surface energy, and calculate excess pressure inside a drop, a liquid bubble, and a soap bubble.
• Explain angle of contact and apply the capillary rise formula $h = 2S\cos\theta/(\rho g a)$, including capillary depression.

Ask the tutor: if any step in the pressure-depth, Bernoulli, Stokes' law, or surface-tension derivations above still feels shaky, ask the tutor to walk through that one worked example again with your own numbers.