Check the motion before assigning zero work to a normal force
A normal force is perpendicular to a contact surface, but it need not be perpendicular to the contact point's velocity. A stationary ideal guide often has zero normal work because motion is tangent to it. A moving support can transfer energy in its normal direction.
Learning outcome: Compare a stationary and an accelerating support, draw normal and weight, and reconcile point-only work with the gravitational-potential boundary.
Starting recap
Use vector dot products for work, gradients for conservative force, definite integrals for accumulation and explicit energy/frame boundaries. Check accessible states before assigning trajectories.
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At a = 2 and t = 1, y = 1 m,v2 m/s,N = 24 N. Normal work24 J equals20 J gravitational gain plus4 J kinetic gain.
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For a = 1 and t = 2, normal work is:
Experiment
Compare a stationary and an accelerating support, draw normal and weight, and reconcile point-only work with the gravitational-potential boundary.
Check whether a constraint actually does zero work
Worked teaching model · synthetic values · Point m2 kg starts at rest on an upward-accelerating horizontal support y=a t²/2 under g10. Contact holds with N=m(g+a). A moving support supplies energy; no physical lift, loading or equipment rating is modeled.
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Forces on the selected recipient
- Solid: Support normal on point; east 0 N, north 22 N.
- Dashed: Gravity on point; 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.
Moving support transfers energy through its normal force
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: N times support displacement (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Moving support transfers energy through its normal force
N times support displacement
| time (s) | support work (J) |
|---|---|
| 0 | 0 |
| 0.04 | 0.0176 |
| 0.08 | 0.0704 |
| 0.12 | 0.1584 |
| 0.16 | 0.2816 |
| 0.2 | 0.44 |
| 0.24 | 0.6336 |
| 0.28 | 0.8624 |
| 0.32 | 1.1264 |
| 0.36 | 1.4256 |
| 0.4 | 1.76 |
| 0.44 | 2.1296 |
| 0.48 | 2.5344 |
| 0.52 | 2.9744 |
| 0.56 | 3.4496 |
| 0.6 | 3.96 |
| 0.64 | 4.5056 |
| 0.68 | 5.0864 |
| 0.72 | 5.7024 |
| 0.76 | 6.3536 |
| 0.8 | 7.04 |
| 0.84 | 7.7616 |
| 0.88 | 8.5184 |
| 0.92 | 9.3104 |
| 0.96 | 10.1376 |
| 1 | 11 |
| 1.04 | 11.8976 |
| 1.08 | 12.8304 |
| 1.12 | 13.7984 |
| 1.16 | 14.8016 |
| 1.2 | 15.84 |
| 1.24 | 16.9136 |
| 1.28 | 18.0224 |
| 1.32 | 19.1664 |
| 1.36 | 20.3456 |
| 1.4 | 21.56 |
| 1.44 | 22.8096 |
| 1.48 | 24.0944 |
| 1.52 | 25.4144 |
| 1.56 | 26.7696 |
| 1.6 | 28.16 |
| 1.64 | 29.5856 |
| 1.68 | 31.0464 |
| 1.72 | 32.5424 |
| 1.76 | 34.0736 |
| 1.8 | 35.64 |
| 1.84 | 37.2416 |
| 1.88 | 38.8784 |
| 1.92 | 40.5504 |
| 1.96 | 42.2576 |
| 2 | 44 |
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 |
|---|---|---|
| Upward displacement | 2 | m |
| Final upward velocity | 2 | m/s |
| Upward normal force | 22 | N |
| Work supplied by moving support | 44 | J |
| Gravitational potential gain | 40 | J |
| Kinetic-energy gain | 4 | J |
| Potential plus kinetic gain | 44 | J |
A normal force is not automatically a zero-work force when its contact point moves.
N−mg=ma fixes N=m(g+a). The point and moving support share upward displacement, so normal work N y is positive when a>0 and t>0. It equals ΔK+ΔUg. For a stationary ideal guide a normal can be perpendicular to velocity and do zero work, but that geometric premise does not apply to this moving support.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
A normal force is not automatically a zero-work force when its contact point moves.
N−mg=ma fixes N=m(g+a). The point and moving support share upward displacement, so normal work N y is positive when a>0 and t>0. It equals ΔK+ΔUg. For a stationary ideal guide a normal can be perpendicular to velocity and do zero work, but that geometric premise does not apply to this moving support.
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Which geometric condition, not force name, makes normal work zero?
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