Pulley systems — mechanical advantage: Worked examples

1.2.12 · D3 · Physics › Newton's Laws & Dynamics › Pulley systems — mechanical advantage

Neeche sab kuch teen facts par tika hua hai jo parent note mein pehle se earn kiye ja chuke hain. Mujhe unhe ek baar mein dobara batane do taaki koi symbol unexplained na rahe:

$T$ = tension, woh pull force jo ek rope ke saath saath carry hoti hai, newtons ( $N$ ) mein measure hoti hai. Ek ideal (massless, frictionless) rope har jagah same $T$ carry karti hai. Dekho Tension in strings.
$W$ = weight kisi hanging object ka $= m g$ , jahan $m$ mass hai kilograms mein aur $g \approx 9.8 m/s^{2}$ gravity ki strength hai. $W$ seedha neeche point karta hai.
$F_{net} = ma$ — Newton's Second Law, Newton's Second Law. Agar kuch accelerate nahi ho raha, $a = 0$ , toh $F_{net} = 0$ (ise equilibrium kehte hain).

Bas yahi poora toolkit hai. Dekho yeh kitna door tak jaata hai.

Scenario matrix

Har pulley problem in case classes mein se ek (ya blend) hota hai. Table mein har cell aur woh example listed hai jo use resolve karta hai.

#	Case class	Yeh kaun sa awkward corner test karta hai	Example
A	Fixed pulley, static	MA jo exactly 1 hai (koi multiplication nahi)	Ex 1
B	Movable pulley, static	MA = 2, aur distance penalty	Ex 2
C	Many strands, static	General $IMA = n$ , bada real number	Ex 3
D	Two masses, dynamic ( $a \neq = 0$ )	Atwood: net force ka sign, motion ki direction	Ex 4
E	Degenerate input $m_{1} = m_{2}$	Limiting case: $a \to 0$ , system balance karta hai	Ex 5
F	Non-ideal pulley (friction)	AMA $<$ IMA, efficiency $η < 1$	Ex 6
G	Movable pulley with acceleration	Constraint $a_{1} + a_{2} = 2 a_{p}$ real mein use hota hai	Ex 7
H	Real-world word problem	Target effort ke liye $n$ choose karna, units	Ex 8
I	Exam-style twist	Effort ek angle par lagaya / hidden strand count	Ex 9

Do limiting behaviours jinse tumhe kabhi surprised nahi hona chahiye:

Recall Limiting behaviours jo pocket mein rakhni hain

Jaise jaise strands ki number $n \to \infty$ , effort $F = W / n \to 0$ — lekin rope jo tumhe kheenchni hai, $d = nh \to \infty$ . Infinite aasani ki cost infinite kheenchna hai. Aur jaise jaise do Atwood masses equality ke paas aate hain, $m_{1} \to m_{2}$ , acceleration $a = (m_{1} - m_{2}) g / (m_{1} + m_{2}) \to 0$ : system balance mein ruk jaata hai.

Cell A — Fixed pulley, static (IMA = 1)

Figure — Pulley systems — mechanical advantage

Step 1 — Weight nikalo. $W = m g = 20 \times 9.8 = 196 N$ . Yeh step kyun? Force problems mein actual newtons chahiye, kilograms nahi. Weight woh load hai jo rope ko hold karna hai.

Step 2 — Movable object ko support karne wale strands gino. Figure mein crate dekho. Sirf ek rope segment (amber strand) usse uthta hai. Pulley khud move nahi karta, toh kuch bhi load "share" nahi karta — ek strand $W$ sab carry karta hai. Yeh step kyun? Parent rule: $IMA = n_{supporting strands}$ . Yahan $n = 1$ .

Step 3 — Crate ka equilibrium. $n T = W \Rightarrow 1 \cdot T = 196 \Rightarrow T = 196 N$ . Tumhara haath usi rope ko pakde hua hai, toh $F_{effort} = T = 196 N$ . Yeh step kyun? Ek rope = ek tension; ek fixed pulley sirf rope ki direction bend karta hai (cyan arrows dekho), kabhi uski magnitude nahi.

Step 4 — Mechanical advantage. $IMA = W / F_{effort} = 196/196 = 1$ .

Verify: Units: $N / N = 1$ (dimensionless, ek ratio ke liye sahi). Sanity: ek fixed pulley ek redirector hai — tum neeche kheenchte ho upar uthane ke liye, jo convenient hai, lekin tumhe poora $196 N$ feel hota hai. Parent ke Example 1 se match karta hai.

Cell B — Movable pulley, static (IMA = 2 + distance penalty)

Step 1 — Weight. $W = 30 \times 9.8 = 294 N$ . Kyun? Same reason — mass ko actual downward force mein convert karo.

Step 2 — Supporting strands gino. Figure mein, do strands (dono amber) movable pulley se uthte hain usse hold karne ke liye. $n = 2$ . Kyun? Kyunki pulley load ke saath move karta hai, rope uske neeche wrap hoti hai, toh ek rope ke do segments dono upar kheenchte hain.

Step 3 — Movable pulley + load ka equilibrium. $n T = W \Rightarrow 2 T = 294 \Rightarrow T = 147 N$ . Tumhara effort $F_{effort} = T = 147 N$ . $IMA = W / T = 294/147 = 2$ . Kyun? Do strands weight equally share karte hain, har ek $147 N$ carry karta hai.

Step 4 — Distance penalty. Load ko $h = 0.5 m$ uthane ke liye, do strands mein se har ek ko $0.5 m$ chota hona chahiye. Woh $2 \times 0.5 = 1.0 m$ slack tumhare haath se guzarta hai: $d_{effort} = nh = 2 \times 0.5 = 1.0 m$ . Kyun? Constraint relations — rope ki total length fixed hai, toh load side par shortening tumhare side par kheenchne mein appear hoti hai.

Verify (energy, Work–Energy Theorem): $W_{in} = F_{effort} \times d_{effort} = 147 \times 1.0 = 147 J$ . $W_{out} = W \times h = 294 \times 0.5 = 147 J$ . ✔ Equal — koi free lunch nahi.

Cell C — Many strands, static (general IMA = n)

Step 1 — Weight. $W = 200 \times 9.8 = 1960 N$ .

Step 2 — Strand count. Diya hua hai $n = 5$ . $IMA = 5$ . Kyun? Sirf movable block se attached (ya wrapping) strands gino — fixed-pulley redirections $n$ mein add nahi hote.

Step 3 — Effort. $F_{effort} = W / n = 1960/5 = 392 N$ . Kyun? Paanch strands $1960 N$ share karte hain ⇒ har ek $= 392 N$ .

Step 4 — Rope pulled. $d = nh = 5 \times 0.4 = 2.0 m$ .

Verify: $W_{in} = 392 \times 2.0 = 784 J$ ; $W_{out} = 1960 \times 0.4 = 784 J$ . ✔ Ek akela insaan ( $\sim 400 N$ push kar sakta hai) ek $2$ -tonne-force load barely manage kar sakta hai — yahi high $n$ ka point hai.

Cell D — Dynamic Atwood ( $a \neq = 0$ )

Step 1 — Ek rope, ek tension; ek string, ek speed. Ideal rope ⇒ dono sides par same $T$ . Inextensible rope ek fixed pulley par ⇒ dono masses same $∣ a ∣$ share karte hain. Dekho Atwood machine. Kyun? Tension in strings + Constraint relations: agar ek side $x$ se neeche jaaye, doosri $x$ se upar jaati hai.

Step 2 — Heavier mass $m_{1}$ ka free-body (woh girta hai, toh iske liye neeche ko positive lo): $m_{1} g - T = m_{1} a \Rightarrow 6 (9.8) - T = 6 a$ . Kyun? $m_{1}$ par net downward force = weight minus tension jo upar kheench rahi hai.

Step 3 — Lighter mass $m_{2}$ ka free-body (woh uthta hai, upar positive lo): $T - m_{2} g = m_{2} a \Rightarrow T - 4 (9.8) = 4 a$ .

Step 4 — Dono equations add karo $T$ cancel karne ke liye: $(m_{1} - m_{2}) g = (m_{1} + m_{2}) a \Rightarrow a = \frac{( 6 - 4 ) ( 9.8 )}{6 + 4} = \frac{2 ( 9.8 )}{10} = 1.96 m/s^{2}$ . Kyun add karo? Add karne se $T$ instantly khatam ho jaata hai, $a$ isolate ho jaata hai.

Step 5 — $T$ ke liye back-substitute karo: $T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 ( 6 ) ( 4 ) ( 9.8 )}{10} = 47.04 N$ .

Verify: $a = 1.96 < g = 9.8$ ✔ (kuch weight rope se held hai, toh yeh free-fall se slower girta hai). Check karo $T$ , $m_{2} g = 39.2 N$ aur $m_{1} g = 58.8 N$ ke beech hai: $39.2 < 47.04 < 58.8$ ✔ (tension light weight se zyada honi chahiye use uthane ke liye, aur heavy weight se kam taaki woh gir sake).

Cell E — Degenerate input ( $m_{1} = m_{2}$ )

Step 1 — Derived formulas use karo (unhe sabhi inputs handle karne chahiye, including equal ones): $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{0 \cdot g}{10} = 0 m/s^{2}$ . Kyun? Jab driving imbalance $(m_{1} - m_{2})$ khatam ho jaata hai, kuch bhi accelerate karne ke liye koi net force nahi.

Step 2 — Tension. $T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 ( 5 ) ( 5 ) ( 9.8 )}{10} = 49 N$ .

Step 3 — Equilibrium se cross-check karo. $a = 0$ ke saath, har side sirf ek hanging weight hai: $T = m g = 5 \times 9.8 = 49 N$ . ✔ Kyun check karo? Ek achhe formula ko degenerate case mein simple static answer par reduce hona chahiye — isi tarah tum use trust karte ho.

Verify: $a = 0$ aur $T = 49 N$ bilkul "do equal weights balance karte hain" wali picture se match karta hai. System kisi bhi position par equilibrium mein hai (neutral balance).

Cell F — Non-ideal pulley (friction, AMA < IMA)

Step 1 — Ideal effort (frictionless). $F_{ideal} = W / IMA = 800/4 = 200 N$ . Kyun? Yeh geometry akele se set kiya hua floor hai.

Step 2 — Actual MA pane ke liye efficiency use karo. $η = \frac{AMA}{IMA} \Rightarrow AMA = η \cdot IMA = 0.75 \times 4 = 3$ . Kyun? Friction tumhare input work ka kuch hissa waste karta hai, toh achieved multiplication ideal se neeche aa jaata hai.

Step 3 — Actual effort. $F_{actual} = W / AMA = 800/3 \approx 266.67 N$ . Kyun? Ab tum same $800 N$ ek smaller effective advantage se support kar rahe ho, toh tumhe harder push karna padta hai.

Verify: $266.67 > 200$ ✔ (friction real machines ko hamesha harder banata hai, kabhi easier nahi). Efficiency cross-check: $η = F_{ideal} / F_{actual} = 200/266.67 = 0.75$ ✔.

Cell G — Movable pulley with acceleration (constraint real mein use hota hai)

Step 1 — Ek movable pulley ke liye constraint yaad karo. $a_{1} + a_{2} = 2 a_{p}$ . Yeh kyun aur "sab equal" kyun nahi? Movable pulley par rope do segments hai; pulley ki position dono ki total par depend karti hai. Yeh "rope length = constant" ko do baar differentiate karne se aata hai — Constraint relations. Ise kabhi guess mat karo.

Step 2 — $a_{p}$ solve karo. $a_{p} = \frac{a _{1} + a _{2}}{2} = \frac{3 + 1}{2} = 2 m/s^{2}$ . Kyun average? Har end equally contribute karta hai ki do-segment loop kitni tezi se short hoti hai, toh pulley unki mean rate par move karta hai.

Verify: $2$ , $1$ aur $3$ ke beech hai ✔ — ek genuine average. Degenerate check: agar $a_{1} = a_{2} = 1$ , toh $a_{p} = 1$ (poori cheez rigidly move karti hai, sensible). Agar ek end fixed hai ( $a_{2} = 0$ ), $a_{p} = a_{1} /2$ — "pull $2 h$ to lift $h$ " distance rule ko do baar differentiate karne se match karta hai. ✔

Cell H — Real-world word problem (n choose karo, units dekho)

Step 1 — Weight. $W = 150 \times 9.8 = 1470 N$ .

Step 2 — Required condition. Use $F_{effort} = W / n \leq F_{max}$ chahiye, yani $n \geq W / F_{max} = 1470/250 = 5.88$ . Kyun $\geq$ ? Zyada strands = kam effort; use kam se kam itna chahiye ki effort uski limit tak aaye.

Step 3 — Poore strand count tak round up karo. $n = 6$ (kyunki $n$ ek whole number hona chahiye aur $5.88$ round up hota hai). Kyun round up, round down kyun nahi? $n = 5$ deta hai $F = 1470/5 = 294 N > 250 N$ — bahut hard. Toh $n = 6$ .

Step 4 — $n = 6$ par effort check karo. $F_{effort} = 1470/6 = 245 N \leq 250 N$ ✔.

Step 5 — Rope pulled. $d = nh = 6 \times 2.5 = 15 m$ .

Verify: Energy: $W_{in} = 245 \times 15 = 3675 J$ ; $W_{out} = 1470 \times 2.5 = 3675 J$ ✔. $d$ ke units: (strands, dimensionless) $\times$ metres $=$ metres ✔.

Cell I — Exam-style twist (angle par effort / hidden count)

Step 1 — Load ke equilibrium se tension nikalo. Do vertical supporting strands load hold karte hain: $2 T = W \Rightarrow T = 120/2 = 60 N$ . Kyun load se, haath se kyun nahi? Strand tension woh hai jo movable pulley ko support karna hai, aur ek ideal fixed guide pulley sirf redirect karta hai — $T$ change nahi karta.

Step 2 — Angle actually kya change karta hai? Ek ideal rope mein tension ki magnitude har jagah same hoti hai, toh tumhare pulling strand ke saath force abhi bhi $60°$ direction chahe ho, $60 N$ hi rehta hai. Angle sirf woh direction change karta hai jis mein tum pull karte ho, tension nahi. Student B sahi hai. Kyun? Ek frictionless guide pulley tension redirect karta hai; woh use kabhi rescale nahi karta (parent Mistake: "tension pulley ke paas se change hoti hai" — nahi hoti).

Step 3 — Pull ki magnitude. $F_{effort} = T = 60 N$ , vertical se $60°$ ki direction mein. $IMA = W / F_{effort} = 120/60 = 2$ — angle se unchanged.

Verify: Tumhare pull ki horizontal component $T sin 60° = 60 \times 0.866 = 51.96 N$ hai aur vertical $T cos 60° = 30 N$ — resultant $51.9 6^{2} + 3 0^{2} = 60 N$ hai ✔, confirming karta hai ki pull magnitude exactly $T$ hai, angle notwithstanding. (Load ka support abhi bhi do vertical strands se aata hai, is angled strand se nahi.)

Recall Saare cells par rapid self-test

Fixed pulley MA ::: $1$ — sirf redirection. Movable pulley MA ::: $2$ — do strands load share karte hain. $n = 5$ strands, $W = 1960 N$ ke liye effort ::: $392 N$ . $6$ aur $4 kg$ ke liye Atwood $a$ ::: $1.96 m/s^{2}$ . $6$ aur $4 kg$ ke liye Atwood $T$ ::: $47.04 N$ . Equal masses Atwood $a$ ::: $0$ (balance). $IMA = 4$ , $η = 0.75$ ⇒ AMA ::: $3$ , toh effort $= W /3$ . Movable-pulley constraint $a_{1} = 3, a_{2} = 1$ ke saath ::: $a_{p} = 2 m/s^{2}$ . Angled effort strand tension change karta hai? ::: Nahi — magnitude unchanged, sirf direction.

Connections

Parent topic — woh core theory jo yeh examples exercise karti hain.
Newton's Second Law — har free-body diagram (Ex 4, 5).
Tension in strings — ek rope, ek tension (Ex 1, 9).
Constraint relations — $a_{1} + a_{2} = 2 a_{p}$ rule (Ex 7).
Atwood machine — Cells D aur E.
Friction — Cell F mein efficiency loss.
Work–Energy Theorem — har energy verification.
Inclined plane mechanical advantage — doosri machine mein same force-for-distance trade.

Worked examples — Pulley systems — mechanical advantage