Key Equations
Scalars & Vectors - OCR A-Level Physics
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Horizontal component of a vector
$$F_x = F\cos\theta$$
- Where:
- $F_x$ = same units as F
- $F$ = magnitude of the vector
- $\theta$ = angle between the vector and the horizontal (degrees or radians)
Use when the angle is measured from the horizontal. The adjacent component uses cos. If the angle is from the vertical, cos gives the vertical component instead.
Vertical component of a vector
$$F_y = F\sin\theta$$
- Where:
- $F_y$ = same units as F
- $F$ = magnitude of the vector
- $\theta$ = angle between the vector and the horizontal (degrees or radians)
Use when the angle is measured from the horizontal. The opposite component uses sin. Always draw the right-angled triangle to confirm which trig function to use.
Resultant of two perpendicular vectors
$$F = \sqrt{F_x^2 + F_y^2}$$
- Where:
- $F$ = resultant magnitude (same units as components)
- $F_x$ = horizontal component
- $F_y$ = vertical component
Pythagoras' theorem applied to perpendicular components. Only valid when the two components are at 90 degrees to each other.
Angle of resultant vector
$$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$$
- Where:
- $\theta$ = degrees or radians
- $F_y$ = component opposite to the angle
- $F_x$ = component adjacent to the angle
Gives the angle measured from the horizontal (if F_x is horizontal and F_y is vertical). Always state the reference direction in your answer.
Weight component parallel to a slope
$$F_{\parallel} = mg\sin\theta$$
- Where:
- $F_{\parallel}$ = N
- $m$ = kg
- $g$ = N kg^-1 (or m s^-2)
- $\theta$ = angle of slope to horizontal
The component of weight that pulls the object down the slope. As the slope gets steeper (theta increases), this component increases. At theta = 90 degrees (vertical), it equals mg.
Weight component perpendicular to a slope
$$F_{\perp} = mg\cos\theta$$
- Where:
- $F_{\perp}$ = N
- $m$ = kg
- $g$ = N kg^-1 (or m s^-2)
- $\theta$ = angle of slope to horizontal
The component of weight pushing the object into the surface. Equals the normal contact force if there is no other perpendicular force. At theta = 0 (flat), this equals mg.