Atwood machine — derivation: Worked examples — Physics deep dive

Before anything, two reminders in plain words:

$a$ (acceleration) means "how fast the speed changes each second", measured in $m/s^{2}$ .
$T$ (tension) means "how hard the rope pulls", measured in newtons ( $N = kg \cdot m/s^{2}$ ).
$g$ is the strength of gravity near Earth, $g \approx 10 m/s^{2}$ (we use $10$ for clean numbers unless told otherwise).

The scenario matrix

Every Atwood problem falls into one of these case classes. The examples below fill every row.

#	Case class	What is special	Example
A	Heavy on left ( $m_{1} > m_{2}$ )	$a > 0$ , motion as assumed	Ex 1
B	Heavy on right ( $m_{1} < m_{2}$ )	$a < 0$ — sign tells you it moves the other way	Ex 2
C	Equal masses ( $m_{1} = m_{2}$ )	Degenerate: $a = 0$ , static	Ex 3
D	One mass zero ( $m_{2} = 0$ )	Degenerate: free fall, $T = 0$	Ex 4
E	One mass huge ( $m_{1} ≫ m_{2}$ )	Limiting value: $a \to g$ , $T \to 2 m_{2} g$	Ex 5
F	Real-world / measure $g$	Nearly equal masses, timing experiment	Ex 6
G	Kinematics follow-up	Use $a$ in $s, v, t$ equations	Ex 7
H	Exam twist: extra applied force	Formula must be re-derived, not memorised	Ex 8

Case A — heavy on the left

Worked example Example 1 (Cell A)

$m_{1} = 6 kg$ (left), $m_{2} = 2 kg$ (right), $g = 10 m/s^{2}$ . Find $a$ and $T$ .

Forecast: Left is heavier, so left falls. Guess: is $a$ closer to $g$ or closer to $0$ ? (The mass ratio is $3 : 1$ , fairly lopsided — expect $a$ well above zero, but not full $g$ .)

Step 1 — Identify the driving force. The net pull is the weight difference $(m_{1} - m_{2}) g = (6 - 2) (10) = 40 N$ . Why this step? From the parent note, gravity on the two sides fights across the pulley; only the difference drives the system.

Step 2 — Identify the inertia. Total mass being moved $= m_{1} + m_{2} = 8 kg$ . Why this step? Both masses must accelerate together, so the rope has to shove $8 kg$ of stuff.

Step 3 — Divide. $a = \frac{40}{8} = 5 m/s^{2} .$ Why this step? $a = \frac{driving force}{total mass}$ is just $F = ma$ for the whole system.

Step 4 — Tension. $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g = \frac{2 ( 6 ) ( 2 )}{8} (10) = \frac{24}{8} (10) = 30 N .$ Why this step? Direct substitution into the boxed formula.

Verify: Check on the light mass $m_{2}$ (moving up): net up force $= T - m_{2} g = 30 - 20 = 10 N$ , and $m_{2} a = 2 (5) = 10 N$ ✓. Note $30 N$ lies between $m_{2} g = 20$ and $m_{1} g = 60$ — as tension always must. Units: $N$ ✓, $m/s^{2}$ ✓.

Case B — heavy on the right (the sign flips)

Worked example Example 2 (Cell B)

Same setup but now $m_{1} = 2 kg$ (left), $m_{2} = 6 kg$ (right). We stubbornly keep the parent's convention: left ( $m_{1}$ ) assumed to go down, positive.

Forecast: We assumed wrong — the heavy side is on the right now. What sign should $a$ come out?

Step 1 — Plug into the formula unchanged. $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 2 - 6 ) ( 10 )}{8} = \frac{- 40}{8} = - 5 m/s^{2} .$ Why this step? We do not re-guess the direction. We trust the algebra to correct us.

Step 2 — Read the sign. $a = - 5 m/s^{2}$ . The minus means " $m_{1}$ actually accelerates up, not down" — exactly opposite to our assumption. Why this step? A negative answer under a chosen positive direction always means "reality points the other way." This is the whole reason we're allowed to guess the direction freely.

Step 3 — Tension is unaffected by the sign. $T = \frac{2 ( 2 ) ( 6 )}{8} (10) = 30 N .$ Why this step? Tension depends on $2 m_{1} m_{2}$ and $m_{1} + m_{2}$ , both symmetric — swapping the masses can't change $T$ . Same $30 N$ as Example 1.

Verify: Magnitude of $a$ is $5 m/s^{2}$ , same as Example 1 — because it's the same two masses, just mirrored. Physics can't care which side you labelled "1". ✓

Case C — equal masses (degenerate: nothing moves)

Case D — one mass zero (degenerate: free fall)

Case E — one mass huge (limiting behaviour)

Worked example Example 5 (Cell E)

$m_{1} = 1000 kg$ , $m_{2} = 1 kg$ , $g = 10$ . What does $a$ approach, and what does $T$ approach?

Forecast: The huge mass basically falls freely and yanks the tiny one up. Predict $a$ and the upward acceleration the small mass feels.

Step 1 — Acceleration. $a = \frac{( 1000 - 1 ) ( 10 )}{1001} = \frac{9990}{1001} \approx 9.98 m/s^{2} .$ Why this step? As $m_{1}$ dwarfs $m_{2}$ , the fraction $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} \to 1$ , so $a \to g$ . The heavy mass is nearly in free fall.

Step 2 — Tension. $T = \frac{2 ( 1000 ) ( 1 )}{1001} (10) = \frac{20000}{1001} \approx 19.98 N .$ Why this step? As $m_{1} \to \infty$ , $\frac{2 m _{1} m _{2}}{m _{1} + m _{2}} \to \frac{2 m _{1} m _{2}}{m _{1}} = 2 m_{2}$ , so $T \to 2 m_{2} g = 2 (1) (10) = 20 N$ .

Step 3 — Interpret the small mass. Net up force on $m_{2}$ : $T - m_{2} g \approx 19.98 - 10 = 9.98 N$ , giving it $\approx 9.98 m/s^{2}$ upward — nearly $g$ upward. It feels a " $2 g$ " world: gravity $g$ down plus tension pulling it up at nearly $2 g$ . Why this step? This makes the limit tangible: an infinitely heavy driver accelerates the tiny mass at essentially $g$ against gravity.

Verify: $a = 9990/1001 \approx 9.980$ , $T = 20000/1001 \approx 19.980$ ; both approach the predicted limits $g$ and $2 m_{2} g$ ✓.

Case F — real-world: measuring $g$

Worked example Example 6 (Cell F)

Atwood's original trick: make masses nearly equal so acceleration is small and easy to time. Take $m_{1} = 0.520 kg$ , $m_{2} = 0.480 kg$ . In an experiment the system falls $h = 0.500 m$ from rest in $t = 1.010 s$ . Estimate $g$ .

Forecast: Will measured $g$ come out near $9.8$ ? What tiny acceleration do we expect?

Step 1 — Get $a$ from the timing (kinematics). From rest, $h = \frac{1}{2} a t^{2}$ , so $a = \frac{2 h}{t ^{2}} = \frac{2 ( 0.500 )}{( 1.010 ) ^{2}} = \frac{1.000}{1.0201} \approx 0.9803 m/s^{2} .$ Why this step? We can measure distance and time directly; acceleration is inferred from them.

Step 2 — Relate $a$ to $g$ via the Atwood formula. Solving $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ for $g$ : $g = a \cdot \frac{m _{1} + m _{2}}{m _{1} - m _{2}} = 0.9803 \cdot \frac{1.000}{0.040} .$ Why this step? The machine amplifies $g$ into a measurable $a$ ; we invert that amplification to recover $g$ .

Step 3 — Compute. $g \approx 0.9803 \times 25 = 24.5 m/s^{2} .$ Hmm — that's far too big! The point: with $\frac{m _{1} + m _{2}}{m _{1} - m _{2}} = 25$ , any small timing error is multiplied by $25$ . Real experiments use much finer timing. With the ideal $g = 9.8$ , the true acceleration would be $a = 9.8/25 = 0.392 m/s^{2}$ , giving $t = 2 h / a = 1.000/0.392 \approx 1.597 s$ . Why this step? This shows the real design trade-off: nearly-equal masses give a slow, timeable fall, but demand precise timing because the amplification factor is huge.

Verify: With ideal values: $a = 0.392 m/s^{2}$ from $g = 9.8$ ; then $t^{2} = 2 h / a = 1.000/0.392 = 2.551$ , $t \approx 1.597 s$ ✓. See Hinglish note for the same idea narrated.

Case G — kinematics follow-up

Case H — exam twist: an extra applied force

Worked example Example 8 (Cell H)

An examiner adds a downward force $F = 10 N$ pulling on the left mass in addition to gravity. Masses $m_{1} = 4 kg$ (left), $m_{2} = 2 kg$ (right), $g = 10$ . Find $a$ .

Forecast: The plain formula no longer applies — why not? You must re-derive. Will $a$ be bigger or smaller than the no- $F$ case?

Step 1 — Write Newton's law for each mass (left assumed down = +). Left: $m_{1} g + F - T = m_{1} a$ . Right: $T - m_{2} g = m_{2} a$ . Why this step? The extra $F$ is a new term the boxed formula never accounted for, so we go back to Newton's Second Law and Free Body Diagrams and rebuild the equations from scratch.

Step 2 — Add to cancel $T$ . $m_{1} g + F - m_{2} g = (m_{1} + m_{2}) a .$ Why this step? Same trick as the parent derivation: $+ T$ and $- T$ cancel on adding.

Step 3 — Solve. $a = \frac{( m _{1} - m _{2} ) g + F}{m _{1} + m _{2}} = \frac{( 4 - 2 ) ( 10 ) + 10}{6} = \frac{20 + 10}{6} = \frac{30}{6} = 5 m/s^{2} .$ Why this step? The driving force now includes $F$ ; the inertia $m_{1} + m_{2}$ is unchanged. Bigger than the $F = 0$ case ( $a = 20/6 \approx 3.33$ ), as expected — you added a helping pull.

Step 4 — Tension. From the right equation, $T = m_{2} (g + a) = 2 (10 + 5) = 30 N$ . Why this step? With $a$ known, plug into either single-mass equation; the simplest is $m_{2}$ 's.

Verify: Left equation: $m_{1} g + F - T = 40 + 10 - 30 = 20 = m_{1} a = 4 (5)$ ✓. The lesson: memorised formulas break the moment a new force appears — always be ready to re-derive.

Recall The one habit that beats every scenario

Guess a positive direction, write $F_{n e t} = ma$ for each mass, add to kill tension. That single procedure solved cells A–H, including the ones where the boxed formula didn't apply. See Constraint Relations for why "same $a$ " is legitimate every time.

Connections

Atwood machine — derivation — the source formulas we stress-tested.
Newton's Second Law — rebuilt every case with $F_{n e t} = ma$ .
Free Body Diagrams — needed for the extra-force twist (Cell H).
Constraint Relations — justifies equal $∣ a ∣$ in every example.
Tension in Strings — why $T$ stayed symmetric in Example 2.
Harmonic Mean — the shape of the tension formula behind Cell E's limit.