Derive the coordinate constraint before eliminating tension
Two virtual hanging masses share a taut inextensible cord over an ideal massless pulley. Choose upward for the first mass and downward for the second as positive directions. Fixed cord length then requires equal signed displacement magnitudes, velocities and accelerations in these paired coordinates; it does not mean their upward Cartesian accelerations have the same sign.
Learning outcome: Retain both recipient diagrams and the differentiated cord constraint. Compare unequal masses in both orders and the equal-mass limit.
Starting recap
Differentiate/integrate with explicit initial conditions; identify each force recipient and the inertial frame. Preserve force dimensions, geometric constraints and the distinction between model and solver error.
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For m1=3,m2=2, acceleration is−2 m/s² in the declared paired coordinates and tension remains24 N. Changing the coordinate sign is not a negative physical tension.
Open the experimentPredict
For m1=2,m2=3,g10, the common cord tension is:
Experiment
Retain both recipient diagrams and the differentiated cord constraint. Compare unequal masses in both orders and the equal-mass limit.
Derive a shared acceleration from a geometric constraint
Worked teaching model · synthetic values · Two virtual hanging point masses joined by one taut massless inextensible cord over an ideal massless frictionless pulley, g10. First upward and second downward coordinates share positive acceleration. Finite pulley inertia and slack are excluded; no physical apparatus instruction.
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Forces on the selected recipient
- Solid: Cord on first mass; east 0 N, north 24 N.
- Dashed: Gravity on first mass; east 0 N, north -20 N.
Both components use the same pixels per newton. Every arrow acts on the named recipient; third-law partners on other bodies are not added here. The net force is the sum, not an extra interaction.
First mass displacement from rest
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: constraint-compatible trajectory (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for First mass displacement from rest
constraint-compatible trajectory
| time (s) | upward displacement (m) |
|---|---|
| 0 | 0 |
| 0.025 | 0.000625 |
| 0.05 | 0.0025 |
| 0.075 | 0.005625 |
| 0.1 | 0.01 |
| 0.125 | 0.015625 |
| 0.15 | 0.0225 |
| 0.175 | 0.030625 |
| 0.2 | 0.04 |
| 0.225 | 0.050625 |
| 0.25 | 0.0625 |
| 0.275 | 0.075625 |
| 0.3 | 0.09 |
| 0.325 | 0.105625 |
| 0.35 | 0.1225 |
| 0.375 | 0.140625 |
| 0.4 | 0.16 |
| 0.425 | 0.180625 |
| 0.45 | 0.2025 |
| 0.475 | 0.225625 |
| 0.5 | 0.25 |
| 0.525 | 0.275625 |
| 0.55 | 0.3025 |
| 0.575 | 0.330625 |
| 0.6 | 0.36 |
| 0.625 | 0.390625 |
| 0.65 | 0.4225 |
| 0.675 | 0.455625 |
| 0.7 | 0.49 |
| 0.725 | 0.525625 |
| 0.75 | 0.5625 |
| 0.775 | 0.600625 |
| 0.8 | 0.64 |
| 0.825 | 0.680625 |
| 0.85 | 0.7225 |
| 0.875 | 0.765625 |
| 0.9 | 0.81 |
| 0.925 | 0.855625 |
| 0.95 | 0.9025 |
| 0.975 | 0.950625 |
| 1 | 1 |
| 1.025 | 1.050625 |
| 1.05 | 1.1025 |
| 1.075 | 1.155625 |
| 1.1 | 1.21 |
| 1.125 | 1.265625 |
| 1.15 | 1.3225 |
| 1.175 | 1.380625 |
| 1.2 | 1.44 |
| 1.225 | 1.500625 |
| 1.25 | 1.5625 |
| 1.275 | 1.625625 |
| 1.3 | 1.69 |
| 1.325 | 1.755625 |
| 1.35 | 1.8225 |
| 1.375 | 1.890625 |
| 1.4 | 1.96 |
| 1.425 | 2.030625 |
| 1.45 | 2.1025 |
| 1.475 | 2.175625 |
| 1.5 | 2.25 |
| 1.525 | 2.325625 |
| 1.55 | 2.4025 |
| 1.575 | 2.480625 |
| 1.6 | 2.56 |
| 1.625 | 2.640625 |
| 1.65 | 2.7225 |
| 1.675 | 2.805625 |
| 1.7 | 2.89 |
| 1.725 | 2.975625 |
| 1.75 | 3.0625 |
| 1.775 | 3.150625 |
| 1.8 | 3.24 |
| 1.825 | 3.330625 |
| 1.85 | 3.4225 |
| 1.875 | 3.515625 |
| 1.9 | 3.61 |
| 1.925 | 3.705625 |
| 1.95 | 3.8025 |
| 1.975 | 3.900625 |
| 2 | 4 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| First upward and second downward acceleration | 2 | m/s² |
| Common cord tension | 24 | N |
| First net upward force | 4 | N |
| Second net downward force | 6 | N |
The geometric cord constraint and two recipient equations jointly determine acceleration and tension.
Let first upward and second downward coordinates be positive. Fixed cord length requires equal displacement magnitudes and accelerations; T−m1g=m1a and m2g−T=m2a. Adding eliminates the internal cord interaction and yields a=g(m2−m1)/(m1+m2). T=2m1m2g/(m1+m2) stays positive. A negative signed acceleration reverses the chosen direction, not the sign of tension; the equality of tensions relies on the massless ideal pulley/cord.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
The geometric cord constraint and two recipient equations jointly determine acceleration and tension.
Let first upward and second downward coordinates be positive. Fixed cord length requires equal displacement magnitudes and accelerations; T−m1g=m1a and m2g−T=m2a. Adding eliminates the internal cord interaction and yields a=g(m2−m1)/(m1+m2). T=2m1m2g/(m1+m2) stays positive. A negative signed acceleration reverses the chosen direction, not the sign of tension; the equality of tensions relies on the massless ideal pulley/cord.
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Which physical change would invalidate a common tension before it invalidates the cord-length relation?
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