Question bank — Inclined planes — with and without friction

False. Sirf gravity ka perpendicular slice surface mein press karta hai, toh $N=mg\cos \theta$, jo kisi bhi real tilt ke liye ($\theta >0$) $mg$ se kam hoti hai.

False. $a=g\sin \theta$ mein koi mass nahi — driving force $mg\sin \theta$ aur inertia $m$ dono mass ke saath scale karte hain aur cancel ho jaate hain, toh sabhi blocks ek hi acceleration share karte hain.

False. Zyada steep tilt ka matlab hai gravity zyada slope ke along jhukti hai aur andar kam, toh $N=mg\cos \theta$ $\theta$ badhne par ghatta hai, vertical wall par $0$ tak pahuncha jaata hai.

False. Friction actual ya impending relative motion ka virodh karta hai. Neeche sliding → friction upar; upar push kiya gaya → friction neeche point karta hai; gentle slope par static hold → friction upar hold karne ke liye.

True (marginally). Yeh exactly woh angle hai jahan slope pull maximum static friction ke barabar hoti hai, $mg\sin \theta ={\mu }_{s}mg\cos \theta$ — sliding shuru hone se pehle rest ka aakhri pal.

False. Slope (sliding) share sine use karta hai: $\theta =0$ par sliding force zero honi chahiye, aur sirf $\sin 0=0$ yeh deliver karta hai. "Slope gets Sine."

False. Kinetic friction sirf sliding ke dauran exist karta hai aur sirf motion ka virodh karta hai — yeh kabhi naye motion ka cause nahi ho sakta, sirf existing motion ke liye resistance.

False. $\tan {\theta }_c={\mu }_s$ mein koi mass nahi — repose angle sirf surfaces ki stickiness par depend karta hai, na ki object kitna bhaari hai.

False. Mirroring sirf "left" aur "right" flip karta hai; $N=mg\cos \theta$ aur $mg\sin \theta$ unchanged hain kyunki yeh tilt magnitude $\theta$ par depend karte hain, na ki hill kis taraf face karti hai. Sirf tumhara positive-along-slope ka labelling mirror ke saath flip hota hai.

Sine aur cosine swap ho gaye. Normal force perpendicular gravity slice ko balance karta hai, toh $N=mg\cos \theta$. Limit test: $\theta =0$ par hume $N=mg$ chahiye, jiske liye $\cos$ chahiye (kyunki $\cos 0=1$).

Galat comparison. Sliding ka decision slope ke along hota hai: yeh tab slide karta hai jab along-slope pull $mg\sin \theta$, static friction ${\mu }_{s}mg\cos \theta$ se zyada ho jaaye, yani jab $\tan \theta >{\mu }_s$. $mg$ ko $N$ se compare karna alag directions mix karta hai.

Friction ka sign galat hai. Upar move karte hue, friction upward motion ka virodh karta hai, toh yeh slope ke neeche point karta hai aur subtract hona chahiye: $F-mg\sin \theta -{\mu }_{k}mg\cos \theta =ma$. Upar jaate samay, gravity aur friction dono tumse ladte hain.

Perpendicular direction balance hoti hai, lekin along-slope direction nahi — $mg\sin \theta$ unopposed hai, toh block accelerate karta hai. Ek axis ko balance karna equilibrium nahi hai.

Yahan negative result ka matlab hai block kabhi chala hi nahi ($\tan \theta <{\mu }_k$), yeh nahi ki kinetic friction use reverse kar raha hai. Kinetic friction sirf slow kar sakta hai, drive nahi.

Un-rotated axes ek real, solvable pair of equations dete hain — lekin $N\ne mg$. Horizontal $x$ ke along: $N\sin \theta =m{a}_x$; vertical $y$ ke along: $N\cos \theta -mg=m{a}_y$. Kyunki block slide bhi karta hai, $a_x$ aur $a_y$ dono nonzero hain, toh $N$ do coupled equations mein appear karta hai aur tumhe unhe simultaneously solve karna hoga. Axes ko rotate karna $N$ ko ek axis par akela bitha deta hai, jisse yeh clean $N=mg\cos \theta$ mil jaata hai — woh convenience hi toh poora point hai.

Taaki normal force purely ek axis par ho aur acceleration purely doosri par — phir sirf gravity ko resolve karna padta hai, teen forces ko ek saath resolve karne ki bajaye.

Kyunki "vertical & horizontal" aur "normal & slope" dono $90^{\circ }$ par lines ke pairs hain. Pehle pair ko $\theta$ se rotate karne par yeh exactly doosre par aa jaata hai, angle ko unchanged carry karte hue — upar wala geometry figure dekho. Toh $mg$ aur $N$ ke beech ka angle $\theta$ ke barabar hota hai.

Driving force $mg\sin \theta$ $m$ ke proportional hai, aur Newton's law usi $m$ (inertia) se divide karta hai. $m$ ke do factors cancel ho jaate hain — exactly free fall ki tarah.

Slipping par, $\tan {\theta }_c={\mu }_s$ — mass, length, aur $g$ sab cancel ho jaate hain. Woh ek angle measure karo jahan sliding shuru hoti hai aur tumhe directly friction coefficient mil jaata hai.

Kinetic friction ek term subtract karta hai: $a=g(\sin \theta -{\mu }_k\cos \theta )$ versus $g\sin \theta$. Jab tak block slide karta hai, woh $-{\mu }_{k}g\cos \theta$ speed-up remove karta hai.

Kyunki friction $N$ ke proportional hai, aur $N=mg\cos \theta$ slope steep hone par shrink karta hai. Surface mein kam "crush" matlab kam grip.

Yeh ek acceleration $g\sin \theta$ deta hai jo ek scaled-down $g$ hai, jo tumhe gravity ke effects ko slow motion mein study karne deta hai full free fall ki jagah.

Sliding force $mg\sin 0=0$ (kuch bhi ise along push nahi karta), aur $N=mg\cos 0=mg$ — surface poora weight carry karta hai. Yeh ordinary flat-ground physics recover karta hai.

$N=mg\cos 90^{\circ }=0$ (kuch bhi wall mein press nahi karta) aur along-slope pull $mg\sin 90^{\circ }=mg$ hai — seedha neeche pure free fall.

Kyunki $\cos (-\theta )=\cos \theta$, normal $N=mg\cos \theta$ identical hai — surface utna hi crush karta hai. Lekin $\sin (-\theta )=-\sin \theta$, toh along-slope pull direction flip karta hai: block doosri taraf slide karna chahta hai. Magnitudes mirrored hill se match karte hain; sirf slide-component ka sign reverse hota hai.

Sirf utni badi jitni pull balance karne ke liye chahiye: $f_s=mg\sin \theta$, na ki iska maximum ${\mu }_{s}mg\cos \theta$. Static friction ek responder hai, demand ko ek ceiling tak match karta hai.

Abhi bhi static, apni maximum value ${\mu }_{s}mg\cos \theta$ par. Kinetic friction sirf sliding shuru hone ke turant baad appear karta hai.

Sirf tab agar ${\mu }_k$ itna bada ho ki $\tan \theta <{\mu }_k$ (toh kinetic friction pull se zyada ho). Warna yeh slide karta rehta hai, kyunki ek baar move karne par sirf (usually chhota) ${\mu }_k$ act karta hai.

Slope block ko $g\sin 30^{\circ }=g/2$ milta hai; free-falling block ko poora $g$. Incline gravity se ek $\sin \theta$ factor "chura" leta hai — energy view ke liye Work-Energy Theorem on Inclines dekho.