Inclined plane simulator
A block sits on a ramp you control. Drag the handle at the ramp's top edge and gravity splits into its two components in front of you; slide the friction coefficients and watch the block decide, live, whether it stays put or slides.
μk can never exceed μs: lowering the static slider drags the kinetic one down with it.
| Quantity | Value |
|---|---|
| Weight mg | 19.60 N |
| mg sin(θ), along the slope | 9.80 N |
| mg cos(θ), into the slope | 16.97 N |
| Normal force N | 16.97 N |
| Static maximum μsN | 10.18 N |
| Friction force f (static, up the slope) | 9.80 N |
| Acceleration a | 0.00 m/s² |
The block slides only when mg sin(θ) beats the static maximum μsmg cos(θ). At the defaults, 9.80 N pulls down the slope and friction can hold up to 10.18 N, so it stays put. Tip the ramp one more degree and watch the verdict flip.
The tilted-axes method
Every inclined plane problem starts the same way: stop using horizontal and vertical axes and tilt them to match the ramp. One axis runs along the slope, the other runs perpendicular to it. The only force that now sits at an awkward angle is gravity, so gravity is the one you decompose:
- mg sin(θ) points along the slope, downhill. This is the component trying to make the block slide.
- mg cos(θ) presses perpendicular into the slope. This is the component the surface has to push back against.
On a bare incline with nothing else pushing on the block, the perpendicular direction is balanced, so the normal force is N = mg cos(θ). Not mg: the surface only supports the part of the weight that presses into it, which is why the normal arrow in the simulator shrinks as you steepen the ramp.
The verdict is a one-line comparison along the slope. Static friction can supply at most μsN = μsmg cos(θ). The block slides only when the pull down the slope wins: mg sin(θ) > μsmg cos(θ). Notice that mass appears on both sides and cancels, which the simulator confirms: change the mass and every arrow rescales, but the verdict and the acceleration never move. Dividing both sides by mg cos(θ) gives the compact version, tan(θ) > μs.
Once the block is moving, kinetic friction μkmg cos(θ) acts up the slope, and Newton's second law along the tilted x-axis gives a = g (sin(θ) - μk cos(θ)), independent of mass. At the default settings (2 kg, 30°, μs = 0.6), the downhill pull is 9.80 N against a static maximum of 10.18 N, so the block holds with 9.80 N of static friction and a = 0. Steepen to 31° and the comparison flips.
What to try
- Find the critical angle. With μs = 0.6, nudge the angle up one degree at a time. The verdict flips at 31°, the first whole degree where tan(θ) beats 0.6.
- Prove mass is a spectator. Set any angle, then sweep mass from 0.5 to 10 kg. Forces scale, acceleration does not.
- Make it frictionless. Drag μs to zero and the acceleration becomes g sin(θ): 4.9 m/s² at 30°, half of g.
For the worked problems behind this simulator, see inclined plane problems and how to find the normal force. To get numbers for a specific setup, use the friction calculator, and to practice drawing the arrows yourself, try the free-body diagram builder.
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