Resolve the weight of an object on a slope into its parallel and normal components, add friction, and find whether it slides, how fast it accelerates and how much force you need to push it up. The slope can be an angle, a rise and run or a grade percentage.
Last updated:
Slope given as
°
%
0 for a frictionless slope
m
For the sliding time, speed at the bottom and work done
m/s²
Enter your values to see the result.
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How to use the Inclined Plane Calculator
Enter the mass in kg or lb.
Give the slope as an angle, a rise and run, or a grade percentage.
Enter the friction coefficients (0 for frictionless).
Optionally add the slope length for time, speed and work.
Read the forces, the slide check and the working, then compare angles in the chart.
Inclined plane formulas
Quantity
Formula
Example (50 kg, 25°, μs 0.4, μk 0.3)
Parallel force
F∥ = m g sin θ
207.2 N
Normal force
N = m g cos θ
444.4 N
Slides if
tan θ > μs
0.466 > 0.4 → yes
Acceleration down
a = g (sin θ − μk cos θ)
1.48 m/s²
Push up (constant speed)
F = m g (sin θ + μk cos θ)
340.5 N
Hold still (minimum)
F = m g (sin θ − μs cos θ)
29.5 N
Ideal mechanical advantage
L / h = 1 / sin θ
2.37
Slope angle, grade and ratio
Angle
Grade
Rise : run
sin θ (share of weight along the slope)
2.86°
5%
1 : 20
5.0%
4.76°
8.3%
1 : 12
8.3%
5.71°
10%
1 : 10
9.95%
14.04°
25%
1 : 4
24.3%
30°
57.7%
1 : 1.73
50%
45°
100%
1 : 1
70.7%
Frequently asked questions
How do you calculate forces on an inclined plane?
Split the weight W = m g into a component along the slope, F∥ = m g sin θ, and one into the slope, N = m g cos θ. Friction is μ × N and acts against the motion (or the tendency to move).
What is the acceleration of an object sliding down a ramp?
a = g (sin θ − μk cos θ). Without friction it is simply g sin θ: about 4.9 m/s² on a 30° slope.
How much force does it take to push something up a ramp?
At constant speed, F = m g (sin θ + μk cos θ) parallel to the slope. A 50 kg box on a 25° ramp with μk = 0.3 needs about 340 N, versus 490 N to lift it straight up.
When will an object slide down by itself?
When m g sin θ exceeds the maximum static friction μs m g cos θ, which is the same as tan θ > μs. With μs = 0.4 that happens above about 21.8°.
What is the mechanical advantage of an inclined plane?
Ideally (no friction) it is length ÷ height = 1 / sin θ. You push with less force over a longer distance; friction lowers the real advantage and the efficiency.
How do I convert a grade or ratio to an angle?
θ = arctan(rise / run). A 10% grade is 5.71°, a 1:12 wheelchair ramp is 4.76° and a 1:4 ramp is 14.04°.
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