Exercises — Angle of friction, angle of repose — derivation

KYA chahiye: wo tilt jis par block abhi slide karne wala ho. YE tool kyun: Angle of repose defined hi hai $\tan {\theta }_r={\mu }_s$ se (verge-of-slipping condition, mass cancel ho jaati hai). ${\theta }_r=\arctan (0.5)=26.57°$ Bas itna hi — koi mass nahi, koi area nahi, sirf coefficient.

KYA chahiye: L1.1 ke relation ko ulta karo. KYU: Pehle slip ka tilt hi angle of repose hai, isliye ${\mu }_s=\tan {\theta }_r$. ${\mu }_s=\tan 40°=0.839$

Normal: floor flat hai, isliye $N=mg=3\times 10=30\text{N}$. Friction (verge par): $f={\mu }_{s}N=0.4\times 30=12\text{N}$. Total reaction (Pythagoras — ye dono perpendicular hain): $R=\sqrt{N^2+f^2}=\sqrt{30^2+12^2}=\sqrt{1044}\approx 32.31\text{N}$

KYA axes: incline ke along aur perpendicular resolve karo (motion, agar ho, to uske along hi hogi). $N=mg\cos \theta =40\times \cos 15°=38.64\text{N}$ $mg\sin \theta =40\times \sin 15°=10.35\text{N (down-slope)}$ KYU friction NEECHE point karti hai: hum block ko upar push kar rahe hain, toh impending motion up-slope hai; friction usse oppose karti hai, isliye wo down-slope point karti hai magnitude $f={\mu }_{s}N$ ke saath. $f=0.5\times 38.64=19.32\text{N}$ Verge-of-sliding-up balance (upar ki forces = neeche ki forces): $P=mg\sin \theta +f=10.35+19.32=29.67\text{N}$

Flat floor par "normal" aur "vertical" ek hi line hai, isliye: $\lambda =\arctan ({\mu }_s)=\arctan (0.4)=21.80°$ L1.3 ke numbers se check karo: $\tan \lambda =f/N=12/30=0.4$. ✓

Repose angle ${\mu }_s$ ke saath badhta hai. Plastic coin ka, jiska smaller ${\mu }_s$ hai, smaller repose angle hoga, isliye wo pehle slip karega: ${\theta }_r^{\text{plastic}}=\arctan (0.3)=16.70°$ Steel coin $\arctan (0.6)=30.96°$ tak tika rehta hai.

$R=\sqrt{N^2+f^2}$ se shuru karo aur $f={\mu }_{s}N$ substitute karo: $R=\sqrt{N^2+({\mu }_{s}N{)}^2}=\sqrt{N^2(1+{\mu }_s^2)}=N\sqrt{1+{\mu }_s^2}$ Flat floor par $N=mg$, isliye $R=mg\sqrt{1+{\mu }_s^2}$. $m=3,{\mu }_s=0.4$ ke liye: $R=30\sqrt{1+0.16}=30\sqrt{1.16}=32.31\text{N}$ — L1.3 se match karta hai. ✓ Iska matlab: $R$ hamesha kam se kam $N$ ke barabar hoga (kyunki $\sqrt{1+{\mu }_s^2}\ge 1$), aur friction badhne ke saath ye bhi badhta hai.

Repose par block equilibrium mein hai, isliye $\vec{R}+m\vec{g}=0$, yaani $\vec{R}=-m\vec{g}$. Gravity vertical hai (neeche), isliye $R$ vertical hona chahiye (upar). Geometry: $R$ (tilted) normal se $\lambda$ angle par hai; vertical usi normal se ${\theta }_r$ angle par hai. Equilibrium $R$ ko vertical par force karta hai, jo sirf tab possible hai jab $\lambda ={\theta }_r$ ho. Ye do angles ke barabar hone ka visual proof hai — figure dekhein: seedha $R$ arrow aur orange weight arrow ek hi line hain.

(a) ${\theta }_r=\arctan (0)=0°$. Frictionless plane par koi bhi tilt block ko slide kar deta hai — "sabse steep safe angle" flat ho jaata hai. (b) Jaise-jaise ${\mu }_s\to \infty$, $\arctan ({\mu }_s)\to 90°$. Infinite grip se block vertical wall par bhi chipka rehta hai; repose angle $90°$ ke paas jaata hai (lekin kabhi pahunchta nahi). Key insight: $\arctan$ function $[0,\infty )$ ko $[0°,90°)$ par map karta hai — toh real repose angle kabhi $90°$ nahi ho sakta aur na hi exceed kar sakta hai. $90°$ se steep ramp, ramp nahi, ceiling hai.

Banked surface par ek parked car bilkul ek incline par block ke jaisi hai. Ye tab neeche slide karta hai jab bank tilt repose angle se zyada ho jaaye: ${\theta }_{\max }=\arctan ({\mu }_s)=\arctan (0.7)=34.99°$ Connection: banking mein, yahi $\arctan \mu$ safe-speed range mein friction ka contribution set karta hai. ${\theta }_r$ se steep bank ek stationary car ko slip karne dega — isliye real banks modest rehte hain aur baaki centripetal force ke liye speed par depend karte hain.

KYA: block neeche slide karne wala hai, isliye friction upar slope ki taraf max value par point karti hai. Horizontal push $P$ ko incline axes par resolve karo:

• incline ke along (up-slope component): $P\cos \theta$
• perpendicular (surface mein): $P\sin \theta$

Perpendicular equilibrium: $N=mg\cos \theta +P\sin \theta$ Along-incline equilibrium (sliding down ke verge par: down-pull = up-push + friction): $mg\sin \theta =P\cos \theta +{\mu }_{s}N$ $N$ substitute karo: $mg\sin \theta =P\cos \theta +{\mu }_s(mg\cos \theta +P\sin \theta )$ $mg\sin \theta -{\mu }_{s}mg\cos \theta =P(\cos \theta +{\mu }_s\sin \theta )$ $P=\frac{mg(\sin \theta -{\mu }_s\cos \theta )}{\cos \theta +{\mu }_s\sin \theta }$ Numbers ($mg=20,\theta =20°,{\mu }_s=0.25$): $P=\frac{20(0.342-0.25\times 0.940)}{0.940+0.25\times 0.342}=\frac{20(0.342-0.235)}{0.940+0.0855}=\frac{2.145}{1.0255}=2.09\text{N}$

Insight: har contact ka apna repose angle hota hai, aur repose angle sirf us contact ke ${\mu }_s$ par depend karta hai (masses cancel ho jaati hain — isliye B ka extra weight A ke plane ke saath repose angle ko change nahi karta).

• B-on-A tab slip karta hai jab tilt $\arctan (0.20)=11.31°$ tak pahunche.
• A-on-plane tab slip karta hai jab tilt $\arctan (0.35)=19.29°$ tak pahunche.

Smaller angle pehle aata hai: B pehle A se slide karta hai, tilt par $\theta =\arctan (0.20)=11.31°$ KYU B ka weight A ko zyada usefully grip nahi karta: haan, B ka weight A ke neeche $N$ badhata hai, lekin A ki driving force $mg\sin \theta$ bhi usi factor se badhti hai — toh A ka threshold apne $\arctan (0.35)$ par hi rehta hai. Dono thresholds independent hain; bas $\mu$'s compare karo.

L4.2 ka result aur uska mirror use karo. $mg=50,\theta =15°,{\mu }_s=0.30$ ke saath: $\sin 15°=0.2588,\cos 15°=0.9659$.

Minimum $P$ — friction UP point karti hai (about to slide down): $P_{\min }=\frac{mg(\sin \theta -{\mu }_s\cos \theta )}{\cos \theta +{\mu }_s\sin \theta }=\frac{50(0.2588-0.30\times 0.9659)}{0.9659+0.30\times 0.2588}$ $=\frac{50(0.2588-0.2898)}{0.9659+0.0776}=\frac{50(-0.0310)}{1.0435}=-1.485\text{N}$ Negative minimum ka matlab hai: $P=0$ ke saath bhi block neeche nahi slide karta (kyunki $15°<\arctan 0.30=16.7°$). Toh physically $P_{\min }=0$ hai — lower end par koi push zaruri nahi.

Maximum $P$ — friction DOWN point karti hai (about to slide up): $P_{\max }=\frac{mg(\sin \theta +{\mu }_s\cos \theta )}{\cos \theta -{\mu }_s\sin \theta }=\frac{50(0.2588+0.2898)}{0.9659-0.0776}$ $=\frac{50(0.5486)}{0.8883}=30.88\text{N}$ Stable range: $0\le P\le 30.88\text{N}$. Isse zyada hone par block hill ke upar dhakeel diya jaata hai.