Week 8 — Fluid properties, pressure, buoyancy

What this week is about

Week 8 is the opening chapter of fluid mechanics. You move from solids (Weeks 1–7) to substances that flow — and the whole week is built around three big ideas:

1. What is a fluid, and how do we describe it? → Density $\rho$, specific weight $\gamma$, specific gravity $SG$, and viscosity $\mu$.
2. How does pressure behave inside a fluid at rest? → Pascal’s law (isotropy) and the hydrostatic equation $\Delta p=\rho g\Delta h$, then manometers as a direct application.
3. Why do things float? → Archimedes’ principle, $F_B=\rho gV$, with $\rho$ the displaced-fluid density.

All three feed each other. Viscosity is the bridge from Week 7’s friction-stress idea; pressure depth-variation is just a force balance on a fluid column; buoyancy falls out of pressure being bigger at the bottom than at the top. See the in-depth note for those connections written out.

Notes in this week

• Lecture summary — the reconstructed lecture with derivations, worked examples 1–5, and tutorial answers.
• Cheatsheet — every formula, table, common mistake, and the quiz (mixed difficulty, reshuffles every visit).
• In-depth analysis — why each result holds, full worked example per topic, and an exam-style sample end-to-end.
• Study guide — what’s directly supported vs inferred, common mistakes, practice questions, confidence report.
• Workshop prep — 5-minute and 20-minute revision plans for Portfolio 7 and Lab Report 2.

Any extra notes you drop into this folder will appear here automatically.

What I need to know before the workshop

• Density, mass, and volume — and how $\rho gh$ generates a pressure
• Free body diagram with three forces (top pressure, bottom pressure, weight)
• How to read a manometer diagram and pick a path from one point to another
• Newton’s law of viscosity $\tau =\mu du/dy$ with linear velocity profile across a thin gap
• Archimedes: $F_B$ equals the weight of displaced fluid
• Stay consistent with $g$ — lecture uses $9.8m/{\mathrm{s}}^2$; some tutorial solutions use $9.81$

Assessment relevance

Week 8 carries three live assessment items:

• Portfolio 7 in the Workshop class (lecture slide 27 — the tutorial deck calls the same item Portfolio 8).
• Tutorial class with the eight problems in Tutorial 8.pdf (worked answers in the solutions PDF).
• Laboratory class — assessed worksheet. Data-collection session that feeds the 10% Lab Report. Bring the worksheet, collect data carefully, submit at the end. The Lab/Practical sits on top of the usual tutorial-and-portfolio rhythm this week, so plan your time accordingly.
• Mastering Physics: Fluids Introduction, Forces on Submerged Bodies.

Hydrostatics and buoyancy are exam staples — expect a manometer path-walking question and at least one Archimedes problem on the paper.

Overview

Week 8 opens fluid mechanics. The lecture distinguishes fluids from solids (a fluid deforms continuously under any shear stress), introduces the key scalar fluid properties (density $\rho$, specific weight $\gamma$, specific gravity $SG$, dynamic viscosity $\mu$, kinematic viscosity $\nu$), then builds up hydrostatics: pressure as a normal stress, Pascal’s law, the hydrostatic equation $\Delta p=\rho g\Delta h$, absolute vs gauge pressure, manometers, and finally buoyancy ($F_B=\rho gV$).

Week 8 learning activities (slide 27) confirm: Mastering Physics (Fluids Introduction, Forces on Submerged Bodies), Wolfson Chapter 15 reading, Portfolio 7 in the Workshop class, tutorial attendance, and an assessed Laboratory Class — “Collect data and submit your worksheet”. The tutorial slide deck (slide 19) refers to the same lab as “Laboratory 2 — Worksheet” and the Portfolio item as Portfolio 8. Lab data collection feeds the 10% Lab Report assessment.

Key concepts

Fluid definition. A fluid is a substance that deforms continuously (flows) when acted on by a shearing stress of any magnitude. At rest the net force on a fluid element is zero. Liquids are nearly incompressible and occupy a fixed volume; gases are easily compressed and fill the container. A free surface is the boundary between a liquid and the gas above it.

Density and its forms.

• Density $\rho =m/V$ in $kg/{\mathrm{m}}^3$. Constant density $\Rightarrow$ incompressible.
• Specific weight (unit weight) $\gamma =\rho g$ — weight per unit volume.
• Specific gravity (relative density) $SG=\rho /{\rho }_{\text{water}}$, dimensionless, water at STP as reference (${\rho }_w=1000kg/{\mathrm{m}}^3$).

Force vs stress. Distributed contact forces are described as stresses. For kinetic friction $F_{\text{fric}}={\mu }_k{F}_N$, and as a stress $F_{\text{fric}}=\tau A$. Replacing the slippery-plate stack with a fluid layer motivates viscosity.

Viscosity. A measure of the importance of friction in a flowing fluid. For a Newtonian fluid the shear stress is proportional to the velocity gradient (Newton’s law of viscosity). Dynamic viscosity $\mu$ has SI units $\mathrm{Pa}\cdot \mathrm{s}$ (equivalently $\mathrm{N}s/{\mathrm{m}}^2$). Kinematic viscosity $\nu =\mu /\rho$.

Pressure and Pascal’s law. Pressure is a normal stress that acts inward on a surface, $P=F/A$ in pascals ($\mathrm{Pa}=N/{\mathrm{m}}^2$). Pascal’s law: at a point in a fluid (at rest or moving) the pressure has the same magnitude in all directions — it is isotropic. Against a solid boundary pressure acts perpendicular to the boundary (distributed normal force).

Pressure varies linearly with depth. Taking a small cylindrical fluid element of cross-sectional area $A$ and height $\Delta z$ at rest, force balance $F_{p,2}-F_{p,1}-F_W=0$ gives $\Delta p=\rho g\Delta z$. Integrating for a homogeneous fluid: $p=\rho gz+p_0$.

Absolute vs gauge pressure.

• Absolute pressure is measured above a perfect vacuum: $p_{\text{abs}}=p_{\text{atm}}+\rho gz$.
• Gauge pressure uses atmospheric pressure as the datum: $p_{\text{gauge}}=\rho gz$. Standard atmosphere $\approx 101.325\mathrm{kPa}$.
• Vacuum (negative gauge) means $p_{\text{abs}}<p_{\text{atm}}$.

Two-point pressure rules. Points at equal depth in a continuous column of the same fluid have equal pressure ($p_A=p_B$). When the connecting path crosses a fluid–fluid boundary, work from first principles, applying $\Delta p=\rho g\Delta h$ on each segment (add when going down, subtract when going up).

Manometers. Barometer (vertical column with vacuum at top) gives $p_{\text{atm}}=\rho gh$ — about $10\mathrm{m}$ of water or $0.8\mathrm{m}$ of mercury for $100\mathrm{kPa}$. A piezometer is the simplest pressure tap. A U-tube manometer uses a heavy manometer fluid (typically mercury, $SG=13.55$) to span larger pressures. A differential manometer measures the pressure difference between two containers.

Buoyancy (Archimedes). Because pressure grows with depth, the bottom of any submerged volume is at higher pressure than the top, producing a net upward buoyant force equal to the weight of the displaced fluid: $F_B=\rho gV$, with $\rho$ the displaced-fluid density and $V$ the displaced volume. Sink/rise/neutral conditions follow from comparing $F_B$ with the object’s weight. For partially submerged floating bodies the displaced volume equals the submerged portion.

Core formulas

Quantity	Formula	Units
Density	$\rho =m/V$	$kg/{\mathrm{m}}^3$
Specific weight	$\gamma =\rho g$	$N/{\mathrm{m}}^3$
Specific gravity	$SG=\rho /{\rho }_{\text{water}}$	dimensionless
Friction as a stress	$F_{\text{fric}}=\tau A$	N
Newton’s law of viscosity	$\tau =\mu \frac{du}{dy}=\mu \frac{\Delta u}{\Delta y}$	Pa
Kinematic viscosity	$\nu =\mu /\rho$	${\mathrm{m}}^2/s$
Pressure	$P=F/A$	$\mathrm{Pa}=N/{\mathrm{m}}^2$
Hydrostatic increment	$\Delta p=\rho g\Delta z=\rho g\Delta h$	Pa
Absolute pressure	$p_{\text{abs}}=p_{\text{atm}}+\rho gz$	Pa
Gauge pressure	$p_{\text{gauge}}=\rho gz$	Pa
Buoyant force	$F_B=\rho gV$	N

Useful reference values from the lecture/notes: ${\rho }_{\text{water}}=1000kg/{\mathrm{m}}^3$, ${\rho }_{\text{air}}\approx 1.3kg/{\mathrm{m}}^3$, ${\rho }_{\text{mercury}}\approx 13550\text{–}13600kg/{\mathrm{m}}^3$, $g=9.8m/{\mathrm{s}}^2$ (lecture) or $9.81$ (some solutions), $p_{\text{atm,std}}=101.325\mathrm{kPa}$.

Worked examples

Example 1 — Viscous drag on a moving plate (slide 11, notes p. 1)

A fluid of viscosity $\mu =0.89\times 10^{-3}\mathrm{N}s/{\mathrm{m}}^2$ fills a gap $\Delta y=5\mathrm{mm}=5\times 10^{-3}\mathrm{m}$ between two parallel plates, contact area $A=0.5{\mathrm{m}}^2$. Find the hanging mass needed to drag the upper plate at $u_{\text{top}}=0.25m/s$.

For the upper plate at constant velocity, $F_T=F_W=mg$. Apply Newton’s law of viscosity with a linear profile from $u=0$ at the floor to $u_{\text{top}}$ at the moving plate:

= \frac{0.25 - 0}{5\times 10^{-3}}\times 0.89\times 10^{-3} = 0.0445\,\mathrm{Pa}.$$ Shear force on the plate equals the required cable tension: $$V = \tau A = 0.0445 \times 0.5 = 0.0223\,\mathrm{N} = F_T.$$ $$m = \frac{F_T}{g} = \frac{0.0223}{9.8} \approx 0.0023\,\mathrm{kg}.$$ ### Pressure variation derivation (notes p. 2) Free body on a static fluid cylinder of area $A$, height $\Delta z$, with $P_2$ pushing up on the bottom and $P_1$ pushing down on the top, weight $F_W = \rho V g = \rho g A\,\Delta z$: $$\sum F_y = 0 \;\Rightarrow\; P_2 A - P_1 A - \rho g A\,\Delta z = 0 \;\Rightarrow\; \boxed{\,\Delta P = \rho g\,\Delta z = \rho g\,\Delta h\,}$$ ### Example 2 — Entrapped air pressure via mixed columns (slide 17, notes p. 3–4) Tanks of water connected by an open-air channel and an oil layer ($SG = 0.8$, so $\rho_\text{oil} = 800\,\mathrm{kg/m^3}$) to a vertical tube of length $1.0\,\mathrm{m}$ ending at entrapped air at point A. $p_\text{atm} = 100\,\mathrm{kPa}$, water depths $0.5\,\mathrm{m}$ and $0.8\,\mathrm{m}$ as labelled. Walking the manometer path E→D→C→B→A (equal-depth and same-fluid points equate, air column neglected): $$p_A = p_E + \rho_w g(0.5) + \rho_w g(0.8) - \rho_\text{oil} g(1.0).$$ $$p_A = 100\,000 + 1000(9.8)(0.5) + 1000(9.8)(0.8) - 800(9.8)(1.0)$$ $$p_A = 104\,900\,\mathrm{Pa} \;\text{(absolute)},\qquad p_{A,\text{gauge}} = 4900\,\mathrm{Pa} \approx 4.9\,\mathrm{kPa}.$$ ### Example 3 — Differential manometer between two oil containers (slide 20, notes p. 5) Oil $SG = 2.5$ in both containers A and B, mercury manometer fluid $SG = 13.55$, $p_\text{atm} = 100\,\mathrm{kPa}$. With $\rho_\text{oil} = 2500\,\mathrm{kg/m^3}$ and $\rho_\text{Hg} = 13\,550\,\mathrm{kg/m^3}$: (a) Pressure at A from the open column ($1.5\,\mathrm{m}$ of oil above): $$p_A = p_\text{atm} + \rho_\text{oil} g (1.5) = 100\,000 + 2500(9.8)(1.5) = 136\,750\,\mathrm{Pa} \approx 136.75\,\mathrm{kPa}.$$ (b) Step through the manometer (A → C up $1.1\,\mathrm{m}$ of oil, C → D down $0.52\,\mathrm{m}$ of mercury, D → B down $2\,\mathrm{m}$ of oil): $$p_C = p_A - \rho_\text{oil} g(1.1) = 109\,800\,\mathrm{Pa},$$ $$p_D = p_C + \rho_\text{Hg} g(0.52) = 178\,850\,\mathrm{Pa},$$ $$p_B = p_D + \rho_\text{oil} g(2.0) = 227\,990\,\mathrm{Pa} \approx 228\,\mathrm{kPa}.$$ ### Example 4 — Inclined-tube mercury manometer (slide 21, notes p. 6) Two water tanks A and B, mercury column geometry shows water layers of height $a$ and $a$ on either side of a mercury slab of height $2a$, $26.8\,\mathrm{cm}$ inclined length at angle $\theta$. Given $p_B - p_A = 20\,\mathrm{kPa}$: $$p_B = p_A + \rho_w g(a) + \rho_\text{Hg} g(2a) - \rho_w g(a)$$ $$\Rightarrow\; p_B - p_A = 2a\,\rho_\text{Hg}\,g$$ $$a = \frac{p_B - p_A}{2\rho_\text{Hg}\,g} = \frac{20\,000}{2(13\,600)(9.8)} = 0.075\,\mathrm{m}.$$ The vertical drop spanned by the inclined segment is $2a$, so $\sin\theta = 2a / 0.268$: $$\theta = \sin^{-1}\!\left(\frac{2(0.075)}{0.268}\right) \approx 34^\circ.$$ ### Example 5 — Cable tension on a submerged concrete block (slide 25, notes p. 7) Block $0.4 \times 0.4 \times 0.3\,\mathrm{m} \Rightarrow V = 0.048\,\mathrm{m^3}$, $\rho_\text{concrete} = 2300\,\mathrm{kg/m^3}$, $\rho_\text{sea} = 1025\,\mathrm{kg/m^3}$. (a) In air, $F_T = W = \rho_c V g = 2300(0.048)(9.8) = 1083\,\mathrm{N}$. (b) Submerged, $F_T = W - F_B = 1083 - \rho_\text{sea} V g = 1083 - 1025(0.048)(9.8) = 600\,\mathrm{N}$. ## Things to practise The tutorial drills the same toolbox. Brief answers from the solutions PDF: - **Tutorial Ex 1.** Trapezoidal-cross-section pool $4\,\mathrm{m}\times 8\,\mathrm{m}$, depths $1\,\mathrm{m}$ and $3\,\mathrm{m}$. Volume $= 64\,\mathrm{m^3}$, mass $= 64\,000\,\mathrm{kg}$. - **Tutorial Ex 2.** Slider bearing, $\mu = 0.271\,\mathrm{Pa\cdot s}$, gap $0.25\,\mathrm{mm}$, $u_w = 3\,\mathrm{mm/s}$, plate $0.5\times 0.5\,\mathrm{m}$. $\tau = 3.252\,\mathrm{Pa}$, $V = 0.813\,\mathrm{N}$, $P = V u_w = 0.00244\,\mathrm{W}$. - **Tutorial Ex 3.** Concentric-cylinder bearing, $r_1 = 1.9\,\mathrm{cm}$, $r_2 = 2.0\,\mathrm{cm}$, $l = 10\,\mathrm{cm}$, $\omega = 500\, \mathrm{rad/s}$, $\mu = 0.104\,\mathrm{Pa\cdot s}$. Use $\Delta u = \omega r_2 = 10\,\mathrm{m/s}$, $A = 2\pi r_2 l = 0.004\pi\,\mathrm{m^2}$. $\tau = 1040\,\mathrm{Pa}$, $V = 13.07\,\mathrm{N}$, $P = 130.7\,\mathrm{W}$. - **Tutorial Ex 4.** Oil ($SG = 0.9$, $0.5\,\mathrm{m}$) over water ($1.2\,\mathrm{m}$). Top of water $p_A = 4410\,\mathrm{Pa}$, bottom $p_B = 16\,170\,\mathrm{Pa}$. - **Tutorial Ex 5.** Multi-chamber water/air container, gauge pressures. Approximate path-walking results: $p_A \approx 12.75\,\mathrm{kPa}$, $p_B \approx -2.94\,\mathrm{kPa}$, $p_C \approx -2.94\,\mathrm{kPa}$ (air column negligible), $p_D \approx -18.63\,\mathrm{kPa}$. - **Tutorial Ex 6.** Iceberg with $\tfrac{1}{7}$ above sea water. (a) $SG_\text{ice} = \rho_\text{ice}/\rho_\text{sw} = 6/7 \approx 0.857$. (b) In pure water $\rho_\text{ice} = 1025(6/7) = 878.57\,\mathrm{kg/m^3}$, submerged fraction $= 87.9\%$, so $12.1\%$ above the surface. - **Tutorial Ex 7.** Hydrometer (cylinder $6\,\mathrm{mm}$ dia $\times 180\,\mathrm{mm}$, mass $0.6\,\mathrm{g}$; sphere $20\,\mathrm{mm}$ dia, mass $6.4\,\mathrm{g}$) in liquid $SG = 0.8$. Total mass $7\times 10^{-3}\,\mathrm{kg}$. Solving $m = \rho(\tfrac{4}{3}\pi r_s^3 + \pi r_c^2 d)$ gives $d = 0.1613\,\mathrm{m} = 161.3\,\mathrm{mm}$. - **Tutorial Ex 8 (challenge).** Differential U-tube manometer, gauge pressures $p_1 = 38\,\mathrm{kPa}$, $p_2 = -50\,\mathrm{kPa}$, mercury $SG = 13.55$, water in the pipe. Equating pressures at the matched level: $R_P = (p_1 - p_2)/[g(\rho_\text{Hg} - \rho_w)] = 88\,000/[9.81(12\,550)] = 0.715\,\mathrm{m} = 715\,\mathrm{mm}$ of Hg. - **Bonus (express $50\,\mathrm{kPa}$ as a column height).** $h = p/(\rho g)$. Mercury: $0.376\,\mathrm{m} = 376.2\,\mathrm{mm}$. Water: $5.097\,\mathrm{m}$. Acetylene tetrabromide ($SG = 2.94$): $1.734\,\mathrm{m}$. ## Common pitfalls - **Sign of $\Delta h$.** Going *down* increases pressure (add $\rho g h$); going *up* decreases it (subtract). Pick a path and be consistent. - **Same fluid + same depth = same pressure** only if the points sit on a continuous path *within that one fluid*. Crossing a fluid–fluid interface (water → mercury → oil) requires the first-principles $\rho g \Delta h$ for each segment. - **Air columns are usually negligible** because $\rho_\text{air} \approx 1.3\,\mathrm{kg/m^3}$ is $\sim$1000× smaller than water — a few Pa over short vertical heights. Drop them unless told otherwise. - **Absolute vs gauge.** $p_\text{abs} = p_\text{atm} + p_\text{gauge}$. Negative gauge pressure (vacuum) is allowed and is *not* an error. Read the question for which the answer should be in. - **Specific gravity is a ratio.** $SG_\text{Hg} = 13.55$ means $\rho_\text{Hg} = 13\,550\,\mathrm{kg/m^3}$ (relative to water at $1000\,\mathrm{kg/m^3}$). Don't plug $SG$ into $\rho g h$ directly. - **Viscosity gap geometry.** $\Delta u$ is the velocity *difference* across the gap, $\Delta y$ is the gap thickness — not the plate thickness or the plate length. - **Buoyancy uses displaced fluid density**, not the object density. For a floating body the displaced volume is the *submerged* volume, not the whole body. - **$g$ value.** Lecture uses $9.8\,\mathrm{m/s^2}$; some tutorial solutions use $9.81$. Stay consistent within a problem. - **Inclined manometers.** Translate the inclined length into a vertical height with $\sin\theta$ before applying $\rho g h$. ## Source citations - Lecture slides — `EGD102-Physics/Lecture8_CTP1.pdf` (slides 1–28), particularly slide 3 (fluid definition), 7 (density forms), 10 (viscosity), 13–16 (pressure, Pascal, depth, two-point rules), 18–19 (manometers), 23–24 (buoyancy), 11/17/20/21/25 (Examples 1–5). - Handwritten worked examples — `EGD102-Physics/EGD102 - Lecture8 - Notes.pdf` pages 1 (Example 1), 2 (pressure-variation derivation, $\Delta P = \rho g \Delta h$), 3–4 (Example 2 walk-through), 5 (Example 3), 6 (Example 4), 7 (Example 5). - Tutorial problems — `EGD102-Physics/Tutorial 8.pdf` (Exercises 1–8). - Tutorial answers — `EGD102-Physics/Tutorial 8_Solutions.pdf` (numerical results quoted in *Things to practise*). - Reference text mentioned on slide 28: Wolfson, R. 2020. *Essential University Physics, Volume 1, Global Edition*, 4th ed. (SI Units), Chapter 15.