Equilibrium of rigid bodies — translational + rotational: Worked examples

1.5.18 · D3 · Physics › Rotational Mechanics › Equilibrium of rigid bodies — translational + rotational

Intuition Yeh page kis liye hai

Parent note ne aapko do rules diye: net force zero aur net torque zero. Lekin rules tabhi pakke hote hain jab aapne unhe har trap mein survive karte dekha ho jo yeh topic set kar sakta hai. Toh neeche hum pehle ek scenario matrix banate hain — ek checklist jisme har tarah ki situation hai jo yeh topic aap par throw kar sakta hai — aur phir har cell ko poori tarah solve karte hain taaki aap kabhi aisi case se na mile jo aapne pehle beat nahi ki ho.

Ek symbol, ek meaning. Is poore page mein hum torque ko $τ = F d$ likhte hain, jahan $d$ hamesha pivot se force ki line of action tak ki perpendicular distance (moment arm) hoti hai. $r$ ko hum pivot se us point tak ki straight-line distance ke liye reserve karte hain jahan force apply hoti hai, aur jab force angled ho toh hum $d$ ko $r$ se $d = r sin θ$ ke zariye banate hain. Har symbol wahan re-define hota hai jahan woh pehli baar aata hai.

Scenario matrix

Har equilibrium problem inhi cells mein se ek (ya mix) hoti hai. Hum sabhi ko cover karenge.

Cell	Kya cheez ise alag banati hai	Example
C1 Torque ka sign	ek force clockwise ghoomta hai, ek counter-clockwise	Ex 1
C2 Force pivot se hokar	uska moment arm zero hai → woh vanish ho jaata hai	Ex 1, Ex 4
C3 Angled force	perpendicular component ( $F sin θ$ ) mein split karna padega	Ex 2
C4 Zero / degenerate input	force pivot se guzarti hai, ya angle = 0/90° hai	Ex 3
C5 Limiting behaviour	kisi angle ko → 0 ya → 90° jaane do, dekho answer blow up hota hai ya vanish	Ex 5
C6 Couple (zero net force, non-zero torque)	force balance hold karta hai, body phir bhi spin karti hai	Ex 6
C7 Friction unknown (inequality)	minimum $μ$ ke liye solve karo	Ex 7
C8 Real-world word problem	words → forces → equations mein translate karo	Ex 8
C9 Exam twist (sirf balance nahi, pivot reaction bhi dhundho)	do support reactions, dono solve karo	Ex 9

Prerequisite refreshers agar koi word shaky lage: Torque, Centre of Mass, Couple and Moment of a Couple, Static Friction, Centre of Gravity vs Centre of Mass, Newton's Laws of Motion.

Example 1 — Unequal seesaw (cells C1, C2)

Forecast: Aage padhne se pehle guess karo — right force $30 N$ se badi hogi ya chhoti? (Woh zyada door hai…)

Figure — Equilibrium of rigid bodies — translational + rotational

Figure dekho: do red arrows applied forces hain; black double-arrows moment arms $0.4 m$ aur $0.6 m$ hain jo rod ke saath $O$ se measure ki gayi hain. Kyunki rod horizontal hai aur forces vertical hain, yeh along-the-rod distances hi perpendicular distances $d$ hain — koi angle correction zaruri nahi.

Step 1 — Pivot $O$ par choose karo. Yeh step kyun? $O$ par support rod ko upar push karta hai, lekin woh push $O$ par hi act karti hai, toh $O$ tak uski perpendicular distance $d = 0$ hai → uska torque zero hai. $O$ choose karne se unknown support force vanish ho jaati hai (cell C2).

Step 2 — Har torque ko sign ke saath likho. Yeh step kyun? Torque magnitude $τ = F d$ hai jahan $d$ pivot se force ki line tak ki perpendicular distance hai. Left force rod ko counter-clockwise ghumaati hai (positive, +), right force clockwise (negative, −) — yeh cell C1 hai.

$\sum τ_{O} = 0 : (30) (0.4) - (F) (0.6) = 0$

Step 3 — Solve karo. $12 = 0.6 F ⟹ F = 20 N .$

Verify: Left twist $= 30 \times 0.4 = 12 N\cdotm$ ; right twist $= 20 \times 0.6 = 12 N\cdotm$ . Equal aur opposite → net torque zero. ✓ Door wali force chhoti hai, jo intuition se match karta hai: zyada distance, kam force chahiye.

Example 2 — Angled force (cell C3)

Forecast: Us $40 N$ ka sirf ek hissa hi rod ko actually twist karega. Kaun sa hissa — aur kya woh sine piece hai ya cosine?

Figure dekho: red arrow poora pull $T$ hai. Rod ke saath black double-arrow $r = 0.5 m$ hai — $O$ se us point tak ki distance jahan rope pakad rahi hai. Chhota black upward arrow perpendicular slice $T sin 3 0^{\circ}$ hai; dotted horizontal piece $T cos 3 0^{\circ}$ rod ke saath saath point karta hai aur kuch nahi karta. Chhota arc $3 0^{\circ}$ angle mark karta hai rope aur rod ke beech.

Step 1 — $T$ ko do pieces mein split karo. Yeh step kyun? Woh force jo rod ke saath point karti hai, rod ko twist nahi kar sakti — door ki edge ke along seedha kheenchna kabhi door nahi kholega. Sirf rod ke perpendicular wala piece use ghumaata hai. Woh perpendicular piece $T sin θ$ hai; along-the-rod piece $T cos θ$ hai aur turning ke liye waste ho jaata hai (cell C3).

Step 2 — Torque likho. Yeh step kyun? Torque = (perpendicular force) × (rod ke saath distance) $= (T sin θ) r$ . Equivalently, moment arm $d = r sin θ$ hai, toh $τ = T d = T r sin θ$ — same cheez.

$τ = T r sin θ = (40) (0.5) sin 3 0^{\circ} .$

Step 3 — Compute karo. $sin 3 0^{\circ} = 0.5$ hone se: $τ = 40 \times 0.5 \times 0.5 = 10 N\cdotm .$

Verify: Units $= N \times m = N\cdotm$ ✓. Sanity check: agar rope seedha upar pull kare ( $θ = 9 0^{\circ}$ , $sin = 1$ ) toh poora $40 \times 0.5 = 20 N\cdotm$ milega — double, jo sahi hai kyunki tab saari force perpendicular hogi.

Example 3 — Degenerate case: force pivot se hokar (cell C4)

Forecast: Kya koi twist hoga?

Step 1 — Perpendicular distance dhundho. Yeh step kyun? Rope ki line of action seedhi $O$ se hokar guzarti hai, toh $O$ se us line tak ki perpendicular distance $d = 0$ hai (cell C4 — degenerate input).

Step 2 — Formula apply karo. $τ = F d = 40 \times 0 = 0 N\cdotm .$

Verify: Equivalently, $r$ ( $O$ se pull point tak) aur $F$ ke beech angle $0^{\circ}$ hai, aur $sin 0^{\circ} = 0$ , toh $τ = F r sin 0^{\circ} = 0$ . ✓ Lesson: koi bhi force jiska line pivot se guzarti hai woh zero torque contribute karta hai — isliye hi smart pivot choices kaam karti hain.

Example 4 — Balanced plank, weight at centre of mass (cells C1, C2)

Forecast: Plank ki apni weight — kya woh person ki help karti hai, ya unse ladd ti hai? Solve karne se pehle guess karo.

Figure dekho: black triangle support hai, left end se $1.5 m$ door. Centre par black arrow plank ka apna weight $W$ hai; left end par red arrow person $P$ hai. Do double-arrows support se moment arms hain: person support ke $1.5 m$ left mein hai, weight $0.5 m$ right mein — woh opposite sides par hain, isliye balance ho sakte hain.

Step 1 — Plank ki weight locate karo. Yeh step kyun? Uniform body ke liye, poora weight centre of mass par act karta hai — geometric centre par, jo yahan left end se $2 m$ door hai (dekho Centre of Gravity vs Centre of Mass). $W$ ko kahi aur rakhne se moment arm galat hoga (parent ka Mistake D).

Step 2 — Pivot support par choose karo. Yeh step kyun? Support ki upward normal force support par hi act karti hai → moment arm $d = 0$ , toh woh disappear ho jaati hai (cell C2). Sirf $W$ aur $P$ bachte hain.

Step 3 — Support se perpendicular distances. Yeh step kyun? Support left end se $1.5 m$ door hai, aur forces vertical hain jabki plank horizontal hai, toh along-the-plank distances hi moment arms $d$ hain.

Plank weight $W$ : centre left end se $2 m$ par; support $1.5 m$ par → $d = 0.5 m$ support ke right mein → clockwise ghumaata hai.
Person $P$ : left end par ( $0 m$ left se) → $d = 1.5 m$ support ke left mein → counter-clockwise ghumaata hai.

Step 4 — Torques balance karo. Yeh step kyun? Counter-clockwise (person) clockwise (plank weight) ke equal hona chahiye — cell C1. $P (1.5) = W (0.5) ⟹ P (1.5) = 60 \times 0.5 = 30.$

Step 5 — Solve karo. $P = \frac{30}{1.5} = 20 N .$

Verify: Person torque $= 20 \times 1.5 = 30 N\cdotm$ CCW; plank torque $= 60 \times 0.5 = 30 N\cdotm$ CW → cancel ho jaate hain. ✓ Plank ki apni weight person ke against fight karti hai (opposite sense), toh person ka weight exactly wahi hai jo ise balance karne ke liye chahiye.

Example 5 — Ladder ka limiting behaviour (cell C5)

Forecast: Near-vertical ladder — ushe bahut zyada friction chahiye ya almost zero?

Figure dekho: red curve $μ_{m i n} = \frac{1}{2} cot θ$ ko ladder angle $θ$ ke against plot karta hai. Note karo ki woh right side par $0$ tak dive karta hai (steep ladder) aur left side par infinity ki taraf shoot karta hai (flat ladder). Do black dots $3 0^{\circ}$ aur $6 0^{\circ}$ ke liye hum jo answers compute karte hain unhe mark karte hain.

Step 1 — $cot θ = cos θ / sin θ$ yaad karo. Yeh step kyun? $cot$ measure karta hai ladder kitna "horizontal" lean karta hai; jab $θ \to 9 0^{\circ}$ toh yeh $0$ tak shrink hota hai, jab $θ \to 0^{\circ}$ toh $\infty$ tak explode karta hai (cell C5).

Step 2 — Do limits lo. $θ \to 9 0^{\circ} : cot θ \to 0 ⟹ μ_{m i n} \to 0.$ $θ \to 0^{\circ} : cot θ \to \infty ⟹ μ_{m i n} \to \infty.$ Yeh step kyun? Yeh batata hai ki vertical ladder ko barely friction chahiye; floor-flat ladder ko impossible friction chahiye — use slip karna hi padega. Everyday sense se match karta hai.

Step 3 — Numbers plug in karo. $θ = 6 0^{\circ} : μ_{m i n} = \frac{1}{2} cot 6 0^{\circ} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{2 3} \approx 0.289.$ $θ = 3 0^{\circ} : μ_{m i n} = \frac{1}{2} cot 3 0^{\circ} = \frac{1}{2} \cdot 3 = \frac{3}{2} \approx 0.866.$

Verify: Flatter ladder ( $3 0^{\circ}$ ) ko $\approx 0.87$ friction chahiye jabki steeper wale ko $\approx 0.29$ — teen guna zyada. ✓ Upar ke limits se consistent hai.

Example 6 — Ek pure couple (cell C6)

Forecast: Do pushes equal aur opposite hain — toh zarur wheel balanced hoga… sahi hai?

Figure dekho: red circle wheel hai. Do black arrows top aur bottom par equal-and-opposite pushes hain. Unke beech dotted line separation $d = 0.35 m$ hai. Dono arrows trace karo: woh dono wheel ko same taraf spin karne ki koshish karte hain, toh unke turning effects cancel hone ki jagah add hote hain.

Step 1 — Force balance check karo. Yeh step kyun? $\sum F = + 8 - 8 = 0$ . Translational equilibrium hold karta hai.

Step 2 — Couple ka torque compute karo. Yeh step kyun? Couple do equal, opposite, parallel forces hote hain jo distance $d$ se separated hote hain (dekho Couple and Moment of a Couple). Iska torque kisi bhi point ke baare mein $F d$ hota hai — dono forces wheel ko same tarah ghumaate hain, toh add hote hain cancel hone ki jagah (cell C6). Yahan $d =$ diameter $= 0.35 m$ . $τ = F d = 8 \times 0.35 = 2.8 N\cdotm .$

Step 3 — Conclusion. Net force $= 0$ lekin net torque $= 2.8 N\cdotm \neq = 0$ → NOT in equilibrium; wheel angularly accelerate karta hai.

Verify: Yeh parent ka "steering wheel" intuition numerically prove hua — second condition genuinely independent hai. ✓

Example 7 — Minimum friction as an inequality (cell C7)

Forecast: $4 5^{\circ}$ lean se badi friction chahiye ya chhoti?

Charon forces (pehle har symbol naam do).

$W = 200 N$ : beam ka weight, neeche, uske centre par.
$N_{f}$ : floor normal force, upar, foot par.
$N_{w}$ : wall normal force, horizontal (wall frictionless hai toh woh sirf seedha bahar push kar sakti hai), top par.
$f$ : floor friction force, horizontal, foot par, wall ki taraf pointing taaki foot bahar slide na kare. Yeh Static Friction se static-friction force hai; yeh koi bhi value le sakti hai ceiling $μ N_{f}$ tak.

Step 1 — Vertical balance. Yeh step kyun? Sirf floor normal $N_{f}$ (upar) aur weight $W$ (neeche) vertical hain: $N_{f} = W = 200 N .$

Step 2 — Horizontal balance. Yeh step kyun? Sirf friction $f$ (wall ki taraf) aur wall push $N_{w}$ (wall se door) horizontal hain, toh unhe cancel karna chahiye: $f = N_{w} .$

Step 3 — Foot par torque. Yeh step kyun? Foot par $N_{f}$ aur $f$ hain — dono wahan act karte hain → moment arm $d = 0$ → disappear ho jaate hain. Sirf $W$ aur $N_{w}$ bachte hain. Unke moment arms: weight beam ke centre par act karta hai, jiska foot se horizontal distance $\frac{L}{2} cos θ$ hai; wall push $N_{w}$ horizontal hai aur top par act karti hai, jiska foot se height $L sin θ$ hai. (In lengths mein $L$ factor hai, beam ki length.) $N_{w} (L sin θ) = W \frac{L}{2} cos θ .$

Step 4 — Length cancel hoti hai; $N_{w}$ ke liye solve karo. Yeh step kyun? Har term mein $L$ ka ek factor hai, toh divide out ho jaata hai — isliye answer ko kabhi numerical length ki zaroorat nahi hoti: $N_{w} = \frac{W}{2} \frac{c o s θ}{s i n θ} = \frac{W}{2} cot θ .$ $cot 5 0^{\circ} \approx 0.8391$ ke saath: $N_{w} = \frac{200}{2} \times 0.8391 = 100 \times 0.8391 \approx 83.9 N .$

Step 5 — $μ_{m i n}$ ke liye static-friction inequality apply karo. Yeh step kyun? Static friction ceiling $μ N_{f}$ tak koi bhi force supply kar sakti hai, yaani isko $f \leq μ N_{f}$ follow karna chahiye (dekho Static Friction). No-slip ke liye zaruri hai ki $f \leq μ N_{f} ⟹ μ \geq \frac{f}{N _{f}} .$ Step 2 se, $f = N_{w} = \frac{W}{2} cot θ$ , aur Step 1 se, $N_{f} = W$ , toh $μ \geq \frac{\frac{W}{2} c o t θ}{W} = \frac{1}{2} cot θ .$ Numerically, $μ_{m i n} = \frac{1}{2} cot 5 0^{\circ} = 0.5 \times 0.8391 \approx 0.42$ .

Verify: $μ_{m i n} = \frac{1}{2} cot 5 0^{\circ} = 0.4196 \approx 0.42$ ✓, aur $N_{w} \approx 83.9 N$ ✓. $4 5^{\circ}$ par yeh $\frac{1}{2} cot 4 5^{\circ} = 0.5$ hota; steeper $5 0^{\circ}$ ko kam friction chahiye ( $0.42$ ). ✓

Example 8 — Real-world: hanging sign (cell C8)

Forecast: Kya $T$ total weight $120 + 180 = 300 N$ se zyada hoga ya kam?

Figure dekho: hinge (black dot) wall par hai; right end par red arrow cable tension $T$ hai, seedha upar kyunki cable vertical hai. Do black downward arrows hain: beam ka weight (centre par, $1.5 m$ ) aur sign ka weight (right end par, $3 m$ ). Kyunki cable vertical hai aur beam horizontal, har moment arm simply hinge se along-the-beam distance hai.

Step 1 — Words ko forces mein translate karo. Yeh step kyun? Beam weight $120 N$ centre par act karta hai ( $d = 1.5 m$ hinge se); sign weight $180 N$ right end par ( $d = 3 m$ ); cable tension $T$ upar right end par ( $d = 3 m$ ); hinge reaction left par ( $d = 0$ ).

Step 2 — Pivot hinge par rakho. Yeh step kyun? Hinge force hinge par act karti hai → $d = 0$ → chali gayi. Sirf $120$ , $180$ , aur $T$ bachte hain.

Step 3 — Torque balance (CCW positive). $T (3) - 120 (1.5) - 180 (3) = 0.$ Yeh step kyun? $T$ lift karta hai (CCW, +); dono weights neeche pull karte hain (CW, −).

Step 4 — Solve karo. $3 T = 180 + 540 = 720 ⟹ T = 240 N .$

Verify: $T = 240 N$ total weight $300 N$ se kam hai — kyunki hinge load share karta hai. Hinge baaki $300 - 240 = 60 N$ upward carry karta hai. Torque check: $240 \times 3 = 720 = 120 \times 1.5 + 180 \times 3 = 180 + 540$ . ✓

Example 9 — Exam twist: DONO support reactions dhundho (cell C9)

Forecast: Kaun sa support zyada load carry karta hai — truck ke nazar aane wala, ya door wala?

Figure dekho: do black triangles supports $A$ (left) aur $B$ (right) hain, $10 m$ apart. Black up-arrows unknown reactions $R_{A}$ aur $R_{B}$ hain. Centre par ( $5 m$ ) black down-arrow beam ka apna weight hai; $A$ se $3 m$ par red down-arrow truck hai. Saari forces vertical hain aur beam horizontal, toh har moment arm sirf jis bhi pivot se choose karo usse along-the-beam distance hai.

Step 1 — $R_{B}$ isolate karne ke liye $A$ par torque lo. Yeh step kyun? $R_{A}$ $A$ par act karti hai → moment arm $d = 0$ → drop out ho jaati hai (cell C2), sirf ek unknown $R_{B}$ bachta hai. Beam weight centre par act karta hai ( $d = 5 m$ ), truck $d = 3 m$ par, aur $R_{B}$ $d = 10 m$ par. Counter-clockwise positive lete hain: $R_{B} (10) - 500 (5) - 2000 (3) = 0.$ Signs kyun? $R_{B}$ right end par upar push karta hai → CCW (+); dono weights neeche pull karte hain → CW (−).

Step 2 — $R_{B}$ ke liye solve karo. $10 R_{B} = 2500 + 6000 = 8500 ⟹ R_{B} = 850 N .$

Step 3 — $R_{A}$ ke liye vertical force balance. Yeh step kyun? Translational equilibrium $\sum F_{y} = 0$ : total upward = total downward. $R_{A} + R_{B} = 500 + 2000 = 2500 ⟹ R_{A} = 2500 - 850 = 1650 N .$

Step 4 — $B$ par torque leke cross-check karo. Yeh step kyun? Kyunki $\sum F = 0$ already hold karta hai, torque kisi bhi pivot ke baare mein same hota hai (parent ka "free pivot" theorem). $B$ lena $R_{A}$ isolate karta hai aur same number dena chahiye: $R_{A} (10) = 500 (5) + 2000 (7) = 2500 + 14000 = 16500 ⟹ R_{A} = 1650 N .$

Verify: Dono routes $R_{A} = 1650 N$ dete hain ✓. Aur $R_{A} = 1650 > R_{B} = 850$ : truck ke nazar wala support zyada load carry karta hai, jaise expect tha. Total $1650 + 850 = 2500 N$ saari weight ko balance karta hai ✓.

Recall Kya har cell cover hui?

C1 signs — Ex 1,4 · C2 force through pivot — Ex 1,4,7,8,9 · C3 angled force — Ex 2 · C4 degenerate/zero — Ex 3 · C5 limiting — Ex 5 · C6 couple — Ex 6 · C7 friction inequality — Ex 7 · C8 word problem — Ex 8 · C9 both reactions — Ex 9. Har cell hit.