Friction — static (maximum), kinetic, rolling: Question bank

1.2.6 · D5 · Physics › Newton's Laws & Dynamics › Friction — static (maximum), kinetic, rolling

0. Symbol recap — pehle yeh padho

Kisi bhi trap se pehle, is page par use hone wale har symbol ko plain words mein define kiya gaya hai aur saath mein figure bhi di gayi hai.

Definition Symbols ki cast

$N$ = normal force: surface ka perpendicular ("normal" matlab 90° par) push jo object ko wapas deta hai. Figure 1 mein red arrow dekho jo floor se seedha bahar point kar raha hai.
$m g$ = weight: object ki mass $m$ times gravity $g$ , seedha neeche pull karta hai (black down-arrow, Figure 1). Flat ground par surface ko $N = m g$ supply karna padta hai taaki object neeche na dhanse.
$f_{s}$ = static friction: along-surface force jo tab act karta hai jab object abhi slide nahi kar raha. Yeh self-adjust karta hai.
$f_{s}^{ma x}$ = maximum static friction: wo ceiling jo $f_{s}$ sliding se ठीक pehle reach kar sakta hai, $f_{s}^{ma x} = μ_{s} N$ . Iske baad bonds snap ho jaate hain aur motion shuru hoti hai.
$f_{k}$ = kinetic friction: along-surface force jab object slide kar raha ho, $f_{k} = μ_{k} N$ , roughly constant.
$μ_{s}, μ_{k}, μ_{r}$ = static, kinetic, rolling friction ke coefficients: pure numbers (koi units nahi) jo surface ki stickiness capture karte hain.
$θ$ = incline ka tilt angle; $θ_{r}$ = angle of repose, wo tilt jis par sliding abhi shuru hoti hai.
$R$ = wheel radius; $a$ = kitna normal force ek rolling wheel par forward shift hota hai.

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

1. Do zaroor dekhne wali pictures

Figure 2 wo incline decomposition hai jis par is page ke har incline trap ka aadhar hai. Figure 3 dikhata hai kyun static-vs-kinetic ek jump hai, slide nahi.

2. True or false — justify karo

Static friction hamesha

μ_{s} N

ke barabar hoti hai.

False.

μ_{s} N

sirf ceiling hai (

f_{s}^{ma x}

). Static friction self-adjust karti hai jitni zaroorat ho sliding rokne ke liye, toh usually

f_{s} < μ_{s} N

Ek stationary block jo level ground par hai aur jis par koi horizontal push nahi hai, usse friction feel hoti hai.

False. Koi cheez isse slide karne ki koshish nahi kar rahi, toh relative motion ki koi tendency nahi, isliye static friction exactly zero hai — yeh sirf driving force oppose karne ke liye appear hoti hai.

Kinetic friction jitna tez slide karo utni zyada strong hoti hai.

False basic model mein.

f_{k} = μ_{k} N

speed se independent hai; sirf surface roughness

μ_{k}

aur load

N

matter karte hain yahaan.

Same weight ka ek wider box zyada friction experience karta hai kyunki zyada surface touch karti hai.

False. Apparent area double karne se pressure per patch half ho jaata hai; dono effects cancel ho jaate hain, isliye friction

N

par depend karta hai, apparent area par nahi.

Friction kisi object ko kabhi speed up nahi kar sakti.

False. Friction contact par relative sliding oppose karta hai, object ki motion nahi. Jab car accelerate karti hai, tyres par static friction forward point karti hai aur car ko drive karti hai.

μ_{s}

1 se zyada ho sakta hai.

True.

μ

forces ka ratio hai, angle ya probability nahi; bahut sticky/rough surfaces (rubber on rubber)

μ_{s} > 1

de sakte hain, matlab isse slide karne ke liye weight se zyada force chahiye.

Ek baar jab

μ_{k}

aur

N

pata ho, kinetic friction fully determined hai.

True. Static (inequality) ke unlike, kinetic friction ek fixed value

μ_{k} N

hai jab sliding shuru ho jaaye — koi self-adjusting nahi.

Ek frictionless incline par block ka normal force phir bhi hota hai.

True. Normal force surface se aata hai jo andar push hone se resist karti hai; yeh

m g cos θ

hai regardless of friction. "Frictionless" sirf along-surface force ko khatam karta hai.

Rolling friction aur kinetic friction ke coefficient ki size same hoti hai.

False.

μ_{r} = a / R

jahan deformation offset

a

millimetre ka fraction hota hai aur radius

R

ke against, toh

μ_{r} ≪ μ_{k}

— yahi reason hai ki wheels dragging se better hote hain.

3. Spot the error

"Box ko 30 N se push kiya gaya hai aur

f_{s}^{ma x} = 50

N hai, toh friction 50 N hai."

Error: yeh slide nahi kar raha, toh static friction sirf 30 N supply karti hai jo push balance karne ke liye chahiye, ceiling value

f_{s}^{ma x}

nahi. Friction yahaan 30 N hai.

"Yeh slide kar raha hai, toh main

f_{k} = μ_{s} N

use karunga."

Error: sliding matlab kinetic law apply hoga,

f_{k} = μ_{k} N

, aur

μ_{k} < μ_{s}

μ_{s}

use karne se moving object ki friction overestimate ho jaati hai.

"Incline par driving force

m g

hai, toh jab

μ_{s} m g < m g

tab slide karta hai."

Error: sirf component along the plane,

m g sin θ

, sliding drive karta hai, aur friction isse

μ_{s} m g cos θ

se oppose karta hai. Sahi condition hai

tan θ > μ_{s}

"Normal force hamesha

m g

ke barabar hoti hai."

Error:

N = m g

sirf horizontal surface ke liye hai jab koi vertical push nahi. Incline par

N = m g cos θ

; downward push

N

badhata hai, upward pull

N

ghatata hai.

"Angle of repose block ki mass par depend karta hai, kyunki bhaari blocks zyada grip karte hain."

Error: mass cancel ho jaata hai —

m g sin θ_{r} = μ_{s} m g cos θ_{r}

tan θ_{r} = μ_{s}

milta hai, mass-free. Repose angle sirf surfaces par depend karta hai.

"Box ko constant velocity par move karte rakhne ke liye

f_{s}^{ma x}

se zyada push karna padta hai."

Error: ek baar move ho jaane ke baad sirf kinetic friction

μ_{k} N

se fight karni hai, jo ceiling

f_{s}^{ma x}

se kam hai.

f_{k}

ke barabar push se constant velocity milti hai (

a = 0

"Driving tyre par static friction backward point karti hai, toh friction car slow karti hai."

Error: tyre surface road ke relative backward slip karne ki tendency rakhta hai, toh uss par static friction forward point karti hai (slip ke opposite, hamare sign convention ke according) — yahi accelerating force hai.

4. Why questions

Static friction ko single value ki jagah inequality se kyun model kiya jaata hai?

Kyunki stationary object ka net force zero hota hai, friction ko exactly jo bhi push apply ho usse cancel karna padta hai — toh yeh breaking-point ceiling

μ_{s} N

tak kai values leta hai.

μ_{k}

μ_{s}

se chhota kyun hota hai?

Static contacts still rehte hain aur adhesive bonds ko fully mature hone dete hain; sliding contacts bonds snap kar dete hain unhe settle hone se pehle, toh average mein kam/weaker bonds resist karte hain, isliye smaller coefficient milta hai.

Friction apparent contact area par kyun depend nahi karta?

Zyada apparent area same load ko zyada patches par spread karta hai, pressure per patch kam karta hai aur har jagah kam bonds banata hai; patches mein increase aur per patch decrease cancel ho jaate hain, aur

f

sirf total

N

se tied rehta hai.

Ek rolling wheel kisi resistance ko experience kyun karta hai agar yeh slide nahi karta?

Wheel aur ground thoda deform hote hain aur perfectly rebound nahi karte (hysteresis), normal force ko centre se

a

aage shift karta hai; wo offset ek backward torque

N a

create karta hai jo ek small force

μ_{r} N

ke equivalent hai.

Plane ko tilt karna eventually kisi bhi block ko slide kyun kara deta hai, regardless of

μ_{s}

Jaise-jaise

θ

badhta hai, driving

m g sin θ

badhta hai jabki friction ceiling

μ_{s} m g cos θ

girti hai; jab

tan θ

μ_{s}

se exceed kar jaata hai tab ceiling drive ko match nahi kar sakti.

Walking kaam kyun karti hai — kya friction motion rok nahi deta?

Tumhara paaon ground ke against backward push karta hai aur usi taraf slip karne ki tendency rakhta hai; static friction us tendency ko oppose karke tumhe forward push karta hai, toh yahi tumhe propel karta hai.

f = μ N

mein weight ki jagah

N

kyun use kiya jaata hai?

Friction is baat se arise hota hai ki surfaces ko kitna hard press kiya ja raha hai, jo perpendicular contact force

N

hai. Weight sirf special flat, unloaded case mein

N

set karta hai.

5. Edge cases

Ek block jo free-fall mein hai ek vertical wall ke saath-saath (wall se press nahi ho raha), us par friction kya hai?

Zero. Wall par block ko press karne wala koi normal force nahi hai (

N = 0

f = μ N = 0

— friction ke liye contact pressure chahiye.

Ek block ek aise incline par baitha hai jahan exactly

tan θ = μ_{s}

hai. Kya hoga?

Yeh exact verge par hai: driving force friction ceiling ke barabar hai. Yeh na accelerate karta hai na koi spare grip hai — ruke rehne aur slide karne ke beech ki boundary.

Tum ek resting box ko exactly

f_{s}^{ma x}

ke barabar force se push karte ho. Kya yeh move karta hai?

Exact ceiling par yeh threshold par hai — friction abhi barely match kar sakti hai, toh acceleration zero hai, lekin koi bhi infinitesimal extra push sliding shuru kar dega.

Ek block slide kar raha hai, phir tum push karna band kar dete ho. Kaisi friction act karti hai?

Jab tak move kar raha hai tab tak usse kinetic friction

μ_{k} N

feel hogi jo uski motion oppose karegi, use decelerate karegi; ruk jaane ke baad, friction jo zaroorat ho usi mein drop ho jaati hai (yahaan zero).

Do identical boxes, ek doosre ke upar stacked, floor par. Kya bottom box par floor friction ceiling double ho jaati hai?

Haan — bottom box ab dono ka weight carry karta hai, toh

N

double ho jaata hai aur

f_{s}^{ma x} = μ_{s} N

double ho jaata hai. Friction

N

ke through scale hua, area ke nahi.

Perfectly smooth (

μ = 0

) incline par block ka acceleration kya hoga?

g sin θ

slope ke neeche. Koi friction nahi toh along-plane sirf weight component

m g sin θ

act karta hai,

a = g sin θ

deta hai.

Kya static friction momentarily

μ_{s} N

se exceed kar sakti hai?

Nahi.

μ_{s} N

bonds ka physical breaking point hai; isse zyada demand karo toh bonds fail ho jaate hain, isliye object slide karna shuru kar deta hai aur kinetic friction le leti hai.

Ek wheel bina slipping ke roll karta hai. Contact point par kaun si friction act karti hai, static ya kinetic?

Static — "without slipping" matlab contact point instantaneously ground ke relative rest mein hai, toh koi sliding nahi hai jise kinetic friction resist kare.

Recall Self-check

Kya har "false" answer ne specific galat assumption identify ki (ceiling vs actual, area vs $N$ , speed independence)? ::: Agar tumne sirf "false" likha bina mechanism ke, toh revisit karo — reasoning hi poora point hai. Figure 2 se, kya tum $m g sin θ$ ko $μ_{s} m g cos θ$ ke against balance karke $tan θ_{r} = μ_{s}$ re-derive kar sakte ho? ::: Agar nahi, toh Section 1 aur Inclined Plane Problems dobara padho.

Yeh bhi dekho: Newton's Second Law · Free Body Diagrams · Rolling without Slipping · Circular Motion · Work-Energy Theorem.