Exercises — Fluid definition — shear stress, no fixed shape

Ek fluid define hota hai continuous deformation under any shear, however small se.

• Steel bar → thoda sa fixed amount tilt karta hai aur ruk jaata hai → solid.
• Air → flows, apna container fill karta hai → fluid (gas ek fluid hai).
• Honey → flows (slowly, high $\mu$) → fluid.
• Rubber → ek fixed amount deform karta hai aur hold karta hai → solid.
• Water → flows → fluid. Fluids: air, honey, water.

Dono "force per area" hain, surface ke relative direction se split ki gayi: $\sigma =\frac{F_n}{A}(\perp \text{, normal}),\tau =\frac{F_t}{A}(\parallel \text{, shear}).$ Shear part woh hai jise rest mein ek fluid resist nahi kar sakta.

Static rest par ek fluid zero shear stress sustain karta hai — yeh uski defining property hai. Koi motion nahi, isliye $du/dy=0$, aur $\tau =\mu \cdot 0=0$. (Wahan phir bhi normal stress hoti hai — pressure — lekin koi shear nahi.)

Kyun linear ⇒ constant gradient: kyunki bottom fluid fixed wall se stick karta hai (speed $0$) aur top fluid plate se stick karta hai (speed $0.6$), aur beech mein kuch bhi line ko bend nahi karta, speed ek steady rate par climb karti hai. Ek straight line har jagah ek slope rakhti hai, aur woh slope exactly $du/dy$ hai. Isliye gradient = (speed difference)/(gap). Pehle convert karo: $3\text{mm}=0.003\text{m}$. $\frac{du}{dy}=\frac{0.6-0}{0.003}=200{\text{s}}^{-1}.$ Kaisa dikhta hai: Figure 1 upar mein, straight amber line $0.003\text{m}$ ki height par $0.6\text{m/s}$ rise karti hai — ek steep, constant slope. Har cyan arrow neeche wale se ek fixed fraction zyada lamba hai.

Kyun $\tau =\mu du/dy$ yahan apply hota hai: glycerine ek Newtonian fluid hai, aur woh plates ke beech move kar raha hai, isliye har layer agle ke upar slide karti hai. Newton's law kehta hai adjacent layers ke beech sideways drag is baat ke proportional hai ki unki speeds per unit height kitni different hain — woh difference-per-height exactly $du/dy$ hai, aur $\mu$ proportionality ka constant hai. Constant gradient ke saath stress pure film mein same hai: $\tau =\mu \frac{du}{dy}=1.5\times 200=300\text{Pa}.$

Kyun area se multiply karte hain: $\tau$ force per unit area hai. Puri plate fluid ko area $A$ par touch karti hai, aur same stress har patch par act karta hai, isliye total tangential force = stress × area. $\tau =F_t/A$ ko invert karke: $F_t=\tau A=300\times 0.25=75\text{N}.$

Kyun hum $\mu$ solve kar sakte hain: Newton's law $\tau =\mu du/dy$ teen quantities ko link karta hai; koi bhi do diye hone par hum teesra nikal sakte hain. Yahan $\tau$ aur $du/dy$ measured hain, isliye $\mu$ bas unka ratio hai — yahi literally lab mein viscosity measure karne ka tarika hai (ek viscometer): $\mu =\frac{\tau }{du/dy}=\frac{48}{120}=0.4\text{Pa⋅s}.$

Yahan profile curved hai, isliye $du/dy$ constant nahi hai — hume differentiate karna padega. Step 1 (kya aur kyun): velocity gradient $u(y)$ ka slope hai, isliye differentiate karo: $\frac{du}{dy}=\frac{d}{dy}\left(500y^2\right)=1000y.$ Step 2: top par evaluate karo, $y=h=0.01$: ${\frac{du}{dy}|}_{y=h}=1000\times 0.01=10{\text{s}}^{-1}.$ Step 3: Newton's law apply karo: $\tau =\mu \frac{du}{dy}=0.02\times 10=0.2\text{Pa}.$ Kaisa dikhta hai: neeche figure mein curved amber profile bottom par almost flat hai (small slope, small $\tau$) aur top par steepest hai (largest slope, largest $\tau$).

$\frac{du}{dy}=1000y$, aur $y=0$ par yeh $1000\times 0=0$ hai. $\tau =\mu \times 0=0\text{Pa}.$ Parabolic profile ka bottom momentarily flat hota hai (zero slope), isliye wahan shear zero hai. Jab profile curved ho toh stress har jagah same nahi hota.

$\tau =\mu du/dy$ se, gradient $du/dy=\tau /\mu$ hai. Same $\tau$, isliye gradient inversely proportional to $\mu$ hai. ${\left(\frac{du}{dy}\right)}_A=\frac{20}{0.8}=25{\text{s}}^{-1},{\left(\frac{du}{dy}\right)}_B=\frac{20}{0.1}=200{\text{s}}^{-1}.$ Fluid B (thinner wala) $200/25=8$ times tez shear karta hai. Ratio = ${\mu }_A/{\mu }_B=8$.

Do different materials ke liye do different laws. Solid ($\tau =G\gamma$): strain angle ek fixed value par pahuncha aur ruk jaata hai — square ek permanent lean $\gamma$ par tilt hota hai aur ise hold karta hai. $\gamma =\frac{\tau }{G}=\frac{5}{2.5\times 10^5}=2\times 10^{-5}\text{(radians, ek fixed tilt)}.$ Fluid ($\tau =\mu \dot{\gamma }$, jahan $\dot{\gamma }=du/dy$): tilt kabhi nahi rukta; yeh steady rate par grow karta hai. $\dot{\gamma }=\frac{\tau }{\mu }=\frac{5}{0.05}=100{\text{s}}^{-1}.$ Interpretation: solid ek permanent, tiny angle $\gamma =2\times 10^{-5}$ rad tilt karta hai aur ise hold karta hai (woh stress ko spring ki tarah store karta hai). Fluid ka tilt $\dot{\gamma }=100$ radians per second ki rate se hamesha grow karta hai — woh flows. Note karo $\gamma$ ek dimensionless angle hai, jabki $\dot{\gamma }$ units ${\text{s}}^{-1}$ carry karta hai aur velocity gradient ke equal hai. Yahi solid aur fluid ke beech poora distinction hai.

(a) Force. Pehle gradient: $du/dy=2/0.005=400{\text{s}}^{-1}$. Phir $\tau =\mu \frac{du}{dy}=0.3\times 400=120\text{Pa},F_t=\tau A=120\times 0.4=48\text{N}.$ (b) Power. Power force times woh speed hai jis par woh move karti hai: $P=F_{t}u=48\times 2=96\text{W}.$ Kyun force × speed? Power kaam karne ki rate hai; kaam force × distance hai, aur distance ÷ time = speed. Isliye jo drag tum feel karte ho woh tumhari $96$ joules effort har second oil ke andar heat mein convert karta hai.

Tall dome hold karne ke liye, slanted outer surfaces ko shear forces support karni padti hain (gravity sides ko neeche aur bahar khichti hai, jo un surfaces ke along ek tangential pull hai).

• Steel: $\tau =G\gamma$ obey karta hai. Woh required shear ko ek tiny finite strain $\gamma$ se balance kar sakta hai, equilibrium reach kar sakta hai, aur dome hold kar sakta hai.
• Water: rest mein woh zero shear sustain kar sakta hai ($G_{\text{static}}=0$, isliye koi bhi residual shear forces $\dot{\gamma }>0$ deti hain). Woh tangential pull balance nahi kar sakta, isliye woh deform karta rehta hai — woh slump aur spread karta hai jab tak sirf normal (vertical, pressure) forces remain na ho jaayein, yaani ek flat puddle. Isliye no fixed shape.

Key insight — kyun dono layers mein stress equal hai (interface par Newton's laws): do layers ke beech boundary par fluid ki thin sheet picture karo. Bottom layer usey stress ${\tau }_1$ (ek force ${\tau }_{1}A$) se pull karti hai, aur top layer stress ${\tau }_2$ (ek force ${\tau }_{2}A$) se, opposite directions mein. Is interface sheet ka essentially koi mass nahi hai. Newton's second law kehta hai force = mass × acceleration; mass $\approx 0$ ke saath, koi bhi unbalanced force infinite acceleration deta — impossible. Isliye forces balance karne chahiye: ${\tau }_{1}A={\tau }_{2}A$, hence ${\tau }_1={\tau }_2=\tau$. Shear stress dono layers mein same hai — exactly jaise same current series mein resistors mein flow karta hai. Step 1: interface speed ko $u_i$ maano. Har layer linear hai, isliye uska gradient (speed jump)/(uski thickness) hai: $\tau ={\mu }_1\frac{u_i-0}{y_1}={\mu }_2\frac{u-u_i}{y_2}.$ Step 2: numbers plug karo ($y_1=0.001$, $y_2=0.002$, $u=1$): $0.2\frac{u_i}{0.001}=0.5\frac{1-u_i}{0.002}\Rightarrow 200u_i=250(1-u_i).$ Step 3: solve karo: $200u_i=250-250u_i\Rightarrow 450u_i=250\Rightarrow u_i=0.5\overline{5}\text{m/s}\approx 0.556\text{m/s}$. Step 4: shear stress: $\tau =200u_i=200\times 0.5556\approx 111.1\text{Pa}.$ (a) $\tau \approx 111.1\text{Pa}$, (b) $u_i\approx 0.556\text{m/s}$. Kaisa dikhta hai: velocity profile ek bent straight line hai — thinner, less-viscous bottom layer mein steeper (bigger slope), thicker, more-viscous top layer mein gentler — lekin dono segments same stress feel karte hain.

Force limit se peeche chalt ke formulas chain karo. $F_t=\tau A=\left(\mu \frac{u}{y}\right)A\le F_{\text{max}}\Rightarrow \mu \le \frac{F_{\text{max}}y}{Au}.$ $\mu \le \frac{30\times 0.004}{0.6\times 1.5}=\frac{0.12}{0.9}\approx 0.133\text{Pa⋅s}.$ $\mu \le 0.133\text{Pa⋅s}$ wala oil choose karo.

• (a) $\mu \to 0$: $\tau =\mu du/dy\to 0$ kisi bhi finite gradient ke liye — ek ideal (inviscid) fluid move karte hue bhi koi shear stress carry nahi karta. (Real fluids ka hamesha small but nonzero $\mu$ hota hai, isliye yeh ek idealisation hai, kabhi exactly reach nahi hoti.)
• (b) $du/dy\to 0$: $\tau =\mu \cdot 0\to 0$. Yeh static case hai — rest mein fluid zero shear sustain karta hai, uski defining property. Layers ke beech koi motion difference nahi matlab unke beech koi drag nahi.
• (c) $\mu \to \infty$: $\tau$ finite rakhne ke liye tumhein $du/dy\to 0$ chahiye, yaani fluid barely shear karta hai — neighbouring layers almost same speed par move karne ko forced hain, isliye material rigid solid ki tarah behave karta hai. Infinite viscosity ≈ solid: woh flow karna band kar deta hai.
• (d) $y\to 0$ fixed plate speed par: linear profile ke saath $du/dy=u/y$, isliye jab gap shrink hota hai $du/dy=u/y\to \infty$, aur isliye $\tau =\mu u/y\to \infty$. Film ko zero thickness tak squeeze karna required shear stress (aur drag force $F_t=\tau A$) ko blow up kar deta hai — yahi exactly reason hai kyun extremely thin lubricating films huge stresses transmit kar sakti hain aur kyun bearings kabhi truly dry nahi run ki jaati.

Upar ki sab cheez do ideas hain baar baar apply ki gayi: shear stress = tangential force per area ($\tau =F_t/A$), aur ek moving Newtonian fluid ke liye, $\tau =\mu du/dy$. Straight profile → gradient speed/gap hai. Curved profile → differentiate karo aur ek point par evaluate karo. Stacked layers → same stress (massless interface par force balance), different slopes. Rest mein → $\tau =0$, jo ise fluid banata hai.