Separate contact into normal reaction and friction
Resolve a surface contact force into perpendicular reaction and tangential friction, and calculate how an angled pull or push changes the reaction.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.1. Paper 1 algebra and trigonometry are assumed. Bodies are modelled as particles: forces act at one point and rotational effects are not considered. Resolve numerically; a scale drawing is not a substitute for calculation. Force diagrams are qualitative unless stated otherwise. Where mass is supplied in this sequence, use the stated g = 10 m s⁻². Check non-negative tension and contact reaction and use F ≤ μR except at limiting friction. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Resolve oblique forces and use equilibrium.
- Calculate a resultant from perpendicular components.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How is normal reaction directed relative to a surface?
What you will learn
- Identify the two components of a rough surface’s contact force.
- Calculate reaction and contact resultant using the supplied applied-force directions.
A rough surface’s total force has two components
Represent contact by a normal reaction R perpendicular to the surface and friction F along it. For maintained non-adhesive contact, R ≥ 0. The combined contact-force magnitude is √(R² + F²), not generally R alone. If F is non-zero, the resultant is inclined to the normal by an angle with tangent |F|/R, when R > 0. The friction direction depends on the tendency for relative sliding.
A 5 kg particle on a horizontal surface is pulled right by 20 N at 30° above the horizontal and held at rest by contact. With g = 10 m s⁻², vertical balance gives R + 20sin30° = 50, so R = 40 N. Horizontal balance requires friction 20cos30° = 10√3 N left. The total contact force has magnitude √(40² + (10√3)²) = 10√19 N. Available friction still has to be sufficient for the proposed equilibrium.
Pulling up and pushing down have opposite normal effects
An upward component of an applied force reduces the normal reaction, whereas a downward component increases it. A 4 kg particle pushed right at 10 N and 30° below the horizontal has R = 40 + 10sin30° = 45 N and requires friction 5√3 N left for rest. Using R = mg in either angled case would miss the vertical contribution and produce the wrong friction limit later.
On an incline, R is perpendicular to the plane, not vertical. Resolve all forces normal to that plane before finding it. If an upward pull calculation gives R < 0, a non-adhesive surface cannot exert the needed downward contact: maintained contact is impossible under that arrangement. At R = 0 the body is at the contact boundary; the dry-friction model gives no friction capacity there. A force balance that needs both negative R and positive friction is not physically valid.
Pause and explain
A horizontal-surface pull has upward component U and weight W. What is R for vertical balance?
Worked example
A 5 kg particle is at rest on a rough horizontal surface under a 20 N pull at 30° above the horizontal. Use g = 10 m s⁻². Find R, required F and the contact-force magnitude.
Show the worked solution
- Vertical balance gives R = 50 − 20sin30° = 40 N.
- Horizontal balance requires F = 20cos30° = 10√3 N left.
- The perpendicular contact components give magnitude √(1600 + 300) = 10√19 N.
Answer R = 40 N; required F = 10√3 N left; contact magnitude 10√19 N ≈ 43.6 N.
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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 4 kg particle is at rest on a rough horizontal surface under a 10 N push at 30° below the rightward horizontal. Use g = 10 m s⁻². Find R and required friction.
Give me a hint
The push adds a downward component to the weight.
Compare my reasoning
- Weight is 40 N and downward push component is 10sin30° = 5 N.
- Vertical balance gives R = 45 N.
- The rightward push component is 10cos30° = 5√3 N, requiring that friction leftwards.
R = 45 N; required friction 5√3 N left.
Look for these in your work
- Use the downward sign of the push component.
- Find the tangential force separately from reaction.
A rough contact has normal component 24 N and tangential friction component 7 N. Find the total contact magnitude and its angle to the normal.
Give me a hint
The components are perpendicular, so use Pythagoras and the correct angle ratio.
Compare my reasoning
- Contact magnitude is √(24² + 7²) = 25 N.
- If β is measured from the normal towards friction, tan β = 7/24.
- Hence β ≈ 16.3°, on the side indicated by the friction direction.
Contact magnitude 25 N; angle 16.3° from the normal towards friction.
Look for these in your work
- Use the normal and tangential components as perpendicular legs.
- Specify the reference direction for the angle.
A 20 kg particle is at rest on a rough horizontal surface under a 100 N pull at 30° above the rightward horizontal. Use g = 10 m s⁻². Find R and required F and explain why friction capacity cannot be certified without μ.
Give me a hint
Balance the upward pull component before calculating the tangential force.
Compare my reasoning
- Weight is 200 N, so R = 200 − 100sin30° = 150 N.
- Horizontal balance needs F = 100cos30° = 50√3 N left.
- Rest requires 50√3 ≤ 150μ; without μ the available friction capacity is not established.
R = 150 N; required F = 50√3 N left; equilibrium requires μ ≥ 1/√3.
Look for these in your work
- Include the normal effect of the pull.
- Distinguish a required friction force from proof that the surface can supply it.
Common mistakes
- R is a contact component, not the full rough-contact resultant.
- Include upward and downward applied components when calculating R.
- Reject negative reaction for a non-adhesive contact.
What can you explain now?
Does an upward angled pull leave R = mg on a horizontal surface?
Compare with the explanation
Not generally; its upward component reduces R.
Vertical balance includes every force component, not only weight and contact.
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Make it stick
Separate rough contact into R and F, resolve the other forces and check both contact signs and any required friction capacity.
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.