Friction — static (maximum), kinetic, rolling: Exercises

1.2.6 · D4 · Physics › Newton's Laws & Dynamics › Friction — static (maximum), kinetic, rolling

Shuru karne se pehle, ek reminder un saare symbols ka jo hum use karte hain, taaki kuch unexplained na lage:

Definition Symbol glossary

$N$ — normal force: woh push jo surface perpendicular (yaani $9 0^{\circ}$ par) apni taraf se deti hai. Socho ek floor ek box ko seedha upar push kar rahi hai.
$f_{s}$ — static friction: tab kaam karta hai jab surfaces abhi slide nahi kar rahi. Self-adjusting hoti hai, $f_{s} \leq μ_{s} N$ .
$f_{k}$ — kinetic friction: slide karte waqt kaam karta hai, $f_{k} = μ_{k} N$ .
$f_{r}$ — rolling friction: ek wheel/ball jo bina slipping ke roll karta hai uske liye choti si resisting force; $f_{r} = μ_{r} N$ jahan $μ_{r}$ bahut chhota hota hai.
$μ_{s}, μ_{k}, μ_{r}$ — pure numbers (koi unit nahi) jo stickiness measure karte hain: static, kinetic, rolling. $μ_{s} > μ_{k} ≫ μ_{r}$ .
$P$ — ek applied push: ek external force jo tum lagate ho, yahan slope ke along directed hai.
$θ$ — ek angle, tilt ki matra, degrees mein measure hoti hai.
$tan θ$ — ek right triangle ka "opposite over adjacent": slope kitna steep hai yeh batata hai.

Level 1 — Recognition

L1.1 — Regime ko identify karo

Ek crate zameen par rakha hai. Tum usse chhoo nahi rahe. Us par kaunsi friction force kaam kar rahi hai, aur woh kitni badi hai?

Recall Solution

KYA: Koi horizontal push nahi matlab slide karne ki koi tendency nahi. KYUN: Static friction self-adjusting hoti hai — yeh sirf utna hi supply karti hai jitna sliding rokne ke liye zaroori ho. Zaroori $= 0$ , isliye yeh $0$ supply karti hai. Answer: $f_{s} = 0 N$ . Friction nahi hoti $μ_{s} N$ yahan; woh sirf ceiling hai, actual value nahi.

L1.2 — Ceiling padhna

Ek $4 kg$ block level floor par, $μ_{s} = 0.6$ . Friction se sabse zyada horizontal force kitni supply ho sakti hai block ke free hone se pehle?

Recall Solution

Step 1 — $N$ nikalo. $N = m g = 4 \times 10 = 40 N$ . Kyun: block vertically nahi hilta, isliye upward normal force neeche wale weight ko exactly balance karni chahiye; koi aur vertical push nahi hai, isliye $N = m g$ . Step 2 — ceiling formula lagao. $f_{s}^{ma x} = μ_{s} N = 0.6 \times 40 = 24 N$ . Kyun: static friction jo sabse zyada supply kar sakti hai woh ceiling $μ_{s} N$ hai, $f_{s}^{ma x}$ ki definition ke hisaab se. Answer: $24 N$ .

Level 2 — Application

L2.1 — Ceiling ke neeche

Ek $10 kg$ box, $μ_{s} = 0.5$ , $μ_{k} = 0.3$ , horizontally $30 N$ se push kiya. Friction force aur acceleration nikalo.

Recall Solution

Step 1 — ceiling. $N = m g = 100 N$ , isliye $f_{s}^{ma x} = 0.5 \times 100 = 50 N$ . Kyun: hum jaanna chahte hain ki static friction kitni zyada supply kar sakti hai box ke break free hone se pehle, isliye pehle ceiling nikalni hogi. Step 2 — compare karo. Push $= 30 N < 50 N$ ⇒ slide nahi karta. Kyun: agar push ceiling se neeche hai, static friction usse still match kar sakti hai, isliye koi sliding nahi hoti. Step 3 — actual friction. Static aur nahi hua ⇒ net force zero ⇒ friction push ko exactly cancel karti hai: $f_{s} = 30 N$ . Kyun: $a = 0$ matlab (Newton's 2nd law) horizontal forces ka sum zero hai, isliye friction push ke barabar honi chahiye. Step 4 — acceleration. $a = 0$ . Kyun: box slide nahi karta aur koi aur cheez use move nahi karti, isliye uski velocity unchanged hai, yaani $a = 0$ .

L2.2 — Ceiling ke upar

Wahi box, ab $70 N$ se push kiya. Friction aur acceleration nikalo.

Recall Solution

Step 1 — ceiling. Abhi bhi $f_{s}^{ma x} = 50 N$ . Kyun: mass, $μ_{s}$ aur $g$ unchanged hain, isliye static ceiling $μ_{s} m g$ pehle jaisi hi hai. Step 2 — compare karo. $70 > 50$ ⇒ box slide karta hai. Kyun: push ab maximum static friction se zyada hai, isliye friction box ko hold nahi kar sakti aur woh break free ho jaata hai. Step 3 — kinetic par switch karo. Ek baar move hone par kinetic law apply hota hai: $f_{k} = μ_{k} N = 0.3 \times 100 = 30 N$ . Kyun: sliding shuru ho gayi, isliye sliding-bond model ( $f_{k} = μ_{k} N$ , chhota) static wale ki jagah le leta hai. Step 4 — Newton's 2nd law. Net $= 70 - 30 = 40 N$ , isliye $a = \frac{40}{10} = 4 m/s^{2}$ . Kyun: friction ab known aur constant hai, isliye leftover (unbalanced) force ko mass se divide karne par acceleration milta hai.

L2.3 — Angle of repose se friction nikalna

Ek coin tab slide karna shuru karta hai jab tray $3 7^{\circ}$ tak tilt ki jaati hai. $μ_{s}$ nikalo. (Lo $tan 3 7^{\circ} = 0.75$ .)

Figure — Friction — static (maximum), kinetic, rolling

Recall Solution

KYUN yeh tool use karein: us tilt par jahan sliding abhi shuru hoti hai, down-slope pull maximum friction ke barabar hoti hai — figure dekho. Red arrow down-slope pull $m g sin θ$ hai; green arrow maximum static friction $f_{s}^{ma x} = μ_{s} m g cos θ$ hai jo slope ke upar point kar rahi hai. Slipping ke verge par yeh equal aur opposite hoti hain. Ek-line reminder ki $tan θ_{r} = μ_{s}$ kyun hota hai. Red $=$ green set karo: $m g sin θ_{r} = μ_{s} m g cos θ_{r}$ . $m g$ cancel karo aur $cos θ_{r}$ se divide karo: $\frac{sin θ _{r}}{cos θ _{r}} = μ_{s}$ , yaani $tan θ_{r} = μ_{s}$ . Step 1: $μ_{s} = tan θ_{r} = tan 3 7^{\circ} = 0.75$ . Kyun: first slip par tilt exactly angle of repose hai, isliye derived condition $tan θ_{r} = μ_{s}$ directly apply hoti hai. Answer: $μ_{s} = 0.75$ .

Level 3 — Analysis

L3.1 — Incline par block: kya yeh slide karega?

Ek $2 kg$ block $3 0^{\circ}$ incline par ruka hua hai, $μ_{s} = 0.7$ . Kya yeh slide karega? Agar nahi, toh us par friction force kitni hai? (Lo $sin 3 0^{\circ} = 0.5$ , $cos 3 0^{\circ} = 0.866$ .)

Recall Solution

Step 1 — gravity resolve karo. $m g$ ko slope along wale part (jo sliding drive karta hai) aur slope mein wale part (jo surface press karta hai) mein split karo. Figure ka decomposition dekho. Kyun: gravity seedhi neeche point karti hai lekin block sirf slope ke along slide kar sakta hai, isliye hum motion drive karne wale part ko surface ke resist karne wale part se alag karte hain.

Along slope: $m g sin θ = 2 \times 10 \times 0.5 = 10 N$ .
Into slope: $m g cos θ = 2 \times 10 \times 0.866 = 17.32 N$ . Step 2 — normal force. $N = m g cos θ = 17.32 N$ . Kyun: block slope ke perpendicular move nahi karta, isliye $N$ sirf gravity ke into-slope part ko balance karta hai. Step 3 — ceiling. $f_{s}^{ma x} = μ_{s} N = 0.7 \times 17.32 = 12.12 N$ . Kyun: yahi sabse zyada friction surface supply kar sakti hai, humein judge karna hai ki block hold karega ya nahi. Step 4 — driving vs ceiling compare karo. Driving $= 10 N < 12.12 N$ ⇒ slide nahi karta. Kyun: down-slope pull friction ceiling se neeche hai, isliye static friction usse poori tarah cancel kar sakti hai. Step 5 — actual friction. Static, koi acceleration nahi ⇒ friction driving pull ko balance karti hai: $f_{s} = 10 N$ (slope ke upar). Kyun: $a = 0$ ⇒ slope ke along forces ka sum zero hai, isliye friction $10 N$ pull ke barabar hai, $12.12 N$ ceiling ke nahi. Answer: ruka rehta hai; $f_{s} = 10 N$ .

L3.2 — Do blocks, kya stack slip karega?

Usi $3 0^{\circ}$ incline par, mass double hokar $4 kg$ ho jaata hai (wahi $μ_{s} = 0.7$ ). Kya ab yeh slide karega? Ek sentence mein explain karo ki answer kyun nahi badal sakta.

Recall Solution

Key insight: driving force $m g sin θ$ aur ceiling $μ_{s} m g cos θ$ dono $m$ ke saath scale karte hain. Unka ratio $tan θ / μ_{s}$ hai, jo mass se independent hai. Check: driving $= 4 \times 10 \times 0.5 = 20 N$ ; ceiling $= 0.7 \times 4 \times 10 \times 0.866 = 24.25 N$ . Abhi bhi $20 < 24.25$ ⇒ koi slide nahi. Kyun unchanged: sliding is baat par depend karti hai ki $tan θ > μ_{s}$ hai ya nahi, jisme koi $m$ nahi hai. Kyunki $tan 3 0^{\circ} = 0.577 < 0.7$ , is slope par koi bhi mass ise slide nahi kara sakta.

Level 4 — Synthesis

L4.1 — Incline ke saath extra push

Ek $5 kg$ block $3 0^{\circ}$ incline par, $μ_{s} = 0.4$ , $μ_{k} = 0.3$ . Akela chhoda jaaye toh yeh slide karta (check karo!). Tum ise surface ke parallel force $P$ se slope upar push karte ho, aur yeh ruka rehta hai. $P$ ka woh range nikalo jo ise static rakhe. (Use karo $sin 3 0^{\circ} = 0.5$ , $cos 3 0^{\circ} = 0.866$ .)

Recall Solution

Figure mein blue arrow push $P$ slope ke upar hai, white arrow gravity ka down-slope pull $m g sin θ$ hai, aur do red arrows friction ke do possible directions dikhate hain — slope ke upar jab $P$ chhota ho (block neeche slide karne ki tendency) aur slope ke neeche jab $P$ bada ho (block upar slide karne ki tendency). Step 0 — sanity check. $tan 3 0^{\circ} = 0.577 > μ_{s} = 0.4$ , isliye bina help ke yeh neeche slide karta hai. Theek hai — akela friction ise rok nahi sakta. Kyun: agar $tan θ > μ_{s}$ toh down-slope pull already friction ceiling se zyada ho jaati hai, isliye push genuinely zaroori hai. Setup. $N = m g cos θ = 5 \times 10 \times 0.866 = 43.3 N$ , isliye $f_{s}^{ma x} = μ_{s} N = 0.4 \times 43.3 = 17.32 N$ ; gravity ka down-slope pull $m g sin θ = 5 \times 10 \times 0.5 = 25 N$ hai. Kyun: do extreme cases test karne se pehle hum fixed quantities (ceiling aur gravity pull) collect karte hain. Case A — sabse chhota $P$ (block neeche slide karne ki kagar). Friction gravity rok ne mein $P$ ki madad karne ke liye poori strength se upar point karti hai: $P_{min} + f_{s}^{ma x} = m g sin θ ⟹ P_{min} = 25 - 17.32 = 7.68 N .$ Kyun: agar $P$ aur chhota hota, toh maximum up-slope friction bhi block ko neeche slide hone se nahi rok sakti. Case B — sabse bada $P$ (block upar slide karne ki kagar). Friction flip hokar poori strength se neeche point karti hai, $P$ ko resist karti hai: $P_{ma x} = m g sin θ + f_{s}^{ma x} = 25 + 17.32 = 42.32 N .$ Kyun: agar $P$ aur bada hota, toh maximum down-slope friction bhi block ko upar slide hone se nahi rok sakti. Answer: block $7.68 N \leq P \leq 42.32 N$ ke liye static rehta hai.

Level 5 — Mastery

L5.1 — Sliding + energy

L4 wala $5 kg$ block usi $3 0^{\circ}$ incline par ( $μ_{k} = 0.3$ ) bina push ke rest se release hota hai. Yeh slope ke along $d = 4 m$ neeche slide karta hai. Nikalo (a) acceleration aur (b) us stretch ke bottom par speed. (Use karo $sin 3 0^{\circ} = 0.5$ , $cos 3 0^{\circ} = 0.866$ .)

Recall Solution

Part (a) — Newton's 2nd law se acceleration.

Neeche driving: $m g sin θ = 25 N$ .
Kinetic friction (motion ko oppose, isliye slope ke upar): $f_{k} = μ_{k} m g cos θ = 0.3 \times 43.3 = 12.99 N$ .
Slope ke along net: $25 - 12.99 = 12.01 N$ .
$a = \frac{12.01}{5} = 2.40 m/s^{2}$ slope ke neeche.

Part (b) — Work-Energy Theorem se speed. Kyun yeh tool: work–energy net force ko distance ke saath directly kinetic energy ke change se link karta hai, isliye hume time ki zaroorat nahi padti final speed ke liye. Kyun yeh (net force)×d tak reduce hota hai: net force yahan constant hai (teeno forces constant hain) aur straight-line motion ke along point karta hai, isliye uska work simply force × distance hai bina kisi angle factor ke. Block rest se shuru karta hai, isliye kinetic energy ka change sirf final $\frac{1}{2} m v^{2}$ hai: $\frac{1}{2} m v^{2} = (net force) \times d = 12.01 \times 4 = 48.05 J$ $v = \frac{2 \times 48.05}{5} = 19.22 = 4.38 m/s .$ Answer: $a \approx 2.40 m/s^{2}$ , $v \approx 4.38 m/s$ .

L5.2 — Rolling vs sliding

Radius $R = 0.25 m$ ka ek wheel bina slipping ke roll karta hai. Normal force ka contact point wheel ke centre se $a = 0.5 mm$ aage hai. Rolling coefficient $μ_{r}$ nikalo aur ise typical kinetic value $μ_{k} = 0.3$ se compare karo.

Recall Solution

Glossary se yaad karo ki $f_{r} = μ_{r} N$ ek rolling wheel par chhoti resisting force hai, aur (parent note se) $μ_{r} = a / R$ , jahan $a$ yeh hai ki wheel aur ground thoda deform hone ki wajah se normal force kitna aage shift hoti hai. Step 1 — $μ_{r} = a / R$ use karo. Kyun: shifted normal force ek backward torque $N a$ banata hai; ise $f_{r} R$ ke saath balance karne par $f_{r} = N a / R$ milta hai, yaani $μ_{r} = a / R$ . $μ_{r} = \frac{a}{R} = \frac{0.5 \times 1 0 ^{- 3} m}{0.25 m} = 0.002.$ Step 2 — compare karo. $\frac{μ _{k}}{μ _{r}} = \frac{0.3}{0.002} = 150$ . Kyun: dono coefficients ko divide karne par pata chalta hai ki sliding, rolling se kitni zyada resist karti hai. Answer: $μ_{r} = 0.002$ — kinetic friction se lagbhag 150 guna chhota. Isliye load ko roll karna, drag karne se behtar hota hai. Dekho bhi Rolling without Slipping.

L5.3 — Self-driving foot

$60 kg$ mass ka ek insaan chalta hai aur $1.5 m/s^{2}$ se forward accelerate karta hai. Uska paon slip nahi karta. Kaunsi force use forward push karti hai, friction kis direction mein kaam karti hai, aur minimum $μ_{s}$ kya hona chahiye jo yeh possible kare?

Recall Solution

Direction: forward accelerate karne ke liye paon zameen ko backward push karta hai; static friction ki wajah se zameen paon ko forward push karti hai. Friction yahan forward driving force hai. Kyun: friction contact par relative sliding ko oppose karti hai, aur paon backward slide karne ki tendency rakhta hai, isliye friction forward point karti hai. Step 1 — required force. $F = ma = 60 \times 1.5 = 90 N$ (yahi static friction hai). Kyun: insaan par sirf yahi horizontal force hai, isliye Newton's 2nd law iska size $ma$ fix karta hai. Step 2 — required $μ_{s}$ . Yeh ceiling $μ_{s} N = μ_{s} m g = μ_{s} \times 600$ ke andar fit honi chahiye. Kyun: static friction $μ_{s} N$ se zyada nahi ho sakti, isliye surface itni sticky honi chahiye ki zaroori $90 N$ supply kar sake. $μ_{s} \times 600 \geq 90 ⟹ μ_{s} \geq 0.15.$ Answer: friction $90 N$ forward supply karti hai; koi bhi $μ_{s} \geq 0.15$ kaam karega.

Recall checkpoint

Recall Quick self-check

Un-pushed box par actual static friction? ::: Zero — yeh sirf utna supply karti hai jitna zaroori ho. Incline par normal force kya carry karta hai? ::: Sirf perpendicular part, $N = m g cos θ$ . Incline ka slide-or-not answer mass-independent kyun hai? ::: Driving aur ceiling dono $m$ ke saath scale karte hain; slide hoga iff $tan θ > μ_{s}$ . Sliding shuru hone ke baad kaunsa coefficient friction govern karta hai? ::: Kinetic wala, $μ_{k}$ ( $μ_{s}$ se chhota). Pushed-incline problem mein static friction ki do limits kyun ho sakti hain? ::: Friction upar ya neeche point kar sakti hai is baat par depend karke ki block kis taraf slide karna chahta hai. Rolling friction itna tiny kyun hai? ::: $μ_{r} = a / R$ jahan deformation offset $a ≪ R$ hota hai.