Inputs
Mass of the body, incline angle of the plane, and the static and kinetic friction coefficients of the material pairing.
Mechanics · Coulomb friction
Friction and self-locking on an inclined plane
Determine whether a body on an inclined plane holds by itself, and what force is needed to hold it, push it up or let it down under control. Static and kinetic friction are treated separately, so breakaway and steady-state forces are distinguished – and the friction angle shows directly how much margin is left before it slides.
Coulomb friction model on a rigid body: friction proportional to the normal force and independent of area and speed. Tipping, rolling resistance, adhesion and dynamic effects are not assessed. The friction coefficients are inputs, because they depend on material pairing, surface and lubrication and scatter considerably.
ρ₀tan α ≤ μ₀
02
Forces and self-locking
Set the mass, incline and friction coefficients.
Inputs and method
Friction on an inclined plane
Check whether a body on a ramp holds by itself, and determine the forces to hold it, push it up and let it down.
Calculation
The weight force FG = m·g resolves into the downhill force FH = FG·sin α and the normal force FN = FG·cos α. Friction acts only on the normal force, so μ₀·FN is available at rest and μ·FN while moving. From that follow the holding force FH − μ₀·FN, the force uphill FH + μ·FN and the force when letting it down FH − μ·FN. The plane is self-locking exactly when tan α ≤ μ₀, i.e. when the incline stays below the friction angle ρ₀ = arctan μ₀.
Example
100 kg on 20° with μ₀ = 0.2 and μ = 0.15: FH = 335.4 N, FN = 921.5 N, available static friction 184.3 N. The friction angle is only 11.3°, so the body slides and needs a holding force of 151.1 N; pushing it up requires 473.6 N at 70.8 % efficiency.
Sources and limits: Classical statics with Coulomb friction, g = 9.80665 m/s². Friction proportional to the normal force and independent of area and speed; tipping, rolling resistance, adhesion and dynamic effects are excluded. The friction coefficients are inputs, as they depend on material pairing, surface, contamination and lubrication.
Friction on an inclined plane in detail
Check whether a body on a ramp holds by itself, and determine the forces to hold it, push it up and let it down.
What happens on an inclined plane?
On an inclined surface gravity no longer pulls a body straight into its support but resolves into two parts: one directed down the slope, which wants to make it slide, and one pressing perpendicular to the surface, which alone produces friction. As the angle rises the first grows and the second shrinks – which is why there is one particular angle at which every body, regardless of weight, begins to slide.
Formula and variables
Self-locking for tan α ≤ μ₀, i.e. α ≤ ρ₀ = arctan μ₀
FH = FG · sin α · FN = FG · cos αHolding force: F = FH − μ₀ · FN (zero when self-locking)Uphill: F = FH + μ · FN · Breakaway: F = FH + μ₀ · FNLetting down: F = FH − μ · FNη = FH / (FH + μ · FN) · a = g · (sin α − μ · cos α)
| Symbol / input | Meaning |
|---|---|
| FG, FH, FN | Weight force and its components along and perpendicular to the plane. |
| μ₀, μ | Static and kinetic friction coefficients of the material pairing. |
| α, ρ₀ | Incline angle of the plane and the friction angle it is compared with. |
| η | Efficiency of the plane when pushing the load up. |
Choose the inputs correctly
Mass of the body, incline angle of the plane, and the static and kinetic friction coefficients of the material pairing.
How to use the calculator
Enter the mass of the body and the inclination of the plane in degrees. Add the static friction coefficient for the body at rest and the kinetic one for the body in motion, both from a source verified for the actual pairing. The calculator first answers the decisive question of whether the plane is self-locking at all, then gives all three load cases: holding, pushing up and letting down.
Worked example
100 kg on 20° with μ₀ = 0.2 and μ = 0.15: FH = 335.4 N, FN = 921.5 N, available static friction 184.3 N. The friction angle is only 11.3°, so the body slides and needs a holding force of 151.1 N; pushing it up requires 473.6 N at 70.8 % efficiency.
Which value is the decisive one?
The most important output is not a force but the comparison of incline angle and friction angle: it decides between holding and sliding and at the same time shows how much margin remains. The forces below answer the follow-on sizing question – the breakaway force governing start-up and the uphill force governing continuous operation. The efficiency puts into context how much of the effort actually goes into lifting work.
Mass in kilograms, angle in degrees, forces in newtons. The friction coefficients are dimensionless. The standard value g = 9.80665 m/s² is used.
Typical applications
Sizing ramps and loading decks, checking whether a load stays put without additional securing, estimating the pull of a winch or thrust cylinder on an inclined guide, and judging the self-locking of wedges and adjustment mechanisms.
Assumptions, limits and common mistakes
Classical statics with Coulomb friction, g = 9.80665 m/s². Friction proportional to the normal force and independent of area and speed; tipping, rolling resistance, adhesion and dynamic effects are excluded. The friction coefficients are inputs, as they depend on material pairing, surface, contamination and lubrication.
Common mistake: Do not enter a percentage slope as an angle: 10 % is 5.7°, not 10°, the relation being α = arctan(slope/100). Do not set the static and kinetic coefficients equal – that removes exactly the difference between breakaway and steady-state force. And do not assume a heavier body slides sooner: the mass cancels out of the sliding criterion completely.
Frequently asked questions
Why does self-locking not depend on the mass?
Because both the downhill force and the friction force are proportional to the weight. The condition FG·sin α ≤ μ₀·FG·cos α can be divided by FG and becomes tan α ≤ μ₀. A piano and a coin start to slide on the same surface at exactly the same angle.
What is the friction angle and why is it so useful?
It is the angle ρ₀ = arctan μ₀ at which the downhill force just reaches the available static friction. Because it has the same unit as the inclination, it can be compared with it directly: the gap between α and ρ₀ is the margin before sliding, without having to compute any forces.
Why is the force when letting it down sometimes negative?
Because the kinetic friction then exceeds the downhill force. The body stops by itself instead of running away and has to be actively pushed to keep moving down at all. The calculator reports that separately, because the direction of the required force reverses.
Why does efficiency fall on shallow ramps?
Because at a shallow inclination the normal force, and hence the friction, is close to the full weight while the lifting work to be done tends to zero. The efficiency FH/(FH + μ·FN) therefore drops as the angle gets smaller – a shallow ramp needs less force but consumes more energy over the longer distance.
Can I rely on self-locking as a means of securing?
Not without further consideration. Friction coefficients scatter widely and fall considerably with moisture, oil, dust or vibration; under oscillating load static friction can break down almost entirely. For safety-relevant retention use a positive mechanical lock and treat self-locking at most as additional margin.