Choose useful axes on an incline
You will be able to: Resolve gravity separately from the free-body diagram and apply perpendicular and parallel force equations.
Why does a steeper ramp produce a greater acceleration?
A block slides down a smooth ramp. Gravity still points straight down, while the support force points perpendicular to the ramp. Only part of gravity acts along the allowed motion.
A useful starting point: Connected objects and the system boundary →
Words and symbols before equations
- Incline angle θ
- Angle of the ramp above the horizontal.
- Parallel component
- Projection of a vector along the ramp.
- Perpendicular component
- Projection normal to the ramp.
- Contact constraint
- The block remains on the straight ramp, so perpendicular acceleration is zero.
What this picture assumes
Smooth straight ramp with maintained contact, g = 10 m/s². The dot diagram shows weight and normal only; resolved weight components are in a separate chart.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Ramp angle 30° above horizontal. N = 17.32 N; downhill gravity component 10 N; a = 5 m/s² downhill.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Draw only actual forces: weight mg downward and normal N perpendicular outward. In a separate calculation resolve weight into mg sin θ downhill and mg cos θ into the plane.
Perpendicular balance gives N − mg cos θ = 0. Along the frictionless ramp, mg sin θ = ma, giving a = g sin θ. Mass cancels because both gravitational force and inertia scale with mass.
N is not generally mg. At zero angle it equals mg; on a steep ramp it is smaller. If another force pulls away from the ramp or perpendicular acceleration is present, the normal equation must change.
A worked example, step by step
A 2 kg block slides on a frictionless 30° ramp. Find normal force and downhill acceleration with g = 10 m/s².
- Weight is 20 N vertically down; normal is perpendicular to the ramp.
- N = 20 cos 30° ≈ 17.32 N.
- Downhill weight component is 20 sin 30° = 10 N.
- a = 10/2 = 5 m/s² downhill; the block’s mass does not affect this ideal acceleration.
Weight components are not extra forces to add to weight in the free-body diagram.
Does increasing mass change downhill acceleration on the same smooth ramp?
Compare with an explanation
No in this model; both gravity and inertia increase in the same proportion.
Predict. Change one thing. Explain.
Keep the mass fixed and increase ramp angle. Compare the actual force diagram with the separate numerical components. Predict both normal force and downhill acceleration.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Ramp angle 30° above horizontal. N = 17.32 N; downhill gravity component 10 N; a = 5 m/s² downhill.
Smooth straight ramp with maintained contact, g = 10 m/s². The dot diagram shows weight and normal only; resolved weight components are in a separate chart.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus 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 questionFor a 4 kg block on a frictionless 30° ramp, label actual forces, find the normal force and acceleration, and compare with a 2 kg block.
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Compare with the answer and four-point rubric
- 1 point: Actual forces are weight 40 N down and normal perpendicular to the plane.
- 1 point: N = 40 cos 30° ≈ 34.64 N.
- 1 point: a = 10 sin 30° = 5 m/s² downhill.
- 1 point: The 2 kg block has the same acceleration, with half the force magnitudes.
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 axes simplify a ramp problem?
Parallel and perpendicular to the ramp.
RECALL 2Why is N not automatically mg?
Only the perpendicular gravity component is balanced in this model.
RECALL 3What remains vertical when the ramp tilts?
The gravitational force.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Choose useful axes on an incline
- N = mg cos θ for this contact-constrained frictionless ramp.
- ΣF_parallel = mg sin θ = ma.
- a = g sin θ; g = 10 m/s² here.
Remember: Weight components are not extra forces to add to weight in the free-body diagram.
Conditions: Smooth straight ramp with maintained contact, g = 10 m/s². The dot diagram shows weight and normal only; resolved weight components are in a separate chart.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.5.A; 2.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.5, objectives 2.5.A; 2.2.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. 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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