Worked examples — Friction — static (maximum), kinetic, rolling

1.2.6 · D3 · Physics › Newton's Laws & Dynamics › Friction — static (maximum), kinetic, rolling

Yeh page friction topic ka stress test hai. Pehle hum har tarah ki situation list karenge jo yeh topic throw kar sakta hai, phir guarantee karenge ki har ek ko kam se kam ek baar solve kiya jaye.

Yahan sab kuch parent note ke teen facts pe tika hua hai. Unhe plain words mein dobara bata lete hain taaki koi bhi symbol bina matlab ke use na ho:

Recall Jin symbols ka hum sahara lenge

• $m$ (mass) object mein "cheez" ki matra hai, kilograms (kg) mein. Iska weight — gravity ka pull — $mg$ hota hai, jahan $g$ gravity ki strength hai.
• $N$ (normal force) surface ke perpendicular, ground ka pushback hai. Units: newtons (N).
• Static friction $f_s$ self-adjusting hai: yeh sirf utna hi supply karta hai jitna sliding rokne ke liye chahiye, lekin ek ceiling $f_s^{max}={\mu }_{s}N$ se zyada kabhi nahi. Isliye yeh ek inequality hai $f_s\le {\mu }_{s}N$.
• Kinetic friction sliding shuru hone ke baad fixed hota hai: $f_k={\mu }_{k}N$, aur ${\mu }_k<{\mu }_s$.
• Rolling friction $f_r={\mu }_{r}N$ bina slipping ke roll karne wale wheel ko resist karta hai, jahan ${\mu }_r\ll {\mu }_k$.

Symbol $\mu$ ("mu") bas ek number hai — bada $\mu$ matlab stickier surfaces. $g=10{\text{m/s}}^2$ poori taraf unless stated.

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The scenario matrix

Har friction problem in case classes mein se ek (ya inka blend) hoti hai. Har row ek alag trap hai. Last column us example ka naam hai jo ise cover karta hai.

#	Case class	Deciding question	Degenerate / limit to watch	Covered by
A	Push below the ceiling	Kya applied force $<f_s^{max}$ hai?	Zero push ⇒ zero friction	Ex 1
B	Push above the ceiling	Slip hone ke baad ${\mu }_k$ use karo	Push exactly $=f_s^{max}$ (verge)	Ex 2
C	Incline, block stays	Kya $\tan \theta <{\mu }_s$ hai?	$\theta =0$ (flat) ⇒ koi driving force nahi	Ex 3
D	Incline, block slides	$\theta >{\theta }_r$: net force down-slope	$\theta =90^{\circ }$ ⇒ free fall	Ex 4
E	Angled force up (lift)	Angle pe push karne se $N$ badalta hai	Upar kheeenchne se $N\to 0$	Ex 5
E'	Angled force down (press)	Neeche angle $N$ badhata hai	Zyada press ⇒ zyada grip	Ex 6
F	Friction as the driver	"Slide karne ki tendency" ka sign flip ho jaata hai	Max grip = ${\mu }_{s}N$	Ex 7
G	Two-body / connected	Shared motion pe friction	Zero friction ⇒ Atwood limit	Ex 8
R	Rolling resistance / grip→roll	${\mu }_r=a/R$, ${\mu }_k$ se bohot chhota	Roll shuru hone se pehle grip limit	Ex 9
H	Real-world word + exam twist	Stopping distance se $\mu$ nikalna	Wet road ⇒ $\mu$ girta hai	Ex 10

Pattern notice karo: har problem ek driving tendency ko ek ceiling ${\mu }_{s}N$ se compare karke shuru hoti hai. Woh comparison sahi karo aur baaki sab Newton's second law hai.

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Example 1 — Case A: push below the ceiling (aur zero-push corner)

Step 1 — Ceiling nikalo. $f_s^{max}={\mu }_{s}N={\mu }_{s}mg=0.5\times 10\times 10=50\text{N}$ Yeh step kyun? Har decision driving force ko is ceiling se compare karta hai. Yahi referee hai.

Step 2 — Case (a), zero push. Koi push nahi toh slide karne ki koi tendency nahi, isliye static friction kuch nahi supply karta: $f_s=0\text{N}$. Yeh step kyun? Yeh Case A ka degenerate corner hai. Friction reactive hai — koi cause nahi, koi effect nahi. Beginners yahan galti se $50\text{N}$ likhte hain.

Step 3 — Case (b), 30 N push. Positive = push direction lete hue compare karo: $30\text{N}<50\text{N}$. Crate move nahi karta, isliye Newton's Second Law ke zariye $a=0$ ke saath friction push ko exactly cancel karta hai: $f_s=30\text{N}$ Yeh step kyun? Static friction self-adjusting hai; yeh ceiling ko nahi balki push ko copy karta hai.

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Example 2 — Case B: push above the ceiling (aur verge corner)

Step 1 — Ceiling yaad karo: $f_s^{max}=50\text{N}$ (Ex 1 se). Yeh step kyun? Same referee.

Step 2 — Case (a), the verge. $50\text{N}=f_s^{max}$. Crate brink pe hai: friction abhi just barely hold kar sakti hai, isliye $a=0$ rehta hai. Yeh step kyun? Inequality $f_s\le {\mu }_{s}N$ mein equality bhi include hai. Ek hair zyada aur yeh free ho jaata hai.

Step 3 — Case (b), it slides. $65>50$, toh yeh move karta hai. Kinetic friction pe switch karo — bonds ab continuously toot rahe hain: $f_k={\mu }_{k}mg=0.3\times 100=30\text{N}$ Yeh step kyun? Sliding hone ke baad kinetic law govern karta hai, aur yeh chhota hota hai. Yahan ${\mu }_s$ rakhna ek classic error hai.

Step 4 — Newton's second law (positive = push direction, friction negative): $a=\frac{F-f_k}{m}=\frac{65-30}{10}=3.5{\text{m/s}}^2$ Yeh step kyun? Friction maloom hone ke baad net force / mass acceleration deta hai.

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Example 3 — Case C: incline, block stays put

Figure dekho. Gravity $mg$ seedha neeche point karti hai. Positive direction = slope ke neeche. Hum gravity ko ramp ke apne directions ke saath do perpendicular pieces mein split karte hain (orange aur teal arrows): ek slope ke neeche (driving piece, $+$) aur ek slope ke andar (jo $N$ set karta hai).

Step 1 — Gravity resolve karo. Figure ki geometry se: $\text{driving}=mg\sin \theta ,N=mg\cos \theta$ Yeh step kyun? Friction surface ke saath react karta hai, isliye humein force component surface ke along chahiye (wahi jo use fight karna hai) aur surface ke perpendicular wala (jo ceiling set karta hai).

Step 2 — Slope ko friction condition se compare karo. Block ruka rehta hai agar driving $\le$ ceiling: $mg\sin \theta \le {\mu }_{s}mg\cos \theta ⟺\tan \theta \le {\mu }_s$ Yeh step kyun? $mg\cos \theta$ se divide karne pe force comparison ek clean angle comparison ban jaata hai — isisi liye incline shortcut $\tan {\theta }_r={\mu }_s$ exist karta hai.

Step 3 — Numbers plug karo. $\tan 20^{\circ }\approx 0.364<0.5$. Yeh ruka rehta hai. Yeh step kyun? Sirf angle wala test sabse fast check hai.

Step 4 — Actual friction value. Kyunki yeh static hai aur $a=0$ hai, friction (up-slope point karta hai, $-$) driving piece ko balance karta hai: $f_s=mg\sin \theta =4\times 10\times \sin 20^{\circ }\approx 13.7\text{N}$ Yeh step kyun? ${\mu }_{s}N$ nahi! Static friction sirf utna hi supply karta hai jitna zarurat hai (Case A logic, ab ramp pe).

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Example 4 — Case D: incline, block slides (aur $90^{\circ }$ limit)

Positive direction = slope ke neeche, jaise agree kiya tha.

Step 1 — Slip test. $\tan 40^{\circ }\approx 0.839>{\mu }_s=0.5$. Driving ceiling ko beat karta hai ⇒ yeh slide karta hai. Yeh step kyun? Same referee as Ex 3, lekin ab inequality ulti hai.

Step 2 — Slide hone ke baad, kinetic friction slope ke upar act karta hai (neeche ki slide ko oppose karta hai, isliye yeh negative hai — figure mein plum arrow dekho): $f_k={\mu }_{k}mg\cos \theta =0.3\times 40\times \cos 40^{\circ }\approx 9.19\text{N}$ Yeh step kyun? Motion ke dauran kinetic law apply hota hai; friction actual sliding direction ko oppose karta hai.

Step 3 — Newton's second law slope ke along (driving $+$, friction $-$): $a=g\sin \theta -{\mu }_{k}g\cos \theta =10\sin 40^{\circ }-0.3\times 10\cos 40^{\circ }\approx 6.43-2.30=4.13{\text{m/s}}^2$ Yeh step kyun? Mass ke upar net down-slope force. Friction subtract hota hai kyunki yeh up-slope point karta hai.

Step 4 — $\theta \to 90^{\circ }$ limit. Tab $\cos \theta \to 0$, isliye $N\to 0$ aur $f_k\to 0$; $a\to g\sin 90^{\circ }=g$. Free fall. Yeh step kyun? Vertical "ramp" block ke against kuch press nahi karta, isliye friction khatam ho jaata hai — yeh sanity check hai ki model boundary pe sahi behave karta hai.

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Example 5 — Case E: upar angle pe kheenchna (aur $N\to 0$ corner)

Positive direction = horizontal drag direction. Rope tension do pieces mein split hoti hai: horizontal piece $T\cos \theta$ (drag karta hai, $+$) aur vertical piece $T\sin \theta$ (upar uthata hai, ground ka load halka karta hai) — teal aur orange arrows dekho.

Step 1 — Vertical balance $N$ set karta hai. Ground ab pura weight nahi carry karta: $N=mg-T\sin \theta$ Yeh step kyun? Upar kheenchne se crate halka lagta hai, isliye $N$ shrink hota hai — aur $N$ poori friction budget control karta hai.

Step 2 — (a) $T=40$ ke saath ceiling. $N=100-40\sin 30^{\circ }=100-20=80\text{N},f_s^{max}=0.5\times 80=40\text{N}$ Yeh step kyun? Lift ne ceiling ko Ex 1 ke 50 N se neeche 40 N kar diya.

Step 3 — Horizontal drive vs ceiling. $T\cos 30^{\circ }=40\times 0.866=34.6\text{N}<40\text{N}\Rightarrow \text{stays.}{f}_s=34.6\text{N}$ Yeh step kyun? Actual pull ko reduced ceiling se compare karo; static friction pull ke saath match karta hai.

Step 4 — (b) $N=0$ corner. $N=0$ set karo: $mg=T\sin \theta \Rightarrow T=\frac{mg}{\sin 30^{\circ }}=\frac{100}{0.5}=200\text{N}$ Yeh step kyun? Is se aage crate ground se bilkul lift ho jaata hai — friction khatam ho jaata hai kyunki press karne ke liye kuch nahi bacha.

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Example 6 — Case E': neeche angle pe dhakkelna (mirror case — zyada grip)

Positive direction = horizontal shove direction. Ab vertical piece $P\sin \theta$ neeche point karta hai, ground ke load mein add ho jaata hai — Ex 5 ka mirror image (orange arrow floor mein press karte hue dekho).

Step 1 — Vertical balance $N$ set karta hai. Neeche ka push load badhata hai: $N=mg+P\sin \theta$ Yeh step kyun? Neeche press karne se ground zyada pushback karta hai, isliye $N$ badhta hai — aur bada $N$ friction ceiling raise karta hai.

Step 2 — (a) $P=40$ ke saath ceiling. $N=100+40\sin 30^{\circ }=100+20=120\text{N},f_s^{max}=0.5\times 120=60\text{N}$ Yeh step kyun? Press ne ceiling ko Ex 1 ke 50 N se upar 60 N kar diya.

Step 3 — Horizontal drive vs ceiling. $P\cos 30^{\circ }=40\times 0.866=34.6\text{N}<60\text{N}\Rightarrow \text{stays.}{f}_s=34.6\text{N}$ Yeh step kyun? Actual shove ko raised ceiling se compare karo; static friction shove ke saath match karta hai.

Step 4 — Mirror comparison. Pull-up ratio tha $34.6/40=0.865$; push-down ratio hai $34.6/60=0.577$. Same horizontal drive, lekin neeche wala version slipping se bohot dur hai. Yeh step kyun? Yahi mirror case ka poora point hai: neeche angle karne se grip milti hai, upar angle karne se grip jaati hai.

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Example 7 — Case F: friction as the DRIVER (walking / accelerating car)

Positive direction = car ka forward motion. Yahan friction villain nahi balki hero hai. Tyre ka contact patch road ke against backward slip karne ki tendency rakhta hai (driven wheel road ko backward push karta hai), isliye static friction tyre ko — aur car ko — forward push karta hai (figure mein orange arrow). Yahi free-body insight hai jo parent note ne flag kiya tha.

Step 1 — Max grip ceiling hai. Forward force static friction ke maximum se zyada nahi ho sakta: $F_{\max }={\mu }_{s}N={\mu }_{s}mg=0.8\times 1200\times 10=9600\text{N}$ Yeh step kyun? Static friction abhi bhi $f_s\le {\mu }_{s}N$ maanta hai; ceiling cap karta hai ki road car ko kitni zyada push kar sakti hai.

Step 2 — Newton's second law. $a_{\max }=\frac{F_{\max }}{m}=\frac{9600}{1200}=8{\text{m/s}}^2$ Yeh step kyun? Grip limit pe, yeh sabse bada acceleration possible hai. Engine ko aur hard push karo aur tyre ${\mu }_{s}N$ se zyada ho jaata hai, kinetic (spinning) friction mein break ho jaata hai, aur grip girta hai.

Step 3 — Degenerate note. Ice pe, ${\mu }_s\to 0.1$, toh $a_{\max }\to 1{\text{m/s}}^2$ — wheels lagbhag immediately spin karte hain. Yeh step kyun? Dikhata hai ki model surface ke saath scale karta hai, everyday experience se match karta hai.

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Example 8 — Case G: connected bodies (aur frictionless limit)

Positive direction = jis direction mein system move karna chahta hai: $A$ pulley ki taraf, $B$ neeche.

Step 1 — Kya yeh move karta hai? Drive ko STATIC ceiling se compare karo. Block $B$ ka weight $m_{B}g$ cord kheenchta hai, aur woh tension hi kuch cheez hai jo $A$ ko table pe drag karne ki koshish karta hai. Koi motion assume karne se pehle hume poochna hoga: kya $A$ pe static friction us pull ko hold kar sakta hai? Iska maximum hai $f_s^{max}={\mu }_s{m}_{A}g=0.25\times 3\times 10=7.5\text{N},m_{B}g=20\text{N}.$ Kyunki $20\text{N}>7.5\text{N}$, drive static ceiling ko beat karta hai ⇒ system move karta hai. Yeh step kyun? Two-body system bhi same referee se govern hota hai: kuch bhi tab tak nahi move karta jab tak driving pull $f_s^{max}={\mu }_s{m}_{A}g$ se zyada na ho. Hum yahan ${\mu }_s$ use karte hain (na ki ${\mu }_k$) kyunki hum test kar rahe hain ki kya yeh start hota hai — kinetic friction tabhi exist karta hai jab sliding already shuru ho chuki ho, isliye ise "does it move?" test mein use karna circular hoga.

Step 2 — Ab jo slide ho raha hai, $A$ pe kinetic friction use karo (pulley ki taraf iske slide ko oppose karta hai, isliye yeh negative hai): $f_k={\mu }_k{m}_{A}g=0.2\times 3\times 10=6\text{N}$ Yeh step kyun? Motion confirm ho gaya, isliye kinetic law ab apply hota hai; friction $A$ ki actual motion se ladhta hai.

Step 3 — Poore system ke liye Newton's second law. Girta weight drive karta hai ($+$), friction resist karta hai ($-$); dono blocks ek acceleration share karte hain aur internal tension cancel ho jaata hai: $a=\frac{m_{B}g-f_k}{m_A+m_B}=\frac{20-6}{5}=\frac{14}{5}=2.8{\text{m/s}}^2$ Yeh step kyun? Pair ko ek mass treat karne se unknown tension ek hi jhatke mein hat jaati hai.

Step 4 — Tension (block $B$ isolate karo: $m_{B}g-T=m_{B}a$): $T=m_B(g-a)=2(10-2.8)=14.4\text{N}$ Yeh step kyun? Internal force nikaalne ke liye ek body isolate karo.

Step 5 — Frictionless limit. ${\mu }_k=0$ set karo: $a=20/5=4{\text{m/s}}^2$, $T=2(10-4)=12\text{N}$ — table-pe-classic Atwood result. Yeh step kyun? Confirm karta hai ki jab friction switch off ho jaata hai tab hamara formula sahi se reduce hota hai.

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Example 9 — Case R: rolling resistance aur grip→roll boundary

Positive direction = cart ka forward motion.

Step 1 — Geometry se rolling coefficient. Parent note ne ${\mu }_r=a/R$ derive kiya tha (normal force $a$ aage shift hoti hai, ek backward torque $Na$ banata hai jo $f_{r}R$ se balance hota hai): ${\mu }_r=\frac{a}{R}=\frac{0.0015}{0.30}=0.005$ Yeh step kyun? Yeh ek chhoti si deformation length ko pure number mein convert karta hai jo resistance scale karta hai.

Step 2 — Rolling-resistance force. $f_r={\mu }_{r}N={\mu }_{r}mg=0.005\times 50\times 10=2.5\text{N}$ Yeh step kyun? Flat ground pe $N=mg$; ${\mu }_r$ se multiply karo gentle backward drag milta hai jo rolling wheel feel karta hai.

Step 3 — (b) Grip ceiling (roll→slip boundary). Jab tak drive static grip limit se neeche rahe contact patch pe, wheel roll karta hai; usse cross karo aur patch slip ho jaata hai: $f_{grip}^{max}={\mu }_{s}N=0.7\times 50\times 10=350\text{N}$ Yeh step kyun? Rolling-without-slipping contact point pe static friction se hold hoti hai — wahi ${\mu }_{s}N$ ceiling. Isse cross karo aur tyre kinetic friction mein skid karta hai.

Step 4 — Dono regimes compare karo. $\frac{f_{grip}^{max}}{f_r}=\frac{350}{2.5}=140$. Yeh step kyun? Numbers mein dikhata hai kyun wheels kamaal hain: everyday drag ($2.5\text{N}$) slip karne se pehle call kar sakte grip se $140\times$ chhota hai.

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Example 10 — Case H: real-world word problem + exam twist

Positive direction = car ka forward motion; friction (backward) negative hai aur decelerate karta hai.

Step 1 — Friction se deceleration. Brakes locked hone par, kinetic friction hi poora force hai: $ma={\mu }_{k}mg\Rightarrow a={\mu }_{k}g$ Yeh step kyun? Mass cancel ho jaata hai — rukna sirf ${\mu }_k$ aur $g$ pe depend karta hai, ek beautiful energy-friendly fact.

Step 2 — Kinematics link. $v^2=u^2-2ad$ use karo final $v=0$ ke saath: $0=u^2-2({\mu }_{k}g)d\Rightarrow {\mu }_k=\frac{u^2}{2gd}$ Yeh step kyun? Measurable distance ko unknown ${\mu }_k$ se connect karta hai.

Step 3 — Plug in (a). ${\mu }_k=\frac{20^2}{2\times 10\times 25}=\frac{400}{500}=0.8$ Yeh step kyun? Direct substitution; ek plausibly grippy dry road.

Step 4 — Twist (b). Wet: ${\mu }_k^{\prime }=0.4$. Kyunki $d=\frac{u^2}{2{\mu }_{k}g}$ hai, ${\mu }_k$ aadha karne se $d$ double ho jaata hai: $d^{\prime }=\frac{20^2}{2\times 0.4\times 10}=\frac{400}{8}=50\text{m}$ Yeh step kyun? Fixed speed pe $d\propto 1/{\mu }_k$ — exam twist ka reward proportionality spot karna hai, na dobara derive karna.

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Recap — woh ek habit jo sab solve kar deti hai