Visual walkthrough — Free body diagrams — systematic drawing technique

1.2.4 · D2 · Physics › Newton's Laws & Dynamics › Free body diagrams — systematic drawing technique

Hum sirf chaar ideas use karte hain, har ek ko use karne se pehle define karte hain: ek force (ek push ya pull, arrow ke roop mein draw ki gayi), mass $m$ (kisi cheez ko speed up karna kitna mushkil hai), weight $mg$ (gravity ka pull), aur acceleration $a$ (speed kitni tezi se change ho rahi hai). Agar inme se koi shaky lage, koi baat nahi — hum inhe neeche se dobara earn karte hain.

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Step 1 — Machine se milte hain

KYA. Do blocks. Block $A$ (mass $m_1$) ek flat, frictionless table par baitha hai. Ek string $A$ se horizontally edge tak jaati hai, pulley (ek chhota wheel) ke upar se mud jaati hai, aur block $B$ (mass $m_2$) tak neeche aati hai jo hawa mein latk raha hai.

PEHLE poora scene kyun draw karte hain? Kisi cheez ko dot mein shrink karne se pehle, hume dekhna hoga kaun kisse touch karta hai. Forces sirf wahan appear hoti hain jahan objects touch karte hain ya jahan gravity pahunchti hai. Ye picture har contact ka map hai.

PICTURE. Red string do blocks ke beech messenger hai. Block $B$ (accent object) latka hua hai, girne aur $A$ ko sideways kheenchne ke liye taiyar.

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Step 2 — Hanging block $B$ ka Free body diagram

KYA. $B$ ko ek dot mein shrink karo. Do forces use touch karti hain: gravity neeche $m_{2}g$ strength ke saath khiinch rahi hai, aur string upar $T$ strength ke saath khiinch rahi hai.

SIRF do hi kyun? $B$ ki boundary scan karo. Isse sirf string touch kar rahi hai (→ tension $T$, upar). Koi aur cheez isse touch nahi karti, isliye koi aur push nahi kar sakti — sivaaye gravity ke, jo bina touch kiye pahunchti hai (→ $m_{2}g$, neeche). Yahi complete list hai.

PICTURE. Up-arrow $T$ (red string) down-arrow $m_{2}g$ se ladta hai. Kyunki $B$ actually girta hai, gravity jeetti hai — down arrow lamba hai.

Ab Newton's second law, Newton's Second Law, vertical direction mein apply karo. Hum neeche ko positive choose karte hain kyunki $B$ actually usi direction mein move karta hai:

$\underset{\text{down pull, wins}}{\underbrace{m_{2}g}}-\underset{\text{up pull, loses}}{\underbrace{T}}=\underset{\text{B’s inertia}}{\underbrace{m_2}}\underset{\text{shared accel}}{\underbrace{a}}$

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Step 3 — Table block $A$ ka Free body diagram

KYA. $A$ ko ek dot mein shrink karo. Teen forces isse touch karti hain: gravity $m_{1}g$ neeche, table ki normal force $N$ upar, aur string ka tension $T$ pulley ki taraf sideways khiinch raha hai.

SIDEWAYS kyun, neeche kyun nahi? $A$ ki side par string horizontally chalti hai, isliye woh sirf horizontally khiinch sakti hai. Table sirf apne aap ke perpendicular push kar sakta hai — seedha upar — kyunki ek frictionless surface ke paas grip karne ke liye kuch nahi hota (dekho Normal Force and Friction).

PICTURE. Vertically, $N$ (upar) exactly $m_{1}g$ (neeche) ko balance karta hai: $A$ kabhi table nahi chhodta. Horizontally, sirf $T$ (red) act karta hai — isliye $A$ ki saari motion horizontal hai.

Do axes mein split karo (isliye parent note axes par insist karta hai — vectors plain arithmetic ban jaate hain):

$y:\underset{\text{table push up}}{\underbrace{N}}-\underset{\text{weight down}}{\underbrace{m_{1}g}}=0\Rightarrow N=m_{1}g$

$=0$ kyun? $A$ table par flat rehta hai — koi vertical acceleration nahi — isliye vertical forces exactly cancel hone chahiye.

$x:\underset{\text{string pull}}{\underbrace{T}}=\underset{\text{A’s inertia}}{\underbrace{m_1}}\underset{\text{shared accel}}{\underbrace{a}}$

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Step 4 — Tension dono ends par same kyun hoti hai

KYA. Humne Step 2 aur Step 3 mein same letter $T$ likha. Yeh ek claim hai, accident nahi. Ek ideal string massless aur inextensible (stretch nahi ho sakti) hoti hai.

SAME $T$ kyun? String ke kisi bhi chhote piece par Newton's second law hai $\sum F=(\text{mass of piece})\times a$. Ek massless string ke liye right side $0$ hai, isliye ek piece mein enter hone wala pull barabar hai piece se baahir jaane wale pull ke — tension rope ke saath chalte waqt nahi badal sakti. Pulley bas us unchanged pull ko corner ke upar redirect karti hai. (Dekho Tension in Strings and Pulleys.)

PICTURE. Red string ko $A$ se follow karo, wheel ke upar, $B$ tak neeche: tension arrows har jagah same length ke hain. Pulley direction mod deti hai, strength nahi.

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Step 5 — Acceleration dono blocks ke liye same kyun hai

KYA. Humne dono equations mein same letter $a$ bhi likha. Yeh "ideal string" ka inextensible wala part hai.

SAME $a$ kyun? String stretch ya slack nahi ho sakti, isliye uski total length fixed hai. Agar $B$ $1$ cm girta hai, toh $A$ ki side par string exactly $1$ cm chhhoti ho jaati hai, $A$ ko exactly $1$ cm pulley ki taraf kheenchti hai. Same distance same time mein $\Rightarrow$ same speed $\Rightarrow$ same acceleration.

PICTURE. Red length jo $A$ ki side se jaati hai exactly equal hai red length ke jo $B$ ki side par pahunchti hai. Distances equal marked $\Rightarrow$ accelerations equal marked.

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Step 6 — Equations add karo, tension khatam karo

KYA. Dono equations line up karo aur add karo. $\begin{array}{rl}\text{(B)}& m_{2}g-T=m_{2}a\\ \text{(A)}& T=m_{1}a\end{array}$

ADD kyun karte hain? $T$ ek internal force hai — yeh system ke andar rehti hai, ek block ko doosre ke through kheenchti hai. Equation (B) mein yeh $-T$ ke roop mein appear hoti hai; equation (A) mein $+T$ ke roop mein. Add karne se woh cancel ho jaati hain: $-T+T=0$. Internal forces hamesha is tarah cancel hoti hain, sirf external driver bachta hai.

PICTURE. Do arrows stack karo: $+T$ aur $-T$ annihilate ho jaate hain, aur jo bachta hai woh driving weight $m_{2}g$ hai ek side par aur combined inertia $m_1+m_2$ doosri side par.

Term by term add karte hain:

$\underset{\text{external driver}}{\underbrace{m_{2}g}}=\underset{\text{total inertia moved}}{\underbrace{(m_1+m_2)}}a$

$(m_1+m_2)$ se dono sides divide karke $a$ solve karo:

$\boxed{a=\frac{m_2}{m_1+m_2}g}$

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Step 7 — Edge aur degenerate cases (kabhi kisi limit se surprise mat hona)

KYA & KYUN. Ek formula jis par tum trust karte ho woh hai jise tumne extreme conditions par stress-test kiya ho. Hum har input ko $0$ ya $\infty$ tak push karte hain aur check karte hain ki answer sane rehta hai.

PICTURE. Teen sliders, teen sanity checks, har ek ke saath block cartoon jo ise obvious banata hai.

Case	Formula deta hai	Kya ye sense banata hai?
$m_2\to 0$ (kuch bhi hang nahi)	$a\to \frac{0}{m_1}g=0$	✓ Koi driver nahi, koi motion nahi.
$m_1\to 0$ (table block weightless)	$a\to \frac{m_2}{m_2}g=g$	✓ $B$ freely girta hai; string massless $A$ ko free mein drag karti hai.
$m_1=m_2=m$	$a=\frac{m}{2m}g=\frac{g}{2}$	✓ Equal partners gravity ka pull barabar share karte hain.
$m_2\to \infty$ (bahut bada hanging block)	$a\to g$	✓ $A$ ki inertia negligible ho jaati hai; near free-fall.

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Ek-picture summary

Poori derivation ek canvas par: do FBDs (left), woh do equations jo woh produce karte hain (middle), aur tension cancel hone ke baad single boxed answer (right). Red string follow karo — yeh har panel ko saath baandhti hai.

Recall Feynman retelling — poori kahani simple words mein

Socho ek block slippery table par hai jisme se ek rope edge ke upar se, ek chhote wheel ke upar se, neeche ek doosre block tak jaati hai jo hawa mein latk raha hai. Latka hua block girna chahta hai, lekin woh apne aalsi dost se table par bandha hai, isliye jab woh girta hai toh table block ko sideways drag karta hai. Pehle mein hanging block ko akele dekhta hoon: gravity use bahut zyada neeche kheenchti hai, rope use thoda upar kheenchti hai, aur woh haarta hai, isliye girta hai — isse mujhe ek equation milti hai. Phir mein table block ko akele dekhta hoon: gravity neeche exactly cancel hoti hai table ke upar push karne se, aur use move karne ke liye sirf rope sideways kheenchti hai — isse mujhe doosri equation milti hai. Clever part yeh hai: rope stretch nahi ho sakti aur kuch nahi weighs, isliye dono blocks same rope-pull feel karte hain aur same rate par speed up karte hain. Iska matlab hai mere do equations mein same do unknowns share hain. Jab mein equations add karta hoon, rope-pull khud ko cancel kar leti hai (yeh andar wali force hai — ek block ko kheenchti hai doosre ko kheench ke), aur mujhe ek beautifully simple kahani milti hai: hanging block par gravity driving karti hai, aur dono blocks' masses load share karte hain. Isliye acceleration hanging weight divided by total weight-to-move hai, times gravity — hamesha real free-fall se thoda kam, kyunki aalsa table block cheezein hold back kar raha hai.

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Active Recall