Coefficients of friction — measurement, material dependence: Worked examples

1.2.7 · D3 · Physics › Newton's Laws & Dynamics › Coefficients of friction — measurement, material dependence

Kisi bhi example se pehle, ek idea bilkul crystal clear hona chahiye, kyunki friction ki aadhi galtiyan isi ko ignore karne se aati hain.

Definition Static friction ek

range hai, ek fixed value nahi Is poore page mein do coefficients aate hain, toh ek baar naam fix kar lete hain:

$μ_{s}$ static-friction coefficient hai — yeh object ke stuck rehne tak maximum grip set karta hai.
$μ_{k}$ kinetic-friction coefficient hai — yeh sliding shuru hone ke baad (roughly constant) drag set karta hai.

Kinetic friction (sliding ke dauran) ka size fixed hota hai: $f_{k} = μ_{k} N$ . Static friction (stuck rehne par) utna hota hai jitna object ko still rakhne ke liye chahiye — $0$ se lekar ek ceiling tak $f_{s, m a x} = μ_{s} N$ . $0 \leq f_{s} \leq μ_{s} N$ Toh KISI BHI problem mein sabse pehla sawaal yeh hai: kya yeh slide karta hai ya nahi? Iska jawab milta hai motion cause karne waali force ko ceiling $μ_{s} N$ se compare karke. Sirf tab jab pushing force ceiling se zyada ho, object move karta hai — aur sirf tab tum $f = μ_{k} N$ likh sakte ho.

Figure 1 — grip curve.

Figure — Coefficients of friction — measurement, material dependence

Red bar dekho: jaise-jaise tum zyada push karte ho (right move karte hue), static friction tumse match karne ke liye badhta hai, diagonal trace karta hua — friction aur push equal hain, kuch nahi hilta. Peak par ( $μ_{s} N$ ) block free ho jaata hai; friction tab drop karke flat kinetic level $μ_{k} N$ par aa jaata hai. Yeh ek picture explain karti hai kyun $μ_{s} > μ_{k}$ cheezein move shuru karte waqt ek "jerk" jaisi feel deta hai. Hum Ex 1 aur Ex 2 mein is curve ki taraf wapas point karenge.

Scenario matrix

Har friction problem inhi cells mein se ek (ya mix) hai. Neeche har example tagged hai us cell/cells se jo woh cover karta hai.

#	Cell (case class)	Kya tricky hai	Example
A	Horizontal pull, kya yeh move bhi karega?	push ko ceiling $μ_{s} N$ se compare karna padega	Ex 1
B	Horizontal pull, yeh move karta hai → $a$ nikalo	$μ_{k}$ use karo, $\sum F = ma$ lagao	Ex 2
C	Angled pull (upar)	pull $N$ ko change karta hai → friction change hoti hai	Ex 3
D	Angle pe neeche push karo	$N$ badhta hai, move karna mushkil ho jaata hai	Ex 4
E	Incline, angle of repose (verge)	$μ_{s} = tan θ_{c}$ , mass cancel ho jaata hai	Ex 5
F	Incline, block slides → $a$ nikalo	gravity-pull aur $μ_{k} N$ ka net	Ex 6
G	Incline, block ruka rehta hai (repose se neeche)	$f_{s}$ ceiling se kam hai	Ex 7
H	Degenerate / limiting: $θ = 0$ , $θ = 9 0^{\circ}$ , $μ = 0$	formulas ko sanity-check karo	Ex 8
I	Real-world word problem	story ko numbers mein translate karo	Ex 9
J	Exam twist: two-surface / trick	ek subtle catch jo examiners ko pasand hai	Ex 10

Hum Free Body Diagrams aur parent note se coordinate decomposition par rely karte hain, aur Newton's Second Law $\sum F = ma$ par bhi. Neeche har example apna free-body diagram (FBD) lekar khulta hai — object par saari forces ki picture — kyunki FBD ke bina friction problems mein safely reason karna impossible hai.

Ex 1 — Kya yeh move bhi karega? (Cell A)

Worked example Ceiling se neeche

Ek $5 kg$ crate floor par rakha hai jahan $μ_{s} = 0.40$ , $μ_{k} = 0.30$ . Tum horizontally $15 N$ push karte ho. Kya yeh move karta hai? Abhi friction force kitni hai?

Forecast: Padhne se pehle guess karo — kya yeh slide karega, aur friction $μ_{s} N$ hogi, $μ_{k} N$ hogi, ya kuch aur?

FBD — Figure 2.

Chaar forces: weight

m g

neeche, normal

N

upar, tumhari push right, friction left.

Step 1 — Normal force. Floor horizontal hai, koi vertical pull nahi, toh vertical balance deta hai $N = m g = 5 \times 9.8 = 49 N$ . Yeh step kyun? Friction hamesha $μ N$ hoti hai; jab tak $N$ pata na ho tab tak friction nahi pata. Dekho Normal Force.

Step 2 — Ceiling compute karo. $f_{s, m a x} = μ_{s} N = 0.40 \times 49 = 19.6 N$ . Yeh step kyun? Yeh available maximum grip hai — Figure 1 ki grip curve ka peak. Agar hamaari push isse beat kar de, toh crate slide karega.

Step 3 — Compare karo. Push $= 15 N < 19.6 N = f_{s, m a x}$ . Toh yeh move nahi karta. Yeh step kyun? Yeh "kya slide karega?" test hai — har static problem ka dil. Figure 1 par hum abhi rising diagonal par hain, peak tak nahi pahunche.

Step 4 — Actual friction batao. Kyunki yeh stuck hai aur accelerate nahi kar raha, friction push ko exactly cancel karti hai: $f_{s} = 15 N$ (na $19.6$ , na $μ_{k} N$ ). Yeh step kyun? Static friction ceiling tak jitna chahiye utna adapt karti hai. Yeh apne available $19.6 N$ mein se sirf $15$ use kar rahi hai — Figure 1 ke diagonal par ek point.

Verify: $\sum F = 15 - 15 = 0 \Rightarrow a = 0$ . ✓ "Not moving" ke saath consistent. Units: sab newtons. ✓

Ex 2 — Yeh move karta hai, acceleration nikalo (Cell B)

Worked example Ceiling ke paar

Wahi crate ( $5 kg$ , $μ_{s} = 0.40$ , $μ_{k} = 0.30$ ). Ab tum $25 N$ se push karte ho. Acceleration nikalo.

Forecast: Kya hum $\sum F = ma$ mein $μ_{s}$ use karenge ya $μ_{k}$ ?

FBD: Figure 2 jaisa hi, lekin ab push arrow friction arrow se lamba hai — ek net force bachi hai.

Step 1 — Sliding test karo. Ceiling $= μ_{s} N = 0.40 \times 49 = 19.6 N$ . Push $25 > 19.6$ , toh yeh slides karta hai. Yeh step kyun? Sirf motion confirm hone ke baad hum kinetic friction par switch kar sakte hain. Figure 1 par hum peak ke paar push kar chuke hain.

Step 2 — Kinetic friction. Move karne ke baad, $f_{k} = μ_{k} N = 0.30 \times 49 = 14.7 N$ . Yeh step kyun? Sliding ke dauran friction ki fixed value hoti hai $μ_{k} N$ — Figure 1 mein flat red level, peak se neeche.

Step 3 — Newton's second law. $\sum F = 25 - 14.7 = 10.3 N$ , toh $a = \frac{10.3}{5} = 2.06 m/s^{2}$ . Yeh step kyun? Net force divided by mass gives acceleration. Dekho Newton's Second Law.

Verify: Units $N / kg = m/s^{2}$ . ✓ Kyunki $μ_{k} < μ_{s}$ , friction potential $19.6$ se $14.7$ tak drop ho gayi — Figure 1 mein "jerk into motion" (drop). ✓

Ex 3 — Angled pull (upar ki taraf) (Cell C)

Worked example Upar kheenchna grip ko change karta hai

Ek $10 kg$ sled, $μ_{s} = 0.50$ . Tum ek rope se $3 0^{\circ}$ horizontal se upar kheenchte ho. Woh tension $T$ kya hogi jo isse just start karaye?

Forecast: Purely horizontal pull se zyada force chahiye ya kam? Socho upward component $N$ ka kya karta hai.

FBD — Figure 3.

Red pull

T

tilt hai upar ki taraf; iska vertical part weight se ladhta hai, iska horizontal part friction se ladhta hai.

Step 1 — Pull decompose karo. $T$ ko horizontal $T cos 3 0^{\circ}$ (motion drive karta hai) aur vertical $T sin 3 0^{\circ}$ (lift karta hai) mein split karo. Yeh step kyun? Alag directions alag kaam karte hain; sirf horizontal part friction se ladhta hai, sirf vertical part $N$ ko change karta hai. Same decomposition idea jaisi Inclined Plane Dynamics mein hai.

Step 2 — Naya normal force. Vertical balance: $N + T sin 3 0^{\circ} = m g$ , toh $N = m g - T sin 3 0^{\circ}$ . Yeh step kyun? Upward pull kuch weight relieve karta hai, toh floor kam push-back karta hai — friction weak hoti hai.

Step 3 — Verge-of-slip condition. $T cos 3 0^{\circ} = μ_{s} N = μ_{s} (m g - T sin 3 0^{\circ})$ . Yeh step kyun? Motion shuru hone ke exact moment par, horizontal pull maximum friction ke equal hoti hai.

Step 4 — $T$ ke liye solve karo. $T cos 3 0^{\circ} + μ_{s} T sin 3 0^{\circ} = μ_{s} m g$ $T = \frac{μ _{s} m g}{c o s 3 0 ^{\circ} + μ _{s} s i n 3 0 ^{\circ}} = \frac{0.50 \times 10 \times 9.8}{0.8660 + 0.50 \times 0.5} = \frac{49}{1.1160} \approx 43.9 N$ Yeh step kyun? $T$ terms collect karo, phir isolate karo.

Verify: Purely horizontal pull ke liye $μ_{s} m g = 49 N$ chahiye hota. Humne sirf $43.9 N$ paaya — kam, confirm karta hai ki upward tilt $N$ cut karke help karta hai. ✓ Units: N. ✓

Ex 4 — Angle pe neeche push karo (Cell D)

Worked example Neeche push karna tumhare against kaam karta hai

Wahi $10 kg$ sled, $μ_{s} = 0.50$ , lekin ab tum $3 0^{\circ}$ horizontal se neeche push karte ho (jaise lawn mower dhakelna). Woh force $F$ kya hogi jo isse just start karaye?

Forecast: Ex 3 se compare karo — ab zyada force chahiye ya kam?

FBD — Figure 4.

Red push

F

neeche tilt hai; iska vertical part weight mein add hota hai, toh

N

badhta hai.

Step 1 — Decompose karo. Horizontal $F cos 3 0^{\circ}$ drive karta hai; vertical $F sin 3 0^{\circ}$ neeche press karta hai. Yeh step kyun? Downward component ab squeeze mein add hota hai.

Step 2 — Normal force. $N = m g + F sin 3 0^{\circ}$ . Yeh step kyun? Neeche push karne ka matlab floor ko zyada push-back karna padta hai → bada $N$ → zyada friction overcome karni padti hai.

Step 3 — Verge condition. $F cos 3 0^{\circ} = μ_{s} (m g + F sin 3 0^{\circ})$ . Yeh step kyun? Motion shuru hone ke exact moment par horizontal drive maximum friction $μ_{s} N$ ke equal hoti hai, aur yahan $N = m g + F sin 3 0^{\circ}$ — toh wahi force $F$ jo drive karta hai woh friction bhi inflate karta hai jise use beat karna hai.

Step 4 — Solve karo. $F (cos 3 0^{\circ} - μ_{s} sin 3 0^{\circ}) = μ_{s} m g$ $F = \frac{0.50 \times 10 \times 9.8}{0.8660 - 0.50 \times 0.5} = \frac{49}{0.6160} \approx 79.5 N$ Yeh step kyun? Ab $μ_{s} sin$ term subtract karta hai, denominator shrink karta hai, $F$ blow up karta hai.

Verify: $79.5 N > 49 N$ (horizontal) $> 43.9 N$ (Ex 3, upward). Neeche push karna sabse bura strategy hai — intuition se match karta hai. ✓

Ex 5 — Angle of repose (Cell E)

Worked example Incline par verge

Ek block tab just slide karna shuru karta hai jab ramp $θ_{c} = 2 7^{\circ}$ reach karta hai. $μ_{s}$ nikalo.

Forecast: Kya tumhe block ki mass chahiye?

FBD — Figure 5.

Weight $m g$ split hota hai $m g sin θ$ (down-slope, red) aur $m g cos θ$ (slope mein); normal $N$ aur friction $f$ unhe balance karte hain.

Step 1 — Incline par verge-of-slip. Parent note se, $μ_{s} = tan θ_{c}$ . Yeh step kyun? Critical angle par, $m g sin θ_{c} = μ_{s} m g cos θ_{c}$ , aur $m g$ cancel ho jaata hai.

Step 2 — Evaluate karo. $μ_{s} = tan 2 7^{\circ} \approx 0.510$ . Yeh step kyun? Jab general result $μ_{s} = tan θ_{c}$ pata ho, toh measured critical angle $2 7^{\circ}$ plug in karne par ek actual number milta hai — woh number hi woh coefficient hai jise hum dhundh rahe the.

Verify: Dimensionless (kisi angle ka tan). ✓ Mass kabhi aayi hi nahi — ramp method ki khoobsoorti yahi hai. ✓

Ex 6 — Incline, block slides, acceleration nikalo (Cell F)

Worked example Steep ramp par neeche slide karna

Ek $2 kg$ block ek $3 5^{\circ}$ incline par $μ_{k} = 0.25$ ke saath. Yeh already neeche slide kar raha hai. Iska acceleration nikalo.

Forecast: Positive (speed up) ya friction kaafi hai isse slow karne ke liye?

FBD: Figure 5 jaisa layout, lekin ab friction $f = μ_{k} N$ apne fixed kinetic size par hai, up-slope point kar rahi hai downhill slide ke against. (Yaad rakho tilted axes: $x$ along-slope, $y$ perpendicular.)

Step 1 — Perpendicular balance. $N = m g cos 3 5^{\circ} = 2 \times 9.8 \times 0.8192 = 16.06 N$ . Yeh step kyun? Surface se off koi motion nahi, toh perpendicular forces cancel ho jaati hain.

Step 2 — Kinetic friction. $f_{k} = μ_{k} N = 0.25 \times 16.06 = 4.01 N$ , slope ke upar point karte hue (downhill motion ke opposing). Yeh step kyun? Kinetic friction hamesha sliding direction ke opposite hoti hai.

Step 3 — Along-slope Newton's law. Downhill gravity pull $= m g sin 3 5^{\circ} = 2 \times 9.8 \times 0.5736 = 11.24 N$ . $\sum F = m g sin 3 5^{\circ} - f_{k} = 11.24 - 4.01 = 7.23 N$ $a = \frac{7.23}{2} = 3.62 m/s^{2} (slope ke neeche)$ Yeh step kyun? Net driving force divided by mass.

Verify: Kyunki $tan 3 5^{\circ} = 0.70 > μ_{k} = 0.25$ , gravity friction ko beat karta hai, toh acceleration positive hai (down-slope). ✓ Units N/kg = m/s². ✓

Ex 7 — Incline, block ruka rehta hai (Cell G)

Worked example Repose angle se neeche

Ek $3 kg$ block ek $1 5^{\circ}$ incline par $μ_{s} = 0.40$ ke saath. Kya yeh slide karta hai? Actual friction force kya hai?

Forecast: Kya yahan friction $μ_{s} N$ ke equal hai?

FBD: Figure 5 jaisa layout, lekin block rest mein hai — static friction $f_{s}$ (up-slope) utni hai jitni chahiye, zaroori nahi ki ceiling tak. (Ex 6 jaisi hi tilted axes.)

Step 1 — Ceiling compute karo. $N = m g cos 1 5^{\circ} = 3 \times 9.8 \times 0.9659 = 28.40 N$ ; $f_{s, m a x} = μ_{s} N = 0.40 \times 28.40 = 11.36 N$ . Yeh step kyun? Humein woh maximum grip chahiye jo surface offer kar sakti hai taaki hum test kar sakein ki downhill pull usse exceed karti hai ya nahi — incline par "does it slide?" test.

Step 2 — Downhill pull. $m g sin 1 5^{\circ} = 3 \times 9.8 \times 0.2588 = 7.61 N$ . Yeh step kyun? Gravity ka yeh along-slope component akela hi block ko slide karne ki koshish karta hai; Step 1 ki ceiling se compare karne ke liye ise jaanna zaroori hai.

Step 3 — Compare karo. Downhill pull $7.61 < 11.36$ ceiling → block ruka rehta hai. Equivalently, $tan 1 5^{\circ} = 0.268 < μ_{s} = 0.40$ . Yeh step kyun? Repose test: yeh sirf tab slide karta hai jab $tan θ > μ_{s}$ .

Step 4 — Actual friction. Kyunki static hai aur still hai, friction downhill pull ko exactly balance karti hai: $f_{s} = 7.61 N$ (slope ke upar), NOT $11.36$ . Yeh step kyun? Static friction sirf utna supply karti hai jitna zaroorat ho — Ex 1 ki echo.

Verify: $\sum F_{along} = 7.61 - 7.61 = 0$ . ✓ Yeh apne $11.36 N$ budget mein se $7.61$ use kar rahi hai. ✓

Ex 8 — Degenerate & limiting cases (Cell H)

Worked example Extremes ko sanity-check karo

Incline formulas ko har boundary par test karo: $θ = 0$ , $θ \to 9 0^{\circ}$ , aur $μ = 0$ .

Case (i) — Flat floor, $θ = 0$ . Repose formula $μ_{s} = tan θ_{c}$ ek formula hai critical tilt angle ke liye, $μ_{s}$ ke liye nahi — material coefficient $μ_{s}$ surface pair ki ek fixed property hai aur sirf isliye $0$ nahi ho jaata kyunki ramp flat hai. $θ = 0$ par gravity ka koi down-slope component nahi hota ( $m g sin 0 = 0$ ), toh sliding drive hi nahi hoti aur repose test trigger hi nahi hota. Formula sirf slipping ke verge par apply hota hai; $θ = 0$ par hum us verge ke kareebi bhi nahi hain, toh $tan 0 = 0$ read off karna meaningless hai, yeh claim nahi ki $μ_{s} = 0$ . Note karo $N = m g cos 0 = m g$ , full weight — woh familiar flat-ground case. ✓

Case (ii) — Vertical wall, $θ \to 9 0^{\circ}$ . $tan 9 0^{\circ} \to \infty$ . Interpretation: block ko is steep incline par hold karne ke liye tumhe infinite $μ_{s}$ chahiye hoga — yaani yeh hamesha slide karta hai. Physically, $9 0^{\circ}$ par normal force $N = m g cos 9 0^{\circ} = 0$ ho jaata hai, toh koi squeeze nahi, hence bilkul friction nahi ( $f_{s, m a x} = μ_{s} N = 0$ ) — kuch bhi block ko hold nahi karta aur yeh free-fall karta hai. ✓

Case (iii) — Frictionless, $μ = 0$ . Ex 6 ke result $a = g (sin θ - μ_{k} cos θ)$ mein $μ_{k} = 0$ daalo, mila $a = g sin θ$ . $θ = 3 0^{\circ}$ par: $a = 9.8 \times 0.5 = 4.9 m/s^{2}$ . Yeh pure-gravity slide hai — us angle par object ki maximum possible acceleration neeche. ✓

Verify: Teeno limits $N = m g cos θ$ aur repose logic se agree karte hain; $θ = 0$ deta hai $N = m g$ , $θ = 9 0^{\circ}$ deta hai $N = 0$ , $μ = 0$ deta hai $a = g sin θ$ . Koi formula break nahi hoti. ✓

Ex 9 — Real-world word problem (Cell I)

Worked example Braking car

Ek car ( $1200 kg$ ) dry asphalt par skid karke ruk jaati hai, $μ_{k} = 0.80$ . Iske wheels locked hain (pure sliding). Iska deceleration kya hai, aur $20 m/s$ se kitni door tak jaata hai?

Forecast: Kya car ki mass stopping distance ko affect karti hai?

FBD: Figure 2 jaisa flat-ground layout — weight $m g$ neeche, normal $N$ upar, aur akela horizontal force hai kinetic friction $f_{k}$ backward point karte hue.

Step 1 — Normal force. Flat road: $N = m g = 1200 \times 9.8 = 11760 N$ . Yeh step kyun? Friction $μ_{k} N$ hai, toh pehle $N$ chahiye; flat ground par yeh full weight ke equal hota hai.

Step 2 — Friction = akela horizontal force. $f_{k} = μ_{k} N = 0.80 \times 11760 = 9408 N$ , backward. Yeh step kyun? Locked wheels matlab pure sliding, toh kinetic friction (not static) car ko rokti hai.

Step 3 — Deceleration. $a = \frac{f _{k}}{m} = \frac{9408}{1200} = 7.84 m/s^{2}$ . Note karo $a = μ_{k} g = 0.80 \times 9.8 = 7.84$ — mass cancel ho jaata hai! Yeh step kyun? $f_{k} = μ_{k} m g$ aur $a = f_{k} / m$ , toh $m$ divide ho jaata hai.

Step 4 — Stopping distance. Kinematics $v^{2} = v_{0}^{2} - 2 a d \Rightarrow 0 = 2 0^{2} - 2 (7.84) d$ , toh $d = \frac{400}{15.68} = 25.5 m$ . Yeh step kyun? Constant deceleration hone par $v^{2} = v_{0}^{2} - 2 a d$ use kar sakte hain.

Verify: Mass $d$ mein kabhi appear nahi karti — ek bhaari truck aur halki car same $μ_{k}$ ke saath same distance mein rukenge. ✓ Units: m. ✓

Ex 10 — Exam-style twist (Cell J)

Worked example "Kaun sa surface pehle break karta hai" wala catch

Ek block ek table par rakha hai; table floor par rakha hai. Table+block ek saath push kiye jaate hain. Block–table $μ_{s} = 0.30$ , table–floor $μ_{s} = 0.50$ . Agar tum table ko horizontally push karo, aur block mass $= 2 kg$ ho, toh system ki maximum acceleration kya hai isse pehle ki block table par slip kare?

Forecast: Kaun sa surface acceleration limit karta hai — upar wala grippier ya slipperier surface?

FBD — Figure 6.

Block ko isolate karo: weight $m g$ neeche, table se normal upar, aur akela horizontal force hai table se red static friction — yeh akela hi block ko accelerate karna chahiye.

Step 1 — Block ko isolate karo. Block par akela horizontal force uske neeche table ki friction hai. Woh friction hi block ki saari acceleration supply karni chahiye. Yeh step kyun? Block ka apna free body diagram draw karo — dekho Free Body Diagrams.

Step 2 — Top surface se max friction. $f_{s, m a x} = μ_{s, top} m g = 0.30 \times 2 \times 9.8 = 5.88 N$ . Yeh step kyun? Yeh sabse bada push-forward hai jo table sirf friction ke zariye block ko de sakta hai.

Step 3 — Max acceleration. $a_{m a x} = \frac{f _{s, m a x}}{m} = \frac{5.88}{2} = 2.94 m/s^{2}$ . Note karo $a_{m a x} = μ_{s, top} g = 0.30 \times 9.8$ . Yeh step kyun? Isse aage, friction block ko table ke saath accelerate nahi rakh sakti → yeh slip kar jaata hai.

Verify: $a_{m a x} = μ_{s} g$ , block mass se independent. ✓ Top ( $0.30$ ) surface limit karta hai, floor ( $0.50$ ) nahi — exam trap yeh sochna hai ki bada $μ$ matter karta hai. Is sawaal ke liye nahi karta. ✓

Recall Self-test: cell ka naam batao

"Tum ek box push karte ho aur yeh nahi hilta; friction nikalo." ::: Cell A — friction tumhari push ke equal hai, $μ_{s} N$ se neeche. "Ek rope crate ko $2 0^{\circ}$ upar pull karti hai; starting tension nikalo." ::: Cell C — $T = μ_{s} m g / (cos θ + μ_{s} sin θ)$ . "Block $4 0^{\circ}$ ramp par neeche slide karta hai; $a$ nikalo." ::: Cell F — $a = g (sin θ - μ_{k} cos θ)$ . "Car skid karti hai; stopping distance nikalo." ::: Cell I — $a = μ_{k} g$ , phir kinematics. "Ramp $θ_{c}$ tak tilt hota hai jahan woh just slip kare." ::: Cell E — $μ_{s} = tan θ_{c}$ .

Connections

Newton's Second Law — har "find $a$ " example $\sum F = ma$ hai.
Static vs Kinetic Friction — "does it move?" branch decide karta hai inme se kaun.
Inclined Plane Dynamics — Ex 5–8 along/perpendicular split use karte hain.
Normal Force — Ex 3–4 is par hinge karte hain ki angled forces $N$ ko kaise change karte hain.
Free Body Diagrams — Ex 10 ke isolation ka tool.
Lubrication & Tribology — kyun $μ_{k}$ surface tables mein alag-alag hota hai.

Worked examples — Coefficients of friction — measurement, material dependence