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Atwood machine: acceleration, tension, and the first metre

SPH4U Grade 12 Physics · Dynamics

In a dynamics lab, a 600 g mass and a 400 g mass hang from the ends of a light string run over a frictionless, massless pulley clamped to a shelf bracket. The pair is released from rest and the heavier mass descends 90.0 cm to a foam pad while a motion sensor watches.

Given
  • m₁ = 600 g — Descending mass
  • m₂ = 400 g — Rising mass
  • d = 90 cm — Drop to the pad
Determine
  1. (a)the acceleration of the system
  2. (b)the tension in the string
  3. (c)the speed of the masses just before the pad
  4. (d)the time the descent takes
Step 1 of 4(a) · solve for Acceleration

Gravity only gets to pull on the 200 g difference, but it has to move the full 1000 g of string-connected hardware — that is the whole formula, difference over sum. The classic wreck is a = g: these masses accelerate at exactly g/5, not g.

Rearranged for a
a=(m1−m2)gm1+m2a = \frac{\left(m_1 - m_2\right) g}{m_1 + m_2}
Your values, in your units
a=((600 g)−(400 g))(9.80665 m/s2)(600 g)+(400 g)a = \frac{\left(\left( 600\ \text{g} \right) - \left( 400\ \text{g} \right)\right) \left( 9.80665\ \text{m/s}^{2} \right)}{\left( 600\ \text{g} \right) + \left( 400\ \text{g} \right)}
Converted to base units
a=((0.6 kg)−(0.4 kg))(9.80665 m/s2)(0.6 kg)+(0.4 kg)a = \frac{\left(\left( 0.6\ \text{kg} \right) - \left( 0.4\ \text{kg} \right)\right) \left( 9.80665\ \text{m/s}^{2} \right)}{\left( 0.6\ \text{kg} \right) + \left( 0.4\ \text{kg} \right)}
Answer
a=1.9613 m/s2a = 1.9613\ \text{m/s}^{2}

Carried onward at full precision, not this rounded figure.

Open the Atwood Machine Acceleration solver →

Step 2 of 4(b) · solve for Rope tension

Isolate the RISING mass: for it to accelerate upward, the string must pull harder than its 3.92 N weight, so T = m₂(g + a). The tension is one number for the whole string — check it against the falling side, T = m₁(g − a), which must and does give the same 4.71 N.

Rearranged for T
T=m(g+a)T = m\left(g + a\right)
1.9613 m/s²carried from step 1
Your values, in your units
T=(400 g)((9.80665 m/s2)+(1.96133 m/s2))T = \left( 400\ \text{g} \right)\left(\left( 9.80665\ \text{m/s}^{2} \right) + \left( 1.96133\ \text{m/s}^{2} \right)\right)
Converted to base units
T=(0.4 kg)((9.80665 m/s2)+(1.96133 m/s2))T = \left( 0.4\ \text{kg} \right)\left(\left( 9.80665\ \text{m/s}^{2} \right) + \left( 1.96133\ \text{m/s}^{2} \right)\right)
Answer
T=4.7072 NT = 4.7072\ \text{N}

Carried onward at full precision, not this rounded figure.

Open the Rope Tension When Lifting a Mass solver →

Step 3 of 4(c) · solve for Final velocity

Dynamics hands its answer to kinematics: v² = v₀² + 2ad with the system's own 1.96 m/s² from step 1, never with 9.8. Feeding in g here is the single most common error in Atwood problems — it answers a different question, namely free fall.

Rearranged for v
v=v02+2adv = \sqrt{v_0^2 + 2 a d}
1.9613 m/s²carried from step 1
Your values, in your units
v=(0 m/s)2+2 (1.96133 m/s2) (90 cm)v = \sqrt{\left( 0\ \text{m/s} \right)^2 + 2 \, \left( 1.96133\ \text{m/s}^{2} \right) \, \left( 90\ \text{cm} \right)}
Converted to base units
v=(0 m/s)2+2 (1.96133 m/s2) (0.9 m)v = \sqrt{\left( 0\ \text{m/s} \right)^2 + 2 \, \left( 1.96133\ \text{m/s}^{2} \right) \, \left( 0.9\ \text{m} \right)}
Answer
v=1.8789 m/sv = 1.8789\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Velocity-Displacement Relation (v² = v₀² + 2ad) solver →

Step 4 of 4(d) · solve for Time

With the final speed and the acceleration both in hand, t = v/a closes the story. Nearly a full second to fall 90 cm — the machine's entire purpose is to dilute free fall until a stopwatch can catch it.

Rearranged for t
t=v−v0at = \tfrac{v - v_0}{a}
1.8789 m/scarried from step 3
1.9613 m/s²carried from step 1
Your values, in your units
t=(1.87893 m/s)−(0 m/s)(1.96133 m/s2)t = \tfrac{\left( 1.87893\ \text{m/s} \right) - \left( 0\ \text{m/s} \right)}{\left( 1.96133\ \text{m/s}^{2} \right)}
Answer
t=957.99 mst = 957.99\ \text{ms}

Carried onward at full precision, not this rounded figure.

Open the Final Velocity (Uniform Acceleration) solver →

Therefore the system accelerates at 1.96 m/s² (exactly g/5), the string carries 4.71 N everywhere along its length, and the heavier mass meets the pad at 1.88 m/s after 0.958 s of descent.

Why this order

The order is Newton first, kinematics second, and the boundary between them is the acceleration. Step 1 is the only place any force reasoning happens: the imbalance (m₁ − m₂)g drives, the total m₁ + m₂ resists, and everything after that is bookkeeping on a uniformly accelerated object. The tension in step 2 is where the physics is usually lost. Students reach for the weight of one mass (3.92 N), the weight of the other (5.88 N), or the difference (1.96 N) — and the true value, 4.71 N, is none of them. It sits strictly between the two weights, because the string has to hold the light mass back less than the heavy one demands, and that in-between-ness is not a coincidence of these numbers; it is what makes the system move at all.

Checking the tension from both ends — m₂(g + a) from the rising side, m₁(g − a) from the falling side — is the free audit this chain is built around, exactly the discipline a free-body diagram is supposed to teach. And the machine itself has a history worth telling: George Atwood published it in 1784 precisely because free fall was too fast for eighteenth-century clocks. With these masses the apparatus scales gravity down five-fold, and a 90 cm drop stretches to nearly a second — a fall that a pendulum clock, or today a $40 photogate, can time to the millisecond. Diluting a phenomenon until your instruments can see it is still a standard experimental move; the Atwood machine is just its most teachable ancestor.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.