Coefficients of friction — measurement, material dependence: Question bank

1.2.7 · D5 · Physics › Newton's Laws & Dynamics › Coefficients of friction — measurement, material dependence

Shuru karne se pehle, ek shared vocabulary reminder taaki har symbol samajh aaye:

Sahi hai ya galat — justify karo

Rubber-on-road

μ

1 se zyada ho sakta hai — kya yeh allowed hai?

Sahi.

μ

sirf ek ratio

f / N

hai; koi cheez friction ko normal force se bada hone se nahi rokti. Soft rubber itni deeply interlock karta hai ki

μ \approx 1

ya zyada real hota hai.

Horizontal floor par rakhe block ki mass double karne se

μ_{s}

double ho jaata hai

Galat. Isse

N

double hota hai (isliye friction force

f = μ_{s} N

bhi), lekin

μ_{s}

ek ratio hai aur fixed rehta hai — yeh surface pair ki property hai, load ki nahi.

Ek brick ko uske chote end par khada karne se, wide face ki jagah,

μ_{s}

badal jaata hai

Galat. Apparent contact area

f = μ N

mein enter nahi karta. Real contact sirf asperity tips par hota hai, jinki total area

N

ke saath track karti hai, footprint ke saath nahi.

Flat table par, jab koi sideways push nahi hai, static friction force exactly

μ_{s} N

hoti hai

Galat. Jab koi sideways force apply nahi hai, friction zero hai — yeh sirf utna hi supply karti hai jitna zaroorat ho. Yeh

μ_{s} N

tak sirf slipping ki verge par pahunchti hai.

Angle-of-repose method mein block ki mass jaanna zaroori hai

Galat. Driving term

m g sin θ

aur normal term

m g cos θ

dono mein

m g

hai, jo cancel ho jaata hai:

μ_{s} = tan θ_{c}

. Ek protractor kaafi hai.

μ_{k}

saari sliding speeds ke liye genuinely constant hota hai

Pehli approximation ke roop mein mostly sahi, lekin exactly nahi — bahut zyada speeds par heating aur material changes isse shift kar dete hain. Introductory dynamics mein hum ise constant maante hain.

Kisi bhi surface pair ke liye,

μ_{s} \geq μ_{k}

hamesha hold karta hai

Almost hamesha sahi — resting asperities cold-weld aur interlock ho jaati hain, free hone ke liye extra force chahiye. Kuch rare engineered pairs nearly tie kar sakte hain, lekin normal materials mein

μ_{s} < μ_{k}

nahi dikh-ta.

Frictionless incline par angle of repose

0^{\circ}

hota hai

Iss sense mein sahi ki

μ_{s} = 0 \Rightarrow tan θ_{c} = 0 \Rightarrow θ_{c} = 0

: block thodi si bhi tilt par slide kar jaata hai, isliye koi non-zero "just holds" angle nahi hota.

Slip angle se neeche ramp par still baitha block ka friction

m g sin θ

ke barabar hota hai

Sahi. Yeh slide nahi kar raha, isliye along-slope forces balance hoti hain: static friction exactly driving component

m g sin θ

match karne tak badhti hai, jo ceiling

μ_{s} N

se kam hai.

Do steel plates ke beech tel daalne se

μ

isliye kam hota hai kyunki isse weight kam hoti hai

Galat. Weight unchanged rehti hai; oil film asperities ko alag kar deta hai taaki woh cold-weld na ho sakein, same

N

ke liye friction force cut ho jaati hai — dekho Lubrication & Tribology.

Error dhundho

"Ek incline par,

N = m g

, isliye

μ_{s} = f / m g

." Galti kahan hai?

Galat

N

. Tilted surface par normal force

N = m g cos θ

hoti hai,

m g

nahi. Sirf

m g cos θ

ramp mein press karta hai; baaki sliding drive karta hai.

"Kyunki

μ_{s} > μ_{k}

, ek sliding block hamesha decelerate karta hai." Kya missing hai?

Applied force. Akela kinetic friction ise decelerate kar deta, lekin agar tum kaafi zyada push karte raho toh net force zero ho sakti hai (constant speed) ya forward bhi.

μ_{s} > μ_{k}

sirf yeh batata hai ki start karna continue karne se mushkil hai.

"Horizontal pull par

μ_{k}

paane ke liye, box ko accelerate karte waqt spring padho." Theek karo.

Ise constant velocity par padho (

a = 0

). Sirf tabhi applied force exactly kinetic friction ke barabar hoti hai; accelerate karte waqt, teri push ka ek hissa

ma

mein jaata hai aur Newton's Second Law ke zariye reading ko pollute karta hai.

"Bada

μ

matlab object bhaari hai." Yeh theek karo.

μ

mein weight ke baare mein koi information nahi hoti; yeh do surfaces kis cheez se bani hain se set hoti hai. Ek halka Teflon puck aur ek bhaari steel block dono ka chota ya bada

μ

ho sakta hai, jo sirf unke surface pair par depend karta hai.

μ

steel ki property hai, isliye steel ka hamesha

μ \approx 0.6

hota hai." Galat kaise?

μ

ek pair ki hoti hai, kabhi ek material ki nahi. Dry steel-on-steel

\sim 0.6

hai, lekin steel-on-ice

\sim 0.03

hai aur steel-on-oiled-steel

\sim 0.1

hai. Hamesha kehna padega "steel on kya".

"Slipping ki verge par,

m g sin θ_{c} = N

." Pakdo ise.

Right side friction ceiling

μ_{s} N

honi chahiye,

N

nahi. Sahi along-slope balance yeh hai:

m g sin θ_{c} = μ_{s} N = μ_{s} m g cos θ_{c}

, jisse

μ_{s} = tan θ_{c}

milta hai.

"Real contact area block ke footprint ke barabar hota hai." Yeh galat kyun hai?

Surfaces sirf microscopic asperity peaks par touch karti hain, isliye real contact footprint ka ek tiny fraction hai aur

N

ke saath badhta hai jab bumps flatten hote hain — exactly isliye apparent area

f = μ N

mein se bahar ho jaata hai.

Kyun wale questions

μ

dimensionless kyun hai?

Yeh ek force divided by ek force hai (

f / N

), isliye saari units cancel ho jaati hain, ek pure comparison number reh jaata hai.

Mass incline method mein cancel kyun ho jaati hai lekin angle nahi?

Har force term

m g

ke proportional hai, isliye along-slope ko perpendicular balance se divide karne par

m g

poori tarah remove ho jaata hai, sirf

tan θ_{c}

reh jaata hai — ek angle-only result.

Static case ke liye friction ko inequality

f_{s} \leq μ_{s} N

se model kyun karna chahiye?

Kyunki static friction jo bhi push cancel karni ho uske hisaab se adapt karti hai; uski ek maximum hai jo exceed nahi ho sakti, lekin usse neeche yeh exactly equilibrium ke liye zaroorat ki value leti hai. Dekho Static vs Kinetic Friction.

Do surfaces ko zyada press karne se grip force kyun badhti hai?

Zyada press karne se zyada asperities contact mein flatten hoti hain, real contact area badhta hai, aur friction usi real area ke saath scale karti hai — isliye

N

ke saath. Yeh

f = μ N

ki microscopic origin hai.

Dono measurement methods se pehle free body diagram kyun draw karte hain?

Har force identify karne ke liye aur axes choose karne ke liye taaki "surface ke perpendicular koi motion nahi" cleanly

N

de sake, aur along-surface balance friction ko isolate kare — iske bina

m g

ka cancellation asaani se galat ho sakta hai.

Microscopic level par

μ_{s}

usually

μ_{k}

se bada kyun hota hai?

Rest par bumps ko settle hone aur ek doosre ki valleys mein cold-weld hone ka waqt milta hai; ek baar motion mein hone par yeh settle hue bina tops par skim karte hain, isliye motion sustain karne ke liye kam force chahiye.

Hum "ice ka coefficient of friction kya hai?" standalone question nahi pooch sakte — kyun?

Friction do surfaces ke interaction se arise hoti hai, isliye akele ek surface ka

μ

nahi hota — meltwater aur adhesion is par depend karte hain ki ice kisse slide kar rahi hai.

Edge cases

Agar ramp angle exactly slip angle ke barabar hai, toh kya block move kar raha hai?

Nahi — yeh slipping ki verge par hai. Static friction

μ_{s} N

par max out hai aur block equilibrium ke last instant mein hai; koi bhi nudge motion shuru kar degi.

Bilkul horizontal, undisturbed surface par rakhe block par friction force kya hai?

Exactly zero. Koi sideways driving force nahi hai jise static friction oppose kare, isliye yeh kuch supply nahi karti, chahe

μ_{s} > 0

ho.

μ_{s} = 0

ke liye (idealised frictionless pair), angle of repose kya hai?

0^{\circ}

se upar koi nahi hai; block kisi bhi tilt ko resist nahi kar sakta, isliye kisi bhi

θ > 0

par slide karta hai. Yeh

tan θ_{c} = μ_{s}

ki degenerate limit hai.

Jab

μ_{s} \to \infty

(imagine perfect grip),

θ_{c}

ka kya hoga?

θ_{c} \to 9 0^{\circ}

: block tab tak chipka rahega jab tak ramp nearly vertical na ho.

tan θ_{c} = μ_{s}

exactly blow up karta hai jab angle right angle ke paas pahunchta hai.

Ek block launch kiye jaane ke baad rough incline par upar slide karta hai — kinetic friction kis direction mein point karti hai?

Slope ke neeche, upar ki motion ko oppose karte hue. Kinetic friction hamesha sliding ki direction ko oppose karti hai, isliye gravity ke along-slope pull ke saath milkar yeh block ko strongly decelerate karti hai — dekho Inclined Plane Dynamics.

Agar ek object free fall mein hai (koi surface contact nahi), toh

N

aur hence

f

kya hai?

Dono zero — koi surface press nahi kar rahi,

N = 0

, isliye

f = μ N = 0

. Koi contact nahi matlab koi friction nahi, chahe

μ

kuch bhi ho.

Exactly constant-velocity slide angle par, ramp ke along net force kya hai?

Zero — constant velocity ka matlab

a = 0

hai, isliye

m g sin θ_{slide} = μ_{k} m g cos θ_{slide}

, jisse

μ_{k} = tan θ_{slide}

milta hai.

Recall Jaane se pehle ek-line self-test

$μ$ ek ratio hai, ek pair ki property hai, apparent area aur weight ko ignore karta hai, static ≥ kinetic, aur ramp par $N$ $m g cos θ$ hoti hai — $m g$ nahi. Agar inme se kuch bhi surprising laga, upar woh trap dobara dekho.

Connections

Parent topic — in traps ke peeche full derivations.
Static vs Kinetic Friction — inequality vs equality ka distinction.
Normal Force — kyun inclines par $N \neq = m g$ hota hai.
Newton's Second Law — kyun $μ_{k}$ padhne ke liye constant velocity zaroori hai.
Inclined Plane Dynamics — ramps par friction ka direction.
Free Body Diagrams — har setup ke peeche ka tool.
Lubrication & Tribology — films $μ$ ko kaise cut karti hain.