Put a crate on a ramp and its weight stops being one number. Gravity still pulls straight down, but the ramp only cares about the two directions it knows: along the slope and into it. So we split the weight into those two components, and everything else in this chapter follows.

$F_{\parallel} = mg\sin\theta$ — read aloud F-parallel equals m g sine theta. $F_{\parallel}$ is the part of the weight pulling the crate DOWN the slope, in newtons; $m$ is the mass in kilograms; $g$ is 9.81 m/s²; and $\theta$ — theta — is the ramp's angle above the horizontal, in degrees. The other half is $N = mg\cos\theta$, the normal force: how hard the crate presses INTO the surface, and therefore how hard the surface presses back. Normal here means perpendicular, not ordinary.

Which trig goes where is the exam's favourite trap, so anchor it with a limit rather than a mnemonic. Flatten the ramp to $\theta = 0$: nothing slides, so the along-slope part must vanish — and $\sin 0 = 0$. Meanwhile the floor still holds the whole weight — and $\cos 0 = 1$. Sine along, cosine into. Check it that way once and you will never need to memorise it.