Use an inequality before assigning a friction value
A point starts at rest on an incline. Resolve its weight into mg sinθ down the slope and mg cosθ normal to the surface. The normal reaction is mg cosθ because there is no normal acceleration or other normal applied force. These components are parts of one weight vector, not extra forces in addition to weight.
Learning outcome: Compare 15°,30°,45° at 2 kg and repeat one state at 1 and 3 kg. Retain actual friction separately from its maximum, and predict which outputs scale with mass.
Starting recap
Draw recipient-specific interactions, resolve signed components and identify the domain of each force/contact model before substituting.
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At 15° and 2 kg, normal≈19.3185 N and required static friction≈5.17638 N. Capacity≈9.65926 N exceeds demand, so actual friction is 5.17638 N and acceleration is zero.
Open the experimentPredict
For a resting object below the static threshold, friction is:
Experiment
Compare 15°,30°,45° at 2 kg and repeat one state at 1 and 3 kg. Retain actual friction separately from its maximum, and predict which outputs scale with mass.
Distinguish static adjustment from kinetic sliding
Worked teaching model · synthetic values · Initially resting point on a straight incline, g=10, static coefficient .5 and kinetic coefficient .3. No added push. Equality at the static bound remains an admissible rest state; beyond it release is downhill. No physical ramp activity.
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Demand and static capacity ratio
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: static demand ratio (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Demand and static capacity ratio
static demand ratio
| angle (degrees) | tan(theta) |
|---|---|
| 0 | 0 |
| 1.5 | 0.0261859 |
| 3 | 0.0524078 |
| 4.5 | 0.0787017 |
| 6 | 0.105104 |
| 7.5 | 0.131652 |
| 9 | 0.158384 |
| 10.5 | 0.185339 |
| 12 | 0.212557 |
| 13.5 | 0.240079 |
| 15 | 0.267949 |
| 16.5 | 0.296213 |
| 18 | 0.32492 |
| 19.5 | 0.354119 |
| 21 | 0.383864 |
| 22.5 | 0.414214 |
| 24 | 0.445229 |
| 25.5 | 0.476976 |
| 27 | 0.509525 |
| 28.5 | 0.542956 |
| 30 | 0.57735 |
| 31.5 | 0.612801 |
| 33 | 0.649408 |
| 34.5 | 0.687281 |
| 36 | 0.726543 |
| 37.5 | 0.767327 |
| 39 | 0.809784 |
| 40.5 | 0.854081 |
| 42 | 0.900404 |
| 43.5 | 0.948965 |
| 45 | 1 |
| 46.5 | 1.05378 |
| 48 | 1.11061 |
| 49.5 | 1.17085 |
| 51 | 1.2349 |
| 52.5 | 1.30323 |
| 54 | 1.37638 |
| 55.5 | 1.45501 |
| 57 | 1.53986 |
| 58.5 | 1.63185 |
| 60 | 1.73205 |
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 |
|---|---|---|
| Normal magnitude | 17.3205 | N |
| Down-slope weight component | 10 | N |
| Maximum static friction | 8.66025 | N |
| Actual up-slope friction | 5.19615 | N |
| Down-slope acceleration | 2.40192 | m/s² |
| Rest branch admissible | 0 | flag |
| Continuous critical angle | 26.5651 | degrees |
The resting branch fails; the stipulated release begins downhill with kinetic friction.
Rest requires mg sinθ≤mu_s mg cosθ, or tanθ≤mu_s. The critical angle is about 26.565°, between this control’s 15° and 30° samples. Equality is admissible rest, not automatic motion. After release, downhill acceleration g(sinθ−mu_k cosθ) is mass independent in this ideal model, while all contact and weight forces scale with mass.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
The resting branch fails; the stipulated release begins downhill with kinetic friction.
Rest requires mg sinθ≤mu_s mg cosθ, or tanθ≤mu_s. The critical angle is about 26.565°, between this control’s 15° and 30° samples. Equality is admissible rest, not automatic motion. After release, downhill acceleration g(sinθ−mu_k cosθ) is mass independent in this ideal model, while all contact and weight forces scale with mass.
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What evidence would challenge the assumption that one pair of friction coefficients describes all states?
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