Visual walkthrough — Pulley systems — mechanical advantage

We build the whole story around one running example: a movable pulley with two strands, then push it to $n$ strands, then check the edge cases.

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Step 1 — What is a rope actually doing? (Tension, from zero)

WHAT. Before any pulley, look at one straight piece of rope with a weight $W$ hanging on the bottom and your hand pulling the top.

WHY. Every pulley result rests on one idea: what number describes "how hard the rope pulls"? That number is called tension, written $T$. We must define it before we use it anywhere.

PICTURE. The rope in the figure pulls up on the weight (arrow up, size $T$) and pulls down on your hand (arrow down, size $T$) — same length arrows, opposite directions.

Because the hanging weight is not moving, the up-pull must exactly cancel the down-weight:

$\underset{\text{rope pulls up}}{\underbrace{T}}=\underset{\text{gravity pulls down}}{\underbrace{W}}$

Here $T$ is what the rope carries, $W=mg$ is the weight ($m$ = mass in kg, $g\approx 9.8{\text{m/s}}^2$). One strand, so you feel the full weight. No advantage yet — that is the baseline we must beat. See Tension in strings.

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Step 2 — Why is tension the SAME all along one rope?

WHAT. Grab a tiny imaginary snippet of the rope and ask: what forces act on it?

WHY. If we can prove the tension doesn't change from point to point, then wherever we grab the rope — near the load, near the ceiling, at your hand — the number $T$ is identical. That single fact is the engine of the whole derivation.

PICTURE. The snippet has the rope to its left tugging left with tension $T_L$, and the rope to its right tugging right with tension $T_R$.

Now apply Newton's Second Law to the snippet:

$\underset{\text{net force on snippet}}{\underbrace{T_R-T_L}}=\underset{\text{its mass}}{\underbrace{m_{\text{snippet}}}}\cdot a$

An ideal rope is massless, so $m_{\text{snippet}}=0$:

$T_R-T_L=0\cdot a=0\Rightarrow T_R=T_L$

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Step 3 — Why doesn't the pulley change $T$ either?

WHAT. Now bend the rope over a wheel — a pulley. Does going around the corner change the tension?

WHY. We are about to use "same $T$ on both sides of the pulley" constantly. If a pulley secretly scaled $T$, the counting trick would collapse. So we settle it now.

PICTURE. The rope comes up the left side (tension $T$), wraps the top of an ideal (frictionless, massless) wheel, and goes down the right side. The wheel is free to spin, so it offers no resistance along the rope — it only bends the direction.

$T_{\text{left of pulley}}=T_{\text{right of pulley}}=T$

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Step 4 — Attach the load to a MOVABLE pulley (the key move)

WHAT. Instead of hanging the weight on a rope end, hang it on the axle of a pulley, and thread one rope: one end tied to the ceiling, over the movable pulley, up to your hand.

WHY. This is the whole trick. Look at the movable pulley: two rope strands leave its top — the ceiling strand and your-hand strand. Both belong to the same rope, so (Steps 2–3) both carry the same $T$. Now two strands share one load.

PICTURE. The movable pulley (carrying weight $W$) is held up by two upward arrows, each of size $T$. Count them: 1, 2.

Draw the free-body of the movable pulley + load (nothing accelerating, static lift):

$\underset{\text{strand 1 up}}{\underbrace{T}}+\underset{\text{strand 2 up}}{\underbrace{T}}=\underset{\text{weight down}}{\underbrace{W}}$

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

You hold one strand, so your effort is $F_{\text{effort}}=T=W/2$. You lift a weight $W$ by pulling only half of it.

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

WHAT. Build a system where $n$ strands of the same rope rise from the movable block (more wheels, same single rope).

WHY. The equilibrium equation just gets more identical up-arrows. Nothing new physically — only the count changes. This is where the clean formula appears.

PICTURE. $n$ equal up-arrows (each $T$) hold the block; one down-arrow $W$. Here we draw $n=4$.

$\underset{n\text{ strands, each }T}{\underbrace{T+T+\cdots +T}}=W\Rightarrow \underset{\text{total up-pull}}{\underbrace{nT}}=W$

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

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Step 6 — The catch: you must pull FARTHER (energy check)

WHAT. Raise the load by height $h$. How much rope must your hand reel in?

WHY. Force never comes free. If you got $n$ times the force for the same pull distance, you'd create energy from nothing — impossible. The geometry forces a price.

PICTURE. When the block rises $h$, every one of the $n$ supporting strands gets shorter by $h$. All that removed slack passes through your one hand.

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

Now check the energy books using Work–Energy Theorem:

$W_{\text{in}}=\underset{W/n}{\underbrace{F_{\text{effort}}}}\times \underset{nh}{\underbrace{d_{\text{effort}}}}=\frac{W}{n}\cdot nh=Wh=\underset{\text{lifting }W\text{ by }h}{\underbrace{W_{\text{out}}}}✓$

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Step 7 — Edge & degenerate cases (never get surprised)

Every scenario the reader can hit, drawn:

• $n=1$ — single fixed pulley. Only one strand holds the load (it hangs from a rope end, the wheel is bolted to the ceiling). $\text{IMA}=1$: no multiplication, pure direction change (pull down → load goes up). Convenience, not power.
• $n=2$ — single movable pulley (Step 4). Half the effort, twice the pull.
• Zero effort $T\to 0$. Then $W=nT\to 0$: with no pull, only a weightless load stays up. Consistent — no free lifting.
• Real pulley (mass + friction). Now Steps 2–3 fail slightly: strands carry different $T$, so actual MA drops. Efficiency $\eta =\text{AMA}/\text{IMA}<1$. See Friction.
• Dynamic case (moving system). Set $a\ne 0$: this is the Atwood machine. With the Constraint relations for one movable pulley, $a_1+a_2=2a_{\text{pulley}}$ — the pulley moves at the average of its two rope ends.

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The one-picture summary

One rope → one tension $T$ (Steps 1–3). Hang the load on a movable pulley so $n$ strands share it → $nT=W$ → $\text{IMA}=n$ (Steps 4–5). Pay for it in distance: $d_{\text{effort}}=nh$, keeping $W_{\text{in}}=W_{\text{out}}$ (Step 6).

Recall Feynman retelling (plain words)

A rope is a lazy messenger: it passes the same pull along its whole length, and a smooth wheel just lets it turn a corner without changing that pull. So if you tie the heavy thing to a moving wheel and loop the rope so that two strands hang the wheel, both strands share the weight — each takes half, and your hand only feels half. Want a quarter? Arrange four strands. The number of strands holding the moving wheel is exactly how many times easier the pull becomes. But the rope is honest: to lift the thing one step, every one of those strands must shorten one step, and all that rope piles up in your hand — so you reel in as many steps as there are strands. Easier pull, longer pull, same total effort. No cheating the universe.

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Connections

• Newton's Second Law — Step 2 uses $F_{net}=ma$ on a massless snippet.
• Tension in strings — the uniform-$T$ backbone (Steps 1–3).
• Work–Energy Theorem — the $W_{in}=W_{out}$ check (Step 6).
• Constraint relations — the movable-pulley average rule (Step 7).
• Atwood machine — the dynamic $a\ne 0$ case.
• Friction — why real AMA $<$ IMA.
• Inclined plane mechanical advantage — same force-for-distance trade in another machine.

Concept Map