Week 8 In-Depth — Why Pressure, Viscosity, and Buoyancy Work

Read the cheatsheet first. This note explains why each result holds, then works one substantial example per topic with every step on the page. The final section assembles an exam-style worked sample end-to-end.

1 — Why a fluid is “a thing that flows under any shear stress”

A solid resists shear finitely: push sideways and the molecules deflect by a small amount, then settle into a new equilibrium. Push hard enough and the solid breaks, but for small shears nothing keeps moving.

A fluid is the opposite. Apply any non-zero shear stress $\tau$ and the fluid element keeps deforming — molecules slip past each other indefinitely. The deformation never reaches equilibrium; the rate of deformation does.

This is the deep reason behind every formula this week:

• At rest, a fluid carries zero shear stress. Otherwise it would flow.
• Pressure (a normal stress) is the only stress an unrestrained fluid at rest can carry, and by Pascal’s argument it must be isotropic.
• Viscosity is the constant of proportionality between shear stress and rate of deformation — Newton’s law of viscosity. Fluids with $\tau$ linear in $du/dy$ are Newtonian; water, air, and most oils are.

Worked example — viscous drag on a plate (Example 1)

A fluid of viscosity $\mu =0.89\times 10^{-3}\mathrm{Pa}\cdot \mathrm{s}$ fills a $5\mathrm{mm}$ gap between two horizontal plates. The bottom plate is stationary. The top plate has area $A=0.5{\mathrm{m}}^2$ and slides at $u_{\text{top}}=0.25m/s$. Find the hanging mass $m$ that holds the cable in tension at constant velocity.

Step 1 — velocity profile. For thin gaps with both plates Newtonian, the steady-state velocity profile is linear (this falls out of solving the Navier–Stokes equations; for now take it as given). So

$$\frac{du}{dy}=\frac{u_{\text{top}}-0}{\Delta y}=\frac{0.25}{5\times 10^{-3}}=50{\mathrm{s}}^{-1}.$$

Step 2 — shear stress. Newton’s law:

$$\tau =\mu \frac{du}{dy}=0.89\times 10^{-3}\times 50=0.0445\mathrm{Pa}.$$

Step 3 — drag force on the plate. The shear stress acts over the wetted area:

$$V_{\text{shear}}=\tau A=0.0445\times 0.5=0.0223\mathrm{N}.$$

Step 4 — at constant velocity, cable tension = drag = weight of the hanging mass:

$$F_T=mg\Rightarrow m=\frac{0.0223}{9.8}\approx 2.3\mathrm{g}.$$

A tiny mass — because the fluid is thin, the gap is narrow, and the plate moves slowly. Try the tutorial slider-bearing problem (Ex 2) where $\mu$ is 300$\times$ larger; the answer comes out in newtons, not millinewtons.

Why the linear profile matters

The linear velocity profile is the simplest geometry where Newton’s law gives a constant shear stress across the gap. If the profile curved, $\tau$ would vary with $y$ and you’d need calculus on the inside, not just at the boundary. In Week 8 every viscosity question assumes linear profiles — your job is just to identify $\Delta u$, $\Delta y$, and $A$.

---

2 — Why pressure grows linearly with depth

Take a small cylindrical fluid element inside a static fluid: cross-section $A$, height $\Delta z$, with its axis vertical. Three vertical forces act:

• Pressure pushing down on the top: $F_1=p_{1}A$ (downward).
• Pressure pushing up on the bottom: $F_2=p_{2}A$ (upward).
• Weight of the element: $F_W=\rho Vg=\rho A\Delta zg$ (downward).

The element is at rest, so $\sum F_y=0$:

$$p_{2}A-p_{1}A-\rho gA\Delta z=0⟹\boxed{\Delta p=p_2-p_1=\rho g\Delta z}.$$

The area cancelled. Two important consequences:

1. The shape and width of the container do not matter. A 1 m column of water generates the same pressure at its base whether it’s a pencil-thin tube or a swimming pool. This is the “hydrostatic paradox” — it confuses everyone the first time.
2. Two points at equal depth in one connected fluid have equal pressure. Same $\rho$, same $\Delta z$, so same $\Delta p$. This is what makes a U-tube work.

Why isotropy (Pascal’s law) holds

Imagine a tiny wedge-shaped fluid element at a point, sides $\delta x$, $\delta y$, hypotenuse $\delta s$, oriented at angle $\theta$. The fluid is at rest. Apply force balance in $x$ and $y$ (the weight is cubic in the small lengths and vanishes faster than the surface forces, which are quadratic). You get $p_x=p_y=p_n$ — the pressure on any face equals the pressure on any other face. Pressure has no preferred direction at a point.

That’s why you can write “pressure at a point” without specifying a surface. And it’s why pressure at $A$ equals pressure at $B$ if they’re at the same depth in one fluid, regardless of the orientation of the wall they sit against.

Worked example — mixed-column pressure (Example 2)

Two open water tanks (depths $0.5\mathrm{m}$ and $0.8\mathrm{m}$ to a connecting channel), an oil layer ($SG=0.8$, ${\rho }_{\text{oil}}=800kg/{\mathrm{m}}^3$), and a vertical tube $1.0\mathrm{m}$ long ending at entrapped air at point $A$. Atmosphere $p_{\text{atm}}=100\mathrm{kPa}$. Find $p_A$.

Walk a path from atmosphere (point $E$) to $A$. Order: $E\to D\to C\to B\to A$. Sketch the geometry, then go segment-by-segment:

• $E\to D$: down $0.5\mathrm{m}$ of water $\Rightarrow$ add ${\rho }_{w}g(0.5)$.
• $D\to C$: same fluid (water), same depth on the connecting channel $\Rightarrow$ no change.
• $C\to B$: down $0.8\mathrm{m}$ of water $\Rightarrow$ add ${\rho }_{w}g(0.8)$.
• $B\to A$: up $1.0\mathrm{m}$ through oil $\Rightarrow$ subtract ${\rho }_{\text{oil}}g(1.0)$.

(Air in the tube has $\rho \approx 1.3kg/{\mathrm{m}}^3$ — its contribution over $\le 1\mathrm{m}$ is a few Pa, ignored.)

$$p_A=p_E+{\rho }_{w}g(0.5)+{\rho }_{w}g(0.8)-{\rho }_{\text{oil}}g(1.0).$$ $$p_A=100000+1000(9.8)(0.5)+1000(9.8)(0.8)-800(9.8)(1.0)$$ $$=100000+4900+7840-7840=104900\mathrm{Pa}\text{(absolute)}.$$

Gauge: $p_{A,\text{gauge}}=4900\mathrm{Pa}\approx 4.9\mathrm{kPa}$.

The pattern is mechanical. Atmosphere $+$ signed sum of $\rho g\Delta h$ along the path.

Worked example — differential manometer (Example 3)

Two containers of oil ($SG=2.5\Rightarrow {\rho }_{\text{oil}}=2500kg/{\mathrm{m}}^3$) joined by a mercury manometer ($SG=13.55\Rightarrow {\rho }_{\text{Hg}}=13550$). Container A has $1.5\mathrm{m}$ of oil above it (open to atmosphere); the manometer between A and B reads: A $\to$ C up $1.1\mathrm{m}$ of oil, C $\to$ D down $0.52\mathrm{m}$ of mercury, D $\to$ B down $2.0\mathrm{m}$ of oil. Atmosphere $100\mathrm{kPa}$. Find $p_A$ and $p_B$.

Pressure at A (open column above A):

$$p_A=p_{\text{atm}}+{\rho }_{\text{oil}}g(1.5)=100000+2500(9.8)(1.5)=136750\mathrm{Pa}.$$

Step through to B:

$$p_C=p_A-{\rho }_{\text{oil}}g(1.1)=136750-26950=109800\mathrm{Pa}.$$ $$p_D=p_C+{\rho }_{\text{Hg}}g(0.52)=109800+69050=178850\mathrm{Pa}.$$ $$p_B=p_D+{\rho }_{\text{oil}}g(2.0)=178850+49000=227850\mathrm{Pa}\approx 228\mathrm{kPa}.$$

(Small rounding to the lecturer’s $\approx 227990$; the difference is in the $g=9.8$ vs $9.81$ choice.)

Why inclined manometers help — Example 4

Two water tanks A, B joined by a mercury manometer, but the mercury sits along $26.8\mathrm{cm}$ of inclined tube at angle $\theta$. The geometry shows water columns $a$ on each side and a mercury slab of total vertical drop $2a$. Given $p_B-p_A=20\mathrm{kPa}$, find $a$ and $\theta$.

Path A $\to$ B walks: down $a$ of water, down $2a$ of mercury, up $a$ of water. The water terms cancel:

$$p_B-p_A=2a{\rho }_{\text{Hg}}g\Rightarrow a=\frac{20000}{2(13600)(9.8)}=0.075\mathrm{m}.$$

The inclined segment has length $0.268\mathrm{m}$ but only contributes a vertical drop of $2a=0.15\mathrm{m}$. The geometry forces:

$$\sin \theta =\frac{2a}{0.268}=\frac{0.15}{0.268}\Rightarrow \theta ={\sin }^{-1}(0.5597)\approx 34^{\circ }.$$

Why this is useful. A short vertical drop of mercury becomes a long inclined column — easier to read precisely. The trade-off is that you must convert inclined length to vertical drop with $\sin \theta$ before applying $\rho gh$.

---

3 — Why buoyancy works — the bottom is at higher pressure

Take a submerged object of volume $V$. Pressure on every surface element points inward, perpendicular to the surface. Pressure on the bottom of the object is higher than pressure on the top, because depth is larger. Integrate pressure over the whole surface and the net horizontal forces cancel by symmetry; the vertical net force is upward and (after the divergence theorem, but it falls out of any simple geometry too) equals:

$$F_B={\rho }_{\text{fluid}}g{V}_{\text{displaced}}.$$

Notice three crucial details:

1. The fluid’s density appears, not the object’s. The buoyant force only knows about the fluid pressing on the surfaces.
2. The displaced volume — for a fully submerged object that’s $V_{\text{object}}$; for a floating object that’s only the submerged portion.
3. No surface integral needed in practice. Once you accept the result, you read off $V$ from the geometry and multiply.

Worked example — cable tension on a submerged block (Example 5)

A concrete block $0.4\times 0.4\times 0.3\mathrm{m}$ ($V=0.048{\mathrm{m}}^3$), ${\rho }_c=2300kg/{\mathrm{m}}^3$, in seawater ${\rho }_{\text{sea}}=1025kg/{\mathrm{m}}^3$. A cable holds the block hanging from a crane.

(a) In air — buoyancy from air is negligible (${\rho }_{\text{air}}\sim 10^{-3}{\rho }_c$):

$$F_T=W={\rho }_{c}Vg=2300\times 0.048\times 9.8=1083\mathrm{N}.$$

(b) Fully submerged in seawater:

$$F_B={\rho }_{\text{sea}}Vg=1025\times 0.048\times 9.8=483\mathrm{N}.$$ $$F_T=W-F_B=1083-483=600\mathrm{N}.$$

The block “loses” $483\mathrm{N}$ of apparent weight to buoyancy. The cable carries the difference.

Floating bodies and the iceberg

For a floating object, only the submerged volume displaces fluid. Equilibrium says weight equals buoyant force:

$${\rho }_{\text{object}}{V}_{\text{object}}g={\rho }_{\text{fluid}}{V}_{\text{submerged}}g.$$

So the submerged fraction is ${\rho }_{\text{object}}/{\rho }_{\text{fluid}}$. An iceberg with $1/7$ above seawater has $6/7$ submerged, so ${\rho }_{\text{ice}}=\frac{6}{7}{\rho }_{\text{seawater}}$ — and in pure water the submerged fraction becomes ${\rho }_{\text{ice}}/{\rho }_{\text{water}}=(6/7)(1025/1000)\approx 0.879$, i.e. about $12.1\%$ above the surface.

Hydrometers

A hydrometer floats vertically; total mass equals mass of displaced fluid:

$$m={\rho }_{\text{fluid}}\left(\frac{4}{3}\pi r_s^3+\pi r_c^{2}d\right),$$

where the bulb (sphere of radius $r_s$) is fully submerged and the cylindrical stem (radius $r_c$) has depth $d$ submerged. Solving for $d$ at a known ${\rho }_{\text{fluid}}$ gives the calibration mark. Tutorial Ex 7 walks through this with $SG=0.8$ and gives $d=161.3\mathrm{mm}$.

---

4 — How these three connect

All three pillars share one ingredient: force balance on a fluid element.

Topic	Force balance setup
Viscosity	Horizontal: shear stress on top face = applied drag, integrated over $A$.
Hydrostatic pressure	Vertical: pressure top + pressure bottom + weight = 0 on a column.
Buoyancy	Vertical: pressure top + pressure bottom (over the whole surface) gives a net upward force equal to displaced weight.

If you can draw a free body of a small fluid element and write $\sum F=0$, you can derive every Week 8 result. The formulas in the cheatsheet are shortcuts; the underlying physics is one equation applied in different geometries.

A second connection worth seeing: viscosity matters for moving fluids, pressure for static fluids, buoyancy for static fluids with submerged objects. Week 9 (fluid dynamics) brings viscosity and pressure together via the Reynolds number and viscous losses in pipes.

---

5 — Exam-style sample, end-to-end

A spherical buoy of radius $r=0.30\mathrm{m}$ and mass $m=50\mathrm{kg}$ is tethered by a vertical cable to the seabed in seawater (${\rho }_{\text{sea}}=1025kg/{\mathrm{m}}^3$). The buoy is fully submerged at the operating depth. (a) Find the buoyant force on the buoy. (b) Find the tension in the cable. (c) The system also acts as an inclined-tube manometer reading the gauge pressure at the buoy’s centre to a surface station; the inclined tube ($\theta =30^{\circ }$) holds mercury (${\rho }_{\text{Hg}}=13600kg/{\mathrm{m}}^3$). If the gauge pressure at the buoy’s centre is $1.5\mathrm{kPa}$, what is the inclined length of mercury required?

Take $g=9.8m/{\mathrm{s}}^2$.

(a) Buoyant force.

Volume of the buoy: $V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (0.30{)}^3=0.1131{\mathrm{m}}^3$.

$$F_B={\rho }_{\text{sea}}Vg=1025\times 0.1131\times 9.8=1136\mathrm{N}.$$

(b) Cable tension.

Free body of the buoy (vertical): up = down. $F_B$ acts up; $W=mg$ acts down; cable tension $T$ pulls down (cable is below the buoy, anchoring it).

$$\sum F_y=0\Rightarrow F_B-W-T=0.$$ $$W=50\times 9.8=490\mathrm{N},T=F_B-W=1136-490=646\mathrm{N}.$$

Positive $T$ confirms the cable is in tension (the buoy wants to rise — buoyancy exceeds weight).

(c) Inclined mercury column for $p_{\text{gauge}}=1.5\mathrm{kPa}$.

For a vertical column, $h_{\text{vert}}=p/({\rho }_{\text{Hg}}g)$:

$$h_{\text{vert}}=\frac{1500}{13600\times 9.8}=0.01125\mathrm{m}=11.25\mathrm{mm}.$$

For an inclined tube at $\theta =30^{\circ }$, the inclined length $L$ satisfies $L\sin \theta =h_{\text{vert}}$:

$$L=\frac{h_{\text{vert}}}{\sin 30^{\circ }}=\frac{0.01125}{0.5}=0.0225\mathrm{m}=22.5\mathrm{mm}.$$

The incline doubles the readable length — that’s the design trade.

Summary box.

Quantity	Value
Buoy volume $V$	$0.1131{\mathrm{m}}^3$
Buoyant force $F_B$	$1136\mathrm{N}$
Weight $W$	$490\mathrm{N}$
Cable tension $T$	$646\mathrm{N}$
Vertical Hg column $h_{\text{vert}}$	$11.25\mathrm{mm}$
Inclined Hg length $L$	$22.5\mathrm{mm}$

The template — list the volumes / densities / depths, write a free body, apply $\sum F=0$ or $\Delta p=\rho g\Delta h$, plug in last — is the template for every Week 8 exam question.

---

Where to look next

• The cheatsheet for fast revision plus the quiz.
• The lecture summary for the full reconstruction (all five worked examples and the tutorial answers).
• Week 9 picks up fluid dynamics — Bernoulli’s equation, viscous losses, the Reynolds number. The hydrostatic results here are the limiting case of “no flow.”