Fluid definition — shear stress, no fixed shape: Worked examples

This is the practice deep-dive for the parent note. We built the law $τ = μ \frac{d u}{d y}$ there. Here we hammer it against every kind of situation the topic can throw at you — including the weird ones (zero gap, non-linear profile, "it's a solid" traps). Read the matrix first, then do each example after forecasting the answer yourself.

Recall Symbols we will use (all earned in the parent note)

$τ$ ::: shear stress = tangential force per area, units Pa. $μ$ ::: dynamic viscosity, units Pa·s ("thickness" of the fluid). $d u / d y$ ::: velocity gradient = how fast the layer speed changes as you go up, units $s^{- 1}$ . $F_{t}$ ::: tangential (shear) force, units N. $A$ ::: contact area, units $m^{2}$ . $γ, \overset{γ}{˙}$ ::: shear strain (an angle), and its rate of change $d u / d y$ .

The scenario matrix

Everything about the fluid definition + Newton's viscosity law reduces to these case-classes. Each worked example below is tagged with the cell it fills.

Cell	Case class	What is tricky	Example
A	Linear profile, straight plug-in	nothing — the baseline	Ex 1
B	Recover force / area from $τ$	multiply back by $A$	Ex 2
C	Solve for the unknown $μ$	rearrange the law	Ex 3
D	Static fluid, $\overset{γ}{˙} = 0$	answer is zero — degenerate	Ex 4
E	Non-linear velocity profile	must use local slope (derivative)	Ex 5
F	Limiting behaviour: gap $\to 0$	gradient blows up	Ex 6
G	Real-world word problem	translate words to $u, y, A, μ$	Ex 7
H	Exam twist: solid-vs-fluid decision	is it $G γ$ or $μ \overset{γ}{˙}$ ?	Ex 8
I	Unit / sanity trap (mm vs m)	convert before dividing	Ex 9

Ex 1 — Baseline linear profile (Cell A)

Ex 2 — Recover the force (Cell B)

Ex 3 — Solve for the viscosity (Cell C)

Ex 4 — The degenerate case: fluid at rest (Cell D)

Worked example Zero rate of strain

A glass of water sits perfectly still on a table. What is the shear stress inside it?

Forecast: the defining property of a fluid should show up here — guess before reading on.

Step 1 — Nothing moves, so every layer has the same speed (zero): $u (y) = 0$ everywhere. Why this step? "At rest" means the velocity field is identically zero.

Step 2 — Therefore the gradient is $d u / d y = 0$ . Why this step? A constant (here, zero) has zero slope.

Step 3 — Apply the law: $τ = μ \times 0 = 0$ . $τ = 0 Pa .$ Why this step? No rate of strain ⇒ no viscous stress.

Verify: This is exactly the definition of a fluid — it sustains zero shear at static equilibrium. If $τ$ were nonzero the water would still be flowing, contradicting "at rest." ✓ (See Hydrostatics — Fluids at Rest — only normal pressure survives.)

Ex 5 — Non-linear profile: use the local slope (Cell E)

Worked example Curved velocity profile

Fluid flows past a wall with the measured profile $u (y) = 50 y - 2500 y^{2}$ (SI units, valid for $0 \leq y \leq 0.01$ m). Viscosity $μ = 0.02$ Pa·s. Find the shear stress right at the wall ( $y = 0$ ) and at $y = 4 mm$ .

Forecast: the profile is steepest at the wall, so shear should be largest at $y = 0$ and smaller higher up.

Step 1 — For a curved profile we cannot use "speed/gap." We need the local slope, i.e. the derivative — that is precisely what $\frac{d u}{d y}$ means. Why this tool and not rise/run? Rise/run only works for straight lines. A derivative gives the slope of the tangent at each point — the honest velocity gradient there. Look at the two tangent lines in the figure: their steepness differs. $\frac{d u}{d y} = 50 - 5000 y .$

Step 2 — At the wall, $y = 0$ : $d u / d y = 50 s^{- 1}$ , so $τ_{wall} = 0.02 \times 50 = 1.0 Pa .$ Why this step? Wall stress uses the slope evaluated at $y = 0$ .

Step 3 — At $y = 4 mm = 0.004$ m: $d u / d y = 50 - 5000 (0.004) = 50 - 20 = 30 s^{- 1}$ , so $τ = 0.02 \times 30 = 0.6 Pa .$ Why this step? Same derivative, evaluated at the new height.

Verify: $1.0 > 0.6$ ✓ — shear drops as we move away from the wall, matching the forecast (profile flattens). Units $Pa\cdots \times s^{- 1} = Pa$ ✓.

Ex 6 — Limiting behaviour: gap shrinks to zero (Cell F)

Worked example What happens as

$y \to 0$ ? Keep plate speed fixed at $u = 0.3$ m/s and viscosity $μ = 0.1$ Pa·s. Compute $τ$ for gaps $y = 2$ mm, $0.5$ mm, $0.1$ mm. What is the trend?

Forecast: thinner film squeezed at the same speed = harder to shear. Guess: $τ$ grows.

Step 1 — Gradient at each gap: $d u / d y = u / y$ .

$y = 0.002$ : $150 s^{- 1}$
$y = 0.0005$ : $600 s^{- 1}$
$y = 0.0001$ : $3000 s^{- 1}$ Why this step? Same speed over a smaller gap → steeper gradient.

Step 2 — Multiply by $μ = 0.1$ : $τ = 15 Pa, 60 Pa, 300 Pa .$ Why this step? Newton's law with the new gradients.

Step 3 — Take the limit: as $y \to 0$ , $\frac{u}{y} \to \infty$ , so $τ \to \infty$ . Why this step? Dividing a fixed speed by a shrinking gap grows without bound.

Verify: halving the gap doubles $τ$ ( $15 \to$ ... $y$ cut by 4 gives $τ \times 4 = 60$ ✓; cut by 20 gives $\times 20 = 300$ ✓). Physically: this is why thin lubricant films resist enormous shear — the basis of bearings.

Ex 7 — Real-world word problem (Cell G)

Worked example Skater on a water film

An ice skater's blade ( $A = 6.0 cm^{2}$ ) glides on a $10 μ m$ film of melt-water ( $μ = 1.8 \times 1 0^{- 3}$ Pa·s) at speed $8$ m/s. What viscous drag force resists the skater?

Forecast: super-thin film + real speed → sizeable gradient, but water is thin and area tiny; expect a fraction of a newton.

Step 1 — Translate words to symbols: $A = 6.0 cm^{2} = 6.0 \times 1 0^{- 4} m^{2}$ , $y = 10 μ m = 1.0 \times 1 0^{- 5}$ m, $u = 8$ m/s. Why this step? Every quantity must be SI before it touches the formula.

Step 2 — Gradient (linear film): $d u / d y = 8/ (1.0 \times 1 0^{- 5}) = 8.0 \times 1 0^{5} s^{- 1}$ . Why this step? Thin film, so large gradient — that's where the drag comes from.

Step 3 — Stress: $τ = μ (d u / d y) = (1.8 \times 1 0^{- 3}) (8.0 \times 1 0^{5}) = 1440$ Pa. Why this step? Newton's law.

Step 4 — Force: $F_{t} = τ A = 1440 \times 6.0 \times 1 0^{- 4} = 0.864$ N. Why this step? Convert stress back to a force via the blade area.

Verify: $\approx 0.86$ N of drag on a fast skater is believable (ice is famously low-friction). Units check: Pa·s × s⁻¹ × m² = N ✓.

Ex 8 — Exam twist: solid or fluid? (Cell H)

Worked example Which law applies?

A block of rubber ( $G = 5 \times 1 0^{5}$ Pa) is pushed by a steady shear $τ = 2 \times 1 0^{4}$ Pa. (a) Find its steady shear strain $γ$ . (b) A cousin insists rubber "flows" and uses $τ = μ \overset{γ}{˙}$ . Who is right?

Forecast: rubber tilts and stops — that's solid behaviour, so it has a finite $γ$ .

Step 1 — Decide the material class. Rubber reaches a fixed deformation and holds it, so it is a solid → use $τ = G γ$ , not the viscosity law. Why this step? The whole topic hinges on this fork: solids store shear ( $G γ$ ), fluids can't ( $μ \overset{γ}{˙}$ ). Choosing the law wrongly is the classic trap.

Step 2 — Solve for strain: $γ = τ / G = (2 \times 1 0^{4}) / (5 \times 1 0^{5}) = 0.04$ (a pure number = radians of tilt). $γ = 0.04.$ Why this step? $τ = G γ \Rightarrow γ = τ / G$ .

Step 3 — Judge the cousin: a fluid would keep tilting ( $\overset{γ}{˙} > 0$ forever). Rubber does not — it settles. So the cousin is wrong; the fluid law does not apply. Why this step? The defining test is "does it keep deforming?" Rubber says no.

Verify: $γ = 0.04 rad \approx 2.3°$ tilt — small and steady, exactly as a solid should behave. Contrast with Stress and Strain in Solids. ✓

Ex 9 — Unit/sanity trap (Cell I)

Worked example Spot the mm-vs-m error

A student computes shear in glycerin ( $μ = 1.4$ Pa·s), plate speed $0.6$ m/s, gap $3$ mm, and writes $d u / d y = 0.6/3 = 0.2 s^{- 1}$ , so $τ = 0.28$ Pa. Find the error and the correct $τ$ .

Forecast: glycerin is very thick; $0.28$ Pa looks suspiciously tiny. Something's off.

Step 1 — Diagnose: the student divided by 3 mm expressed as 3, forgetting to convert to metres. Why this step? $d u / d y$ must be (m/s)/(m); mixing m/s with mm silently multiplies the answer by 1000-error.

Step 2 — Correct gap: $y = 3 mm = 0.003$ m, so $d u / d y = 0.6/0.003 = 200 s^{- 1}$ . Why this step? Consistent SI units.

Step 3 — Correct stress: $τ = 1.4 \times 200 = 280$ Pa. $τ = 280 Pa .$ Why this step? Newton's law with the fixed gradient.

Verify: the true answer is exactly $1000 \times$ the wrong one ( $0.28 \to 280$ ) — the signature of a mm/m slip. And $280$ Pa for thick glycerin is reasonable, unlike $0.28$ Pa. ✓

Recall Quick self-test (cover the right side)

Linear profile, which formula for the gradient? ::: $d u / d y = u / y$ . Curved profile, which tool for the gradient? ::: the derivative $d u / d y$ evaluated locally. Fluid at rest, shear stress? ::: exactly zero (defining property). As gap $y \to 0$ at fixed speed, $τ$ does what? ::: grows without bound ( $\to \infty$ ). Rubber under steady shear — which law? ::: $τ = G γ$ (it's a solid, finite strain). First thing to check before plugging numbers? ::: convert mm/cm/µm to metres.

Connections

Velocity Profile and No-Slip Condition — where curved profiles like Ex 5 come from.
Viscosity and Newtonian vs Non-Newtonian Fluids — deeper on $μ$ used throughout.
Hydrostatics — Fluids at Rest — the $\overset{γ}{˙} = 0$ world of Ex 4.
Stress and Strain in Solids — the $τ = G γ$ side of Ex 8.
Pressure in Fluids — the normal-stress companion to shear.