Worked examples — Inclined planes — with and without friction

1.2.11 · D3 · Physics › Newton's Laws & Dynamics › Inclined planes — with and without friction

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The scenario matrix

Har incline problem inhi cells mein se kisi ek mein aata hai. Last column uss example ka naam deta hai jo ise cover karta hai.

Ex 1 — Dono degenerate limits (Cells A & B)

Step 1 — Flat case, $\theta =0^{\circ }$. $mg\sin 0^{\circ }=5(9.8)(0)=0\text{N}N=mg\cos 0^{\circ }=5(9.8)(1)=49\text{N}$ Yeh step kyun? Hum formula ko uski edge par test karte hain. $\sin 0=0$ ka matlab hai koi along-slope pull nahi — kuch bhi block ko slide karne ki koshish nahi karta, bilkul wahi jo "flat" ka matlab hona chahiye. $\cos 0=1$ ka matlab hai surface poora weight carry karti hai, $N=mg=49\text{N}$.

Step 2 — Vertical case, $\theta =90^{\circ }$. $mg\sin 90^{\circ }=5(9.8)(1)=49\text{N}N=mg\cos 90^{\circ }=5(9.8)(0)=0\text{N}$ Yeh step kyun? $90^{\circ }$ par "slope" ek vertical wall hai. Poora weight $49\text{N}$ along-slope pull ban jaata hai (block free fall mein hai), aur $N=0$ — ek vertical wall kisi cheez par press nahi karti jo uske against tiki ho. Right figure dekho: blue normal arrow zero tak shrink ho gaya hai.

Ex 2 — Frictionless slide, mass cancel ho jaata hai (Cell C)

Step 1 — $a=g\sin \theta$ use karo. $a=9.8\sin 37^{\circ }=9.8(0.6018)=5.90{\text{m/s}}^2\text{ down the slope}$ Yeh step kyun? Frictionless ka matlab kuch bhi along-slope pull ka oppose nahi karta, toh slope ke along Newton's 2nd law simply $mg\sin \theta =ma$ hai; $m$'s divide ho jaate hain.

Step 2 — Mass double karo. $m$ kabhi bhi $a=g\sin \theta$ mein appear nahi karta, toh $a=5.90{\text{m/s}}^2$ phir se same. Yeh step kyun? Yahi parent se "beautiful" cancellation hai — driving force $\propto m$, inertia $\propto m$, toh mass drop out ho jaata hai. Ek pankh aur ek patthar tie karte hain.

Ex 3 — Rough but stays put; REAL friction nikalo (Cell D)

Step 1 — $\tan \theta$ vs ${\mu }_s$ se slide test karo. $\tan 20^{\circ }=0.364\le {\mu }_s=0.60$ Yeh step kyun? Block tabhi slip karta hai jab along-slope pull max friction ko beat kare, yaani $\tan \theta >{\mu }_s$. Yahan $0.364\le 0.60$, toh yeh rest mein rehta hai.

Step 2 — Actual friction = jo pull ko balance kare. Kyunki yeh static hai aur move nahi kar raha, along-slope forces cancel hote hain: $f_s=mg\sin \theta =3(9.8)\sin 20^{\circ }=3(9.8)(0.342)=10.05\text{N (up the slope)}$ Yeh step kyun? Static friction automatically ${\mu }_{s}N$ nahi hota. Yeh sirf utna supply karta hai jitna block ko hold karne ke liye zaroori ho — yahan sirf itna ki $10.05\text{N}$ pull cancel ho sake. Maximum available hai ${\mu }_{s}N=0.60\times 3(9.8)\cos 20^{\circ }=16.58\text{N}$, comfortably zyada.

Ex 4 — Rough, actually slides down (Cell E)

Step 1 — Confirm karo ki motion possible hai. $\tan 40^{\circ }=0.839>{\mu }_k=0.25$, toh ek baar move karne par yeh slide karta rehta hai (pull kinetic friction ko beat karti hai). Yeh step kyun? Agar $\tan \theta <{\mu }_k$ hota, toh friction term pull ko swamp kar deti aur formula ek nonsensical "negative" acceleration deta jiska matlab "yeh slide nahi karta." Hum pehle check karte hain.

Step 2 — Sliding-down formula apply karo. $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(\sin 40^{\circ }-0.25\cos 40^{\circ })$ $=9.8(0.6428-0.25\times 0.766)=9.8(0.6428-0.1915)=9.8(0.4513)=4.42{\text{m/s}}^2$ Yeh step kyun? Kinetic friction up the slope act karta hai (downhill motion ko oppose karta hai), toh yeh pull se ${\mu }_{k}g\cos \theta$ subtract karta hai.

Ex 5 — Reverse: slip angle se ${\mu }_s$ nikalo (Cell F)

Step 1 — Angle of repose use karo. ${\mu }_s=\tan {\theta }_c=\tan 31^{\circ }=0.601$ Yeh step kyun? Slipping ki verge par, $mg\sin \theta ={\mu }_{s}mg\cos \theta$. Mass aur $g$ cancel ho jaate hain, leaving ${\mu }_s=\tan {\theta }_c$ — ek stickiness number jo seedha ek protractor se padha jaata hai. Yeh parent se elegant measurement trick hai, jo Friction — Static and Kinetic se linked hai.

Ex 6 — Slope ke upar push kiya: dono forces tumse ladte hain (Cell G)

Step 1 — Har along-slope force ko uske sign ke saath list karo. Up = positive. Gravity ka slope-pull neeche hai; friction upward motion ko oppose karta hai, toh woh bhi neeche point karta hai: $F-mg\sin \theta -{\mu }_{k}mg\cos \theta =ma$ Yeh step kyun? Upar jaate waqt, nature (gravity) aur friction dono downhill point karte hain — figure ke dono down-arrows dekho tumhare single up-arrow ke against.

Step 2 — Numbers plug karo. $mg\sin 25^{\circ }=4(9.8)(0.4226)=16.57\text{N}$ ${\mu }_{k}mg\cos 25^{\circ }=0.30\times 4(9.8)(0.9063)=10.66\text{N}$ $a=\frac{40-16.57-10.66}{4}=\frac{12.77}{4}=3.19{\text{m/s}}^2\text{ up the slope}$ Yeh step kyun? Net up-slope force $=F$ minus dono opponents, phir mass se divide karo (Newton's Second Law).

Ex 7 — Real-world word problem: ramp par kinematics (Cell H)

Step 1 — Slope ke along acceleration. $a=g\sin 30^{\circ }=9.8(0.5)=4.9{\text{m/s}}^2$ Yeh step kyun? Frictionless hai, toh $a=g\sin \theta$; mass irrelevant hai.

Step 2 — $v^2=u^2+2as$ se final speed. Rest se shuru ($u=0$), $s=3\text{m}$ travel karta hai: $v=\sqrt{2as}=\sqrt{2(4.9)(3)}=\sqrt{29.4}=5.42\text{m/s}$ Yeh step kyun? Humein distance aur acceleration pata hai lekin abhi time nahi, toh time-free kinematic equation natural choice hai.

Step 3 — $v=u+at$ se time. $t=\frac{v}{a}=\frac{5.42}{4.9}=1.11\text{s}$ Yeh step kyun? Ab $v$ pata hai, toh time tak pahunchne ka yahi sabse quick rasta hai.

Ex 8 — Exam twist: block ko slide hone se rokne ke liye minimum force (Cell I)

Step 1 — Verge-of-sliding-down balance set up karo. Neeche slide karne ki verge par, dono $F$ aur maximum static friction upar point karte hain, gravity ka pull neeche point karta hai: $F+{\mu }_{s}mg\cos \theta =mg\sin \theta$ Yeh step kyun? Hum sabse chhota $F$ chahte hain; friction yahan ek free ally hai, toh hum ise poori strength ${\mu }_{s}N$ par upar work karne dete hain. Is $F$ se kam kuch bhi block ko neeche slip kara dega.

Step 2 — $F$ ke liye solve karo. $mg\sin 35^{\circ }=2(9.8)(0.5736)=11.24\text{N}$ ${\mu }_{s}mg\cos 35^{\circ }=0.30\times 2(9.8)(0.8192)=4.82\text{N}$ $F=11.24-4.82=6.42\text{N}$ Yeh step kyun? Applied force sirf gravity ke pull aur jo friction akele hold kar sakti hai, uske beech ke shortfall ko pura karne ki zaroorat hai.

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