Angle of friction, angle of repose — derivation: Worked examples

1.2.8 · D3 · Physics › Newton's Laws & Dynamics › Angle of friction, angle of repose — derivation

Yeh page parent derivation ke liye drill ground hai. Humne derive kiya tha ki friction surface ko ek tilted reaction deta hai, aur ek ramp jo $arctan (μ_{s})$ tak tip ho jaaye woh slipping ke verge par hoti hai. Ab hum us ek idea ko har us case class mein run karte hain jisme woh appear ho sakta hai — small angles, exact $45°$ pivot, steep angles, zero-friction limit, push-along-the-floor problem, aur ek exam twist jahan answer simply $arctan (μ)$ nahi hota.

Koi bhi symbol aane se pehle: $μ_{s}$ (padho "mew-sub-ess") static coefficient of friction hai, ek plain number jo measure karta hai ki ek surface pair kitna "grippy" hai — dekho Static and Kinetic Friction. $arctan (x)$ (padho "arc-tangent of $x$ ") yeh sawaal hai "kaun sa angle hai jiska tangent $x$ ke barabar hai?". Kisi angle ka $tan (θ)$ us angle ke right triangle ki (opposite side) $/$ (adjacent side) ka ratio hai — angle jitna steep hoga, yeh ratio utna bada hoga.

Scenario matrix

Is topic ke har problem ka answer in cells mein se ek mein aata hai. Neeche ke worked examples mein har ek par cell(s) ka label laga hai.

Cell	Isko alag kya banata hai	Covered by
A. Small angle ( $θ_{r} < 45°$ , so $μ_{s} < 1$ )	gentle slope, "typical" surface	Ex 1
B. The pivot ( $θ_{r} = 45°$ , $μ_{s} = 1$ )	tangent exactly one ke barabar	Ex 2
C. Steep angle ( $θ_{r} > 45°$ , $μ_{s} > 1$ )	bahut grippy (rubber, rough rock)	Ex 3
D. Degenerate: $μ_{s} = 0$	frictionless limit, angle $\to 0$	Ex 4
E. Reaction geometry (horizontal floor, find $R$ and $λ$ )	$N$ aur $f$ ka vector sum	Ex 5
F. Below vs at vs above repose (kya yeh move kar raha hai?)	inequality reasoning, sirf equality nahi	Ex 6
G. Real-world word problem	ek kahani ko $arctan (μ)$ mein translate karo	Ex 7
H. Exam twist: extra applied force	answer plain $arctan (μ)$ NAHI hai	Ex 8

Case A — Small angle

Worked example Ex 1 (Cell A) — Tilting book par coin

Ek coin slide karna shuru karta hai jab ek book ko slowly $θ_{r} = 20°$ tak tilt kiya jaata hai. $μ_{s}$ find karo, aur predict karo ki yeh surface $μ_{s} = 1$ se "grippier" hai ya "more slippery".

Forecast: $20°$ kaafi $45°$ se neeche hai. Andaza lagao: kya $μ_{s}$ $1$ se upar hoga ya neeche?

Step 1. Repose condition likhte hain $tan θ_{r} = μ_{s}$ . Yeh step kyun? Verge par, down-slope pull $m g sin θ$ exactly friction limit $μ_{s} m g cos θ$ ke barabar hota hai; divide karne par $tan θ_{r} = μ_{s}$ milta hai ( $m g$ cancel ho jaata hai).

Step 2. $μ_{s} = tan 20° \approx 0.364$ . Yeh step kyun? $arctan$ aur $tan$ ek doosre ko undo karte hain; hum sirf measured tilt ka tangent evaluate karte hain.

Verify: $0.364 < 1$ , toh yeh pivot case se more slippery hai — ek chhoti si tilt par sliding hona iske consistent hai. Feeding it back: $arctan (0.364) = 20°$ . ✓

Case B — $45°$ par pivot

Case C — Steep angle

Case D — Degenerate frictionless limit

Case E — Horizontal floor par reaction geometry

Figure — Angle of friction, angle of repose — derivation

Worked example Ex 5 (Cell E) — Pushed block ka total reaction

Ek $2 kg$ block horizontal floor par rakha hai, $μ_{s} = 0.5$ , aur horizontally push kiya jaata hai jab tak yeh slide karne ke just verge par na ho. $g = 10 m/s^{2}$ lo. Find karo (a) total contact reaction $R$ ka angle $λ$ vertical se, aur (b) $R$ ki magnitude.

Forecast: Kya $R$ weight $20 N$ se bada hoga ya chhota? Computing se pehle guess karo.

Step 1. Normal balances weight: $N = m g = 2 \times 10 = 20 N$ . Yeh step kyun? Vertically block accelerate nahi karta, isliye $N = m g$ (equilibrium).

Step 2. Maximum friction: $f = μ_{s} N = 0.5 \times 20 = 10 N$ . Yeh step kyun? "Just about to slide" matlab friction apni ceiling $μ_{s} N$ tak pahunch chuka hai.

Step 3. Reaction ka angle: $λ = arctan (f / N) = arctan (0.5) \approx 26.57°$ . Yeh step kyun? Figure dekho — $N$ seedha upar point karta hai ( $λ$ ke adjacent), $f$ sideways point karta hai ( $λ$ ke opposite); tangent hai opposite/adjacent $= f / N$ .

Step 4. Magnitude: $R = N^{2} + f^{2} = 2 0^{2} + 1 0^{2} = 500 \approx 22.36 N$ . Yeh step kyun? $N$ aur $f$ perpendicular hain, isliye unka vector sum hypotenuse hai — Pythagoras (dekho Resolving Vectors into Components).

Verify: $R = N 1 + μ_{s}^{2} = 20 1.25 \approx 22.36 N$ — same number, toh shortcut aur direct sum agree karte hain. Units: poora newtons mein. ✓ Aur $R > 20 N$ (friction ek sideways slice add karta hai), forecast se match karta hai.

Case F — Below / at / above repose

Worked example Ex 6 (Cell F) — Teen alag tilts par kya yeh move karta hai?

Ek ramp ka $μ_{s} = 0.577$ hai (toh $θ_{r} = 30°$ ). Ek block uske upar rakha hai. Decide karo ki yeh (i) $θ = 20°$ , (ii) $θ = 30°$ , (iii) $θ = 40°$ par ruka rehega ya nahi.

Forecast: Kaam karne se pehle har ek ka outcome guess karo.

Step 1. Block ruka tabhi aur tabhi rehega jab required friction $m g sin θ$ ceiling $μ_{s} m g cos θ$ se exceed nahi karta, yaani $tan θ \leq μ_{s}$ . Yeh step kyun? Equilibrium ke liye friction ko poora down-slope pull supply karna padta hai; lekin friction zyaada se zyaada $μ_{s} N$ supply kar sakta hai. Yeh ek inequality hai, sirf verge par equality nahi.

Step 2 (i). $θ = 20°$ : $tan 20° \approx 0.364 < 0.577$ . Ruka rehega (friction spare capacity hai). Kyun? Required grip ceiling se neeche hai.

Step 3 (ii). $θ = 30°$ : $tan 30° \approx 0.577 = μ_{s}$ . Verge par — exact repose angle. Kyun? Required grip exactly ceiling ke barabar hai; thodi si bhi extra tilt ise tod degi.

Step 4 (iii). $θ = 40°$ : $tan 40° \approx 0.839 > 0.577$ . Slide karta hai (balance karna impossible hai). Kyun? Demand friction ceiling se zyaada hai; surplus $m g (sin θ - μ_{s} cos θ)$ block ko down-slope accelerate karta hai.

Verify: Teen verdicts stay / verge / slide, $θ_{r} = 30°$ ko exactly straddle karte hain jaise matrix predict karta hai. Figure mein, red down-slope arrow green friction cap ko sirf $30°$ ke baad overtake karta hai. ✓

Case G — Real-world word problem

Worked example Ex 7 (Cell G) — Grain silo cone

Dry sand zaamin par daalne par naturally ek cone banata hai jiske sides horizontal se $34°$ banate hain — yeh literally uska angle of repose hai. Sand grains ke beech effective $μ_{s}$ estimate karo, aur hence $10 m$ base radius wale cone ki height nikalo.

Forecast: Pile $34°$ se zyaada steep hone se mana karta hai — yeh grain-on-grain grip ke baare mein kya batata hai?

Step 1. $μ_{s} = tan θ_{r} = tan 34° \approx 0.675$ . Yeh step kyun? Ek settled pile apne steepest stable slope par hoti hai — woh hi angle of repose hai, isliye wahi $tan θ_{r} = μ_{s}$ relation loose grains par bhi apply hota hai.

Step 2. Cone height $h = r tan θ_{r} = 10 \times tan 34° \approx 6.75 m$ . Yeh step kyun? Cone ka slant base se $θ_{r}$ angle banata hai; height/radius exactly $tan θ_{r}$ hai (us slope triangle ka opposite/adjacent).

Verify: $h / r = 6.75/10 = 0.675 = μ_{s} = tan 34°$ . ✓ Units: metres. Wahi $arctan (μ)$ idea jo ek ramp ko limit karta hai woh sand pile kitni tall ho sakti hai yeh bhi cap karta hai.

Case H — Exam twist (answer plain arctan μ NAHI hai)

Worked example Ex 8 (Cell H) — Extra downward push ke saath repose

Ek rough incline ( $μ_{s} = 0.5$ ) par ek block ko ek extra force $P$ se slope ke against press kiya jaata hai jo surface ke perpendicular (uske andar) directed hai. Sliding ke verge par, kya critical tilt abhi bhi $arctan (0.5)$ hai? General mein condition work out karo aur comment karo.

Forecast: Ek inward push add karne se block ko hold karna zyaada mushkil hoga ya aasaan? Guess karo ki repose angle kaise shift hota hai.

Step 1. Perpendicular balance mein ab $P$ bhi shaamil hai: $N = m g cos θ + P$ . Yeh step kyun? Extra push block ko surface mein press karta hai, isliye woh normal force mein add hota hai (dekho Block on an Inclined Plane).

Step 2. Verge condition (down-slope pull = friction ceiling): $m g sin θ = μ_{s} N = μ_{s} (m g cos θ + P)$ . Yeh step kyun? Sirf gravity down-slope drive karti hai ( $P$ ka along-slope component zero hai kyunki woh perpendicular hai), jabki friction badhti hai kyunki $N$ badh gayi.

Step 3. Rearrange karo: $tan θ = μ_{s} (1 + \frac{P}{m g cos θ}) > μ_{s}$ . Yeh step kyun? Extra $P$ term strictly positive hai, isliye critical $tan θ$ , $μ_{s}$ se zyaada hota hai — block ek steeper tilt tolerate kar sakta hai.

Step 4 (special check). Agar $P = 0$ , yeh collapse ho jaata hai $tan θ = μ_{s}$ mein, ordinary repose angle $arctan (0.5) \approx 26.57°$ . Yeh step kyun? Ek achha general formula tab known case mein reduce hona chahiye jab naya ingredient vanish ho jaaye.

Verify: $P = 0$ ke saath, $arctan (0.5) = 26.57°$ standard result recover karta hai. ✓ $P > 0$ ke saath, pressing force ramp ko steeper jaane deta hai — intuition se match karta hai ki block ko slope mein push karna usse grip karne mein help karta hai. Naive "answer hamesha $arctan μ$ hota hai" trap exactly wahi hai jo yeh cell test karta hai.

Recall Checkpoint

Repose se neeche, kya friction apni maximum value par hoti hai? ::: Nahi — friction sirf utna supply karti hai jitna zaroori hai ( $m g sin θ$ ), apni ceiling $μ_{s} N$ se neeche rehte hue verge tak. Ex 5 mein $R$ weight se bada kyun nikla? ::: Kyunki $R$ , $N$ aur sideways friction $f$ ka hypotenuse hai; ek perpendicular slice add karne se resultant hamesha lamba ho jaata hai. Ex 8 mein $P$ down-slope driving term mein kyun appear nahi karta? ::: $P$ surface ke perpendicular hai, isliye slope ke along uska zero component hai — yeh sirf $N$ ko boost karta hai.

Connections

Static and Kinetic Friction — $μ_{s}$ kahan se aata hai.
Block on an Inclined Plane — Ex 6 aur 8 mein reuse kiya gaya resolution setup.
Resolving Vectors into Components — Pythagoras aur $sin / cos$ splitting.
Newton's First Law (Equilibrium) — har case ki $\sum F = 0$ backbone.
Banking of Roads — safe cornering speed ke liye wahi $arctan (μ)$ recipe.