Atwood machine — derivation: Exercises — Physics deep dive

The two engines you will use everywhere below, from Newton's Second Law applied to each mass:

The figure below shows the two free-body diagrams you will draw for every plain-Atwood problem — one tension arrow up on each mass, one weight arrow down, opposite motion arrows.

What the figure shows (read this alongside it): A cyan circle at the top is the frictionless, massless pulley. A single white rope runs over it, down the left to the amber block $m_{1}$ (the heavier mass) and down the right to the cyan block $m_{2}$ (the lighter mass). On each block there is a cyan arrow pointing up labelled $T$ — these two arrows are the same length on purpose, because it is one rope over a frictionless pulley, so the tension is a single value $T$ at both ends. On each block there is also an amber arrow pointing down: $m_{1} g$ under the left block, $m_{2} g$ under the right block. Finally, two white "motion" arrows show the outcome: $m_{1}$ accelerates down with magnitude $a$ , and $m_{2}$ accelerates up with the same magnitude $a$ (opposite directions, equal size — the inextensible-string rule). Because the two cyan $T$ arrows are identical, adding the two Newton equations cancels $T$ — that is the whole trick, visible right there in the picture.

Level 1 — Recognition

L1.1 — Which quantity is the denominator of BOTH the acceleration and the tension formula, and what does it physically represent?

Recall Solution — L1.1

The denominator is the ==total mass $m_{1} + m_{2}$ ==. Physically it is the inertia of the whole system — all the "stuff" that must be set moving by the net force. Both $a$ and $T$ divide by it, because both are ultimately governed by how sluggish the coupled system is.

L1.2 — Without computing anything, state whether the tension $T$ is bigger than $m_{2} g$ , smaller than $m_{1} g$ , or both. Give the one-line reason.

Recall Solution — L1.2

Both. $T$ lies strictly between $m_{2} g$ and $m_{1} g$ .

On the rising mass $m_{2}$ : it accelerates up, so the net force is up, so $T > m_{2} g$ .
On the falling mass $m_{1}$ : it accelerates down, so the net force is down, so $T < m_{1} g$ . One tension, sandwiched: $m_{2} g < T < m_{1} g$ .

L1.3 — For $m_{1} = 4 kg$ , $m_{2} = 4 kg$ , write down $a$ and $T$ by inspection (no algebra).

Recall Solution — L1.3

Equal masses ⇒ balanced ⇒ $a = 0$ . With no acceleration each side is in equilibrium, so the rope simply holds a weight up: $T = m g = 4 \times 10 = 40 N$ . (Check with the formula: $T = \frac{2 \cdot 4 \cdot 4}{8} \cdot 10 = 40 N$ ✓.)

Level 2 — Application

L2.1 — $m_{1} = 6 kg$ , $m_{2} = 2 kg$ . Find $a$ and $T$ .

Recall Solution — L2.1

Why the net force is $m_{1} g - m_{2} g$ . Draw the two free bodies (figure above). Treat the rope-coupled pair as one system moving together, calling the direction " $m_{1}$ down, $m_{2}$ up" positive. Write Newton's Second Law for each and add: $on m_{1} m_{1} g - T + on m_{2} T - m_{2} g = m_{1} a + m_{2} a .$ The two tension terms $+ T$ and $- T$ cancel (same rope, same $T$ ), leaving $m_{1} g - m_{2} g$ on the left. So the net driving force on the whole system is the weight difference $= (6 - 2) \times 10 = 40 N$ , resisted by total inertia $= 8 kg$ . $a = \frac{( 6 - 2 ) 10}{6 + 2} = \frac{40}{8} = 5 m/s^{2}$ $T = \frac{2 \cdot 6 \cdot 2}{8} \cdot 10 = \frac{24}{8} \cdot 10 = 30 N$ Check on $m_{2}$ : $T - m_{2} g = 30 - 20 = 10 = m_{2} a = 2 \times 5$ ✓. And $20 = m_{2} g < 30 = T < 60 = m_{1} g$ ✓.

L2.2 — $m_{1} = 0.55 kg$ , $m_{2} = 0.45 kg$ . Find $a$ . Then find the time to descend $h = 0.8 m$ from rest.

Recall Solution — L2.2

$a = \frac{( 0.55 - 0.45 ) 10}{0.55 + 0.45} = \frac{0.10 \times 10}{1.0} = 1 m/s^{2}$ > The acceleration is constant, so ordinary kinematics (constant-$a$ motion) applies. Starting from rest, $h = \frac{1}{2} a t^{2}$ : $t = \frac{2 h}{a} = \frac{2 \times 0.8}{1} = 1.6 \approx 1.265 s$

L2.3 — Same masses as L2.2. What is the velocity of the falling mass just as it has dropped $0.8 m$ ?

Recall Solution — L2.3

Use $v^{2} = u^{2} + 2 ah$ with $u = 0$ : $v = 2 ah = 2 \times 1 \times 0.8 = 1.6 \approx 1.265 m/s$ (Numerically equal to the time here only because $a = 1$ ; that is a coincidence of the numbers, not a rule.)

Level 3 — Analysis

L3.1 — A student measures $a = 2 m/s^{2}$ and knows the total mass is $m_{1} + m_{2} = 10 kg$ . Find $m_{1}$ and $m_{2}$ ( $g = 10$ ).

Recall Solution — L3.1

From $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ : the weight difference is $m_{1} - m_{2} = \frac{a ( m _{1} + m _{2} )}{g} = \frac{2 \times 10}{10} = 2 kg .$ Now solve the pair $m_{1} + m_{2} = 10$ , $m_{1} - m_{2} = 2$ : $m_{1} = \frac{10 + 2}{2} = 6 kg, m_{2} = \frac{10 - 2}{2} = 4 kg .$ Check: $a = \frac{( 6 - 4 ) 10}{10} = 2$ ✓.

L3.2 — Suppose you assumed the wrong direction: you guessed $m_{2}$ falls when in fact $m_{1}$ (heavier) falls. You wrote $m_{2} g - T = m_{2} a$ and $T - m_{1} g = m_{1} a$ . Solve and interpret the sign. Use $m_{1} = 6, m_{2} = 2$ .

Recall Solution — L3.2

Add the two (wrong-direction) equations: $m_{2} g - m_{1} g = (m_{1} + m_{2}) a$ , so $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}} = \frac{( 2 - 6 ) 10}{8} = - 5 m/s^{2} .$ The negative sign is the machine correcting your guess: it means the system actually accelerates opposite to the direction you assumed positive — i.e. $m_{1}$ really falls, at magnitude $5 m/s^{2}$ . This is exactly the L2.1 answer. Lesson: a wrong direction guess is safe; the algebra self-corrects, you just read the sign.

L3.3 — Two students argue. Student A: "Bigger $m_{1}$ always means bigger tension." Student B: "Not necessarily." Who is right? Test with $m_{2} = 2$ fixed, comparing $m_{1} = 6$ vs $m_{1} = 100$ .

Recall Solution — L3.3

$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g = \frac{2 \cdot 2 \cdot m _{1}}{m _{1} + 2} \cdot 10 = \frac{40 m _{1}}{m _{1} + 2}$ .

$m_{1} = 6$ : $T = \frac{240}{8} = 30 N$ .
$m_{1} = 100$ : $T = \frac{4000}{102} \approx 39.2 N$ . Tension did increase, but notice it is saturating: as $m_{1} \to \infty$ , $T \to 2 m_{2} g = 40 N$ and never exceeds it. So Student A is right that it increases, but Student B captures the deeper truth — tension is bounded by $2 m_{2} g$ , no matter how huge $m_{1}$ gets. Do not confuse this tension ceiling with acceleration: the acceleration of the light mass tops out at $g$ (as $m_{1} \to \infty$ , $a \to g$ ), not $2 g$ . The reason the tension reaches $2 m_{2} g$ is that the rope must supply $m_{2} g$ to hold $m_{2}$ 's weight plus another $m_{2} g$ to accelerate it upward at $g$ — two contributions of $m_{2} g$ , giving a force of $2 m_{2} g$ but an acceleration of only $g$ .

Level 4 — Synthesis

L4.1 — Atwood on an inclined side. Mass $m_{2} = 3 kg$ sits on a frictionless incline whose slope makes angle $θ = 3 0^{\circ}$ with the horizontal. Its rope runs over a pulley at the top of the incline down to a hanging mass $m_{1} = 4 kg$ . Find $a$ and $T$ . ( $g = 10$ , $sin 3 0^{\circ} = 0.5$ .)

Recall Solution — L4.1

This blends the Atwood method with inclined-plane resolution. For a mass on a frictionless incline whose slope makes angle $θ$ with the horizontal, the component of gravity along the slope (the part that pulls the block back down the ramp) is $m_{2} g sin θ$ — see the figure below, where only the along-slope part fights the rope's pull. Everything else — one rope, one tension, one shared $a$ — is unchanged.

Let $m_{1}$ fall (down), pulling $m_{2}$ up the slope.

$m_{1}$ (down +): $m_{1} g - T = m_{1} a$
$m_{2}$ (up-slope +): $T - m_{2} g sin θ = m_{2} a$

Add to cancel $T$ : $m_{1} g - m_{2} g sin θ = (m_{1} + m_{2}) a$ $a = \frac{( m _{1} - m _{2} s i n θ ) g}{m _{1} + m _{2}} = \frac{( 4 - 3 \times 0.5 ) 10}{7} = \frac{( 4 - 1.5 ) 10}{7} = \frac{25}{7} \approx 3.571 m/s^{2} .$ Tension from $m_{1}$ 's equation: $T = m_{1} (g - a) = 4 (10 - 3.571) = 4 \times 6.429 \approx 25.71 N$ . Check on $m_{2}$ : $T - m_{2} g sin θ = 25.71 - 15 = 10.71 = m_{2} a = 3 \times 3.571$ ✓. Notice the incline replaces the light side's full weight $m_{2} g$ with only its slope component $m_{2} g sin θ$ . The steeper the ramp, the closer $sin θ$ gets to $1$ and the more the light side behaves like a fully hanging mass; a vertical wall ( $θ = 9 0^{\circ}$ , $sin θ = 1$ ) recovers the plain Atwood exactly, while a flat table ( $θ = 0^{\circ}$ , $sin θ = 0$ ) leaves nothing to oppose $m_{1}$ so $a = \frac{m _{1} g}{m _{1} + m _{2}}$ .

The figure makes the key move visible: on the slope, gravity splits into an along-slope part $m_{2} g sin θ$ (the only part that opposes the pull) and an into-slope part that the surface cancels.

L4.2 — Energy cross-check. For the plain Atwood of L2.1 ( $m_{1} = 6, m_{2} = 2$ ), the heavy mass drops $h = 0.9 m$ . Use energy conservation to find the final speed, and confirm it matches the kinematics $v = 2 ah$ .

Recall Solution — L4.2

Energy view: $m_{1}$ falls $h$ (loses PE $m_{1} g h$ ), $m_{2}$ rises $h$ (gains PE $m_{2} g h$ ). The net PE released becomes the shared kinetic energy of both masses (they move at the same speed $v$ ): $(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2}$ $v^{2} = \frac{2 ( m _{1} - m _{2} ) g h}{m _{1} + m _{2}} = \frac{2 ( 6 - 2 ) ( 10 ) ( 0.9 )}{8} = \frac{144}{8} = 18 \Rightarrow v = 18 \approx 4.243 m/s .$ Kinematics cross-check: $a = 5 m/s^{2}$ (from L2.1), $v = 2 ah = 2 \times 5 \times 0.9 = 9 \cdot 2 = 18 \approx 4.243 m/s$ ✓. The two methods agree because $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ is exactly the factor buried in the energy equation.

Level 5 — Mastery

L5.1 — Prove the tension bound. Show that for fixed $m_{2}$ and any $m_{1} > 0$ , the tension satisfies $T < 2 m_{2} g$ , and that this limit is approached as $m_{1} \to \infty$ .

Recall Solution — L5.1

Write $T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}$ . Divide numerator and denominator by $m_{1}$ : $T = \frac{2 m _{2} g}{1 + \frac{m _{2}}{m _{1}}} .$ Since $m_{1}, m_{2} > 0$ , the term $\frac{m _{2}}{m _{1}} > 0$ , so the denominator is $> 1$ , hence $T < 2 m_{2} g$ . As $m_{1} \to \infty$ , $\frac{m _{2}}{m _{1}} \to 0$ , so the denominator $\to 1$ and $T \to 2 m_{2} g$ . This is the limiting behaviour flagged in the parent note: an infinitely heavy $m_{1}$ pulls $m_{2}$ up with a net force $\to m_{2} g$ (net accel $g$ ), so the rope must supply weight $m_{2} g$ plus the force $m_{2} g$ to accelerate it — total $2 m_{2} g$ . The bound is never reached, only approached. This is the harmonic-mean structure of $\frac{2 m _{1} m _{2}}{m _{1} + m _{2}}$ showing its ceiling.

L5.2 — Find the mass ratio for a target acceleration. For what ratio $r = m_{1} / m_{2}$ is the acceleration exactly $a = g /3$ ?

Recall Solution — L5.2

Divide the acceleration formula's top and bottom by $m_{2}$ : $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( r - 1 ) g}{r + 1} .$ Set equal to $g /3$ and cancel $g$ : $\frac{r - 1}{r + 1} = \frac{1}{3} .$ Cross-multiply and solve: $3 (r - 1) = r + 1 \Rightarrow 3 r - 3 = r + 1 \Rightarrow 2 r = 4 \Rightarrow r = 2.$ So $m_{1} = 2 m_{2}$ . Check: $a = \frac{( 2 - 1 ) g}{2 + 1} = \frac{g}{3}$ ✓. Notice acceleration depends only on the ratio, never the absolute masses — doubling both masses leaves $a$ untouched.

L5.3 — Design / degenerate case. You want to measure $g$ à la Atwood with an acceleration of only $a = 0.05 g$ so it is slow enough to time. If the total mass must be $m_{1} + m_{2} = 2 kg$ , find $m_{1}$ and $m_{2}$ . Then state what happens to $a$ and $T$ in the fully degenerate limit $m_{2} \to 0$ .

Recall Solution — L5.3

From $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = 0.05 g$ , the fraction $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} = 0.05$ , so $m_{1} - m_{2} = 0.05 \times (m_{1} + m_{2}) = 0.05 \times 2 = 0.1 kg .$ With $m_{1} + m_{2} = 2$ : $m_{1} = \frac{2 + 0.1}{2} = 1.05 kg$ , $m_{2} = \frac{2 - 0.1}{2} = 0.95 kg$ . Check: $a = \frac{( 1.05 - 0.95 ) g}{2} = \frac{0.1 g}{2} = 0.05 g$ ✓.

Degenerate limit $m_{2} \to 0$ : the light side vanishes.

$a = \frac{( m _{1} - 0 ) g}{m _{1} + 0} = g$ — the heavy mass is in free fall (nothing holds it back).
$T = \frac{2 m _{1} \cdot 0}{m _{1} + 0} g = 0$ — the rope goes slack, there is nothing on the other end to pull against. Both agree perfectly with intuition: no partner, no tension, pure gravity.

Recall Feynman recap — the whole ladder in one breath

Every problem here is the same two equations from Newton's Second Law — heavy side $m_{1} g - T = m_{1} a$ , light side $T - (opposing force) = m_{2} a$ — added together to kill $T$ . Because it is one massless rope over a frictionless pulley, the tension is a single value, so $+ T$ and $- T$ cancel on adding, leaving $(driving force) = (m_{1} + m_{2}) a$ . Solve that one line for $a$ , then back-substitute into either equation for $T$ . What changes across levels is only the "opposing force": full weight $m_{2} g$ on the flat machine, slope component $m_{2} g sin θ$ on an incline, zero when the partner vanishes. Master that one move — write two, add to cancel $T$ , solve, back-substitute — and every Atwood variant collapses into arithmetic.

Connections

Atwood machine — derivation (index 1.2.10) — the parent derivation these exercises drill.
Newton's Second Law — the single law behind every solution.
Free Body Diagrams — isolate each mass before writing equations.
Tension in Strings — why one tension serves both sides.
Constraint Relations — the inextensible-string equal- $a$ rule used throughout.
Inclined Plane Problems — the L4.1 slope variant.
Pulley Systems & Mechanical Advantage — where multi-rope generalisations live.
Harmonic Mean — the ceiling structure proved in L5.1.