Preview

Balancing Act

Q: Place a 2kg mass at 1m left. Where must a 4kg mass go to balance?

You can work it out this way: use τ = mgd: 2×1 = 4×d.

Q: Can two equal masses at equal distances ever be unbalanced?

You can work it out this way: think about what torque = r × F means when r and F are equal.

Discover torque and rotational equilibrium

Two children walk toward a seesaw: a big kid and a small kid. Everyone on the playground already knows the ending — the big kid's side slams down. Except today the big kid sits close to the middle, the small kid scoots all the way out to the end, and the plank floats level. The playground's physics is wrong, and four hundred years of engineering says so: a crane hoists a four-tonne load at the end of a long boom while a squat block of concrete, parked barely a meter behind the pivot, holds the whole thing still. The secret is that rotation doesn't price weight — it prices weight TIMES distance, τ = mgd, and it keeps two separate books at once. Book one: the torques must cancel, clockwise against counterclockwise, or the beam turns. Book two: the forces must cancel too, which is why the pivot quietly pushes up with the sum of every weight aboard — 29.4 newtons here, more the moment you add mass — doing its hardest work precisely when the beam looks most peaceful. Master the product and the seesaw stops being a mystery: double the mass, halve the distance, nothing moves. In this lab you'll catch a 2 kg mass balancing a 1 kg mass with your own eyes, walk the three presets from playground symmetry to crane arithmetic, and then deliberately unbalance the beam — and fix it with the distance slider alone, which is the moment the weight-only instinct finally lets go.

What you'll be able to do

Falsify the weight-only rule with the panel's own numbers: at the data defaults (1 kg at 2.0 m left vs 2 kg at 1.0 m right — the right side is TWICE as heavy) the rows read τ Left 19.60 = τ Right 19.60 N·m, Net Torque 0.00, BALANCED — because the heavier side's lever arm is exactly half. Torque is mgd, not mg: 'the greater the lever arm, the greater the magnitude of the torque' (OpenStax §10.6, verified)
Apply the law of moments m₁d₁ = m₂d₂ across the preset tour: Simple Seesaw ties 5 kg at 2.0 m both ways (98.00 = 98.00 N·m), Unequal Mass balances 10 kg at 1.0 m against 5 kg at 2.0 m (98.00 = 98.00 — the heavier side does NOT sink), and Crane Counterweight holds 4 kg at 3.0 m with 15 kg at just 0.8 m (117.60 = 117.60) — mass and distance trade in exact inverse proportion, which is why OpenStax's torque balance 'may be used to measure mass' with g cancelling out (§12.2 Example 12.3, verified)
Keep both equilibrium books at once: at defaults the Pivot Reaction row reads 29.4 N = (1+2)·9.8 N — the first condition ΣF = 0 holding while the second Στ = 0 holds ('when rotational and translational equilibrium conditions hold simultaneously in one frame… they also hold in any other inertial frame', §12.1 verified). Slide masses anywhere: the reaction never leaves (m₁+m₂)g, balance or not — the pivot works hardest exactly when the beam is balanced

Formulas

\tau = r \times F

Torque

\sum \tau = 0

Rotational equilibrium condition

\tau = m \cdot g \cdot d

Torque from weight

m_1 d_1 = m_2 d_2

Law of moments (balance condition)

Make a prediction

Default state: Mass 1 is 1.0 kg parked 2.0 m left of the pivot; Mass 2 is 2.0 kg parked 1.0 m right of it. The right side carries TWICE the mass. What does the Net Torque row read — and what does the beam do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Replay the prediction live from the defaults: slide Mass 2 up from 2.0 to 4.0 kg and watch Net Torque swing to −19.60 N·m as the beam leans right — then restore the balance using ONLY the Distance 2 slider: 4.0 kg at 0.5 m ties 19.60 = 19.60 again. Mass broke it, distance fixed it — the trade is the lesson.
Hit the Unequal Mass preset and read all three torque rows slowly: 10 kg at 1.0 m against 5 kg at 2.0 m — τ Left 98.00 = τ Right 98.00, Net 0.00, BALANCED, with the heavier side plainly aboard (F1 row 98.0 N vs F2 49.0 N). This is the predict gate's wrong answer losing on preset autoplay.
Hit Crane Counterweight and run the tipping audit: 15 kg at 0.8 m holds 4 kg at 3.0 m (117.60 = 117.60). Now slide Mass 1 down and find exactly where the crane tips — at 10 kg the left side brings only 78.40 N·m and the beam goes right with a 39.20 N·m deficit. The margin is m₁·0.8 = 4·3; the counterweight's address matters more than its size.
Audit the pivot's invisible labor: from the defaults (Pivot Reaction 29.4 N), add mass anywhere or tour the presets — Seesaw 98.0 N, Unequal 147.0 N, Crane 186.2 N — and watch the row hold (m₁+m₂)·g exactly, tilted or level. ΣF = 0 never sleeps; the pivot pushes hardest precisely when the beam balances.

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Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: 0.00 N·m — the beam stays level: the heavier side's lever arm is exactly half (1.0 m vs 2.0 m), so τ = mgd ties at 19.60 = 19.60 N·m and the BALANCED indicator lights

Torque prices force AND lever arm together: τ = r⊥F (OpenStax §10.6 — 'the greater the lever arm, the greater the magnitude of the torque'). Left side: 1 kg × 9.8 m/s² × 2.0 m = 19.60 N·m counterclockwise. Right side: 2 kg × 9.8 m/s² × 1.0 m = 19.60 N·m clockwise. The books tie, Net Torque reads 0.00, and the second equilibrium condition Στ = 0 (OpenStax §12.1) says the beam does not turn. Option B is the weight-only rule — the strategy Siegler's balance-scale studies documented children using (compare the weights, ignore the distances) and the exact misconception PhET built Balancing Act to defeat ('Predict how objects of various masses can be used to make a plank balance'). It half-knows the physics: mass really does buy torque, 2 kg pulls twice as hard as 1 kg — but the distance slider sells it back at the same exchange rate, and here the heavier side's arm is exactly half. Option C makes the mirror-image error, crowning distance alone — it would tip the beam left, but 2.0 m doesn't beat 1.0 m any more than 2 kg beats 1 kg; only the PRODUCT competes. The playground isn't crazy: equal arms are just the special case where the weight comparison happens to be the whole story.

Quiz (0/3)

Place a 2kg mass at 1m left. Where must a 4kg mass go to balance?

Can two equal masses at equal distances ever be unbalanced?

Where is the center of mass of three unequal masses?

You can now

Falsify the weight-only rule with the panel's own numbers: at the data defaults (1 kg at 2.0 m left vs 2 kg at 1.0 m right — the right side is TWICE as heavy) the rows read τ Left 19.60 = τ Right 19.60 N·m, Net Torque 0.00, BALANCED — because the heavier side's lever arm is exactly half. Torque is mgd, not mg: 'the greater the lever arm, the greater the magnitude of the torque' (OpenStax §10.6, verified)
Apply the law of moments m₁d₁ = m₂d₂ across the preset tour: Simple Seesaw ties 5 kg at 2.0 m both ways (98.00 = 98.00 N·m), Unequal Mass balances 10 kg at 1.0 m against 5 kg at 2.0 m (98.00 = 98.00 — the heavier side does NOT sink), and Crane Counterweight holds 4 kg at 3.0 m with 15 kg at just 0.8 m (117.60 = 117.60) — mass and distance trade in exact inverse proportion, which is why OpenStax's torque balance 'may be used to measure mass' with g cancelling out (§12.2 Example 12.3, verified)
Keep both equilibrium books at once: at defaults the Pivot Reaction row reads 29.4 N = (1+2)·9.8 N — the first condition ΣF = 0 holding while the second Στ = 0 holds ('when rotational and translational equilibrium conditions hold simultaneously in one frame… they also hold in any other inertial frame', §12.1 verified). Slide masses anywhere: the reaction never leaves (m₁+m₂)g, balance or not — the pivot works hardest exactly when the beam is balanced