Drag the apex to change the incline angle directly
Controls
Readings
- Weight, mg
- 19.6N
- Weight’s parallel component, mg sinθ
- 9.8N
- Normal force, N
- 17.0N
- Maximum static friction, μsN
- 6.8N
- Current friction force, f
- 5.1N
- Net force
- 4.7N
- Acceleration, a
- 2.35m/s²
- Speed, v
- 0.00m/s
- Distance travelled, s
- 0.00m
- Time, t
- 0.0s
- Angle of repose, θᵣ
- 21.8°
How to use this simulation
- Release the block at a shallow angle, say 10°: mg sinθ is small, friction easily balances it, the block stays put, and both the net force and acceleration read zero.
- Raise the angle slowly and note exactly where the block first starts to move; that angle should match the "angle of repose" reading.
- Drag the apex directly to change the angle: both the arc and the angle label update immediately.
- Turn off "show weight components" to see only the weight, N and f, then turn it back on to see how the two dashed arrows are just the weight, split in two.
- Set both friction coefficients to zero: with no friction at all, the block slides at any angle, however small, with acceleration simply a = g sinθ.
Why moving a fridge uses a ramp, not a straight lift
Lifting a heavy fridge straight up into a truck is hard work, but rolling the same fridge up a tilted board takes far less force. The height gained is exactly the same either way, so why does it feel easier? Because on a slope, only part of the weight pushes the object along the surface — the rest presses it into the surface instead.
Pour sand onto a growing pile and it keeps its shape only up to a point: past a certain steepness, the top layer simply slides off, and the pile settles at a fairly consistent angle. Nobody measured that angle on purpose — the physics of friction fixed it, and that fixed angle is exactly what this page calls the angle of repose.
Mountain roads often carry signs reading "10% grade" or "steep hill", warning drivers about exactly this slope. A steeper grade means a bigger share of a vehicle’s weight is pulling it downhill, which is why brakes and tyre friction matter so much more on the way down — the same accounting this page runs for a block.
Starting from zero: what an inclined plane is, and why the weight gets split
An inclined plane is any flat surface set at an angle to the horizontal. A body resting on one feels two main forces: its weight mg, always pointing straight down, and the normal force N from the surface, always perpendicular to the surface. If there is friction, a third force acts too, along the surface.
The trouble is that weight points straight down, while any motion on the slope happens along the surface, not straight down. So mg cannot be compared to the surface directly — it has to be split into two parts: one along the slope (which pushes the body along it), and one perpendicular to the slope (which presses against N).
Geometry gives the split directly: if the surface makes an angle θ with the horizontal, the component along the slope is mg sinθ and the component perpendicular to it is mg cosθ. As θ grows, sinθ grows and cosθ shrinks — a steeper slope means more of the weight tries to push the body down it, and less presses it into the surface. That is exactly why climbing a steep hill is harder than walking on a nearly flat road.
With no other vertical force holding the body down or lifting it, the surface must fully support the perpendicular component, so the normal force is exactly N = mg cosθ. This N matters because friction depends directly on it.
Key terms at a glance
Get the vocabulary straight before the formulas; definition questions in exams come straight from this table.
| Term | Symbol | What it means | SI unit |
|---|---|---|---|
| Angle of incline | θ | The angle between the surface and the horizontal | degree (°) |
| Normal force | N | The force the surface exerts on the body, perpendicular to it | newton (N) |
| Coefficient of static friction | μs | How much friction can resist, up to the moment motion begins | no unit |
| Coefficient of kinetic friction | μk | How much friction resists once the body is already moving | no unit |
| Maximum static friction | μsN | The largest force friction can supply just before motion starts | newton (N) |
| Angle of repose | θr | The angle at which a body just begins to slide; tanθr = μs | degree (°) |
| Net force | ΣF | The sum of all forces along the surface | newton (N) |
Where the formulas come from: deriving the angle of repose and the sliding acceleration
The line between staying still and sliding is exactly the angle of repose. Two short derivations show where it comes from.
The angle of repose, θr: deriving tanθr = μs
For a body to stay still, friction must exactly balance the component pushing it downhill, mg sinθ. Friction can grow only as far as its limit, μsN = μs mg cosθ. So the condition to stay still is mg sinθ ≤ μs mg cosθ.
Dividing both sides by mg cosθ (safe as long as cosθ is not zero) gives sinθ/cosθ ≤ μs, that is, tanθ ≤ μs. The angle at which this turns into an equality, tanθr = μs, is the angle of repose θr — beyond it, the body cannot remain still.
Notice what is missing: mass. It cancels from both sides, since both the pushing component and the normal force scale with m in exactly the same way. So the angle of repose depends only on the surfaces involved, never on how heavy the body is.
tanθr = μsθr = tan⁻¹μs
The sliding acceleration, a = g(sinθ − μkcosθ)
Once the angle passes the angle of repose, the block starts sliding, and friction is no longer at whatever value is needed to balance things — it follows the kinetic friction law instead: f = μkN = μk mg cosθ, always opposing the motion (here, pointing back up the slope).
Applying Newton’s second law along the surface, the net force is ΣF = mg sinθ − f = mg sinθ − μk mg cosθ. Dividing by mass gives the acceleration, a = ΣF/m = g sinθ − μk g cosθ = g(sinθ − μkcosθ). Mass cancels here too, so heavy and light blocks on the same slope with the same friction accelerate identically — just like free fall.
With no friction (μk = 0), this collapses to the familiar a = g sinθ for a smooth incline.
a = g(sinθ − μkcosθ)acceleration along the slope while sliding
Static friction versus kinetic friction: why they are different numbers
A common misconception is that friction is a single fixed number. In reality, the friction while a body is at rest (static friction) and the friction once it starts moving (kinetic friction) are usually different, and almost always μk < μs.
Static friction is not a constant — it adjusts to whatever is needed, up to its ceiling μsN. As long as the applied force stays below that ceiling, static friction matches it exactly, holding the body in balance. That is why the simulation’s friction reading matches mg sinθ exactly at shallow angles, not μsN.
The moment the limit is crossed, friction stops adjusting and drops to a fixed value, μkN. μk is usually smaller than μs because once motion begins, the tiny interlocking points between the two surfaces have already been broken free, so keeping a moving body going takes less force than starting it moving in the first place. That is exactly why the first push to slide a heavy cupboard is the hardest part — once it is moving, keeping it moving is easier.
Try these in the simulation
Guess each result before you run it, then check.
- Set the angle to 30°, μs = 0.40 and μk = 0.30 — the simulation’s own defaults. Work out by hand whether the block will slide, then run it and check.
- Lower the angle gradually and watch exactly when the status text switches from "sliding down" to "at rest". That angle should match the "angle of repose" reading.
- At the same angle, raise μs until the block becomes static again — the angle of repose has grown past the incline’s angle.
- Lower μk while keeping μs fixed: the angle at which the block starts sliding never changes (μs decides that), but once it slides, the acceleration grows, since a = g(sinθ − μkcosθ) has a smaller μk.
- Double the mass. Neither the acceleration nor the decision to slide changes, since mass cancels in both formulas — only the weight, normal force and friction force in newtons double.
Solved problems
Write down what is given, then the formula, then substitute. That order earns full marks in an exam too.
Problem 1: resolving the weight into components
A 5 kg crate sits on a 30° incline. Find its weight, and the components along and perpendicular to the slope.
mg = 5 × 9.8 = 49.0 N. mg sinθ = 49.0 × sin30° = 24.5 N. mg cosθ = 49.0 × cos30° = 42.4 N.
Problem 2: will the crate stay put?
For the crate in Problem 1, if μs = 0.50, find the maximum static friction and decide whether the crate stays at rest.
Maximum static friction = μsN = 0.50 × 42.4 = 21.2 N. Since mg sinθ = 24.5 N exceeds that, the crate cannot stay at rest — it slides.
Problem 3: finding the angle of repose
A surface has a static friction coefficient of μs = 0.35. Find its angle of repose.
tanθr = μs = 0.35, so θr = tan⁻¹(0.35) = 19.3°. Any incline steeper than this angle will let a body start sliding on its own.
Problem 4: comparing acceleration with and without friction
A block slides down a 40° incline. Find its acceleration with no friction, and with μk = 0.20.
Frictionless: a = g sinθ = 9.8 × sin40° = 6.30 m/s². With friction: a = g(sinθ − μkcosθ) = 9.8 × (sin40° − 0.20 × cos40°) = 4.80 m/s². Friction slowed it down, but did not stop it.
Problem 5: the simulation defaults, worked in full
The simulation’s own values: angle 30°, mass 2 kg, μs = 0.40, μk = 0.30, ramp length 6 m. Will the block stay still, and if not, how long does it take to reach the bottom, and at what speed?
mg = 19.6 N, mg sinθ = 9.8 N, N = 17.0 N, maximum static friction = μsN = 6.8 N. Since 9.8 > 6.8, the block slides.
Kinetic friction = μkN = 5.1 N. Acceleration, a = (mg sinθ − friction) / m = (9.8 − 5.1) / 2 = 2.35 m/s². Time to cover 6 m, t = √(2s/a) = 2.26 s, and speed v = √(2as) = 5.31 m/s. Run the simulation and check these exact readings.
Problem 6: a crate that does NOT slide
A 3 kg crate sits on a 25° incline with μs = 0.60. Does it stay at rest?
This surface’s angle of repose is θr = tan⁻¹(0.60) = 31.0°, which is greater than the given angle of 25°, so the crate stays at rest. Check: N = 29.4 × cos25° = 26.6 N, maximum static friction = 0.60 × 26.6 = 16.0 N, and mg sinθ = 29.4 × sin25° = 12.4 N, which is smaller.
Problem 7: speed on a smooth incline
A block is released from rest on a frictionless 45° incline. Find its speed after sliding 10 m.
a = g sinθ = 9.8 × sin45° = 6.93 m/s². v = √(2as) = √(2 × 6.93 × 10) = 11.77 m/s.
Problem 8: stopping distance for a block sliding up the slope
A block is given a push of 8 m/s up a 20° incline with μk = 0.25. How far does it travel before stopping?
Going up, both gravity and friction act downhill (friction now points down the slope, opposing the upward motion): a = g(sinθ + μkcosθ) = 9.8 × (sin20° + 0.25 × cos20°) = 5.65 m/s². From v² = u² − 2as, 0 = 8² − 2 × 5.65 × s, so s = 5.66 m.
Problem 9: the μk that gives a constant sliding speed
A block slides down a 35° incline at exactly constant velocity (zero acceleration). Find μk.
Zero acceleration means mg sinθ = μk mg cosθ, so μk = tanθ = tan35° = 0.70. Notice this looks exactly like the angle-of-repose formula — here kinetic friction is meeting the constant-speed condition, rather than a static balance.
At a glance: friction coefficients of familiar materials
These rounded, illustrative kinetic-friction values show how much materials can differ. Ice on ice is almost frictionless; rubber on concrete is about as grippy as everyday surfaces get.
| Material pair | Approximate μk |
|---|---|
| Ice on ice | 0.03 |
| Teflon on steel | 0.05 |
| Wood on wood | 0.30 |
| Steel on steel | 0.55 |
| Rubber on concrete (a tyre) | 0.70 |
Common mistakes
Avoiding these keeps marks from slipping away on inclined-plane problems.
- Mixing up mg sinθ and mg cosθ. An easy check: at a shallow angle (nearly flat), the along-slope component should be small, and sinθ stays small for small angles while cosθ stays near 1.
- Assuming N = mg. On an incline, N = mg cosθ, not mg — N equals mg only on a flat horizontal surface.
- Treating static friction as always equal to μsN. In reality static friction adjusts to whatever is needed, only reaching μsN right at the point motion begins.
- Treating μk and μs as the same number. μk is usually smaller than μs, which is why a body already sliding is easier to keep moving than to start moving.
- Assuming the angle of repose depends on mass. tanθr = μs has no mass in it at all — heavy and light bodies on the same surface start sliding at the same angle.
- Forgetting the cosθ in the friction term of the acceleration formula. Friction depends on N, and N = mg cosθ, so cosθ has to appear wherever friction does.
Inclined planes in real life
These calculations are not confined to the exam hall — the same rules run quietly behind many everyday slopes.
- Wheelchair ramps: building codes limit how steep a ramp may be, because past a certain angle mg sinθ grows large enough that pushing a wheelchair up becomes genuinely hard.
- Mountain-road grade signs: a steeper percentage grade means a bigger mg sinθ pulling a vehicle downhill, which is exactly why brakes matter more on steep descents.
- Grain silos and sand piles: grain or sand can only pile up to its angle of repose before the top layer slides off — a direct, visible application of tanθr = μs.
- Ski slopes: a steeper slope gives a skier more acceleration (a = g sinθ, since friction on snow is very low), which is why steep runs demand more skill.
- Conveyor belts: an inclined belt carrying goods upward is kept below a maximum angle, or the goods lose their grip on the belt and slide back down.
- Staircases: a staircase is really a series of very steep inclined steps, so the height-to-depth ratio of each step is tied to the same physics as a ramp.
Exam corner
Definition questions typically ask for the angle of repose, and for the difference between static and kinetic friction, stated precisely. Numerical questions almost always start with resolving the weight into mg sinθ and mg cosθ before anything else, and a wrong first step there ruins every answer after it. The animation on Newton’s laws already showed friction opposing a push on a flat floor; here the same friction is doing its work on a tilted one — reading both pages together makes the idea click faster.
A typical structured question
Scenario: a 2 kg crate is placed on a 30° incline with μs = 0.40 and μk = 0.30.
(a) Define the angle of repose. (b) State the difference between static and kinetic friction. (c) Show numerically whether the crate stays at rest. (d) "Increasing the incline’s angle increases the block’s acceleration linearly" — discuss whether this is correct.
Answer to (c): N = 17.0 N, maximum static friction = 6.8 N, and mg sinθ = 9.8 N exceeds it, so the crate does not stay at rest. Answer to (d): a = g(sinθ − μkcosθ) is a combination of sinθ and cosθ, not a linear function of θ — the statement is incorrect; acceleration grows with angle, but not in a straight-line way.
Revision: the last-minute summary
One pass over this list the night before the exam is usually enough.
- Weight components: along the slope, mg sinθ; perpendicular to it, mg cosθ = N (with no other vertical force).
- Condition to stay at rest: mg sinθ ≤ μsN, that is, tanθ ≤ μs.
- Angle of repose: tanθr = μs, θr = tan⁻¹μs — independent of mass.
- Sliding acceleration: a = g(sinθ − μkcosθ); with no friction, a = g sinθ.
- μk is usually smaller than μs, so friction drops once motion begins.
- Constant-velocity sliding condition: μk = tanθ, giving zero acceleration.
Frequently asked questions
What is an inclined plane?
An inclined plane is any flat surface set at an angle to the horizontal. A body resting on it has its weight analysed as two components: one along the surface and one perpendicular to it.
What are the formulas for the weight components on an incline?
The component along the slope is mg sinθ, which pushes the body downhill. The component perpendicular to the slope is mg cosθ, which equals the normal force N when no other vertical force is present.
What is the angle of repose?
The angle of repose (θr) is the angle at which a body on an incline just begins to slide. Its formula is tanθr = μs, where μs is the coefficient of static friction.
What is the difference between static and kinetic friction?
Static friction acts while a body is at rest and adjusts up to a maximum of μsN. Kinetic friction acts once the body is moving and usually takes a fixed value μkN, where μk is smaller than μs.
What is the formula for acceleration while sliding down an incline?
With friction, a = g(sinθ − μkcosθ). Without friction, this simplifies to a = g sinθ.
Does the angle of repose depend on the mass of the body?
No. The formula tanθr = μs contains no mass term, because mass cancels equally from both the downhill component and the normal force. Heavy and light bodies on the same surface start sliding at the same angle.
Is the normal force on an incline always equal to mg?
No. On an incline, the normal force is N = mg cosθ, which is less than mg whenever θ is greater than zero. N equals mg only on a flat, horizontal surface.
Does a heavier object slide more easily on an incline?
No. In the condition mg sinθ ≤ μsN, both mg and N scale with mass in the same way, so mass cancels out entirely. Whether an object slides depends only on the angle and the friction coefficients, not on how heavy it is.
What condition lets a block slide down an incline at a constant speed?
Acceleration must be zero, which means mg sinθ = μk mg cosθ, so μk = tanθ. At that exact friction coefficient, the friction force matches the downhill component perfectly.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
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