Normal Force on an Incline (N = mg cos θ)
Worked example: 10 kg on a 60° ramp → 49.033 N — press Try an example to run it live, then adjust anything.
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Grade 12Grade 12 Physics
Resolving on the incline →
UniversityEngineering Mechanics
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Normal Force on an Incline (N = mg cos θ) explained
On a slope, gravity still pulls straight down, but the surface can only push back perpendicular to itself — so it carries just the cos θ share of the weight: N = mg cos θ, with g = 9.80665 m/s². A 10 kg block on a 60° ramp presses in with only 10 × 9.80665 × cos 60° ≈ 49 N, half its 98 N weight. Level ground (θ = 0°) recovers N = mg, and a vertical wall (θ = 90°) gives N = 0, which is why nothing rests on a wall.
This is the quiet half of every incline problem, because friction is proportional to N: as the slope steepens, the down-slope pull grows while the friction budget shrinks, and the object eventually lets go. The classic error is using the full weight for N on a ramp, which overestimates friction and predicts that boxes stay put when they will actually slide.
Normal Force on an Incline (N = mg cos θ) formula
- = Normal force (N)
- = Mass (kg)
- = Incline angle (°)