Resolving forces on a ramp
You will be able to: Resolve weight along and perpendicular to a frictionless incline.
Why does only part of gravity speed a block down a ramp?
A block on a gently tilted board accelerates less strongly than one dropped straight down. The board prevents motion through its surface, so it changes which part of gravity accelerates the block.
A useful starting point: Vector components →
Words and symbols before equations
- Incline angle θ
- Angle the ramp makes with the horizontal, in degrees here.
- Component
- Projection of a vector along a chosen axis.
- Parallel / perpendicular
- Along the surface / at a right angle to the surface.
What this picture assumes
Straight frictionless ramp, g=10 m/s². No acceleration perpendicular to the ramp. Component diagram is separate from the real-force diagram.
Connect the picture to the physics
Choose one axis down the ramp and one away from it. Weight still points vertically down. Its along-ramp component is mg sinθ; its into-ramp component is mg cosθ.
For a straight fixed ramp, the block has no acceleration perpendicular to it. With no other perpendicular forces, N=mg cosθ. Along the ramp, no friction means ma=mg sinθ, giving a=g sinθ.
Check extremes before trusting trigonometry. At θ=0°, the along-ramp component must vanish. As θ approaches 90°, it approaches mg. The two components replace the weight in a component sum; do not add weight again.
A worked example, step by step
A 2 kg block slides on a frictionless 30° ramp. Use g=10 m/s², sin30°=0.5 and cos30°≈0.866. Find its normal force and acceleration.
- Weight = 20 N vertically down.
- Along-ramp weight = 20×0.5 = 10 N down the slope.
- Normal force = 20×0.866 ≈17.3 N, balancing the perpendicular component.
- a = 10/2 = 5 m/s² down the ramp. Mass cancels for this frictionless model.
The normal force is not mg on this incline. It balances only the perpendicular component of weight.
Would doubling the mass double the frictionless acceleration?
Compare with an explanation
No. Both gravitational force and inertia double, so a=g sinθ stays the same.
Predict. Change one thing. Explain.
Change the angle from 0° to 30° to 60°, keeping mass fixed. Compare the down-slope force and normal force. Explain why one grows while the other shrinks.
Ramp angle 30°. Down-ramp acceleration 5 m/s²; normal force 17.32 N. All force arrows use 4.75 drawing units/N. Components replace weight in the calculation; they are not extra forces. Ramp guide is schematic.
Straight frictionless ramp, g=10 m/s². No acceleration perpendicular to the ramp. Component diagram is separate from the real-force diagram.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 3 kg block is on a frictionless 30° ramp. (a) Draw weight and normal force, (b) find the parallel component of weight, (c) find N, and (d) predict acceleration for a 6 kg block.
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Compare with the answer and four-point rubric
- 1 point: Weight vertical down; normal perpendicular away from the ramp.
- 1 point: 15 N down-ramp.
- 1 point: 30 cos30°≈26 N.
- 1 point: 5 m/s² down-ramp, unchanged because mass cancels.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which way does weight point on a ramp?
Vertically downward.
RECALL 2Why choose tilted axes?
They separate the constrained and free-motion directions.
RECALL 3When does N=mg cosθ apply?
When there are no other perpendicular forces and no perpendicular acceleration.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Resolving forces on a ramp
- F_parallel = mg sinθ; N = mg cosθ here.
- a_down = g sinθ for a frictionless fixed ramp.
Remember: The normal force is not mg on this incline. It balances only the perpendicular component of weight.
Conditions: Straight frictionless ramp, g=10 m/s². No acceleration perpendicular to the ramp. Component diagram is separate from the real-force diagram.
Refresh Kid · Unit 2 · Objectives 2.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.5, objectives 2.5.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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