On a ramp, the motion runs along the slope, so the smart choice of axes tilts with it: $x$ along the slope, $y$ perpendicular to it. The normal force then lies along $y$, and only gravity needs to be split into components.
Key idea
On a ramp at angle $\theta$, gravity splits into $mg\sin\theta$ along the slope (downhill) and $mg\cos\theta$ into the slope. Without friction, $F_N = mg\cos\theta$ and $a = g\sin\theta$ down the slope.
Why it works
The angle between the weight (straight down) and the perpendicular to the ramp equals the ramp angle $\theta$, because both lines are turned by the same amount from the vertical and the ramp from the horizontal. So the into-the-slope part is adjacent to $\theta$ (cosine) and the along-the-slope part is opposite (sine). Check the extremes: on flat ground ($\theta = 0$), nothing pulls along the surface; on a vertical wall ($\theta = 90^\circ$), all of gravity is along it.
Worked examples
Worked example: Acceleration down a ramp
A block slides down a frictionless ramp at $20^\circ$. Find its acceleration.
- $a = g\sin\theta = 9.8\sin 20^\circ$.
- $a \approx 3.35\ \text{m/s}^2$ down the slope, for any mass.
Answer: about $3.35\ \text{m/s}^2$
Worked example: Holding a cart
How hard must you push, parallel to a frictionless $15^\circ$ ramp, to hold a $40\ \text{kg}$ cart still? What is the normal force?
- Along the slope: push $= mg\sin\theta = 392\sin 15^\circ \approx 101\ \text{N}$.
- Into the slope: $F_N = mg\cos\theta = 392\cos 15^\circ \approx 379\ \text{N}$.
Answer: about $101\ \text{N}$; normal force about $379\ \text{N}$
Common mistake
Swapping sine and cosine for the two parts of the weight. Check with a nearly flat ramp: the along-the-slope part should be tiny and the normal force almost $mg$. Since $\sin 0^\circ = 0$ and $\cos 0^\circ = 1$, the slope part is $mg\sin\theta$ and the normal part is $mg\cos\theta$.
Try it in the Science Lab: Forces at Work has a block on a ramp you can tilt.