Pulley systems — mechanical advantage: Worked examples

Everything below rests on three facts already earned in the parent note. Let me restate them in one breath so no symbol is unexplained:

$T$ = tension, the pull force carried along a rope, measured in newtons ( $N$ ). One ideal (massless, frictionless) rope carries the same $T$ everywhere. See Tension in strings.
$W$ = weight of a hanging object $= m g$ , where $m$ is mass in kilograms and $g \approx 9.8 m/s^{2}$ is the strength of gravity. $W$ points straight down.
$F_{net} = ma$ — Newton's Second Law, Newton's Second Law. If nothing accelerates, $a = 0$ , so $F_{net} = 0$ (this is called equilibrium).

That is the whole toolkit. Watch how far it goes.

The scenario matrix

Every pulley problem is one (or a blend) of these case classes. The table lists each cell and the example that resolves it.

#	Case class	The awkward corner it tests	Example
A	Fixed pulley, static	MA that is exactly 1 (no multiplication)	Ex 1
B	Movable pulley, static	MA = 2, and the distance penalty	Ex 2
C	Many strands, static	General $IMA = n$ , big real number	Ex 3
D	Two masses, dynamic ( $a \neq = 0$ )	Atwood: sign of net force, direction of motion	Ex 4
E	Degenerate input $m_{1} = m_{2}$	Limiting case: $a \to 0$ , system balances	Ex 5
F	Non-ideal pulley (friction)	AMA $<$ IMA, efficiency $η < 1$	Ex 6
G	Movable pulley with acceleration	Constraint $a_{1} + a_{2} = 2 a_{p}$ used for real	Ex 7
H	Real-world word problem	Choosing $n$ to hit a target effort, units	Ex 8
I	Exam-style twist	Effort applied at an angle / hidden strand count	Ex 9

The two limiting behaviours you must never be surprised by:

Recall Limiting behaviours to keep in your pocket

As the number of strands $n \to \infty$ , the effort $F = W / n \to 0$ — but the rope you must pull, $d = nh \to \infty$ . Infinite ease costs infinite pulling. And as the two Atwood masses approach equality, $m_{1} \to m_{2}$ , the acceleration $a = (m_{1} - m_{2}) g / (m_{1} + m_{2}) \to 0$ : the system stalls in balance.

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

Figure — Pulley systems — mechanical advantage

Step 1 — Find the weight. $W = m g = 20 \times 9.8 = 196 N$ . Why this step? Force problems need actual newtons, not kilograms. Weight is the load the rope must hold.

Step 2 — Count the strands supporting the movable object. Look at the crate in the figure. Only one rope segment (the amber strand) rises from it. The pulley itself does not move, so nothing "shares" the load — a single strand carries all of $W$ . Why this step? The parent rule: $IMA = n_{supporting strands}$ . Here $n = 1$ .

Step 3 — Equilibrium of the crate. $n T = W \Rightarrow 1 \cdot T = 196 \Rightarrow T = 196 N$ . Your hand holds that same rope, so $F_{effort} = T = 196 N$ . Why this step? One rope = one tension; a fixed pulley only bends the rope's direction (see cyan arrows), never its magnitude.

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

Verify: Units: $N / N = 1$ (dimensionless, correct for a ratio). Sanity: a fixed pulley is a redirector — you pull down to lift up, which is convenient, but you feel the full $196 N$ . Matches the parent's Example 1.

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

Step 1 — Weight. $W = 30 \times 9.8 = 294 N$ . Why? Same reason — convert mass to the actual downward force.

Step 2 — Count supporting strands. In the figure, two strands (both amber) rise from the movable pulley to hold it up. $n = 2$ . Why? Because the pulley moves with the load, the rope wraps under it, so two segments of the one rope both pull up.

Step 3 — Equilibrium of the movable pulley + load. $n T = W \Rightarrow 2 T = 294 \Rightarrow T = 147 N$ . Your effort $F_{effort} = T = 147 N$ . $IMA = W / T = 294/147 = 2$ . Why? The two strands share the weight equally, each carrying $147 N$ .

Step 4 — Distance penalty. To raise the load by $h = 0.5 m$ , each of the two strands must shorten by $0.5 m$ . That $2 \times 0.5 = 1.0 m$ of slack passes through your hand: $d_{effort} = nh = 2 \times 0.5 = 1.0 m$ . Why? Constraint relations — the rope's total length is fixed, so shortening on the load side must appear as pulling on your side.

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 — no free lunch.

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

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

Step 2 — Strand count. Given $n = 5$ . $IMA = 5$ . Why? Count only strands attached to (or wrapping) the movable block — fixed-pulley redirections don't add to $n$ .

Step 3 — Effort. $F_{effort} = W / n = 1960/5 = 392 N$ . Why? Five strands share $1960 N$ ⇒ each $= 392 N$ .

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

Verify: $W_{in} = 392 \times 2.0 = 784 J$ ; $W_{out} = 1960 \times 0.4 = 784 J$ . ✔ A single person (can push $\sim 400 N$ ) can just manage a $2$ -tonne-force load — that's the point of high $n$ .

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

Step 1 — One rope, one tension; one string, one speed. Ideal rope ⇒ same $T$ both sides. Inextensible rope over a fixed pulley ⇒ both masses share the same $∣ a ∣$ . See Atwood machine. Why? Tension in strings + Constraint relations: if one side descends by $x$ , the other rises by $x$ .

Step 2 — Free-body the heavier mass $m_{1}$ (it falls, so take down as positive for it): $m_{1} g - T = m_{1} a \Rightarrow 6 (9.8) - T = 6 a$ . Why? Net downward force on $m_{1}$ = weight minus tension pulling up.

Step 3 — Free-body the lighter mass $m_{2}$ (it rises, take up positive): $T - m_{2} g = m_{2} a \Rightarrow T - 4 (9.8) = 4 a$ .

Step 4 — Add the two equations to cancel $T$ : $(m_{1} - m_{2}) g = (m_{1} + m_{2}) a \Rightarrow a = \frac{( 6 - 4 ) ( 9.8 )}{6 + 4} = \frac{2 ( 9.8 )}{10} = 1.96 m/s^{2}$ . Why add? Adding kills $T$ instantly, isolating $a$ .

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

Verify: $a = 1.96 < g = 9.8$ ✔ (some weight is held by the rope, so it falls slower than free-fall). Check $T$ lies between $m_{2} g = 39.2 N$ and $m_{1} g = 58.8 N$ : $39.2 < 47.04 < 58.8$ ✔ (tension must exceed the light weight to lift it, and be less than the heavy weight to let it fall).

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

Step 1 — Use the derived formulas (they must handle all inputs, including equal ones): $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{0 \cdot g}{10} = 0 m/s^{2}$ . Why? When the driving imbalance $(m_{1} - m_{2})$ vanishes, there is no net force to accelerate anything.

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 — Cross-check against equilibrium. With $a = 0$ , each side is just a hanging weight: $T = m g = 5 \times 9.8 = 49 N$ . ✔ Why check? A good formula must reduce to the simple static answer in the degenerate case — this is how you trust it.

Verify: $a = 0$ and $T = 49 N$ matches the naive "two equal weights balance" picture exactly. The system is in equilibrium at any position (neutral balance).

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

Step 1 — Ideal effort (frictionless). $F_{ideal} = W / IMA = 800/4 = 200 N$ . Why? This is the floor set by geometry alone.

Step 2 — Use efficiency to get actual MA. $η = \frac{AMA}{IMA} \Rightarrow AMA = η \cdot IMA = 0.75 \times 4 = 3$ . Why? Friction wastes part of your input work, so the achieved multiplication drops below the ideal.

Step 3 — Actual effort. $F_{actual} = W / AMA = 800/3 \approx 266.67 N$ . Why? You now support the same $800 N$ with a smaller effective advantage, so you must push harder.

Verify: $266.67 > 200$ ✔ (friction always makes real machines harder, never easier). Efficiency cross-check: $η = F_{ideal} / F_{actual} = 200/266.67 = 0.75$ ✔.

Cell G — Movable pulley with acceleration (constraint used for real)

Step 1 — Recall the constraint for one movable pulley. $a_{1} + a_{2} = 2 a_{p}$ . Why this and not "all equal"? The rope over the movable pulley is two segments; the pulley's position depends on the total of both. This comes from differentiating "rope length = constant" twice — Constraint relations. Never guess it.

Step 2 — Solve for $a_{p}$ . $a_{p} = \frac{a _{1} + a _{2}}{2} = \frac{3 + 1}{2} = 2 m/s^{2}$ . Why the average? Each end contributes equally to how fast the two-segment loop shortens, so the pulley moves at their mean rate.

Verify: $2$ lies between $1$ and $3$ ✔ — a genuine average. Degenerate check: if $a_{1} = a_{2} = 1$ , then $a_{p} = 1$ (whole thing moves rigidly, sensible). If one end is fixed ( $a_{2} = 0$ ), $a_{p} = a_{1} /2$ — matching the "pull $2 h$ to lift $h$ " distance rule differentiated twice. ✔

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

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

Step 2 — Required condition. She needs $F_{effort} = W / n \leq F_{max}$ , i.e. $n \geq W / F_{max} = 1470/250 = 5.88$ . Why $\geq$ ? More strands = smaller effort; she needs at least enough to bring effort down to her limit.

Step 3 — Round up to a whole strand count. $n = 6$ (since $n$ must be a whole number and $5.88$ rounds up). Why round up, not down? $n = 5$ gives $F = 1470/5 = 294 N > 250 N$ — too hard. So $n = 6$ .

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

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

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

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

Step 1 — Find the tension from the load's equilibrium. The two vertical supporting strands hold the load: $2 T = W \Rightarrow T = 120/2 = 60 N$ . Why from the load, not the hand? The strand tension is set by what the movable pulley must support, and an ideal fixed guide pulley only redirects — it does not change $T$ .

Step 2 — What does the angle actually change? The magnitude of tension in one ideal rope is the same everywhere, so the force along your pulling strand is still $60 N$ regardless of the $60°$ direction. The angle changes only the direction you pull, not the tension. Student B is right. Why? A frictionless guide pulley redirects tension; it never rescales it (parent Mistake: "tension changes across a pulley" — it doesn't).

Step 3 — The pull magnitude. $F_{effort} = T = 60 N$ , directed at $60°$ from vertical. $IMA = W / F_{effort} = 120/60 = 2$ — unchanged by the angle.

Verify: The horizontal component of your pull is $T sin 60° = 60 \times 0.866 = 51.96 N$ and the vertical is $T cos 60° = 30 N$ — the resultant is $51.9 6^{2} + 3 0^{2} = 60 N$ ✔, confirming the pull magnitude is exactly $T$ , angle notwithstanding. (The support of the load still comes from the two vertical strands, not from this angled strand.)

Recall Rapid self-test across all cells

Fixed pulley MA ::: $1$ — redirection only. Movable pulley MA ::: $2$ — two strands share the load. Effort for $n = 5$ strands, $W = 1960 N$ ::: $392 N$ . Atwood $a$ for $6$ and $4 kg$ ::: $1.96 m/s^{2}$ . Atwood $T$ for $6$ and $4 kg$ ::: $47.04 N$ . Equal masses Atwood $a$ ::: $0$ (balance). $IMA = 4$ , $η = 0.75$ ⇒ AMA ::: $3$ , so effort $= W /3$ . Movable-pulley constraint with $a_{1} = 3, a_{2} = 1$ ::: $a_{p} = 2 m/s^{2}$ . Angled effort strand changes tension? ::: No — magnitude unchanged, only direction.

Connections

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

Worked examples — Pulley systems — mechanical advantage