Tension in inextensible strings — Hinglish notes

1.2.9 · Physics › Newton's Laws & Dynamics

WHAT is tension?

WHY is tension the same throughout a massless string?

Yeh derive kiya jaata hai, assume nahi.

HOW the inextensibility constraint works (the 80/20 idea)

Derivation for two masses over a pulley: Maano $x_{1}, x_{2}$ string ki lengths hain pulley se neeche har mass tak. Total length: $L = x_{1} + x_{2} + (wrap over pulley) = constant$ Time ke saath differentiate karo: $\overset{x}{˙}_{1} + \overset{x}{˙}_{2} = 0 \Rightarrow \overset{x}{¨}_{1} = - \overset{x}{¨}_{2}$

Toh magnitudes equal hain, $∣ a_{1} ∣ = ∣ a_{2} ∣ = a$ . Jab ek mass neeche jaata hai, doosra utni hi amount upar jaata hai. Yahi saara raaz hai.

Figure — Tension in inextensible strings

Worked Example 1 — Two masses, frictionless pulley (Atwood machine)

Masses $m_{1} > m_{2}$ ek ideal pulley par latke hain. $a$ aur $T$ nikalo.

Step 1 — Har mass ka FBD. Kyun? Newton's 2nd law har body par alag se apply hoti hai; tension har ek par ek external force hai.

$m_{1}$ ke liye (neeche $a$ ke saath accelerate karta hai): $m_{1} g - T = m_{1} a$ Is sign ka kyun? Net force = (neeche weight) − (upar tension), aur motion downward hai, toh hum is mass ke liye down ko positive lete hain.

$m_{2}$ ke liye (upar same $a$ ke saath accelerate karta hai, inextensibility ki wajah se): $T - m_{2} g = m_{2} a$ Same $a$ kyun? String inextensible hai ⟹ equal acceleration magnitudes.

Step 2 — Equations add karo $T$ eliminate karne ke liye: $m_{1} g - m_{2} g = (m_{1} + m_{2}) a \Rightarrow a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$

Step 3 — $T$ ke liye back-substitute karo: $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ Yeh step kyun? $a$ ko kisi bhi equation mein plug karo; yahi actual answer hai jo question maangta hai.

Worked Example 2 — Block on table connected to hanging block

$m_{1}$ ek frictionless table par hai, string edge par ek pulley ke upar hai, hanging mass $m_{2}$ hai.

Step 1 — FBD. $m_{2}$ (girta hai): $m_{2} g - T = m_{2} a$ $m_{1}$ (horizontally slide karta hai): $T = m_{1} a$ Kyun? Table par, horizontally sirf $T$ hi act karta hai; weight aur normal vertically cancel ho jaate hain.

Step 2 — Add karo: $m_{2} g = (m_{1} + m_{2}) a \Rightarrow a = \frac{m _{2} g}{m _{1} + m _{2}}, T = \frac{m _{1} m _{2} g}{m _{1} + m _{2}}$

Worked Example 3 — Connected blocks pulled by a force

Do blocks $m_{1}, m_{2}$ ek frictionless floor par hain, unke beech string hai, force $F$ $m_{2}$ ko kheenchti hai (toh $m_{1}$ peeche se drag hota hai).

Step 1 — System pehle ( $a$ nikalo): $F = (m_{1} + m_{2}) a \Rightarrow a = \frac{F}{m _{1} + m _{2}}$ . Ek system ki tarah kyun treat karein? Woh $a$ share karte hain; internal tension puri system ke liye cancel ho jaati hai.

Step 2 — $m_{1}$ ko isolate karo $T$ nikalne ke liye: sirf $T$ hi $m_{1}$ ko pull karta hai, toh $T = m_{1} a = \frac{m _{1} F}{m _{1} + m _{2}}$ $m_{2}$ nahi, $m_{1}$ kyun? Tension hi sirf woh force hai jo $m_{1}$ ko accelerate karti hai, toh yeh directly $m_{1} a$ ke barabar hai.

Common Mistakes (Steel-manned)

Active Recall

Recall Quick self-test (answers cover karo!)

Massless string mein tension uniform kyun hoti hai? → $T_{2} - T_{1} = (d m) a$ , aur $d m = 0$ .
"Inextensible" mathematically kya deta hai? → equal acceleration magnitudes ( $\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}$ ).
Kya string push kar sakti hai? → Nahi, sirf apni length ke along pull kar sakti hai.
Atwood tension formula? → $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ .

Recall Feynman: explain to a 12-year-old

Socho do bachche ek tree branch par rope ke opposite ends par hain. Rope stretch nahi ho sakti, toh agar ek bachcha 1 metre neeche gire, doosra zaroor 1 metre upar jaayega — unhe same amount move karne par majboor kiya jaata hai. Rope bas pull unke beech pass karti hai; yeh kabhi kisi ko dhakka nahi deti, sirf kheenchti hai. Agar ek bachcha bhaari ho, toh woh "apni side mein zyada rope-pull" nahi banata — balki woh tug jeetta hai aur neeche slide karta hai, halke bachche ko upar kheenchta hai. Pull (tension) rope ke along har jagah same hoti hai jab tak rope khud kuch bhi nahi waznti.

Connections

Newton's Second Law — har tension equation wahi $F = ma$ hai jo ek body par apply hoti hai.
Free Body Diagrams — woh tool jo tension ko visible banata hai.
Constraint Relations — "inextensible ⟹ same $a$ " ka generalization.
Atwood Machine — canonical tension problem.
Frictionless Pulleys vs Pulleys with Inertia — jab $T$ dono sides par same nahi hoti.
Normal Force — doosra contact-force cousin (pull ki jagah push karta hai).

Tension string ke relative kis direction mein act karta hai?

String ke along, attached body ko string ki taraf kheenchta hai (kabhi push nahi karta).

Ideal string mein tension same kyun hoti hai throughout?

Yeh massless hai, toh

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

kisi bhi finite

a

ke liye.

"Inextensible" connected masses ke liye kya imply karta hai?

Unke equal acceleration magnitudes hote hain,

∣ a_{1} ∣ = ∣ a_{2} ∣

Atwood machine acceleration formula?

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

Atwood machine tension formula?

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

Acceleration

a

ke saath string par girte hue single mass ke liye

T

kya hai?

T = m (g - a)

m g

se kam.

Ideal pulley ke dono sides par tension same kyun hoti hai?

Pulley massless aur frictionless hai, toh uske paas koi rotational inertia nahi jo difference create kare.

Frictionless table par block (

m_{1}

) aur hanging mass (

m_{2}

): tension?

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