1.2.8
Jab ek surface kisi block ko push karti hai, woh actually do components se push karti hai: perpendicular ==normal force N aur parallel friction force f total contact reaction force R milti hai. Angle of friction bas woh angle hai jo yeh total reaction, normal ke saath banati hai. Angle of repose ek ramp ka sabse steep tilt hai jis par ek block abhi bhi just slip hone wala hota hai. Sundar punchline yeh hai: yeh dono angles same number hain, aur dono arctan ( μ )
Definition Angle of friction (λ Total contact reaction R aur normal N ke beech ka angle, us waqt jab friction apne maximum par ho (f = μ s N
Definition Angle of repose (θ r α maximum inclination jis par rakha hua body equilibrium mein rehe (just sliding ki verge par).
YEH YAHAN SE KYU SHURU KAREN? Kyunki surface do forces wapas deti hai; unke resultant ko samajhna sabse clean first-principles route hai.
Ek block ek horizontal rough surface par ruka hua hai. Surface exert karti hai:
Normal force N
Friction force f
Yeh dono ek reaction mein combine ho jaate hain:
R = N 2 + f 2
Yeh step kyun? Do perpendicular vectors Pythagoras se add hote hain.
R N λ tan λ = normal ke along component surface ke along component = N f
Yeh step kyun? tan
Slipping ki verge par , friction maximum hoti hai: f = μ s N tan λ = N μ s N = μ s
λ = tan − 1 ( μ s )
ALAG SETUP KYU? Ab hum plane ko tilt karte hain aur poochhte hain: kis tilt par block apne weight ke neeche slide karna shuru karta hai?
Block ko θ m g parallel aur perpendicular axes mein incline ke according resolve karo:
Incline ke along (down-slope, slide karne ke liye pull karta hai): m g sin θ
Perpendicular (surface mein press karta hai): m g cos θ
Is tarah resolve kyun karte hain? Kyunki motion (agar koi ho) incline ke along hoti hai, isliye hum driving force aur normal force ko clean alag axes par chahte hain.
Perpendicular equilibrium (surface se koi motion nahi):
N = m g cos θ
Sliding ki verge par , down-slope pull max static friction ke barabar hoti hai:
m g sin θ = f m a x = μ s N = μ s m g cos θ
Yeh step kyun? "Just slip hone wala hai" ka matlab hai driving force bilkul friction limit tak badh gayi hai — na zyada, na kam.
Dono sides ko m g cos θ tan θ = μ s
Toh critical angle (angle of repose) hai:
θ r = tan − 1 ( μ s )
Dono derivations tan ( angle ) = μ s λ = θ r Angle of friction = Angle of repose. Geometrically: plane ko repose tak tilt karne se true vertical weight bilkul total reaction R λ θ r
Worked example Example 1 — Repose experiment seμ 30° μ s
Step 1: Repose par, tan θ r = μ s Kyun? Yeh wahi verge-of-slipping condition hai jo humne derive ki.
Step 2: μ s = tan 30° = 3 1 ≈ 0.577
Dhyan do: koi mass nahi, koi area nahi chahiye — bas angle. Yahi cancellation ki power hai.
Worked example Example 2 —μ μ s = 1.0
Step 1: θ r = arctan ( μ s ) = arctan ( 1.0 ) = 45° Kyun? tan 45° = 1 1 : 1 45°
Worked example Example 3 — Reaction force directionEk 2 kg μ s = 0.5 g = 10
Step 1: λ = arctan ( 0.5 ) = 26.57° Step 2: N = m g = 20 N f = μ s N = 10 N Step 3: R = 2 0 2 + 1 0 2 = 500 ≈ 22.4 N Yeh step kyun? Total reaction N f
Common mistake "Bhaari blocks chhote angle par slip hote hain."Yeh sahi kyun lagta hai: Bhaari objects zyada "stuck" lagte hain, toh surely mass matter karta hoga?
Fix: Driving force (m g sin θ aur friction limit (μ m g cos θ m g sirf μ s
Common mistake Repose angle ke liyeμ k Yeh sahi kyun lagta hai: Block move hone wala hai, toh lagta hai motion friction use hogi.
Fix: "Move hone wala hai" abhi bhi static hai — μ s μ k < μ s
Common mistake Yeh sochna kiλ θ r Yeh sahi kyun lagta hai: Woh equal numbers hain, toh phir kyun nahi?
Fix: λ normal se measure ki jaati hai; θ r plane ka horizontal se tilt hai. Woh value mein equal hain lekin alag geometric contexts mein define hain. Equality ek result hai, definition nahi.
Recall Feynman: 12-saal ke bache ko explain karoSocho tum ek slide par khade ho. Jab yeh almost flat hoti hai, tum slide nahi hote — slide tumhare pair pakad leti hai. Ise thoda aur tilt karo, thoda aur... ek special steepness par, whoosh , tum slide karna shuru kar dete ho! Woh special steepness angle of repose hai. Cool baat yeh hai, koi farak nahi padta ki tum chhote bache ho ya bhaari adult — tum dono usi tilt par slide karna shuru karte ho, kyunki slide bhaare logon ko bhi usi hisaab se zyada pakadti hai. "Grip strength" number μ μ
"Repose = Reaction = arctan(μ)." Teeno R's usi angle ki taraf point karte hain. Aur: "Friction ka angle aur resting ramp twins hain — dono ka tan mu hai."
Recall Active recall checkpoint
Repose derivation mein mass cancel kyun hota hai?
tan λ N Verge-of-slipping ek static situation hai ya kinetic?
Define angle of friction Angle between total contact reaction R N f = μ s N
Define angle of repose Rough incline ka maximum tilt jis par body equilibrium mein rahe (just slide hone wali ho).
Formula for angle of friction λ λ = arctan ( μ s ) tan λ = f / N = μ s
Formula for angle of repose θ r θ r = arctan ( μ s ) m g sin θ = μ s m g cos θ
Relation between angle of friction and angle of repose Dono equal hain: λ = θ r = arctan ( μ s )
Why does mass not affect the angle of repose? Driving force m g sin θ μ s m g cos θ m g
On an incline, what is N N = m g cos θ
Which coefficient governs the angle of repose, μ s μ k μ s
If μ s = 1 45° arctan ( 1 ) = 45°
Magnitude of total reaction R
Static and Kinetic Friction — μ s μ k Block on an Inclined Plane — weight resolve karne ka wahi setup reuse hua.Resolving Vectors into Components — m g sin θ m g cos θ Newton's First Law (Equilibrium) — isliye ∑ F = 0 Banking of Roads — same arctan μ
tan lambda equals f over N
Perpendicular mg cos theta
Angle of repose equals arctan mu
