Inclined Plane Simulator

When a block on a slope slides, and how fast

A block on an inclined plane slides when the pull of gravity down the slope, mg sin θ, is more than static friction can hold, μs mg cos θ, which happens once tan θ is larger than μs. It then accelerates down the slope at a = g(sin θ − μk cos θ). Everything follows from resolving the weight along the slope and into it: mg sin θ drives the block, and mg cos θ presses it into the surface, which pushes back with the normal force N = mg cos θ that both kinds of friction are proportional to.

Mass appears in every one of those forces, so it cancels from the acceleration, and a heavy block and a light one of the same material slide together. A push changes that, because a push is not proportional to the mass: the acceleration is still the net force along the slope divided by the mass, as the force calculator works it out for F = ma, but the same push now accelerates a heavy block less than a light one.

The free-body diagram of a block on a slope

Three forces act on a block on a slope: its weight mg, straight down; the normal force N, at right angles to the surface; and friction f, along the surface, against the way the block moves or would move. A push F along the slope makes four. The arrows on the block are that diagram, all drawn to one scale, so a longer arrow is a larger force.

The two dashed arrows are the weight resolved into mg sin θ down the slope and mg cos θ into it. They are the weight, not extra forces, and each lies on the same line as the force it is weighed against: mg cos θ opposite the normal force, which it exactly matches, and mg sin θ on the line friction acts along, so with no push the difference between the two is the net force.

Using the simulator

Drag anywhere on the scene to tilt the slope, or select it and use the arrow keys. The sliders set the mass, the two friction coefficients, a push along the slope (positive pushes up it), the length of the slope and gravity. Play runs the slide from rest, and the scrubber steps through it by hand.

Every force the diagram draws is also read out, so the simulator doubles as an inclined plane calculator. The readouts give the acceleration and the net force, the friction and the normal force, the part of the weight along the slope, the critical angle and the range of pushes that hold the block still. For a block that slides they also give the time and the speed at the end of the slope, and the energy over the whole slope is split into the potential energy released, the work done by the push, the kinetic energy at the end and the heat from friction.

Worked example: a 2 kg block on a 30° wooden slope

The simulator opens on this case: a 2 kg block on a 5 m slope at 30°, wood on wood, with μs = 0.5 and μk = 0.3, and g = 9.81 m/s².

• Weight: mg = 2 × 9.81 = 19.62 N.
• Along the slope: mg sin 30° = 9.81 N. Into it: mg cos 30° = 16.99 N, which is the normal force.
• The most static friction can give is μs N = 0.5 × 16.99 = 8.50 N, less than the 9.81 N pulling the block down, so it slides. Put the other way, tan 30° = 0.577 is above μs.
• Kinetic friction then acts up the slope: μk N = 0.3 × 16.99 = 5.097 N, leaving a net force of 9.81 − 5.097 = 4.713 N down it.
• Acceleration: the net force over the mass, which is a = g(sin 30° − 0.3 cos 30°) = 9.81 × 0.2402 = 2.356 m/s².
• From rest over 5 m: t = √(2 × 5 / 2.356) = 2.06 s and v = √(2 × 2.356 × 5) = 4.854 m/s, as the kinematics calculator gives for u = 0, a = 2.356 m/s² and s = 5 m.

The energy balances to the joule. The block drops 5 × sin 30° = 2.5 m, which releases mgh = 19.62 × 2.5 = 49.05 J. Of that, ½mv² = 23.56 J is kinetic energy at the bottom and the heat is μk N d = 5.097 × 5 = 25.49 J, and 23.56 + 25.49 = 49.05. Those are the energy readouts at the opening settings, where the push does no work.

Static friction is only as strong as it needs to be

Set the same block on a 20° slope and it stays put, because tan 20° = 0.364 is below μs = 0.5. The static friction readout then shows 6.71 N, exactly the pull down the slope, mg sin 20°, although static friction could give up to μs N = 9.218 N. μs N is the ceiling, not the value: static friction takes whatever size up to that ceiling keeps the block still, and points whichever way that needs, which is down the slope when a push up it is stronger than the pull of gravity along it.

The same is true on level ground, and at an angle of 0 the simulator reproduces Example 6.10 in OpenStax University Physics: a 20.0 kg crate with μs = 0.700 and μk = 0.600, with g = 9.80 m/s². Pushes of 20, 30 and 120 N leave it still, met by static friction of exactly 20, 30 and 120 N, because the most static friction can give is 0.700 × 196 = 137.2 N. At 180 N the crate slides, friction drops to 0.600 × 196 = 117.6 N, and it accelerates at (180 − 117.6) / 20.0 = 3.12 m/s². The book prints 3.10 m/s² because it rounds the friction to 118 N first.

Finding μs from the angle where it slips

With no push the block stays put up to the critical angle, where tan θ = μs, and slides above it. Mechanics texts also call this angle the angle of repose. Neither the mass nor gravity comes into it, since the pull down the slope and the limit of static friction are both proportional to mg, which is why tilting a board until a block slips is the standard way to measure a coefficient of static friction. The critical angle readout is arctan μs: 26.57° for μs = 0.5 and 45° for μs = 1.

Just past that angle the block does not creep away. The friction holding it drops from μs N to μk N the moment it moves, so it sets off with an acceleration of about g cos θ (μs − μk). At 26.5° the default block is held; at 26.6° it slides at 1.761 m/s². The friction plot shows the same thing as a step. The rule tan θ = μk describes something else, a block that keeps sliding at constant speed once it has been started, which a block starting from rest never does.

Pushing a block up the slope

A push up the slope holds the block still anywhere between mg(sin θ − μs cos θ) and mg(sin θ + μs cos θ), which for the opening block is 1.314 N to 18.31 N, the range the push readout gives. To start the block moving up the slope, the push has to beat the top of that range. Once it is moving, kinetic friction takes over, pointing down the slope, and a push of mg(sin θ + μk cos θ) = 9.81 + 5.097 = 14.91 N keeps it going at a steady speed. The simulator starts every block from rest, so at 14.9 N it shows the block held, with 5.09 N of static friction pointing down the slope.

Push harder than 18.31 N and the block accelerates up the slope. At 20 N, a = 20/2 − g(sin 30° + 0.3 cos 30°) = 10 − 7.454 = 2.546 m/s², and over the 5 m the push does 100 J of work: 49.05 J becomes potential energy, 25.46 J kinetic energy and 25.49 J heat.

With no friction at all, the push that holds the block, or keeps it moving up at a steady speed, is mg sin θ = mg h/L, where h is the height the slope rises over its length L. That is why a ramp counts as a simple machine: it trades distance for force, with an ideal mechanical advantage of L/h, which is 2 for a 30° slope. Friction takes some of that back, and the opening block needs 14.91 N rather than 9.81 N to keep it moving up.

Where the energy goes

With no push, the potential energy released as the block slides down becomes kinetic energy and heat: mgh = ½mv² + μk N d. A push adds its work, F times the distance, when it points along the motion and takes it away when it points against it, the same force times distance the work and power calculator computes. The energy plot draws each term against the distance slid. Every line is straight, because every force is constant, and each gradient is a force: mg sin θ for the potential energy, μk N for the heat and the net force for the kinetic energy.

That is also why friction costs speed. On a frictionless slope the whole 49.05 J would become kinetic energy and the block would reach √(2gh) = 7.00 m/s, the speed of a straight drop of 2.5 m, which the gravitational potential energy calculator gives from the height alone. With wood on wood it reaches 4.854 m/s, and a little over half the energy has gone into warming the block and the slope.

What this model leaves out

• A block that is launched. The block always starts from rest. One sent up the slope slows with friction pointing down the slope, stops, and then either stays or slides back with friction pointing up it: two accelerations rather than one. A block already sliding down and speeding up has the acceleration shown here with μs set equal to μk.
• A push at an angle. The push here acts along the slope. Tilt it by an angle φ away from the surface and the normal force falls to mg cos θ − F sin φ, taking both frictions down with it.
• Rolling. A ball rolls rather than slides, and a solid ball rolling without slipping accelerates at (5/7) g sin θ, losing no energy to friction at all.
• Real surfaces. The coefficients are constants, the same at any speed and for any area of contact, which is the Amontons and Coulomb model. Lubricated or very fast contacts depart from it, and a real μs varies from place to place along a board.
• A curved track, air resistance and tipping. The slope is straight, there is no drag, and the block never topples, although a tall, narrow one would tip before it slid.

Common mistakes

• Writing N = mg on a slope. The slope only pushes back against the part of the weight pressing into it, mg cos θ. Using mg overstates friction by about 15 percent at 30°.
• Deciding whether it starts to slide with μk. A block at rest starts to move only when the pull beats μs N. μk matters once it is moving.
• Treating static friction as μs N every time. μs N is the most it can give. On a 20° slope the 2 kg block needs, and gets, only 6.71 N.
• Taking tan θ = μk as the angle where it starts to slide. That is the slope on which a moving block keeps a constant speed. The start is set by tan θ = μs.
• Giving friction a fixed direction. Friction opposes the motion, or the motion it is preventing. Push the block up the slope and friction points down it.
• Expecting a heavier block to slide faster. Without a push the mass cancels, and every block of the same material has the same acceleration.
• Leaving the calculator in radians. sin 30 in radians is −0.988, not 0.5.