Infer a missing force mass or friction coefficient
Rearrange a Newton equation using measured acceleration, then reject parameters that contradict the stated motion and contact model.
Section 4.4 · Newton’s laws of motion
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Write Newton equations for horizontal and inclined sliding.
- Rearrange linear equations and interpret negative results.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 5 kg particle accelerates at 4 m s⁻² under drive 30 N and opposing resistance Q. Which equation is correct?
What you will learn
- Recover an unknown force, mass or coefficient from a specified motion.
- Check positive mass, non-negative friction and the assumed force directions.
Measured acceleration supplies the required resultant
A 5 kg particle driven horizontally by 30 N with forward acceleration 4 m s⁻² needs resultant 20 N. If the only opposition is a constant resistance Q, then 30 − Q = 20, so Q = 10 N. Do not treat Q = ma: the resistance is only one part of the resultant. For an unknown mass under drive 42 N and resistance 6 N with acceleration 3, 42 − 6 = 3m gives m = 12 kg.
For downhill sliding on a rough slope, mg sin θ − μmg cos θ = ma. A 4 kg particle with sin θ = 3/5, cos θ = 4/5 and downward a = 2 gives 24 − 32μ = 8, so μ = 1/2. Substituting it back recovers friction 16 N and resultant 8 N downwards. The inferred coefficient applies to the specified friction model, not automatically to all real motion conditions.
An algebraic result must satisfy the model
If that same gravity-only downhill particle is reported to accelerate at 8 m s⁻², the equation gives 24 − 32μ = 32 and μ = −1/4. A negative coefficient is incompatible with a resistive friction model. The reported data require a different force model, corrected measurements or corrected information; taking the absolute value would silently change the equation and would not reproduce the acceleration.
Negative signed acceleration itself is often valid; it can mean a particle moving in the positive direction is slowing. Negative mass or a negative friction magnitude is a different kind of issue. Distinguish a signed component from a non-negative physical parameter and state the interval for which the assumed direction of sliding remains true. Never fix a failed model merely by removing a minus sign.
Pause and explain
A downhill rough-slope calculation gives μ = −1/4. What is the appropriate conclusion?
Worked example
A 4 kg particle slides down a slope with sin θ = 3/5 and cos θ = 4/5. Its downward acceleration is 2 m s⁻² and only weight, reaction and sliding friction act. Find μ.
Show the worked solution
- Reaction is R = 40 × 4/5 = 32 N and downslope weight is 24 N.
- Downward Newton equation is 24 − 32μ = 4 × 2 = 8.
- Thus μ = 16/32 = 1/2; substituting gives the required 8 N resultant.
Answer μ = 1/2; friction 16 N up the plane.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 5 kg particle has a forward drive 30 N, an opposing resistance Q and acceleration 4 m s⁻² forward. Find Q.
Give me a hint
First calculate the resultant required by the mass and acceleration.
Compare my reasoning
- The required resultant is 5 × 4 = 20 N.
- The force equation is 30 − Q = 20.
- Therefore Q = 10 N, opposing motion as stated.
Q = 10 N.
Look for these in your work
- Use ma for the resultant.
- Retain the opposing sign for resistance.
A particle of unknown mass m has a 42 N drive and 6 N opposing resistance, giving forward acceleration 3 m s⁻². Find m and test its sign.
Give me a hint
The driving and opposing forces determine the net force before dividing by acceleration.
Compare my reasoning
- The resultant is 42 − 6 = 36 N.
- Newton’s law gives 36 = 3m, so m = 12 kg.
- The positive mass is admissible and reproduces acceleration 36/12 = 3.
m = 12 kg.
Look for these in your work
- Subtract the supplied resistance.
- Check positive mass and substitute into the original equation.
A 4 kg particle slides downhill with sin θ = 3/5 and cos θ = 4/5. Only weight, reaction and resistive sliding friction act, but its reported downhill acceleration is 8 m s⁻². Infer μ and test the report.
Give me a hint
Compare the reported acceleration with the gravity-only downslope acceleration before accepting a parameter.
Compare my reasoning
- Reaction is 32 N and downslope weight is 24 N.
- The equation 24 − 32μ = 4 × 8 gives μ = −1/4.
- That coefficient is invalid for the stated resistive model; an additional driving force or corrected data would be needed.
Inferred μ = −1/4 is invalid; the report is incompatible with the stated model.
Look for these in your work
- Quantify the incompatible parameter.
- Keep the conclusion limited to the supplied force model rather than altering the sign.
Common mistakes
- The unknown resistance is not usually the whole resultant.
- Reject a negative coefficient in the stated resistive model.
- A negative acceleration can still be physically valid.
What can you explain now?
Can you replace a calculated negative friction coefficient with its absolute value?
Compare with the explanation
No; that changes the equation and fails to reproduce the supplied acceleration.
The negative coefficient signals incompatibility with the stated force model.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Infer the parameter from the full resultant equation, then substitute and check its physical constraints.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.