Use smooth contact on inclined planes and vertical walls
Set friction to zero, resolve weight in surface directions and test whether the remaining forces can support equilibrium.
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 vertical weight parallel and normal to an incline.
- Use two perpendicular component equilibrium equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For an incline θ above the horizontal, what is the weight’s downslope component?
What you will learn
- Solve smooth-contact equilibrium with forces along or oblique to a plane.
- Explain what the smooth-particle model omits and when it cannot support rest.
Smooth does not mean no contact force
A smooth surface supplies normal reaction but no friction. On a plane inclined at θ to the horizontal, weight W has downslope component Wsin θ and into-plane component Wcos θ. If a taut string parallel to the plane supports the particle, equilibrium gives T = Wsin θ and R = Wcos θ. A 50 N weight on a smooth 30° plane therefore needs T = 25 N and R = 25√3 N.
A smooth particle on a non-horizontal plane without another upslope force cannot remain at rest, because Wsin θ is unbalanced. Setting R = W would not fix that missing parallel force. The smooth model neglects tangential resistance; a real surface may not satisfy it exactly. It is useful only while the stated contact and applied-force assumptions remain appropriate, and the particle model omits possible tipping of an extended object.
An oblique support force also changes the reaction
A horizontal force P pushing a particle towards the uphill side of a smooth plane has upslope component Pcos θ and into-plane component Psin θ. Rest requires Pcos θ = Wsin θ, so P = Wtan θ; the normal reaction is R = Wcos θ + Psin θ = W/cos θ. With W = 30 N and θ = 30°, P = 10√3 N and R = 20√3 N. The extra horizontal push increases the contact reaction.
At a smooth vertical wall, its normal reaction is horizontal and cannot support vertical weight. A 40 N particle pressed against the wall horizontally by 15 N has R = 15 N, but still needs a vertical upward force of 40 N, such as a taut vertical string. Without that support it cannot stay at rest in this model. Contact alone is not a guarantee of equilibrium in all directions.
Pause and explain
What can a smooth vertical wall’s reaction balance?
Worked example
A 50 N particle is held on a smooth 30° incline by a string parallel to the plane. Find T and R.
Show the worked solution
- Resolve weight: 50sin30° = 25 N downslope and 50cos30° = 25√3 N into the plane.
- Parallel balance gives T = 25 N upslope.
- Normal balance gives R = 25√3 N away from the plane; smooth contact supplies no friction.
Answer T = 25 N; R = 25√3 N ≈ 43.3 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.
An 80 N particle is at rest on a smooth 30° incline with a string parallel to the plane. Find tension and reaction.
Give me a hint
Balance independently along and normal to the plane.
Compare my reasoning
- Downslope weight is 80sin30° = 40 N.
- Hence tension is 40 N upslope.
- Normal reaction is 80cos30° = 40√3 N; no friction is included.
T = 40 N; R = 40√3 N ≈ 69.3 N.
Look for these in your work
- Use the correct weight components.
- Apply the zero-friction assumption in the parallel equation.
A 30 N particle rests on a smooth 30° incline under a horizontal force P towards the uphill side. Find P and R.
Give me a hint
The horizontal force has an upslope component and an into-plane component.
Compare my reasoning
- Parallel balance gives Pcos30° = 30sin30°, so P = 10√3 N.
- Normal balance gives R = 30cos30° + Psin30°.
- Substitution gives R = 15√3 + 5√3 = 20√3 N.
P = 10√3 N; R = 20√3 N.
Look for these in your work
- Resolve the horizontal force in both plane directions.
- Check that the push increases the normal reaction.
A 40 N particle rests against a smooth vertical wall under a horizontal force 15 N into the wall and a taut vertical string. Find R and T, and explain why removing the string prevents rest.
Give me a hint
Separate the wall-normal horizontal balance from the vertical support balance.
Compare my reasoning
- Horizontal balance gives R = 15 N away from the wall.
- Vertical balance gives T = 40 N upwards.
- Without the string, weight is unbalanced because smooth contact provides no friction and its normal force is horizontal.
R = 15 N; T = 40 N; without the string the smooth-wall model cannot support rest.
Look for these in your work
- Use the correct direction of the wall reaction.
- Explain the missing upward force under the changed model.
Common mistakes
- Smooth contact removes friction but not normal reaction.
- Weight’s plane components are Wsin θ along and Wcos θ normal to the incline.
- A wall’s horizontal normal cannot balance vertical weight.
What can you explain now?
Can a smooth inclined plane alone hold a particle at rest when θ > 0 and W > 0?
Compare with the explanation
No: the downslope component Wsin θ is unbalanced.
The normal reaction has no component along the plane, and friction is absent.
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Make it stick
Set friction to zero, resolve every force along and normal to a surface and test whether both directions can balance under a smooth-contact model.
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.