Visual walkthrough — Angle of friction, angle of repose — derivation

1.2.8 · D2 · Physics › Newton's Laws & Dynamics › Angle of friction, angle of repose — derivation

Humein teen ideas chahiye honge jo doosri jagah se aate hain, aur jab-jab kaam aayenge tab-tab re-explain karenge: Static and Kinetic Friction se pushing forces, Block on an Inclined Plane se tilted-plane picture, aur Resolving Vectors into Components se ek arrow ko todne ki trick.

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Step 1 — Ek surface DO directions mein push karta hai

KYA. Ek block ko flat table par rakho aur apni ungli se use sideways slide karne ki koshish karo. Table do alag tareekon se ek saath resist karta hai.

KYUN. Isse pehle ki hum kisi force ke angles ki baat karein, humein pehle jaanna hai ki kaun sa force hai. Table pull nahi kar sakta, sirf push kar sakta hai. Picture dekho: yeh seedha upar push karta hai (block ko dhasnay se rokta hai) aur yeh sideways push karta hai (block ko grip karta hai taaki woh slide na ho). Hum har push ko ek naam dete hain:

• $N$ = normal force, seedha upar push. "Normal" ek purana word hai perpendicular ke liye — yeh surface se $90^{\circ }$ par point karta hai.
• $f$ = friction force, sideways grip. Yeh hamesha ulti direction mein point karta hai jisme block slide karne ki koshish kar raha hai.

PICTURE. Upar wala kala arrow $N$ hai. Horizontal kala arrow (tumhari ungli ki push ke ulti direction mein) $f$ hai. Do arrows, ek corner share karte hue — block ka contact point.

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Step 2 — Dono pushes ko ek arrow $R$ mein jodo

KYA. Ek point se do arrows ko ek single arrow se replace kiya ja sakta hai: woh diagonal jo un dono ke rectangle se banta hai. Us single arrow ko $R$ bolte hain.

KYUN. Surface ko nahi pata ki woh do forces de raha hai — woh split humara bookkeeping tha. Physically ek hi total shove hai. $N$ aur $f$ ko $R$ (total contact reaction) mein combine karne se hum ek saaf question puch sakte hain: surface actually kis direction mein push kar raha hai, aur woh seedhe-upar se kitna door hai?

PICTURE. Lal arrow $R$ hai, woh diagonal. Kyunki $N$ (upar) aur $f$ (sideways) right angle par milte hain, yeh ek right-angled rectangle ki do sides banate hain, aur $R$ uski diagonal hai.

$R=\sqrt{\underset{\text{upar wali side}}{\underbrace{N^2}}+\underset{\text{sideways side}}{\underbrace{f^2}}}$

Pythagoras kyun, simple addition kyun nahi? Agar do arrows same direction mein point karte to hum sirf lengths add kar lete. Yeh $90^{\circ }$ par point karte hain, isliye shortcut hypotenuse rule hai — woh tool jo exactly right angles ke liye bana hai.

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Step 3 — Angle $\lambda$ define karo, aur triangle se padho

KYA. Lal arrow $R$ seedhe-upar direction $N$ se door jhuk jaata hai. Yeh jhukav ek angle hai. Ise $\lambda$ (Greek letter "lambda") naam do.

KYUN. Hum ek number chahte hain jo capture kare "surface ka total push kitna tilted hai." $R$ aur normal $N$ ke beech ka angle woh number hai. Yahi ==angle of friction $\lambda$== hai.

PICTURE. Lal right triangle dekho: seedhi side $N$ hai, flat side $f$ hai, aur tirhki side $R$ hai. Angle $\lambda$ neeche wale corner par baith hai, seedhe $N$ aur diagonal $R$ ke beech.

"Kitna tilted hai" ko arithmetic mein convert karne ke liye hum tangent use karte hain:

$\tan \lambda =\frac{\text{side opposite to }\lambda }{\text{side next to }\lambda }=\frac{f}{N}$

Term by term:

• $f$ woh side hai jo angle $\lambda$ ke saamne hai ("opposite" side).
• $N$ woh side hai jo angle $\lambda$ ko chhoo rahi hai seedha upar ki taraf ("adjacent" side).
• Inका ratio $f/N$ steepness measure karta hai: same $N$ ke liye zyada sideways grip $f$ matlab zyada tilted $R$, matlab bada $\lambda$.

$\tan$ kyun, $\sin$ ya $\cos$ kyun nahi? $\sin$ ek side ko diagonal $R$ se compare karta hai; $\cos$ bhi. Lekin hum $f$ aur $N$ seedha jaante hain aur inhi ka ratio chahte hain. Tangent woh ek trig ratio hai jo triangle ki do legs se bana hai — $R$ compute karne ki zaroorat hi nahi.

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Step 4 — Verge tak push karo: $N$ cancel ho jaate hain

KYA. Apni ungli ki push badhao jab tak block abhi slide hone wala ho lekin abhi tak hua na ho. Friction ab apni absolute maximum par hai.

KYUN. Friction aalsee hai — yeh utni hi grip supply karta hai jitni zaroorat ho, ek ceiling tak. Woh ceiling (Static and Kinetic Friction se) hai:

$f_{\max }={\mu }_{s}N$

Yahan ${\mu }_s$ (coefficient of static friction) ek single number hai jo batata hai ki do surfaces kitne grippy hain. Rubber-on-concrete grippy hai (bada ${\mu }_s$); ice-on-ice slippery hai (chota ${\mu }_s$).

PICTURE. Jaise $f$ apni ceiling ${\mu }_{s}N$ ki taraf badhta hai, lal arrow $R$ vertical se door-door tilt hota jaata hai — jab tak apni steepest lean par pahunch jaaye. Woh steepest lean hi $\lambda$ hai.

Step 3 ke ratio mein ceiling substitute karo:

$\tan \lambda =\frac{f_{\max }}{N}=\frac{{\mu }_{s}N}{N}={\mu }_s$

Cancellation term by term dekho: upar ka $N$ aur neeche ka $N$ same normal force hain, isliye divide karke $1$ ho jaata hai. Jo bachta hai woh pure ${\mu }_s$ hai — block ka weight, uska size, kuch bhi matter nahi karta.

$\boxed{\lambda =\arctan ({\mu }_s)}$

Symbol $\arctan$ (jo ${\tan }^{-1}$ bhi likha jaata hai) ulta sawaal puchta hai: "kaun sa angle is tangent wala hai?" Agar $\tan \lambda ={\mu }_s$ hai, toh $\lambda =\arctan ({\mu }_s)$ woh angle hai jo jawab deta hai.

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Step 5 — Naya scene: poora plane tilt karo

KYA. Ab finger bhool jao. Same block lo aur surface ko hi dheere-dheere tilt karo, jaise kisi kitab ka ek sira uthao. $\theta$ tilt ko horizontal se measure karo.

KYUN. Hum doosra angle dhundh rahe hain — angle of repose, woh steepest tilt jis par block abhi bhi slide karne se mana karta hai. Yeh Block on an Inclined Plane wala setup hai. Ab gravity woh "pushing" kar rahi hai jo pehle hamari finger karti thi.

PICTURE. Lal arrow block ka weight $mg$ hai, seedha neeche point karta hua (gravity kabhi tilt nahi hoti). Plane $\theta$ se tilt hai. Notice karo weight ab plane ke axes ke saath aligned nahi hai — yeh neeche point karta hai jabki surface slanted hai. Yahi mismatch hai jise hum resolve karna chahte hain.

• $m$ = block ki mass, $g$ = gravity ki strength ($\approx 10{\text{m/s}}^2$). Toh $mg$ = block ka weight.

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Step 6 — Weight ko ramp ke saath split karo

KYA. Single weight arrow $mg$ ko do arrows mein todo: ek slope ke saath, ek slope mein.

KYUN. Motion, agar hoti hai, ramp ke neeche chalti hai. Toh natural axes hain along-the-ramp aur perpendicular-to-the-ramp (yeh Resolving Vectors into Components hai). $mg$ ko is tarah split karna us hisse ko alag kar deta hai jo sliding drive karta hai us hisse se jo surface mein press karta hai.

PICTURE. Lal weight arrow se, do components nikalo:

• Slope ke neeche (sliding drive karta hai): $mg\sin \theta$.
• Slope mein (surface press karta hai): $mg\cos \theta$.

Down-slope part ke liye $\sin$ kyun? Jaise zyada tilt karte ho (bada $\theta$), zyada weight ramp ke saath jhuk jaata hai — aur $\sin \theta$ $0$ se $1$ ki taraf badhta hai jaise $\theta$ $0^{\circ }\to 90^{\circ }$ jaata hai. Yeh driving force ko perfectly track karta hai. Meanwhile $\cos \theta$ $1$ se $0$ ki taraf ghat-ta hai — pressing force weak hoti jaati hai jaise ramp steep hota hai, bilkul intuition ke saath match karta hai (near-vertical wall bahut kam press karta hai).

Perpendicular direction balance mein hai (block na uda jaata hai na dhasnay hai):

$N=mg\cos \theta$

• $N$ = surface ka normal push, ab sirf pressing part $mg\cos \theta$ ko match karna hai, pure weight ko nahi.

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Step 7 — Verge condition: driving force friction ki ceiling se milti hai

KYA. Tilt karte raho jab tak down-slope pull just surface ki maximum friction ke barabar ho jaaye. Wahi tilt angle of repose hai.

KYUN. Is tilt se neeche, friction chupchap pull ko match karta hai aur kuch nahi hiltaa (Newton's First Law (Equilibrium) — forces balance hote hain, toh motion nahi). Special tilt par, pull friction ceiling tak pahunch jaata hai. Ek aur nudge aur friction haar jaata hai.

PICTURE. Do lal arrows ramp ke saath, nose to nose: down-slope pull $mg\sin \theta$ versus uphill friction $f_{\max }={\mu }_{s}N$. Repose par yeh equal length hain — ek perfect standoff.

$\underset{\text{driving pull}}{\underbrace{mg\sin \theta }}=\underset{\text{friction ceiling}}{\underbrace{{\mu }_{s}N}}={\mu }_s(mg\cos \theta )$

Humne Step 6 se $N=mg\cos \theta$ use kiya. Ab dono sides ko $mg\cos \theta$ se divide karo aur cheezein gayab hote dekho:

$\frac{mg\sin \theta }{mg\cos \theta }=\frac{{\mu }_{s}mg\cos \theta }{mg\cos \theta }$

• Dono sides ka $mg$ cancel ho jaata hai — mass chali gayi (bhaari aur halke dono blocks ek hi tilt par slip karte hain).
• Left par, $\frac{\sin \theta }{\cos \theta }$ by definition $\tan \theta$ hai.
• Right par, $mg\cos \theta$ apne aap se divide ho ke $1$ ho jaata hai, ${\mu }_s$ bachta hai.

$\tan {\theta }_r={\mu }_s⟹\boxed{{\theta }_r=\arctan ({\mu }_s)}$

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Step 8 — Edge & degenerate cases (koi gap mat chodo)

KYA. Extremes check karo taaki koi reader us scene par na atke jo humne skip kiya.

PICTURE. Teen ramps badhte tilt ke saath, flat se steep tak, reaction arrow ke saath.

• Frictionless surface, ${\mu }_s=0$. Toh ${\theta }_r=\arctan (0)=0^{\circ }$. Block kisi bhi tilt par slide kar jaata hai — sirf thodi si slope par bhi. Sense banta hai: bilkul grip nahi. Lal reaction $R$ seedha upar point karta hai ($\lambda =0$), kyunki friction ke bina sirf $N$ hai.
• Bilkul flat, $\theta =0^{\circ }$. Toh $mg\sin 0=0$: zero driving pull, isliye block apne aap kabhi slide nahi karta. Repose sirf ${\mu }_s=0$ ke liye flatness par pahuncha jaata hai, kabhi exceed nahi hota.
• Bahut grippy, ${\mu }_s=1$. Toh ${\theta }_r=\arctan (1)=45^{\circ }$. Exactly $1:1$ slope. Rubber on dry concrete yahan hota hai.
• Grip $1$ se zyada, ${\mu }_s>1$. Bahut sticky surfaces ke liye possible hai (silicone, kuch rubbers): $\arctan$ $45^{\circ }$ ke aage jaata hai lekin kabhi $90^{\circ }$ nahi pahunchta — kyunki $\tan 90^{\circ }$ infinite hai. Physically: koi ordinary friction ek truly vertical wall par block ko sirf grip se nahi rok sakti. Formula honestly predict karta hai: ${\theta }_r\to 90^{\circ }$ sirf tab jab ${\mu }_s\to \infty$.

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Ek-picture summary

Ek image, poori kahani. Left par, flat-surface triangle: $N$ upar, $f$ sideways, lal $R$ normal se $\lambda$ jhuka hua, $\tan \lambda =f/N={\mu }_s$ ke saath. Right par, tilted ramp: weight $mg\sin \theta$ aur $mg\cos \theta$ mein split, lal reaction ramp ke normal se ${\theta }_r$ jhuka hua, $\tan {\theta }_r={\mu }_s$ ke saath. Same ${\mu }_s$ dono ko feed karta hai — isliye dono jhuke angles ek hi hain, $\lambda ={\theta }_r=\arctan ({\mu }_s)$.

Recall Feynman: poora walkthrough simple words mein

Table sirf push kar sakta hai. Woh upar push karta hai tumhe hold karne ke liye (woh hai $N$) aur sideways push karta hai tumhe grip karne ke liye (woh hai $f$). Un dono pushes ko ek tilted arrow $R$ mein glue karo; woh seedhe-upar se kitna tilt hai woh hamara pehla angle hai, $\lambda$. Tilt grip-over-hold ke barabar hai, jo breaking point par exactly grippiness number $\mu$ hai. Toh $\lambda =\arctan \mu$.

Ab ek kitab tilt karo jab tak ek coin slide karne lage. Gravity coin ko seedha neeche khichti hai, lekin slope par us pull ka sirf ek hissa ramp ke neeche jaata hai ($mg\sin \theta$) jabki baaki press karta hai ($mg\cos \theta$). Coin usi waqt slip karta hai jab down-ramp pull grip ceiling ke barabar ho jaaye — aur jab tum yeh likhte ho, coin ka weight dono sides se cancel ho jaata hai, sirf $\tan \theta =\mu$ bachta hai. Toh woh tilt jahan yeh slip karta hai, angle of repose, bhi $\arctan \mu$ hai. Do alag experiments, ek grippiness number, ek answer. Angle of friction aur angle of repose twins hain.

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Connections

• Static and Kinetic Friction — jahan se ${\mu }_s$ aur ceiling $f_{\max }={\mu }_{s}N$ aate hain.
• Block on an Inclined Plane — Steps 5–7 ka tilted setup.
• Resolving Vectors into Components — Step 6 ka weight-splitting tool.
• Newton's First Law (Equilibrium) — kyun balanced forces matlab repose se neeche motion nahi.
• Banking of Roads — same $\arctan \mu$ logic safe cornering speeds ke liye.