Tension in inextensible strings: Worked examples — Hinglish deep dive

1.2.9 · D3 · Physics › Newton's Laws & Dynamics › Tension in inextensible strings

Scenario matrix

Kuch bhi solve karne se pehle, aao territory map karte hain. Neeche har row ek case class hai — ek genuinely alag situation jo yeh topic produce kar sakta hai. Jo examples baad mein aate hain woh us cell ke saath tagged hain, aur milke woh har row fill karte hain.

#	Case class	Kya naya hai / kya galat ho sakta hai	Example
A	Dono masses hang karte hain (Atwood) — unequal	kaun sa side neeche jaata hai? $a$ ka sign	Ex 1
B	Balanced case $m_{1} = m_{2}$ — degenerate	$a = 0$ , static, $T = m g$	Ex 2
C	Ek mass zero — degenerate limit	free-fall check $a \to g$ , $T \to 0$	Ex 2
D	Horizontal + vertical (table + hanger)	weight table body par cancel ho jaata hai	Ex 3
E	Force ek chain of blocks ko pull karti hai	pehle system, phir $T$ ke liye isolate	Ex 4
F	Limiting behaviour ( $m \to \infty$ )	kya $T$ ek static value approach karta hai?	Ex 4
G	Do hanging masses ko ek external upward force se lift kiya	tension $m g$ se zyada ho sakti hai	Ex 5
H	Incline + hanger — $sin θ$ ka sign	system kis taraf slide karega?	Ex 6
I	Real-world word problem (elevator + person)	apparent weight = cable mein tension	Ex 7
J	Exam twist — teen masses, beech wali tension dhundo	tension segments ke beech alag hoti hai	Ex 8

Example 1 — Cell A · Atwood, unequal masses

Forecast: Bhaari side ( $m_{1}$ ) zaroor giregi, halki side zaroor uthegi. Kya tumhara anumaan hai ki $T$ , $m_{2} g = 20 N$ se zyada hoga ya kam? (Padhne se pehle guess karo — sochो: string ko $m_{2}$ ko lift bhi karna hai aur speed up bhi, toh...)

Figure padhna: picture mein dono masses upar lavender pulley se latke hue hain. Coral arrows follow karo — woh weights $m_{1} g$ aur $m_{2} g$ hain, dono seedhe neeche point karte hain. Lavender arrows tension $T$ hain, har block ko upar string ki taraf kheenchte hain. Bahar ke chhote slate arrows actual motion dikhate hain: $m_{1}$ neeche jaata hai, $m_{2}$ upar, same acceleration magnitude $a$ ke saath.

Figure — Tension in inextensible strings

Step 1 — Har mass ke liye free body diagram banao (figure mein jaisa hai). Yeh step kyun? Newton's law ek baar mein ek body par apply hota hai; har body sirf apna weight (neeche) aur tension (upar) feel karti hai.

$m_{1}$ ke liye, neeche ko positive chuno (yeh girta hai): $m_{1} g - T = m_{1} a$ $m_{2}$ ke liye, upar ko positive chuno (yeh uthta hai): $T - m_{2} g = m_{2} a$ Har body ke liye alag "positive" kyun? Kyunki inextensible string unhe same amount $a$ par accelerate karti hai lekin opposite directions mein. Har body ki apni motion ko positive lene se hum dono ke liye ek number $a$ use kar sakte hain.

Step 2 — Dono equations add karo $T$ cancel karne ke liye: $m_{1} g - m_{2} g = (m_{1} + m_{2}) a \Rightarrow a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 6 - 2 ) \cdot 10}{8} = 5 m/s^{2}$ Add kyun kiya? $- T$ aur $+ T$ equal aur opposite hain; add karne se woh unknown khatam ho jaata hai jo abhi chahiye nahi.

Step 3 — General tension derive karo, phir plug in karo. $T - m_{2} g = m_{2} a$ mein back-substitute karo: $T = m_{2} (g + a) = m_{2} (g + \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}) = m_{2} \cdot \frac{( m _{1} + m _{2} ) g + ( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ Yahan algebra kyun ki? Yahan se famous Atwood tension formula aata hai — yeh magic nahi hai, yeh bas $T = m_{2} (g + a)$ hai jisme $a$ substitute karke simplify kiya gaya. Numbers plug in karo: $T = \frac{2 \cdot 6 \cdot 2}{8} \cdot 10 = 30 N$

Verify: $T = 30 N > m_{2} g = 20 N$ ✓ — forecast se match (string $m_{2}$ ko lift aur accelerate karti hai). Direct route: $T = m_{2} (g + a) = 2 (10 + 5) = 30 N$ ✓ — same answer, derived formula confirm.

Example 2 — Cells B & C · Degenerate Atwood

Forecast: (a) mein dono sides perfectly matched hain — equals ke beech ek tug-of-war. $a$ kya hona chahiye? (b) mein ek side par kuch bhi nahi hai; doosri side kya kar rahi hai?

Step 1 — General formulas use karo jo humne Example 1 mein derive ki thi: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ Unhe reuse kyun kiya? Degenerate cases ko general result mein plug in karke check karna best hai — yahi kisi bhi formula ko sanity-check karne ka tarika hai.

Step 2 — Case (a), balanced: $a = \frac{( 4 - 4 ) \cdot 10}{8} = 0, T = \frac{2 \cdot 4 \cdot 4}{8} \cdot 10 = 40 N$ $a = 0$ kyun? Koi net driving difference nahi, toh kuch bhi accelerate nahi karta — system static hai. Har side sirf apna weight hold karti hai: $T = m g = 4 \cdot 10 = 40 N$ ✓.

Step 3 — Case (b), ek side khali ( $m_{2} = 0$ ): $a = \frac{( 4 - 0 ) \cdot 10}{4} = 10 = g, T = \frac{2 \cdot 4 \cdot 0}{4} \cdot 10 = 0$ $a = g$ aur $T = 0$ kyun? Pull karne ke liye kuch nahi toh $m_{1}$ free fall mein hai; ek slack, force-free string zero tension carry karti hai.

Verify: (a) $a = 0$ , $T = m g = 40 N$ ✓ (static balance). (b) $a = g = 10$ , $T = 0$ ✓ (free fall). Dono degenerate limits physically sahi behave karte hain.

Example 3 — Cell D · Table par block + hanging block

Figure padhna: lavender block $m_{1}$ slate table line par hai; string flat jaati hai butter-coloured pulley tak edge par, phir neeche coral hanging block $m_{2}$ tak bend karti hai. $m_{1}$ par mint arrow dhyan se dekho — woh normal force $N$ hai, exactly $m_{1}$ ka weight cancel karti hai taaki table par vertically kuch nahi hota. $m_{1}$ par sirf un-cancelled force hai lavender tension arrow jo horizontally pulley ki taraf point karta hai. $m_{2}$ par, coral weight arrow (neeche) lavender tension arrow (upar) ko beat karta hai, toh $m_{2}$ neeche accelerate karta hai — slate arrow labelled $a$ .

Forecast: Hanging mass poore system ko pull karti hai. Kya tension $m_{2} g = 30 N$ ke equal, zyada, ya kam hogi? (Hint: $m_{2}$ neeche accelerate kar raha hai, toh upar wala $T$ uska weight pura match nahi kar sakta.)

Step 1 — $m_{2}$ ka FBD (girta hai, neeche positive): $m_{2} g - T = m_{2} a$ $m_{1}$ ka FBD (horizontally slide karta hai, pull ki direction positive): $T = m_{1} a$ $m_{1}$ ke liye weight kyun nahi aaya? Table par uska weight normal force se cancel ho jaata hai (figure mein mint arrow); sirf horizontal force $T$ hai.

Step 2 — $T = m_{1} a$ pehli equation mein substitute karo: $m_{2} g - m_{1} a = m_{2} a \Rightarrow a = \frac{m _{2} g}{m _{1} + m _{2}} = \frac{3 \cdot 10}{5} = 6 m/s^{2}$ Substitute kyun? Isse $T$ eliminate hota hai taaki pehle $a$ solve kar sako.

Step 3 — $T$ nikalo: $T = m_{1} a = 2 \cdot 6 = 12 N$

Verify: $T = 12 N < m_{2} g = 30 N$ ✓ (forecast se match: falling mass ke liye $T < m g$ ). Cross-check: boxed formula $T = \frac{m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 \cdot 3 \cdot 10}{5} = 12 N$ ✓.

Example 4 — Cells E & F · Force se kheencha gaya chain of blocks, aur ek limit

Forecast: String ko sirf $m_{1}$ (peeche wala block) accelerate karna hai. Toh kya $T$ poora $12 N$ hona chahiye, ya sirf ek fraction?

Step 1 — Dono blocks ko ek system maano: $F = (m_{1} + m_{2}) a \Rightarrow a = \frac{F}{m _{1} + m _{2}} = \frac{12}{6} = 2 m/s^{2}$ System view kyun? Internal tension combined body ke liye cancel ho jaati hai (dono ends par equal aur opposite), sirf external $F$ bachti hai.

Step 2 — $m_{1}$ ko isolate karo (string uski sirf horizontal force hai): $T = m_{1} a = 4 \cdot 2 = 8 N$ $m_{2}$ nahi, $m_{1}$ kyun? $m_{1}$ ke liye, $T$ akela accelerating force hai, toh seedha $T = m_{1} a$ . $m_{2}$ ke liye $F - T = m_{2} a$ hoga (ek extra force), jo messier hai.

Step 3 — Limit (Cell F): $F$ fixed ke saath, $T = m_{1} a = m_{1} \cdot \frac{F}{m _{1} + m _{2}} = \frac{m _{1} F}{m _{1} + m _{2}} m_{2} \to \infty 0$ Zero kyun? Bahut bhaari front block barely accelerate karta hai ( $a \to 0$ ), toh string ko $m_{1}$ ko almost koi force se nahi kheenchna padta.

Verify: $a = 2$ , $T = 8 N$ . $m_{2}$ ki apni equation check karo: $F - T = 12 - 8 = 4 = m_{2} a = 2 \cdot 2 = 4$ ✓. Limit $T \to 0$ jab $m_{2} \to \infty$ ✓.

Example 5 — Cell G · Do masses ko upward force se lift karna

Forecast: Poori assembly upar accelerate karti hai. Kya lower tension $T_{1}$ ( $m_{1}$ ko pakde) $m_{1} g = 10 N$ se zyada hogi ya kam?

Step 1 — System (dono masses) $a$ nikalne ke liye: $F - (m_{1} + m_{2}) g = (m_{1} + m_{2}) a \Rightarrow a = \frac{F}{m _{1} + m _{2}} - g = \frac{30}{2} - 10 = 5 m/s^{2} (upar)$ Pehle system kyun? Internal string tension cancel ho jaati hai; sirf $F$ aur total weight bachti hai.

Step 2 — $m_{1}$ ko isolate karo (neeche weight feel karta hai aur upar lower tension $T_{1}$ ): $T_{1} - m_{1} g = m_{1} a \Rightarrow T_{1} = m_{1} (g + a) = 1 (10 + 5) = 15 N$ $m_{1}$ ke liye up-positive kyun? Yeh upar accelerate karta hai, toh upar iska natural positive direction hai.

Verify: $T_{1} = 15 N > m_{1} g = 10 N$ ✓ (upar accelerate karne ke liye extra upward pull chahiye). $m_{2}$ check karo: forces hain $F$ upar, $T_{1}$ neeche, weight neeche: $F - T_{1} - m_{2} g = 30 - 15 - 10 = 5 = m_{2} a = 1 \cdot 5$ ✓.

Example 6 — Cell H · Incline + hanging mass ( $sin θ$ ka sign)

Figure padhna: lavender wedge incline hai; uska base angle slate-labelled $3 0^{\circ}$ hai. Mint block $m_{1}$ slope par rakhaa hai, uski string slope ke upar jaati hai butter pulley tak, phir neeche coral hanging block $m_{2}$ tak. Slope par key coral arrow poora weight nahi hai — sirf along-slope component $m_{1} g sin θ$ hai, gravity ka woh part jo actually $m_{1}$ ko downhill slide karne ki koshish karta hai. Lavender arrows tension $T$ hain: ek $m_{1}$ ko slope ke upar kheenchta hai, ek $m_{2}$ ko upar pakde rakhta hai. System kis taraf move karta hai yeh decide karne ke liye coral down-slope pull ko $m_{2}$ ke full weight se compare karo.

Forecast: Incline $m_{1}$ ko slope ke neeche sirf component $m_{1} g sin θ = 2 \cdot 10 \cdot 0.5 = 10 N$ se kheenchti hai, jabki $m_{2}$ apne full weight $30 N$ se kheenchta hai. System kis taraf move karta hai — kya $m_{2}$ neeche aata hai, ya $m_{1}$ neeche slide hoke $m_{2}$ ko upar kheechta hai?

Step 1 — $m_{1}$ ka weight slope ke along resolve karo. Resolve kyun? Incline par gravity ke do parts hain: slope ke along ( $m_{1} g sin θ$ , jo $m_{1}$ ko ramp ke neeche kheench sakta hai) aur slope ke andar ( $m_{1} g cos θ$ , normal force se balanced). Sirf along-slope part string se compete karta hai.

Assume karo $m_{2}$ neeche jaata hai (toh $m_{1}$ ramp par chadtha hai), dono acceleration $a$ ke saath:

$m_{2}$ (neeche positive): $m_{2} g - T = m_{2} a$
$m_{1}$ (slope ke upar positive): $T - m_{1} g sin θ = m_{1} a$

Step 2 — $T$ cancel karne ke liye add karo: $m_{2} g - m_{1} g sin θ = (m_{1} + m_{2}) a \Rightarrow a = \frac{( m _{2} - m _{1} s i n θ ) g}{m _{1} + m _{2}} = \frac{( 3 - 2 \cdot 0.5 ) \cdot 10}{5} = 4 m/s^{2}$ Numerator ka sign check kyun karte hain? Agar $m_{2} - m_{1} sin θ$ negative aata, toh humara assumed direction galat tha aur block incline ke neeche slide karta. Yahan yeh $+ 2 > 0$ hai, toh $m_{2}$ sach mein neeche aata hai. ✓

Step 3 — $m_{2} g - T = m_{2} a$ se tension: $T = m_{2} (g - a) = 3 (10 - 4) = 18 N$

Verify: $a = 4$ , $T = 18 N$ . $m_{1}$ check karo: $T - m_{1} g sin θ = 18 - 10 = 8 = m_{1} a = 2 \cdot 4 = 8$ ✓. Aur $T = 18 < m_{2} g = 30$ ✓ (descending mass).

Worked example Degenerate edge check — perfectly balanced incline (

$m_{2} = m_{1} sin θ$ ) Statement: $m_{1} = 2 kg$ , $θ = 3 0^{\circ}$ rakho (toh $m_{1} sin θ = 1$ ), lekin ab exactly $m_{2} = 1 kg$ set karo. Kya hoga? Step 1 — General acceleration mein plug in karo: $a = \frac{( m _{2} - m _{1} s i n θ ) g}{m _{1} + m _{2}} = \frac{( 1 - 2 \cdot 0.5 ) \cdot 10}{3} = \frac{0}{3} = 0$ Numerator zero kyun hota hai? Hanging weight $m_{2} g$ exactly down-slope pull $m_{1} g sin θ$ ke barabar hai; dono tugs cancel ho jaate hain, toh kuch bhi accelerate nahi karta. Step 2 — Is static case mein tension: $a = 0$ ke saath, $T = m_{2} (g - a) = m_{2} g = 1 \cdot 10 = 10 N$ . Aur slope par $T = m_{1} g sin θ = 10 N$ — dono sides agree. Verify: $a = 0$ , $T = 10 N = m_{2} g = m_{1} g sin θ$ ✓. Yeh balanced Atwood (Cell B) ka incline version hai: numerator $m_{2} - m_{1} sin θ$ ka zero se guzarna exactly "system ek taraf slide karta hai" aur "doosri taraf" ke beech ki boundary hai.

Example 7 — Cell I · Real-world: elevator mein ek insaan

Forecast: Upar accelerate karte hue, kya insaan apne usual $600 N$ se bhaari feel karega ya halka? Aur kya cable total static weight $5000 N$ se zyada carry karegi?

Step 1 — Akele insaan ka FBD. Do forces act karti hain: scale ka normal force $N$ (upar) aur weight $m g$ (neeche). Insaan upar $a$ ke saath accelerate karta hai, toh upar positive lo: $N - m g = ma \Rightarrow N = m (g + a) = 60 (10 + 2) = 720 N$ Scale reading $N$ kyun hai? Scale woh force read karta hai jo insaan use neeche press karta hai; Newton's third law se woh scale ke upar push karne wale $N$ ke equal hai. Toh "reading" hai hi $N$ .

Step 2 — Poore elevator system ka FBD. Elevator + person ko ek body $M = 500 kg$ maano; cable tension $T$ upar kheenchti hai, total weight $M g$ neeche, aur poori cheez upar $a$ ke saath accelerate karti hai: $T - M g = M a \Rightarrow T = M (g + a) = 500 (10 + 2) = 6000 N$ Cable ke liye person + elevator kyun bundle kiya? Cable andar ki har cheez hold karti hai; combined body ke liye floor-on-person aur person-on-floor ke internal forces cancel ho jaate hain, sirf $T$ aur total weight bachte hain.

Verify: $N = 720 N > 600 N$ ✓ (upar accelerate karte hue bhaari feel hota hai — yeh "apparent weight" hai). $T = 6000 N > 5000 N$ ✓ (cable static hang se zyada mehnat karti hai). Units: $kg \cdot (m/s^{2}) = N$ ✓.

Example 8 — Cell J · Exam twist: teen masses, beech wali tension dhundo

Forecast: Do alag strings hain, toh dono tensions equal hona zaruri nahi. Tumhara anumaan kya hai — pull ke paas wali string zyada hogi ya peeche wali?

Step 1 — $a$ ke liye pehle system: $F = (m_{1} + m_{2} + m_{3}) a \Rightarrow a = \frac{12}{6} = 2 m/s^{2}$ System view kyun? Teeno ek acceleration share karte hain; poori train ke liye internal tensions cancel ho jaate hain, sirf $F$ bachta hai.

Step 2 — Peeche wala block $m_{3}$ isolate karo (sirf $T_{23}$ use kheenchta hai): $T_{23} = m_{3} a = 3 \cdot 2 = 6 N$ Akele $m_{3}$ kyun? Last string sirf $m_{3}$ ko accelerate karti hai, toh seedha $T_{23} = m_{3} a$ .

Step 3 — Group ${m_{2} + m_{3}}$ ko saath isolate karo (string $T_{12}$ dono ko accelerate karti hai): $T_{12} = (m_{2} + m_{3}) a = (2 + 3) \cdot 2 = 10 N$ $m_{2}$ aur $m_{3}$ ko group kyun kiya? Middle string ke peeche ki har cheez us string se kheenchi jaati hai, toh $T_{12}$ dono trailing blocks ka load carry karta hai.

Verify: $T_{12} = 10 N > T_{23} = 6 N$ ✓ — front string zyada carry karti hai, jaise forecast kiya tha. $m_{1}$ check karo: $F - T_{12} = 12 - 10 = 2 = m_{1} a = 1 \cdot 2$ ✓. $m_{2}$ check karo: $T_{12} - T_{23} = 10 - 6 = 4 = m_{2} a = 2 \cdot 2$ ✓. Yeh confirm karta hai ki tension strings ke beech alag hoti hai — directly kab tensions equal hote hain aur kab nahi se link.

Active Recall

Recall Kaun se cells

$T > m g$ vs $T < m g$ produce karte hain? Girna / descend karna → $T < m g$ (Ex 1, 3, 6). Upar accelerate karna → $T > m g$ (Ex 5, 7). Static → $T = m g$ (Ex 2a, incline edge check).

Recall Kya batata hai ki tumhara assumed direction galat tha?

Negative acceleration numerator (Ex 6: $m_{2} - m_{1} sin θ$ ). Agar exactly zero hai, toh system balanced aur static hai ( $a = 0$ ).

Recall Kya do alag strings guaranteed equal tension carry karti hain?

Nahi. Ek massless string ke andar tension uniform hoti hai, lekin do alag strings (Ex 8) jo bhi unka trailing load demand kare woh carry karti hain.

Connections

Newton's Second Law — upar har step $F = ma$ hai ek chosen body par.
Free Body Diagrams — woh drawing jo har equation obvious bana deti hai.
Constraint Relations — isliye ek $a$ saare connected bodies ke liye kaam karta hai.
Atwood Machine — Cells A, B, C.
Frictionless Pulleys vs Pulleys with Inertia — woh limit jahan Cell J ke "different tensions" ek single pulley mein bhi ghus jaate hain.
Normal Force — Cells D aur I mein use hone wala vertical partner.