Tension in inextensible strings: Question bank — Hinglish deep dive

1.2.9 · D5 · Physics › Newton's Laws & Dynamics › Tension in inextensible strings

Neeche har jagah yeh notation use hogi: $T$ = tension (string ke saath kheenchne wali force), $g$ = gravitational acceleration ( $\approx 9.8 m/s^{2}$ neeche ki taraf), $a$ = acceleration ka magnitude, $m$ = mass. Saari strings ideal hain (massless + inextensible) aur saare pulleys ideal hain (massless + frictionless) jab tak question explicitly kuch aur na kahe.

Teen pictures jinse har trap aata hai

Traps se pehle, in teen diagrams ko dekho jinka ye questions baar baar reference karte hain. Ek baar jab tum inhe visualize kar lo, toh verbal traps bilkul obvious ho jaate hain.

Figure 1 — ek mass ka free-body diagram (FBD). Ek free-body diagram ka matlab hai: sirf uss ek object ko draw karo, aur uske upar har force ka arrow, aur kuch nahi. Ek string par latke mass ke liye, bilkul do arrows kaam karte hain: gravity $m g$ neeche ki taraf kheenchti hai, tension $T$ upar ki taraf kheenchta hai (body se door, string ki taraf). Inhi do arrows ka net hi $ma$ ke barabar hota hai.

Figure — Tension in inextensible strings

Arrow ki lengths dekho: jab mass girta hai (neeche accelerate karta hai), toh neeche wala arrow $m g$ lamba hona chahiye, isliye $T < m g$ . Yeh ek picture "tension = weight" wale trap ko khatam kar deti hai.

Figure 2 — string element (tension uniform kyun hoti hai). String ka ek chhota sa tukda zoom karke dekho. Iske mass ko $d m$ kaho (yahan " $d$ " ka matlab hai "bahut chhota sa piece"). Yeh chhota sa tukda khud ek body hai, isliye iska apna FBD banta hai: baayein wali string ise $T_{1}$ se baayi taraf kheenchti hai, daayein wali string ise $T_{2}$ se daayein taraf kheenchti hai. Is tukde ke liye Newton ka law hai $T_{2} - T_{1} = (d m) a$ .

Ek massless string ke liye $d m = 0$ hai, toh kisi bhi finite $a$ ke liye right side zero ho jaata hai, aur $T_{2} = T_{1}$ maanna padta hai. Isliye tension har jagah same hoti hai — yeh $d m \to 0$ ka consequence hai, koi assumption nahi.

Figure 3 — inextensibility constraint (accelerations match kyun karte hain). Ek coordinate set karo: $x_{1}$ = string ki length pulley se mass 1 tak neeche, aur $x_{2}$ = pulley se mass 2 tak neeche. Har side ke liye positive $x$ neeche ki taraf hai (uss side par lambi string = mass neeche). Rope ki total length fixed hai: $x_{1} + x_{2} + (fixed wrap over pulley) = L = constant .$ Ek baar time ke saath differentiate karo: $\overset{x}{˙}_{1} + \overset{x}{˙}_{2} = 0$ (agar ek side badhti hai, doosri utni hi ghatti hai). Phir se differentiate karo: $\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}$ . Equal magnitude, opposite sign — mass 1 bilkul utni hi tezi se neeche jaata hai jitni tezi se mass 2 upar jaata hai.

"Inextensible" ka poora content yahi hai; general rule ke liye Constraint Relations dekho. Ab traps.

Sahi ya galat — justify karo

Ek string kisi block ko push kar sakti hai agar tum ise tezi se compress karo.

Galat. String floppy hoti hai; compression mein yeh buckle ho jaati hai (socho ek dhage ko push karna — woh fold ho jaata hai). Yeh sirf apni length ke saath force sustain karti hai pulling direction mein, yaani Figure 1 ke upar wale arrow ki direction mein.

m_{1} \neq = m_{2}

wali Atwood Machine mein, bhaari side ka tension zyada hota hai.

Galat. Figure 2 ek single

T

dikhata hai jo pulley ke upar se guzarti hai; ideal string + ideal pulley ke liye yeh dono sides par identical hoti hai. Masses alag hain, isliye accelerations alag hain — tension nahi.

Agar do masses ek pulley ke upar balanced hain (

m_{1} = m_{2}

), toh string mein tension zero hoti hai.

Galat. Kuch accelerate nahi karta, lekin har Figure-1 diagram mein upar wala arrow abhi bhi neeche wale arrow ko cancel karta hai, isliye

T = m g

0

nahi.

Neeche ki taraf

a

acceleration ke saath girte mass ke liye, tension uske weight

m g

ke barabar hoti hai.

Galat. Figure 1 mein neeche wala arrow lamba hai, isliye

m g - T = ma

, jo deta hai

T = m (g - a) < m g

. Tension sirf tab weight ke barabar hoti hai jab

a = 0

ho (dono arrows equal hों).

String ki inextensibility connected masses ko same velocity vector rakhne par majboor karti hai.

Galat. Figure 3 deta hai

\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}

: equal magnitudes, opposite signs. Directions alag hain — ek upar jaata hai jab doosra neeche jaata hai.

Ek ideal string kaatne se tension turant zero ho jaati hai.

Sahi. Tension material ke through transmit hoti hai (Figure 2 ke

T_{1}, T_{2}

arrows); material kaat do aur kheenchne ke liye kuch nahi bachta, isliye cut complete hote hi

T = 0

ho jaati hai.

Agar Atwood machine mein dono masses double kar do, toh acceleration double ho jaata hai.

Galat.

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

ek ratio hai — dono ko double karne par yeh unchanged rehta hai. Sirf tension double hoti hai.

Ek massless string ek mass wale pulley par — phir bhi dono sides par equal tension hogi.

Galat. Inertia wale pulley ko spin up karne ke liye net torque chahiye, jiske liye iske across alag

T_{1}, T_{2}

chahiye (Figure 2 ka zero-difference argument toot jaata hai). Dekho Frictionless Pulleys vs Pulleys with Inertia.

Ek real (bhaari) rope mein tension har jagah same hoti hai.

Galat. Figure 2 mein,

d m \neq = 0

, isliye

T_{2} - T_{1} = (d m) a

nonzero hai aur

T

vary karta hai. Uniform tension sirf massless idealization ka consequence hai.

Error pakdo

"Girte mass ke liye maine

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

likha, upar ko positive lete hue."

Mass neeche jaata hai, isliye uske Figure-1 diagram mein neeche wala arrow jeet ta hai; net force neeche ki taraf hai. Sahi equation hai

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

(is body ke liye down positive).

"Dono masses accelerate kar rahe hain, toh maine globally 'up is positive' leke dono FBD equations ko add kiya."

Figure 3 mein dono masses opposite directions mein move karte hain, isliye ek global sign ek tension term ko flip kar deta hai aur

T

cancel nahi hota. Fix: har body ke liye motion ki direction ko positive lo, phir add karne par

T

saaf cancel ho jaata hai.

"Latka hua mass

m_{2} g

force se kheenchta hai, toh table par rakhe block ko

T = m_{2} g

feel hota hai."

Yeh sirf tab sahi hai jab kuch accelerate na ho. Yahan system accelerate kar raha hai, isliye

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

. Static shortcut nahi, har Figure-1 diagram par $F=ma$ equations use karo.

"Force

F

front block

m_{2}

ko kheenchti hai,

m_{1}

ko peeche se kheechthi hai, isliye

T = F

F

poore system ko accelerate karti hai; uss force ka sirf ek hissa string ke through

m_{1}

tak pahunchta hai.

m_{1}

ka FBD isolate karne par milta hai

T = m_{1} a = \frac{m _{1} F}{m _{1} + m _{2}} < F

"Pulley string ko redirect karta hai isliye tension badal jaati hai."

Ek ideal (massless, frictionless) pulley sirf string ki direction redirect karta hai; Figure 2 dikhata hai

T_{2} - T_{1} = (d m) a = 0

, isliye yeh tension mein na kuch add karta hai na kuch remove.

"Tension uniform hai, isliye Atwood machine mein dono masses par same net force lagti hai."

Same tension ≠ same net force. Har Figure-1 diagram mein apna alag weight arrow bhi hota hai, isliye net forces (aur is tarah acceleration directions) alag hote hain. Uniform

T

har body par sirf do mein se ek arrow hai.

Why questions

Tension hamesha us body se door kyun hoti hai jis par woh act karta hai?

String sirf pull kar sakti hai, aur pull karne ka matlab hai body ko string ki taraf kheenchna — isliye Figure 1 mein body par tension ka arrow string ke saath, body se door string ki taraf point karta hai.

"Inextensible" equal acceleration magnitudes mein kaise translate hota hai?

Figure 3: total length

L

constant hai, isliye

\overset{x}{˙}_{1} + \overset{x}{˙}_{2} = 0

aur phir differentiate karne par

\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}

milta hai. Equal magnitude, opposite sign — dekho Constraint Relations.

Do connected blocks ko

a

find karne ke liye ek system ki tarah kyun treat kar sakte hain?

Internal tension Figure 2 ke equal-and-opposite

T

arrows ki tarah appear hoti hai, isliye poore system ke liye yeh cancel ho jaati hai; sirf external forces common acceleration set karte hain.

"Same

T

throughout" ke liye massless assumption kyun zaroori hai?

Figure 2 ki string-element equation,

T_{2} - T_{1} = (d m) a

T_{2} = T_{1}

sirf tab force karti hai jab

d m = 0

ho. Koi bhi mass tension ko string ke saath vary karne deta hai.

Atwood tension

T < m_{1} g

aur

T > m_{2} g

kyun satisfy karta hai jab

m_{1} > m_{2}

ho?

Bhaare mass ke Figure-1 diagram mein weight arrow jeet ta hai (

T < m_{1} g

); halke mass ke diagram mein tension arrow jeet ta hai (

T > m_{2} g

). Single

T

dono weights ke beech mein hota hai.

Ek frictionless table par sliding block ki horizontal equation mein Normal Force kyun nahi aati?

Normal force vertical hai aur block ke FBD mein uske weight ko cancel karti hai; horizontally sirf tension arrow act karta hai, isliye sirf

T = m_{1} a

bachta hai.

Edge cases

m_{2} = 0

wali Atwood machine (ek side khaali):

a

aur

T

ka kya hoga?

a = g

(bacha hua mass free fall mein hai) aur

T = 0

(doosri side mein kuch nahi jo pull create kare).

Atwood machine mein

m_{1} \to \infty

jab

m_{2}

fixed ho:

a

kya approach karega?

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

. Bahut bada mass essentially free-fall karta hai, chhote wale ko lagbhag

g

se upar kheenchta hai.

Block-on-table with hanging mass, jaise

m_{1} \to \infty

T

kya approach karega?

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

. Bahut bada table block barely move karta hai, isliye latka hua mass lagbhag statically

T = m_{2} g

par latakta hai.

Ek string slack ho jaati hai (jaise ek mass string se tezi se upar phenka gaya ho):

T

kya hai?

T = 0

. Ek slack string koi force exert nahi karti; tension kabhi negative nahi ho sakti, isliye yeh zero ho jaati hai jab tak string phir se taut na ho.

Do blocks frictionless floor par force

F

se ek saath push kiye gaye (contact, koi string nahi). Kya internal contact force "tension" hai?

Nahi — blocks ke beech contact push kar sakta hai (yeh ek normal force hai). Tension sirf kisi cheez mein exist kart hai jo kheenchi ja rahi ho, jaise ek connecting string.

Massless string, lekin pulley mein friction hai jo use grip karta hai: kya tension abhi bhi dono sides par equal hogi?

Nahi. Friction pulley ko slipping resist karne deta hai, isliye yeh apne across ek tension difference support kar sakta hai — ideal "same

T

" rule toot jaata hai.

Recall

Recall Top traps ke liye one-line rescues

Tension = weight? → Sirf jab $a = 0$ ho; girte mass ke liye $T = m (g - a)$ .
Bhaari side, zyada tension? → Nahi — same $T$ , alag accelerations.
String ko push karo? → Kabhi nahi; strings sirf pull karti hain.
Kisi bhi pulley par same $T$ ? → Sirf ideal (massless, frictionless) pulleys par.

Connections

Newton's Second Law — yahan har trap ko har body par $F = ma$ likhne se solve kiya jaata hai.
Free Body Diagrams — woh tool jo Figures 1–2 mein arrows draw karta hai.
Constraint Relations — Figure 3 ki length equation ka general form.
Atwood Machine — tension traps ka canonical source.
Frictionless Pulleys vs Pulleys with Inertia — jab "same $T$ " fail ho jaata hai.
Normal Force — pushing cousin, pulling tension se contrast mein.