Atwood machine — derivation: Visual walkthrough

Step 1 — Draw the machine and name every part

WHAT. Two lumps of stuff hang from a single rope. The rope goes up, bends over a wheel (a pulley), and comes back down. We give the left lump the name $m_{1}$ and the right lump the name $m_{2}$ . The letter $m$ just means "how much stuff is in this lump" — its mass, measured in kilograms.

WHY. Before any physics, we must be able to point at each object and each force with a name. If we skip this, every later equation is just letters floating in air.

PICTURE. In the figure below, the two lumps are blue (left, $m_{1}$ ) and orange (right, $m_{2}$ ). The rope is the gray line bending over the gray wheel. Notice: it is one continuous rope — that single fact is what ties the whole problem together.

Step 2 — The word "tension" and why the same arrow appears on both sides

WHAT. A rope can only pull, never push. Wherever a rope is attached, it pulls that object along the rope, toward the pulley. We call the strength of this pull the tension and give it the name $T$ (in newtons, N — a unit of force).

WHY this idea and not "the rope holds each weight". A beginner guesses the left rope pulls with $m_{1} g$ and the right with $m_{2} g$ . But it is one rope, and a massless rope must have the same tension throughout (proved in Tension in Strings) — otherwise a tiny piece of rope with zero mass would feel a net force and need infinite acceleration. So there is exactly one number $T$ , pulling up on the left lump and up on the right lump.

PICTURE. Two equal-length green arrows labelled $T$ , both pointing up (toward the pulley) — same length because same number.

Step 3 — Add gravity, and commit to a direction of motion

WHAT. Every lump is pulled straight down by gravity with a force called its weight, $m g$ . Here $g \approx 10 m/s^{2}$ is the strength of gravity (how hard Earth pulls per kilogram). So the left lump feels $m_{1} g$ down, the right feels $m_{2} g$ down.

Now we guess which way the system moves. Say $m_{1} > m_{2}$ (left is heavier). Then the heavy left side sinks and the light right side rises. We declare:

for $m_{1}$ : down is the positive direction,
for $m_{2}$ : up is the positive direction.

WHY. Newton's law needs signs, and signs need a chosen positive direction. Choosing "positive = the way it actually moves" for each lump keeps every acceleration a positive number. If we guessed the heavy side wrong, the final $a$ just comes out negative and politely tells us so — nothing breaks.

PICTURE. Blue down-arrow $m_{1} g$ (long, heavy side), orange down-arrow $m_{2} g$ (short), green up-arrows $T$ on both. A curved motion arrow shows left-down / right-up. These are the two free body diagrams — see Free Body Diagrams.

Step 4 — Newton's law on the heavy lump $m_{1}$

WHAT. Newton's Second Law says: (net force in the positive direction) $=$ (mass) $\times$ (acceleration). For $m_{1}$ , positive is down. Down forces: weight $m_{1} g$ (helps, $+$ ). Up forces: tension $T$ (opposes, $-$ ).

$weight of m_{1} pulls DOWN (+) m_{1} g - rope pull pulls UP (-) T = its mass m_{1} its accel. (down, +) a (1)$

WHY. We isolate one lump so only two forces appear. The acceleration is called $a$ — the same letter we will reuse for $m_{2}$ , because the inextensible rope forces equal acceleration magnitudes (Step 6 proves it visually).

PICTURE. Left lump alone. Down $m_{1} g$ (long), up $T$ (short), so the leftover downward stub is the net force $m_{1} g - T$ that drives $m_{1} a$ .

Step 5 — Newton's law on the light lump $m_{2}$

WHAT. Now $m_{2}$ , whose positive direction is up. Up forces: tension $T$ ( $+$ ). Down forces: weight $m_{2} g$ ( $-$ ).

$rope pull pulls UP (+) T - weight of m_{2} pulls DOWN (-) m_{2} g = its mass m_{2} its accel. (up, +) a (2)$

WHY. Same $T$ (Step 2), same $a$ (Step 6). Notice the signs flipped compared to $(1)$ : here tension helps (it lifts $m_{2}$ ) and gravity fights it. That sign-flip is exactly why adding the equations will cancel $T$ .

PICTURE. Right lump alone. Up $T$ (longer), down $m_{2} g$ (shorter), leftover upward stub is the net force $T - m_{2} g$ that drives $m_{2} a$ .

Step 6 — Why both accelerations are the same $a$ (the constraint)

WHAT. If the left lump moves down by a distance $x$ , the rope that vanished on the left must reappear on the right (the rope can't stretch or bunch up), so the right lump rises by the same $x$ . Same distance in the same time ⇒ same speed ⇒ same acceleration magnitude $a$ .

WHY we needed this. In Steps 4 and 5 we wrote the same $a$ in both. This step is the licence to do that. It is the constraint relation (Constraint Relations): one rope, one length, so the two motions are locked together, opposite in direction but equal in size.

PICTURE. A "before/after" strip: rope moves, left drops $x$ , right rises $x$ . The equal blue and orange displacement brackets show why $a$ is shared.

Step 7 — Add the two equations to kill $T$ , and solve for $a$

WHAT. Stack $(1)$ and $(2)$ and add them straight down:

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

The $- T$ and $+ T$ cancel. Solve:

$a = \frac{( m _{1} - m _{2} ) extra weight, in masses g}{total stuff being moved m _{1} + m _{2}}$

WHY add, not subtract? Because $T$ appears once as $- T$ and once as $+ T$ ; addition is the one move that deletes the unknown we don't yet want. What survives is a single equation in the single unknown $a$ .

Read the formula physically: the top $(m_{1} - m_{2}) g$ is the net driving force — only the difference in weight pushes the system (the shared part just balances). The bottom $(m_{1} + m_{2})$ is the total inertia — all the mass has to be shoved, both lumps. More difference → faster; more total mass → slower.

PICTURE. A balance-beam cartoon: the difference weight $(m_{1} - m_{2}) g$ sits on the "push" pan; the total mass $(m_{1} + m_{2})$ sits on the "resist" pan; their ratio (times $g$ ) is $a$ .

Step 8 — Back-substitute to get the tension $T$

WHAT. Put $a$ into equation $(2)$ , $T = m_{2} (g + a)$ :

= m_2 \cdot \frac{(m_1+m_2)g + (m_1-m_2)g}{m_1+m_2} = \boxed{\,\frac{2 m_1 m_2}{m_1+m_2}\,g\,}$$ **WHY equation (2) and not (1)?** Either works — both must give the same $T$ (that's a free consistency check). We use $(2)$ because $T$ is already alone on the left there. **Read it:** the result is **symmetric** in $m_1$ and $m_2$ (swap them, nothing changes) — correct, because *the rope has no idea which side you named "1"*. The quantity $\frac{2m_1m_2}{m_1+m_2}$ is the [[Harmonic Mean]] combination of the two masses, so $T$ always lands **between** the two weights $m_2 g$ and $m_1 g$. **PICTURE.** A number line from $m_2 g$ to $m_1 g$ with $T$ marked strictly inside — tension is never as big as the heavy weight nor as small as the light one. ![[deepdives/dd-physics-1.2.10-d2-s08.png]] --- ## Step 9 — Every edge case, checked on one picture **WHAT.** A formula you trust must survive its extremes. Feed the boxed results the special inputs: | Input | $a$ | $T$ | Meaning | |---|---|---|---| | $m_1 = m_2 = m$ | $0$ | $mg$ | Balanced — nothing moves; rope just holds each weight | | $m_2 = 0$ | $g$ | $0$ | Left lump in **free fall**; nothing on the right to pull | | $m_1 \to \infty$ | $\to g$ | $\to 2m_2 g$ | Huge left mass yanks $m_2$ up at $\approx 2g$ net | | $m_1 < m_2$ | negative | still $+$ | Our guess was backwards; sign of $a$ self-corrects | **WHY.** These are the scenarios a real problem might hand you. The last row matters most: if you assumed the wrong heavy side, **$a$ comes out negative** and simply means "it moves the other way" — the derivation never breaks. **PICTURE.** Three mini-panels: (a) equal masses, motion frozen; (b) $m_2=0$, free fall arrow $=g$; (c) $m_1$ huge, $m_2$ flung upward with a $2g$-length arrow. ![[deepdives/dd-physics-1.2.10-d2-s09.png]] > [!example] Plug-in check (matches the parent's Example 1) > $m_1 = 5$, $m_2 = 3$, $g = 10$: > $$a = \frac{(5-3)(10)}{5+3} = \frac{20}{8} = 2.5\ \text{m/s}^2, \qquad T = \frac{2(5)(3)}{8}(10) = 37.5\ \text{N}.$$ > Cross-check on $m_2$: $T - m_2 g = 37.5 - 30 = 7.5 = 3(2.5) = m_2 a$ ✓. And $T=37.5$ sits between $m_2g=30$ and $m_1g=50$ ✓. --- ## The one-picture summary Everything above, compressed: two free bodies → two Newton equations → add to kill $T$ → the difference-over-sum for $a$ and double-product-over-sum for $T$. ![[deepdives/dd-physics-1.2.10-d2-s10.png]] > [!mnemonic] The two boxes in one breath > **"DIFFERENCE over SUM for $a$; DOUBLE-PRODUCT over SUM for $T$."** Both denominators are $m_1+m_2$ — the total inertia always sits on the bottom. > [!recall]- Feynman: the whole walkthrough in plain words > Two buckets hang on one light rope over a smooth wheel. Because the rope is light and the wheel is smooth, the rope pulls *up* on both buckets with the exact same strength — call it $T$. Gravity pulls each bucket down by its own weight. Pick the heavy bucket; write "its weight minus the rope pull equals its mass times how fast it speeds up." Pick the light bucket; write "the rope pull minus its weight equals its mass times the same speed-up" — *same* speed-up because one rope can't stretch, so both ends move together. Add those two sentences and the mysterious rope pull $T$ vanishes, leaving: the *extra* weight (the difference) is what does the pushing, and *all* the stuff (the sum) is what has to be pushed. Divide them, sprinkle in gravity $g$, and that's the acceleration. Slide it back into either sentence to find the rope pull, which always lands somewhere between the two weights. Try equal buckets (nothing moves), or an empty bucket (the other free-falls), and the formulas agree — so you can trust them. > [!recall]- Quick self-test > Why can we write the same $a$ for both masses? ::: The rope is inextensible — a constraint that locks their displacements (hence accelerations) equal in size, opposite in direction. > Why does adding equations $(1)$ and $(2)$ help? ::: $T$ appears as $-T$ and $+T$; adding cancels it, leaving one equation in $a$. > Why must $T$ lie between $m_2g$ and $m_1g$? ::: It is the harmonic-mean combination $\frac{2m_1m_2}{m_1+m_2}g$, which is always between the two weights. > If $m_1<m_2$, what does the formula do? ::: Gives a negative $a$, meaning the system moves opposite to the guessed direction. --- ## Connections - [[Newton's Second Law]] — $F_{net}=ma$ applied per lump (Steps 4–5). - [[Free Body Diagrams]] — isolating each mass (Steps 3–5). - [[Tension in Strings]] — why one $T$ everywhere (Step 2). - [[Constraint Relations]] — inextensible rope ⇒ shared $a$ (Step 6). - [[Inclined Plane Problems]] — same recipe with a gravity component. - [[Pulley Systems & Mechanical Advantage]] — many-pulley generalisation. - [[Harmonic Mean]] — the shape of the tension result (Step 8).