Fluids I: pressure, floating and flow

Push a book sideways and it slides as one piece: it keeps its shape. Push water sideways and it simply flows. That is the whole difference. A fluid is anything that flows. Liquids and gases are both fluids.

Why do fluids flow? A solid can resist a sideways (shearing) push by deforming a little and pushing back. A fluid at rest cannot. Even a tiny sideways force makes its layers slide over each other. So a fluid at rest can push on a surface only at right angles to it. This one fact explains a lot of the chapter.

• Liquids have a fixed volume but no fixed shape. They are nearly incompressible, so in this chapter we treat their density as constant.
• Gases fill whatever container they are in and are easily compressed.

A sharp knife cuts but a blunt one does not, even though your hand pushes just as hard. A camel walks on sand without sinking because its feet are wide. School-bag straps are made wide so they do not dig into your shoulders. In each case what matters is not just the force but how much area it is spread over. That quantity is pressure.

A fluid at rest cannot sustain a sideways shear stress. On every little surface inside it, the neighbouring fluid pushes normally. At one point the pressure has the same value whichever way that surface faces; changing its orientation changes the direction of force, not the scalar pressure.

Three facts about pressure in a fluid at rest. All three are tested:

1. Pressure is a scalar. It has no direction. What has a direction is the force the fluid exerts on a surface, and that force is always normal (perpendicular) to the surface.
2. At a point, pressure is the same in every direction. A tiny probe at one point feels the same push from above, below and the sides. (NCERT shows this with a small triangular wedge of fluid: the forces balance only if \(P_a = P_b = P_c\).)
3. Pressure increases with depth. Deeper water has more water above it pushing down.

Imagine a thin horizontal cylinder of liquid of end area A. Its weight acts downward, so it supplies no horizontal force. The pressure forces on its two ends are P₁A and P₂A in opposite directions. Horizontal equilibrium requires P₁A = P₂A, hence P₁ = P₂. If pressures differed, the liquid would accelerate instead of staying at rest.

This comparison applies within the same connected fluid in equilibrium. It does not equate the pressures in two unrelated sealed vessels merely because their points are at the same height. Across a flat liquid–liquid interface, pressure is continuous when interfacial curvature effects are negligible.

A lower layer of water supports the layers above it. Pressure therefore increases downward. We can derive the amount of that increase by balancing forces on a small vertical column; we do not need to guess from the total amount of liquid in the vessel.

Imagine a vertical cylinder of liquid with cross-section \(A\) and height \(h\), its top at the surface. It is in equilibrium, so the upward push on its bottom must hold up the downward push on its top plus its own weight:

\( P A = P_0 A + mg = P_0 A + (\rho A h) g \;\Rightarrow\; P = P_0 + \rho g h \).

Note that \(A\) cancels. Pressure depends only on vertical depth, not on how wide the vessel is, what shape it has, or how much liquid there is.

Fill three differently shaped vessels with the same base area to the same height. The pressure at the bottom, and so the force on the base, is the same in all three, even though they hold very different amounts of water. How can the base of the narrow-necked vessel feel more than the weight of the water in it? Because its sloping walls push down on the water. In the flared vessel the walls hold some of the water up.

The simple product ρgh assumes constant density. If density is ρ(z), divide the liquid into thin horizontal slices. A slice of thickness dz adds pressure dP = ρ(z)g dz. Add the contributions by integrating. Here z measures depth downward from the free surface:

\[P(h)-P_0=g\int_0^h\rho(z)\,dz.\]

For separate uniform layers, integration becomes a sum: P − P₀ = g(ρ₁h₁ + ρ₂h₂ + …), where ρᵢ and hᵢ are each layer's density and vertical thickness. At a given depth, include all the layers above that point.

Two upright vessels are connected at their bottoms by a pipe. Initially, the taller column gives a greater pressure at its end of the pipe. Water flows toward the shorter column. With the same liquid and equal air pressure above both surfaces, the final state at rest has equal surface heights, because the bottom pressures must match.

Equal heights do not necessarily mean equal volumes. A wider cylinder holds more water at the same height. To find the final height, conserve the total liquid volume. Assume both vessels have constant cross-sectional areas and a common bottom level, and the pipe's volume is negligible.

A uniform water column of height h and area A has mass ρAh. Its centre of mass is h/2 above the bottom, so its gravitational potential energy is U = ρgAh²/2. Add this energy for the two vessels before and after the transfer.

Gravity does positive work as the water redistributes: W_g = U_initial − U_final. During transfer some energy becomes kinetic; as real water settles, viscosity turns it into internal energy. In a perfectly lossless model, the liquid would keep oscillating instead of settling. This is why equalisation does not imply conservation of the initial gravitational potential energy.

We live at the bottom of an ocean of air several hundred kilometres deep. Its weight presses on everything with about 10⁵ N on every square metre, roughly the weight of a 10-tonne truck on each m². We don't get crushed because the fluids inside our body push back just as hard.

You use this every day. When you drink through a straw, you don't "pull" the juice up. You lower the pressure inside the straw, and the atmosphere pressing on the juice in the glass pushes it up. A suction hook sticks because you squeeze the air out from behind it and the atmosphere then holds it against the wall.

Pour oil into one arm of a U-tube containing water. The oil floats on the water and pushes the water up the other arm. To solve these, use one rule: in the same liquid, at the same level, at rest, the pressure is the same. Choose the level of the oil–water interface. Below it, both arms contain water.

\(P_0 + \rho_1 g h_1 = P_0 + \rho_2 g h_2 \;\Rightarrow\; \rho_1 h_1 = \rho_2 h_2\), with both heights measured from the interface level.

Squeeze the end of a toothpaste tube and paste comes out at the nozzle, far from your fingers. Press one side of a water-filled balloon and it bulges everywhere else. An enclosed liquid passes any extra push on to every part of itself. That is Pascal's law, and it lets a mechanic lift a car with one hand.

Two pistons of areas \(A_1\) (small) and \(A_2\) (large) sit on the same enclosed liquid. Push the small one with \(F_1\). That adds pressure \(P = F_1/A_1\), which reaches the large piston unchanged, so it is pushed up with \(F_2 = P A_2\).

Hydraulic brakes. A light push on the brake pedal moves a piston in a small master cylinder. The pressure travels through the brake oil to larger pistons at each wheel, which press the brake pads with a much larger force. The added pressure is transmitted through the brake oil. The force at each piston equals this pressure increase times its piston area; actual front and rear braking forces can differ by design.

In a swimming pool you can lift a friend with one arm. A steel ship weighing thousands of tonnes floats, while a steel nail sinks. A bucket of water feels lighter while it is still under the surface of the well. All of these come from one upward push, the buoyant force (upthrust).

Where does it come from? Think of a cube held under water. Pressure pushes on all six faces. The side pushes cancel. But the bottom face is deeper than the top face, so it is pushed harder: \(F_B = (P_{bottom} - P_{top})A = \rho_l g h A = \rho_l g V\). The net push is upward and equals the weight of a cube of liquid the same size. The material of the body does not appear anywhere in this result.

Use one idea: a floating object displaces its own weight of water; a sunk object displaces only its own volume. For anything denser than water, displacing its weight takes more water than displacing its volume.

A sealed hollow metal ball can float even if its metal is denser than water. Its weight depends on the volume of metal, while the water displaced depends on the outer submerged volume, including the sealed cavity. Use the total mass divided by the outer volume to decide whether it can float.

For outer radius R, cavity radius a and material density ρ_s, ignoring the mass of enclosed air, \(M=\rho_s\frac{4\pi}{3}(R^3-a^3)\). The mean density is \(\bar\rho=M/(4\pi R^3/3)=\rho_s[1-(a/R)^3]\). Here M is the mass in kg, R and a are radii in m, and both densities are in kg/m³.

“Just floats” in the ideal question means fully immersed neutral equilibrium: mean density equals liquid density ρ_l. Hence \((a/R)^3=1-\rho_l/\rho_s\). The same ratio applies to cavity diameter divided by outer diameter. If water enters an open cavity, do not treat that water-filled space as the same sealed displaced volume.

In Test-05 Q13, the sphere's density varies with radial distance r from its centre. An outer shell contains more volume than an equally thick inner shell: for a thin shell, \(dV=4\pi r^2dr\). Therefore a simple average of centre and surface densities is wrong.

For sphere radius R and density \(\rho(r)=\rho_0(1-r/R)\), where ρ₀ is its central density, add the shell masses:

\[M=\int_0^R\rho(r)\,4\pi r^2dr=4\pi\rho_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=\frac{\pi\rho_0R^3}{3}.\]

Dividing by total volume \(4\pi R^3/3\) gives \(\bar\rho=\rho_0/4\). Use this mean density in the ordinary floating relation \(V_\text{sub}/V=\bar\rho/\rho_l\). Here V is the sphere volume and V_sub is its submerged volume. This calculation averages the body's density; it differs from integrating a liquid's pressure through depth.

Let a uniform floating candle have current length ℓ and submerged fraction f = ρ_c/ρ_l, where ρ_c and ρ_l are candle and liquid densities. Its immersed length is fℓ. If burning decreases total length at rate b, that immersed length decreases at rate fb. Relative to a nearly fixed water surface, the bottom therefore rises at fb, while the top falls at (1 − f)b.

For a half-submerged candle, f = 1/2. Burning at b = 4 cm/hour makes the bottom rise at 2 cm/hour and the top fall at 2 cm/hour. The large vessel in Q6 (D ≫ d, with D and d the vessel and candle diameters) makes the change of outside liquid level negligible. The model assumes uniform density and cross-section and no retained wax changing the mass balance.

• Extra load to sink a floater (Q11): extra mass needed = ρ_l × (unsubmerged volume). A 1 m³ cube half under water needs 1000 × 0.5 = 500 kg more.
• Burning candle (Q6): the fraction submerged stays ρ_c/ρ_w. If the candle shortens at rate r, the submerged length shrinks at (ρ_c/ρ_w)r, so the bottom rises at that rate. Half-submerged, burning 4 cm/h → bottom rises 2 cm/h.
• Hollow and variable-density bodies (Q9, Q13): use the average density = total mass ÷ total volume. "Just floats" (fully submerged at rest) means ρ_avg = ρ_l. See the shell-mass calculation above and Problem patterns.
• Body on a spring, fully submerged (Q10): if it hangs from a spring, kx = mg − F_B. If it is tied to the bottom by a spring (a floater held under), kx = F_B − mg.
• Accelerating container: in a lift accelerating up with a, replace g by g + a. Weight and buoyancy both scale, so the fraction submerged does not change. In free fall there is no buoyancy at all.
• Rising body (like a bubble, Q14): with drag neglected, net upward force F_B − W = ma gives a = (ρ_l/ρ_b − 1)g. For a = g/100, ρ_l/ρ_b = 1.01.

Watch smoke rising from an incense stick. For the first few centimetres it rises as a smooth, steady thread. Higher up it suddenly breaks into swirls and eddies. A slow river glides smoothly; white-water rapids churn. These are the two kinds of flow: streamline (smooth) and turbulent (chaotic).

Streamline. In steady flow, the path a particle follows is a streamline. Every particle passing through a given point follows the same streamline. The tangent to a streamline at any point gives the direction of the velocity there.

• Two streamlines never cross. At a crossing a particle would have two velocities at once, which is impossible.
• Crowded streamlines mean fast flow; widely spaced streamlines mean slow flow. (You will see this in the continuity and aerofoil sims.)
• Tube of flow: a bundle of streamlines. No fluid crosses its side walls, so whatever mass enters one end must leave the other. That gives the equation of continuity.

Put your thumb over the end of a garden hose and the water shoots out much faster. A river flows lazily where it is wide and races through a narrow gorge. The water has nowhere else to go: the same amount must pass every section each second, so where the passage is narrow the water must move faster.

Derivation. In time Δt, the fluid entering section 1 occupies a length v₁Δt, so its mass is ρ₁A₁v₁Δt. The same mass must leave through section 2 (no fluid crosses the walls of the tube of flow, and none piles up in steady flow): ρ₁A₁v₁ = ρ₂A₂v₂. For an incompressible liquid ρ₁ = ρ₂.

Why a falling stream of tap water narrows (Test-05 Q25). Once it leaves the tap, water speeds up under gravity: \(v^2 = v_0^2 + 2gh\). Since Av is constant, a larger v means a smaller A, so the stream gets thinner as it falls. This is continuity, not surface tension.

Hold a strip of paper just below your lips and blow over it: it rises. Hang two sheets of paper close together and blow between them: they come together, not apart. A shower curtain gets sucked inward when the shower runs. In each case fast-moving air has lower pressure. Bernoulli's principle explains why, and it is just energy conservation for a flowing fluid.

Follow a small slug of fluid of volume ΔV and mass Δm = ρΔV along a tube of flow, from section 1 (pressure P₁, speed v₁, height h₁) to section 2 (P₂, v₂, h₂).

1. The fluid behind pushes it forward: work done = P₁A₁Δx₁ = P₁ΔV. The fluid ahead pushes back: work = −P₂ΔV. Net work by pressure = (P₁ − P₂)ΔV.
2. Work–energy theorem: this net work = change in KE + change in PE = ½Δm(v₂² − v₁²) + Δm g(h₂ − h₁).
3. Divide by ΔV (with Δm/ΔV = ρ) and rearrange: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂.

In a horizontal pipe, P + ½ρv² = const. P is the static pressure (what a gauge moving with the fluid would read). ½ρv² is the dynamic pressure. Where the fluid is brought to rest, at the nose of an object or the mouth of a tube facing the flow, all of it turns into pressure. The stagnation pressure there is P + ½ρv².

An L-shaped (pitot) tube with its open end facing the current therefore holds a column h = v²/2g higher than a plain vertical tube (Test-05 Q39).

Poke holes at different heights in a plastic bottle full of water. The lowest hole squirts hardest, because it has the most water above it. But it does not necessarily throw the water farthest: a hole lower down also has less height to fall from. That trade-off is the most-asked graph and range question in this chapter.

Derivation. Apply Bernoulli between a point on the free surface (pressure P₀, speed ≈ 0 because the tank is much wider than the hole) and the hole at depth h below the surface (pressure P₀, since it opens to air, speed v):

\(P_0 + 0 + \rho g h = P_0 + \tfrac12\rho v^2 + 0 \;\Rightarrow\; v = \sqrt{2gh}\).

This is the same speed a stone gains falling freely through h. Here h is measured down from the free surface, not up from the bottom.

How do you measure how much water flows through a pipe each second without putting anything in its way? Narrow the pipe smoothly to a throat and back again. In the throat the water speeds up (continuity), so its pressure drops (Bernoulli). Measure that pressure drop with a manometer and you can calculate the flow rate. That device is a venturimeter, and it uses both laws together.

Aeroplanes, spinning cricket balls and a roof lifted by a storm all involve a pressure difference created by moving air. Start with the pressure: a surface feels a greater push from the side at higher pressure. In the ideal, horizontal-flow model, Bernoulli relates that difference to the air speeds.

For speeds \(v_\text{top}\) and \(v_\text{bottom}\), measured relative to the wing, subtract the two Bernoulli equations using the same upstream conditions. Faster flow over the top gives lower pressure there, so the bottom pushes up more strongly than the top pushes down.

View a moving ball in its own frame: air flows past it. Viscosity couples nearby air to its rotating surface. Spin changes the flow around opposite sides and makes the pressure distribution asymmetric. The resulting force is transverse to the ball's motion, bending its path. Reverse the spin while keeping the motion the same and the deflection reverses.

In the model above, air travels left to right and the ball moves left relative to the air. When its top surface moves with the air, the faster upper flow corresponds to lower pressure and an upward force. With the opposite spin, the faster side and force reverse. Always draw the relative airflow and rotation before deciding the direction.

For steady level flight, the vertical lift balances aircraft weight Mg, where M is its mass in kg and g is gravitational acceleration. If the lower airflow speed is v and the upper speed is v + δv, then \(\Delta P=\tfrac12\rho[(v+\delta v)^2-v^2]=\rho v\delta v+\tfrac12\rho(\delta v)^2\).

If δv is much smaller than v, neglect the squared term and use \(Mg=\Delta P A\). This gives the fractional increase \(\delta v/v\approx Mg/(\rho Av^2)\). Here δv is the speed difference in m/s, v is the lower speed in m/s, ρ is air density in kg/m³ and A is effective wing area in m². For Test-05 Q23, 900 km/h = 250 m/s, so the fraction is 0.064, about 6.4%. Without the small-difference approximation, solve \(\delta v/v=\sqrt{1+2Mg/(\rho Av^2)}-1\), which gives about 6.2%.

If air inside a building is approximately at rest while air rushes across its roof, outside pressure can fall below inside pressure. The estimated upward force is \(F=\tfrac12\rho v^2A\). If the wind speed doubles, this ideal estimate grows fourfold.

An atomiser uses a fast stream to lower pressure near a tube opening. Higher pressure over the liquid reservoir then pushes liquid up the tube. We often say the liquid is “sucked up”, but the pressure difference is what moves it.

Water in a bus rises at the rear when the bus accelerates forward. The liquid is at rest relative to the bus after settling, but the bus frame is accelerating. In that frame include an inertial force opposite the acceleration: the effective gravity is \(\boldsymbol g_\text{eff}=\boldsymbol g-\boldsymbol a\).

The pressure increases in the direction of effective gravity. The free surface is perpendicular to it, since points on that surface are at the same atmospheric pressure. For horizontal acceleration, the surface is no longer horizontal.