Inclined Plane Calculator
Calculate effort force, mechanical advantage, and friction loss on an inclined plane (ramp).
Useful for loading docks, ADA ramps, and physics homework.
Inclined Plane
A ramp lets you lift a heavy object using less force, in exchange for moving it a longer distance. The flatter the ramp, the smaller the force — but the longer the path.
Ideal Mechanical Advantage (IMA)
IMA = ramp length / ramp height = 1 / sin(θ)
| Slope | IMA |
|---|---|
| 5° | 11.5:1 |
| 10° | 5.76:1 |
| 15° | 3.86:1 |
| 30° | 2:1 |
| 45° | 1.41:1 |
| 60° | 1.15:1 |
| 90° (vertical) | 1:1 |
Effort to Push Up the Ramp (frictionless)
F_effort = m × g × sin(θ)
For a 100 kg crate on a 15° ramp:
- F_effort = 100 × 9.81 × sin(15°) = 254 N (≈ 26 kgf)
That is far easier than the full 981 N (100 kgf) of vertical lifting.
Adding Friction
If μ is the kinetic coefficient of friction between the load and the ramp surface:
F_effort = m × g × (sin(θ) + μ × cos(θ))
For the same crate with μ = 0.20:
- F_effort = 100 × 9.81 × (0.2588 + 0.20 × 0.9659) = 444 N
Friction can easily double the required effort.
Common Friction Coefficients
| Surfaces | μ_kinetic |
|---|---|
| Steel on dry steel | 0.4–0.6 |
| Wood on wood | 0.2–0.5 |
| Rubber tire on asphalt | 0.6–0.8 |
| Skis on snow | 0.04 |
| Greased surfaces | 0.05 |
| ADA-compliant ramp + casters | 0.05–0.10 |
ADA / Building Code Ramps
US ADA accessibility ramps are limited to 1:12 slope (about 4.76°), giving IMA ≈ 12:1. A 200 kg loaded wheelchair plus user (1962 N) requires only ~163 N (~17 kgf) of horizontal push — well within human capability.
Conservation of Energy
Total work = m × g × h, regardless of the ramp angle. The ramp redistributes that work over a longer distance — you trade peak force for total path length. Friction adds extra heat-dissipated work, which is why steeper ramps with friction are far less efficient.
Worked Example — Loading a Pickup
Pushing a 50 kg motorcycle up a 1.8 m ramp into a 0.6 m bed:
- Angle: arcsin(0.6 / 1.8) = 19.5°
- Frictionless effort: 50 × 9.81 × sin(19.5°) ≈ 164 N (~17 kgf)
- With μ = 0.15: 50 × 9.81 × (0.334 + 0.15 × 0.943) ≈ 234 N (~24 kgf)
- Vertical lift would need 491 N (~50 kgf) — twice as hard.