Atwood machine — derivation: Exercises — Hinglish deep dive

1.2.10 · D4 · Physics › Newton's Laws & Dynamics › Atwood machine — derivation

Neeche diye gaye do engines hain jo tum har jagah use karoge, Newton's Second Law se har mass par apply karke:

Neeche ki figure woh do free-body diagrams dikhati hai jo tum har plain-Atwood problem mein banaaoge — har mass par ek tension arrow upar, ek weight arrow neeche, opposite motion arrows.

Figure kya dikhati hai (iske saath padhna): Top par ek cyan circle hai jo frictionless, massless pulley hai. Ek single white rope uske upar se chalti hai, left side neeche amber block $m_{1}$ (heavier mass) tak aur right side neeche cyan block $m_{2}$ (lighter mass) tak. Har block par ek cyan arrow upar ki taraf point karta hai jis par $T$ likha hai — ye do arrows purposely same length ke hain, kyunki ek hi rope ek frictionless pulley par hai, isliye tension dono ends par ek hi value $T$ hai. Har block par ek amber arrow bhi neeche ki taraf hai: left block ke neeche $m_{1} g$ , right block ke neeche $m_{2} g$ . Aakhir mein, do white "motion" arrows outcome dikhate hain: $m_{1}$ magnitude $a$ ke saath neeche accelerate karta hai, aur $m_{2}$ same magnitude $a$ ke saath upar (opposite directions, equal size — inextensible-string rule). Kyunki dono cyan $T$ arrows identical hain, do Newton equations jodhne par $T$ cancel ho jaata hai — yahi poora trick hai, seedha picture mein dikhayi deti hai.

Level 1 — Recognition

L1.1 — Kaun si quantity DONO acceleration aur tension formula ka denominator hai, aur wo physically kya represent karta hai?

Recall Solution — L1.1

Denominator ==total mass $m_{1} + m_{2}$ == hai. Physically yeh poore system ki inertia hai — wo saari "cheez" jise net force se move karwana hai. $a$ aur $T$ dono isse divide hote hain, kyunki dono ultimately is baat se govern hote hain ki coupled system kitna sluggish hai.

L1.2 — Kuch compute kiye bina, batao ki tension $T$ kya $m_{2} g$ se bada hai, $m_{1} g$ se chota hai, ya dono. Ek line ka reason do.

Recall Solution — L1.2

Dono. $T$ strictly $m_{2} g$ aur $m_{1} g$ ke beech mein hota hai.

Rising mass $m_{2}$ par: woh upar accelerate karta hai, isliye net force upar hai, isliye $T > m_{2} g$ .
Falling mass $m_{1}$ par: woh neeche accelerate karta hai, isliye net force neeche hai, isliye $T < m_{1} g$ . Ek tension, beech mein sandwiched: $m_{2} g < T < m_{1} g$ .

L1.3 — $m_{1} = 4 kg$ , $m_{2} = 4 kg$ ke liye, inspection se $a$ aur $T$ likho (koi algebra nahi).

Recall Solution — L1.3

Equal masses ⇒ balanced ⇒ $a = 0$ . Koi acceleration nahi hone par har side equilibrium mein hai, toh rope sirf ek weight ko upar rakhti hai: $T = m g = 4 \times 10 = 40 N$ . (Formula se check karo: $T = \frac{2 \cdot 4 \cdot 4}{8} \cdot 10 = 40 N$ ✓.)

Level 2 — Application

L2.1 — $m_{1} = 6 kg$ , $m_{2} = 2 kg$ . $a$ aur $T$ dhundho.

Recall Solution — L2.1

Net force $m_{1} g - m_{2} g$ kyun hai. Do free bodies banao (upar ki figure). Rope-coupled pair ko ek system ki tarah treat karo jo saath move karta hai, " $m_{1}$ neeche, $m_{2}$ upar" direction ko positive maano. Har ek ke liye Newton's Second Law likho aur jodhao: $on m_{1} m_{1} g - T + on m_{2} T - m_{2} g = m_{1} a + m_{2} a .$ Do tension terms $+ T$ aur $- T$ cancel ho jaate hain (same rope, same $T$ ), left side par $m_{1} g - m_{2} g$ bachta hai. Toh poore system par net driving force weight difference hai $= (6 - 2) \times 10 = 40 N$ , jo total inertia $= 8 kg$ se resist hoti hai. $a = \frac{( 6 - 2 ) 10}{6 + 2} = \frac{40}{8} = 5 m/s^{2}$ $T = \frac{2 \cdot 6 \cdot 2}{8} \cdot 10 = \frac{24}{8} \cdot 10 = 30 N$ Check $m_{2}$ par: $T - m_{2} g = 30 - 20 = 10 = m_{2} a = 2 \times 5$ ✓. Aur $20 = m_{2} g < 30 = T < 60 = m_{1} g$ ✓.

L2.2 — $m_{1} = 0.55 kg$ , $m_{2} = 0.45 kg$ . $a$ dhundho. Phir rest se $h = 0.8 m$ neeche jaane ka time dhundho.

Recall Solution — L2.2

$a = \frac{( 0.55 - 0.45 ) 10}{0.55 + 0.45} = \frac{0.10 \times 10}{1.0} = 1 m/s^{2}$ > Acceleration constant hai, isliye ordinary kinematics (constant-$a$ motion) apply hoti hai. Rest se shuru karke, $h = \frac{1}{2} a t^{2}$ : $t = \frac{2 h}{a} = \frac{2 \times 0.8}{1} = 1.6 \approx 1.265 s$

L2.3 — L2.2 wale same masses. Falling mass ki velocity kya hogi jab woh theek $0.8 m$ drop kar chuka ho?

Recall Solution — L2.3

$v^{2} = u^{2} + 2 ah$ use karo, $u = 0$ ke saath: $v = 2 ah = 2 \times 1 \times 0.8 = 1.6 \approx 1.265 m/s$ (Yahan numerically time ke barabar sirf isliye hai kyunki $a = 1$ hai; yeh numbers ka coincidence hai, koi rule nahi.)

Level 3 — Analysis

L3.1 — Ek student $a = 2 m/s^{2}$ measure karta hai aur jaanta hai ki total mass $m_{1} + m_{2} = 10 kg$ hai. $m_{1}$ aur $m_{2}$ dhundho ( $g = 10$ ).

Recall Solution — L3.1

$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ se: weight difference hai $m_{1} - m_{2} = \frac{a ( m _{1} + m _{2} )}{g} = \frac{2 \times 10}{10} = 2 kg .$ Ab pair solve karo $m_{1} + m_{2} = 10$ , $m_{1} - m_{2} = 2$ : $m_{1} = \frac{10 + 2}{2} = 6 kg, m_{2} = \frac{10 - 2}{2} = 4 kg .$ Check: $a = \frac{( 6 - 4 ) 10}{10} = 2$ ✓.

L3.2 — Maano tumne galat direction assume ki: tumne guess kiya ki $m_{2}$ girta hai jabki asliyat mein $m_{1}$ (heavier) girta hai. Tumne $m_{2} g - T = m_{2} a$ aur $T - m_{1} g = m_{1} a$ likha. Solve karo aur sign interpret karo. $m_{1} = 6, m_{2} = 2$ use karo.

Recall Solution — L3.2

Do (galat-direction) equations jodhao: $m_{2} g - m_{1} g = (m_{1} + m_{2}) a$ , toh $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}} = \frac{( 2 - 6 ) 10}{8} = - 5 m/s^{2} .$ Negative sign machine ka tumhara guess correct karna hai: matlab system actually us direction ke opposite accelerate karta hai jise tumne positive assume kiya tha — yaani $m_{1}$ waaqai girta hai, magnitude $5 m/s^{2}$ ke saath. Yahi L2.1 ka answer hai. Lesson: galat direction guess karna safe hai; algebra khud correct kar leta hai, bas sign padho.

L3.3 — Do students argue karte hain. Student A: "Bada $m_{1}$ hamesha bada tension deta hai." Student B: "Zaruri nahi." Kaun sahi hai? $m_{2} = 2$ fixed rakhke, $m_{1} = 6$ vs $m_{1} = 100$ compare karo.

Recall Solution — L3.3

$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g = \frac{2 \cdot 2 \cdot m _{1}}{m _{1} + 2} \cdot 10 = \frac{40 m _{1}}{m _{1} + 2}$ .

$m_{1} = 6$ : $T = \frac{240}{8} = 30 N$ .
$m_{1} = 100$ : $T = \frac{4000}{102} \approx 39.2 N$ . Tension badi zaroor hui, lekin dekho yeh saturate ho rahi hai: jaise $m_{1} \to \infty$ , $T \to 2 m_{2} g = 40 N$ aur kabhi isse zyada nahi hoti. Toh Student A sahi hai ki yeh badhti hai, lekin Student B deeper truth pakad raha hai — tension $2 m_{2} g$ se bounded hai, chahe $m_{1}$ kitna bhi bada ho jaaye. Ye tension ceiling ko acceleration se confuse mat karo: light mass ki acceleration $g$ tak ja sakti hai (jaise $m_{1} \to \infty$ , $a \to g$ ), $2 g$ nahi. Tension $2 m_{2} g$ isliye reach karti hai kyunki rope ko $m_{2} g$ supply karni hoti hai $m_{2}$ ka weight rakhne ke liye plus usse $g$ par upar accelerate karne ke liye aur $m_{2} g$ — do contributions of $m_{2} g$ , jo force $2 m_{2} g$ deti hai lekin acceleration sirf $g$ .

Level 4 — Synthesis

L4.1 — Inclined side par Atwood. Mass $m_{2} = 3 kg$ ek frictionless incline par hai jiska slope horizontal se angle $θ = 3 0^{\circ}$ banata hai. Uski rope incline ke top par ek pulley ke upar se hanging mass $m_{1} = 4 kg$ tak jaati hai. $a$ aur $T$ dhundho. ( $g = 10$ , $sin 3 0^{\circ} = 0.5$ .)

Recall Solution — L4.1

Yeh Atwood method ko inclined-plane resolution ke saath blend karta hai. Ek frictionless incline par ek mass ke liye jiska slope horizontal se $θ$ angle banata hai, gravity ka component slope ke along (woh part jo block ko ramp neeche kheenchta hai) $m_{2} g sin θ$ hota hai — neeche ki figure mein dekho, jahan sirf along-slope part rope ki pull ko oppose karta hai. Baaki sab kuch — ek rope, ek tension, ek shared $a$ — unchanged hai.

Maano $m_{1}$ girta hai (down), $m_{2}$ ko slope upar kheenchta hai.

$m_{1}$ (down +): $m_{1} g - T = m_{1} a$
$m_{2}$ (up-slope +): $T - m_{2} g sin θ = m_{2} a$

$T$ cancel karne ke liye jodhao: $m_{1} g - m_{2} g sin θ = (m_{1} + m_{2}) a$ $a = \frac{( m _{1} - m _{2} s i n θ ) g}{m _{1} + m _{2}} = \frac{( 4 - 3 \times 0.5 ) 10}{7} = \frac{( 4 - 1.5 ) 10}{7} = \frac{25}{7} \approx 3.571 m/s^{2} .$ $m_{1}$ ki equation se Tension: $T = m_{1} (g - a) = 4 (10 - 3.571) = 4 \times 6.429 \approx 25.71 N$ . Check $m_{2}$ par: $T - m_{2} g sin θ = 25.71 - 15 = 10.71 = m_{2} a = 3 \times 3.571$ ✓. Dhyan karo ki incline light side ka full weight $m_{2} g$ sirf uske slope component $m_{2} g sin θ$ se replace karta hai. Ramp jitna steep hoga, $sin θ$ utna hi $1$ ke paas hoga aur light side utni hi fully hanging mass jaisi behave karegi; vertical wall ( $θ = 9 0^{\circ}$ , $sin θ = 1$ ) plain Atwood exactly recover karta hai, jabki flat table ( $θ = 0^{\circ}$ , $sin θ = 0$ ) $m_{1}$ ko oppose karne ke liye kuch nahi chhoda toh $a = \frac{m _{1} g}{m _{1} + m _{2}}$ .

Figure mein key move visible hai: slope par, gravity along-slope part $m_{2} g sin θ$ (sirf woh part jo pull ko oppose karta hai) aur into-slope part mein split hoti hai jise surface cancel kar deti hai.

L4.2 — Energy cross-check. L2.1 ke plain Atwood ke liye ( $m_{1} = 6, m_{2} = 2$ ), heavy mass $h = 0.9 m$ drop karta hai. Energy conservation use karke final speed dhundho, aur confirm karo ki yeh kinematics $v = 2 ah$ se match karta hai.

Recall Solution — L4.2

Energy view: $m_{1}$ $h$ girta hai (PE $m_{1} g h$ loose karta hai), $m_{2}$ $h$ uthta hai (PE $m_{2} g h$ gain karta hai). Net released PE dono masses ki shared kinetic energy ban jaati hai (woh same speed $v$ par move karte hain): $(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2}$ $v^{2} = \frac{2 ( m _{1} - m _{2} ) g h}{m _{1} + m _{2}} = \frac{2 ( 6 - 2 ) ( 10 ) ( 0.9 )}{8} = \frac{144}{8} = 18 \Rightarrow v = 18 \approx 4.243 m/s .$ Kinematics cross-check: $a = 5 m/s^{2}$ (L2.1 se), $v = 2 ah = 2 \times 5 \times 0.9 = 9 \cdot 2 = 18 \approx 4.243 m/s$ ✓. Dono methods agree karte hain kyunki $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ wahi factor hai jo energy equation mein chhupa hua hai.

Level 5 — Mastery

L5.1 — Tension bound prove karo. Dikhao ki fixed $m_{2}$ aur kisi bhi $m_{1} > 0$ ke liye, tension $T < 2 m_{2} g$ satisfy karta hai, aur yeh limit $m_{1} \to \infty$ par approach hoti hai.

Recall Solution — L5.1

$T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}$ likho. Numerator aur denominator ko $m_{1}$ se divide karo: $T = \frac{2 m _{2} g}{1 + \frac{m _{2}}{m _{1}}} .$ Kyunki $m_{1}, m_{2} > 0$ hain, term $\frac{m _{2}}{m _{1}} > 0$ hai, isliye denominator $> 1$ hai, hence $T < 2 m_{2} g$ . Jaise $m_{1} \to \infty$ , $\frac{m _{2}}{m _{1}} \to 0$ , toh denominator $\to 1$ aur $T \to 2 m_{2} g$ . Yahi limiting behaviour parent note mein flag kiya gaya hai: infinitely heavy $m_{1}$ $m_{2}$ ko upar net force $\to m_{2} g$ ke saath kheenchta hai (net accel $g$ ), toh rope ko weight $m_{2} g$ supply karni hoti hai plus usse accelerate karne ke liye force $m_{2} g$ — total $2 m_{2} g$ . Bound kabhi reach nahi hoti, sirf approach hoti hai. Yahi $\frac{2 m _{1} m _{2}}{m _{1} + m _{2}}$ ka harmonic-mean structure hai jo apna ceiling dikha raha hai.

L5.2 — Target acceleration ke liye mass ratio dhundho. Kis ratio $r = m_{1} / m_{2}$ par acceleration exactly $a = g /3$ hogi?

Recall Solution — L5.2

Acceleration formula ke top aur bottom ko $m_{2}$ se divide karo: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( r - 1 ) g}{r + 1} .$ $g /3$ ke barabar set karo aur $g$ cancel karo: $\frac{r - 1}{r + 1} = \frac{1}{3} .$ Cross-multiply aur solve karo: $3 (r - 1) = r + 1 \Rightarrow 3 r - 3 = r + 1 \Rightarrow 2 r = 4 \Rightarrow r = 2.$ Toh $m_{1} = 2 m_{2}$ . Check: $a = \frac{( 2 - 1 ) g}{2 + 1} = \frac{g}{3}$ ✓. Dhyan karo acceleration sirf ratio par depend karta hai, kabhi absolute masses par nahi — dono masses double karne par $a$ unchanged rehta hai.

L5.3 — Design / degenerate case. Tum Atwood ki tarah $g$ measure karna chahte ho sirf $a = 0.05 g$ acceleration se taaki timing ke liye slow ho. Agar total mass $m_{1} + m_{2} = 2 kg$ honi chahiye, toh $m_{1}$ aur $m_{2}$ dhundho. Phir batao ki fully degenerate limit $m_{2} \to 0$ mein $a$ aur $T$ ka kya hoga.

Recall Solution — L5.3

$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = 0.05 g$ se, fraction $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} = 0.05$ hai, toh $m_{1} - m_{2} = 0.05 \times (m_{1} + m_{2}) = 0.05 \times 2 = 0.1 kg .$ $m_{1} + m_{2} = 2$ ke saath: $m_{1} = \frac{2 + 0.1}{2} = 1.05 kg$ , $m_{2} = \frac{2 - 0.1}{2} = 0.95 kg$ . Check: $a = \frac{( 1.05 - 0.95 ) g}{2} = \frac{0.1 g}{2} = 0.05 g$ ✓.

Degenerate limit $m_{2} \to 0$ : light side disappear ho jaati hai.

$a = \frac{( m _{1} - 0 ) g}{m _{1} + 0} = g$ — heavy mass free fall mein hai (kuch hold nahi kar raha).
$T = \frac{2 m _{1} \cdot 0}{m _{1} + 0} g = 0$ — rope slack ho jaati hai, doosri end par kuch nahi hai pull karne ko. Dono perfectly intuition se agree karte hain: koi partner nahi, koi tension nahi, pure gravity.

Recall Feynman recap — poori ladder ek saans mein

Yahan ki har problem wahi same do equations hain Newton's Second Law se — heavy side $m_{1} g - T = m_{1} a$ , light side $T - (opposing force) = m_{2} a$ — jodhao $T$ ko khatam karne ke liye. Kyunki yeh ek massless rope hai ek frictionless pulley par, tension ek single value hai, toh jodhne par $+ T$ aur $- T$ cancel ho jaate hain, $(driving force) = (m_{1} + m_{2}) a$ bachta hai. Uss ek line ko $a$ ke liye solve karo, phir kisi bhi equation mein back-substitute karke $T$ nikalo. Jo cheez levels mein change hoti hai woh sirf "opposing force" hai: flat machine par full weight $m_{2} g$ , incline par slope component $m_{2} g sin θ$ , zero jab partner vanish ho jaaye. Ek woh move master karo — do likho, jodhao $T$ cancel karne ke liye, solve karo, back-substitute karo — aur har Atwood variant arithmetic mein collapse ho jaayega.

Connections

Atwood machine — derivation (index 1.2.10) — parent derivation jise ye exercises drill karti hain.
Newton's Second Law — woh single law jo har solution ke peeche hai.
Free Body Diagrams — equations likhne se pehle har mass ko isolate karo.
Tension in Strings — kyun ek tension dono sides serve karti hai.
Constraint Relations — inextensible-string equal- $a$ rule jo poori jagah use hoti hai.
Inclined Plane Problems — L4.1 slope variant.
Pulley Systems & Mechanical Advantage — jahan multi-rope generalisations rehte hain.
Harmonic Mean — L5.1 mein prove ki gayi ceiling structure.