Exercises — Fluid definition — shear stress, no fixed shape

A fluid is defined by continuous deformation under any shear, however small.

• Steel bar → tilts a tiny fixed amount and stops → solid.
• Air → flows, fills its container → fluid (a gas is a fluid).
• Honey → flows (slowly, high $\mu$) → fluid.
• Rubber → deforms a fixed amount and holds → solid.
• Water → flows → fluid. Fluids: air, honey, water.

Both are "force per area", split by direction relative to the surface: $\sigma =\frac{F_n}{A}(\perp \text{, normal}),\tau =\frac{F_t}{A}(\parallel \text{, shear}).$ The shear part is the one a fluid at rest cannot resist.

At static rest a fluid sustains zero shear stress — that is its defining property. With no motion, $du/dy=0$, so $\tau =\mu \cdot 0=0$. (There is still normal stress — pressure — but no shear.)

Why linear ⇒ constant gradient: because the bottom fluid sticks to the fixed wall (speed $0$) and the top fluid sticks to the plate (speed $0.6$), and nothing between bends the line, the speed climbs at a steady rate. A straight line has one slope everywhere, and that slope is exactly $du/dy$. So the gradient = (speed difference)/(gap). First convert: $3\text{mm}=0.003\text{m}$. $\frac{du}{dy}=\frac{0.6-0}{0.003}=200{\text{s}}^{-1}.$ What it looks like: in Figure 1 above, the straight amber line rises $0.6\text{m/s}$ over a height of $0.003\text{m}$ — a steep, constant slope. Every cyan arrow is a fixed fraction longer than the one below it.

Why $\tau =\mu du/dy$ applies here: glycerine is a Newtonian fluid, and it is moving between the plates, so each layer slides over the next. Newton's law says the sideways drag between adjacent layers is proportional to how fast their speeds differ per unit height — that difference-per-height is exactly $du/dy$, and $\mu$ is the constant of proportionality. With a constant gradient the stress is the same throughout the film: $\tau =\mu \frac{du}{dy}=1.5\times 200=300\text{Pa}.$

Why multiply by area: $\tau$ is force per unit area. The whole plate touches the fluid over area $A$, and the same stress acts on every patch, so the total tangential force is stress × area. Inverting $\tau =F_t/A$: $F_t=\tau A=300\times 0.25=75\text{N}.$

Why we can solve for $\mu$: Newton's law $\tau =\mu du/dy$ links three quantities; given any two we get the third. Here $\tau$ and $du/dy$ are measured, so $\mu$ is just their ratio — this is literally how viscosity is measured in a lab (a viscometer): $\mu =\frac{\tau }{du/dy}=\frac{48}{120}=0.4\text{Pa⋅s}.$

Here the profile is curved, so $du/dy$ is not constant — we must differentiate. Step 1 (what & why): the velocity gradient is the slope of $u(y)$, so differentiate: $\frac{du}{dy}=\frac{d}{dy}\left(500y^2\right)=1000y.$ Step 2: evaluate at the top, $y=h=0.01$: ${\frac{du}{dy}|}_{y=h}=1000\times 0.01=10{\text{s}}^{-1}.$ Step 3: apply Newton's law: $\tau =\mu \frac{du}{dy}=0.02\times 10=0.2\text{Pa}.$ What it looks like: the curved amber profile in the figure below is nearly flat at the bottom (small slope, small $\tau$) and steepest at the top (largest slope, largest $\tau$).

$\frac{du}{dy}=1000y$, and at $y=0$ this is $1000\times 0=0$. $\tau =\mu \times 0=0\text{Pa}.$ The bottom of a parabolic profile is momentarily flat (zero slope), so the shear there is zero. The stress is not the same everywhere when the profile curves.

From $\tau =\mu du/dy$, the gradient is $du/dy=\tau /\mu$. Same $\tau$, so the gradient is inversely proportional to $\mu$. ${\left(\frac{du}{dy}\right)}_A=\frac{20}{0.8}=25{\text{s}}^{-1},{\left(\frac{du}{dy}\right)}_B=\frac{20}{0.1}=200{\text{s}}^{-1}.$ Fluid B (the thinner one) shears $200/25=8$ times faster. Ratio = ${\mu }_A/{\mu }_B=8$.

Two different laws for two different materials. Solid ($\tau =G\gamma$): the strain angle reaches a fixed value and stops — the square tilts to a permanent lean $\gamma$ and holds it. $\gamma =\frac{\tau }{G}=\frac{5}{2.5\times 10^5}=2\times 10^{-5}\text{(radians, a fixed tilt)}.$ Fluid ($\tau =\mu \dot{\gamma }$, where $\dot{\gamma }=du/dy$): the tilt never stops; it grows at a steady rate. $\dot{\gamma }=\frac{\tau }{\mu }=\frac{5}{0.05}=100{\text{s}}^{-1}.$ Interpretation: the solid tilts by a permanent, tiny angle $\gamma =2\times 10^{-5}$ rad and holds it (it stores the stress like a spring). The fluid's tilt grows at $\dot{\gamma }=100$ radians per second forever — it flows. Note $\gamma$ is a dimensionless angle, while $\dot{\gamma }$ carries units ${\text{s}}^{-1}$ and equals the velocity gradient. This is the whole distinction between a solid and a fluid.

(a) Force. First the gradient: $du/dy=2/0.005=400{\text{s}}^{-1}$. Then $\tau =\mu \frac{du}{dy}=0.3\times 400=120\text{Pa},F_t=\tau A=120\times 0.4=48\text{N}.$ (b) Power. Power is force times the speed it moves at: $P=F_{t}u=48\times 2=96\text{W}.$ Why force × speed? Power is the rate of doing work; work is force × distance, and distance ÷ time = speed. So the drag you feel converts $96$ joules of your effort into heat every second inside the oil.

To hold a tall dome, the slanted outer surfaces must support shear forces (gravity pulls the sides down and out, which is a tangential pull along those surfaces).

• Steel: obeys $\tau =G\gamma$. It can balance the required shear with a tiny finite strain $\gamma$, reach equilibrium, and hold the dome.
• Water: at rest it can sustain zero shear ($G_{\text{static}}=0$, so any residual shear forces $\dot{\gamma }>0$). It cannot balance the tangential pull, so it keeps deforming — it slumps and spreads until only normal (vertical, pressure) forces remain, i.e. a flat puddle. Hence no fixed shape.

Key insight — why the stress is equal in both layers (Newton's laws on the interface): picture the thin sheet of fluid right at the boundary between the two layers. The bottom layer pulls on it with stress ${\tau }_1$ (a force ${\tau }_{1}A$), and the top layer pulls with stress ${\tau }_2$ (a force ${\tau }_{2}A$), in opposite directions. This interface sheet has essentially no mass. Newton's second law says force = mass × acceleration; with mass $\approx 0$, any unbalanced force would give infinite acceleration — impossible. So the forces must balance: ${\tau }_{1}A={\tau }_{2}A$, hence ${\tau }_1={\tau }_2=\tau$. The shear stress is the same in both layers — exactly like the same current flowing through resistors in series. Step 1: let the interface speed be $u_i$. Each layer is linear, so its gradient is (speed jump)/(its thickness): $\tau ={\mu }_1\frac{u_i-0}{y_1}={\mu }_2\frac{u-u_i}{y_2}.$ Step 2: plug numbers ($y_1=0.001$, $y_2=0.002$, $u=1$): $0.2\frac{u_i}{0.001}=0.5\frac{1-u_i}{0.002}\Rightarrow 200u_i=250(1-u_i).$ Step 3: solve: $200u_i=250-250u_i\Rightarrow 450u_i=250\Rightarrow u_i=0.5\overline{5}\text{m/s}\approx 0.556\text{m/s}$. Step 4: the shear stress: $\tau =200u_i=200\times 0.5556\approx 111.1\text{Pa}.$ (a) $\tau \approx 111.1\text{Pa}$, (b) $u_i\approx 0.556\text{m/s}$. What it looks like: the velocity profile is a bent straight line — steeper (bigger slope) in the thinner, less-viscous bottom layer, gentler in the thicker, more-viscous top layer — but both segments feel the same stress.

Chain the formulas backwards from the force limit. $F_t=\tau A=\left(\mu \frac{u}{y}\right)A\le F_{\text{max}}\Rightarrow \mu \le \frac{F_{\text{max}}y}{Au}.$ $\mu \le \frac{30\times 0.004}{0.6\times 1.5}=\frac{0.12}{0.9}\approx 0.133\text{Pa⋅s}.$ Choose an oil with $\mu \le 0.133\text{Pa⋅s}$.

• (a) $\mu \to 0$: $\tau =\mu du/dy\to 0$ for any finite gradient — an ideal (inviscid) fluid carries no shear stress even while moving. (Real fluids always have small but nonzero $\mu$, so this is an idealisation, never exactly reached.)
• (b) $du/dy\to 0$: $\tau =\mu \cdot 0\to 0$. This is the static case — a fluid at rest sustains zero shear, its defining property. No motion difference between layers means no drag between them.
• (c) $\mu \to \infty$: to keep $\tau$ finite you need $du/dy\to 0$, i.e. the fluid barely shears — neighbouring layers are forced to move at almost the same speed, so the material behaves like a rigid solid. Infinite viscosity ≈ solid: it stops flowing.
• (d) $y\to 0$ at fixed plate speed: with a linear profile $du/dy=u/y$, so as the gap shrinks $du/dy=u/y\to \infty$, and therefore $\tau =\mu u/y\to \infty$. Squeezing the film to zero thickness makes the required shear stress (and drag force $F_t=\tau A$) blow up — this is exactly why extremely thin lubricating films can transmit huge stresses and why bearings are never run truly dry.

Everything above is two ideas applied again and again: shear stress = tangential force per area ($\tau =F_t/A$), and for a moving Newtonian fluid, $\tau =\mu du/dy$. Straight profile → gradient is speed/gap. Curved profile → differentiate and evaluate at a point. Stacked layers → same stress (force balance on a massless interface), different slopes. At rest → $\tau =0$, which is what makes it a fluid.