Mechanical Properties of FluidsPressure
Press a sharp needle against your skin and it goes in. Press the back of a spoon with the same force and nothing happens. An elephant standing on a person's chest would crack the ribs, but a circus performer survives a heavy load placed on a wide, strong plank across the chest. Force alone does not decide what happens; the area the force is spread over matters just as much. The smaller the area, the bigger the effect. This effect is called pressure.
At the Dussehra mela, Ishaan the performer, who weighs 600 N, lies down on a bed of blunt nails. Each nail tip touches his back over about 1 mm². Skin gives way at roughly 3 MPa (3 N on each square millimetre). How many nails does the bed need?
Number of nails: n =
The purple curve is the pressure under each tip, 600 N ÷ (n × 1 mm²). Double the nails and the pressure halves. The red band covers every pressure above 3 MPa, where the skin is pierced. With 200 nails or more, each tip presses at 3 MPa or less, and the bed is safe to lie on. A bed of nails is a trick of area, not of courage.
Force Per Unit Area
When a body is submerged in a fluid at rest, the fluid pushes on every part of its surface, and that push is always normal (perpendicular) to the surface. Suppose the fluid pushed partly along the surface. By Newton's third law the surface would push back on the fluid along itself, and the fluid, which cannot resist a sideways force, would start to flow. A fluid at rest is not flowing, so the push it exerts must be perpendicular to any surface it touches. The same holds for the walls of its container.
We can imagine measuring this push with a small device: an evacuated chamber closed by a light piston of area A, with a calibrated spring behind it. Put it anywhere in the fluid. The fluid pushes the piston inwards with a normal force F, the spring pushes it outwards, and the spring's squeeze tells us F. The average pressure on the piston is the normal force per unit area:
Making the piston smaller and smaller gives the pressure at a point:
Pressure is a scalar. The F in these formulas is the size of the normal force, not a vector, so pressure has no direction of its own. Its dimensions are [ML⁻¹T⁻²] and its SI unit is N m⁻², named the pascal (Pa) after
Tip
Pressure is not a property of walls only. It exists at every point inside a fluid, and a tiny element of the fluid stays still because the pushes on all its faces balance.
Density
The second quantity we need for fluids is the density ρ (rho). For a mass m of fluid filling a volume V,
Its dimensions are [ML⁻³] and its SI unit is kg m⁻³; it is a positive scalar. A liquid is nearly incompressible, so its density hardly changes with pressure. A gas is the opposite: squeeze it and its density goes up a lot.
The relative density of a substance is its density divided by the density of water at 4 °C, which is 1.0 × 10³ kg m⁻³. Being a ratio of two densities, it has no units. Copper has a relative density of 8.9, so its density is
Table 9.1 — densities of some common fluids at STP (0 °C and 1 atm)
| Fluid | ρ (kg m⁻³) |
|---|---|
| Water | 1.00 × 10³ |
| Sea water | 1.03 × 10³ |
| Mercury | 13.6 × 10³ |
| Ethyl alcohol | 0.806 × 10³ |
| Whole blood | 1.06 × 10³ |
| Air | 1.29 |
| Oxygen | 1.43 |
| Hydrogen | 9.0 × 10⁻² |
| Interstellar space | ≈ 10⁻²⁰ |
Look at the jump between the liquids and the gases in Table 9.1: water is about 800 times denser than air. Mercury, 13.6 times denser than water, is the heavyweight among common liquids, which is why it turns up in barometers.
Example 9.1: Standing on Two Bones
Example 9.1 When Riya's grandfather stands, his two thigh bones (femurs) carry the upper part of his body, of mass 48 kg. Each femur has a cross-sectional area of 12 cm². Estimate the average pressure the femurs bear. (Take g = 10 m s⁻².)
Solution The two bones share the load, so the total area is A = 2 × 12 cm² =
The average pressure is
which is
Pascal's Law
The French scientist Blaise Pascal noticed something simple and powerful: in a fluid at rest, the pressure is the same at all points at the same height. It does not matter which way you face a pressure gauge, either. Test both ideas with the probe below, the spring-and-piston device from before. Drag the probe around the tank and turn its face with the chips.
The purple arrow is the water's push on the piston face. Turning the face only turned the arrow; its size, the reading, did not change. Sliding sideways along one level did not change it either. Going deeper did: twice as deep, twice the reading above the air's pressure. Pressure at a point acts equally in all directions, and it is the same everywhere along a horizontal level.
Why Every Direction Agrees
Take a tiny element of fluid at rest, shaped like a right-angled prism. It is so small that gravity's effect across it can be ignored (it is drawn large only for clarity). The rest of the fluid pushes on its three rectangular faces with normal forces: one on face a, one on the slanted face b and one on face c, called Fa, Fb and Fc. The faces have areas Aa, Ab and Ac, and the slant makes the angle θ with face a.
Fig. 9.2 — a small prism-shaped element of a fluid at rest, seen end on.
The element is in equilibrium, so the parts of Fb along and across face a balance the other two forces:
Geometry gives the same two factors for the areas, because face c and face a are the projections of face b:
Divide each force equation by the matching area equation:
The prism could be turned any way, so pressure in a fluid at rest is the same in all directions. Like other kinds of stress, it cannot be given a direction.
Now picture a thin horizontal bar of the fluid with equal end areas. It is in equilibrium, so the horizontal forces on its two ends balance, and the pressures at its ends are equal. If the pressure were different at two points on one level, the fluid between them would feel a net force and flow. In a fluid that is not flowing, the pressure is the same at every point of a horizontal plane.
Variation of Pressure with Depth
Now let gravity in. Take a vertical cylinder of liquid inside a container at rest, with base area A and height h, its top at point 1 and its bottom at point 2. Sideways, the pushes on its curved side cancel. Up and down, three forces act on it: the push P₁A down on its top, the push P₂A up on its base, and the weight of the liquid column itself, mg. The column is at rest, so
Fig. 9.3 — a column of liquid of height h between point 1 and point 2.
The mass of the column is its density times its volume, m = ρV = ρhA. The area A cancels:
The difference in pressure depends only on the vertical distance h, the density ρ and g. Move point 1 up to the open surface, where the pressure is that of the air, Pₐ, and call the pressure at depth h simply P:
So at a depth h below an open surface, the pressure is more than the atmospheric pressure by ρgh. This excess, P − Pₐ = ρgh, is the gauge pressure at that depth. Notice what is missing from Eq. (9.7): the area of the container and its shape. Only the depth counts.
Sina Dives
Sina has a waterproof pressure gauge and four tanks, each 13 m deep, filled with the liquids of Table 9.1. On the right, the depth axis is shared with the tank, so the graph shows the pressure at Sina's depth. Drag Sina down and switch liquids with the chips.
Each line is Eq. (9.7). They all start at 1 atm at the surface, because the air pushes on the top of every tank alike, and the purple line for the chosen liquid climbs by ρg for every metre. Water gives 2 atm at about 10.3 m; mercury, 13.6 times denser, gets there at only 0.76 m.
Example 9.2: A Diver in a Lake
Example 9.2 Kabir dives 15 m below the surface of Bhimtal lake. What is the pressure on him there? (Take Pₐ = 1.01 × 10⁵ Pa, ρ = 1000 kg m⁻³ and g = 10 m s⁻².)
Solution From Eq. (9.7), P = Pₐ + ρgh. The water column adds ρgh = 1000 × 10 × 15 =
So P = 1.01 × 10⁵ Pa + 1.5 × 10⁵ Pa =
At only 15 m the pressure on Kabir has become two and a half times its value at the surface. At a depth of 1 km the extra pressure is about 100 atm, which is why submarines have such thick hulls.
The Hydrostatic Paradox
Three glass vessels of very different shapes, A, B and C, are joined at the bottom by a horizontal pipe, and water is poured in. A holds a lot of water, B very little, and C is a funnel that widens upwards. At the bottom, where the pipe joins them, which vessel has the largest pressure?
Fig. 9.4 — three vessels A, B and C joined at the bottom.
The water settles at the same level in all three, and the pressure at the bottom points is Pₐ + ρgh with the same h for each. If it were larger under one vessel, water would flow from there through the pipe until the levels evened out. That the pressure ignores the amount of liquid and the shape of the vessel is called the hydrostatic paradox.
Atmospheric Pressure and Gauge Pressure
The pressure of the atmosphere at any point equals the weight of a column of air of unit cross-section reaching from that point to the top of the atmosphere. At sea level it is 1.013 × 10⁵ Pa, or 1 atm. The Italian scientist Evangelista Torricelli was the first to measure it. He filled a long glass tube, closed at one end, with mercury and turned it upside down into a trough of mercury. This is the mercury barometer.
Mercury runs out of the tube until the column is a certain height h. The space at the top holds only a trace of mercury vapour, whose pressure is so small that we can take the pressure at the top point A as zero. Point B inside the tube and point C on the open surface are at the same level, so they have the same pressure, and at C that is the air's pressure Pₐ. Applying Eq. (9.6) to the mercury column:
where ρ is the density of mercury.
Fig. 9.5 — (a) the mercury barometer, (b) the open-tube manometer with its points A and B and height difference h.
Put in ρ = 13.6 × 10³ kg m⁻³, g = 9.8 m s⁻² and Pₐ = 1.013 × 10⁵ Pa: the column stands h ≈
Doctors use mm of Hg for blood pressure, and weather scientists use the bar and the millibar: 1 bar = 10⁵ Pa.
An open-tube manometer measures pressure differences. It is a U-tube holding a suitable liquid: a light liquid such as oil for small differences, a dense one such as mercury for large ones. One arm is open to the air, and the other is connected to the system whose pressure P we want. A and B are at the same level in the same liquid, so they have the same pressure, and the liquid above B gives P − Pₐ = ρgh. A manometer, like a tyre gauge or a blood-pressure gauge, reads the gauge pressure, which is proportional to h.
Tip
In every pressure question, decide first: is the pressure asked absolute (P = Pₐ + ρgh) or gauge (P − Pₐ = ρgh)? The force on a window or a hull is set by the pressure difference across it.
Example 9.3: How High Is the Air?
Example 9.3 On a mild day the air at sea level has a density of about 1.2 kg m⁻³. Suppose the air kept this density all the way up. How high would the atmosphere reach? (Take Pₐ = 1.01 × 10⁵ Pa and g = 9.8 m s⁻².)
Solution The whole column of air must supply the sea-level pressure, so by Eq. (9.7) ρgh = Pₐ, and h = 1.01 × 10⁵ ÷ (1.2 × 9.8) ≈
Real air is not like that. It thins out as we go up, because the air below is squeezed by the weight of the air above it, and g itself falls a little. The graph compares the two.
Drag the height and take Sina up.
Height:
The dashed line is the uniform-density atmosphere, which runs out of air at 8.6 km. The solid curve is the real atmosphere: half the pressure is gone by about 5.5 km, but there is still about 0.3 atm on top of Everest, and the air thins on for more than 100 km. Sea-level pressure is not fixed either: when a barometer falls by 10 mm of Hg or more, a storm is often on its way.
Example 9.4: A Submarine's Window
Example 9.4 A research submarine works 800 m down in the ocean. Inside, the air is kept at sea-level atmospheric pressure. Take the density of sea water as 1.03 × 10³ kg m⁻³, g = 10 m s⁻² and Pₐ = 1.01 × 10⁵ Pa. (a) What is the absolute pressure outside? (b) What is the gauge pressure there? (c) What force acts on a window 30 cm × 20 cm?
Solution (a) From Eq. (9.7), P = Pₐ + ρgh = 1.01 × 10⁵ + 1.03 × 10³ × 10 × 800 =
(b) The gauge pressure is P − Pₐ = ρgh =
(c) Outside, the pressure is Pₐ + ρgh; inside, it is Pₐ. The net pressure on the window is therefore the gauge pressure, and with A = 0.30 m × 0.20 m = 0.06 m² the force is F = ρgh × A =
Hydraulic Machines
What happens when we change the pressure on a fluid in a closed vessel? Picture a horizontal cylinder of liquid with a piston at one end and three open vertical tubes standing on it at different places. The liquid stands at the same height in all three tubes, because the pressure in the cylinder is the same everywhere. Push the piston in and the liquid rises in every tube, and again it reaches the same height in each. The extra pressure from the piston reached every part of the liquid equally. This is the second form of Pascal's law: a change in the pressure applied to an enclosed fluid is passed on undiminished to every point of the fluid and to the walls of its container.
A hydraulic lift puts this to work. Two pistons, a small one of area A₁ and a large one of area A₂, close the two ends of a vessel full of oil. A force F₁ on the small piston creates a pressure P = F₁/A₁, which reaches the large piston undiminished and pushes it up with a force P × A₂. Help Sina lift the car.
The force on the big piston is
so the push on the small piston is multiplied by A₂/A₁, the mechanical advantage of the machine. With r = 1.5 cm against 15 cm, A₂/A₁ = 100, and Sina's 147 N holds up 14 700 N. Nothing is free: the oil is incompressible, so the volume pushed out of the small cylinder, L₁A₁, equals the volume that enters the large one, L₂A₂. Each 20 cm stroke raised the car by only 0.2 cm. The work F₁L₁ that Sina does equals the work F₂L₂ done on the car.
Example 9.5: Two Syringes
Example 9.5 In the school lab, Meera joins two water-filled syringes (needles removed) with a tight rubber tube full of water. The pistons have diameters 1.0 cm and 2.5 cm. (a) She pushes the small piston with a force of 8.0 N. What force does the large piston exert? (b) She pushes the small piston in by 5.0 cm. How far does the large piston move out?
Solution (a) The pressure passes undiminished through the water. The force grows by the ratio of the areas, which is the square of the ratio of the diameters: A₂/A₁ = (2.5/1.0)² =
So F₂ = 6.25 × 8.0 N =
(b) Water is practically incompressible, so the volume the small piston pushes in equals the volume the large one lets out: L₁A₁ = L₂A₂. Then L₂ = L₁ × A₁/A₂ = 5.0 cm ÷ 6.25 =
Example 9.6: The Car Lift
Example 9.6 In a service-station car lift, compressed air pushes with a force F₁ on a small piston of radius 4.0 cm. The pressure reaches a large piston of radius 20 cm. The car on it has a mass of 1200 kg. Find F₁ and the air pressure needed. (Take g = 9.8 m s⁻².)
Solution The pressure is passed on undiminished, so
which gives F₁ ≈
Hydraulic brakes in cars work the same way. A gentle push on the brake pedal drives a small piston into the master cylinder. The pressure travels through the brake oil to pistons of larger area at the wheels, which press the brake shoes against the brake lining with a large force. And because the pressure is the same everywhere in the oil, all four wheels are braked equally.
Speed Round: Pressure
P = F/A; pressure at a point acts equally in all directions and is the same along a level; P = Pₐ + ρgh; gauge pressure is P − Pₐ; F₂ = F₁A₂/A₁.
Get five right to move on.