Pulley systems — mechanical advantage: Question bank

1.2.12 · D5 · Physics › Newton's Laws & Dynamics › Pulley systems — mechanical advantage

Ye bank un traps ko thokta hai jo Pulley systems — mechanical advantage ke andar rehte hain. Shuru karne se pehle, neeche diya vocabulary pakka kar lo — har trap answer inhi exact symbols use karta hai.

Definition Is page ke symbols (pehle padho)

$F_{load}$ — wo force jo machine ko support ya lift karni hai, yaani jis object ko tum utha rahe ho uska weight, newtons ( $N$ ) mein. Neeche figures mein orange block ka downward pull socho.
$F_{effort}$ — wo force jo tumhara haath rope ke free end par lagata hai, woh bhi newtons mein. Wo red down-arrow socho jise tum kheenchte ho.
$Q$ (load ka weight ek number ke roop mein) — jab hume " $n$ strands load share karte hain" jaisi equation mein load ke weight ke liye ek single letter chahiye, hum $Q$ likhte hain; yeh exactly $F_{load}$ hai, bas ek chhota naam. (Hum yahan deliberately $W$ letter reuse nahi karte — next box mein dekho kyun.)
$E_{in}$ , $E_{out}$ (energy / work) — wo energy jo tum haath se daaalte ho ( $E_{in}$ ) aur wo energy jo load tak pahunchti hai ( $E_{out}$ ), dono force × distance ke barabar hain, joules ( $J$ ) mein measure hoti hain.
$T$ — rope mein tension (kheench), newtons mein.
$n$ — movable block ko support karne wale rope strands ki sankhya.
MA (Mechanical Advantage) — force-multiplying factor: load force tumhari effort force se kitni baar badi hai. $MA = \frac{F _{load}}{F _{effort}}$ Ise aise padho: "apni $F_{effort}$ ke pratyek newton ke liye main kitna bhaari load ( $F_{load}$ ) thaam sakta hoon?" MA $4$ ka matlab hai $10 N$ ki pull $40 N$ ka load support karti hai.

Visual warm-up — count karne se pehle strands dekho

Intuition Figure 3 — easy pull zyada lambi pull kyun maangta hai (trade-off, drawn)

Yeh topic ka 3Blue1Brown heart hai. Panel by panel, movable pulley utha ke load $h$ height tak charhta hai. Do green segments ko dekho, har ek $h$ se chhota hota hai (do chhote brackets). Woh saari chhoti hui rope kahin jaani chahiye — woh tumhare haath se reel in hoti hai. Do brackets add karo: tumhara haath $d_{effort} = 2 h$ travel karta hai. Force aadhi, distance double: shaded blue "energy rectangle" (force × distance) input aur output dono side par same area hai. Woh barabar area hi energy conservation hai, drawn.

Alt-text: left panel load lifted by h with two strands each shortening by h; right panel two equal-area rectangles, one tall-thin (full weight × small h) and one short-wide (half weight × 2h), shown equal to depict $E_{in} = E_{out}$ .

Movable-pulley constraint derive karna (traps mein referenced)

Neeche kai traps $a_{1} + a_{2} = 2 a_{pulley}$ rule par hinge karte hain. Ise kabhi guess nahi karna — yahan se yeh aata hai, taaki tum ise defend kar sako.

Formula Teen honest steps mein derivation

WHAT (length likho): pulley ke neeche har segment ki length $(y_{p} - y_{end})$ hai, aur do hain, isliye $L = (y_{p} - y_{1}) + (y_{p} - y_{2}) + const = 2 y_{p} - y_{1} - y_{2} + const .$ WHY differentiate: $L$ constant hai, isliye iska rate of change zero hai — ek constant differentiate karne se $0$ milta hai, jo ek length fact ko velocity fact mein badal deta hai. Accelerations tak pahunchne ke liye do baar karo (velocity ka rate of change). Ek baar differentiate karke: $0 = 2 \overset{y}{˙}_{p} - \overset{y}{˙}_{1} - \overset{y}{˙}_{2} .$ Phir se: $0 = 2 \overset{y}{¨}_{p} - \overset{y}{¨}_{1} - \overset{y}{¨}_{2} \Rightarrow a_{1} + a_{2} = 2 a_{pulley} .$ KAISA LAGTA HAI (Figure 4): agar tum ek end fast neeche kheencho aur doosra still rakho, pulley tumhari speed ki half par sink karti hai — yeh hamesha do ends ke average par move karti hai. Isliye " $2$ " wahan hai: do segments har metre mein share karte hain jo pulley travel karta hai.

True or false — justify karo

True or false: Ek single fixed pulley tumhe 1 se zyada mechanical advantage deta hai.

False. Sirf ek strand load support karta hai, isliye

n = 1

aur

IMA = 1

; ek fixed pulley force redirect karta hai (neeche kheencho, upar lift karo) lekin kabhi multiply nahi karta.

True or false: Ek ideal system mein, rope path mein doosra fixed pulley add karne se mechanical advantage badh jaata hai.

False. Ek fixed pulley sirf redirection add karta hai — woh movable block ko support karne wala strand nahi badhata.

IMA = n

supporting strands par depend karta hai, total pulley count par nahi.

True or false: Agar ek pulley system ka

IMA = 4

hai, toh tumhe input se four times energy output milti hai.

False. Energy conserved hai:

E_{in} = E_{out}

. Tumhe 4× force milti hai, lekin tum 4× distance kheencho, isliye energy product (force × distance) unchanged hai — koi extra energy nahi aati.

True or false: Ek continuous ideal rope mein, tension rope jis bhi pulley se guzarti hai uske dono sides par same hoti hai.

True. Rope ka ek tiny piece lo: Newton's 2nd law kehta hai

F_{net} = ma = 0

(mass zero hai), isliye uske left pull aur right pull bilkul equal hone chahiye — matlab

T

poore rope mein same hai. Ek frictionless pulley sirf direction change karta hai, magnitude nahi.

True or false: Pulley ko heavier banana actual mechanical advantage badhata hai.

False — yeh ise ghatata hai. Ek massive pulley khud ko lift aur spin karne ke liye extra tension demand karta hai, isliye real force saving shrink ho jaati hai:

AMA

drop hota hai, aur efficiency

η = AMA / IMA

1 se neeche girti hai.

True or false:

m_{1} = m_{2}

wale Atwood machine mein, rope mein tension ek block ke weight ke barabar hoti hai.

True. Equal masses ke saath

a = 0

, isliye har block equilibrium mein hai: tension exactly ek weight balance karti hai,

T = m g

. System balanced hai, accelerate nahi kar raha.

True or false: Ek system ka ideal mechanical advantage uski efficiency batata hai.

False.

IMA

sirf geometry se set hota hai; efficiency actual ko ideal se compare karti hai,

η = AMA / IMA

, aur sirf real pulleys (friction/mass ke saath) ke liye 1 se neeche jaati hai. Ek ideal system ka

IMA

bada ho sakta hai phir bhi

η = 1

True or false: Movable-pulley setup mein, load aur tumhara haath hamesha same distance move karte hain.

False. Agar 2 strands load support karte hain, tumhara haath twice utna door move karta hai jitna load utha, kyunki dono strands lift height

h

se chhote hone chahiye, isliye tum

d = 2 h

rope reel in karte ho (Figure 3 dekho).

True or false: Movable pulley ke liye, pulley ka acceleration do rope-end accelerations ka sum ke barabar hota hai.

False. Yeh average ke barabar hai:

a_{pulley} = \frac{1}{2} (a_{1} + a_{2})

, Figure 4 mein constant rope length se derived (Constraint relations).

Error dhundho

Ek student likhta hai: "Do masses ek fixed pulley par, isliye heavy side par tension light side se zyada hai." Error dhundho.

Ek ideal rope matlab ek tension: massless-rope piece ka

F_{net} = 0

hai, dono sides par equal pull force karta hai. Net force heavy side par zyada hai (uska weight jeetta hai), lekin string tension

T

har jagah identical hai.

Ek student ek Atwood machine setup karta hai aur likhta hai

m_{1} g - T = m_{1} a

aur

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

, phir conclude karta hai

a = (m_{1} - m_{2}) g

. Error dhundho.

Unhone total mass se divide karna bhool gaye. Equations add karne par

T

cancel ho jaata hai aur milta hai

(m_{1} - m_{2}) g = (m_{1} + m_{2}) a

, isliye

a = (m_{1} - m_{2}) g / (m_{1} + m_{2})

hai,

(m_{1} - m_{2}) g

nahi.

Ek student claim karta hai: "Ek movable pulley ka

IMA = 2

hai, isliye agar main

100 N

ka load attach karun, toh mujhe

200 N

tension feel hogi." Error dhundho.

Ulta hai.

IMA = 2

matlab do strands weight split karte hain, isliye tera effort tension aadha hai:

T = Q /2 = 50 N

, double nahi.

Ek student block-and-tackle ke liye strands count karta hai: "5 pulleys hain, isliye

IMA = 5

." Error dhundho.

IMA

pulley count nahi hai. Sirf woh rope segments count karo jo movable block ko support karte hain; fixed pulleys redirection add karte hain, strands nahi.

Ek student kehta hai: "Load

1 m

utha, isliye main kisi bhi pulley system mein

1 m

rope apne haath se pull karta hun." Error dhundho.

Sirf tab true hai jab

n = 1

ho.

n

supporting strands ke liye, sab

n

1 m

se chhota hona hai, isliye tum ==

n

metres== rope kheenchte ho (

n = 2

case Figure 3 mein dikha hai).

Ek student movable pulley ke liye constraint

a_{1} = a_{2} = a_{pulley}

likhta hai. Error dhundho.

Accelerations equal hain yeh guess karna galat hai. Constant rope length se milta hai

L = 2 y_{p} - y_{1} - y_{2} + const

; do baar differentiate karne par ==

a_{1} + a_{2} = 2 a_{pulley}

== milta hai — pulley do ends ke average par move karti hai (Figure 4).

Ek student kehta hai: "Kyunki MA mujhe kam force mein zyada lift karne deta hai, ek pulley system mere favor mein energy conservation violate karta hai." Error dhundho.

Koi violation nahi — tum force ko distance se trade karte ho. Work force × distance hai; zyada rope jo tum reel in karte ho (

n

times lambi) exactly force reduction (

1/ n

) cancel kar deti hai, isliye

E_{in} = E_{out}

(Figure 3 mein equal blue areas).

Why questions

Ek single massless rope mein tension har jagah same kyun hoti hai?

Koi bhi tiny segment lo: Newton's 2nd law deta hai

F_{net} = ma

, aur

m = 0

ke saath yeh

F_{net} = 0

ho jaata hai, isliye har side ka pull equal hona chahiye. Ek frictionless pulley sirf us pull ko redirect karta hai, uski size fixed rakhta hai.

Mechanical advantage supporting strands ki sankhya ke barabar kyun hota hai?

Kyunki

n

equal-tension strands weight share karte hain: unke upward pulls load balance karte hain,

n T = Q

, isliye har ek

Q / n

carry karta hai. Tera effort ek strand ka tension hai

F_{effort} = T = Q / n

, jo deta hai

IMA = Q / T = n

Mechanical advantage zyada hone par zyada door kyun kheenchna padta hai?

Kyunki energy

=

force × distance conserved hai. Us product ko fixed rakhne ke liye jab force factor

n

se drop ho, distance ko ==factor

n

se badhna== chahiye — exactly Figure 3 ke do equal-area rectangles.

Do Atwood equations ko subtract karne ki jagah add kyun karte hain?

Add karne se ==unknown tension

T

instantly eliminate== ho jaata hai — yeh ek equation mein

+ T

aur doosre mein

- T

ke roop mein aata hai, isliye cancel ho jaate hain, ek single equation for

a

bachti hai.

Ek fixed pulley phir bhi

n = 1

kyun count hota hai

n = 0

ki jagah?

Load rope ke ek strand se latakta hai, jo uska poora weight support karta hai (Figure 1). Woh single supporting strand

n = 1

deta hai; pulley bas uski direction change karta hai.

Ek real pulley ki efficiency 100% se neeche kyun hoti hai?

Kuch input work friction overcome karne aur pulley ki apni mass lift karne mein jaati hai, isliye useful output input se kam hai:

E_{out} < E_{in}

, jo

η = E_{out} / E_{in} < 1

banata hai.

Movable pulley — fixed nahi — force multiplication ka source kyun hai?

Sirf movable pulley mein load hold karne ke liye multiple strands hoti hain (uske up-arrows

Q

share karte hain, Figure 2); ek fixed pulley sirf rope ko upar se guzarne deta hai, ek strand phir bhi poora load carry karta hai.

Movable-pulley constraint

a_{1} + a_{2} = 2 a_{pulley}

mein "2" kyun aata hai?

Kyunki do rope segments movable pulley tak neeche jaate hain (Figure 4). Length

L = 2 y_{p} - y_{1} - y_{2} + const

; do baar differentiate karne par pulley ke acceleration par factor

2

aa jaata hai.

Edge cases

Edge case: Agar tum ek fixed pulley ke upar load lift karne ke liye effort rope seedha neeche kheenchte ho, aur

n = 1

hai, toh mechanical advantage kya hai?

IMA = 1

: koi multiplication nahi, aur tumhe poore weight ke barabar force se kheenchna hoga — lekin kam se kam tum neeche kheench ke apna body weight conveniently use kar sakte ho.

Edge case: Atwood machine mein,

m_{2} \to 0

hone par acceleration ka kya hota hai?

a = (m_{1} - m_{2}) g / (m_{1} + m_{2}) \to g

. Doosri side kuch nahi hone par,

m_{1}

free fall mein hai — rope slack ho jaati hai, tension

\to 0

Edge case: Atwood machine mein,

m_{1} \to m_{2}

hone par kya hota hai?

a \to 0

aur

T \to m g

. System balanced hai; koi bhi position stable rahti hai (ideal case mein neutral equilibrium).

Edge case: Ek movable pulley ka

IMA

kya hai agar tum free end neeche kheencho versus upar — kya answer change hoga?

Nahi.

IMA

supporting strands count karta hai (

n = 2

), jo ek geometric fact hai, independent of jis direction mein tum free end kheenchte ho.

Edge case: Agar rope ka significant mass ho, toh kya tension phir bhi uniform hai?

Nahi. Ek massive rope ko khud accelerate karne ke liye net force chahiye, isliye

F_{net} \neq = 0

aur tension uske saath vary karti hai; "one rope, one tension" rule sirf ideal, massless case mein hold karta hai.

Edge case: Ek movable pulley aise hold ki jaaye ki load move na kare (

a = 0

). Kya phir bhi mechanical advantage hai?

Haan.

IMA

ek static/geometric property hai: equilibrium mein bhi,

n

strands weight share karte hain, isliye

F_{effort} = Q / n

Edge case: Agar load

n

strands ke saath

h

neeche uttara jaaye lift karne ki jagah, toh effort distance kya hai?

Tum

nh

rope bahar nikalte ho (same magnitude, opposite direction). Distance trade-off symmetric hai; energy bookkeeping phir bhi balance hoti hai.

Recall Ek-line survival summary

Supporting strands count karo $IMA = n$ ke liye; ek ideal rope mein ek tension; higher MA matlab proportionally lambi pull; energy hamesha conserved hai. Question :::- Agar koi claim distance penalty ignore karta hai ya ek ideal pulley ke across extra tension invent karta hai, woh galat hai.

Connections

Parent: Pulley systems — mechanical advantage — wo full derivation jise ye traps test karte hain.
Newton's Second Law — kyun massless rope $F_{net} = 0$ force karta hai.
Tension in strings — one-rope-one-tension rule.
Work–Energy Theorem — $E_{in} = E_{out}$ bookkeeping har "no free lunch" answer ke peeche.
Constraint relations — $a_{1} + a_{2} = 2 a_{pulley}$ constraint (Figure 4).
Atwood machine — dynamic edge cases ka source.
Friction — kyun real efficiency 1 se neeche girti hai.
Inclined plane mechanical advantage — wahin force-for-distance trade kahin aur.