On a slope at angle theta to the horizontal
Scalars & Vectors - OCR A-Level Physics
- On a slope at angle theta to the horizontal, resolve the weight into two components.
- The component parallel to the slope (causing the object to slide down) = mg sin(theta).
- The component perpendicular to the slope (into the surface) = mg cos(theta).
- The normal contact force equals the perpendicular component: $N = mg \cos(\theta)$.
- The net force down the $slope = mg \sin(\theta) minus friction$.
$$F_{\parallel} = mg\sin\theta$$
$$F_{\perp} = mg\cos\theta$$
Common Mistake
MEDIUM
Wrong: Using mg cos(theta) for the component down the slope.
Right: The component DOWN the slope is mg sin(theta). The component INTO the slope is mg cos(theta). Remember: as the slope gets steeper (theta increases), sin(theta) increases (more force down the slope), which makes physical sense.
Right: The component DOWN the slope is mg sin(theta). The component INTO the slope is mg cos(theta). Remember: as the slope gets steeper (theta increases), sin(theta) increases (more force down the slope), which makes physical sense.
Examiner Tips and Tricks
- A quick check: when theta = 0 (flat surface), sin(0) = 0 so no force along the slope, and cos(0) = 1 so normal $force = mg$.
- When theta = 90 (vertical), sin(90) = 1 so the full weight acts downward (as expected).
- If your answer fails this check, you have sin and cos swapped.