Atwood machine — derivation: Visual walkthrough — Hinglish deep dive

1.2.10 · D2 · Physics › Newton's Laws & Dynamics › Atwood machine — derivation

Step 1 — Machine draw karo aur har part ko naam do

KYA. Do chunke ek hi rope se latke hue hain. Rope upar jaati hai, ek wheel (pulley) ke upar se mod kaati hai, aur neeche aa jaati hai. Left waale chunke ko hum naam dete hain $m_{1}$ aur right waale ko $m_{2}$ . Letter $m$ bas itna matlab hai "is chunke mein kitna stuff hai" — iska mass, kilograms mein measure hota hai.

KYO. Kisi bhi physics se pehle, humein point karke har object aur har force ko naam se identify karna hoga. Agar hum yeh skip kar dein, toh baad ki har equation sirf letters hogi jaise hawa mein tairti hoon.

PICTURE. Neeche diye figure mein, do chunke blue (left, $m_{1}$ ) aur orange (right, $m_{2}$ ) hain. Rope woh gray line hai jo gray wheel ke upar se mod kaati hai. Dhyan do: yeh ek continuous rope hai — yahi ek fact hai jo poore problem ko ek saath bandhta hai.

Step 2 — "Tension" word aur kyun same arrow dono sides par dikhta hai

KYA. Rope sirf pull kar sakti hai, push nahi. Jahan bhi rope attached hai, woh us object ko rope ke saath, pulley ki taraf kheenchti hai. Is pull ki strength ko hum tension kehte hain aur ise naam dete hain $T$ (newtons mein, N — force ki unit).

YEH IDEA KYUN, "rope har weight ko thaamti hai" nahi. Beginner log guess karte hain ki left rope $m_{1} g$ se kheenchti hai aur right wali $m_{2} g$ se. Lekin yeh ek rope hai, aur ek massless rope mein tension poore mein same honi chahiye (prove hota hai Tension in Strings mein) — warna zero mass wale rope ke chhote tukde par net force hogi aur usse infinite acceleration chahiye hogi. Toh exactly ek number $T$ hai, left chunke ko upar kheenchta hai aur right chunke ko bhi upar kheenchta hai.

PICTURE. Do equal-length green arrows $T$ label kiye, dono upar ki taraf pointing (pulley ki taraf) — same length kyunki same number hai.

Step 3 — Gravity add karo, aur motion ki direction decide karo

KYA. Har chunk seedha neeche gravity se kheencha jaata hai ek force ke saath jise iska weight, $m g$ kehte hain. Yahan $g \approx 10 m/s^{2}$ gravity ki strength hai (Earth per kilogram kitna kheenchti hai). Toh left chunk par $m_{1} g$ neeche feel hota hai, right par $m_{2} g$ neeche.

Ab hum guess karte hain system kis taraf move karega. Maano $m_{1} > m_{2}$ (left wala bhaari hai). Toh bhaari left side neecha jaata hai aur halka right side upar uthta hai. Hum declare karte hain:

$m_{1}$ ke liye: neeche positive direction hai,
$m_{2}$ ke liye: upar positive direction hai.

KYO. Newton's law ko signs chahiye, aur signs ke liye ek chosen positive direction chahiye. "Positive = jis taraf actually move kare" choose karna har lump ke liye acceleration ko ek positive number rakhta hai. Agar humne bhaari side galat guess ki, toh final $a$ negative aayega aur politely bata dega — kuch toot nahi jaata.

PICTURE. Blue down-arrow $m_{1} g$ (lamba, bhaari side), orange down-arrow $m_{2} g$ (chhota), dono par green up-arrows $T$ . Ek curved motion arrow left-neeche / right-upar dikhata hai. Yeh dono free body diagrams hain — dekho Free Body Diagrams.

Step 4 — Bhaari chunk $m_{1}$ par Newton's law

KYA. Newton's Second Law kehta hai: (positive direction mein net force) $=$ (mass) $\times$ (acceleration). $m_{1}$ ke liye, positive neeche hai. Neeche ki forces: weight $m_{1} g$ (help karta hai, $+$ ). Upar ki forces: tension $T$ (oppose karta hai, $-$ ).

$weight of m_{1} pulls DOWN (+) m_{1} g - rope pull pulls UP (-) T = its mass m_{1} its accel. (down, +) a (1)$

KYO. Hum ek chunk ko isolate karte hain toh sirf do forces dikhti hain. Acceleration ko $a$ kehte hain — wahi letter jo hum $m_{2}$ ke liye bhi use karenge, kyunki inextensible rope equal acceleration magnitudes force karti hai (Step 6 isse visually prove karta hai).

PICTURE. Left chunk akela. Neeche $m_{1} g$ (lamba), upar $T$ (chhota), toh bacha hua downward stub net force $m_{1} g - T$ hai jo $m_{1} a$ drive karta hai.

Step 5 — Halke chunk $m_{2}$ par Newton's law

KYA. Ab $m_{2}$ , jiska positive direction upar hai. Upar ki forces: tension $T$ ( $+$ ). Neeche ki forces: weight $m_{2} g$ ( $-$ ).

$rope pull pulls UP (+) T - weight of m_{2} pulls DOWN (-) m_{2} g = its mass m_{2} its accel. (up, +) a (2)$

KYO. Same $T$ (Step 2), same $a$ (Step 6). Dhyan do ki signs flip ho gaye $(1)$ se compare karke: yahan tension help karta hai ( $m_{2}$ ko uthata hai) aur gravity isse fight karti hai. Yahi sign-flip hai kyun equations add karne se $T$ cancel ho jaayega.

PICTURE. Right chunk akela. Upar $T$ (lamba), neeche $m_{2} g$ (chhota), bacha hua upward stub net force $T - m_{2} g$ hai jo $m_{2} a$ drive karta hai.

Step 6 — Kyun dono accelerations same $a$ hain (constraint)

KYA. Agar left chunk distance $x$ neeche move kare, toh rope jo left par gayab hui woh right par reappear karni chahiye (rope stretch ya bunch up nahi ho sakti), toh right chunk same $x$ upar uthta hai. Same distance same time mein ⇒ same speed ⇒ same acceleration magnitude $a$ .

YEH KYUN CHAHIYE THA. Steps 4 aur 5 mein hum dono mein same $a$ likhte hain. Yeh step us kaam ki license hai. Yeh constraint relation hai (Constraint Relations): ek rope, ek length, toh dono motions ek saath locked hain, direction mein opposite lekin size mein equal.

PICTURE. Ek "before/after" strip: rope move karti hai, left $x$ girta hai, right $x$ uthta hai. Equal blue aur orange displacement brackets dikhate hain kyun $a$ shared hai.

Step 7 — Dono equations add karo $T$ hatane ke liye, aur $a$ solve karo

KYA. $(1)$ aur $(2)$ stack karo aur seedha neeche add karo:

m_{1} g - T T - m_{2} g m_{1} g - m_{2} g = = = m_{1} a m_{2} a (m_{1} + m_{2}) a

$- T$ aur $+ T$ cancel ho jaate hain. Solve karo:

$a = \frac{( m _{1} - m _{2} ) extra weight, in masses g}{total stuff being moved m _{1} + m _{2}}$

ADD KYUN, SUBTRACT NAHI? Kyunki $T$ ek baar $- T$ ki tarah aur ek baar $+ T$ ki tarah aata hai; addition woh ek move hai jo us unknown ko delete kar deta hai jo hum abhi nahi chahte. Jo bachta hai woh ek equation hai ek unknown $a$ mein.

Formula ko physically padho: upar $(m_{1} - m_{2}) g$ net driving force hai — sirf weight ka difference system ko push karta hai (shared part balance ho jaata hai). Neeche $(m_{1} + m_{2})$ total inertia hai — saari mass ko dhakka dena padta hai, dono chunks. Zyada difference → tezi; zyada total mass → dhimai.

PICTURE. Ek balance-beam cartoon: difference weight $(m_{1} - m_{2}) g$ "push" pan par; total mass $(m_{1} + m_{2})$ "resist" pan par; unka ratio (times $g$ ) $a$ hai.

Step 8 — Back-substitute karke tension $T$ nikalo

KYA. $a$ ko equation $(2)$ mein dalo, $T = m_{2} (g + a)$ :

= m_2 \cdot \frac{(m_1+m_2)g + (m_1-m_2)g}{m_1+m_2} = \boxed{\,\frac{2 m_1 m_2}{m_1+m_2}\,g\,}$$ **EQUATION (2) KYUN, (1) NAHI?** Dono kaam karte hain — dono ko same $T$ dena chahiye (yeh ek free consistency check hai). Hum $(2)$ use karte hain kyunki $T$ wahan already left par akela hai. **Isko padho:** result $m_1$ aur $m_2$ mein **symmetric** hai (unhe swap karo, kuch nahi badalta) — sahi hai, kyunki *rope ko pata nahi kaunsa side tumne "1" naam diya*. Quantity $\frac{2m_1m_2}{m_1+m_2}$ do masses ka [[Harmonic Mean]] combination hai, isliye $T$ hamesha $m_2 g$ aur $m_1 g$ ke **beech** aata hai. **PICTURE.** $m_2 g$ se $m_1 g$ tak ka ek number line jisme $T$ strictly andar mark kiya hai — tension kabhi bhi bhaare weight jaisa bada ya halke weight jaisa chhota nahi hota. ![[deepdives/dd-physics-1.2.10-d2-s08.png]] --- ## Step 9 — Har edge case, ek picture par check kiya **KYA.** Jo formula tum trust karte ho woh apni extremes mein survive karna chahiye. Boxed results mein special inputs daalo: | Input | $a$ | $T$ | Meaning | |---|---|---|---| | $m_1 = m_2 = m$ | $0$ | $mg$ | Balanced — kuch move nahi hota; rope bas har weight thaamti hai | | $m_2 = 0$ | $g$ | $0$ | Left chunk **free fall** mein; right par kheenchne ke liye kuch nahi | | $m_1 \to \infty$ | $\to g$ | $\to 2m_2 g$ | Bhaari left mass $m_2$ ko $\approx 2g$ net par upar kheenchta hai | | $m_1 < m_2$ | negative | phir bhi $+$ | Humara guess galat tha; $a$ ka sign self-correct kar leta hai | **KYO.** Yeh woh scenarios hain jo ek real problem tumhare haath mein de sakti hai. Aakhri row sabse zyada important hai: agar tumne galat bhaari side assume ki, **$a$ negative aata hai** aur bas itna matlab hai "woh doosri taraf move karta hai" — derivation kabhi nahi toot'ti. **PICTURE.** Teen mini-panels: (a) equal masses, motion frozen; (b) $m_2=0$, free fall arrow $=g$; (c) $m_1$ huge, $m_2$ upar $2g$-length arrow ke saath flung. ![[deepdives/dd-physics-1.2.10-d2-s09.png]] > [!example] Plug-in check (parent ke Example 1 se match karta hai) > $m_1 = 5$, $m_2 = 3$, $g = 10$: > $$a = \frac{(5-3)(10)}{5+3} = \frac{20}{8} = 2.5\ \text{m/s}^2, \qquad T = \frac{2(5)(3)}{8}(10) = 37.5\ \text{N}.$$ > $m_2$ par cross-check: $T - m_2 g = 37.5 - 30 = 7.5 = 3(2.5) = m_2 a$ ✓. Aur $T=37.5$, $m_2g=30$ aur $m_1g=50$ ke beech hai ✓. --- ## Ek-picture summary Upar sab kuch, compress karke: do free bodies → do Newton equations → $T$ hatane ke liye add karo → $a$ ke liye difference-over-sum aur $T$ ke liye double-product-over-sum. ![[deepdives/dd-physics-1.2.10-d2-s10.png]] > [!mnemonic] Do boxes ek saansh mein > **"$a$ ke liye DIFFERENCE over SUM; $T$ ke liye DOUBLE-PRODUCT over SUM."** Dono denominators $m_1+m_2$ hain — total inertia hamesha neeche baith'ti hai. > [!recall]- Feynman: poora walkthrough plain words mein > Do buckets ek halki rope par ek smooth wheel ke upar latke hain. Kyunki rope halki hai aur wheel smooth hai, rope *dono* buckets ko exactly same strength se *upar* kheenchti hai — ise $T$ bolo. Gravity har bucket ko uske apne weight se neeche kheenchti hai. Bhaari bucket lo; likho "uska weight minus rope pull equals uska mass times kitni tezi se speed up hota hai." Halka bucket lo; likho "rope pull minus uska weight equals uska mass times same speed-up" — *same* speed-up kyunki ek rope stretch nahi ho sakti, toh dono ends saath move karte hain. Woh do sentences add karo aur mysterious rope pull $T$ gayab ho jaata hai, bacha yeh: *extra* weight (difference) hi push karta hai, aur *saara* stuff (sum) ko push karna padta hai. Unhe divide karo, gravity $g$ laga do, aur woh acceleration hai. Ise kisi bhi sentence mein slide karke rope pull nikalte hain, jo hamesha dono weights ke beech kahin aata hai. Equal buckets try karo (kuch nahi move karta), ya empty bucket (doosra free-fall mein), aur formulas agree karte hain — toh tum unhe trust kar sakte ho. > [!recall]- Quick self-test > Hum dono masses ke liye same $a$ kyun likh sakte hain? ::: Rope inextensible hai — ek constraint jo unke displacements (isliye accelerations) ko size mein equal, direction mein opposite lock karta hai. > Equations $(1)$ aur $(2)$ add karna kyun help karta hai? ::: $T$ $-T$ aur $+T$ ki tarah aata hai; add karne se yeh cancel ho jaata hai, ek equation $a$ mein bach jaati hai. > $T$ hamesha $m_2g$ aur $m_1g$ ke beech kyun hona chahiye? ::: Yeh harmonic-mean combination $\frac{2m_1m_2}{m_1+m_2}g$ hai, jo hamesha dono weights ke beech hota hai. > Agar $m_1<m_2$, toh formula kya karta hai? ::: Negative $a$ deta hai, matlab system guessed direction ke opposite move karta hai. --- ## Connections - [[Newton's Second Law]] — $F_{net}=ma$ har chunk par apply hota hai (Steps 4–5). - [[Free Body Diagrams]] — har mass ko isolate karna (Steps 3–5). - [[Tension in Strings]] — kyun har jagah ek $T$ (Step 2). - [[Constraint Relations]] — inextensible rope ⇒ shared $a$ (Step 6). - [[Inclined Plane Problems]] — same recipe gravity component ke saath. - [[Pulley Systems & Mechanical Advantage]] — many-pulley generalisation. - [[Harmonic Mean]] — tension result ki shape (Step 8).