Pulley systems — mechanical advantage: Exercises

1.2.12 · D4 · Physics › Newton's Laws & Dynamics › Pulley systems — mechanical advantage

Yahan sab kuch do ideas par tikaa hai jo parent note ne scratch se earn kiye the:

Ek ideal rope ⇒ ek tension $T$ (ek massless rope jisme $F_{net} = 0$ ho, aur ek frictionless pulley sirf direction badalta hai).
$IMA = n$ , jahan $n$ un rope strands ki sankhya hai jo movable block ko support karte hain.

Symbols aane se pehle, words ka ek reminder:

$W$ = load ka weight (ek force, newtons mein, = mass $\times g$ ).
$T$ = tension = rope mein carry hone wali pull force (ek ideal rope par har jagah same).
$F_{effort}$ = woh force jo tumhara haath lagata hai. Tum ek strand pakdte ho, isliye $F_{effort} = T$ .
$h$ = load kitna upar uthta hai; $d$ = tumhara haath kitni rope reel in karta hai.

L1 — Recognition

(Kya tum diagram padh ke sahi se count kar sakte ho?)

Neeche wali figure mein dono L1 configurations ek saath dikhaye gaye hain — baayi panel Problem 1.1 ka fixed pulley hai, daayeen panel Problem 1.2 ka movable pulley hai. Jo problem solve kar rahe ho, us labeled panel ko dekho.

Figure — Pulley systems — mechanical advantage

Problem 1.1

Ek rope ek fixed pulley ke upar se guzarti hai jo ceiling mein bolted hai. Tum ek end ko neeche kheenchte ho; load doosre end se latka hua hai. Ideal mechanical advantage kya hai?

Recall Solution 1.1

Sirf ek hi sawaal poochho jo matter karta hai: load ko upar ki taraf kitne rope strands kheench rahe hain? Upar wali figure ka baayana panel dekho — ek single rope segment load se seedha pulley tak jaata hai. Yeh ek supporting strand hai. $n T = W \Rightarrow 1 \cdot T = W \Rightarrow IMA = n = 1.$ Ek fixed pulley force ko $1$ se multiply karta hai — yeh sirf tumhari pull ko redirect karta hai (neeche ka upar ho jaata hai). Convenient hai, powerful nahi.

Problem 1.2

Ek load ek single movable pulley se latka hua hai. Rope ka ek end ceiling se tied hai; doosra end tumhare haath mein hai. Movable block ko kitne strands support karte hain, aur IMA kya hai?

Recall Solution 1.2

Movable pulley ke neeche rope wrap hoti hai: ek segment ceiling anchor tak jaata hai, doosra tumhare haath tak jaata hai. Dono same rope ke hisse hain, isliye dono same tension $T$ carry karte hain, aur dono movable block ko upar kheenchte hain. Count = $2$ . $2 T = W \Rightarrow IMA = 2.$ Upar wali figure ke daayeen panel mein dono orange strands dekho.

Problem 1.3

Sahi ya galat: "Rope ko kisi corner se redirect karne ke liye ek doosra fixed pulley add karne se mechanical advantage badhta hai."

Recall Solution 1.3

Galat. Ek fixed pulley rope ki direction badalta hai lekin movable block ko support karne wale zero strands add karta hai. MA sirf woh strands count karta hai jo movable load ko lift kar rahe hain. Ek pure redirector $n$ mein kuch contribute nahi karta.

L2 — Application

(Forces aur distances mein geometry daalo.)

Problem 2.1

Ek block-and-tackle mein movable block ko support karne wale 4 strands hain. Usme $50 kg$ ka ek load latka hua hai. Dhundho (a) use hold karne ke liye zaroorat wali effort force, aur (b) load ko $0.8 m$ upar uthane ke liye tumhe kitni rope kheenchni padegi.

Recall Solution 2.1

(a) Weight: $W = m g = 50 \times 9.8 = 490 N$ . Char strands ise barabar baantte hain: $4 T = W \Rightarrow T = \frac{490}{4} = 122.5 N .$ Tum ek strand pakdte ho, isliye $F_{effort} = 122.5 N$ . (b) $d = nh$ kyun? Rope inextensible hai (Constraint relations): uski total length kabhi nahi badlti. Jab load (aur uska movable pulley) $h$ se upar uthta hai, har ek $n$ supporting strands $h$ se chhoti ho jaati hai. Yeh sara removed slack — total mein $n \times h$ — kahin na kahin se bahar aana chahiye, aur ek hi free end tumhare haath mein hai. Isliye tumhara haath $d = nh$ travel karta hai. Yahan: $d = nh = 4 \times 0.8 = 3.2 m .$ Energy check: $W_{in} = 122.5 \times 3.2 = 392 J$ , $W_{out} = 490 \times 0.8 = 392 J$ . ✔ (Work–Energy Theorem)

Problem 2.2

Ek ideal pulley system mein $60 N$ ki effort $300 N$ ka load uthati hai. Iske kitne supporting strands hain?

Recall Solution 2.2

$IMA = \frac{W}{F _{effort}} = \frac{300}{60} = 5.$ Kyunki IMA supporting strands ki sankhya ke barabar hai, $n = 5$ .

Problem 2.3

Ek single movable pulley ek $12 kg$ ki bucket uthata hai. Use stationary hold karne ke liye kitni effort force chahiye, aur use $1.5 m$ upar uthane ke liye tum kitna pull karte ho?

Recall Solution 2.3

$W = 12 \times 9.8 = 117.6 N$ . Movable pulley ⇒ $n = 2$ . $F_{effort} = \frac{W}{2} = 58.8 N, d = 2 \times 1.5 = 3.0 m .$

L3 — Analysis

(Ab motion, constraints, aur real losses aate hain.)

Problem 3.1 (Atwood machine)

Ek single fixed ideal pulley ke upar se do masses latkay hue hain: $m_{1} = 6 kg$ aur $m_{2} = 4 kg$ . System ki acceleration aur rope mein tension dhundho. (Atwood machine)

Recall Solution 3.1

Ek rope ⇒ dono sides par ek tension $T$ . Inextensible rope ⇒ same speed aur same $∣ a ∣$ : jab $m_{1}$ , $x$ se neeche gire, $m_{2}$ , $x$ se upar jaaye. Upar wale sign convention ka use karte hue (girti side $m_{1}$ ke liye neeche positive lo, uthti side $m_{2}$ ke liye upar positive lo), har mass ka free-body banao (Newton's Second Law):

Bhaari side $m_{1}$ girta hai: $m_{1} g - T = m_{1} a$ .
Halki side $m_{2}$ uthta hai: $T - m_{2} g = m_{2} a$ .

Equations add karo taaki $T$ khatam ho jaaye: $(m_{1} - m_{2}) g = (m_{1} + m_{2}) a \Rightarrow a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{2 \times 9.8}{10} = 1.96 m/s^{2} .$ Phir doosri equation se: $T = m_{2} (g + a) = 4 (9.8 + 1.96) = 47.04 N .$ Closed form kahan se aata hai: boxed $a$ ko $T = m_{2} (g + a)$ mein wapas substitute karo: $T = m_{2} (g + \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}) = m_{2} g \frac{( m _{1} + m _{2} ) + ( m _{1} - m _{2} )}{m _{1} + m _{2}} = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} .$ Yeh ek derived result hai, yaad kiya hua nahi. Check: $\frac{2 \cdot 6 \cdot 4 \cdot 9.8}{10} = 47.04 N$ . ✔

Problem 3.2 (real efficiency)

Ek pulley system jisme $IMA = 4$ hai, use $500 N$ ka load uthane ke liye $150 N$ ki effort chahiye (friction aur pulley ka weight kuch kaam chura leta hai). Actual mechanical advantage (AMA) aur efficiency $η$ dhundho.

Recall Solution 3.2

Actual MA real forces use karta hai: $AMA = \frac{W}{F _{effort}} = \frac{500}{150} = 3.333 \dots$ $η = \frac{AMA}{IMA} = \frac{3.333}{4} = 0.8333 \dots \approx 83.3%.$ Friction (Friction) aur pulley weight ne tumhara $\approx 16.7%$ kaam kha liya.

Problem 3.3 (movable-pulley constraint)

Ek movable pulley ek rope par latka hua hai. Rope ka ek end $a_{1} = 3 m/s^{2}$ acceleration se neeche kheencha jaata hai (us end par rope lamba ho raha hai), doosra end fixed hai ( $a_{2} = 0$ ). Movable pulley ki acceleration kya hai? (Constraint relations)

Recall Solution 3.3

Apne sign convention ka use karte hue, hum kisi end ki rate ko positive tab lete hain jab uska segment lamba ho (woh end rope feed kare). Movable pulley ke upar rope do segments se bani hai, isliye pulley average par move karta hai: $a_{pulley} = \frac{a _{1} + a _{2}}{2} = \frac{3 + 0}{2} = 1.5 m/s^{2} .$ Average kyun? Rope ki total length constant: ek end jo bhi slack supply kare, pulley position ko do segments mein absorb karna hoga, isliye woh summed end-rate ki aadhi shift karta hai. Yeh sign kabhi guess mat karo — constant length se derive karo.

L4 — Synthesis

(Ideas ko combine karo: energy, motion, aur ek twist.)

Problem 4.1

Problem 2.3 wale single movable pulley ka use karte hue ( $m = 12 kg$ , $n = 2$ ), maano sirf hold karne ki jagah tum rope ka apna end $0.4 m/s$ ki constant speed se neeche kheench rahe ho. Load ki upar jaane wali speed aur tumhari deliver ki gayi power kya hai? (Ideal maano, isliye $F_{effort} = W /2$ .)

Recall Solution 4.1

Speed: tumhara haath $0.4 m/s$ par rope reel in kar raha hai. Load ko uthane ke liye do strands ko saath mein chota hona hoga, isliye load tumhari haath-speed ki aadhi speed par uthta hai: $v_{load} = \frac{v _{hand}}{n} = \frac{0.4}{2} = 0.2 m/s .$ Yeh wahi $d = nh$ trade hai, ab per second. Power (tumhara input): $F_{effort} = W /2 = 117.6/2 = 58.8 N$ , $0.4 m/s$ par move karte hue: $P_{in} = F_{effort} v_{hand} = 58.8 \times 0.4 = 23.52 W .$ Output check: $P_{out} = W v_{load} = 117.6 \times 0.2 = 23.52 W$ . ✔ Power in = power out (ideal).

Problem 4.2

Ek block-and-tackle ( $IMA = 3$ ) ki efficiency $η = 0.9$ hai. $270 N$ ka load uthane ke liye kitni effort chahiye, aur tumhara kitna input work heat ban jaata hai?

Recall Solution 4.2

Definition $η = \frac{AMA}{IMA}$ se shuru karo (Problem 3.2 se). Actual MA isolate karne ke liye dono sides ko $IMA$ se multiply karo: $AMA = η \times IMA = 0.9 \times 3 = 2.7.$ Ab $AMA = \frac{W}{F _{effort}}$ use karo, effort ke liye rearrange karke: $F_{effort} = \frac{W}{AMA} = \frac{270}{2.7} = 100 N .$ Wasted fraction $= 1 - η = 1 - 0.9 = 0.1 = 10%$ input work friction/pulley weight mein kho jaata hai.

L5 — Mastery

(Constraint + dynamics ke saath poora multi-body system.)

Problem 5.1

Ek load $M = 10 kg$ ek movable pulley se latka hua hai. Iske upar wali rope ka ek end ceiling se fixed hai; doosra end ek fixed pulley ke upar se jaata hai aur ek hanging counterweight $m = 8 kg$ seedha neeche kheenchta hai. Sab kuch ideal hai. Load $M$ ki acceleration aur tension $T$ dhundho.

Setup: do strands (har ek tension $T$ ) $M$ waale movable pulley ko upar kheenchte hain; ek single rope end $m$ ko upar kheenchta hai.

Recall Solution 5.1

Step 1 — Constraint (kaun kitni speed par move karta hai). Movable pulley (jisme $M$ hai) ke $n = 2$ strands hain. Agar $m$ neeche girta hai aur rope ko acceleration $a_{m}$ par kheenchta hai, toh load $M$ us rate ki aadhi par uthta hai: $a_{M} = \frac{a _{m}}{2} .$ Apne sign convention ko follow karte hue, maano load ki upar wali acceleration $a$ hai ( $M$ ke liye up positive, jisko hum uthte dekhte hain), isliye $m$ neeche $2 a$ acceleration par jaata hai ( $m$ ke liye down positive, jisko hum girte dekhte hain).

Step 2 — $M$ ka FBD (uske movable pulley ke saath). Do strands upar kheenchte hain ( $2 T$ ), gravity neeche kheenchti hai ( $M g$ ). Up positive aur acceleration $a$ ke saath: $2T - Mg = M a. \tag{i}$

Step 3 — Counterweight $m$ ka FBD. Gravity $m g$ neeche, tension $T$ upar, down positive aur acceleration $2 a$ ke saath: $mg - T = m(2a). \tag{ii}$

Step 4 — Solve karo. (ii) se: $T = m g - 2 ma = m (g - 2 a)$ . (i) mein substitute karo: $2 m (g - 2 a) - M g = M a$ $2 m g - 4 ma - M g = M a$ $(2 m - M) g = (M + 4 m) a$ $a = \frac{( 2 m - M ) g}{M + 4 m} .$ $M = 10$ , $m = 8$ daalo: $a = \frac{( 16 - 10 ) \times 9.8}{10 + 32} = \frac{6 \times 9.8}{42} = \frac{58.8}{42} = 1.4 m/s^{2} (load rises) .$ Tension: $T = m (g - 2 a) = 8 (9.8 - 2.8) = 8 \times 7.0 = 56 N .$ Sanity check: $2 T = 112 N > M g = 98 N$ , isliye net upward force $M$ ko uthata hai — $a > 0$ ke saath consistent hai. ✔

Problem 5.2

Usi system ke liye, kaun se counterweight mass $m_{⋆}$ par load $M = 10 kg$ equilibrium mein rehta hai ( $a = 0$ )? Mechanical advantage ke zariye interpret karo.

Recall Solution 5.2

$a = \frac{( 2 m - M ) g}{M + 4 m}$ mein $a = 0$ set karo: numerator zero hona chahiye: $2 m_{⋆} - M = 0 \Rightarrow m_{⋆} = \frac{M}{2} = \frac{10}{2} = 5 kg .$ MA interpretation: counterweight ka weight yahan tumhari effort hai. $n = 2$ wale movable pulley se $10 kg$ ke load ko balance karne ke liye tumhe sirf load ke weight ki aadhi chahiye, yani ek $5 kg$ ka counterweight. Yeh exactly $IMA = 2$ hai: effort $= W /2$ .

Connections

Newton's Second Law — upar ke har FBD mein.
Tension in strings — ek rope, ek $T$ .
Constraint relations — $a_{M} = a_{m} /2$ aur $d = nh$ links.
Atwood machine — L3.1 baseline.
Work–Energy Theorem — energy/power checks.
Friction — L3.2 / L4.2 mein efficiency losses.
Inclined plane mechanical advantage — wahi force-for-distance trade.