Tension in inextensible strings: Visual walkthrough

1.2.9 · D2 · Physics › Newton's Laws & Dynamics › Tension in inextensible strings

Step 1 — Scene, bina kisi assumption ke

KYA. Do objects ek hi rope ke do siron se latke hue hain. Rope upar jaake ek wheel (pulley) ke upar se guzarti hai jo use direction badalne deti hai. Hum objects ki "stuff" ki matra (unke masses) ko $m_{1}$ aur $m_{2}$ bolte hain, aur hum agree karte hain ki $m_{1}$ bhaari wala hai ( $m_{1} > m_{2}$ ).

KYUN. Kisi bhi maths se pehle, hume players ko dekhna hoga. $m_{1}$ jaisa symbol tab tak meaningless hai jab tak use picture mein pin nahi kiya jaata. Left weight ko dekho — woh $m_{1}$ hai. Right wale ko dekho — woh $m_{2}$ hai.

PICTURE. Bhaari block left pe baitha hai; dono seedhe neeche latke hain kyunki unhe sideways kheenchne wali cheez sirf rope hogi, aur rope yahan seedha upar kheenchti hai.

Figure — Tension in inextensible strings

Step 2 — Har force ko ek arrow ki tarah draw karo (the Free Body Diagram)

KYA. Hum dono blocks ko alag karte hain aur, har ek ke liye akele, usp par act karne wala har push ya pull ek arrow ke roop mein draw karte hain. Yeh picture ek Free Body Diagram (FBD) hai.

KYUN. Newton's Second Law — woh law $F = ma$ jo hum use karne wale hain — ek body par ek baar apply hoti hai. Agar hum do bodies ki forces mix karte hain, toh law toot jaati hai. Isliye hum har block ko isolate karte hain aur poochte hain: use kya touch kar raha hai, use kya pull kar raha hai?

Har block par do forces act karti hain:

Gravity, seedha neeche kheenchti hai, size $m_{1} g$ left block par aur $m_{2} g$ right par.
Tension $T$ , rope ka pull, seedha upar rope mein jaata hai dono blocks ke liye.

PICTURE. Neeche ke arrows weight hain; upar ke arrows tension hain. Notice karo ki dono upar ke arrows ka label ek hi $T$ hai — parent note ne prove kiya tha: ek massless rope ideal pulley ke upar har jagah equal strength se kheenchti hai.

KYA. Rope stretch nahi ho sakti. Toh agar left block $d$ distance neeche girta hai, toh exactly utni hi rope right side par aani chahiye, right block ko usi $d$ se upar uthate hue. Same distance same time mein ⟹ same speed ⟹ same acceleration magnitude, jise hum $a$ bolenge.

KYUN. Yeh woh constraint hai jo do unknown motions ko ek mein badal deta hai. Iske bina hamare paas do accelerations hote aur hum solve nahi kar paate. "Inextensible" ek hidden gift hai.

PICTURE. Coloured markers dekho: left wala ek grid square neeche slide karta hai exactly jab right wala ek grid square upar slide karta hai. Woh ek doosre ke saath locked hain.

Formula The constraint — with an explicit coordinate choice

"Same distance" ko signs mein badalne ke liye, pehle hume axes choose karne honge. Har block ko ek single coordinate se measure karo jo ek chosen direction mein badhti hai:

Let $x_{1}$ = left block apne start se kitna neeche gaya (toh $x_{1}$ badhta hai jab $m_{1}$ descend karta hai, matlab down positive hai $x_{1}$ ke liye).
Let $x_{2}$ = pulley se right block tak rope ki length (toh $x_{2}$ badhta hai jab $m_{2}$ descend karta hai, matlab down positive hai $x_{2}$ ke liye bhi).

Total rope length fixed hai. Jab left block $x_{1}$ se girta hai, exactly $x_{1}$ rope pulley ke upar se kheenchi jaati hai, toh right side usi amount se chhoti ho jaati hai: $x_{1} + x_{2} = constant .$

$x_{1} + x_{2}$ :: dono sides par latki total rope, jo change nahi ho sakti (inextensible).

Time ke respect mein do baar differentiate karo (pehle velocity, phir acceleration): $\overset{x}{¨}_{1} + \overset{x}{¨}_{2} = 0 ⟹ \overset{x}{¨}_{1} = - \overset{x}{¨}_{2} .$

$\overset{x}{¨}_{1}$ :: left block ki acceleration down positive measure karke;
$\overset{x}{¨}_{2}$ :: right block ki acceleration down positive measure karke;
minus sign :: hamare axis choice se forced hai — kyunki dono coordinates down ko positive count karte hain, ek block ka downward gain doosre ka downward loss hona chahiye (matlab ek upward move). Ek neeche jaata hai, doosra upar.

Sizes lene par, dono accelerations ka same magnitude hai: $∣ \overset{x}{¨}_{1} ∣ = ∣ \overset{x}{¨}_{2} ∣ = a .$ Woh single number $a$ , jo upar define kiya gaya, wahi hai jo agle steps solve karte hain.

Step 4 — Heavy block par Newton's law ( $m_{1}$ )

KYA. $m_{1}$ bhaari hai, isliye woh jeetta hai aur neeche move karta hai. Hum is block ke liye down ko positive direction choose karte hain, aur $F = ma$ likhte hain.

KYUN. Hum "positive = jis taraf woh actually move karta hai" choose karte hain taaki acceleration positive nikle — cleaner bookkeeping. Newton's Second Law kehta hai net force (saare arrows signs ke saath add kiye gaye) mass times acceleration ke barabar hoti hai.

$weight, down (+) m_{1} g - tension, up (-) T = mass \times accel m_{1} a$

$m_{1} g$ :: neeche gravity pull, positive kyunki down yahan hamara chosen positive hai;
$- T$ :: rope ka upar pull, negative kyunki yeh motion ke against hai;
$m_{1} a$ :: bacha hua unbalanced force, ise neeche drive karta hai.

PICTURE. Down-arrow ( $m_{1} g$ ) up-arrow ( $T$ ) se lamba draw kiya gaya hai: unke beech ka farq hi block ko accelerate karta hai.

Step 5 — Light block par Newton's law ( $m_{2}$ )

KYA. $m_{2}$ upar kheencha jaata hai. Hum is block ke liye up ko positive choose karte hain, aur phir $F = ma$ likhte hain.

KYUN. Same law, lekin ek alag sign choice — har body ko apna "positive = uski direction of motion" milta hai. Parent note ne mixing global signs ko ek classic mistake bataya tha; local conventions milke cleanly add hote hain.

$tension, up (+) T - weight, down (-) m_{2} g = mass \times accel m_{2} a$

$T$ :: rope pull, ab positive kyunki up is block ka positive direction hai;
$- m_{2} g$ :: gravity, negative kyunki yeh upward motion se ladhti hai;
$m_{2} a$ :: net upward force jo light block ko utha rahi hai.

PICTURE. Yahan up-arrow ( $T$ ) lamba wala hai: tension is block ke weight ko outpull karti hai, toh yeh utha jaata hai.

Step 6 — $T$ ko khatam karne ke liye dono equations add karo

KYA. Hamare paas do equations hain jo same $T$ aur same $a$ share karti hain. Unhe add karne se $- T$ aur $+ T$ cancel ho jaate hain.

KYUN. $T$ ek unknown hai jo hum abhi nahi chahte. Kyunki humne har block ki positive direction uski apni motion ke saath choose ki, tension terms ke opposite signs hain aur add karne par gayab ho jaate hain — ek equation mein ek unknown, $a$ , bach jaata hai.

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

$a$ ke liye solve karo:

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

$m_{1} - m_{2}$ :: weight mein imbalance — bhaari side ka "extra" pull;
$m_{1} + m_{2}$ :: total mass jo push ho rahi hai (dono blocks move karte hain);
toh $a$ hai "unbalanced pull ÷ total mass moved" — exactly Newton's law disguise mein.

PICTURE. Bar chart do weights dikhata hai; sirf overhang (burnt-orange slice) system ko drive karta hai, aur use dono masses ko accelerate karna hota hai.

Step 7 — Tension $T$ dhundhne ke liye $a$ wapas daalo

KYA. Ab $a$ ko $m_{2}$ equation mein use karo ( $T - m_{2} g = m_{2} a$ ) taaki $T$ mile.

KYUN. $T$ woh doosri cheez thi jo hum chahte the. Do mein se koi bhi equation kaam karti hai; hum substitute karte hain aur simplify karte hain.

$T = m_{2} g + m_{2} a = m_{2} (g + \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}) = m_{2} \cdot \frac{( m _{1} + m _{2} ) + ( m _{1} - m _{2} )}{m _{1} + m _{2}} g$

Upar $2 m_{1}$ ban jaata hai, isliye:

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

$2 m_{1} m_{2}$ :: dono masses mein symmetric — unhe swap karo aur $T$ unchanged rehta hai (rope ko koi farq nahi ki kaun sa side bhaari hai);
$m_{1} + m_{2}$ :: total mass phir se;
poori cheez $m_{2} g$ aur $m_{1} g$ ke beech baithti hai — rope tension ek compromise hai, kabhi bhi bhaari weight jaisi badi nahi, kabhi bhi halki weight jaisi chhoti nahi.

PICTURE. Ek number line $T$ ko exactly dono weights $m_{2} g$ aur $m_{1} g$ ke beech land karta dikhata hai.

Step 8 — Saare edge cases (koi bhi scenario unshown mat chhodte)

KYA. Hum formulas ko extremes par test karte hain yeh pakka karne ke liye ki woh sahi behave kar rahe hain.

KYUN. Jis formula ko tum sanity-check nahi kar sakte, woh ek aisa formula hai jis par tum trust nahi kar sakte. Har limit apne aap mein ek picture hai.

Case	$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$	$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$	Matlab
$m_{1} = m_{2} = m$	$0$	$m g$	Balanced — kuch nahi hila, har side bas apna weight hold karti hai.
$m_{2} = 0$	$g$	$0$	Left block free fall mein; right pe kuch nahi, koi pull nahi.
$m_{1} ≫ m_{2}$	$\to g$	$\to 2 m_{2} g$	Heavy side almost free-fall karta hai; light side yanked hoti hai — tension uske apne weight ke double ke paas!

PICTURE. Teen mini-scenes: balanced still-life, free-fall, aur violent yank — har ek mein arrows size match karte hain.

Ek-picture summary

Upar sab kuch ek single annotated map mein compress kiya gaya: do FBDs (Steps 4–5) do $F = ma$ equations feed karte hain; constraint (Step 3) ek shared $a$ force karta hai; add karne se $T$ cancel hota hai aur $a$ milta hai (Step 6); back-substitution se $T$ milta hai (Step 7). Figure algebra ko arrows ke saath rakhta hai taaki tum har symbol ko ek picture tak trace kar sako.

Recall Feynman retelling — ek dost ko batao

Do blocks ek rope ke siron se latke hain jo ek wheel ke upar se slung hai. Main har block ko akele dekhta hoon aur uske arrows draw karta hoon: gravity use neeche kheenchti hai, rope use upar kheenchti hai. Rope stretch nahi ho sakti, isliye dono blocks ek hi distance same time mein move karne par majboor hain — ek shared acceleration $a$ . Main har block ke liye "net force = mass × a" likhta hoon, is baat ka dhyan rakhte hue ki uski apni motion ki direction ko positive kahunga. Jab main do chhoti equations add karta hoon, rope ka tug ( $T$ ) cancel ho jaata hai aur main bach jaata hoon: weights ka farq masses ka sum drive karta hai. Woh hai $a = (m_{1} - m_{2}) g / (m_{1} + m_{2})$ . Phir main $a$ ko ek equation mein slide karta hoon aur tension $T = 2 m_{1} m_{2} g / (m_{1} + m_{2})$ nikalta hai — ek value jo hamesha politely dono weights ke beech baithti hai. Aakhir mein main silly extremes test karta hoon: equal masses → kuch nahi hila; right side empty → left block free-falls; giant left mass → chhota block apne weight se almost double rope-pull ke saath upar whip hota hai. Har extreme sense banata hai, isliye main formulas par trust karta hoon.

Connections

Newton's Second Law — har boxed step ek block par $F = ma$ hai.
Free Body Diagrams — Steps 2, 4, 5 hain hi FBDs.
Constraint Relations — Step 3, "same $a$ " lock.
Atwood Machine — yeh poora page uski derivation hai.
Frictionless Pulleys vs Pulleys with Inertia — jahan "equal $T$ " hold karna band ho jaata hai.
Normal Force — tension ka pushing cousin.