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

2.2.1 · D3 · Physics › Fluid Mechanics › Fluid definition — shear stress, no fixed shape

Yeh practice deep-dive hai parent note ke liye. Wahan humne law $τ = μ \frac{d u}{d y}$ banaya tha. Yahan hum ise har us situation ke against maarte hain jo topic throw kar sakta hai — including weird wale (zero gap, non-linear profile, "yeh solid hai" wale traps). Pehle matrix padho, phir har example karo apna answer forecast karne ke baad.

Recall Symbols jo hum use karenge (sab parent note mein earn kiye hain)

$τ$ ::: shear stress = tangential force per area, units Pa. $μ$ ::: dynamic viscosity, units Pa·s (fluid ki "thickness"). $d u / d y$ ::: velocity gradient = layer ki speed kitni tezi se badal rahi hai jaise upar jaate hain, units $s^{- 1}$ . $F_{t}$ ::: tangential (shear) force, units N. $A$ ::: contact area, units $m^{2}$ . $γ, \overset{γ}{˙}$ ::: shear strain (ek angle), aur uska rate of change $d u / d y$ .

The scenario matrix

Fluid definition + Newton's viscosity law ke baare mein sab kuch in case-classes tak reduce hota hai. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Kya tricky hai	Example
A	Linear profile, seedha plug-in	kuch nahi — yeh baseline hai	Ex 1
B	$τ$ se force / area recover karna	$A$ se multiply karo wapas	Ex 2
C	Unknown $μ$ solve karna	law ko rearrange karo	Ex 3
D	Static fluid, $\overset{γ}{˙} = 0$	answer hai zero — degenerate case	Ex 4
E	Non-linear velocity profile	local slope (derivative) use karna padega	Ex 5
F	Limiting behaviour: gap $\to 0$	gradient blow up hota hai	Ex 6
G	Real-world word problem	words ko $u, y, A, μ$ mein translate karo	Ex 7
H	Exam twist: solid-vs-fluid decision	kya yeh $G γ$ hai ya $μ \overset{γ}{˙}$ ?	Ex 8
I	Unit / sanity trap (mm vs m)	divide karne se pehle convert karo	Ex 9

Ex 1 — Baseline linear profile (Cell A)

Ex 2 — Force recover karo (Cell B)

Ex 3 — Viscosity solve karo (Cell C)

Worked example Unknown

$μ$ Ek unknown oil ke upar $A = 0.5 m^{2}$ area ki plate drag karte hain, gap $2 mm$ , plate speed $0.3$ m/s, aur measured force $F_{t} = 7.5$ N lagti hai. $μ$ find karo.

Forecast: oils water se "thicker" hote hain, toh expect karo $μ$ tenths of Pa·s ke around, $1 0^{- 3}$ nahi.

Step 1 — Measured force se stress nikalo: $τ = F_{t} / A = 7.5/0.5 = 15$ Pa. Yeh step kyun? Hum force measure karte hain, lekin law stress mein likha hai — pehle convert karo.

Step 2 — Gradient: $d u / d y = 0.3/0.002 = 150 s^{- 1}$ . Yeh step kyun? Linear profile, phir se rise/run.

Step 3 — $μ$ ke liye law rearrange karo: $μ = \frac{τ}{d u / d y} = \frac{15}{150} = 0.1 Pa\cdots .$ Yeh step kyun? $τ = μ (d u / d y)$ , toh $μ$ isolate karne ke liye stress ko gradient se divide karo.

Verify: $0.1$ Pa·s water se 100× zyada hai — "oil" se consistent. Plug back karo: $0.1 \times 150 = 15$ Pa ✓.

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

Worked example Zero rate of strain

Paani ka ek glass table par bilkul still rakha hai. Uske andar shear stress kya hai?

Forecast: fluid ki defining property yahan dikhni chahiye — aage padhne se pehle guess karo.

Step 1 — Kuch move nahi kar raha, toh har layer ki speed same (zero) hai: $u (y) = 0$ har jagah. Yeh step kyun? "At rest" ka matlab hai velocity field identically zero hai.

Step 2 — Isliye gradient hai $d u / d y = 0$ . Yeh step kyun? Ek constant ka (yahan, zero) slope zero hota hai.

Step 3 — Law apply karo: $τ = μ \times 0 = 0$ . $τ = 0 Pa .$ Yeh step kyun? Rate of strain nahi ⇒ viscous stress nahi.

Verify: Yeh bilkul fluid ki definition hai — yeh static equilibrium par zero shear sustain karta hai. Agar $τ$ nonzero hota toh paani abhi bhi flow kar raha hota, jo "at rest" ka contradict karta. ✓ (Dekho Hydrostatics — Fluids at Rest — sirf normal pressure bachti hai.)

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

Worked example Curved velocity profile

Fluid ek wall ke paas measured profile $u (y) = 50 y - 2500 y^{2}$ ke saath flow karta hai (SI units, valid hai $0 \leq y \leq 0.01$ m ke liye). Viscosity $μ = 0.02$ Pa·s. Shear stress find karo bilkul wall par ( $y = 0$ ) aur $y = 4 mm$ par.

Forecast: profile wall par sabse steep hai, toh shear sabse zyada $y = 0$ par hona chahiye aur upar jaane par chhhota.

Step 1 — Curved profile ke liye hum "speed/gap" use nahi kar sakte. Hume local slope chahiye, yaani derivative — yahi $\frac{d u}{d y}$ ka matlab hai. Yeh tool kyun, rise/run nahi? Rise/run sirf straight lines ke liye kaam karta hai. Derivative har point par tangent ka slope deta hai — wahan ka honest velocity gradient. Figure mein do tangent lines dekho: unki steepness alag hai. $\frac{d u}{d y} = 50 - 5000 y .$

Step 2 — Wall par, $y = 0$ : $d u / d y = 50 s^{- 1}$ , toh $τ_{wall} = 0.02 \times 50 = 1.0 Pa .$ Yeh step kyun? Wall stress $y = 0$ par evaluate kiya hua slope use karta hai.

Step 3 — $y = 4 mm = 0.004$ m par: $d u / d y = 50 - 5000 (0.004) = 50 - 20 = 30 s^{- 1}$ , toh $τ = 0.02 \times 30 = 0.6 Pa .$ Yeh step kyun? Same derivative, naye height par evaluate kiya.

Verify: $1.0 > 0.6$ ✓ — wall se door jaane par shear girta hai, forecast se match karta hai (profile flat hota jaata hai). Units $Pa\cdots \times s^{- 1} = Pa$ ✓.

Ex 6 — Limiting behaviour: gap zero tak sirakta hai (Cell F)

Worked example Jab

$y \to 0$ hota hai toh kya hota hai? Plate speed $u = 0.3$ m/s aur viscosity $μ = 0.1$ Pa·s fixed rakho. Gaps $y = 2$ mm, $0.5$ mm, $0.1$ mm ke liye $τ$ compute karo. Trend kya hai?

Forecast: same speed par thinner film squeeze karna = shear karna mushkil. Guess: $τ$ badhega.

Step 1 — Har gap par gradient: $d u / d y = u / y$ .

$y = 0.002$ : $150 s^{- 1}$
$y = 0.0005$ : $600 s^{- 1}$
$y = 0.0001$ : $3000 s^{- 1}$ Yeh step kyun? Same speed chhote gap par → steeper gradient.

Step 2 — $μ = 0.1$ se multiply karo: $τ = 15 Pa, 60 Pa, 300 Pa .$ Yeh step kyun? Naye gradients ke saath Newton's law.

Step 3 — Limit lo: jab $y \to 0$ , $\frac{u}{y} \to \infty$ , toh $τ \to \infty$ . Yeh step kyun? Fixed speed ko shrinking gap se divide karna bina bound ke badhta hai.

Verify: gap ko half karne par $τ$ double hota hai ( $15 \to$ ... $y$ 4 se cut ⇒ $τ \times 4 = 60$ ✓; 20 se cut ⇒ $\times 20 = 300$ ✓). Physically: isliye thin lubricant films enormous shear resist karte hain — bearings ki yahi basis hai.

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

Worked example Skater paani ki film par

Ek ice skater ka blade ( $A = 6.0 cm^{2}$ ) $10 μ m$ melt-water film ( $μ = 1.8 \times 1 0^{- 3}$ Pa·s) par $8$ m/s ki speed se glide karta hai. Skater ko resist karne wala viscous drag force kya hai?

Forecast: super-thin film + real speed → sizeable gradient, lekin water patla hai aur area tiny hai; expect karo ek newton ka fraction.

Step 1 — Words ko symbols mein translate karo: $A = 6.0 cm^{2} = 6.0 \times 1 0^{- 4} m^{2}$ , $y = 10 μ m = 1.0 \times 1 0^{- 5}$ m, $u = 8$ m/s. Yeh step kyun? Formula ko touch karne se pehle har quantity SI honi chahiye.

Step 2 — Gradient (linear film): $d u / d y = 8/ (1.0 \times 1 0^{- 5}) = 8.0 \times 1 0^{5} s^{- 1}$ . Yeh step kyun? Thin film, toh large gradient — drag wahan se aata hai.

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

Step 4 — Force: $F_{t} = τ A = 1440 \times 6.0 \times 1 0^{- 4} = 0.864$ N. Yeh step kyun? Blade area ke zariye stress ko wapas force mein convert karo.

Verify: ek fast skater par $\approx 0.86$ N ka drag believable hai (ice famously low-friction hai). Units check: Pa·s × s⁻¹ × m² = N ✓.

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

Worked example Kaun sa law apply hota hai?

Rubber ka ek block ( $G = 5 \times 1 0^{5}$ Pa) steady shear $τ = 2 \times 1 0^{4}$ Pa se push kiya jaata hai. (a) Uska steady shear strain $γ$ find karo. (b) Ek dost insist karta hai ki rubber "flows" aur $τ = μ \overset{γ}{˙}$ use karta hai. Kaun sahi hai?

Forecast: rubber tilt hoke ruk jaata hai — yeh solid behaviour hai, toh iska finite $γ$ hoga.

Step 1 — Material class decide karo. Rubber fixed deformation tak pahunchta hai aur use hold karta hai, toh yeh ek solid hai → $τ = G γ$ use karo, viscosity law nahi. Yeh step kyun? Pura topic is fork par depend karta hai: solids shear store karte hain ( $G γ$ ), fluids nahi kar sakte ( $μ \overset{γ}{˙}$ ). Law galat choose karna classic trap hai.

Step 2 — Strain solve karo: $γ = τ / G = (2 \times 1 0^{4}) / (5 \times 1 0^{5}) = 0.04$ (ek pure number = tilt ke radians). $γ = 0.04.$ Yeh step kyun? $τ = G γ \Rightarrow γ = τ / G$ .

Step 3 — Dost ko judge karo: ek fluid hamesha tilt karta rahta ( $\overset{γ}{˙} > 0$ forever). Rubber nahi karta — woh settle ho jaata hai. Toh dost galat hai; fluid law apply nahi hota. Yeh step kyun? Defining test hai "kya yeh deform karna jaari rakhta hai?" Rubber kehta hai nahi.

Verify: $γ = 0.04 rad \approx 2.3°$ tilt — chhota aur steady, bilkul waisa jaise ek solid behave kare. Stress and Strain in Solids se contrast karo. ✓

Ex 9 — Unit/sanity trap (Cell I)

Worked example mm-vs-m error dhundho

Ek student glycerin ( $μ = 1.4$ Pa·s) mein shear compute karta hai, plate speed $0.6$ m/s, gap $3$ mm, aur likhta hai $d u / d y = 0.6/3 = 0.2 s^{- 1}$ , toh $τ = 0.28$ Pa. Error dhundho aur sahi $τ$ batao.

Forecast: glycerin bahut thick hai; $0.28$ Pa suspiciously tiny lagta hai. Kuch gadbad hai.

Step 1 — Diagnose karo: student ne 3 mm ko 3 express karke divide kiya, metres mein convert karna bhool gaya. Yeh step kyun? $d u / d y$ hona chahiye (m/s)/(m); m/s ko mm ke saath mix karna silently answer mein 1000 ka error daal deta hai.

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

Step 3 — Correct stress: $τ = 1.4 \times 200 = 280$ Pa. $τ = 280 Pa .$ Yeh step kyun? Fixed gradient ke saath Newton's law.

Verify: sahi answer exactly galat answer ka $1000 \times$ hai ( $0.28 \to 280$ ) — yeh mm/m slip ki signature hai. Aur thick glycerin ke liye $280$ Pa reasonable hai, unlike $0.28$ Pa. ✓

Recall Quick self-test (right side cover karo)

Linear profile, gradient ke liye kaun sa formula? ::: $d u / d y = u / y$ . Curved profile, gradient ke liye kaun sa tool? ::: derivative $d u / d y$ locally evaluate kiya hua. Fluid at rest, shear stress? ::: bilkul zero (defining property). Fixed speed par gap $y \to 0$ hone par, $τ$ kya karta hai? ::: bina bound ke badhta hai ( $\to \infty$ ). Rubber steady shear ke under — kaun sa law? ::: $τ = G γ$ (yeh solid hai, finite strain). Numbers plug karne se pehle sabse pehle kya check karo? ::: mm/cm/µm ko metres mein convert karo.

Connections

Velocity Profile and No-Slip Condition — jahan Ex 5 jaisi curved profiles aati hain.
Viscosity and Newtonian vs Non-Newtonian Fluids — throughout use kiye gaye $μ$ par deeper dive.
Hydrostatics — Fluids at Rest — Ex 4 ki $\overset{γ}{˙} = 0$ duniya.
Stress and Strain in Solids — Ex 8 ka $τ = G γ$ side.
Pressure in Fluids — shear ka normal-stress companion.