Atwood machine — derivation: Worked examples — Hinglish deep dive

1.2.10 · D3 · Physics › Newton's Laws & Dynamics › Atwood machine — derivation

Shuru karne se pehle, do reminders plain words mein:

$a$ (acceleration) ka matlab hai "speed kitni tezi se badal rahi hai har second", measured in $m/s^{2}$ .
$T$ (tension) ka matlab hai "rope kitna strong pull kar rahi hai", measured in newtons ( $N = kg \cdot m/s^{2}$ ).
$g$ Earth ke paas gravity ki strength hai, $g \approx 10 m/s^{2}$ (hum $10$ use karte hain clean numbers ke liye jab tak kuch aur na bata jaaye).

Scenario matrix

Har Atwood problem in case classes mein se ek mein aata hai. Neeche ke examples har row ko fill karte hain.

#	Case class	Kya special hai	Example
A	Left par heavy ( $m_{1} > m_{2}$ )	$a > 0$ , motion jaisi assume ki thi	Ex 1
B	Right par heavy ( $m_{1} < m_{2}$ )	$a < 0$ — sign batata hai ki yeh doosri taraf move karta hai	Ex 2
C	Equal masses ( $m_{1} = m_{2}$ )	Degenerate: $a = 0$ , static	Ex 3
D	Ek mass zero ( $m_{2} = 0$ )	Degenerate: free fall, $T = 0$	Ex 4
E	Ek mass huge ( $m_{1} ≫ m_{2}$ )	Limiting value: $a \to g$ , $T \to 2 m_{2} g$	Ex 5
F	Real-world / $g$ measure karna	Nearly equal masses, timing experiment	Ex 6
G	Kinematics follow-up	$a$ ko $s, v, t$ equations mein use karo	Ex 7
H	Exam twist: extra applied force	Formula ko re-derive karna padega, yaad nahi karna	Ex 8

Case A — left par heavy

Worked example Example 1 (Cell A)

$m_{1} = 6 kg$ (left), $m_{2} = 2 kg$ (right), $g = 10 m/s^{2}$ . $a$ aur $T$ nikalo.

Forecast: Left heavy hai, toh left giregi. Guess karo: kya $a$ , $g$ ke kareeb hai ya $0$ ke kareeb? (Mass ratio $3 : 1$ hai, kaafi lopsided — expect karo ki $a$ zero se kaafi upar hogi, lekin full $g$ nahi.)

Step 1 — Driving force identify karo. Net pull weight difference hai $(m_{1} - m_{2}) g = (6 - 2) (10) = 40 N$ . Yeh step kyun? Parent note se, dono sides ki gravity pulley ke across fight karti hai; sirf difference system ko drive karta hai.

Step 2 — Inertia identify karo. Move hone wala total mass $= m_{1} + m_{2} = 8 kg$ . Yeh step kyun? Dono masses ko saath accelerate karna hota hai, isliye rope ko $8 kg$ cheez ko push karna padta hai.

Step 3 — Divide karo. $a = \frac{40}{8} = 5 m/s^{2} .$ Yeh step kyun? $a = \frac{driving force}{total mass}$ poore system ke liye sirf $F = ma$ hai.

Step 4 — Tension. $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g = \frac{2 ( 6 ) ( 2 )}{8} (10) = \frac{24}{8} (10) = 30 N .$ Yeh step kyun? Boxed formula mein direct substitution.

Verify: Light mass $m_{2}$ par check karo (upar ja rahi hai): net up force $= T - m_{2} g = 30 - 20 = 10 N$ , aur $m_{2} a = 2 (5) = 10 N$ ✓. Note karo ki $30 N$ , $m_{2} g = 20$ aur $m_{1} g = 60$ ke beech hai — jaisa tension hamesha hona chahiye. Units: $N$ ✓, $m/s^{2}$ ✓.

Case B — right par heavy (sign flip ho jaata hai)

Worked example Example 2 (Cell B)

Same setup lekin ab $m_{1} = 2 kg$ (left), $m_{2} = 6 kg$ (right). Hum ziddi hokar parent ka convention rakhte hain: left ( $m_{1}$ ) assume kiya hai ki down jaayegi, positive.

Forecast: Humne galat assume kiya — heavy side ab right par hai. $a$ ka kya sign aana chahiye?

Step 1 — Formula mein bina badle plug karo. $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 2 - 6 ) ( 10 )}{8} = \frac{- 40}{8} = - 5 m/s^{2} .$ Yeh step kyun? Hum direction dobara guess nahi karte. Hum algebra par trust karte hain ki woh hamein correct kare.

Step 2 — Sign padho. $a = - 5 m/s^{2}$ . Minus ka matlab hai " $m_{1}$ actually upar accelerate kar rahi hai, neeche nahi" — humari assumption se bilkul opposite. Yeh step kyun? Chosen positive direction ke under negative answer ka matlab hamesha hota hai "reality doosri taraf point kar rahi hai." Yahi reason hai ki hum direction freely guess karne allowed hain.

Step 3 — Tension sign se affect nahi hoti. $T = \frac{2 ( 2 ) ( 6 )}{8} (10) = 30 N .$ Yeh step kyun? Tension $2 m_{1} m_{2}$ aur $m_{1} + m_{2}$ par depend karti hai, dono symmetric hain — masses ko swap karna $T$ nahi badal sakta. Example 1 jaisi hi $30 N$ .

Verify: $a$ ki magnitude $5 m/s^{2}$ hai, Example 1 jaisi hi — kyunki yeh same do masses hain, sirf mirror image. Physics ko parwah nahi ki tumne kaunsi side ko "1" label kiya. ✓

Case C — equal masses (degenerate: kuch nahi hilta)

Case D — ek mass zero (degenerate: free fall)

Case E — ek mass huge (limiting behaviour)

Worked example Example 5 (Cell E)

$m_{1} = 1000 kg$ , $m_{2} = 1 kg$ , $g = 10$ . $a$ kis value ke paas jaata hai, aur $T$ kis value ke paas jaati hai?

Forecast: Huge mass basically freely girta hai aur tiny mass ko upar yank karta hai. $a$ predict karo aur woh upward acceleration jo small mass feel karti hai.

Step 1 — Acceleration. $a = \frac{( 1000 - 1 ) ( 10 )}{1001} = \frac{9990}{1001} \approx 9.98 m/s^{2} .$ Yeh step kyun? Jaise $m_{1}$ , $m_{2}$ se bahut bada ho jaata hai, fraction $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} \to 1$ , toh $a \to g$ . Heavy mass nearly free fall mein hai.

Step 2 — Tension. $T = \frac{2 ( 1000 ) ( 1 )}{1001} (10) = \frac{20000}{1001} \approx 19.98 N .$ Yeh step kyun? Jaise $m_{1} \to \infty$ , $\frac{2 m _{1} m _{2}}{m _{1} + m _{2}} \to \frac{2 m _{1} m _{2}}{m _{1}} = 2 m_{2}$ , toh $T \to 2 m_{2} g = 2 (1) (10) = 20 N$ .

Step 3 — Small mass ko interpret karo. $m_{2}$ par net up force: $T - m_{2} g \approx 19.98 - 10 = 9.98 N$ , use $\approx 9.98 m/s^{2}$ upward acceleration deta hai — nearly $g$ upward. Yeh ek " $2 g$ " world feel karta hai: gravity $g$ down plus tension nearly $2 g$ par upar pull kar rahi hai. Yeh step kyun? Yeh limit ko tangible banata hai: infinitely heavy driver tiny mass ko essentially $g$ se gravity ke against accelerate karta hai.

Verify: $a = 9990/1001 \approx 9.980$ , $T = 20000/1001 \approx 19.980$ ; dono predicted limits $g$ aur $2 m_{2} g$ ke paas jaate hain ✓.

Case F — real-world: $g$ measure karna

Worked example Example 6 (Cell F)

Atwood ka original trick: masses ko nearly equal banao taaki acceleration small ho aur time karna aasaan ho. Lo $m_{1} = 0.520 kg$ , $m_{2} = 0.480 kg$ . Ek experiment mein system $h = 0.500 m$ rest se $t = 1.010 s$ mein girta hai. $g$ estimate karo.

Forecast: Kya measured $g$ , $9.8$ ke paas aayega? Hum kitni tiny acceleration expect karte hain?

Step 1 — Timing se $a$ nikalo (kinematics). Rest se, $h = \frac{1}{2} a t^{2}$ , toh $a = \frac{2 h}{t ^{2}} = \frac{2 ( 0.500 )}{( 1.010 ) ^{2}} = \frac{1.000}{1.0201} \approx 0.9803 m/s^{2} .$ Yeh step kyun? Hum distance aur time directly measure kar sakte hain; acceleration unse infer ki jaati hai.

Step 2 — $a$ ko Atwood formula ke zariye $g$ se relate karo. $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ se $g$ solve karte hue: $g = a \cdot \frac{m _{1} + m _{2}}{m _{1} - m _{2}} = 0.9803 \cdot \frac{1.000}{0.040} .$ Yeh step kyun? Machine $g$ ko ek measurable $a$ mein amplify karti hai; hum woh amplification invert karte hain $g$ recover karne ke liye.

Step 3 — Compute karo. $g \approx 0.9803 \times 25 = 24.5 m/s^{2} .$ Hmm — yeh bahut zyada bada hai! Point yeh hai: $\frac{m _{1} + m _{2}}{m _{1} - m _{2}} = 25$ ke saath, koi bhi chota timing error $25$ se multiply ho jaata hai. Real experiments mein kaafi finer timing use hoti hai. Ideal $g = 9.8$ ke saath, true acceleration $a = 9.8/25 = 0.392 m/s^{2}$ hoti, giving $t = 2 h / a = 1.000/0.392 \approx 1.597 s$ . Yeh step kyun? Yeh real design trade-off dikhata hai: nearly-equal masses ek slow, timeable fall dete hain, lekin precise timing demand karte hain kyunki amplification factor huge hai.

Verify: Ideal values ke saath: $a = 0.392 m/s^{2}$ $g = 9.8$ se; phir $t^{2} = 2 h / a = 1.000/0.392 = 2.551$ , $t \approx 1.597 s$ ✓. Same idea narrated ke liye Hinglish note dekho.

Case G — kinematics follow-up

Case H — exam twist: ek extra applied force

Worked example Example 8 (Cell H)

Ek examiner ek downward force $F = 10 N$ add karta hai jo left mass par gravity ke alawa pull karta hai. Masses $m_{1} = 4 kg$ (left), $m_{2} = 2 kg$ (right), $g = 10$ . $a$ nikalo.

Forecast: Plain formula ab apply nahi hota — kyun nahi? Tumhe re-derive karna padega. Kya $a$ , no- $F$ case se bada hoga ya chhota?

Step 1 — Har mass ke liye Newton's law likho (left assumed down = +). Left: $m_{1} g + F - T = m_{1} a$ . Right: $T - m_{2} g = m_{2} a$ . Yeh step kyun? Extra $F$ ek naya term hai jiska boxed formula mein koi account nahi tha, isliye hum Newton's Second Law aur Free Body Diagrams par waapas jaate hain aur equations scratch se rebuild karte hain.

Step 2 — Add karo taaki $T$ cancel ho jaaye. $m_{1} g + F - m_{2} g = (m_{1} + m_{2}) a .$ Yeh step kyun? Parent derivation jaisi hi trick: add karne par $+ T$ aur $- T$ cancel ho jaate hain.

Step 3 — Solve karo. $a = \frac{( m _{1} - m _{2} ) g + F}{m _{1} + m _{2}} = \frac{( 4 - 2 ) ( 10 ) + 10}{6} = \frac{20 + 10}{6} = \frac{30}{6} = 5 m/s^{2} .$ Yeh step kyun? Driving force mein ab $F$ bhi include hai; inertia $m_{1} + m_{2}$ unchanged hai. $F = 0$ case se bada ( $a = 20/6 \approx 3.33$ ), jaisa expect tha — tumne ek helping pull add ki.

Step 4 — Tension. Right equation se, $T = m_{2} (g + a) = 2 (10 + 5) = 30 N$ . Yeh step kyun? $a$ known hone par, kisi bhi single-mass equation mein plug karo; sabse simple $m_{2}$ ki hai.

Verify: Left equation: $m_{1} g + F - T = 40 + 10 - 30 = 20 = m_{1} a = 4 (5)$ ✓. Lesson: Yaad kiye hue formulas tab break ho jaate hain jab koi naya force aata hai — hamesha re-derive karne ke liye ready raho.

Recall Ek aisi aadat jo har scenario beat karta hai

Ek positive direction guess karo, har mass ke liye $F_{n e t} = ma$ likho, tension kill karne ke liye add karo. Usi ek procedure ne cells A–H solve kiye, including woh bhi jahan boxed formula apply nahi hota tha. "Same $a$ " kyun har baar legitimate hai, yeh jaanne ke liye Constraint Relations dekho.

Connections

Atwood machine — derivation — woh source formulas jinhein humne stress-test kiya.
Newton's Second Law — $F_{n e t} = ma$ se har case rebuild kiya.
Free Body Diagrams — extra-force twist ke liye zaroori tha (Cell H).
Constraint Relations — har example mein equal $∣ a ∣$ justify karta hai.
Tension in Strings — kyun $T$ Example 2 mein symmetric rahi.
Harmonic Mean — Cell E ki limit ke peeche tension formula ki shape.