Any single vector can be split into two perpendicular components
Scalars & Vectors - OCR A-Level Physics
Key Definition
Resolving a vector
Splitting a single vector into two perpendicular componentsThe two perpendicular parts that a single vector can be split into. that, added back together, give the original vector. The reverse process of vector addition.
Splitting a single vector into two perpendicular componentsThe two perpendicular parts that a single vector can be split into. that, added back together, give the original vector. The reverse process of vector addition.
Diagram pending
Right-angled triangle. Vector $\mathbf{F}$ along the hypotenuse, drawn from origin at angle $\theta$ above the horizontal. Adjacent side (horizontal) labelled $F_x = F\cos\theta$. Opposite side (vertical) labelled $F_y = F\sin\theta$. Small right-angle square at the corner.
Will be replaced with a GeoGebra SVG in stream 2.
- Any single vector can be split into two perpendicular components. Choose the axes to suit the problem: horizontal and vertical for projectile motion, parallel and perpendicular to a slope for inclined-plane problems.
- The component adjacent to the angle uses $\cos\theta$. The component opposite the angle uses $\sin\theta$.
- The angle $\theta$ is measured between the vector and the axis you project onto.
- Always draw the right-angled triangle with the vector as the hypotenuse and its two components as the other sides. Label which angle is $\theta$ before substituting numbers.
$$F_x = F\cos\theta \qquad F_y = F\sin\theta$$
- Pythagoras check: $F_x^2 + F_y^2 = F^2\cos^2\theta + F^2\sin^2\theta = F^2(\cos^2\theta + \sin^2\theta) = F^2$. The components always recombine to the original magnitude.
- Example: a $50 \text{ N}$ force at $37^\circ$ above the horizontal gives $F_x = 50\cos 37^\circ = 39.9 \text{ N}$ and $F_y = 50\sin 37^\circ = 30.1 \text{ N}$. Check: $\sqrt{39.9^2 + 30.1^2} = 50 \text{ N}$.
- Sign convention: choose a positive direction (e.g. right is $+x$, up is $+y$). Components pointing the other way are written as negative numbers when you later sum them.
Common Mistake
HIGH
Wrong: Swapping $\sin$ and $\cos$. Using $F\sin\theta$ for the component adjacent to the angle when $\theta$ is measured from the horizontal.
Right: The component adjacent to the angle uses $\cos$. The component opposite the angle uses $\sin$. Draw the right-angled triangle, mark $\theta$, and identify "adjacent" and "opposite" before writing the trig function.
Right: The component adjacent to the angle uses $\cos$. The component opposite the angle uses $\sin$. Draw the right-angled triangle, mark $\theta$, and identify "adjacent" and "opposite" before writing the trig function.
Examiner Tips and Tricks
- If the angle is given from the vertical (not the horizontal), $\cos$ and $\sin$ swap roles: $F\cos\theta$ then gives the vertical component and $F\sin\theta$ the horizontal.
- Limiting check: at $\theta = 0$, $\cos\theta = 1$ and $\sin\theta = 0$, so the vector lies entirely along the axis it makes the angle with.
- On inclined-plane problems, resolve along the slope and perpendicular to it, not horizontal and vertical, so the normal force lies along one axis.