Visual walkthrough — Equilibrium of rigid bodies — translational + rotational

1.5.18 · D2 · Physics › Rotational Mechanics › Equilibrium of rigid bodies — translational + rotational

Hum assume karte hain ki tum sirf itna jaante ho: ek force ek push ya pull hai (ek arrow), aur objects ka weight unhe neeche khichta hai. Baaki sab — torque, moment arm, couple, pivot-independence — hum page pe hi grow karenge.

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Step 1 — Ek force ek arrow hai, aur yeh DO alag cheezein kar sakti hai

Hum yahan se kyun start karte hain. Sabka pehla instinct yeh hota hai ki "arrows ko add karo, agar woh cancel ho jayein toh kaam khatam." Yeh instinct sirf effect (a) ke baare mein hai — sliding. Yeh poori tarah effect (b) ko ignore karta hai — spinning. Poori wajah ki ek rigid body ko do rules ki zaroorat hoti hai woh yeh hai ki ek arrow secretly do tarah ki motion control karta hai. Hume effect (b) ko visible banana hoga pehle kuch aur karne se pehle.

PICTURE. Neeche, wahi blue arrow do baar draw kiya gaya hai. Left side par hum uski sliding effect track karte hain (door ka centre kahan jaata hai). Right side par hum uski spinning effect track karte hain (woh red hinge ke baare mein kitna ghoomti hai).

Dhyaan do: arrow ki size shove ko govern karti hai. Lekin spin kuch aisi cheez par depend karta hai jo size akele nahi batati — hinge se kitni door woh act karti hai, aur kis slant par. Hume iske liye ek number chahiye. Aage chalo.

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Step 2 — "Turning power" banana: the moment arm

Yeh number kyun, aur sirf "force $\times$ pivot tak distance" kyun nahi. Picture dekho: agar tum arrow ki apni line ke saath slide karo, spinning effect kabhi nahi badlti. Toh spin full distance $r$ par depend nahi kar sakti — jahan se tum grab karte ho wahan se pivot tak — yeh sirf pivot aur line of action (woh infinite line jis par arrow rehta hai) ke beech ki shortest (perpendicular) gap par depend karni chahiye. Woh perpendicular gap moment arm hai, jise $d$ likhte hain.

PICTURE. Green segment $d$ hai. Dekho kaise $d=r\sin \theta$ seedha right triangle se nikalta hai: $\vec{r}$ (pivot → grab point) hypotenuse hai, $\theta$ $\vec{r}$ aur force ke beech ka angle hai, aur $\theta$ ke opposite wali side exactly perpendicular drop $d$ hai.

$\sin \theta$ kyun aur $\cos$ ya raw angle kyun nahi? Hume ek aisa factor chahiye jo 1 ho jab force $\vec{r}$ ke perpendicular ho (best possible spin) aur 0 ho jab force seedha $\vec{r}$ ke along point kare (koi spin nahi — tum pivot ki taraf ya usse door pull kar rahe ho). Woh plain-trig function jo $90^{\circ }$ par $1$ aur $0^{\circ }$ par $0$ ho, woh $\sin$ hai. Yahi poori wajah hai ki woh wahan appear karta hai. Zyada ke liye Torque dekho.

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Step 3 — Do arrows jo force ke roop mein cancel ho jaate hain lekin SPIN karte hain paaglon ki tarah (the couple)

Yeh step crucial kyun hai. Yeh single picture proof hai ki sirf force balance equilibrium nahi hai. Yeh woh counterexample hai jo ek doosra rule zaroor banata hai. Is configuration ka ek naam hai: couple (dekho Couple and Moment of a Couple).

PICTURE. Top arrow right point karta hai, bottom left point karta hai, $d$ se separated. Arrows ki tarah: $+F$ aur $-F$ add hokar $0$ dete hain. Spins ki tarah: top door ki far side ko ek taraf push karta hai, bottom near side ko doosri taraf push karta hai, aur — yahi magic hai — woh do pushes body ko same rotational sense mein rotate karte hain. Unke torques add ho jaate hain.

Toh: ek body ka $\sum \vec{F}=0$ ho sakta hai aur phir bhi woh spin up ho rahi hoti hai. Force balance kaafi nahi hai. Hum majboor hain ek doosri, independent cheez demand karne ke liye: torque balance.

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Step 4 — Yeh do rules aate kahan se hain asliye mein? (Newton, promoted)

Particles par sum kyun karte hain. Ek rigid body laakhon particles ka ek lock-shuda collection hai. Unke beech internal forces equal-opposite pairs mein aate hain (Newton's 3rd law), toh jab hum sab forces add karte hain, internal wale annihilate ho jaate hain — sirf external forces bachti hain.

PICTURE. Body ek dots ka cloud hai; blue internal force pairs arrow-to-arrow cancel ho jaate hain; sirf bahar wale (external) arrows rehte hain.

Sab particles par Newton's law $\vec{F}=m\vec{a}$ add karne se poori body ke statements milte hain:

$\underset{\text{total bahari push}}{\underbrace{\sum {\vec{F}}_{\text{ext}}}}=\underset{\text{total mass}}{\underbrace{M}}\underset{\begin{array}{c}\text{centre of mass}\\ \text{ki accel.}\end{array}}{\underbrace{{\vec{a}}_{cm}}}\underset{\text{total bahari twist}}{\underbrace{\sum {\vec{\tau }}_{\text{ext}}}}=\underset{\begin{array}{c}\text{moment of}\\ \text{inertia}\end{array}}{\underbrace{I}}\underset{\begin{array}{c}\text{angular}\\ \text{acceleration}\end{array}}{\underbrace{\vec{\alpha }}}$

Left equation centre-of-mass motion hai; right uska rotational twin hai, jahan $I$ Moment of Inertia hai. Body ko sliding start nahi karne ke liye hume ${\vec{a}}_{cm}=0$ chahiye, aur spinning start nahi karne ke liye hume $\vec{\alpha }=0$ chahiye:

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Step 5 — Pivot freely choose karo (the beautiful fact)

Yeh kaam kyun karta hai — picture argument. Naye point $B$ ke baare mein torque, purane point $A$ ke baare mein torque hai, plus ek correction jo total force ke proportional hai. Agar total force zero hai, correction gayab ho jaata hai — toh dono answers agree karne chahiye.

PICTURE. Point $A$ aur point $B$, vector $\vec{d}$ se offset. Har particle ka arm same shift $\vec{d}$ se badal jaata hai. Woh common shift $\sum {\vec{F}}_i=0$ se multiply hota hai, toh woh disappear ho jaata hai.

${\vec{\tau }}_B=\sum ({\vec{r}}_i-\vec{d})\times {\vec{F}}_i=\underset{={\vec{\tau }}_A}{\underbrace{\sum {\vec{r}}_i\times {\vec{F}}_i}}-\vec{d}\times \underset{=0}{\underbrace{\sum {\vec{F}}_i}}={\vec{\tau }}_A.$

Isko padhte hain: ${\vec{r}}_i$ = $A$ se particle $i$ tak arm; $\vec{d}$ = $A$ se $B$ tak shift; doosra term mar jaata hai kyunki $\sum {\vec{F}}_i=0$. Conclusion: ${\vec{\tau }}_B={\vec{\tau }}_A$. Pivot choice free hai. ∎

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Step 6 — Sign convention: ise arithmetic mein badalna

Sign ki zaroorat hi kyun hai. Do forces opposite senses mein twist kar sakti hain; signs ke bina woh galat tarike se add ho jaayenge. Signs opposite spins ko cancel karne dete hain — exactly woh jo balance require karta hai.

PICTURE. Ek green curved arrow (CCW, $+$) aur ek red curved arrow (CW, $-$) same pivot ke baare mein. Balance ke liye, total CCW twisting exactly total CW twisting ke equal honi chahiye.

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Step 7 — Degenerate & edge cases (kabhi surprise mat ho)

Hume corner cases cover karne chahiye taaki koi scenario tumhe dhoka na de.

Case A — Pivot se guzarti force ⇒ zero torque. Agar kisi force ki line of action pivot se guzarti hai, toh $d=0$, toh $\tau =F\cdot 0=0$. Yahi wajah hai ki jab hum wahan pivot karte hain toh seesaw ki support force aur ladder ki floor reactions gayab ho jaati hain.

Case B — $\vec{r}$ ke parallel force ($\theta =0$ ya $180^{\circ }$) ⇒ zero torque. Yahan $\sin \theta =0$, toh $\tau =Fr\sin \theta =0$. Pivot ki taraf ya usse seedha door pull karna body ko kabhi spin nahi karata.

Case C — Perpendicular force ($\theta =90^{\circ }$) ⇒ maximum torque. $\sin 90^{\circ }=1$, toh $\tau =Fr$, sabse zyada faayda.

Case D — Balanced forces, unbalanced couple (the trap). $\sum \vec{F}=0$ lekin body phir bhi spin karti hai (Step 3). Equilibrium fail hoti hai. Yeh woh case hai jise log bhool jaate hain.

PICTURE. Chaar mini-panels, ek har case ke liye, moment arm $d$ aur resulting $\tau$ label kiye gaye.

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Ek picture mein sab kuch

Upar sab kuch, compressed: couple ke do arrows (force cancel, twist survive) do boxed rules aur free-pivot fact ke saath side by side.

Recall Feynman retelling — poora walkthrough simple shabdon mein

Ek floating door socho jis par ek red hinge-dot hai. Step 1: ek push do kaam karta hai — poori door ko slide karta hai AUR use dot ke baare mein spin karta hai. Step 2: spin-power (torque) push ki size times dot se push ki line tak ki perpendicular distance ke barabar hai — kyunki line ke saath slide karne se kuch nahi badlta. Step 3: ab dono edges ko equally opposite directions mein push karo: door slide nahi karti (pushes cancel ho jaate hain) lekin ghoomti hai — proof ki "pushes cancel" poori kahani nahi hai. Step 4: Newton's $F=ma$ ko poori body tak promote karne se do twin laws milte hain — ek sliding ke liye, ek spinning ke liye — toh "mat hilo" ka matlab hai ki total push AUR total twist dono zero hone chahiye. Step 5: aur yahan ek gift hai — ek baar pushes already cancel ho jaayein, total twist same rehta hai chahe tum use kisi bhi dot ke baare mein measure karo, toh apna dot wahin rakh do jis force ki tumhe parwah nahi taaki woh gayab ho jaaye. Step 6: anticlockwise twists ko $+$ aur clockwise ko $-$ kaho taaki opposites cancel ho sakein. Yahi poori theory hai: forces sliding ko rokti hain, torques spinning ko tame karte hain.