Pulley systems — mechanical advantage: Visual walkthrough

1.2.12 · D2 · Physics › Newton's Laws & Dynamics › Pulley systems — mechanical advantage

Hum poori kahani ek running example ke around banate hain: ek movable pulley jisme do strands hain, phir use $n$ strands tak le jaate hain, phir edge cases check karte hain.

Step 1 — Rope actually kya kar rahi hai? (Tension, zero se shuru)

KYA. Kisi bhi pulley se pehle, ek seedhi rope ka piece dekho jisme neeche ek weight $W$ latka hai aur upar tumhara haath kheench raha hai.

KYUN. Har pulley result ek idea par tikaa hai: "rope kitni zor se kheenchti hai" — yeh number kya hai? Uss number ko tension kehte hain, likha jaata hai $T$ . Isse kahin bhi use karne se pehle define karna zaroori hai.

PICTURE. Figure mein rope weight ko upar kheenchti hai (arrow upar, size $T$ ) aur tumhare haath ko neeche kheenchti hai (arrow neeche, size $T$ ) — same length ke arrows, opposite directions.

Figure — Pulley systems — mechanical advantage

Kyunki latka hua weight hil nahi raha, upar ki pull bilkul neeche ki gravity ko cancel karni chahiye:

$rope upar kheenchti hai T = gravity neeche kheenchti hai W$

Yahan $T$ woh hai jo rope carry karti hai, $W = m g$ weight hai ( $m$ = mass kg mein, $g \approx 9.8 m/s^{2}$ ). Ek strand hai, toh tum poora weight feel karte ho. Abhi koi advantage nahi — yeh woh baseline hai jise hume beat karna hai. Dekho Tension in strings.

Step 2 — Tension ek rope ke poore length mein SAME kyun hoti hai?

KYA. Rope ka ek chhota sa imaginary tukda lo aur pucho: uski kya forces act kar rahi hain?

KYUN. Agar hum prove kar sakein ki tension ek point se doosre point tak change nahi hoti, toh rope ko jahan bhi pakdo — load ke paas, ceiling ke paas, tumhare haath par — woh number $T$ same rahega. Yahi ek fact poori derivation ka engine hai.

PICTURE. Tukde par baayi taraf ki rope tension $T_{L}$ se baayi taraf kheenchti hai, aur daayeen taraf ki rope tension $T_{R}$ se daayeen taraf kheenchti hai.

Ab tukde par Newton's Second Law apply karo:

$tukde par net force T_{R} - T_{L} = uska mass m_{tukda} \cdot a$

Ideal rope massless hoti hai, toh $m_{tukda} = 0$ :

$T_{R} - T_{L} = 0 \cdot a = 0 \Rightarrow T_{R} = T_{L}$

Step 3 — Pulley bhi $T$ ko change kyun nahi karti?

KYA. Ab rope ko ek wheel ke upar se mode karo — ek pulley. Kya corner se guzarne par tension change hoti hai?

KYUN. Hum aage bar bar "pulley ke dono sides par same $T$ " use karne wale hain. Agar pulley secretly $T$ ko scale kar deti, toh counting trick fail ho jaati. Toh ab ise settle karte hain.

PICTURE. Rope baayi side se upar aati hai (tension $T$ ), ek ideal (frictionless, massless) wheel ke upar wrap hoti hai, aur daayeen side se neeche jaati hai. Wheel freely spin kar sakta hai, isliye rope ke saath koi resistance nahi — sirf direction bend hota hai.

$T_{pulley ke baayi taraf} = T_{pulley ke daayeen taraf} = T$

Step 4 — Load ko MOVABLE pulley par lagao (yahi hai poora trick)

KYA. Weight ko rope ke end par latkaane ki bajay, ek pulley ke axle par latkaao, aur ek rope daalo: ek end ceiling se baandho, movable pulley ke upar se, phir tumhare haath tak.

KYUN. Yahi poora trick hai. Movable pulley dekho: iske upar se do rope strands nikalti hain — ceiling strand aur tumhare-haath wali strand. Dono ek hi rope ki hain, isliye (Steps 2–3) dono same $T$ carry karti hain. Ab do strands ek load share karti hain.

PICTURE. Movable pulley (jo weight $W$ carry kar raha hai) do upar ke arrows se tika hai, har ek size $T$ ka. Gino: 1, 2.

Movable pulley + load ka free-body diagram banao (kuch accelerate nahi kar raha, static lift):

$strand 1 upar T + strand 2 upar T = weight neeche W$

$2 T = W \Rightarrow T = \frac{W}{2}$

Tum ek strand pakde ho, isliye tumhari effort hai $F_{effort} = T = W /2$ . Tum weight $W$ ko sirf uska aadha kheenchkar uthaa rahe ho.

Step 5 — Generalise karo: $n$ strands ⇒ IMA $= n$

KYA. Aisa system banao jahan ek hi rope ke $n$ strands movable block se upar uthein (zyada wheels, same ek rope).

KYUN. Equilibrium equation mein bas aur identical upar-waale arrows aa jaate hain. Physically kuch naya nahi — sirf count badalta hai. Yahin clean formula saamne aata hai.

PICTURE. $n$ equal upar-ke arrows (har ek $T$ ) block ko pakde hain; ek neeche-ka arrow $W$ . Yahan hum $n = 4$ draw karte hain.

$n strands, har ek T T + T + \dots + T = W \Rightarrow kul upar-ki pull n T = W$

$T = \frac{W}{n} \Rightarrow IMA = \frac{W}{T} = n$

Step 6 — Catch yeh hai: tumhe ZYADA door kheenchna padega (energy check)

KYA. Load ko height $h$ tak uthao. Tumhare haath ko kitni rope reel in karni padegi?

KYUN. Force kabhi free nahi milti. Agar same pull distance ke liye $n$ guna force milti, toh tum energy kuch nahi se bana rahe hote — yeh impossible hai. Geometry ek price force karti hai.

PICTURE. Jab block $h$ uthta hai, $n$ supporting strands mein se har ek $h$ se chhoti ho jaati hai. Woh saari removed slack tumhare ek haath se guzarti hai.

$d_{effort} = strands n \times load rise h = nh$

Ab Work–Energy Theorem se energy check karo:

$W_{in} = W / n F_{effort} \times nh d_{effort} = \frac{W}{n} \cdot nh = W h = W ko h tak uthana W_{out} ✓$

Step 7 — Edge & degenerate cases (kabhi surprise mat ho)

Har woh scenario jo reader face kar sakta hai, drawn:

$n = 1$ — single fixed pulley. Sirf ek strand load ko pakde hai (woh rope end se latka hai, wheel ceiling se bolted hai). $IMA = 1$ : koi multiplication nahi, pure direction change (neeche kheencho → load upar jaata hai). Convenience hai, power nahi.
$n = 2$ — single movable pulley (Step 4). Aadha effort, double pull.
Zero effort $T \to 0$ . Tab $W = n T \to 0$ : bina pull ke, sirf ek weightless load tika rehta hai. Consistent hai — koi free lifting nahi.
Real pulley (mass + friction). Ab Steps 2–3 thoda fail karte hain: strands alag $T$ carry karte hain, isliye actual MA girta hai. Efficiency $η = AMA / IMA < 1$ . Dekho Friction.
Dynamic case (moving system). $a \neq = 0$ set karo: yahi Atwood machine hai. Ek movable pulley ke liye Constraint relations ke saath, $a_{1} + a_{2} = 2 a_{pulley}$ — pulley apne dono rope ends ke average par move karta hai.

Ek picture mein summary

Ek rope → ek tension $T$ (Steps 1–3). Load ko movable pulley par latkaao taaki $n$ strands use share karein → $n T = W$ → $IMA = n$ (Steps 4–5). Distance mein payment karo: $d_{effort} = nh$ , $W_{in} = W_{out}$ banaye rakho (Step 6).

Recall Feynman retelling (seedhe words mein)

Ek rope ek aalsi messenger hai: woh apni poori length mein same pull pass karti hai, aur ek smooth wheel bas ise corner mude deta hai bina uss pull ko change kiye. Toh agar tum bhaari cheez ko ek moving wheel par baandhte ho aur rope ko is tarah loop karte ho ki do strands wheel ko latkaayein, toh dono strands weight share karte hain — har ek aadha leta hai, aur tumhara haath sirf aadha feel karta hai. Chautha chahiye? Chaar strands arrange karo. Moving wheel ko pakde hue strands ki sankhya bilkul utni hi baar easier pull ban jaati hai. Lekin rope imaandaar hai: cheez ko ek kadam uthane ke liye, un strands mein se har ek ko ek kadam chhota hona hoga, aur woh saari rope tumhare haath mein pile hoti hai — toh tum utne hi kadam reel in karte ho jitne strands hain. Aasaan pull, lambaa pull, same total effort. Universe ko cheat karne ka koi raasta nahi.

Connections

Newton's Second Law — Step 2 mein $F_{n e t} = ma$ ek massless snippet par use hota hai.
Tension in strings — uniform- $T$ ki backbone (Steps 1–3).
Work–Energy Theorem — $W_{in} = W_{o u t}$ check (Step 6).
Constraint relations — movable-pulley average rule (Step 7).
Atwood machine — dynamic $a \neq = 0$ case.
Friction — kyun real AMA $<$ IMA.
Inclined plane mechanical advantage — ek aur machine mein same force-for-distance trade.

Visual walkthrough — Pulley systems — mechanical advantage