Atwood machine — derivation: Question bank — Physics deep dive

Before we start, one reminder of the symbols so nothing here is unearned:

$m_{1}, m_{2}$ — the two hanging masses (kg). We take $m_{1} \geq m_{2}$ unless told otherwise.
$g$ — gravitational acceleration ( $\approx 9.8 m/s^{2}$ ), the pull of Earth per kg.
$a$ — the common acceleration magnitude both masses share (they move at the same rate because one rope ties them).
$T$ — the tension, the pulling force the rope exerts, the same at every point of a massless rope.
The two results we lean on: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ and $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ .

See it first — the picture every trap comes from

Before the traps, look at the two free-body diagrams below. Every question on this page is really about these two arrows and the rope that ties them together.

Notice, in the figure: on each mass there are exactly two arrows — weight $m g$ pulling down (chalk-pink) and the same tension $T$ pulling up (chalk-blue). The rope over the pulley just bends that single blue force around the wheel; it is the same length arrow on both sides because the string is massless. That one fact — one blue arrow, two identical copies — is what makes the tension uniform.

Now watch the two Newton equations fall straight out of the arrows. On the heavy side, down wins, so writing down-as-positive gives $m_{1} g - T = m_{1} a$ . On the light side, up wins, so writing up-as-positive gives $T - m_{2} g = m_{2} a$ . Add them and the blue $T$ arrows cancel:

The second figure is the whole algebra as a picture: the two blue tension arrows are equal and opposite in the sum, so they annihilate, leaving only the pink weight difference $m_{1} g - m_{2} g$ pushing the total mass $m_{1} + m_{2}$ . That is exactly $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} .$ Put this $a$ back into the light-side equation, $T = m_{2} (g + a)$ , and simplify to $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g .$

The last picture shows why the answers behave the way they do — how $a$ and $T$ slide as one mass grows:

Keep these three figures in your mind's eye; each trap below is just misreading one of these arrows.

True or false — justify

Both masses have the same acceleration magnitude.

True — the string is inextensible, so if one end drops by

x

the other rises by exactly

x

; equal displacement over equal time means equal speed and equal acceleration (Constraint Relations).

The tension is different on the two sides of the pulley.

False — the string and pulley are massless and the pulley is frictionless, so the rope only redirects the force;

T

is one single value everywhere (Tension in Strings).

The tension equals the weight of the lighter mass.

False — that would only hold in equilibrium (

a = 0

). Here

T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g

, which sits strictly between

m_{2} g

and

m_{1} g

whenever the masses differ.

If you double both masses, the acceleration doubles.

False —

a = \frac{( 2 m _{1} - 2 m _{2} ) g}{2 m _{1} + 2 m _{2}}

has the factor of 2 cancel top and bottom, so

a

is unchanged. Acceleration depends on the ratio of masses, not their absolute size.

The heavier mass falls with acceleration

g

False — it falls with

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} < g

, because the rope holds part of its weight back; only if

m_{2} = 0

does it reach free fall.

Gravity on the two masses adds up to drive the system.

False — because the rope routes over the pulley,

m_{2}

's weight opposes

m_{1}

's motion. The net driving force is the difference

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

, never the sum.

Tension is symmetric in

m_{1}

and

m_{2}

True —

\frac{2 m _{1} m _{2}}{m _{1} + m _{2}}

is unchanged if you swap the labels, which makes sense: the rope cannot tell which side you decided to call "1".

m_{1} = m_{2}

the rope goes slack.

False — the system is balanced so

a = 0

, but the rope is taut, carrying

T = m g

to hold each equal weight up against gravity.

Spot the error

"For

m_{1}

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

with up as positive."

Error — the sign convention is wrong. If

m_{1}

accelerates down, you must take down as positive for it, so the equation is correct only when down is

+

; stated with "up as positive" the signs of

a

don't match its motion.

"Add the equations to get

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

Error — you added the weights instead of subtracting. Correct addition is

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

, giving

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

; the tensions cancel and the weights subtract.

"Since each mass weighs

m g

, the rope must pull each up with

m g

, so

T = m_{1} g

on the heavy side."

Error — that assumes equilibrium. The masses are accelerating, so Newton's second law gives net force

\neq = 0

; a single tension

T < m_{1} g

acts on the heavy side.

"Each side has its own acceleration because they move in opposite directions."

Error — opposite direction does not mean different magnitude. The inextensible string forces equal

∣ a ∣

; only the sign (up vs down) differs.

"On the light mass:

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

, so if

a > 0

then

T < m_{2} g

Error — rearrange:

T = m_{2} (g + a)

, and since

a > 0

this gives

T > m_{2} g

. The light mass accelerates upward, so the rope must pull with more than its weight; the extra

m_{2} a

is what lifts it.

"The pulley adds an extra upward force, so it should appear in the equations."

Error — an ideal (massless, frictionless) pulley stores and adds no force; it only changes the rope's direction. It never enters the two Newton equations.

Why questions

Why do we add the two Newton equations rather than subtract them?

Because the unknown tension appears as

- T

in one and

+ T

in the other; adding cancels

T

and leaves a single equation in

a

alone.

Why is the total mass

(m_{1} + m_{2})

in the denominator of both formulas?

Because the whole rope-coupled system moves as one — the total mass is the inertia that resists the driving force, so it always sets the scale of the response.

Why did Atwood build the machine with nearly equal masses?

Nearly equal masses give a tiny weight difference, so

a

is a small fraction of

g

. A slow acceleration is easy to time by hand, letting you measure

g

accurately (Newton's Second Law).

Why does tension come out less than the heavy mass's weight?

The heavy mass accelerates downward, so its net force must point down:

m_{1} g - T > 0

, forcing

T < m_{1} g

Why does tension come out more than the light mass's weight?

The light mass accelerates upward, so its net force points up:

T - m_{2} g > 0

, forcing

T > m_{2} g

Why must we draw a separate free-body diagram for each mass?

Because Newton's second law applies to one body at a time; isolating each mass shows exactly the two forces (weight down, tension up) acting on it (Free Body Diagrams).

Why is the assumption of a massless string essential for "

T

is uniform"?

A massive rope would need its own net force to accelerate, so tension would differ end-to-end; only a massless rope needs zero net force, keeping

T

the same everywhere.

Why does the tension expression resemble a harmonic-mean combination?

Because

\frac{2 m _{1} m _{2}}{m _{1} + m _{2}}

is exactly the harmonic mean of

m_{1}

and

m_{2}

; it is symmetric and weighted toward the smaller mass, reflecting that the lighter side limits the shared force (Harmonic Mean).

Edge cases

What happens to

a

and

T

when

m_{1} = m_{2} = m

a = 0

(balanced, nothing accelerates) and

T = \frac{2 m ^{2}}{2 m} g = m g

(the rope simply holds each weight up).

What happens as

m_{2} \to 0

(one side empty)?

a \to g

(the lone mass free-falls) and

T \to 0

(there is nothing on the other end to pull, so the rope carries no force).

What happens as

m_{1} \to \infty

with

m_{2}

fixed?

a \to g

(the huge mass falls essentially freely) and

T \to 2 m_{2} g

— the light mass then feels a net upward force of

T - m_{2} g \to m_{2} g

, giving it an upward acceleration approaching

g

If our sign guess is wrong and we assume the lighter side falls, what tells us?

The formula returns a negative

a

. That negative sign self-corrects, meaning the true motion is opposite to our guess — the physics is unharmed.

Can the tension ever exceed both weights

m_{1} g

and

m_{2} g

at once?

No —

T

always lies strictly between

m_{2} g

and

m_{1} g

(equal to both only when

m_{1} = m_{2}

). It can never be larger than the heavier weight in an ideal Atwood machine.

If the pulley had mass (real world), would

T

still be the same on both sides?

No — a real pulley needs a net torque to angularly accelerate, so tensions differ across it. That case leaves ideal-Atwood territory and needs rotational dynamics (Pulley Systems & Mechanical Advantage).

Recall One-line summary of every trap

The rope makes the masses share $a$ , makes $T$ uniform, and makes the drive the difference of weights over the sum of masses. Break any one idealisation (mass on rope, mass on pulley, friction) and the neat symmetry breaks with it.

Connections

Atwood machine — derivation — the parent derivation these traps test.
Newton's Second Law — the engine behind every "why" here.
Free Body Diagrams — the tool that resolves most sign traps.
Tension in Strings — why $T$ is uniform.
Constraint Relations — why the accelerations are equal.
Pulley Systems & Mechanical Advantage — what changes when the pulley is real.
Harmonic Mean — the shape of the tension formula.