What it does
A block on a ramp is the classic physics problem for splitting a force into components. This calculator takes the mass, the angle and (optionally) friction and a push up the slope. It finds every force, decides whether static friction can hold the block, and gives the acceleration. It also shows how hard you'd have to push to move the block up at a steady speed and the ramp's mechanical advantage.
How it's calculated
Gravity pulls straight down with the weight mg, using standard gravity g = 9.80665 m/s2. Tilt the axes to line up with the slope and the weight splits into two parts:
Normal force N = mg cos θ
Static friction holds if |Fpush − mg sin θ| ≤ μs N
Otherwise a = (|Fpush − mg sin θ| − μk N) ÷ m
Ideal mechanical advantage = 1 ÷ sin θ = length ÷ height
Friction always opposes motion (or the tendency to move). If you leave μs empty, the calculator uses μk for both.
Worked example
A 10 kg box sits on a 30° ramp with μk = 0.2 and no push. Weight = 10 × 9.80665 = 98.07 N. Along the slope: 98.07 × sin 30° = 49.03 N. Normal force: 98.07 × cos 30° = 84.93 N. Friction can supply at most 0.2 × 84.93 = 16.99 N, less than 49.03 N, so the box slides. Net force = 49.03 − 16.99 = 32.05 N, and a = 32.05 ÷ 10 = 3.20 m/s2 down the slope.
To push it up at a steady speed you'd need 49.03 + 16.99 = 66.02 N. Lifting it straight up takes 98.07 N, so the ramp's actual mechanical advantage is 98.07 ÷ 66.02 = 1.49; without friction it would be 1 ÷ sin 30° = 2.
Angles and ideal mechanical advantage
| Angle | sin θ | cos θ | Ideal MA |
|---|---|---|---|
| 5° | 0.0872 | 0.9962 | 11.47 |
| 10° | 0.1736 | 0.9848 | 5.76 |
| 15° | 0.2588 | 0.9659 | 3.86 |
| 30° | 0.5 | 0.8660 | 2 |
| 45° | 0.7071 | 0.7071 | 1.414 |
| 60° | 0.8660 | 0.5 | 1.155 |
Limitations
- The block is treated as a point: no rolling, tipping or air resistance. A rolling ball or wheel accelerates differently because some energy goes into spinning.
- Friction coefficients depend on both surfaces and their condition. Measure them if you need a reliable answer.
- The push is assumed to be parallel to the slope. A push at another angle also changes the normal force.