Normal Force on an Incline
Block on an incline with adjustable angle; normal force decreases as cosine of the angle
Objective
Verify that the normal force on an inclined surface equals N = m·g·cosθ — not the full weight — by reading it as a function of angle. On a frictionless incline the block slides down under the unbalanced along-slope component F∥ = m·g·sinθ (acceleration a = g·sinθ, independent of mass), while the normal force stays equal to the perpendicular projection of weight, N = m·g·cosθ — unchanged by the motion. The block is idealized as a point mass.
Setup
- Set the angle slider to θ = 0° and the mass slider to m = 5 kg. Press Start and note the Normal Force N readout — it should equal the Weight W readout (both ≈ 49.05 N). This is the flat-surface baseline.
- Press Reset. Drag the angle slider to θ = 30° (leave mass at 5 kg). Before pressing Start, read the N readout: it should show ≈ 42.48 N. Record this as your prediction, then press Start to confirm the animated result.
- Press Reset. Drag the angle slider to θ = 60°. Predict N = m·g·cos60° ≈ 24.53 N. Press Start; the block slides down and the N readout holds at this value.
- Press Reset. Drag the angle slider to θ = 80° and mass to m = 10 kg. Predict N ≈ 17.03 N, W = 98.10 N. Press Start and verify the readouts match the analytic values within 0.1 N.
- Sweep the mass slider from 1 kg to 10 kg at a fixed angle (e.g. θ = 45°) without pressing Start — watch W and N both scale while the ratio N/W stays constant at cos45° ≈ 0.707. This confirms that the cosine relationship is mass-independent.
Analytical Prediction
The normal force is the component of weight perpendicular to the incline surface: N = m·g·cosθ. Weight is W = m·g regardless of angle. With m = 5 kg, g = 9.81 m/s², and θ = 30°:
At θ = 60°:
The ratio N(30°)/N(60°) = cos30°/cos60° = 0.866/0.500 ≈ 1.73, so the 30° case carries 73% more normal force than the 60° case — a large difference the bar chart makes immediate. The Pythagorean identity N² + F∥² = W² always holds: at 30°, 42.48² + 24.53² ≈ 2406 ≈ 49.05².
Results Analysis
After pressing Start at each test angle, the block slides down the frictionless incline while the Normal Force N readout stays constant (N does not depend on position). Compare the N readout to the analytic prediction. At θ = 30°, m = 5 kg, N reads 42.48 N (tolerance ± 0.05 N). At θ = 60°, it reads ≈ 24.53 N. The Weight W readout remains 49.05 N throughout — confirming W is angle-independent while N is not. The bar chart secondary view makes the trade-off visual: as θ increases, the sky-blue N bar shrinks and the amber F∥ bar grows, always satisfying N² + F∥² = W². The amber F∥ vector is the unbalanced net force that drives the slide.
Source of Error
This simulation models the block as a rigid point mass on a perfectly smooth, rigid incline with no friction, no air drag, and no rotational degrees of freedom. The analytical prediction N = m·g·cosθ assumes the same idealizations — a point mass on a frictionless incline under uniform gravity. The block slides under the unbalanced along-slope component F∥ = m·g·sinθ, but the normal force N is the geometric perpendicular projection of weight, computed directly from the angle rather than integrated — so it equals m·g·cosθ exactly throughout the slide, independent of any integration drift in the block's position. Any gap between the N readout and the analytic value is purely decimal rounding.
Further Exploration
- Set θ = 0° and m = 5 kg — does N equal W exactly? Now slowly drag θ toward 90°. At what angle does N drop below half of W? (Hint: cos θ = 0.5 at θ = 60°.) Does the bar chart confirm this?
- Fix θ = 45° and sweep mass from 1 kg to 10 kg. Does the ratio N/W stay constant? What does this tell you about the mass-independence of the cosine relationship?
- Compare θ = 30° and θ = 60°. The angles differ by 30°, but N at 30° is about 1.73 times N at 60° — not 1.5 times. Why? (Hint: the cosine function is not linear — equal angle steps produce unequal normal-force steps.)
- Set θ = 80° and m = 10 kg. The normal force N ≈ 17 N while W ≈ 98 N. If you placed a book on this near-vertical ramp, what force would need to hold it in place along the surface, and why is it so much larger than N?