Laws of Motion: Class 11 Physics Chapter 4 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

100 min
29 practice questions
30 MCQs

Is chapter mein aap kya seekhenge

Galileo ki reasoning se samjhao ki bina kisi net force ke uniform motion kyun chalta rehta hai, aur Newton ka first law batao
Momentum aur impulse ki definition do, aur F = dp/dt = ma ko vector aur component form mein apply karo
Newton ka third law batao, action–reaction pairs ko sahi pehchano, aur conservation of momentum derive karo
Force triangles aur components ki madad se particle ke equilibrium ki condition apply karo
Mechanics ke common forces samjhao aur static, kinetic aur rolling friction ke laws use karo
Level aur banked roads par circular motion analyse karo, aur optimum speed aur speeds ki safe range nikalo
Free-body diagrams ke saath connected aur accelerating bodies ke problems step by step solve karo

NCERT Physics Class 11, Chapter 4, pp. 49–70 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

4.1 Introduction

Chapters 2 aur 3 mein humne motion describe kiya: body kahan hai, kitni tez chalti hai aur uski velocity kaise badalti hai. Humne kabhi yeh nahi poocha ki woh kyun waise chalti hai. Yahi sawaal, yaani motion ke causes, is chapter ka subject hai, aur iska jawab poori mechanics ki neenv hai (गति के नियम, laws of motion).

Roz ka experience ek pehla jawab deta hai. Carrom ki goti tabhi chalti hai jab striker usse takrata hai, aur rolling ball tab rukti hai jab fielder use rokta hai. Kabhi push ya pull ke liye contact chahiye, jaise striker mein; kabhi yeh khaali space ke paar kaam karta hai, jaise jab aam daal se girta hai ya magnet pin ko apni taraf kheenchta hai. To shuru karna aur rokna, dono ke liye bahar se ek force chahiye. Lekin us goti ka kya jo pehle se achhe powder wale board par steadily fisal rahi hai? Kya kisi ko use lagataar push karte rehna padega?

4.2 Aristotle ki galti

Greek philosopher Aristotle (384–322 BCE) ka jawab tha haan. Unke hisaab se body tabhi tak chalti hai jab tak koi cheez us par act karti rehti hai; mover ko hata do to motion khatam. Unhone yeh bhi kaha ki udte teer ke peeche ghusti hawa use aage dhakelti rehti hai. Seedhe shabdon mein, Aristotle ka rule kehta hai: body ko chalte rakhne ke liye, us par bahar se force lagataar act karna chahiye.

Yeh samajhna aasaan hai ki yeh itna obvious kyun laga. Sabzi wale ka thela tabhi tak chalta hai jab tak woh use push karta rehta hai; jaise hi woh chhodta hai, thela dheema hokar ruk jaata hai. Earth par akeli chhodi gayi cheezein hamesha ruk jaati lagti hain.

Galti yeh hai ki thela kabhi akela chhoda hi nahi jaata. Road aur axles us par friction lagate hain, aur hawa us par drag lagati hai. Jab thela steady speed se chalta hai, to sabzi wale ka push sirf in opposing forces ko cancel karta hai, isliye thele par net force zero hota hai (Fig. 4.1). Jab woh chhod deta hai, to sirf friction hi horizontal force bachta hai, aur woh thele ko dheema karta hai. Friction aur drag ko poori tarah hata do, to thela bina kisi push ke chalta rahega. Aristotle ke observations sahi the; unke hisaab-kitaab mein ek force chhoot gaya tha. Breakthrough tab aaya jab kisi ne poocha ki agar woh chhoota hua force band kar diya jaaye to kya hoga, aur yeh poochne wale the Galileo.

1/3Pull aur friction barabar aur opposite hain: net force zero, velocity constant.

Fig. 4.1Push kiye gaye thele ko chalte rehne ke liye force kyun chahiye lagta hai: push sirf friction ko cancel karta hai.NCERT p. 50

Floor par do crates le jaate thele ka side view. Step 1: right ki taraf ek blue pull arrow aur left ki taraf ek orange friction arrow ki length barabar hai, aur ek green velocity arrow right ki taraf point karta hai; label kehta hai steady pull, net force zero. Step 2: pull arrow gayab hai, thela aage badh gaya hai, sirf friction act karta hai aur velocity arrow aadha lamba hai; label kehta hai friction use dheema karta hai. Step 3: floor ek ideal smooth surface ki tarah bana hai, koi friction arrow aur koi pull nahi hai, aur velocity arrow phir se poori length ka hai; label kehta hai woh chalta rehta hai.

4.3 The law of inertia

Seventeenth century ki shuruaat mein Galileo Galilei ne smooth, halki dhalaan wale tracks par rolling balls ka time naapa aur ek pattern pakda. Dhalaan par neeche jaate hue ball ki speed badhti hai; upar jaate hue speed ghatti hai; aur track jitna steep ho, change utna tez. To jo track kisi bhi taraf dhalaan na ho, us par koi change nahi hona chahiye, agar friction hataya ja sake: ball constant velocity se chalti rahegi.

Aamne-saamne do ramps wala ek thought experiment yahi baat dikhata hai (Fig. 4.2). Left ramp par kisi height se ball chhodo. Woh neeche aati hai, bottom paar karti hai aur right ramp par tab tak chadhti hai jab tak lagbhag apne starting level par wapas na aa jaaye; ramps jitne smooth, woh utna kareeb pahunchti hai, isliye frictionless ball theek usi level tak pahunchegi. Ab right ramp ko neeche jhukao. Same level dobara paane ke liye ball ko us par aur door jaana padta hai. Right ramp ko flat kar do, to woh level kabhi nahi milega, isliye ball bina ruke chalti rahegi. Asli friction use aakhir mein rok deta hai, lekin yeh trend ideal case ke baare mein zyada shak nahi chhodta.

2/3Halki slope: wahi height, lekin kaafi lambe raaste ke baad.

Fig. 4.2Galileo ka double slope (frictionless ideal): ball hamesha apni starting height dobara paane ki koshish karti hai.NCERT Fig. 4.1 · p. 50

Left mein ek fixed slope jismein release point par ball ki dashed outline hai, aur poore figure ke paar starting height dikhati ek dashed horizontal line hai. Right mein doosri slope apne paaye par hinged hai. Step 1: doosri slope steep hai aur ball thodi chadhai ke baad starting height par ruk jaati hai. Step 2: doosri slope halke angle tak neeche ki gayi hai aur ball phir se starting height par rukti hai, lekin kaafi aage. Step 3: doosri slope flat padi hai aur ball us par aage ki taraf arrow ke saath chalti hai, label kehta hai ki woh height kabhi nahi milti.

Isne do hazaar saal ki soch ulat di. Forces ke hisaab se, ek jagah ruke rehna aur steadily fisalna ek hi state hai: dono mein body par forces ka total zero hota hai. Force ki zaroorat velocity ke change ke liye hoti hai. Section 4.2 mein thele wala sirf friction cancel karne ke liye push karta hai. Har body jo kar rahi hai wahi karte rehna chahti hai, ya to ruki rehti hai ya steady speed se seedhi chalti rehti hai; is tendency ko inertia (जड़त्व) kehte hain.

Definition

Inertia

Body ki apni velocity ko na badalne ki tendency: agar ruki hai to ruki rahe, ya agar chal rahi hai to steady speed se seedhi chalti rahe. Zyada mass matlab zyada inertia.

ExampleEk loaded truck ko start ya stop karna scooter se kahin zyada mushkil hai.

Try karke dekho

Tumbler aur kaagaz ki patti

Tumhe chahiye

ek khaali steel tumbler
lagbhag 30 cm lambi kaagaz ki patti
ek smooth table

Pehle guess karo

Tum kaagaz ko jhatke se kheench lete ho. Tumbler kya karta hai?

Woh kaagaz ke saath ghasit jaata haiWoh lagbhag wahin rehta hai jahan thaWoh tumhari taraf gir jaata hai

Kaagaz ki patti table par aise rakho ki lagbhag 10 cm kinare se bahar latka ho, aur patti ke doosre sire par tumbler seedha khada karo.
Latakte sire ko pakdo aur patti ko ek hi tez movement mein jhatke se neeche aur door kheencho.
Sab kuch wapas rakho aur ab patti ko dheere kheencho. Compare karo ki tumbler kya karta hai.

History seNCERT p. 51

Ancient Indian science mein motion ke ideas

Vaisesika school ke thinkers ne motion ke causes ko alag-alag kinds mein baanta. Nodana ek lagataar push tha, jaise ek dhaara tairte lakkad ko steadily bahati hai; abhighata ek achanak chot thi, jaise mallet ka khoonti par maarna. Kaha jaata tha ki chalti body vega le kar chalti hai, yaani seedhi line mein chalte rehne ki ek andar ki tendency, jo doosri cheezon, hawa samet, ke contact se ghisti jaati hai. Vega shaayad inertia ke sabse kareeb ka ancient idea hai, aur uska dheere-dheere khatam hona friction aur air resistance jaisa hai.

Unhone yeh bhi pehchana ki poori body ka motion, chahe girna ho, ghoomna ho ya vibrate karna, uske particles ki movements se banta hai, aur woh space mein ek point ko teen directions ke along distances se locate karte the. 1150 CE mein mathematician Bhaskara II ne tatkalika gati describe ki, yaani ek single instant par motion, jo us idea ke kareeb hai jise aaj calculus instantaneous velocity kehta hai.

4.4 Newton's first law of motion

Galileo ne dikha diya tha ki purani mechanics mein kya galat tha; nayi mechanics zyaadatar Isaac Newton (1642–1727) ne banayi, jinki Principia (1687) mein motion ke teen laws diye gaye. Newton ne Galileo ke law of inertia ko apna first law banaya.

LawNewton's first law of motion

Ek body ruki rehti hai, ya constant speed se seedhi line mein chalti rehti hai, jab tak us par koi net external force act na kare. Doosre shabdon mein, zero net external force ka matlab zero acceleration, aur sirf net external force hi body ko accelerate kar sakta hai.

Hum is law ko do taraf se use karte hain. Kabhi humein pata hota hai ki forces cancel ho rahe hain aur hum nateeja nikalte hain ki acceleration zero hai: deep space mein, har star se door, tairta dhool ka kan lagbhag koi force mehsoos nahi karta, isliye uski jo bhi velocity hai woh wahi rakhta hai. Zyaadatar hum pehle se har force ki list nahi bana sakte, lekin hum dekh sakte hain ki body accelerate nahi kar rahi, aur tab law batata hai ki us par forces ka total zero hona chahiye. Earth par gravity hamesha act karti hai, aur chalti cheezon ko friction aur drag milta hai, isliye ruki hui ya uniform motion wali body forces se free nahi hoti; uske forces bas cancel ho jaate hain.

Desk par rakha ek phone lo. Us par do forces act karte hain: neeche ki taraf uska weight $W$ aur upar ki taraf desk ka push, yaani normal force $N$ . Kisi ne $N$ pehle se nahi naapa; desk jitna push chahiye utna de deti hai, aur load ke hisaab se khud ko adjust kar leti hai. Sahi reasoning sirf ek hi direction mein chalti hai: phone ruka hua dikhta hai, isliye first law se net force zero hai, isliye $N = W$ . Yeh kehna ki phone ruka hai kyunki $N$ barabar hai $W$ ke, logic ko ulta kar deta hai.

Ab ek car socho jo traffic light se chalti hai, phir seedhi road par steady speed se cruise karti hai. Jab tak uski speed badhti hai woh accelerate karti hai, isliye us par ek net external force act karna chahiye. Uska engine aur gears internal hain: akele woh car ko utna hi accelerate kar sakte hain jitna tum apne joote ke feete kheench kar khud ko utha sakte ho. Engine wheels ko ghumata hai, tyres road ko peeche push karte hain, aur road friction (Section 4.9) ke zariye car ko aage push karti hai. Jab car steadily cruise karti hai, to woh aage wala friction air drag aur rolling resistance ko theek balance karta hai, aur net force phir se zero ho jaata hai.

Inertia bus mein lagne wale jaane-pehchane jhatke ko samjhati hai (Fig. 4.3). Jab khade passenger ki bus achanak chalti hai, to floor ka friction pairon ko bus ke saath aage kheenchta hai, lekin upar ki body wahin rehna chahti hai, isliye passenger bus ke relative peeche jhuk jaata hai jab tak muscles body ko saath nahi le aate. Jab bus zor se brake lagati hai, to friction pairon ko rok deta hai jabki upar ki body aage chalti rehti hai, aur passenger aage ki taraf jhatka khaata hai. Handrail pakadna madad karta hai kyunki tab haath woh force de sakta hai jo upar ki body ko chahiye.

2/3Friction pairon ko aage kheenchta hai; upar ki body peeche reh jaati hai aur passenger peeche jhuk jaata hai.

Fig. 4.3Friction pairon ko bus ke saath chalata hai; inertia baaki body ko wahi karte rakhti hai jo woh kar rahi thi.NCERT p. 52

Bus ke andar ka side view jismein beech mein ek passenger khada hai. Step 1: bus ruki hai aur passenger seedha khada hai. Step 2: ek blue acceleration arrow aage, right ki taraf point karta hai; pairon par ek orange friction arrow bhi aage point karta hai, aur passenger ki upar ki body peeche jhuki hai, label kehta hai body peeche reh jaati hai. Step 3: acceleration arrow peeche point karta hai, pairon par friction arrow peeche point karta hai, aur upar ki body aage jhuki hai, label kehta hai body aage chalti rehti hai.

Example 4.1NCERT Example 4.1 · p. 52

Probe ka peeche chhoda gaya package

Kisi bhi star ya planet se door ek space probe apne thrusters chalata hai aur steady $3.0 m / s^{2}$ se accelerate karta hai. Uska robotic arm ek chhota instrument package chhod deta hai. (a) Chhodne ke theek baad package ka acceleration kya hai? (b) $4.0 s$ baad probe aur package kitni door hain, aur woh kitni tezi se alag ho rahe hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.5 Newton's second law of motion

First law zero net force wala special case cover karta hai. Second law baaki har case sambhalta hai: yeh batata hai ki koi net external force body ke motion ko kitna badalta hai. Ise state karne ke liye humein ek nayi quantity chahiye.

Momentum

Body ke mass ko uski velocity se multiply karo to tumhe uska momentum (संवेग) milta hai:

Momentum

𝐩 = m 𝐯

(4.1)

jahan

$𝐩$: momentum, $k g m s^{- 1}$ mein
$m$: mass
$𝐯$: velocity

Kyunki velocity ek vector hai, momentum bhi vector hai, aur yeh velocity ki direction mein hi point karta hai. Yeh khaas product naam dene laayak kyun hai? Roz ke kai observations iski taraf ishaara karte hain:

Mass matter karta hai. Loaded tempo ko walking speed tak push karne mein khaali scooter ko usi speed tak le jaane se kahin zyada mehnat, ya kahin zyada time lagta hai. Tempo ko rokna bhi utna hi mushkil hai.
Speed matter karti hai. Dheere se uchhali tennis ball ko hatheli se rokna aasaan hai; wahi ball $40 m / s$ se aaye to chubhti hai. Ek hi mass ke liye, zyada speed ka matlab hai ki use diye gaye time mein rokne ke liye zyada force chahiye.
Time matter karta hai. Deewar se sand pit par koodo to mushkil se kuch mehsoos hota hai; concrete par akde pairon se utro to ghutno mein jhatka lagta hai. Momentum ka change same hai, lekin sand aur mude ghutne use lambe time mein phaila dete hain, isliye force chhota hota hai.
Same force same time tak lage to momentum ka change same hota hai. Ek halki trolley aur ek bhaari trolley, dono rest se shuru, par ek second ke liye ek steady force lagao. Halki trolley zyada tez ho jaati hai, lekin dono ka momentum aakhir mein bilkul same hota hai. Yahi second law ka sabse important clue hai.

Momentum sirf direction mein bhi badal sakta hai. Throw se pehle hammer (wire par steel ki ball) ghumata athlete thodi der ke liye uski speed steady rakhta hai, phir bhi uski direction lagataar mudti rehti hai, aur use modne ke liye athlete ko wire par andar ki taraf lagataar kheenchna padta hai. Ise aur tez ghumao, ya chhoti wire par, to zaroori pull badh jaata hai. Momentum ka tez change, chahe size mein ho ya direction mein, bade force ki maang karta hai. Is pull par hum Section 4.10 mein wapas aayenge.

LawNewton's second law of motion

Body ka momentum kitni tezi se badalta hai, yeh us par lage net external force ke proportional hota hai, aur momentum usi direction mein badalta hai jismein woh force act karta hai.

DerivationSecond law se F = ma tak

Maan lo ek net force $𝐅$ chhote time $Δ t$ ke liye act karta hai, aur momentum ko $Δ 𝐩$ se badal deta hai. Law kehta hai $𝐅 \propto Δ 𝐩 / Δ t$ , isliye kisi constant $k$ ke liye $𝐅 = k Δ 𝐩 / Δ t$ . $Δ t \to 0$ karne par yeh ratio time ke respect mein $𝐩$ ka derivative ban jaata hai:

𝐅 = k \frac{d 𝐩}{d t}

(4.2)

Jis body ka mass nahi badalta, uske liye $m$ derivative ke bahar aa jaata hai:

\begin{aligned} \frac{d 𝐩}{d t} & = \frac{d}{d t} (m 𝐯) \\ = m \frac{d 𝐯}{d t} = m 𝐚 \end{aligned}

(4.3)

isliye law aisa padhta hai

𝐅 = k m 𝐚

(4.4)

Humne abhi tak force ki unit define nahi ki hai, isliye hum $k$ chunne ke liye free hain. Sabse simple choice $k = 1$ hai, jo isi equation se force ki unit ko define karti hai:

Newton ka second law

𝐅 = \frac{d 𝐩}{d t} = m 𝐚

(4.5)

Force ki SI unit, newton, woh force hai jo $1 k g$ mass ko $1 m / s^{2}$ ka acceleration deta hai: $1 N = 1 k g m s^{- 2}$ . Ek average apple ka weight lagbhag $1 N$ hota hai.

Use karne se pehle second law ki chaar baatein dhyaan dene laayak hain.

Isme first law shaamil hai. (4.5) mein $𝐅 = 0$ rakho to $𝐚 = 0$ milta hai.
Yeh ek vector law hai. Iska matlab teen alag equations hain, har axis ke liye ek ((4.6)). Force sirf apni direction wala velocity component badalta hai; right angle wala component waisa hi rehta hai. Projectile ke liye gravity vertically act karti hai, isliye horizontal velocity constant rehti hai. Force hamesha acceleration ki direction mein hota hai, lekin woh velocity ki direction mein, uske against, uske right angle par ya kisi bhi angle par point kar sakta hai.
Yeh systems par bhi lagta hai. Rigid body, ya particles ke kisi bhi group ke liye, $𝐅$ ka matlab total external force aur $𝐚$ ka matlab centre of mass ka acceleration hai (Chapter 6). System ke parts ke beech ke forces internal hain aur chhod diye jaate hain.
Yeh time mein local hai. Kisi instant par acceleration usi instant aur jagah ke force se tay hota hai. Body ko pehle lage forces ki koi yaad nahi rehti (Fig. 4.4).

Components mein second law

\begin{aligned} F_{x} & = \frac{d p_{x}}{d t} = m a_{x} \\ F_{y} & = \frac{d p_{y}}{d t} = m a_{y} \\ F_{z} & = \frac{d p_{z}}{d t} = m a_{z} \end{aligned}

(4.6)

2/3Chhodte hi sirf gravity ka force bachta hai: koi horizontal force nahi, koi horizontal acceleration nahi.

Fig. 4.4Chhodne ke baad packet us instant wali horizontal velocity rakhta hai, lekin helicopter ka acceleration nahi (air resistance ignore kiya gaya).NCERT Fig. 4.5 · p. 55

Ek helicopter blue acceleration arrow ke saath right ki taraf udta hai. Step 1: ek yellow packet uske hook se latka hai aur uske paas bhi wahi blue acceleration arrow hai, label kehta hai pakda hua, saath accelerate karta hai. Step 2: hook gayab hai; packet ke paas ab sirf neeche point karta orange gravity arrow aur right point karta green velocity arrow hai, label kehta hai chhoda gaya, sirf gravity act karti hai. Step 3: helicopter kaafi aage nikal gaya hai, jabki packet ek dashed curved path par gira hai; uske paas ab bhi wahi green horizontal velocity arrow aur orange gravity arrow hai, label kehta hai packet ab peeche reh gaya.

Example 4.2NCERT Example 4.2 · p. 55

Keel thokna

$1.2 k g$ mass ka hammer head $6.0 m / s$ se ek keel par lagta hai, aur keel ke lakdi ke beam mein $1.5 c m$ ghusne ke baad dono saath ruk jaate hain. Retarding force ko constant aur keel ke mass ko negligible maano. Nikalo (a) rukte waqt hammer head par average force aur (b) rukne ka time.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.3NCERT Example 4.3 · p. 55

Position–time law se force

$2.0 k g$ ki trolley $x$ -axis ke along aise chalti hai ki SI units mein uski position $x = 4.0 + 3.0 t - 0.50 t^{2}$ hai. (a) Us par force nikalo. (b) Woh kab ek pal ke liye rest mein hoti hai, aur us instant par force kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Impulse

Bat ball ko sirf kuch milliseconds ke liye maarta hai, aur us dauraan force aise badhta-ghatta hai jise record karna bahut mushkil hai. Lekin pehle aur baad mein ball ki velocity naapna aasaan hai, aur second law dono ko jodta hai: poore contact mein force × time ko jodne se momentum ka change milta hai. Is product ko force ka impulse kehte hain:

Impulse

𝐉 = 𝐅_{av} Δ t = Δ 𝐩

(4.7)

jahan

$𝐅_{av}$: contact time $Δ t$ par average force
$Δ 𝐩$: momentum ka change, final minus initial

Jab force badalta hai, to impulse force–time graph ke neeche ka area hota hai. Same area ek lamba, patla spike bhi ho sakta hai ya ek neecha, chauda hump bhi (Fig. 4.5): isi wajah se landing par ghutne modna, ball aate waqt cricketer ka haathon ko peeche le jaana, car ka crumple zone ya helmet ki foam lining, sab time ko lamba karke force kam karte hain.

3/3Dono areas barabar hain: dono landings mein momentum ka change same hai.

Fig. 4.5Force–time graph ke neeche ka area impulse hai. Same impulse ko lambe time mein phailane se peak force kam ho jaata hai.NCERT p. 56

Ek graph jismein vertical axis par force aur horizontal axis par time hai. Step 1: ek lamba, patla orange spike, label kehta hai akde pair, chhota aur bada. Step 2: spike fade hota hai aur ek neecha, chauda blue hump dikhta hai, label kehta hai mude ghutne, lamba aur halka. Step 3: dono shapes saath dikhaye gaye hain aur label hai barabar area, barabar impulse, kyunki koodne wale ka momentum change dono tarah se same hai.

Jo force bahut bada ho lekin sirf ek pal ke liye rahe, use impulsive force kehte hain. Yeh koi alag kind ka force nahi hai: apne size aur chhoti duration ke alawa, second law ise bilkul ek halke push ki tarah treat karta hai. Kyunki contact time itna chhota hota hai, iske act karte waqt body mushkil se hilti hai, isliye impact ke dauraan hum uski position fixed maan sakte hain.

Example 4.4NCERT Example 4.4 · p. 56

Deewar se takrakar lautti football

$0.43 k g$ ki football $8.0 m / s$ se deewar se seedhi takrati hai aur $6.0 m / s$ se seedhi wapas lautti hai. Contact $20 m s$ tak rehta hai. Ball par impulse aur us par deewar ka average force nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.6 Newton's third law of motion

Second law batata hai ki force kya karta hai; third law batata hai ki forces aate kahan se hain. Body par har force kisi doosri body ka lagaya hota hai, aur yeh asar hamesha dono taraf chalta hai. Chhoti naav se jetty par utro to jaise tum aage badhte ho, naav peeche khisak jaati hai. Jo bodies ek doosre ko chhooti nahi, unmein yeh dono-taraf wala nature dekhna mushkil hai. Earth girte nariyal ko neeche kheenchti hai; Newton ka daava hai ki nariyal bhi Earth ko utne hi zor se upar kheenchta hai. $1.2 k g$ ke nariyal ke liye yeh force lagbhag $12 N$ hai, jo $6 \times 10^{24} k g$ mass wali Earth ko sirf lagbhag $2 \times 10^{- 24} m / s^{2}$ ka acceleration deta hai. Isi liye humein yeh kabhi pata nahi chalta.

To force kabhi ek-tarfa maamla nahi hota; yeh do bodies ke beech ka interaction hai, aur iske dono sides hamesha size mein barabar aur direction mein opposite hote hain. Newton ne ise aise kaha:

Har Action ke against hamesha ek barabar Reaction hota hai.

Isaac Newton, Law IIIPrincipia (1687), Andrew Motte ke English translation (1729) mein

LawNewton's third law of motion

Forces hamesha pairs mein hote hain. Agar body B body A par force lagati hai, to A bhi usi instant B par same magnitude ka force opposite direction mein lagati hai.

𝐅_{AB} = - 𝐅_{BA}

(4.8)

jahan

$𝐅_{AB}$: B dwara A par force
$𝐅_{BA}$: A dwara B par force

Newton ke shabd itne chhote hain ki unhe aksar galat padha jaata hai. Teen baatein zyaadatar galtiyon se bachati hain:

Dono forces alag bodies par act karte hain. Jab tum sirf A ka motion study karte ho, to sirf $𝐅_{AB}$ matter karta hai, isliye A ki equation of motion mein yeh pair ek doosre ko kabhi cancel nahi kar sakta. Sirf jab A aur B ko saath mein ek system maana jaata hai, tab dono internal forces ban jaate hain jinka sum zero hai. Isi tarah, body ke andar ke saare internal forces pairs mein cancel ho jaate hain, aur isi wajah se second law poori body par kaam karta hai.
Koi pehle nahi aata. Dono forces saath paida hote hain aur saath khatam hote hain; ek doosre ka cause nahi hai. Kisi ko bhi action kaha ja sakta hai.
Action aur reaction bas forces hain. Yeh shabd aur kuch nahi jodte. Har force ko "B dwara A par force" ke naam se bulana sabse clear hai.

Fig. 4.6 law ko kaam karte dikhata hai. Smooth ice par do skaters hatheliyan milakar push karte hain. Har ek doosre ko same force se push karta hai, isliye halka skater, jiski inertia kam hai, zyada acceleration paata hai. Woh opposite directions mein fisal kar alag hote hain, aur kyunki forces same time tak act karte hain, unka momentum barabar aur opposite hota hai: $40 k g$ wala skater $60 k g$ wale se $1.5$ guna tez chalta hai.

2/3B, A ko theek utne hi zor se push karta hai jitna A, B ko, opposite direction mein.

Fig. 4.6Push ek hi interaction hai: dono skaters par usi instant barabar aur opposite forces.NCERT p. 57

Ice par do skaters aamne-saamne khade hain: left mein A, label 60 kg, aur right mein chhota B, label 40 kg. Step 1: unke haath ek doosre se lage hain aur dono rest mein hain. Step 2: haathon se barabar length ke do orange force arrows shuru hote hain, ek A par left ki taraf aur ek B par right ki taraf. Step 3: skaters alag ho gaye hain; A ke paas left point karta chhota green velocity arrow hai, label dheema, aur B ke paas right point karta lamba arrow hai, label tez.

Example 4.5NCERT Example 4.5 · p. 57

Smooth floor se takrakar uchhalti ball

$60 g$ ki ball smooth horizontal floor par $10 m / s$ se takrati hai, uski velocity vertical se $60^{\circ}$ ka angle banati hai. Smooth floor sirf apne right angle par hi push kar sakta hai. Ball floor ko $3.0 m / s$ ke upward velocity component ke saath chhodti hai. Nikalo (a) ball par impulse, (b) floor par force ki direction aur (c) rebound ki speed aur direction.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek book table par rakhi hai. Third law se, book par Earth ke pull (uske weight) ka partner hai

table ka book par upar ki taraf pushbook ka table par neeche ki taraf pushbook ka Earth par gravitational pullEarth ka table par upar ki taraf push

Answer dekho

Answer (C) book ka Earth par gravitational pull

Weight Earth dwara book par force hai, isliye uska partner book dwara Earth par force hai. Table ka push bhi book par act karta hai, isliye woh partner nahi ho sakta, bhale hi yahan woh barabar ho.

4.7 Conservation of momentum

Second aur third laws ko saath rakho to ek powerful result nikalta hai. Ek cannon chalao. Jab tak barood jalta hai, cannon ball ko force $𝐅$ se aage push karti hai, aur ball cannon ko $- 𝐅$ se peeche push karti hai, usi chhote time $Δ t$ ke liye. Ball ka momentum $𝐅 Δ t$ se badalta hai aur cannon ka $- 𝐅 Δ t$ se. Dono rest se shuru hue the, isliye baad mein ball ka momentum aur cannon ka recoil momentum jud kar zero hote hain, theek shot se pehle ki tarah.

LawLaw of conservation of momentum

Agar interact karte particles ke group (ek isolated system) par koi net external force act na kare, to unka total momentum kabhi nahi badalta.

DerivationCollision mein momentum

Do bodies A aur B, momenta $𝐩_{A}$ aur $𝐩_{B}$ ke saath, takrati hain aur momenta $𝐩_{A}^{'}$ aur $𝐩_{B}^{'}$ ke saath alag hoti hain. Contact time $Δ t$ ke dauraan har ek sirf doosre ka force mehsoos karti hai. Second law se:

\begin{aligned} 𝐅_{AB} Δ t & = 𝐩_{A}^{'} - 𝐩_{A} \\ 𝐅_{BA} Δ t & = 𝐩_{B}^{'} - 𝐩_{B} \end{aligned}

Jodne par left side zero hai kyunki $𝐅_{AB} = - 𝐅_{BA}$ ((4.8)). Rearrange karne par:

Momentum conserve hota hai

𝐩_{A}^{'} + 𝐩_{B}^{'} = 𝐩_{A} + 𝐩_{B}

(4.9)

Is argument mein kuch bhi is par depend nahi karta ki energy ka kya hota hai, isliye result tab bhi sahi hai jab bodies uchhal kar alag hoti hain (elastic collision) aur tab bhi jab pichak kar chipak jaati hain (inelastic collision). Elastic collision mein total kinetic energy bhi conserve hoti hai (Chapter 5). Bahut saare particles ke liye bhi internal forces pair by pair cancel hote hain, isliye jab tak koi external force act na kare, total momentum nahi badal sakta.

3/4Elastic bounce: A aakhir mein $- \frac{1}{3} m / s$ par, B $\frac{14}{3} m / s$ par. Total: 4.

Fig. 4.7

3.0 m / s

se chalti

2.0 k g

ki trolley aur opposite taraf se

2.0 m / s

se aati

1.0 k g

ki trolley ke beech collision. Momentum ek se doosri mein jaata hai, lekin total

4.0 k g m s^{- 1}

hi rehta hai.NCERT p. 58

2 kg ki trolley A aur 1 kg ki trolley B track par chalti hain; unke neeche A, B aur total ke momentum ke horizontal bars hain, aur ek dashed line total dikhati hai. Step 1, pehle: A lambe arrow ke saath right chalti hai aur B left; bar A right mein 6 units hai, bar B left mein 2 units aur total bar 4 tak pahunchta hai. Step 2, contact mein: barabar aur opposite orange force arrows; dono trolleys ki ek hi velocity hai, bars A aur B 2.7 aur 1.3 units hain aur total ab bhi 4 hai. Step 3, uchhal kar alag: A dheere left chalti hai aur B tez right; bar A thoda negative hai, bar B 4.7 hai aur total 4 hai. Step 4, chipak jaati hain: dono saath right chalti hain; bars 2.7 aur 1.3 hain aur total phir se 4 hai.

Example

Railway wagons ko jodna

Goods yard mein $1.5 m / s$ se chalta $20 t$ ka wagon ek ruke hue $30 t$ wagon se takrata hai, aur dono jud jaate hain. Chhote impact ke dauraan friction ignore karke unki common speed aur har wagon ko mila impulse nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

3.0 m / s

se chalti

0.20 k g

ki ball ek ruki hui

0.30 k g

ball se seedhi takrati hai aur wahin ruk jaati hai. Doosri ball kitni tezi se chalti hai, m/s mein?

m/s

Answer dekho

Answer 2 m/s

0.20 \times 3.0 = 0.30 v

, isliye

v = 2.0 m / s

4.8 Particle ka equilibrium

Particle equilibrium (साम्यावस्था) mein hota hai jab us par net external force zero ho. First law se tab woh ya to rest mein hota hai ya constant velocity se chal raha hota hai. Do forces ke saath iska seedha matlab hai ki woh barabar aur opposite hain:

𝐅_{1} = - 𝐅_{2}

(4.10)

Ek point par act karte teen forces ke saath, equilibrium ke liye unka vector sum zero hona chahiye:

𝐅_{1} + 𝐅_{2} + 𝐅_{3} = 0

(4.11)

Geometry ki nazar se dekho to iska matlab hai ki kinhi do forces ka resultant (parallelogram law se) teesre ke barabar aur opposite hai. Doosre tareeke se, agar hum teeno forces ko head to tail banayein, to woh band hokar ek triangle bana lete hain (Fig. 4.8). Yeh idea kitne bhi forces par lagta hai: $n$ forces ke under particle tabhi equilibrium mein hai jab unke arrows, head to tail jode jaane par, ek closed polygon banayein.

3/3Aakhri arrow wahin khatam hota hai jahan pehla shuru hua tha: vector sum zero hai.

Fig. 4.8Equilibrium wale forces, head to tail banane par, ek closed triangle banate hain. Yahan ki tensions Example 4.6 wali hain.NCERT Fig. 4.7 · p. 58

Left mein ek knot jise teen forces kheench rahe hain: upar aur left ki taraf ek lamba blue arrow, upar aur right ki taraf ek chhota dashed orange arrow, aur seedha neeche ek vermilion arrow. Step 1 mein sirf yeh teen forces hain. Step 2: blue aur orange arrows ki copies right panel mein khisakti hain aur head to tail judti hain, label hai head to tail. Step 3: neeche wale arrow ki copy khisak kar aati hai aur uska head theek starting point par girta hai, triangle band karta hai, label kehta hai band, sum zero hai.

Calculations ke liye hum aksar iske bajaye axes ke along resolve karte hain. Tab (4.11) har axis ke liye ek scalar equation mein bant jaata hai:

\begin{aligned} F_{1 x} + F_{2 x} + F_{3 x} & = 0 \\ F_{1 y} + F_{2 y} + F_{3 y} & = 0 \\ F_{1 z} + F_{2 z} + F_{3 z} & = 0 \end{aligned}

(4.12)

Extended body ke liye yeh sirf aadhi kahani hai. Body par net force zero ho phir bhi woh ghoomna shuru kar sakti hai, isliye uske equilibrium ke liye net torque bhi zero hona chahiye, jise hum Chapter 6 mein padhenge. Particle ka koi size nahi hota, isliye uske liye zero net force kaafi hai.

Example 4.6NCERT Example 4.6 · p. 58

Do cords par latki lantern

$5.0 k g$ ki lantern ek knot K se latki hai. Ek cord K se ceiling tak vertical se $30^{\circ}$ par jaati hai; doosri vertical se $60^{\circ}$ par deewar tak jaati hai. Har cord mein tension nikalo. $g = 10 m / s^{2}$ lo aur cords ko massless maano.

Fig. 4.9Set-up aur knot ka free-body diagram.

Left: ek ceiling aur ek wall bracket, jinse do cords ek knot K par milti hain; ceiling wali cord ek dashed vertical line se 30 degrees ka angle banati hai aur deewar wali cord usse 60 degrees ka. Knot ke neeche 5.0 kilogram ki lantern latki hai. Right: knot ka free-body diagram, jismein pehli cord ke along tension T1, doosri cord ke along ek chhota dashed tension T2 aur neeche point karta weight W hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.9 Mechanics ke common forces

Gravity khaas hai: yeh un bodies ke beech act karti hai jo ek doosre ko chhoo nahi rahi, khaali space ke paar. Earth ke paas har object ise mehsoos karta hai, aur yahi Moon ko Earth ke chaaron taraf aur planets ko Sun ke chaaron taraf ghumati hai. (Charged aur magnetised bodies bhi door se electric aur magnetic forces mehsoos karti hain; unhe hum yahan chhod dete hain.)

Mechanics mein milne wala baaki har force ek contact force hai: yeh tabhi tak hota hai jab tak ek object doosre ko chhoota hai, chahe solid ho ya fluid, aur yeh third-law pairs mein aata hai. Jahan do solid surfaces ek doosre ko dabati hain, wahan contact force ko do parts mein todna useful hai (Fig. 4.10). Surfaces ke perpendicular wala part normal reaction hai; unke along wala part friction hai. Fluids bhi contact forces lagate hain: paani ya hawa mein object ek upward buoyant force mehsoos karta hai jo uske hataye fluid ke weight ke barabar hota hai, aur fluid mein chalti body ek viscous drag mehsoos karti hai, jise hawa mein hum air resistance kehte hain.

2/3Halka push: friction aata hai, isliye contact force push ke against peeche jhuk jaata hai.

Fig. 4.10Ek contact force, do components: normal (floor ke perpendicular) aur friction (floor ke along). Crate abhi fisalna shuru nahi hua.NCERT Fig. 4.9 · p. 59

Floor par ek crate khada hai, uske centre se neeche point karta ek orange weight arrow aur floor ke contact force ke liye ek purple arrow hai. Step 1: koi push nahi; contact force seedha upar point karta hai. Step 2: left se ek chhota blue push arrow act karta hai; purple contact force ab push ke against peeche jhuka hai, aur dashed lines use ek vertical normal part aur floor ke along ek horizontal friction part mein todti hain. Step 3: ek lamba push arrow; contact force aur peeche jhukta hai aur uska friction part lamba hai, jabki normal part same rehta hai.

Do aur contact forces lagbhag har problem mein aate hain. Kheenchi ya dabayi gayi spring apni natural length ki taraf wapas push ya pull karti hai. Chhote deformations ke liye restoring force natural length se displacement $x$ ke proportional hota hai:

F = - k x

jahan

$k$: spring constant (force constant), $N / m$ mein
$x$: extension (positive) ya compression (negative)

Minus sign batata hai ki force hamesha displacement ka virodh karta hai. String ek bahut bade $k$ wali spring ki tarah behave karti hai: yeh lagbhag bilkul nahi khinchti, aur iske lagaye pull ko tension kehte hain. Problems mein hum aksar string ko massless aur inextensible maante hain. Tab tension uski poori length mein same rehta hai, aur ek smooth, halki pulley sirf pull ki direction badalti hai, uska size nahi.

Physicists chaar fundamental forces maante hain: gravitational, electromagnetic, strong aur weak. Strong aur weak forces sirf atomic nuclei ke andar act karte hain, isliye mechanics mein sirf gravity aur electromagnetism matter karte hain. Kamaal ki baat yeh hai ki har contact force, normal reaction aur friction se lekar tension, spring forces aur drag tak, asal mein electrical hai: yeh dono surfaces ke atoms ke charged electrons aur nuclei ke beech ke forces se aata hai. Inhe atom by atom nikalna bahut hi complicated hoga, isliye mechanics har ek ko experiment se mile ek simple law se describe karti hai.

4.9.1 Friction

Floor par khadi steel ki almirah par jhuk kar zor lagao: kuch nahi hota. Aur zor lagao: phir bhi kuch nahi. Kyunki almirah accelerate nahi karti, first law kehta hai ki us par horizontal forces balance hone chahiye, isliye floor use tumhare force ke bilkul barabar force se wapas push kar raha hai. Woh force static friction $f_{s}$ hai. Koi push nahi to static friction bilkul nahi; jaise-jaise tum zyada push karte ho, yeh badh kar tumhare barabar ho jaata hai. Yeh impending motion ka virodh karta hai, yaani woh fisalna jo floor frictionless hota to hota.

Definition

Static friction

Do surfaces ke beech ka friction force jab woh ek doosre par fisal nahi rahi. Yeh fisalne ki kisi bhi tendency ka virodh karne ke liye khud ko adjust karta hai, ek maximum value $μ_{s} N$ tak.

Static friction bina limit ke nahi badh sakta: kaafi zor se push karo to almirah fisalne lagti hai. Experiments dikhate hain ki sabse bada possible static friction contact ke area par depend nahi karta aur, achhe approximation mein, normal force ke proportional hota hai:

Limiting static friction

(f_{s})_{\max} = μ_{s} N

(4.13)

Constant $μ_{s}$ , yaani coefficient of static friction, sirf contact wali surfaces ke pair par depend karta hai. Isliye static friction ka rule ek inequality hai:

Static friction ka law

f_{s} \leq μ_{s} N

(4.14)

Fisalna shuru hote hi friction force static maximum se neeche gir jaata hai. Yeh kinetic friction (ya sliding friction) bhi contact area par depend nahi karta, speed se lagbhag independent hai, aur ek milte-julte law ko follow karta hai:

Kinetic friction

f_{k} = μ_{k} N

(4.15)

Definition

Kinetic friction

Ek doosre par fisalti do surfaces ke beech ka friction force, $f_{k} = μ_{k} N$ , relative motion ke against. Surfaces ke kisi pair ke liye $μ_{k}$ $μ_{s}$ se kam hota hai.

Kyunki $μ_{k} < μ_{s}$ , body ko fisalna shuru karwane mein use fisalte rakhne se zyada force lagta hai (Fig. 4.11). Chalne ke baad us par net force $F - μ_{k} N$ hai, isliye block shuru karne wale push se chhote push mein bhi steady speed se fisal sakta hai, aur push rukne par $μ_{k} g$ se dheema hokar ruk jaata hai. Neeche friction lab mein dono regimes try karo (Fig. 4.12).

2/3Peak par static friction apni limit $μ_{s} N$ tak pahunchta hai.

Fig. 4.11Push badhne par block par friction. Static friction

μ_{s} N

tak push ke barabar rehta hai; fisalne ke baad kinetic friction

μ_{k} N

ke aas-paas rehta hai.NCERT Fig. 4.10 · p. 60

Ek graph jismein horizontal axis par applied force aur vertical axis par friction hai. Ek blue line origin se 45 degrees par badhti hai, ek shaded region se jiska label rest mein hai, ek peak tak; ek dashed girawat neeche ek orange horizontal line tak le jaati hai, ek region mein jiska label fisal raha hai. Step 1: ek dot badhti line par hai, jahan friction push ke barabar hai. Step 2: dot peak par hai, yaani limiting static friction. Step 3: fisalne wala region chamakta hai aur dot neeche wali horizontal line par hai, yaani constant kinetic friction.

InteractiveFriction lab

Do modes wala ek interactive block-aur-friction lab. Push mode mein ek block level floor par rakha hai; ek slider applied force set karta hai, aur arrows push, friction, normal force aur weight dikhate hain. Tilt mode mein ek slider board ko jhukata hai, aur arrows board ke perpendicular normal force, slope par upar ki taraf friction aur vertical weight dikhate hain. Sliders mass aur dono friction coefficients set karte hain. Ek graph push ya tilt angle ke saath friction plot karta hai, abhi ki state ke liye ek dot ke saath, aur readouts normal force, friction aur acceleration batate hain. Ek status line batati hai ki block rest mein hai ya fisal raha hai, aur ek baar fisalne par woh tab tak fisalta rehta hai jab tak push ya tilt kinetic limit se neeche na gir jaaye.

Fig. 4.12Level floor par block ko push karo, ya uske neeche ka board jhukao. Dekho ki static friction push ke barabar kaise rehta hai jab tak woh

μ_{s} N

tak na pahunch jaaye.

Table 4.1Coefficients of friction ki typical values (sookhi surfaces; asli values safaai aur roughness ke saath kaafi badalti hain)

Surfaces	$μ_{s}$	$μ_{k}$
Sookhe concrete par rubber	1.0	0.8
Steel par steel	0.74	0.57
Wood par wood	0.25–0.5	0.2
Ice par ice	0.1	0.03
Teflon par Teflon	0.04	0.04

Newton ke law of gravitation ki tarah, yeh friction rules nature ke gehre laws nahi hain. Yeh measurements se nikale rules of thumb hain, aur asli surfaces inse kaafi alag ho sakti hain. Phir bhi, zyaadatar calculations ke liye humein bas inhi ki zaroorat hai.

Dhyaan do ki friction asal mein kiska virodh karta hai: surfaces ke beech relative motion ka, motion ka nahi. Delivery truck ke khule bed par rakha crate truck ke saath speed badhata hai. Crate par sirf ek horizontal force hai, truck bed ka friction, aur woh aage ki taraf, motion ki direction mein point karta hai. Agar bed frictionless hota, to truck crate ke neeche se nikal jaata. Static friction us impending slip ka virodh karta hai, aur aisa karke crate ko truck ke saath aage kheenchta hai.

Example 4.7NCERT Example 4.7 · p. 61

Truck bed par crate

Ek crate truck ke flat bed par rakha hai. Unke beech friction ke coefficients $μ_{s} = 0.35$ aur $μ_{k} = 0.28$ hain. (a) Bina crate fisle truck ka sabse zyada acceleration kitna ho sakta hai? (b) Driver $4.0 s$ mein rest se $54 k m / h$ tak jaata hai. Kya crate fisalta hai, aur agar haan, to truck ke relative uska acceleration kya hai? $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Ruler se coefficient of friction naapo

Tumhe chahiye

ek hardback book ya ek smooth phatta
ek eraser aur ek sikka
ek ruler

Pehle guess karo

Tum pehle eraser par ek doosra waisa hi eraser rakhte ho. Ek eraser ke comparison mein, yeh stack kis angle par fisalna shuru karta hai?

zyada angle parsame angle parkam angle par

Book ko flat rakho aur eraser ko uske door wale sire ke paas rakho.
Us sire ko bahut dheere uthao jab tak eraser bas fisalna shuru na kare. Book ko wahin rok kar rakho.
Uthaye gaye sire ki table se height $h$ aur neeche wale kinare se uthaye gaye sire ke neeche wale point tak horizontal distance $d$ naapo.
Tab $\tan θ = h / d$ , aur fisalne wale angle par yeh $μ_{s}$ ke barabar hai. Teen baar repeat karo aur average nikalo, phir sikke aur do erasers ke stack ke saath try karo.

Table 4.2Fisalne ke angle ke results

Cheez	$h$ (cm)	$d$ (cm)	$μ_{s} = h / d$
Ek eraser
Do erasers
Sikka

Tumhari entries apne aap save hoti hain.

Example 4.8NCERT Example 4.8 · p. 61

Jhuke board par tiffin box

Ek steel ka tiffin box $1.20 m$ lambe lakdi ke board par rakha hai. Board ka ek sira dheere uthaya jaata hai, aur box tab fisalna shuru karta hai jab woh sira doosre se $0.50 m$ ooncha hota hai. (a) $μ_{s}$ nikalo. (b) Agar $μ_{k} = 0.30$ hai, to board ko usi angle par rakhne par fisalte box ka acceleration kya hai? $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.9NCERT Example 4.9 · p. 61

Ek block aur latakta pan

Horizontal table par rakha $4.0 k g$ ka block ek halki string se, jo table ke kinare par smooth pulley ke upar se jaati hai, aazaadi se latakte $1.5 k g$ ke pan se bandha hai. Block aur table ke beech $μ_{s} = 0.30$ aur $μ_{k} = 0.20$ hai. System ko rest se chhoda jaata hai. Kya yeh chalta hai? Agar haan, to acceleration aur tension nikalo. $g = 9.8 m / s^{2}$ lo.

Fig. 4.13Motion ki direction ke along forces.

Table par 4.0 kilogram ka block rakha hai. Usse ek string horizontally table ke right kinare ki chhoti pulley tak jaati hai aur phir neeche 1.5 kilogram ke latakte pan tak. Ek green arrow block ko right ki taraf kheenchta tension dikhata hai, ek orange arrow block par left point karta friction dikhata hai, aur blue arrows block ko right ki taraf aur pan ko neeche ki taraf accelerate karte dikhate hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Rolling friction

Floor par ek eent ghaseeto aur phir utne hi weight ki ball rolling karo: ball kahin zyada der tak chalti rehti hai. Ideal case mein, sakht road par rolling karta sakht wheel use sirf ek point par chhoota hai jo road par ek pal ke liye rest mein hota hai, isliye kuch nahi fisalta aur sliding friction bilkul nahi hota. Asli wheels phir bhi dheeme hote hain, kyunki tyre thoda pichakta hai aur road thodi dabti hai. Ab woh ek point ke bajaye ek chhote patch par chhoote hain, aur us patch par contact force thoda peeche jhuk jaata hai (Fig. 4.14). Iska peeche wala component rolling friction hai. Same load ke liye yeh aam taur par sliding friction se sau se hazaar guna tak chhota hota hai, isi liye wheels ya rollers par load ghaseetne ke comparison mein kahin aasaani se chalta hai.

2/3Dono surfaces deform hoti hain; patch par contact force peeche jhukta hai, rolling ka virodh karta hai.

Fig. 4.14Rolling friction deformation se aata hai. Ball bearings shaft aur uske housing ke beech fisalne ko rolling se badal dete hain.NCERT Fig. 4.13 · p. 62

Step 1: ek ideal rigid wheel flat road ko sirf ek marked point par chhoota hai, label hai point contact. Step 2: tyre neeche se pichka hai aur road thodi dabi hai, isliye woh ek contact patch par milte hain; ek green arrow wheel ko right ki taraf rolling karte dikhata hai, aur patch se ek purple contact-force arrow peeche, left ki taraf jhukta hai. Step 3: wheel ke paas ek ball bearing dikhta hai: ek outer ring, ek central shaft aur unke beech aath balls ki ring, label hai ball bearing.

Friction aksar ek musibat hai. Machine mein fisalte parts power ko heat ki tarah barbaad karte hain aur ek doosre ko ghisa dete hain. Engineers isse lubricating oil se, fisalne ko rolling mein badalne wale ball bearings se, aur surfaces ke beech hawa ki patli gaddi se ladte hain, jaise hovercraft ya air-hockey table mein.

Utni hi baar friction bilkul wahi hai jiski humein zaroorat hai. Tum chal sakte ho sirf isliye kyunki zameen ka friction tumhare pair ko aage push karta hai; insaanon aur jaanwaron ko bhi accelerate karne ke liye external force chahiye, bilkul carts ki tarah. Road ka wahi friction car ki speed badhata hai, use mod par ghumata hai aur, brakes ke zariye, use rokta hai. Oil ya ice se dhaki road par car inmein se kuch bhi nahi kar sakti.

Compare

Friction: dushman bhi, dost bhi

Ise kam karo

Engines aur gearboxes: oil films
Cycle hubs aur fans: ball bearings
Hovercraft: hawa ki gaddi
Polished, achhe se fit moving parts

Iski zaroorat hai

Chalna aur daudna
Tyres ki pakad, steering aur braking
Keelein, screws aur gaanthein jo pakad kar rakhti hain
Pen ya pencil se likhna

4.10 Circular motion

Chapter 3 mein humne dekha ki radius $R$ ke circle par constant speed $v$ se chalti body centre ki taraf accelerate karti hai, magnitude $v^{2} / R$ ke saath. Second law se, ek net force bhi centre ki taraf point karna chahiye. Uska size hai

Centripetal force

f_{c} = \frac{m v^{2}}{R}

(4.16)

Is andar ki taraf wale net force ko centripetal force (अभिकेंद्र बल) kehte hain. Yeh naam ek kaam batata hai, koi naya kind ka force nahi. Har circular motion mein koi asli force, jo kisi asli object ne lagaya ho, yeh kaam karta hai: ghumaye patthar ke liye string ka tension, planet ke liye Sun ki gravity, Moon ya satellite ke liye Earth ki gravity, flat mod par car ke liye friction, aur banked track par car ya maut ka kuan (wall of death) mein rider ke liye normal force ka ek part. Inke saath "centripetal force" ko kabhi ek extra arrow ki tarah mat banao; yeh centre ki taraf inka resultant hi hai.

Definition

Centripetal force

Mass $m$ ki body ko radius $R$ ke circle mein speed $v$ se chalte rakhne ke liye centre ki taraf zaroori net force; iska magnitude $m v^{2} / R$ hai. Ise hamesha tension, gravity, friction ya normal force jaise asli forces dete hain.

Level road par car ka motion

DerivationFlat mod par maximum speed

Car par teen forces act karte hain: uska weight $m g$ , normal force $N$ aur road ka friction $f$ . Vertically kuch accelerate nahi karta, isliye $N - m g = 0$ :

N = m g

(4.17)

Sirf ek horizontal force hai, road ka tyres par friction, jise poora centripetal force dena hai. Tyres side mein nahi fisalte, isliye yeh static friction hai, jo car ke bahar ki taraf impending skid ka virodh karta hai. Static friction (4.14) se limited hai:

\begin{aligned} \frac{m v^{2}}{R} & = f_{s} \\ \leq μ_{s} N = μ_{s} m g \end{aligned}

isliye $v^{2} \leq μ_{s} R g$ . Mass cancel ho jaata hai, aur mod par sabse zyada safe speed hai

Maximum speed, level road

v_{\max} = \sqrt{μ_{s} R g}

(4.18)

Same tyres aur road par loaded truck aur khaali truck ki limit same hoti hai. Baarish ya ret $μ_{s}$ ko kam karti hai aur uske saath $v_{\max}$ ko bhi; tez mod (chhota $R$ ) bhi ise kam karta hai.

Example 4.10NCERT Example 4.10 · p. 64

Sookhi aur geeli road par scooter

Ek scooter rider radius $25 m$ ka flat mod $36 k m / h$ se leta hai. Kya mod safe hai (a) $μ_{s} = 0.45$ wali sookhi road par aur (b) baarish ke baad, jab $μ_{s} = 0.20$ ho? (c) Geeli road par safe limit kya hai? $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Banked road par car ka motion

Sirf friction par bharosa karna risky hai, isliye highway ke mod aur racetracks banked hote hain: road ka bahari kinara andar wale kinare se ooncha banaya jaata hai. Tab normal force, road ke perpendicular hone ki wajah se, mod ke centre ki taraf jhuk jaata hai, aur uska horizontal component centripetal force dene mein madad karta hai (Fig. 4.15).

DerivationBanked road par maximum speed

Maan lo road angle $θ$ par banked hai. Tez speed par car upar aur bahar ki taraf skid karna chahti hai, isliye friction $f$ slope par neeche ki taraf act karta hai. Vertically koi acceleration nahi hai:

\begin{aligned} N \cos θ & = m g \\ + f \sin θ \end{aligned}

(4.19a)

Horizontally, centre ki taraf $N$ aur $f$ ke components centripetal force dete hain:

\begin{aligned} N \sin θ & + f \cos θ \\ = \frac{m v^{2}}{R} \end{aligned}

(4.19b)

Kyunki $f \leq μ_{s} N$ , speed sabse zyada tab hoti hai jab $f = μ_{s} N$ :

\begin{aligned} N \cos θ & - μ_{s} N \sin θ \\ = m g \end{aligned}

(4.20a)

\begin{aligned} N \sin θ & + μ_{s} N \cos θ \\ = \frac{m v_{\max}^{}}{R} \end{aligned}

(4.20b)

(4.20b) ko (4.20a) se divide karo. $N$ aur $m$ dono cancel ho jaate hain:

\begin{aligned} \frac{v_{\max}^{}}{R g} \\ = \frac{\sin θ + μ_{s} \cos θ}{\cos θ - μ_{s} \sin θ} \end{aligned}

Fraction ke upar aur neeche ko $\cos θ$ se divide karne par result milta hai:

Maximum speed, banked road

\frac{v_{\max}^{}}{R g} = \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}

(4.21)

$θ = 0$ ke saath yeh (4.18) ban jaata hai; koi bhi banking limit ko badha deti hai. Bilkul friction na ho ( $μ_{s} = 0$ ) to sirf ek hi speed chalti hai:

Optimum speed

v_{0} = \sqrt{R g \tan θ}

(4.22)

Optimum speed $v_{0}$ par jhuka hua normal force akela hi $m v^{2} / R$ de deta hai, isliye tyres ko side ka koi friction nahi chahiye aur woh sabse kam ghiste hain. $v_{0}$ se tez chalao to friction ko slope par neeche act karna padta hai; dheere chalao to car bank par neeche fisalna chahti hai, isliye friction slope par upar act karta hai. $f$ ko ulta karke derivation dohrane se sabse kam safe speed milti hai:

Minimum speed, banked road

v_{\min}^{} = R g \frac{\tan θ - μ_{s}}{1 + μ_{s} \tan θ}

Agar $\tan θ \leq μ_{s}$ ho to right-hand side zero ya negative hai, isliye koi lower limit nahi hai: car ko bank par bina neeche fisle park bhi kiya ja sakta hai.

2/4Banked, $v_{0}$ par: normal force aur weight jud kar ek horizontal net force dete hain; koi friction nahi chahiye.

Fig. 4.15

m v^{2} / R

kaun deta hai? Level road par friction; banked road par jhuka hua normal force, aur friction ek ya doosri direction mein madad karta hai.NCERT Fig. 4.14 · p. 63

Road par ek car peeche se dikhti hai; top left mein ek arrow mod ke centre ki taraf point karta hai. Step 1, level road: blue normal force seedha upar, vermilion weight seedha neeche, aur tyres par ek green friction arrow centre ki taraf point karta hai. Step 2, ideal speed par banked road: road aur car bahar ki taraf upar jhuke hain; normal force centre ki taraf jhukta hai, koi friction arrow nahi hai, aur ek dashed black arrow net force ko horizontally centre ki taraf point karte dikhata hai. Step 3, banked aur tez: ek green friction arrow slope par neeche point karta hai. Step 4, banked aur dheemi: friction arrow slope par upar point karta hai.

InteractiveBanked road lab

Road par ek car ka rear view, jiska bank angle tum ek slider se set karte ho. Sliders speed, mod ka radius aur coefficient of static friction bhi set karte hain. Arrows road ke perpendicular normal force, vertical weight aur zaroori friction dikhate hain, jo slope par neeche, upar point karta hai ya gayab ho jaata hai; jab tyres jitna de sakte hain usse zyada friction chahiye, to yeh orange ho jaata hai. Readouts optimum speed, sabse kam aur sabse zyada safe speeds aur centripetal acceleration dikhate hain, aur ek status line batati hai ki car pakad bana kar rakhti hai, bahar skid karti hai ya andar fisalti hai. Buttons speed ko optimum par set karte hain, road ko flat karte hain ya use icy banate hain.

Fig. 4.16Radius, bank angle, friction aur speed set karo. Lab nikalta hai ki tyres ko kitna friction dena hai aur kya woh de sakte hain.

Example 4.11NCERT Example 4.11 · p. 64

Highway ka mod design karna

Radius $120 m$ ka highway mod $10^{\circ}$ par banked hai, aur tyres aur road ke beech $μ_{s} = 0.30$ hai. Nikalo (a) optimum speed, (b) maximum safe speed aur (c) kya car mod par bina neeche fisle ruk kar intezaar kar sakti hai. $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.11 Mechanics ke problems solve karna

Teen laws ke saath ab tum bahut saare problems sambhal sakte ho. Ek typical problem mein kai bodies hoti hain, jo strings se judi hain, ek doosre par rakhi hain ya supports ko daba rahi hain, aur har ek ko gravity bhi kheench rahi hai. Sabse badi aazaadi yeh hai: tum laws of motion ko arrangement ke kisi bhi part par apply kar sakte ho, jise hum system kehte hain, bas shart yeh hai ki baaki sab cheezein, yaani environment, us par jo bhi forces lagati hain unhe shaamil karo. Ab tak ke worked examples ne chupchaap yahi idea use kiya hai. Yeh raha ek routine ki tarah:

Poore arrangement ka sketch banao: bodies, strings, pulleys aur supports.
Ek convenient part ko apna system chuno.
Use akela banao, bahar se us par act karte har force ke saath: contact wali bodies se, strings se aur gravity se. System jo forces doosri cheezon par lagata hai unhe chhod do. Yeh ek free-body diagram hai; iska matlab yeh nahi ki net force zero hai.
Jo forces tumhe pakka pata hain unhe mark karo (weight seedha neeche, tension string ke along, normal force surface ke perpendicular) aur baaki ko unknowns label karo. Har force ko "Y dwara X par" naam do: friction, tension, thrust, buoyancy aur weight sab bas forces hain.
Agar aur equations chahiye, to doosra system chuno. Agar A ke diagram mein B dwara force $𝐅$ hai, to B ke diagram mein A dwara $- 𝐅$ hona chahiye.
Har system ke liye convenient axes ke along $𝐅 = m 𝐚$ likho, solve karo, aur units aur ek limiting case check karo.

3/4Scale par: floor ka push $N_{2}$ upar, student ka push $N_{1}$ neeche aur uska apna chhota weight $W_{2}$ .

Fig. 4.17Wahi lift teen alag systems ki tarah. System ke parts ke beech ke forces cancel hokar uske diagram se hat jaate hain (scale ka weight saaf dikhne ke liye bada dikhaya gaya hai).NCERT Fig. 4.15 · p. 65

Left: cable se latka ek lift cabin, bagal mein upar point karta blue acceleration arrow; andar ek student bathroom scale par khada hai. Step 1: right panel kehta hai system chuno, uske forces banao. Step 2, sirf student: ek box jismein upar ek lamba green arrow, scale ka normal force, aur neeche ek chhota vermilion arrow, student ka weight. Step 3, sirf scale: ek flat box jismein floor se upar ek lamba purple arrow, student se neeche ek green arrow aur scale ke apne weight ke liye neeche ek chhota sa vermilion arrow. Step 4, student plus scale: floor se upar ek lamba purple arrow aur total weight ke liye neeche ek vermilion arrow; student aur scale ke beech ke forces ab nahi dikhte.

Example 4.12NCERT Example 4.12 · p. 65

Lift mein bathroom scale

$50 k g$ ka student $2.0 k g$ ke bathroom scale par khada hai jo ek lift ke floor par hai, aur lift $1.5 m / s^{2}$ se upar accelerate karti hai. Nikalo (a) student aur scale ke beech force, (b) scale aur floor ke beech force, aur (c) action–reaction pairs ki list banao. $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yeh example dikhata hai ki jaana-pehchana result "normal force weight ke barabar hai" dhyaan se kyun use karna chahiye. Yeh sirf tab sahi hai jab body equilibrium mein ho (desk par, ya constant velocity se chalti lift mein), aur iska third law se koi lena-dena nahi. Free-body diagrams aise jaal saaf dikha dete hain; neeche ki exercises inhe aadat bana dengi.

Quick check

3 mein se 1

50 k g

ka ek insaan lift mein scale par khada hai jo

2.0 m / s^{2}

se neeche accelerate kar rahi hai. Scale padhta hai (

g = 10 m / s^{2}

)

600 N

500 N

400 N

zero

Answer dekho

Answer (C) $400 N$

N = m (g - a) = 50 \times 8.0 = 400 N

600 N

upward acceleration ke liye hota; zero sirf free fall mein.

Recap

Summary aur key points

Aristotle sochte the ki chalti body ko lagataar force chahiye. Asal mein force sirf friction aur drag ko balance karne ke liye chahiye; zero net force ke saath motion bina badle chalta rehta hai (Section 4.2).
Galileo ke slope experiments se law of inertia mila: rest aur seedhi line mein uniform motion, zero net force ki barabar states hain (Section 4.3).
First law: body ki velocity sirf tab badalti hai jab us par net external force act kare (Section 4.4).
Momentum $𝐩 = m 𝐯$ hai. Second law: $𝐅 = d 𝐩 / d t = m 𝐚$ , jahan $1 N = 1 k g m s^{- 2}$ ((4.5)).
Second law ek vector law hai (har axis ke liye ek equation), systems par lagta hai jahan $𝐅$ total external force hai, aur local hai: abhi ka force abhi ka acceleration tay karta hai (chaar baatein).
Impulse $= 𝐅_{av} Δ t = Δ 𝐩$ , force–time graph ke neeche ka area; lamba contact time matlab chhota force (impulse).
Third law: forces ek saath, barabar aur opposite pairs mein aate hain jo alag bodies par act karte hain, isliye ek body ke liye woh kabhi cancel nahi hote (Section 4.6).
Isolated system ka total momentum conserve hota hai, elastic aur inelastic dono collisions mein ((4.9)).
Particle equilibrium mein hota hai jab us par forces ka sum zero ho: head to tail banane par woh ek polygon band karte hain (Section 4.8).
Mechanics mein sirf gravity non-contact force hai; normal reaction, friction, tension, spring forces aur drag electrical origin ke contact forces hain (Section 4.9).
Static friction $μ_{s} N$ tak khud ko adjust karta hai; kinetic friction $μ_{k} N$ hai jahan $μ_{k} < μ_{s}$ ; rolling friction kahin chhota hai (Section 4.9.1).
Centripetal force $m v^{2} / R$ asli forces dete hain. Level road par $v_{\max} = \sqrt{μ_{s} R g}$ ; banked road par $v_{0} = \sqrt{R g \tan θ}$ aur $v_{\max}$ zyada hai (Section 4.10).
Free-body diagrams se problems solve karo: system chuno, bahar se us par har force banao, third-law pairs ko consistent rakho, phir $𝐅 = m 𝐚$ apply karo (Section 4.11).

Flash cards · 10

Formula revision

1 / 10

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 4.1	Practice Q1
NCERT Exercises 4.2	Practice Q2
NCERT Exercises 4.3	Practice Q3
NCERT Exercises 4.4	Practice Q4
NCERT Exercises 4.5	Practice Q5
NCERT Exercises 4.6	Practice Q6
NCERT Exercises 4.7	Practice Q7
NCERT Exercises 4.8	Practice Q8
NCERT Exercises 4.9	Practice Q9
NCERT Exercises 4.10	Practice Q10
NCERT Exercises 4.11	Practice Q11
NCERT Exercises 4.12	Practice Q12
NCERT Exercises 4.13	Practice Q13
NCERT Exercises 4.14	Practice Q14
NCERT Exercises 4.15	Practice Q15
NCERT Exercises 4.16	Practice Q16
NCERT Exercises 4.17	Practice Q17
NCERT Exercises 4.18	Practice Q18
NCERT Exercises 4.19	Practice Q19
NCERT Exercises 4.20	Practice Q20
NCERT Exercises 4.21	Practice Q21
NCERT Exercises 4.22	Practice Q22
NCERT Exercises 4.23	Practice Q23

Quick check: MCQs

4.3 The law of inertia3 questions
4.4 Newton's first law of motion3 questions
Impulse4 questions
4.6 Newton's third law of motion3 questions
4.7 Conservation of momentum3 questions
4.8 Particle ka equilibrium3 questions
Rolling friction4 questions
Banked road par car ka motion4 questions
4.11 Mechanics ke problems solve karna3 questions

Glossary · 15

Key terms (English + Hindi)

centripetal forceअभिकेंद्र बलabhikendra bal: Centre ki taraf net force, size $m v^{2} / R$ , jo body ko circle mein chalte rakhta hai.Chapter mein pehli baar yahan aaya →
coefficient of frictionघर्षण गुणांकgharshan gunaank: Surfaces ke kisi pair ke liye limiting friction aur normal force ka ratio: static friction ke liye $μ_{s}$ aur sliding ke liye $μ_{k}$ .Chapter mein pehli baar yahan aaya →

equilibriumसाम्यावस्थाsaamyaavastha: Particle ki woh state jismein us par net external force zero ho, taaki woh rest mein ho ya constant velocity se chale.Chapter mein pehli baar yahan aaya →

forceबलbal: Ek body par doosri body ka lagaya push ya pull, jise newtons mein naapa jaata hai. Net force body ki velocity badalta hai.Chapter mein pehli baar yahan aaya →
free-body diagramमुक्त पिंड आरेखmukt pind aarekh: Ek chuni hui body ya system ka akele sketch, jo us par act karte har external force ko dikhata hai.Chapter mein pehli baar yahan aaya →
frictionघर्षणgharshan: Contact force ka woh component jo chhooti surfaces ke along hota hai. Yeh unke beech asli ya impending fisalne ka virodh karta hai.Chapter mein pehli baar yahan aaya →

impulseआवेगaaveg: Force aur uske act karne ke time ka product, jo uske paida kiye momentum change ke barabar hai; $N s$ mein naapa jaata hai.Chapter mein pehli baar yahan aaya →
inertiaजड़त्वjadatva: Body ki apni rest ya uniform seedhi motion ki state mein kisi bhi change ka virodh karne ki tendency; mass ise naapta hai.Chapter mein pehli baar yahan aaya →

kinetic frictionगतिज घर्षणgatij gharshan: Fisalti surfaces ke beech friction, $f_{k} = μ_{k} N$ , relative motion ke against.Chapter mein pehli baar yahan aaya →

momentumसंवेगsanveg: Mass aur velocity ka product, $𝐩 = m 𝐯$ , ek vector jo $k g m s^{- 1}$ mein naapa jaata hai.Chapter mein pehli baar yahan aaya →

newtonन्यूटनnyootan: Force ki SI unit: woh force jo $1 k g$ mass ko $1 m / s^{2}$ ka acceleration deta hai.Chapter mein pehli baar yahan aaya →
normal forceअभिलंब बलabhilamb bal: Contact force ka woh part jo contact wali surfaces ke perpendicular hai. Yeh ek surface ko doosri mein ghusne se rokne ke liye khud ko adjust karta hai.Chapter mein pehli baar yahan aaya →

rolling frictionलोटनिक घर्षणlotanik gharshan: Wheel aur jis surface par woh rolling karta hai, unke halke deformation se hone wala rolling ka virodh; sliding friction se kahin chhota.Chapter mein pehli baar yahan aaya →

static frictionस्थैतिक घर्षणsthaitik gharshan: Na fisal rahi surfaces ke beech friction; yeh maximum $μ_{s} N$ tak khud ko adjust karta hai.Chapter mein pehli baar yahan aaya →

tensionतनावtanaav: Kheenchi hui string, rassi ya cable ka lagaya pull. Massless string mein yeh poori length mein same hota hai.Chapter mein pehli baar yahan aaya →

Laws of Motion: Class 11 Physics Chapter 4 Hinglish mein

Is chapter mein aap kya seekhenge

4.1 Introduction

4.2 Aristotle ki galti

4.3 The law of inertia

4.4 Newton's first law of motion

4.5 Newton's second law of motion

Momentum

Impulse

4.6 Newton's third law of motion

4.7 Conservation of momentum

4.8 Particle ka equilibrium

4.9 Mechanics ke common forces

4.9.1 Friction

Rolling friction

4.10 Circular motion

Level road par car ka motion

Banked road par car ka motion

4.11 Mechanics ke problems solve karna

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye