Frictionless 30°
α = 30°, μ = 0 → a = g/2.
Enter angle and μ — get acceleration along the ramp. If a ≤ 0 the model says it will not slide.
Theory and problem: inclined plane. Ideal frictionless ramp: incline (no friction). Friction force T = μN: friction force.
Enter values — the result shows up here.
With kinetic friction, acceleration along the slope is a = g(sin α − μ cos α). When μ cos α ≥ sin α (μ ≥ tan α), a ≤ 0 — from rest the object does not slide.
Checks: α = 30°, μ = 0 → a = g/2 ≈ 4.905 m/s². With μ = 0.2: a = g(0.5 − 0.2·√3/2) ≈ 3.206 m/s². At α = 10°, μ = 0.3: a ≤ 0.
This is the real slide. The ideal ramp (μ = 0) stays a separate long-tail tool with a = g sin α.
With L and a > 0: t = √(2L/a), vₖ = √(2aL) from rest. Friction work W ≈ T·L is omitted in v1 (needs mass).
Chart: a vs μ at your angle — watch a fall to zero.
a = g(sinα − μ cosα)
If a ≤ 0 — no slide from rest.
With L and a > 0: t = √(2L/a), vk = √(2a L)
α = 30°, μ = 0 → a = g/2.
α = 30°, μ = 0.2 — a ≈ 3.21 m/s².
α = 10°, μ = 0.3 → a ≤ 0.
α = 45°, μ = 0.2, L = 12 m.
α = 20°, μ = 0.35.
α = 15°, μ = 0.7 — often a ≤ 0.
α = 25°, μ = 0.15, L = 20 m.
α = 30°, μ = 0.1, g = 1.62 m/s².
α = 35°, μ = 0.25, L = 50 m.
When a ≤ 0, i.e. μ ≥ tan α (from rest in this kinetic-friction model).
Without friction a = g sin α (for α > 0). Here μ reduces a and can drive it to zero.
Sliding uses μₖ. The breakaway angle ties to μₛ = tan α.
Not in v1 — W ≈ T·L needs mass. Here: a, optional t and vₖ.
Out of scope for v1.
Blank = 9.81 m/s². Moon/Mars — type it in.
Header length (m / ft). Angle in degrees.
Yes in the formula — we treat it as “does not slide” down the ramp.
Raise α until it starts: μₛ ≈ tan α. Or measure a and invert the formula.
T = μ N with N = mg cos α — separate friction-force tool.
Compare with the ideal incline or return to F = m·a.