Atwood Machine Tension

Equivalent forms

T = \mu_{harmonic} \cdot g

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

A 1784 classroom toy that still teaches Newton's second law more cleanly than free fall.

Symbol	Name	SI unit	Dimension	Range
$T$	Rope tensionoutput Tension in the inextensible rope	N	MLT^-2	0 – 1000
$m1$	Mass 1 Mass on left side	kg	M	0.1 – 50
$m2$	Mass 2 Mass on right side	kg	M	0.1 – 50

Unit systems

Symbol	SI	CGS	Imperial
$T$	N	dyn	lbf
$m1$	kg	g	slug
$m2$	kg	g	slug

Where it holds

Massless, inextensible rope; massless, frictionless pulley; non-relativistic.

Dimensional analysis

T \to [MLT^{-2}]

(M\cdot M\cdot LT^{-2})/M = [MLT^{-2}]

Newtons.

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Discovery

George Atwood · 1784

Atwood invented this device to make gravity 'tractable' enough to verify Newton's second law with eighteenth-century pendulum clocks.

Try this

Two masses on a frictionless pulley — what's the tension in the string?

Masses m₁ = 4 kg and m₂ = 6 kg hang over an ideal pulley. Find the rope tension.

Research status: stable

Real-world applications

Elevator counterweight systems
Crane and lift mechanics
Theatrical fly rigs
Force-measurement standards

Common misconceptions

Tension equals weight of the heavier mass — it doesn't
Tension differs on each side of the pulley — only true if pulley has mass/friction
The masses fall at g — they fall at $(m_{2}-m_{1})g/(m_{1}+m_{2})$

Experimental verification

Used in undergraduate labs to measure g to 1% precision; modern variants use photogate timers and precision pulleys.

Derivation

For each mass:

T - m_i g = \mp m_i

a (sign depending on which mass falls).

Adding gives

(m_{1}+m_{2})

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

.

Substituting back:

T = m_{1}(g

+a

) = 2\,\mathrm{m}_{1} m_{2} g / (m_{1}+m_{2})

.

Limiting cases

m_{1} = m_{2}

⟶

T = m g

Balanced: no acceleration; tension equals either weight.

m_{2}

≫

m_{1}

⟶

T \to 2\,\mathrm{m}_{1} g

Light mass falls down; rope tension approaches twice

m_{1} g

.

m_{1} \to 0

⟶

T \to 0

Effectively free fall for

m_{2}

; rope goes slack.

What if…

What if the pulley has mass?

Tensions on the two sides differ; $add I \alpha = (T_{2}-T_{1}) R$ .

What if

m_{1} = m_{2}

?

No acceleration; $T = mg$ . Classic 'static equilibrium' test.

What if rope stretches?

System becomes a coupled oscillator with spring constant k_rope.

1

4 kg vs 6 kg

Given ·

m1:: 4
m2:: 6

Find · T

Steps

Numerator: $2\cdot 4\cdot 6\cdot 9.8 = 470.4$
Denominator: $4 + 6 = 10$
$T = 47.04\,\mathrm{N}$ (note: between $4\cdot g=39.2$ and $6\cdot g=58.8)$

Result ·

T = 2\cdot 4\cdot 6\cdot 9.8 / 10 = 47.04\,\mathrm{N}

2

Near balance

Given ·

m1:: 5
m2:: 5.1

Find · T

Steps

$T = 2\cdot 5\cdot 5.1\cdot 9.8 / 10.1$
$T \approx 49.49\,\mathrm{N}$ — almost equal to $mg = 49\,\mathrm{N}$

Result ·

T \approx 49.49\,\mathrm{N}

Newton's Second Law→Newton's Law of Universal Gravitation→