Fluids at Rest

Fluids at Rest

Part 1 of 2 — Density, Pressure & Pascal's Principle

Where this fits. This lesson starts the fluids-and-waves sequence with fluids that are not moving at all: density and pressure, Pascal's principle, buoyancy and surface tension. The next lesson, Fluids in Motion, sets the fluid flowing with continuity, Bernoulli, viscosity and blood flow.

A fluid is any substance that flows: liquids and gases. A fluid at rest cannot hold a sideways (shear) force, so it can only push perpendicular to any surface it touches. That one fact is the root of everything about pressure.

Density and Specific Gravity

Density $\rho$ (Greek "rho") is mass per unit volume:

$\rho = \frac{m}{V}$

The SI unit is $\text{kg/m}^3$; labs and clinics often use $\text{g/cm}^3$, which is the same as g/mL. Water is the anchor: $1\;\text{g/cm}^3 = 1000\;\text{kg/m}^3 = 1\;\text{kg/L}$. To convert a density from $\text{g/cm}^3$ to $\text{kg/m}^3$, multiply by 1000.

Specific gravity (SG) is a substance's density divided by the density of water:

$SG = \frac{\rho_{\text{substance}}}{\rho_{\text{water}}}$

It is a ratio of two densities, so it has no units. $SG = 0.8$ means 80% as dense as water, or $800\;\text{kg/m}^3$. Numerically, SG equals the density written in $\text{g/cm}^3$.

Memorize water and mercury; the others are useful for estimates and are usually given in a passage.

Two more facts you will use constantly. Weight follows from density: $W = mg = \rho V g$. And liquids barely compress, so the MCAT treats them as incompressible (constant density); gases are compressible, and their density rises with pressure.

Pressure

Pressure is force per unit area, with the force taken perpendicular to the surface:

$P = \frac{F}{A}$

The SI unit is the pascal: $1\;\text{Pa} = 1\;\text{N/m}^2$. Pressure is a scalar: at any point in a fluid at rest it pushes equally in all directions, and the force it produces on a surface always acts perpendicular to that surface. The same force on a smaller area gives a larger pressure, which is why a needle tip pierces skin.

Pressure versus depth. Picture a column of liquid with cross-sectional area $A$ and height $h$ resting on a point. Its weight is $\rho (Ah) g$, spread over area $A$, so it adds $\rho g h$ of pressure. Add whatever presses on the liquid's surface, $P_0$ (usually atmospheric pressure):

$P = P_0 + \rho g h$

Here $h$ is the depth below the surface and $g \approx 10\;\text{m/s}^2$.

• Depth, not shape. Only $\rho$, $g$ and $h$ appear. A narrow tube and a wide lake have the same pressure at the same depth; the container's shape and the total volume of liquid do not matter.
• Same level, same pressure. In one connected fluid at rest, every point at the same depth has the same pressure.
• Linear in depth. Doubling the depth doubles $\rho g h$, but not the total $P$, because $P_0$ does not double.
• Rule of thumb. In water, $\rho g h = (1000)(10)(10) = 10^5\;\text{Pa}$ for every $10\;\text{m}$: about one extra atmosphere per 10 m of depth.

Gauge vs. absolute pressure. Absolute pressure is the total $P$. Gauge pressure is the amount above atmospheric: $P_{\text{gauge}} = P - P_{\text{atm}}$, which for an open liquid is just $\rho g h$. Tire gauges and blood-pressure cuffs report GAUGE pressure, and it can be negative: the thin fluid layer between the lungs and the chest wall sits at about $-4\;\text{mmHg}$, 4 mmHg below atmospheric.

A millimetre of mercury (mmHg) is literally the pressure under a 1 mm column of mercury: $\rho g h = (13{,}600)(9.8)(0.001) \approx 133\;\text{Pa}$.

Practice: Density & Pressure 🎯

Barometers, Manometers and Blood Pressure

Barometer (measures atmospheric pressure). Fill a long tube with mercury, close it, and turn it upside down in a dish of mercury. The column falls until the pressure at the bottom of the column, $\rho_{\text{Hg}} g h$, matches the atmosphere pushing on the dish.

• The space above the mercury is nearly a vacuum, so the reading is ABSOLUTE. At sea level $h = 760\;\text{mm}$, the origin of "760 mmHg".
• A water barometer would need a column 13.6 times taller, about $10.3\;\text{m}$, which is why mercury is used.
• At high altitude the air above is thinner and the column is shorter.

Open-tube manometer (measures gauge pressure). A U-shaped tube of liquid has one arm connected to the gas or vessel and the other arm open to the air. The liquid is pushed DOWN on the higher-pressure side, and the height difference $\Delta h$ between the two surfaces gives $P - P_{\text{atm}} = \rho g \Delta h$. Because the open arm already carries atmospheric pressure, the reading is a gauge pressure.

Blood pressure. A reading of "120/80" means a peak (systolic) pressure of 120 mmHg and a minimum (diastolic) pressure of 80 mmHg, both measured ABOVE atmospheric. The cuff is kept at heart level because blood is a fluid and obeys $\rho g h$ too:

• In a standing adult, arteries in the feet sit about $1.2\;\text{m}$ below the heart, so (taking blood's density as about $1000\;\text{kg/m}^3$) their pressure is higher by roughly $\rho g h = (1000)(10)(1.2) = 1.2 \times 10^4\;\text{Pa} \approx 90\;\text{mmHg}$.
• Arteries in the head, above the heart, have LOWER pressure than at heart level, for the same reason.
• An IV bag works by hydrostatics: hung above the arm, the column of fluid in the line makes the pressure at the needle higher than the pressure in the vein, so fluid flows in.
• Pressure also falls ALONG the circulation as blood flows, but that is resistance, not gravity: see the Fluids in Motion lesson.

Pascal's Principle

Pascal's principle: a pressure change applied to an enclosed, incompressible fluid is transmitted undiminished to every part of the fluid and to the walls of its container.

The hydraulic lift. Two pistons of areas $A_1$ (small) and $A_2$ (large) sit on the same enclosed fluid at the same height. Pressing the small piston raises the pressure everywhere by the same amount, so

$\frac{F_1}{A_1} = \frac{F_2}{A_2} \quad\Rightarrow\quad F_2 = F_1\frac{A_2}{A_1}$

• Force is multiplied by the area ratio $A_2/A_1$.
• Distance is divided by the same ratio. The fluid volume pushed out of the small cylinder equals the volume entering the large one: $A_1 d_1 = A_2 d_2$, where $d$ is the distance each piston moves.
• Work is conserved: $F_1 d_1 = F_2 d_2$. A hydraulic lift, like a lever, trades distance for force and never creates energy.
• If the area ratio is given as a ratio of DIAMETERS (or radii), square it: double the diameter means four times the area.

Pascal's principle and $\rho g h$ work together. Pushing on a piston raises the pressure everywhere by the same amount; points deeper in the fluid still have an extra $\rho g h$ on top. Hydraulic brakes, dentists' chairs and syringes all rely on the same idea.

Worked Example — Hydraulic Lift

Problem. A mechanic pushes down with $100\;\text{N}$ on a piston of area $0.01\;\text{m}^2$. The output piston has area $0.1\;\text{m}^2$.

• (a) What pressure does the mechanic create in the fluid?
• (b) What force does the fluid exert on the large piston?
• (c) How far must the mechanic's piston move to raise the load $0.05\;\text{m}$?
• (d) How much work goes in, and how much comes out?

(a) Pressure created. $P = \dfrac{F_1}{A_1} = \dfrac{100}{0.01} = 10^4\;\text{Pa}$

(b) Force on the large piston. The same $P = 10^4\;\text{Pa}$ from (a) acts there: $F_2 = PA_2 = (10^4)(0.1) = 1000\;\text{N}$. A tenfold area ratio gives a tenfold force.

(c) Distance. The fluid volume pushed out equals the volume pushed in, $A_1 d_1 = A_2 d_2$, so $d_1 = d_2 \dfrac{A_2}{A_1} = (0.05)(10) = 0.5\;\text{m}$.

(d) Work check. With $F_2 = 1000\;\text{N}$ from (b) and $d_1 = 0.5\;\text{m}$ from (c): in, $(100)(0.5) = 50\;\text{J}$; out, $(1000)(0.05) = 50\;\text{J}$. Force ×10, distance ÷10, work unchanged.

Worked Example — Hanging an IV Bag

Problem. A bag of saline ($\rho \approx 1000\;\text{kg/m}^3$) hangs so that the fluid surface in the bag is $1.0\;\text{m}$ above the needle in a patient's arm. Use $g = 10\;\text{m/s}^2$ and $1\;\text{mmHg} \approx 133\;\text{Pa}$.

• (a) With the line clamped and the fluid at rest, what is the gauge pressure at the needle?
• (b) What is that pressure in mmHg?
• (c) Pressure in a peripheral vein is only about 10 mmHg (gauge). Which way does fluid flow when the clamp opens?
• (d) Does a wider or fuller bag at the same height change the needle pressure? What if the bag is lowered to $0.5\;\text{m}$, or below the arm?

(a) Gauge pressure at the needle. $\rho g h = (1000)(10)(1.0) = 1.0 \times 10^4\;\text{Pa}$.

(b) Convert. $\dfrac{10^4}{133} \approx 75\;\text{mmHg}$.

(c) Compare. The needle pressure, about 75 mmHg from (b), is far above the vein's 10 mmHg, so when the clamp opens, saline flows into the vein.

(d) Only the height counts. No: only the HEIGHT of the fluid surface above the needle enters $\rho g h$, so a wider or fuller bag changes nothing. At $0.5\;\text{m}$ the pressure halves to about 38 mmHg, still above venous pressure, so saline flows in more slowly. Below the arm, the needle pressure drops under venous pressure and blood backs up into the line.

Practice: Instruments, Hydraulics & the Body 🎯

Key Takeaways

• Density and specific gravity. $\rho = m/V$; water is $1000\;\text{kg/m}^3$, and $SG = \rho/\rho_{\text{water}}$ has no units (mercury 13.6).
• Pressure grows with depth. $P = P_0 + \rho g h$ depends on depth, never container shape: about +1 atm per 10 m of water.
• Gauge versus absolute. Barometers read absolute pressure; manometers and blood pressure read gauge; absolute = gauge + 760 mmHg.
• Pascal's principle. $F_1/A_1 = F_2/A_2$: a hydraulic lift multiplies force, divides distance and conserves work.

Common traps

• Gauge or absolute. Check which one is asked: absolute pressure is the gauge reading PLUS atmospheric pressure.
• Wider container, more pressure. A wider or fuller container adds nothing at the same depth; only $\rho$, $g$ and $h$ count.
• Scaling the total pressure. Doubling the depth doubles only the $\rho g h$ part, not the total $P$, since $P_0$ stays fixed.
• Pistons and diameters. Pascal transmits PRESSURE, not force, and a diameter ratio must be squared to give the area ratio.

Fluids at Rest

Part 2 of 2 — Buoyancy, Archimedes & Surface Tension

Earlier in this lesson we saw that pressure grows with depth: $P = P_0 + \rho g h$. That single fact explains why things float.

Archimedes' Principle

Where buoyancy comes from. Hold a block of height $h$, with top and bottom faces of area $A$, under water. Its bottom face is deeper than its top face by $h$, so the water pushes up on the bottom harder than it pushes down on the top. The pressure difference is $\Delta P = \rho_{\text{fluid}} g h$, and the net upward force is

$\Delta P \cdot A = \rho_{\text{fluid}}(Ah)g = \rho_{\text{fluid}} V g$

The sideways pushes on the vertical faces cancel. This net upward force from the fluid is the buoyant force, $F_B$.

Archimedes' principle: the buoyant force on an object equals the weight of the fluid it displaces:

$F_B = \rho_{\text{fluid}} \, V_{\text{sub}} \, g$

where $V_{\text{sub}}$ is the submerged volume of the object, which is the volume of fluid pushed aside.

• It uses the FLUID's density, not the object's: equal submerged volumes in the same fluid feel equal buoyant forces, whether the object is lead or cork.
• Only the submerged volume counts.
• For a fully submerged object it does not depend on depth: a liquid is incompressible, so the top-to-bottom pressure difference is the same at any depth.

Float, hover or sink? Compare the object's average density (its total mass divided by its total volume, including any air inside) with the fluid's density.

• A steel ship floats although steel is about 8 times denser than water: its hull encloses air, so its average density is below water's.
• A fish adjusts the gas volume in its swim bladder to change its average density and hover.
• With full lungs most people's average density is slightly below water's, so they float; after a full exhalation many sink.

Floating objects. A floating object is in equilibrium, so the buoyant force equals its weight: $\rho_{\text{fluid}} V_{\text{sub}} g = \rho_{\text{object}} V g$. Rearranging,

$\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$

In water, the fraction submerged equals the object's specific gravity. Keep the two cases apart: a FLOATING object displaces its own WEIGHT of fluid; a fully SUBMERGED object displaces its own VOLUME of fluid.

Apparent weight. An object hung from a scale while fully submerged reads lighter by the buoyant force: $W_{\text{apparent}} = W - F_B$. This is why patients recovering muscle strength often exercise in a pool: the water supports part of their weight.

Practice: Archimedes' Principle 🎯

Specific Gravity Reasoning

Weighing in air and in water. The buoyant force on a fully submerged object equals the weight of an equal volume of water. The ratio of an object's weight to the weight of an equal volume of water is exactly its specific gravity, so

$SG = \frac{W}{F_B} = \frac{W}{W - W_{\text{apparent}}}$

This is how Archimedes could test a crown for gold without melting it down.

Hydrometer. A hydrometer is a weighted float with a scale on its narrow stem. Its weight is fixed, so it always displaces the same WEIGHT of liquid. In a denser liquid that weight is a smaller volume, so the hydrometer rides HIGHER; the liquid's SG is read where the surface meets the stem. Clinically, a higher urine specific gravity means more concentrated urine, as in dehydration.

• Normal urine specific gravity runs from roughly 1.005 to 1.030.

One object, two fluids. The submerged fraction is $\rho_{\text{object}}/\rho_{\text{fluid}}$, so the same object rides higher in a denser fluid: a swimmer floats more easily in seawater (SG about 1.03) than in a pool. An object whose density lies between two liquids that do not mix, such as oil floating on water, comes to rest at the boundary, partly in each layer.

Melting ice. An ice cube floating in a glass of water displaces exactly its own weight of water. When it melts it becomes exactly that weight of water, so the water level does not change.

Surface Tension, Cohesion and Adhesion

• Cohesion is the attraction between molecules of the same substance (water to water, through hydrogen bonds). Adhesion is the attraction between a liquid and a different material (water to glass).
• A molecule inside a liquid is pulled equally in every direction; a molecule at the surface has neighbours only below and beside it, so it feels a net inward pull. The surface behaves like a stretched elastic skin.
• Surface tension $\gamma$ (Greek "gamma") measures that skin's strength as a force per unit length along the surface (N/m), or equivalently the energy needed to create a unit area of new surface. Water's is high because of hydrogen bonding.
• Consequences: droplets pull into spheres (the shape with the least surface area); a water strider or a carefully placed steel needle rests ON the surface, held up by surface tension rather than buoyancy; soaps and other surfactants (surface-active agents) lower $\gamma$.

Capillary action. In a narrow tube, the balance of adhesion and cohesion decides what the liquid does.

The narrower the tube, the larger the rise (or fall). Capillary action draws a drop of blood into a thin glass tube and water into paper towels.

Surface tension in the lungs. The alveoli (tiny air sacs of the lung) are lined with a thin layer of fluid. Its surface tension pulls inward and tends to collapse each alveolus, and the effect is stronger the smaller the alveolus. Pulmonary surfactant, secreted by type II alveolar cells, lowers that surface tension and keeps small alveoli open. Premature infants who cannot yet make enough surfactant develop respiratory distress syndrome.

• The pressure needed to keep a spherical bubble inflated against surface tension is $\Delta P = 2\gamma/r$ (the law of Laplace), so smaller bubbles need more pressure.
• The capillary rise in a tube of radius $r$ is $h = 2\gamma\cos\theta/(\rho g r)$ (Jurin's law), where $\theta$ is the contact angle.

Worked Example — The Block That Floats 60% Submerged

Problem 1. A block of wood floats in water with 60% of its volume submerged.

• (a) What are the block's density and specific gravity?
• (b) What fraction of the block is submerged when it floats in oil ($SG = 0.8$)?
• (c) In what fluids does the block sink instead of floating?

(a) Density from floating. Floating means $F_B = W$:

$\rho_{\text{water}}(0.6V)g = \rho_{\text{wood}}Vg \;\Rightarrow\; \rho_{\text{wood}} = 0.6\,\rho_{\text{water}} = 600\;\text{kg/m}^3$

The submerged fraction IS the specific gravity: 0.6.

(b) Move it to oil. It must still displace its own weight, now of a less dense fluid, so it sinks deeper. With SG 0.6 from (a), the fraction is $0.6/0.8 = 0.75$, or 75% submerged.

(c) The sinking limit. In any fluid with SG below 0.6, even full submersion cannot displace enough weight, and the block sinks.

Problem 2. The same block ($600\;\text{kg/m}^3$) has a volume of $0.01\;\text{m}^3$ (10 L) and is held completely under water. Use $g = 10\;\text{m/s}^2$.

• (a) What buoyant force acts on the block?
• (b) What does the block weigh?
• (c) What downward push holds it under, and does that push change with depth?

(a) Buoyant force. Fully submerged in water, $F_B = (1000)(0.01)(10) = 100\;\text{N}$.

(b) Weight. $W = (600)(0.01)(10) = 60\;\text{N}$.

(c) Pushing it under. With $F_B = 100\;\text{N}$ from (a) and $W = 60\;\text{N}$ from (b), holding it under takes a downward push of $100 - 60 = 40\;\text{N}$. That push is the same at any depth, because the buoyant force on a fully submerged object does not change with depth.

Worked Example — Weighing an Ore Sample

Problem. A geologist's rock sample weighs $50\;\text{N}$ in air and $40\;\text{N}$ when hung fully submerged in water. Use $g = 10\;\text{m/s}^2$.

• (a) What buoyant force acts on the submerged sample?
• (b) What is the sample's volume?
• (c) What is its density?
• (d) Does the shortcut $SG = W/F_B$ agree, and could the sample be gold (SG about 19)?

(a) Buoyant force. $F_B = W - W_{\text{apparent}} = 50 - 40 = 10\;\text{N}$.

(b) Volume. With $F_B = 10\;\text{N}$ from (a), $V = \dfrac{F_B}{\rho_{\text{water}} g} = \dfrac{10}{(1000)(10)} = 1.0 \times 10^{-3}\;\text{m}^3$ (1 L).

(c) Density. Mass $= 50/10 = 5.0\;\text{kg}$. With $V = 1.0 \times 10^{-3}\;\text{m}^3$ from (b), $\rho = 5.0/(1.0 \times 10^{-3}) = 5000\;\text{kg/m}^3$.

(d) Shortcut check. $SG = W/F_B = 50/10 = 5$, the same answer in one line. Gold's SG is about 19, so this is not gold.

Practice: Specific Gravity & Surface Effects 🎯

Key Takeaways

• Archimedes' principle. $F_B = \rho_{\text{fluid}} V_{\text{sub}} g$, the displaced fluid's weight, whatever the depth or material.
• Average density decides. Less dense than the fluid floats, with submerged fraction $\rho_{\text{object}}/\rho_{\text{fluid}}$; equal hovers; denser sinks.
• Weighing under water. Apparent weight $= W - F_B$, and $SG = W/F_B$, which is how a crown or an ore sample is tested.
• Surface tension. Cohesion makes a surface skin; water rises in glass by adhesion; lung surfactant lowers tension so alveoli stay open.

Common traps

• Wrong density for buoyancy. $F_B$ uses the FLUID's density; the object's density only decides whether it sinks or floats.
• Deeper means more buoyancy. A fully submerged object's buoyant force does not grow with depth or with a denser object.
• Submerged versus above. $\rho_{\text{object}}/\rho_{\text{fluid}}$ is the fraction BELOW the surface; the part above is the rest.
• Apparent weight in SG. Divide the TRUE weight by $F_B$ to get the specific gravity, not the apparent weight.

Low-yield extras

• the normal urine specific-gravity range; the law of Laplace formula $\Delta P = 2\gamma/r$; Jurin's law for capillary rise