Inclined planes, friction and connected particles · Planos inclinados, atrito e partículas conectadas
| English | Português |
|---|---|
| limiting friction/ˈlɪmɪtɪŋ ˈfrɪkʃn/ | atrito 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? · Qual descrição define corretamente o atrito limite?
The maximum static friction before slipping, equal to μ times the normal reaction in the model. · O atrito estático máximo antes do deslizamento, igual a μ vezes a reação normal no 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, atrito e partículas conectadas
Resolve parallel and perpendicular to the plane · Resolva paralelo e perpendicular ao plano
Compare the model with the worked case and explain one change. · Compare o modelo com o caso resolvido e explique uma mudança.
Find the downhill component of weight for m=5 kg,θ=30°,g=9.8. · Encontre a componente descendente do peso para m=5 kg, θ=30°, g=9.8.
Resolve weight along the slope: 5×9.8×sin30°=24.5 N. · Resolva o peso ao longo da inclinação: 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. · Encontre o atrito limite quando μ=0.6 e R=40 N.
Limiting friction=μR=0.6×40=24 N. · Atrito limite=μR=0.6×40=24 N.
Static friction is always exactly μR even when the body is not about to slip. · O atrito estático nem sempre é exatamente μR, mesmo quando o corpo não está prestes a deslizar.
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. · O atrito se opõe ao movimento ou à tendência de movimento, nem sempre à direção do sistema de coordenadas. Uma corda leve, inextensível e tensa sobre uma polia lisa fornece tensão igual e magnitude de aceleração comum; cada suposição tem uma função.
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. · Se uma partícula de 2 kg tem força resultante de 7 N, encontre sua aceleração.
Use resultant force: a=7/2=3.5 m/s². · Use a força resultante: a=7/2=3.5 m/s².
Match each part of a complete solution to its purpose. · Associe cada parte de uma solução completa ao seu propósito.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Uma suposição justifica o modelo; uma verificação testa o resultado; a interpretação conecta-o à pergunta.
Use this in your course
- edexcel IAL 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.