Inclined planes, friction and connected particles · Planos inclinados, fricción y partículas conectadas
| English | Español |
|---|---|
| limiting friction/ˈlɪmɪtɪŋ ˈfrɪkʃn/ | fricción límite |
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? · ¿Qué descripción define correctamente la fricción límite?
The maximum static friction before slipping, equal to μ times the normal reaction in the model. · La fricción estática máxima antes del deslizamiento, igual a μ veces la reacción normal en el modelo.
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 · Planos inclinados, fricción y partículas conectadas
Resolve parallel and perpendicular to the plane · Resuelve las fuerzas paralelas y perpendiculares al plano
Compare the model with the worked case and explain one change. · Compara el modelo con el caso resuelto y explica un cambio.
Find the downhill component of weight for m=5 kg,θ=30°,g=9.8. · Encuentra la componente paralela al plano para m=5 kg, θ=30°, g=9.8.
Resolve weight along the slope: 5×9.8×sin30°=24.5 N. · Resuelve el peso a lo largo de la pendiente: 5×9.8×sin(30°)=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. · Encuentra la fricción límite cuando μ=0.6 y R=40 N.
Limiting friction=μR=0.6×40=24 N. · Fricción límite = μR = 0.6×40 = 24 N.
Static friction is always exactly μR even when the body is not about to slip. · La fricción estática no siempre es exactamente μR cuando el cuerpo no está a punto de resbalar.
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. · La fricción se opone al movimiento o a la tendencia a moverse, no necesariamente a la dirección del sistema de coordenadas. Una cuerda ligera tensa e inextensible sobre una polea lisa proporciona tensión igual y magnitud de aceleración común; cada suposición tiene una función específica.
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 una partícula de 2 kg tiene una fuerza neta de 7 N, encuentra su aceleración.
Use resultant force: a=7/2=3.5 m/s². · Usa la fuerza resultante: a = 7/2 = 3.5 m/s².
Match each part of a complete solution to its purpose. · Emparejar cada parte de una solución completa con su propósito.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Una suposición justifica el modelo; una comprobación verifica el resultado; la interpretación lo conecta con la pregunta.
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.