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.
Section 4.1 · Forces and 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.