If you know the horizontal and vertical components
Scalars & Vectors - OCR A-Level Physics
Key Definition
Combining perpendicular components
If a vector has been resolved into perpendicular components $F_x$ and $F_y$, the original magnitude $F$ is recovered using Pythagoras' theorem and the angle is recovered using inverse tan.
If a vector has been resolved into perpendicular components $F_x$ and $F_y$, the original magnitude $F$ is recovered using Pythagoras' theorem and the angle is recovered using inverse tan.
Diagram pending
Right-angled triangle. Horizontal leg labelled $F_x$, vertical leg labelled $F_y$, hypotenuse labelled $F$ as the resultant. Angle $\theta$ marked at the bottom-left corner between $F_x$ and $F$. Right-angle square at the bottom-right corner.
Will be replaced with a GeoGebra SVG in stream 2.
- If you know two perpendicular components, the magnitude of the resultant is found with Pythagoras' theoremFor a right-angled triangle: the square of the hypotenuse equals the sum of the squares of the other two sides..
- The direction (the angle the resultant makes with one axis) is found with trigonometryThe relationships between angles and sides of triangles (sin, cos, tan)., specifically $\tan^{-1}$.
- This only works directly when the two components are perpendicular. For non-perpendicular vectors use the parallelogram cosine-rule form.
$$F = \sqrt{F_x^2 + F_y^2} \qquad \theta = \tan^{-1}\!\left(\frac{F_y}{F_x}\right)$$
- Always state the direction of the resultant relative to a stated reference (e.g. "$35^\circ$ above the horizontal", "N $30^\circ$ E", or "$23^\circ$ to the right of vertical"). An angle without a reference loses the mark.
- Worked example: $F_x = 6.0 \text{ N}$ east, $F_y = 8.0 \text{ N}$ north. Magnitude $F = \sqrt{6.0^2 + 8.0^2} = \sqrt{100} = 10.0 \text{ N}$. Direction $\theta = \tan^{-1}(8.0/6.0) = 53^\circ$ north of east.
- Calculator check: $\tan^{-1}$ on most calculators returns an angle between $-90^\circ$ and $+90^\circ$. If your resultant points into the second or third quadrant ($F_x$ negative), add $180^\circ$ to the calculator answer.
Common Mistake
MEDIUM
Wrong: Quoting magnitude as $F = F_x + F_y$, or writing $\tan\theta = F_x/F_y$ when you meant $F_y/F_x$, or quoting an angle without naming the reference axis.
Right: Magnitude uses Pythagoras: $F = \sqrt{F_x^2 + F_y^2}$. The angle measured from the $x$-axis is $\tan^{-1}(F_y/F_x)$ (opposite over adjacent). Always state the reference: "$\theta$ above the horizontal" or similar.
Right: Magnitude uses Pythagoras: $F = \sqrt{F_x^2 + F_y^2}$. The angle measured from the $x$-axis is $\tan^{-1}(F_y/F_x)$ (opposite over adjacent). Always state the reference: "$\theta$ above the horizontal" or similar.
Examiner Tips and Tricks
- Multi-force problems: resolve every force into $x$ and $y$ components, sum each axis independently, then recombine the totals with Pythagoras and $\tan^{-1}$.
- For projectile motion, components are independent. Treat horizontal and vertical motion separately, then combine speeds at the end if needed.