1.2.11
Ek inclined plane basically gravity ka tilted version hai. Gravity ek block ko "seedha zameen mein" kheenchne ke bajaye, ab uska ek hissa block ko slope ke saath neeche kheenchta hai (jisse woh slide karta hai) aur doosra hissa usse slope mein andar dabata hai (jo normal force aur friction create karta hai). Hamara poora kaam hai ki gravity ko do perpendicular pieces mein split karo jo slope ke aligned hon. Baaki sab khud aa jaata hai.
Intuition Aisi axes chuno jo motion se match karti honFlat ground par hum horizontal/vertical axes use karte hain. Slope par, block incline ke saath move karta hai , horizontally nahi. Agar hum flat axes rakhein, toh motion aur normal force dono ke x y apna coordinate system rotate karte hain taaki:
x slope ke neeche point kare (possible sliding ki direction),y slope ke perpendicular point kare (normal force ki direction).
Ab N y x gravity ko split karna padta hai.
Maano incline horizontal se θ W = m g neeche point karta hai. Hum isse new (tilted) axes ke along resolve karte hain.
θ m g θ equal hote hain.
Intuition Yeh accelerate kyun karta haiAlong-slope pull ko kuch nahi rokta, isliye block slope ke neeche accelerate karna padta hai. Perpendicular direction mein koi motion nahi, isliye wahan forces cancel ho jaate hain.
Newton ka 2nd law, axis by axis:
Perpendicular (y koi acceleration nahi ⇒ N − m g cos θ = 0 N = m g cos θ
Along slope (x m g sin θ = ma a = g sin θ
a = g sin θ Mass cancel ho jaata hai. Ek pankh aur ek boulder frictionless slope par same acceleration se slide karte hain (bilkul free fall ki tarah, sin θ g
Definition Friction forceFriction slope ke along, relative motion (ya tendency) ko oppose karte hue kaam karta hai. Uski maximum static value aur kinetic value:
f s ma x = μ s N , f k = μ k N N = m g cos θ
Slope pull m g sin θ f s μ s m g cos θ ruka rehta hai jab tak
m g sin θ ≤ μ s m g cos θ ⇒ tan θ ≤ μ s
Kinetic friction slope ke upar kaam karta hai (downward motion ko oppose karta hai):
Along slope: m g sin θ − f k = ma f k = μ k m g cos θ ma = m g sin θ − μ k m g cos θ a = g ( sin θ − μ k cos θ )
Frictionless a = g sin θ μ k g cos θ subtract karta hai . Agar sin θ < μ k cos θ tan θ < μ k reverse nahi kar sakta; iska matlab bas yeh hai ki woh kabhi move hi nahi hua (jo 2a se consistent hai).
Worked example Worked: 2 kg block,θ = 30° μ k = 0.20 g = 9.8 N N N = m g cos θ = 2 ( 9.8 ) cos 30° = 16.97 N Yeh step kyun? Ab a a = g ( sin θ − μ k cos θ ) = 9.8 ( 0.5 − 0.2 × 0.866 ) = 9.8 ( 0.5 − 0.173 ) = 3.20 m/s 2 4.9
Worked example Worked: Angle of reposeEk coin ek book se tab slide hona shuru hoti hai jab book ko 26.6° μ s Yeh kyun kaam karta hai: slip angle par tan θ c = μ s μ s = tan 26.6° ≈ 0.50 Yeh step kyun? Koi mass/length nahi chahiye — sirf angle. ✓
Worked example Worked: Block ko slope ke UPAR push karnaForce F dono gravity-component aur friction slope ke neeche point karte hain (friction upward motion ko oppose karta hai).
F − m g sin θ − μ k m g cos θ = ma Dono negative kyun? Upar jaate waqt, "natural" tendency (gravity) neeche ki taraf hai AUR friction motion ko resist karta hai jo upar ki taraf hai ⇒ friction neeche ki taraf hai. Dono tumhare khilaaf hain.
Common mistake Steel-man: "N m g Kyun sahi lagta hai: Flat tables par saalon se N = m g Kyun galat hai: Slope par sirf gravity ka perpendicular part surface mein dabta hai. Slope normal force ke roop mein weight ka kam hissa "carry" karta hai.
Fix: Hamesha project karo. N = m g cos θ N < m g N → 0
Common mistake Steel-man:sin cos Kyun sahi lagta hai: Dono "bas trig" hain, pressure mein flip karna aasaan hai.
Kyun galat hai aur fix: Limit test use karo. θ → 0 nahi honi chahiye, isliye along-slope term vanish hona chahiye ⇒ yeh sin sin 0 = 0 m g cos cos 0 = 1 Letters nahi, limits yaad karo.
Common mistake Steel-man: bhool jaana ki friction kisi bhi direction mein point kar sakta haiKyun sahi lagta hai: "Friction hamesha peeche ki taraf point karta hai."
Fix: Friction actual ya impending relative motion ko oppose karta hai. Neeche slide karna ⇒ friction upar. Upar push karna ⇒ friction neeche. Ek block gentle slope par static raha ⇒ friction upar point karta hai usse hold karne ke liye.
Recall Feynman: 12-saal ke bacche ko explain karoEk toy car ko ramp par imagine karo. Gravity hamesha seedha neeche kheenchti hai, lekin ramp car ko seedha neeche nahi jaane deta — woh sirf ramp ke along roll kar sakti hai. Toh gravity ki pull "share" ho jaati hai: thodi si car ko neeche roll karaati hai (m g sin θ m g cos θ
"S lope ko S ine milta hai; C rush ko C osine milta hai."
Woh force jo isse S lide karaata hai = m g S sin θ C rush karti hai (normal) = m g C cos θ
Incline par hum gravity ko kaun se do pieces mein split karte hain? Along-slope m g sin θ m g cos θ
Inclined plane par axes rotate kyun karte hain? Taaki N
Angle θ N = m g cos θ m g
Frictionless incline ke neeche acceleration? a = g sin θ
a = g sin θ Driving force aur inertia dono m
Block ke rest mein rehne ki condition (static)? tan θ ≤ μ s m g sin θ ≤ μ s m g cos θ
Angle of repose kya hai aur uska formula kya hai? Woh tilt angle jis par sliding abhi shuru hoti hai; tan θ c = μ s
Rough incline par slide karte block ki acceleration? a = g ( sin θ − μ k cos θ )
Jab block neeche slide kare toh kinetic friction ki direction? Slope ke upar (downward motion ko oppose karta hai).
Rough slope par block ko UPAR push karte waqt force F F − m g sin θ − μ k m g cos θ = ma
Sin/cos swap se bachne ka limit test? θ = 0 = 0 sin = m g cos
no friction, drives motion
net = mg sin theta minus f
Tilt axes to match motion
Frictionless: a = g sin theta
Mass-independent acceleration
