Inclined planes, friction and connected particles · Plans inclinés, frottement et particules liées
| English | Français |
|---|---|
| limiting friction/ˈlɪmɪtɪŋ ˈfrɪkʃn/ | frottement limite |
Is friction already at its maximum?
- A crate on a slope is at rest. Friction can adjust to balance the force down the slope; it need not already be at its maximum.
- This lesson studies limiting friction 极限摩擦力: The maximum static friction before slipping, equal to μ times the normal reaction in the model.
Choose the mathematical structure
- Resolve parallel and perpendicular to the plane. Weight components are mg sinθ and mg cosθ. Static friction satisfies F≤μR and equals μR only at the limiting case. Write separate equations for connected particles.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines limiting friction? · Quelle description définit correctement le frottement limitant ?
The maximum static friction before slipping, equal to μ times the normal reaction in the model. · La friction statique maximale avant glissement, égale à μ fois la réaction normale dans le modèle.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
A 5 kg crate on a 30° slope has R=5×9.8×cos30°≈42.44 N. The downhill weight component is 24.5 N. With μ=0.6, maximum friction≈25.46 N, so equilibrium is possible with friction 24.5 N.
Inclined planes, friction and connected particles · Plans inclinés, frottement et particules liées
Resolve parallel and perpendicular to the plane · Résoudre les composantes parallèle et perpendiculaire au plan
Compare the model with the worked case and explain one change. · Compare le modèle avec l'exemple résolu et explique un changement.
Find the downhill component of weight for m=5 kg,θ=30°,g=9.8. · Trouver la composante descendante du poids pour m=5 kg, θ=30°, g=9.8.
Resolve weight along the slope: 5×9.8×sin30°=24.5 N. · Décomposer le poids selon la pente : 5×9.8×sin30°=24.5 N.
Test a tempting shortcut
- Friction opposes motion or the tendency to move, not always the coordinate direction. A taut light inextensible string over a smooth pulley gives equal tension and a common acceleration magnitude; each assumption has a job.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Static friction is always exactly μR even when the body is not about to slip. This claim is false. Explain which definition or assumption it violates.
Find limiting friction when μ=0.6 and R=40 N. · Trouver la force de frottement limite lorsque μ=0.6 et R=40 N.
Limiting friction=μR=0.6×40=24 N. · Frottement limitant = μR = 0.6 × 40 = 24 N.
Static friction is always exactly μR even when the body is not about to slip. · Le frottement statique n'est pas toujours exactement égal à μR lorsque le corps n'est pas sur le point de glisser.
Friction opposes motion or the tendency to move, not always the coordinate direction. A taut light inextensible string over a smooth pulley gives equal tension and a common acceleration magnitude; each assumption has a job. · Le frottement s'oppose au mouvement ou à la tendance au mouvement, et non nécessairement à la direction du repère. Une corde légère, inextensible et tendue passant sur une poulie lisse assure une tension égale et une même grandeur d'accélération ; chaque hypothèse a sa justification.
Interpret a new situation
- Start with force diagrams and a proposed direction of motion. If the calculated direction conflicts with the friction assumption, revisit the model instead of keeping inconsistent signs.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
If a 2 kg particle has net force 7 N, find its acceleration. · Si une particule de 2 kg subit une force nette de 7 N, trouver son accélération.
Use resultant force: a=7/2=3.5 m/s². · Utiliser la force résultante : a = 7 / 2 = 3.5 m/s².
Match each part of a complete solution to its purpose. · Associer chaque partie d'une solution complète à son but.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Une hypothèse justifie le modèle ; une vérification teste le résultat ; l'interprétation le relie à la question.
Use this in your course
- edexcel IAL further mathematics; official unit M1. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The maximum static friction before slipping, equal to μ times the normal reaction in the model. Choose the relationship, show the method, check its assumptions and interpret the result.