Foundations — Angle of friction, angle of repose — derivation

1.2.8 · D1 · Physics › Newton's Laws & Dynamics › Angle of friction, angle of repose — derivation

Is result ko derive karne se pehle — jo parent topic mein hai — tumhe har woh letter aur picture samajhni hogi jo derivation chupke se assume karti hai. Yeh page unhe ek ek karke, bilkul zero se build karta hai. Agar koi symbol parent mein aata hai, toh woh pehle yahan define kiya gaya hai.

1. "Force" ka matlab, aur woh arrow jo ise draw karta hai

Figure 1 dekho. Block sirf floor par baitha hai, phir bhi teen arrows uس par act kar rahe hain. Hume har arrow padhna seekhna hoga in sabko add karne se pehle.

Force ki unit newton hai, likha jaata hai $\text{N}$. Ek newton roughly tumhare haath mein rakhhe ek chhote seb ka weight hai.

2. Weight $mg$ — gravity ka downward arrow

• $m$ = mass (kg)
• $g$ = gravitational field strength $\approx 10\text{N/kg}$ (hum easy arithmetic ke liye $10$ use karte hain; actual value $9.8$ hai)
• $mg$ = weight, ek arrow jo seedha neeche point karta hai, hamesha.

Topic ko yeh kyun chahiye: ek tilted ramp par, gravity hi akela woh cheez hai jo block ko slide karane ki koshish karti hai. Derivation mein har "driving force" $mg$ se aati hai.

3. Normal force $N$ — surface ka perpendicular pushback

Figure 1 mein cyan arrow jis par $N$ likha hai woh floor se seedha upar point karta hai. Ek tilted ramp par yeh seedha ramp se bahar point karta hai — abhi bhi surface ke perpendicular, toh yeh ramp ke saath lean karta hai (figure 4).

Topic ko yeh kyun chahiye: $N$ measure karta hai ki block aur surface kitni tightly squeeze ho rahi hain — aur friction ki strength directly $N$ par built hai.

4. Friction force $f$ — sideways grip

Ise dhyaan se padho: friction ek fixed number nahin hai. Yeh lazy hai — sirf utni hi grip provide karti hai jitni zaroorat hai, right up to maximum ${\mu }_{s}N$ tak. Isse zyada push karo toh block free ho jaata hai.

Topic ko yeh kyun chahiye: poori derivation us moment par jiiti hai jab $f={\mu }_{s}N$. Woh akela equation dono results ka engine hai.

5. Coefficient of friction ${\mu }_s$ — "grip number"

${\mu }_s$ limit equation se hi define hota hai: ${\mu }_s=\frac{f_{\max }}{N}$ Yeh ratio hai "maximum sideways grip ÷ kitna tightly press ho rahe ho". Kyunki yeh do forces ka ratio hai, newtons cancel ho jaate hain aur ${\mu }_s$ sirf ek bare number reh jaata hai.

Topic ko yeh kyun chahiye: poore topic ka punchline hai $\text{angle}=\arctan ({\mu }_s)$. Sab kuch is ek number mein funnel ho jaata hai.

6. Right triangle, aur $\tan$ kya measure karta hai

Derivation forces ko ek angle mein badal deti hai. "Do arrows" aur "ek angle" ke beech ka bridge right triangle hai — toh hume ise aur iske steepness padhne wale tool ko define karna hoga.

Figure 2 se ek key fact note karo: jaise jaise $\theta$ $0°$ se $90°$ ki taraf badhta hai, opposite side stretch hoti hai jabki adjacent same rehti hai, toh $\tan \theta$ $0$ se upar without bound climb karta hai. Har steepness ka apna tangent hota hai — woh one-to-one link hi hume reverse jaane deta hai.

7. $\arctan$ — tangent ko undo karna

Ise is sawaal ki tarah socho: "kis angle ka yeh tangent hai?"

Topic ko yeh kyun chahiye: derivation $\tan (\text{angle})={\mu }_s$ par land karti hai. Angle nikalne ke liye, hume tangent ko undo karna hoga — woh exactly $\arctan$ hai. Yeh final key hai.

8. Vector ko components mein resolve karna

Gravity seedha neeche point karti hai, lekin ek tilted ramp par ramp ki "along" aur "perpendicular" directions bhi tilted hoti hain. Unhe use karne ke liye, hum single weight arrow ko un tilted axes par do arrows mein split karte hain.

Angle $\theta$ par tilted ramp par, downward weight $mg$ split ho jaata hai:

• Slope ke along (block ko neeche slide karane ki koshish karte hue): $mg\sin \theta$
• Slope ke perpendicular (ramp mein press karte hue): $mg\cos \theta$

Topic ko yeh kyun chahiye: repose derivation $mg\sin \theta$ (slide-driver) ko $mg\cos \theta$ (press) se bani friction se compare karti hai. Resolving ke bina, hum woh do forces likh hi nahin sakte.

9. Equilibrium — balanced-arrows condition

Ramp par still baitha ek block ke liye, equilibrium har axis par alag alag hume deta hai:

• Perpendicular: $N=mg\cos \theta$ (surface push pressing component ko balance karta hai)
• Slope ke along, verge par: $mg\sin \theta =f_{\max }={\mu }_{s}N$

Topic ko yeh kyun chahiye: equilibrium woh licence hai jo "yeh forces equal hain" likhne deta hai. Derivation mein har equation asal mein ek equilibrium statement hai.

10. Sab ek saath jodna — do named angles

Ab har symbol exist karta hai, toh hum state kar sakte hain (abhi fully derive nahin) jo parent prove karta hai:

Figure 4 total reaction $R$ dikhata hai — woh single arrow jo $N$ aur $f$ add karke milta hai — angle $\lambda$ se normal se lean karta hua. Woh woh picture hai jo parent ka Pythagoras step build karta hai.

Wahi idea, $\arctan (\mu )$, Banking of Roads mein curved road ke safe-speed tilt ke liye wapas aata hai — toh ise yahan master karna do jagah pay off karta hai.

Prerequisite map

Equipment checklist

Khud ko test karo — right side cover karo aur reveal karne se pehle jawaab do.