Atwood machine — derivation: Question bank — Hinglish deep dive

1.2.10 · D5 · Physics › Newton's Laws & Dynamics › Atwood machine — derivation

Shuru karne se pehle, symbols ka ek reminder taaki yahan kuch bhi unclear na rahe:

$m_{1}, m_{2}$ — do hanging masses (kg). Hum $m_{1} \geq m_{2}$ lete hain jab tak aur na bataya jaye.
$g$ — gravitational acceleration ( $\approx 9.8 m/s^{2}$ ), Earth ka pull har kg par.
$a$ — acceleration magnitude jo dono masses share karte hain (woh ek hi rate se chalte hain kyunki ek rope unhe baandh ke rakhti hai).
$T$ — tension, woh pulling force jo rope exert karti hai, ek massless rope ke har point par same hoti hai.
Do results jinpar hum rely karte hain: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ aur $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ .

Pehle dekho — woh picture jisse har trap aata hai

Traps se pehle, neeche diye gaye do free-body diagrams dekho. Is page ka har sawaal asal mein inhi do arrows aur unhe jodhne wali rope ke baare mein hai.

Figure mein notice karo: har mass par exactly do arrows hain — weight $m g$ neeche ki taraf kheench rahi hai (chalk-pink) aur wahi tension $T$ upar ki taraf kheench rahi hai (chalk-blue). Pulley ke upar se rope bas us ek blue force ko wheel ke around mod deti hai; woh dono taraf same length ka arrow hai kyunki string massless hai. Woh ek fact — ek blue arrow, do identical copies — hi tension ko uniform banata hai.

Ab dekho kaise do Newton equations seedha arrows se nikal aati hain. Heavy side par, neeche jeetता hai, toh neeche-as-positive likhne par $m_{1} g - T = m_{1} a$ milta hai. Light side par, upar jeetता hai, toh upar-as-positive likhne par $T - m_{2} g = m_{2} a$ milta hai. Inhe add karo aur blue $T$ arrows cancel ho jaate hain:

Doosri figure poori algebra ko ek picture ke roop mein dikhati hai: sum mein do blue tension arrows equal aur opposite hain, isliye woh annihilate ho jaate hain, sirf pink weight difference $m_{1} g - m_{2} g$ bachta hai jo total mass $m_{1} + m_{2}$ ko push karta hai. Yahi exactly hai $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} .$ Is $a$ ko light-side equation mein wapas daalo, $T = m_{2} (g + a)$ , aur simplify karo: $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g .$

Aakhri picture dikhati hai kyun answers waise behave karte hain jaise karte hain — $a$ aur $T$ kaise slide karte hain jab ek mass badhta hai:

Ye teen figures apne dimaag mein rakhna; har trap neeche inhi arrows mein se kisi ek ko galat padhne ka case hai.

True ya false — justify karo

Dono masses ka acceleration magnitude same hota hai.

True — string inextensible hai, toh agar ek end

x

neeche jata hai toh doosra exactly

x

upar aata hai; equal time mein equal displacement matlab equal speed aur equal acceleration (Constraint Relations).

Pulley ke dono taraf tension alag hoti hai.

False — string aur pulley massless hain aur pulley frictionless hai, toh rope sirf force ko redirect karti hai;

T

har jagah ek hi value hoti hai (Tension in Strings).

Tension lighter mass ke weight ke barabar hoti hai.

False — yeh sirf equilibrium (

a = 0

) mein hold hoga. Yahan

T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g

hai, jo strictly

m_{2} g

aur

m_{1} g

ke beech hota hai jab bhi masses alag hon.

Agar dono masses double kar do, acceleration double ho jata hai.

False —

a = \frac{( 2 m _{1} - 2 m _{2} ) g}{2 m _{1} + 2 m _{2}}

mein 2 ka factor upar aur neeche cancel ho jata hai, toh

a

unchanged rehta hai. Acceleration masses ke ratio par depend karta hai, unke absolute size par nahi.

Heavier mass

g

acceleration se girta hai.

False — woh

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} < g

se girta hai, kyunki rope uske weight ka ek hissa rok leti hai; sirf jab

m_{2} = 0

ho tab woh free fall tak pahunchta hai.

Dono masses par gravity milkar system ko drive karti hai.

False — kyunki rope pulley ke upar se route hoti hai,

m_{2}

ka weight

m_{1}

ki motion ko oppose karta hai. Net driving force difference

(m_{1} - m_{2}) g

hota hai, kabhi bhi sum nahi.

Tension

m_{1}

aur

m_{2}

mein symmetric hai.

True —

\frac{2 m _{1} m _{2}}{m _{1} + m _{2}}

labels swap karne par unchanged rehta hai, jo sahi bhi lagta hai: rope nahi bata sakti ki tumne kaunhi side ko "1" decide kiya.

Agar

m_{1} = m_{2}

ho toh rope slack ho jaati hai.

False — system balanced hota hai toh

a = 0

, lekin rope taut rehti hai,

T = m g

carry karti hai taaki har equal weight ko gravity ke against hold kar sake.

Error dhundho

m_{1}

ke liye:

m_{1} g - T = m_{1} a

with up as positive."

Error — sign convention galat hai. Agar

m_{1}

neeche accelerate karta hai, toh uske liye down ko positive lena hoga; equation tab sahi hai jab down

+

ho; "up as positive" ke saath

a

ke signs uski motion se match nahi karte.

"Equations add karo aur

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

milta hai."

Error — tumne weights add kar diye subtract karne ki jagah. Sahi addition hai

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

, jisse

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

milta hai; tensions cancel ho jaate hain aur weights subtract ho jaate hain.

"Kyunki har mass

m g

weigh karta hai, rope ko har ek ko

m g

se upar kheenchna chahiye, toh heavy side par

T = m_{1} g

hai."

Error — yeh equilibrium assume karta hai. Masses accelerate kar rahe hain, toh Newton's second law net force

\neq = 0

deta hai; heavy side par ek single tension

T < m_{1} g

act karti hai.

"Har side ka apna acceleration hai kyunki woh opposite directions mein chalte hain."

Error — opposite direction ka matlab different magnitude nahi hota. Inextensible string equal

∣ a ∣

force karti hai; sirf sign (upar vs neeche) alag hota hai.

"Light mass par:

T - m_{2} g = m_{2} a

, toh agar

a > 0

hai toh

T < m_{2} g

hai."

Error — rearrange karo:

T = m_{2} (g + a)

, aur

a > 0

hone se

T > m_{2} g

milta hai. Light mass upar accelerate karta hai, toh rope ko uske weight se zyada pull karna padta hai; extra

m_{2} a

hi use utha ta hai.

"Pulley ek extra upward force add karti hai, toh use equations mein aana chahiye."

Error — ek ideal (massless, frictionless) pulley koi force store ya add nahi karti; woh sirf rope ki direction change karti hai. Woh kabhi bhi do Newton equations mein nahi aati.

Why questions

Hum do Newton equations ko add kyun karte hain, subtract kyun nahi?

Kyunki unknown tension ek mein

- T

aur doosre mein

+ T

ke roop mein appear hoti hai; add karne par

T

cancel ho jaata hai aur sirf

a

mein ek equation bachti hai.

Dono formulas ke denominator mein total mass

(m_{1} + m_{2})

kyun hai?

Kyunki poora rope-coupled system ek ki tarah move karta hai — total mass woh inertia hai jo driving force ko resist karti hai, toh woh hamesha response ka scale set karti hai.

Atwood ne machine nearly equal masses ke saath kyun banaya?

Nearly equal masses ek tiny weight difference dete hain, toh

a

ek choti fraction of

g

hoti hai. Slow acceleration ko haath se time karna aasaan hai, jisse

g

accurately measure kar sakte hain (Newton's Second Law).

Tension heavy mass ke weight se kam kyun aata hai?

Heavy mass neeche accelerate karta hai, toh uska net force neeche point karna chahiye:

m_{1} g - T > 0

, jo force karta hai

T < m_{1} g

Tension light mass ke weight se zyada kyun aata hai?

Light mass upar accelerate karta hai, toh uska net force upar point karta hai:

T - m_{2} g > 0

, jo force karta hai

T > m_{2} g

Hume har mass ke liye alag free-body diagram kyun banana chahiye?

Kyunki Newton's second law ek body par ek baar apply hota hai; har mass ko isolate karne se exactly do forces (weight neeche, tension upar) dikhte hain jo uspar act kar rahe hain (Free Body Diagrams).

T

uniform hai" ke liye massless string ka assumption kyun essential hai?

Ek massive rope ko accelerate karne ke liye apna net force chahiye hoga, toh tension ek end se doosre end tak alag hogi; sirf massless rope ko zero net force chahiye, jo

T

ko har jagah same rakhta hai.

Tension expression harmonic-mean combination jaisi kyun lagti hai?

Kyunki

\frac{2 m _{1} m _{2}}{m _{1} + m _{2}}

exactly

m_{1}

aur

m_{2}

ka harmonic mean hai; yeh symmetric hai aur smaller mass ki taraf weighted hai, jo reflect karta hai ki lighter side shared force ko limit karti hai (Harmonic Mean).

Edge cases

Jab

m_{1} = m_{2} = m

ho toh

a

aur

T

ka kya hoga?

a = 0

(balanced, kuch accelerate nahi karta) aur

T = \frac{2 m ^{2}}{2 m} g = m g

(rope bas har weight ko upar hold karti hai).

Jab

m_{2} \to 0

ho (ek side khaali) toh kya hoga?

a \to g

(akela mass free-fall karta hai) aur

T \to 0

(doosre end par kuch nahi hai pull karne ke liye, toh rope koi force carry nahi karti).

Jab

m_{1} \to \infty

aur

m_{2}

fixed ho toh kya hoga?

a \to g

(bahut bada mass essentially freely girta hai) aur

T \to 2 m_{2} g

— tab light mass ko

T - m_{2} g \to m_{2} g

ka net upward force feel hota hai, jisse uska upward acceleration

g

ke kareeb pahunchta hai.

Agar humara sign guess galat ho aur hum assume karein ki lighter side giregi, toh kya batayega?

Formula negative

a

return karega. Woh negative sign self-correct karta hai, matlab ki actual motion humari guess ke opposite hai — physics harm nahi hoti.

Kya tension dono weights

m_{1} g

aur

m_{2} g

se ek saath zyada ho sakti hai?

Nahi —

T

hamesha strictly

m_{2} g

aur

m_{1} g

ke beech rehti hai (dono ke equal sirf tab jab

m_{1} = m_{2}

ho). Ideal Atwood machine mein yeh heavier weight se kabhi bada nahi ho sakta.

Agar pulley ka mass hota (real world), toh kya

T

dono sides par still same hoti?

Nahi — ek real pulley ko angularly accelerate karne ke liye net torque chahiye, toh tensions uske across alag honge. Woh case ideal-Atwood territory se bahar jata hai aur rotational dynamics ki zaroorat padti hai (Pulley Systems & Mechanical Advantage).

Recall Har trap ki ek-line summary

Rope masses ko $a$ share karati hai, $T$ ko uniform banati hai, aur drive ko masses ke sum par difference of weights banati hai. Kisi ek idealisation ko todo (rope par mass, pulley par mass, friction) aur neat symmetry uske saath toot jaati hai.

Connections

Atwood machine — derivation — woh parent derivation jise ye traps test karte hain.
Newton's Second Law — har "why" ke peeche ka engine.
Free Body Diagrams — woh tool jo zyaadatar sign traps solve karta hai.
Tension in Strings — kyun $T$ uniform hai.
Constraint Relations — kyun accelerations equal hain.
Pulley Systems & Mechanical Advantage — kya badalta hai jab pulley real hoti hai.
Harmonic Mean — tension formula ki shape.