Inclined Plane Force Calculator
Work out inclined plane force instantly with clear inputs, formula shown and shareable results.
About Inclined Plane Force Calculator
On a slope, weight splits into mg sin theta down the incline and mg cos theta into it. Friction is proportional to that second component, so it fades as the slope steepens while the driving component grows. A 50 kg mass on 25 degrees with mu = 0.2 has 207 N pulling it down and 89 N of friction, so 118 N must be held.
Formulas
Components of weight
along slope = mg sin(theta); into slope = mg cos(theta)
Holding force
F = mg sin(theta) - mu.mg cos(theta)
Worked example
With the default values below, Inclined Plane Force Calculator returns Force Needed To Stop It Sliding of 118.345. Change any input above and every figure here is recalculated in your browser.
Inputs used
- Mass on the slopeminimum 0 kg
- 50 kg
- Slope angleaccepts 0–89.9 deg
- 25 deg
- Coefficient of frictionminimum 0
- 0.2
Results
- Force Needed To Stop It Sliding
- 118.345 N
- Gravity Component Along The Slope
- 207.224 N
- Normal Force On The Slope
- 444.392 N
- Friction Resisting Motion
- 88.878 N
Step by step
- 1Values used
- 2Components of weight
- 3Holding force
- 4Force Needed To Stop It Sliding= 118.345 N
- 5Gravity Component Along The Slope= 207.224 N
- 6Normal Force On The Slope= 444.392 N
- 7Friction Resisting Motion= 88.878 N
Assumptions and limits
- The result is derived only from the 3 inputs listed above. Anything not among them is not part of the calculation.
- In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
- Figures are rounded for display to 3 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.
How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.
Frequently Asked Questions
At what angle does it slide on its own?
When tan theta exceeds mu. For mu = 0.2 that is about 11.3 degrees, so 25 degrees is well past it.
Why does friction fall on steeper slopes?
Because the surface is pressed together by mg cos theta, which shrinks as the slope steepens. Vertically there is no normal force and no friction at all.