Pulley systems — mechanical advantage: Question bank

This bank hammers the traps that live inside Pulley systems — mechanical advantage. Before starting, lock in the vocabulary below — every trap answer uses these exact symbols.

Definition The symbols this page uses (read first)

$F_{load}$ — the force the machine has to support or lift, i.e. the weight of the object you are raising, measured in newtons ( $N$ ). Picture the orange block's downward pull in the figures below.
$F_{effort}$ — the force your hand supplies on the free end of the rope, also in newtons. Picture the red down-arrow you tug on.
$Q$ (weight of the load as a number) — when we need a single letter for the load's weight inside an equation like " $n$ strands share the load", we write $Q$ ; it is exactly $F_{load}$ , just a shorter name. (We deliberately do not reuse the letter $W$ here — see the next box for why.)
$E_{in}$ , $E_{out}$ (energy / work) — the energy you feed in with your hand ( $E_{in}$ ) and the energy delivered to the load ( $E_{out}$ ), each equal to force × distance, measured in joules ( $J$ ).
$T$ — the tension (pull) in the rope, in newtons.
$n$ — the number of rope strands supporting the movable block.
MA (Mechanical Advantage) — the force-multiplying factor: how many times bigger the load force is than your effort force. $MA = \frac{F _{load}}{F _{effort}}$ Read it as: "how much heavier a load ( $F_{load}$ ) can I hold for each newton of my own pull ( $F_{effort}$ )?" An MA of $4$ means a $10 N$ pull supports a $40 N$ load.

Visual warm-up — see the strands before you count them

Intuition Figure 3 — why easy pull costs a long pull (the trade-off, drawn)

This is the 3Blue1Brown heart of the topic. Panel by panel, the movable pulley is raised so the load climbs by a height $h$ . Watch the two green segments each shorten by $h$ (the two little brackets). All that shortened rope has to go somewhere — it is reeled in through your hand. Add the two brackets: your hand travels $d_{effort} = 2 h$ . Force halved, distance doubled: the shaded blue "energy rectangle" (force × distance) is the same area on the input and output side. That equal area is energy conservation, 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 $E_{in} = E_{out}$ .

Deriving the movable-pulley constraint (referenced in the traps)

Several traps below hinge on the rule $a_{1} + a_{2} = 2 a_{pulley}$ . It is never to be guessed — here is where it comes from, so you can defend it.

Formula The derivation in three honest steps

WHAT (write the length): each segment below the pulley has length $(y_{p} - y_{end})$ , and there are two of them, so $L = (y_{p} - y_{1}) + (y_{p} - y_{2}) + const = 2 y_{p} - y_{1} - y_{2} + const .$ WHY differentiate: $L$ is constant, so its rate of change is zero — differentiating a constant gives $0$ , which turns a length fact into a velocity fact. Do it twice to reach accelerations (rate of change of velocity). Differentiating once: $0 = 2 \overset{y}{˙}_{p} - \overset{y}{˙}_{1} - \overset{y}{˙}_{2} .$ Again: $0 = 2 \overset{y}{¨}_{p} - \overset{y}{¨}_{1} - \overset{y}{¨}_{2} \Rightarrow a_{1} + a_{2} = 2 a_{pulley} .$ WHAT IT LOOKS LIKE (Figure 4): if you pull one end down fast and hold the other still, the pulley sinks at half your speed — it always moves at the average of the two ends. That is why the " $2$ " is there: two segments share every metre the pulley travels.

True or false — justify

True or false: A single fixed pulley gives you mechanical advantage greater than 1.

False. Only one strand supports the load, so

n = 1

and

IMA = 1

; a fixed pulley redirects force (pull down to lift up) but never multiplies it.

True or false: In an ideal system, adding a second fixed pulley in the rope path increases the mechanical advantage.

False. A fixed pulley adds only redirection — it doesn't add a strand supporting the movable block.

IMA = n

depends on supporting strands, not total pulley count.

True or false: If a pulley system has

IMA = 4

, you get four times the energy out compared to what you put in.

False. Energy is conserved:

E_{in} = E_{out}

. You get 4× the force, but you pull 4× the distance, so the energy product (force × distance) is unchanged — no extra energy appears.

True or false: In one continuous ideal rope, the tension is the same on both sides of every pulley it crosses.

True. Take a tiny piece of massless rope: Newton's 2nd law says

F_{net} = ma = 0

(mass is zero), so the pull on its left must exactly equal the pull on its right — meaning

T

is the same all along. A frictionless pulley only changes direction, never magnitude.

True or false: Making the pulley heavier increases the actual mechanical advantage.

False — it decreases it. A massive pulley demands extra tension just to lift and spin itself, so the real force you save shrinks:

AMA

drops, and efficiency

η = AMA / IMA

falls below 1.

True or false: In an Atwood machine with

m_{1} = m_{2}

, the tension in the rope equals the weight of one block.

True. With equal masses

a = 0

, so each block is in equilibrium: the tension exactly balances one weight,

T = m g

. The system is balanced, not accelerating.

True or false: The ideal mechanical advantage of a system tells you its efficiency.

False.

IMA

is set by geometry alone; efficiency compares actual to ideal,

η = AMA / IMA

, and only drops below 1 for real pulleys with friction/mass. An ideal system has

IMA

large yet

η = 1

True or false: In a movable-pulley setup, the load and your hand always move the same distance.

False. If 2 strands support the load, your hand moves twice as far as the load rises, because both strands must shorten by the lift height

h

, so you reel in

d = 2 h

(see Figure 3).

True or false: For a movable pulley, the pulley's acceleration equals the sum of the two rope-end accelerations.

False. It equals the average:

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

, derived from constant rope length in Figure 4 (Constraint relations).

Spot the error

A student writes: "Two masses over a fixed pulley, so tension on the heavy side is bigger than on the light side." Find the error.

One ideal rope means one tension: a massless-rope piece has

F_{net} = 0

, forcing equal pull on both sides. The net force is bigger on the heavy side (its weight wins), but the string tension

T

is identical everywhere.

A student sets up an Atwood machine and writes

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

and

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

, then concludes

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

. Spot the error.

They forgot to divide by total mass. Adding the equations cancels

T

and gives

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

, so

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

, not

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

A student claims: "A movable pulley has

IMA = 2

, so if I attach a load of

100 N

, the tension I feel is

200 N

." Spot the error.

Backwards.

IMA = 2

means the two strands split the weight, so your effort tension is halved:

T = Q /2 = 50 N

, not doubled.

A student counts strands for a block-and-tackle: "There are 5 pulleys, so

IMA = 5

." Spot the error.

IMA

is not the pulley count. Count only the rope segments supporting the movable block; fixed pulleys add redirection, not strands.

A student says: "The load rises

1 m

, so I pull

1 m

of rope through my hand in any pulley system." Spot the error.

Only true when

n = 1

. For

n

supporting strands, all

n

must shorten by

1 m

, so you pull ==

n

metres== of rope (Figure 3 shows the

n = 2

case).

A student writes the constraint for a movable pulley as

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

. Spot the error.

Guessing accelerations are equal is wrong. Constant rope length gives

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

; differentiating twice yields ==

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

== — the pulley moves at the average of the two ends (Figure 4).

A student states: "Since MA lets me lift more with less force, a pulley system violates conservation of energy in my favour." Spot the error.

No violation — you trade force for distance. Work is force × distance; the extra rope you reel in (

n

times longer) exactly cancels the force reduction (

1/ n

), so

E_{in} = E_{out}

(the equal blue areas in Figure 3).

Why questions

Why is the tension the same everywhere in a single massless rope?

Take any tiny segment: Newton's 2nd law gives

F_{net} = ma

, and with

m = 0

this is

F_{net} = 0

, so the pull on each side must be equal. A frictionless pulley only redirects that pull, keeping its size fixed.

Why does mechanical advantage equal the number of supporting strands?

Because

n

equal-tension strands share the weight: their upward pulls balance the load,

n T = Q

, so each carries

Q / n

. Your effort is one strand's tension

F_{effort} = T = Q / n

, giving

IMA = Q / T = n

Why must you pull farther when the mechanical advantage is higher?

Because energy

=

force × distance is conserved. To keep that product fixed while the force drops by factor

n

, the distance must ==rise by factor

n

== — exactly the two equal-area rectangles in Figure 3.

Why do we add the two Atwood equations instead of subtracting?

Adding ==eliminates the unknown tension

T

== instantly — it appears as

+ T

in one equation and

- T

in the other, so they cancel, leaving a single equation for

a

Why does a fixed pulley still count as

n = 1

rather than

n = 0

The load hangs from one strand of the rope, which supports its full weight (Figure 1). That single supporting strand gives

n = 1

; the pulley just changes its direction.

Why does a real pulley have efficiency below 100%?

Some input work goes into overcoming friction and lifting the pulley's own mass, so the useful output is less than the input:

E_{out} < E_{in}

, making

η = E_{out} / E_{in} < 1

Why is the movable pulley — not the fixed one — the source of force multiplication?

Only the movable pulley has multiple strands leaving it to hold the load (its up-arrows share

Q

, Figure 2); a fixed pulley just has the rope pass over it, one strand still carrying the whole load.

Why does the "2" appear in the movable-pulley constraint

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

Because two rope segments run down to the movable pulley (Figure 4). The length

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

; differentiating twice puts a factor

2

on the pulley's acceleration.

Edge cases

Edge case: What is the mechanical advantage if you pull the effort rope straight down over a fixed pulley to lift a load, and

n = 1

IMA = 1

: no multiplication, and you must pull with force equal to the full weight — but at least you can use your body weight conveniently by pulling down.

Edge case: In an Atwood machine, what happens to the acceleration as

m_{2} \to 0

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

. With nothing on the other side,

m_{1}

is in free fall — the rope goes slack, tension

\to 0

Edge case: In an Atwood machine, what happens as

m_{1} \to m_{2}

a \to 0

and

T \to m g

. The system is balanced; any position stays put (neutral equilibrium in the ideal case).

Edge case: What is the

IMA

of a movable pulley if you pull the free end downward versus upward — does the answer change?

No.

IMA

counts supporting strands (

n = 2

), a geometric fact independent of the direction you happen to pull the free end.

Edge case: If the rope has significant mass, is the tension still uniform?

No. A massive rope needs a net force to accelerate itself, so

F_{net} \neq = 0

and tension varies along it; the "one rope, one tension" rule holds only in the ideal, massless case.

Edge case: A movable pulley is held so the load does not move (

a = 0

). Is there still mechanical advantage?

Yes.

IMA

is a static/geometric property: even in equilibrium,

n

strands share the weight, so

F_{effort} = Q / n

Edge case: What is the effort distance if the load is lowered instead of raised by

h

with

n

strands?

You feed out

nh

of rope (same magnitude, opposite direction). The distance trade-off is symmetric; energy bookkeeping still balances.

Recall One-line survival summary

Count supporting strands for $IMA = n$ ; one ideal rope has one tension; higher MA means proportionally longer pull; energy is always conserved. Question :::- If a claim ignores the distance penalty or invents extra tension across an ideal pulley, it is wrong.

Connections

Parent: Pulley systems — mechanical advantage — the full derivation these traps test.
Newton's Second Law — why massless rope forces $F_{net} = 0$ .
Tension in strings — the one-rope-one-tension rule.
Work–Energy Theorem — the $E_{in} = E_{out}$ bookkeeping behind every "no free lunch" answer.
Constraint relations — the $a_{1} + a_{2} = 2 a_{pulley}$ constraint (Figure 4).
Atwood machine — source of the dynamic edge cases.
Friction — why real efficiency drops below 1.
Inclined plane mechanical advantage — the same force-for-distance trade elsewhere.