Last lesson you combined two arrows into one. This lesson runs the projector backwards: take any single vector pointing at an awkward angle and split it into a horizontal piece and a vertical piece. Two clean perpendicular numbers are far easier to work with than one diagonal, and this one trick powers projectile motion, inclined planes and every 2D force problem you will ever meet.
A force of $50\ \text{N}$ points at exactly $45°$ above the horizontal.
Before reading: do you think the horizontal part and the vertical part of this force are equal, or is one bigger? Write your prediction and one sentence on why.
Any diagonal vector is the sum of a horizontal part and a vertical part
A single vector pointing diagonally is really doing two jobs at once, pushing sideways and pushing up, and resolving simply names those two jobs separately.
Picture the vector as the hypotenuse of a right-angled triangle. The horizontal leg is the $x$-component and the vertical leg is the $y$-component. Because the two legs are perpendicular, they are independent, changing one does not change the other.
Any vector can be resolved into two perpendicular components that lie along chosen axes. The components are independent and recombine to the original by Pythagoras: $F = \sqrt{F_x^2 + F_y^2}$.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: the horizontal and vertical components of a vector are perpendicular, so neither affects the other.
Using $\cos$ and $\sin$ to find each component
We just saw that any vector is the hypotenuse of a right triangle. That raises a question: how do we calculate the length of each leg? This card answers it → trigonometry, with $\cos\theta$ for the adjacent leg and $\sin\theta$ for the opposite leg.
For an angle measured from the $x$-axis, the side next to the angle uses cosine and the side opposite uses sine, exactly the right-triangle ratios from maths.
For a vector of magnitude $F$ at angle $\theta$ to the $x$-axis:
$$F_x = F\cos\theta \qquad\qquad F_y = F\sin\theta$$
A rope pulls a sled with a force of $100\ \text{N}$ at $30°$ above the horizontal. Find the horizontal and vertical components of this force.
For a vector $F$ at angle $\theta$ to the $x$-axis: $F_x = F\cos\theta$ (adjacent) and $F_y = F\sin\theta$ (opposite). Each component is never larger than $F$, and they recombine as $F = \sqrt{F_x^2 + F_y^2}$.
Add the highlighted formulas to your notes before the check below.
Quick check: A velocity of $20\ \text{m s}^{-1}$ points at $60°$ above the horizontal. What is its horizontal component?
Pattern A, Calculate then verify
A force of $80\ \text{N}$ acts at $25°$ above the horizontal. In your book: (1) sketch the vector and its component triangle; (2) calculate $F_x$ and $F_y$; (3) verify your answer by checking that $\sqrt{F_x^2 + F_y^2}$ returns $80\ \text{N}$.
Two truths, one lie: which statement about resolving a vector is the lie?
Dot point: Resolve vectors into components
Turning one 2D problem into two 1D problems
We just learned to calculate each component. That raises a question: why bother splitting a vector at all? This card answers it → because perpendicular components can be handled separately, so a hard 2D problem becomes two easy 1D problems.
Once a diagonal vector is split into horizontal and vertical parts, you can analyse each direction on its own, and that is the entire strategy behind projectiles and inclined planes.
Because the components are independent, you can add all the $x$-components together and all the $y$-components together separately, then recombine at the end. To choose between $\sin$ and $\cos$, always identify which axis the angle is measured from: the component adjacent to the angle uses $\cos$, the component opposite uses $\sin$.
Resolving converts a 2D problem into two independent 1D problems, handle the $x$-direction and $y$-direction separately, then recombine. Always check which axis the angle is measured from: adjacent uses $\cos$, opposite uses $\sin$.
Pause, copy the highlighted strategy into your book before the check below.
Last check: Decide if the statement is true or false.
A component of a vector can be larger than the vector's full magnitude.
Odd one out: for a vector $V$ at angle $\theta$ to the $x$-axis, three of these are correct. Which is not?
Dot point: Resolve vectors into components
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 4(4 marks) 1. A child pulls a wagon with a force of $60\ \text{N}$ directed at $40°$ above the horizontal. Resolve this force into its horizontal and vertical components, and state what each component does to the wagon.
1 mark: $F_x = 60\cos 40°$ · 1 mark: $F_x \approx 46\ \text{N}$ · 1 mark: $F_y = 60\sin 40° \approx 39\ \text{N}$ · 1 mark: $F_x$ drives the wagon forward, $F_y$ lifts it (reducing the normal force)
ApplyBand 5(3 marks) 2. A force of $50\ \text{N}$ acts at $30^\circ$ above the horizontal. Resolve it into horizontal and vertical components, and verify your components recombine to the original magnitude.
1 mark: $F_x = 50\cos 30^\circ = 43\ \text{N}$ · 1 mark: $F_y = 50\sin 30^\circ = 25\ \text{N}$ · 1 mark: check $\sqrt{43^2 + 25^2} \approx 50\ \text{N}$
Dot point: Resolve vectors into components
UnderstandBand 4(3 marks) 3. Explain why resolving a vector into perpendicular components makes two-dimensional problems easier to solve, and state what must be true of the two chosen directions.
1 mark: perpendicular components are independent · 1 mark: each direction can be treated as a separate one-dimensional problem · 1 mark: the two directions must be at right angles
Dot point: Resolve vectors into components
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): The angle is measured from the horizontal, so the horizontal component uses cosine: $F_x = 60\cos 40° = 46.0\ \text{N}$. The vertical component uses sine: $F_y = 60\sin 40° = 38.6\ \text{N}$. The horizontal component $F_x$ (about $46\ \text{N}$) acts along the ground and drives the wagon forward. The vertical component $F_y$ (about $39\ \text{N}$) acts upward, partly lifting the wagon and reducing the normal force between the wagon and the ground.
Q2 (3 marks): Resolving along the axes gives $$F_x = F\cos\theta = 50\cos 30^\circ = 43\ \text{N}$$ and $$F_y = F\sin\theta = 50\sin 30^\circ = 25\ \text{N}$$ To verify, recombine by Pythagoras: $\sqrt{43^2 + 25^2} = \sqrt{1849 + 625} = \sqrt{2474} = 49.7 \approx 50\ \text{N}$, which returns the original magnitude. Note the components sum to $68\ \text{N}$ arithmetically, which is not the magnitude, because components add as vectors and not as numbers.
Q3 (3 marks): Resolving replaces one vector acting in an awkward direction with two vectors acting along chosen axes. Because those axes are perpendicular, the two components are independent: changing one does not affect the other. That lets a two-dimensional problem be split into two separate one-dimensional problems, each solved with ordinary algebra, and the results recombined at the end by Pythagoras and trigonometry. The requirement is that the two chosen directions are at right angles to each other; if they were not, the components would interfere and could not be handled separately.
You can now resolve any vector into perpendicular components and recombine them. With vectors fully mastered, the next lessons apply these tools to real journeys: distance, displacement, speed and velocity.
↩ Back to focus-area hubAt the start you considered a $50\ \text{N}$ force at $45°$. The components are equal at $45°$, because $\cos 45° = \sin 45°$, giving $50\cos 45° = 50\sin 45° = 35.4\ \text{N}$ each. Below $45°$ the horizontal part is bigger; above $45°$ the vertical part is bigger. Revisit Card 2 if this surprised you.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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