Pulley systems — mechanical advantage — Physics notes

WHY does MA appear? (Derive it from first principles)

The key insight: a single ideal (massless, frictionless) rope has the SAME tension everywhere along it.

Now count rope segments supporting the load.

Consider a load hanging from a movable pulley, with $n$ rope segments rising from it. Each segment pulls up with the same tension $T$ . The load (weight $W$ ) is in equilibrium:

$\sum F_{\text{up}} = \sum F_{\text{down}} \;\Rightarrow\; nT = W$

You hold one of those segments, so your effort is $F_{\text{effort}} = T$ . Therefore:

$F_{\text{effort}} = \frac{W}{n} \quad\Rightarrow\quad \boxed{\text{IMA} = \frac{W}{T} = n}$

The distance trade-off (energy bookkeeping)

If the load rises by height $h$ , each of the $n$ supporting segments must shorten by $h$ . That slack has to be pulled in through your hand, so you pull a length:

$d_{\text{effort}} = n\,h$

Check energy: $W_{\text{in}} = F_{\text{effort}}\,d_{\text{effort}} = \frac{W}{n}\cdot nh = W h = W_{\text{out}} \checkmark$

Figure — Pulley systems — mechanical advantage

HOW to solve any pulley problem (method)

One rope → one tension. Label each independent rope with its own $T$ .
Constraint of inextensible string: total rope length is constant. Differentiate twice ⇒ relation between accelerations.
Free-body each mass, write $F_{net}=ma$ along motion.
Solve the system. For "MA" of a static system, set $a=0$ .

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine lifting a heavy bag with a rope thrown over a hook. Hard. Now use a moving hook attached to the bag, with the rope going up, around, down. Now two rope-strings hold the bag, so each only carries half its weight — pulling feels half as heavy! But there's a catch: to lift the bag 1 step, you must reel in rope from both strings, so you pull twice as much rope. You don't get strength for free — you trade pulling farther for pulling easier. Want it 4× easier? Make 4 strings hold the bag.

Flashcards

Mechanical advantage definition

\text{MA}=F_{load}/F_{effort}

, the factor a machine multiplies your input force.

Ideal MA of a pulley system equals

the number of rope strands supporting the movable block,

n

Why is tension uniform in an ideal rope?

Massless rope ⇒

F_{net}=ma=0

, and a frictionless pulley only changes direction, not magnitude.

Force on effort for

n

supporting strands

F_{effort}=W/n

(since

nT=W

Distance you must pull to raise load by

h

(n strands)

d=nh

, so

W_{in}=(W/n)(nh)=Wh=W_{out}

MA of a single fixed pulley

1 (redirects force only, no multiplication).

MA of a single movable pulley

String constraint for one movable pulley

a_1+a_2=2a_{pulley}

(pulley moves at average of the two ends).

Atwood machine acceleration

a=(m_1-m_2)g/(m_1+m_2)

Atwood machine tension

T=2m_1 m_2 g/(m_1+m_2)

Efficiency of a real pulley

\eta = AMA/IMA = W_{out}/W_{in}

Why must you pull farther with high MA?

Energy conservation: less force over more distance keeps

Fd

constant.

Connections

Newton's Second Law — every FBD here uses $F_{net}=ma$ .
Tension in strings — uniform tension assumption.
Work–Energy Theorem — the $W_{in}=W_{out}$ bookkeeping.
Constraint relations — differentiating fixed rope length.
Atwood machine — dynamic two-mass pulley.
Friction — source of real-world efficiency loss.
Inclined plane mechanical advantage — same force-for-distance trade.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, pulley ka funda bilkul simple hai: ye force ko distance ke saath trade karta hai. Agar tumhe heavy load uthana hai par zor kam lagana hai, to pulley use karo — lekin tumhe rope zyada lambi kheechni padegi. Free mein kuch nahi milta, energy in = energy out hamesha. Bas geometry ka khel hai.

Sabse important rule: ek hi ideal rope mein tension har jagah same hota hai (rope massless, pulley frictionless maan lo). Kyun? Kyunki rope ka mass zero hai, to $F_{net}=ma=0$ , aur frictionless pulley sirf direction badalta hai, magnitude nahi. Ab movable block ko jitne rope ke strands upar se support kar rahe hain, woh number hi tumhara Mechanical Advantage (MA) hai. $n$ strands ⇒ $nT=W$ ⇒ effort $=W/n$ . Bas strands gino!

Distance ka penalty mat bhoolna: agar load $h$ upar uthana hai, to har strand ko $h$ chhota hona padega, isliye tum $nh$ lambi rope kheechoge. Check karo: $W_{in}=(W/n)(nh)=Wh=W_{out}$ . Perfect conservation. Isliye 4 MA matlab 4 guna aasaan, par 4 guna lambi pull.

Problems solve karne ka tareeka: ek rope = ek tension label karo, string inextensible constraint se accelerations relate karo ( $a_1+a_2=2a_{pulley}$ movable pulley ke liye), phir har mass ka FBD likho aur $F=ma$ lagao. Atwood machine ke liye yaad rakho $a=(m_1-m_2)g/(m_1+m_2)$ . Galti se kabhi mat sochna ki "zyada pulley = zyada MA" — sirf supporting strands count karo, fixed pulley toh sirf direction change karta hai.

Pulley systems — mechanical advantage

WHAT is mechanical advantage?

WHY does MA appear? (Derive it from first principles)

The distance trade-off (energy bookkeeping)

HOW to solve any pulley problem (method)

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Newton's Laws & Dynamics

Connections