Exercises — Inclined planes — with and without friction

WHAT hum choose karte hain: Frictionless incline par along-slope ek hi force hai $mg\sin \theta$. Newton's Second Law se, $ma=mg\sin \theta$, toh mass cancel ho jaata hai: $a=g\sin \theta$ Yeh answer (b) hai. WHY $\cos$ nahin? Master figure dekho — $\cos$ slope mein andar point karta hai (green), jo woh direction hai jisme koi motion nahin hai; yeh sliding cause nahin kar sakta. WHY $\tan$ nahin? $\tan$ yahan force balance mein kabhi nahin aata — yeh baad mein sirf do forces ke ratio ke roop mein aata hai (angle of repose). Compute: $a=9.8\times \sin 30°=9.8\times 0.5=4.9{\text{m/s}}^2$.

WHAT: Perpendicular direction mein koi acceleration nahin hai, toh $N$ sirf crush component ko balance karta hai (green arrow). $N=mg\cos \theta =4\times 9.8\times \cos 30°=4\times 9.8\times 0.8660=33.9\text{N}$ WHY $mg=39.2\text{N}$ se kam? Kyunki slope weight ka ek part apne andar le leta hai; sirf perpendicular part andar press karta hai. Kisi bhi tilt ke liye $N<mg$ hota hai.

WHY pehle $N$ find karo: friction ko $N$ chahiye. $N=mg\cos \theta =2\times 9.8\times \cos 30°=16.97\text{N}$ WHAT aage: block neeche move kar raha hai, toh kinetic friction (orange) upar slope ki taraf point karta hai, motion ko oppose karte hue. Along-slope Newton's law: $ma=mg\sin \theta -{\mu }_{k}mg\cos \theta$ $m$ se divide karo (cancel ho jaata hai): $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(0.5-0.20\times 0.8660)=9.8(0.5-0.1732)=3.20{\text{m/s}}^2$ Sanity: frictionless $4.9{\text{m/s}}^2$ se kam, jaise friction slow karta hai. ✓

WHY yeh kaam karta hai: angle of repose par down-slope pull maximum static friction ke barabar ho jaata hai: $mg\sin {\theta }_c={\mu }_{s}mg\cos {\theta }_c\Rightarrow {\mu }_s=\tan {\theta }_c$ Mass, $g$, aur coin size sab cancel ho jaate hain — yeh ek pure-angle measurement hai. ${\mu }_s=\tan 26.6°=0.50$

WHAT: Humne pehle hi $a=3.20{\text{m/s}}^2$ (constant) find kar liya. Kinematic relation $v^2=u^2+2as$ use karo jisme $u=0$: $v=\sqrt{2as}=\sqrt{2\times 3.20\times 5}=\sqrt{32.0}=5.66\text{m/s}$ WHY yahan kinematics: acceleration constant hai (slope ke saath saath sari forces constant hain), toh standard constant-$a$ equations apply hoti hain. Yeh wahi idea hai jo Work-Energy Theorem on Inclines deta — energy in, kinetic energy out.

WHAT pehle test karein: driving pull ko maximum static friction se compare karo.

• Down-slope pull: $mg\sin \theta =3\times 9.8\times \sin 25°=12.43\text{N}$.
• Max static friction: ${\mu }_{s}mg\cos \theta =0.60\times 3\times 9.8\times \cos 25°=0.60\times 26.64=15.99\text{N}$. Kyunki $12.43<15.99$, pull friction ko overcome nahin kar sakta ⇒ yeh slide NAHIN karega. Equivalent quick test: $\tan 25°=0.466<{\mu }_s=0.60$ ✓ (angle of repose se neeche hai). WHAT friction actually hai: jab static hota hai toh friction maximum par nahin hoti. Yeh sirf utna supply karti hai jitna balance ke liye zaroori ho. Toh actual friction (UP slope pointing) down-slope pull ke barabar hai: $f_s=mg\sin \theta =12.43\text{N}(\text{not }15.99\text{N})$

WHAT badalta hai: block UPAR move kar raha hai, toh kinetic friction (orange) flip hokar slope ke neeche point karta hai — friction hamesha actual motion ko oppose karta hai. Ab gravity ka pull aur friction dono slope ke neeche point karte hain, dono upward motion se lad rahe hain. $ma=-mg\sin \theta -{\mu }_{k}mg\cos \theta$ Deceleration ki magnitude: $a=g(\sin \theta +{\mu }_k\cos \theta )=9.8(\sin 20°+0.30\cos 20°)$ $=9.8(0.3420+0.30\times 0.9397)=9.8(0.3420+0.2819)=9.8\times 0.6239=6.11{\text{m/s}}^2$ WHY yahan plus sign hai (vs. Problem 2.1 mein minus): friction ki direction flip ho gayi. Neeche jaate waqt, friction pull se subtract karta hai; upar jaate waqt, friction gravity mein add ho jaata hai kyunki dono ab down-slope point karte hain. Master figure mein orange arrow ki dono possible directions compare karo.

WHAT forces hain (block upar move kar raha hai): applied $F$ upar; gravity component $mg\sin \theta$ neeche; kinetic friction ${\mu }_{k}mg\cos \theta$ neeche (upward motion ko oppose karta hai). Slope ke saath Newton's law, up-slope ko positive maante hue: $ma=F-mg\sin \theta -{\mu }_{k}mg\cos \theta$ Pieces compute karo:

• $mg\sin \theta =5\times 9.8\times 0.5=24.5\text{N}$
• $mg\cos \theta =5\times 9.8\times 0.8660=42.44\text{N}$, toh friction $=0.25\times 42.44=10.61\text{N}$ $ma=60-24.5-10.61=24.89\text{N}$ $a=\frac{24.89}{5}=4.98{\text{m/s}}^2\text{up the slope}$ WHY dono subtractions: upar jaate waqt, block DONO gravity (naturally downhill) aur friction (uphill motion ko resist karta hai) se ladhta hai. Dono push ke relative "negative" hain.

WHAT sliding-back decide karta hai: top par block momentarily rest mein hai. Static friction ab ise hold karne ki koshish karta hai. Yeh tabhi wapas slide karta hai jab down-slope pull maximum static friction ko beat kar de. Problem sirf ${\mu }_k$ deta hai; check ke liye assume karo ${\mu }_s\approx {\mu }_k=0.25$.

• Down-slope pull: $mg\sin 30°=24.5\text{N}$
• Max static friction: ${\mu }_{s}mg\cos 30°=0.25\times 42.44=10.61\text{N}$ Kyunki $24.5>10.61$, haan yeh wapas neeche slide karega. (Equivalently $\tan 30°=0.577>0.25$.) Slide-back acceleration: ab neeche move karte hue, friction flip hokar UPAR point karta hai: $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(0.5-0.25\times 0.8660)=9.8(0.5-0.2165)=2.78{\text{m/s}}^2\text{down}$ WHY up-deceleration se choti hai: neeche jaate waqt, friction gravity ke against help karta hai (subtract karta hai), toh net down-slope force upward trip se ladhne wali net force se chhoti hoti hai.

WHAT invert karein: hum sliding-down formula jaante hain aur ${\mu }_k$ chahiye. Start karo yahan se: $a=g(\sin \theta -{\mu }_k\cos \theta )$ ${\mu }_k$ ke liye solve karo: ${\mu }_k=\frac{g\sin \theta -a}{g\cos \theta }=\frac{9.8\times 0.6018-3.0}{9.8\times 0.7986}=\frac{5.898-3.0}{7.826}=\frac{2.898}{7.826}=0.370$ WHY yeh mastery hai: humne usual direction reverse ki — ek measured motion se ek surface property tak. Exactly isi tarah experimentalists ${\mu }_k$ measure karte hain.

WHAT energy accounting hai: slope distance $d$ par, gravity positive work karta hai $mg\sin \theta \cdot d$ (motion ke saath component), friction negative work karta hai $-{\mu }_{k}mg\cos \theta \cdot d$. Net work kinetic energy gain ke barabar hoti hai (rest se start): $\frac{1}{2}m{v}^2=mg\sin \theta d-{\mu }_{k}mg\cos \theta d$ Mass cancel ho jaata hai: $v^2=2gd(\sin \theta -{\mu }_k\cos \theta )$ Compute: $\sin 45°=\cos 45°=0.7071$. $v^2=2\times 9.8\times 3\times (0.7071-0.40\times 0.7071)=58.8\times (0.7071-0.2828)=58.8\times 0.4243=24.94$ $v=\sqrt{24.94}=4.99\text{m/s}$ Kinematics se cross-check: $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8\times 0.4243=4.158{\text{m/s}}^2$; phir $v=\sqrt{2ad}=\sqrt{2\times 4.158\times 3}=\sqrt{24.95}=4.99\text{m/s}$ ✓. Dono routes agree karte hain — energy aur force ek hi physics ke do views hain.

WHAT hum compute karte hain: move hone ke baad net down-slope force hai $F_{\text{net}}=mg\sin \theta -{\mu }_{k}mg\cos \theta =mg\cos \theta (\tan \theta -{\mu }_k)$ $\theta ={\theta }_c$ par hume $\tan {\theta }_c={\mu }_s={\mu }_k$ milta hai, toh $\tan {\theta }_c-{\mu }_k=0\Rightarrow F_{\text{net}}=0$. ${\theta }_c$ se aage: $\tan \theta$ badhta hai ($0$ se $90°$ tak rising function hai), toh $\tan \theta -{\mu }_k>0\Rightarrow F_{\text{net}}>0$: block accelerate karta hai. Interpretation: angle of repose ek knife-edge hai. Bilkul usi par, block neutral balance mein hai (constant velocity agar nudge kiya jaaye, ${\mu }_k={\mu }_s$ ke saath); thoda bhi steeper aur yeh genuinely accelerate karta hai. Isi liye ${\theta }_c$ "stays put" aur "runs away" ke beech ka sharp threshold hai. Note karo $mg\cos \theta$ factor $\theta <90°$ ke liye hamesha positive hai, toh $F_{\text{net}}$ ka sign poori tarah $(\tan \theta -{\mu }_k)$ se decide hota hai.

Moving DOWN ::: friction points UP → $a=g(\sin \theta -{\mu }_k\cos \theta )$. Moving UP ::: friction points DOWN → deceleration $=g(\sin \theta +{\mu }_k\cos \theta )$. At rest, not slipping ::: friction = jo bhi pull balance kare, max ${\mu }_{s}mg\cos \theta$ tak.