Exercises — Coefficients of friction — measurement, material dependence

KYA: hum friction ÷ normal ka ratio chahte hain. KYUN: woh ratio hi coefficient ki definition hai — aur kuch chahiye nahi. ${\mu }_s=\frac{f_{s,\max }}{N}=\frac{20}{50}=0.40$ Dhyaan do: koi unit nahi. Force ÷ force = pure number. Isliye $\mu$ ko dimensionless kaha jaata hai.

KYA: "jis angle par yeh slip karna shuru karta hai" woh angle of repose hai. WOH TOOL KYUN: incline par, $mg\sin \theta$ sliding drive karta hai aur $mg\cos \theta$ surfaces ko press karta hai; verge par dono satisfy karte hain $mg\sin {\theta }_c={\mu }_{s}mg\cos {\theta }_c$, aur mass cancel ho jaata hai, bacha rehta hai ${\mu }_s=\tan {\theta }_c=\tan 30^{\circ }\approx \mathrm{0.577.}$ Yahan ek protractor force meter se zyada kaam ka hai — koi scale nahi, koi mass nahi, sirf angle chahiye.

Zyada force → zyada coefficient (dono same $N$ se divide hote hain), isliye ${\mu }_s>{\mu }_k$. KYUN expected: rest mein microscopic bumps settle hoke ek doosre ki valleys mein "cold-weld" kar lete hain, isliye break free karne mein keep skimming se zyada force laagta hai. Yahi rule hai ==${\mu }_s\ge {\mu }_k$== — "Static Sticks, Kinetic Skips".

Step 1 — $N$ nikalo. KYUN $N=mg$: floor horizontal hai aur crate ka koi vertical acceleration nahi, isliye vertical force balance se upward normal force downward weight ko exactly cancel karni chahiye — vertically aur kuch act nahi karta. Isliye: $N=mg=6\times 9.8=58.8\text{N}.$ Step 2 — starting force = verge of slip. KYUN ${\mu }_s=F_{\max }/N$: $21\text{N}$ jo just start karne ke liye chahiye woh maximum static friction $f_{s,\max }$ hai (isse kam mein woh move hi nahi hota). Isse seedha definition $\mu =f/N$ mein daalo: ${\mu }_s=\frac{f_{s,\max }}{N}=\frac{21}{58.8}\approx \mathrm{0.357.}$ Step 3 — constant speed matlab $a=0$. KYUN reading $f_k$ ke barabar hai: zero acceleration ke saath, Newton's Second Law kehta hai net horizontal force zero hai, isliye applied $15\text{N}$ kinetic friction ko exactly balance karta hai — reading ko confuse karne ke liye koi leftover force nahi. Phir ${\mu }_k=f_k/N$: ${\mu }_k=\frac{15}{58.8}\approx \mathrm{0.255.}$ Sanity check: ${\mu }_s>{\mu }_k$. ✓

Step 1 — ceiling. KYUN $f_{s,\max }={\mu }_{s}N$ ceiling hai: surface jo sabse zyada static friction supply kar sakti hai woh definition ko rearrange karne se milti hai, $f_{s,\max }={\mu }_{s}N$ — yahi maximum grip available hai slip hone se pehle. $N=mg=2\times 9.8=19.6\text{N}$ ke saath: $f_{s,\max }={\mu }_{s}N=0.5\times 19.6=9.8\text{N}.$ Step 2 — push ko ceiling se compare karo: tumhara push $6\text{N}<9.8\text{N}$ hai, isliye demand ceiling se neeche rehti hai aur book ruki rehti hai. Step 3 — actual friction. KYUN yeh push ke barabar hai: kyunki book move nahi ho rahi, uska horizontal acceleration zero hai, isliye friction jo bhi tum apply kar rahe ho use exactly cancel karta hai — na zyada, na kam. Friction khud ko $6\text{N}$ par adjust kar leta hai ($9.8\text{N}$ nahi!). Yahi L1 trap action mein hai.

KYUN ${\mu }_k$ na ki ${\mu }_s$: block already moving hai, isliye kinetic friction in play hai. Constant velocity → $a=0$, isliye slope ke saath driving pull kinetic friction ke barabar hai: $mg\sin \theta ={\mu }_{k}mg\cos \theta .$ KYUN mass cancel hota hai: exactly same factor $mg$ dono sides ko multiply karta hai — driving term $mg\sin \theta$ aur friction term ${\mu }_{k}mg\cos \theta$ dono ko. Dono sides ko $mg$ se divide karne par woh completely hat jaata hai, isliye mass kabhi matter nahi karta: ${\mu }_k=\tan \theta =\tan 18^{\circ }\approx \mathrm{0.325.}$ Same tangent trick — mass phir cancel ho gaya.

Step 1 — weight decompose karo (figure dekho: slate $mg$ arrow ek red along-slope part $mg\sin \theta$ mein split hota hai jo block ko ramp se neeche kheenchta hai, aur ek mint into-slope part $mg\cos \theta$ mein jo ise surface ke against press karta hai; lavender arrow friction $f$ hai jo ise up-slope hold karti hai).

• Driving (down-slope) part: $mg\sin \theta =3\times 9.8\times \sin 20^{\circ }\approx 10.05\text{N}$.
• Pressing part: $N=mg\cos \theta =3\times 9.8\times \cos 20^{\circ }\approx 27.63\text{N}$.

Step 2 — friction ceiling: $f_{s,\max }={\mu }_{s}N=0.6\times 27.63\approx 16.58\text{N}.$ Step 3 — compare karo. Jo pull ise slide karne ki koshish kar rahi hai ($10.05\text{N}$) ceiling ($16.58\text{N}$) se kam hai, isliye woh slide nahi karta. (b) Kyunki woh static aur stationary hai, friction exactly wahi hai jo down-slope pull cancel kare: $f_s=mg\sin \theta \approx 10.05\text{N}(\text{up the slope}).$ $16.58\text{N}$ nahi — woh ceiling unused hai.

Alternative one-line test: $\tan \theta =\tan 20^{\circ }\approx 0.364$ ko ${\mu }_s=0.6$ se compare karo. Kyunki $\tan \theta <{\mu }_s$ hai, koi slip nahi. Faster hai, aur mass kabhi appear nahi karta.

Angle of repose → static: ${\mu }_s=\tan 25^{\circ }\approx 0.466$. Constant-speed slide → kinetic: ${\mu }_k=\tan 21^{\circ }\approx 0.384$. Consistency: ${\theta }_{\text{slide}}<{\theta }_c$ matlab ${\mu }_k<{\mu }_s$ ✓ — ek baar glide karna shuru hone par shallower tilt hold kar sakte ho, exactly wahi "sticks vs skips" wali baat.

Forecast: unchanged, $25^{\circ }$. Mass kabhi ${\mu }_s=\tan {\theta }_c$ mein enter hi nahi kiya. Verify: mass $2m$ ke saath, verge-of-slip balance hai $2mg\sin {\theta }_c={\mu }_s(2mg\cos {\theta }_c).$ Poora factor $2mg$ divide ho jaata hai: $\tan {\theta }_c={\mu }_s.$ Same equation, same angle. Isliye akela protractor ${\mu }_s$ measure karta hai — koi scale required nahi.

KYUN ab Newton's Second Law: woh accelerate kar raha hai, isliye hum slope ke saath $\sum F=ma$ chahte hain, balance nahi. Dekho Newton's Second Law. Figure padho: block $35^{\circ }$ ramp par hai; red arrow $mg\sin \theta$ ise down-slope kheenchta hai, lavender friction arrow $f_k={\mu }_{k}mg\cos \theta$ up-slope point karta hai (motion oppose karta hai), aur mint arrow $mg\cos \theta$ normal press set karta hai. Red minus lavender ka net hi ise accelerate karta hai. Along-slope forces (neeche positive):

• Driving: $mg\sin \theta =3\times 9.8\times \sin 35^{\circ }\approx 16.86\text{N}$ (neeche).
• Kinetic friction motion oppose karta hai, isliye up slope point karta hai: $f_k={\mu }_{k}N={\mu }_{k}mg\cos \theta =0.45\times 3\times 9.8\times \cos 35^{\circ }\approx 10.83\text{N}$.

Newton's Second Law: $ma=mg\sin \theta -{\mu }_{k}mg\cos \theta .$ $m$ se divide karo (mass phir cancel!): $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(\sin 35^{\circ }-0.45\cos 35^{\circ })\approx 2.01{\text{m/s}}^2\text{(down slope)}.$ Positive → woh sach mein neeche accelerate karta hai, "it slides" ke consistent hai.

Step 1 — static coefficient se slip test: $\tan \theta$ ko ${\mu }_s$ se compare karo. $\tan 35^{\circ }\approx 0.700,{\mu }_s=\mathrm{0.55.}$ Step 2 — interpret: $\tan \theta >{\mu }_s$ matlab down-slope pull static ceiling exceed kar raha hai, isliye block break free hoke rest se sliding shuru karta hai. Step 3 — phir coefficients switch karo: ek baar move karne par, kinetic friction (${\mu }_k$) take over karta hai, L4.1 mein mila acceleration deta hai. Yahi poora chain hai: ${\mu }_s$ se test karo → agar slip hoti hai, ${\mu }_k$ se motion compute karo.

Step 1 — Method A se ${\mu }_s$ nikalo: ${\mu }_s=\tan 30^{\circ }\approx 0.577$. Step 2 — Method B ki reading predict karo: flat ground par $N=mg=5\times 9.8=49\text{N}$, aur slip ke verge par $F_{\max }={\mu }_{s}N=0.577\times 49\approx 28.3\text{N}.$ Agar spring scale lagbhag $28\text{N}$ read kare, toh do independent methods agree karte hain — ek achha internal check ki $\mu$ surface pair ki genuine property hai, apparatus ki nahi.

(a) ${\theta }_c=0^{\circ }$: ${\mu }_s=\tan 0^{\circ }=0$. Ek flat table jo block ko zero tilt par slip hone de matlab surfaces perfectly slippery hain — bilkul koi grip nahi. Sensible. (b) ${\theta }_c\to 90^{\circ }$: $\tan {\theta }_c\to \infty$. Ek vertical wall jis se block chipka rahe usse infinite grip chahiye. Reality mein ordinary materials ke liye ${\theta }_c$ kabhi $90^{\circ }$ tak nahi pahunchta, isliye ${\mu }_s$ finite rehta hai — formula warn karta hai ki near-vertical repose ke liye enormous friction chahiye. Limit ke roop mein sensible. (c) Frictionless (${\mu }_s=0$): $\tan {\theta }_c=0$ solve karne par ${\theta }_c=0^{\circ }$ milta hai — block slightest tilt par slip karta hai. Intuition se match karta hai: zero friction ke saath flat se upar koi stable resting angle nahi hai.

Microscopic sach: surfaces sirf tiny asperity tips par touch karti hain. Real contact area $A_{\text{real}}$ apparent face area se kaafi chhota hai, aur yeh surfaces ko kitna dabaaya ja raha hai uspe depend karta hai: $A_{\text{real}}\propto N$. Zyada press → tips flatten → zyada true contact. Consequence: friction $\propto A_{\text{real}}\propto N$, isliye $f=\mu N$ jahan $\mu$ apparent face se independent hai. Brick ko end par khada karne par same weight chhote apparent area par concentrate hota hai, lekin real contact area (aur isliye friction) unchanged rehta hai. Incline test: brick ${\theta }_c$ par slip karna shuru karega chahe apne face par ho ya end par — tangent, aur isliye ${\mu }_s$, nahi badlega. Yahi parent note ka "Why area doesn't appear" callout hai, testable bana diya.

Dry: ${\theta }_c=\arctan 0.6\approx 31.0^{\circ }$. Oiled: ${\theta }_c=\arctan 0.1\approx 5.71^{\circ }$. Oiled surface bahut shallow tilt par slip karta hai — kaafi kam grip. Physical meaning: patla oil film asperity tips ko cold-weld aur interlock hone se rokta hai; woh ek doosre mein ghusne ki jagah fluid layer par ride karte hain. $\mu$ number isliye drop karta hai kyunki interface change ho gaya — yeh reinforce karta hai ki $\mu$ pair plus jo kuch unke beech hai describe karta hai, sirf metal ko nahi. Dekho Lubrication & Tribology.

Starting force se static $\mu$ ::: flat ground par ${\mu }_s=F_{\max }/(mg)$. Incline par one-line slip test ::: $\tan \theta$ ko ${\mu }_s$ se compare karo; slips if $\tan \theta >{\mu }_s$. Sliding shuru hone ke baad, kaun sa coefficient acceleration set karta hai ::: ${\mu }_k$, via $a=g(\sin \theta -{\mu }_k\cos \theta )$. Kya mass double karne se repose angle change hota hai ::: Nahi — ${\mu }_s=\tan {\theta }_c$ mein $mg$ cancel ho jaata hai. Kya resting incline par friction hamesha ${\mu }_{s}N$ hoti hai ::: Nahi, yeh $mg\sin \theta$ ke barabar hoti hai slip ke verge tak.