Inclined plane calculator
Force along the slope, friction and acceleration.
Fill in the fields and the result will appear here automatically.
Find the forces on a body already sliding down a straight ramp. The weight is split into components parallel and normal to the surface, then sliding friction is subtracted. Downhill is positive: positive acceleration means speeding up, negative means slowing until it stops. A stationary body needs a separate static-friction model; this tool does not decide when sliding begins.
How it works
Formula and logic
With g = 9.80665 m/s²: F∥ = mg sin α, N = mg cos α, Ff = μN, Fnet = F∥ − Ff, a = Fnet/m. μ has no unit. The model assumes downhill sliding, constant μ, no applied pull, rolling or air drag. At 90°, N = 0 is a limiting case.
Example
50 kg, 30°, μ = 0.2: F∥ = 245.17 N, N = 424.64 N, friction = 84.928 N, net force = 160.24 N, a = 3.205 m/s². On a horizontal surface with the same mass and μ, a = −1.961 m/s²: a moving body slows; this is not a stability margin.
Fields and units
- Body mass — kg
- Slope angle — °
- Kinetic friction coefficient — 1
How to use
- — Enter mass in kilograms and the angle to the horizontal in degrees, not a percent gradient.
- — Use the kinetic friction coefficient for the actual surfaces; 0.2 is a sample input, not a universal material value.
- — Read signs with downhill positive. Negative acceleration applies while motion remains downhill; the model changes at rest.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: kinetic and static friction JCGM/BIPM: conventional standard acceleration 9.80665 m/s²
- Limitation
- A downhill-sliding model, not a load-securing or structural-stability calculation. Surface condition changes friction.
FAQ
Why does mass cancel from acceleration?
Both forces contain m, giving a = g(sin α − μ cos α). With the same surfaces, extra mass increases the forces but not this acceleration.
Will a crate begin to slide from rest?
This tool cannot establish that. Starting requires mg sin α > μs mg cos α using static μs. The entered coefficient describes sliding, and static friction need not equal μN.
Can I use it for uphill motion or rolling?
No. During uphill motion friction reverses and both forces act downhill; rolling also needs rotational dynamics.
How do I turn a percent gradient into the angle input?
For a rise Δh over horizontal run L, gradient p = 100Δh/L and angle α = arctan(p/100). A 100% gradient means 45°, not 90°. Enter the converted angle in degrees.