On a slope at angle theta to the horizontal
Scalars & Vectors - OCR A-Level Physics
Key Definition
Weight on an inclined plane
On a slope at angle $\theta$ to the horizontal, weight $W = mg$ acts vertically downward. It is resolved into two perpendicular components: one parallel to the slope (causing the object to slide) and one perpendicular to the slope (pressing into the surface).
On a slope at angle $\theta$ to the horizontal, weight $W = mg$ acts vertically downward. It is resolved into two perpendicular components: one parallel to the slope (causing the object to slide) and one perpendicular to the slope (pressing into the surface).
Diagram pending
Box on a ramp inclined at angle $\theta$ to the horizontal. Weight $W = mg$ drawn vertically downward from the centre of the box. Two component arrows along the slope axes: $mg\sin\theta$ pointing down the slope, $mg\cos\theta$ pointing into the slope. Normal contact force $N$ drawn perpendicular to the slope, equal and opposite to $mg\cos\theta$ when there is no perpendicular acceleration. Angle $\theta$ marked at the base of the ramp.
Will be replaced with a GeoGebra SVG in stream 2.
- On a slope at angle $\theta$ to the horizontal, resolve the weight $W = mg$ into two components: parallel to the slope (down the slope) and perpendicular to the slope (into the surface).
- The component parallel to the slope (which would slide the object down) is $mg\sin\theta$.
- The component perpendicular to the slope (pressing into the surface) is $mg\cos\theta$.
- If there is no acceleration perpendicular to the slope, the normal contact force balances the perpendicular weight component: $N = mg\cos\theta$.
- The net force along the slope is $mg\sin\theta - f$, where $f$ is the friction force. If $f \geq mg\sin\theta$, the object does not slide.
$$F_{\parallel} = mg\sin\theta \qquad F_{\perp} = mg\cos\theta$$
Common Mistake
HIGH
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 $mg\cos\theta$. As the slope steepens ($\theta$ increases), $\sin\theta$ increases (more force down the slope, faster sliding), which matches intuition.
Right: The component DOWN the slope is $mg\sin\theta$. The component INTO the slope is $mg\cos\theta$. As the slope steepens ($\theta$ increases), $\sin\theta$ increases (more force down the slope, faster sliding), which matches intuition.
Examiner Tips and Tricks
- Limiting check: at $\theta = 0^\circ$ (flat), $\sin 0 = 0$ so no force along the slope, and $\cos 0 = 1$ so $N = mg$ (full normal force).
- At $\theta = 90^\circ$ (vertical), $\sin 90^\circ = 1$ so the full weight acts down the "slope", and $\cos 90^\circ = 0$ so $N = 0$ (no contact).
- If your answer fails this limiting check, you have $\sin$ and $\cos$ swapped.
- $\theta$ in these formulae is measured between the slope and the horizontal, not between the weight and the slope.