2.2 Components analysis

Achievable AP Physics 1

2. Vectors and their analysis

Our AP Physics 1 course is currently in development and is a work-in-progress.

Components analysis

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This sub-chapter builds on the Basics of vectors section by looking more closely at how to break vectors into components and switch between different representations. These skills show up constantly in physics, especially when you’re resolving forces or motion along convenient directions.

In this sub-chapter, we’ll cover:

Review and transition
Conversion between polar and rectangular forms
Decomposition along standard axes
Resolving vectors in rotated or non-standard axes
Projection of vectors
Applications and example problems

Review and transition

Here’s a quick recap of the key ideas we’ll use throughout this sub-chapter:

Vector basics: A vector is a quantity with both magnitude and direction. You can represent vectors graphically (arrows) and algebraically (components), and you can add or subtract them.
Magnitude of a vector: For a vector expressed as

$v = v_{x} \hat{i} + v_{y} \hat{j},$

the magnitude is

$∣ v ∣ = v_{x}^{2} + v_{y}^{2} .$

With those fundamentals in place, we can focus on two core tasks:

breaking a vector into components along chosen axes
converting between polar and rectangular forms

Conversion between polar and rectangular forms

Vectors are commonly written in two equivalent ways:

Polar form: A magnitude $V$ and an angle $θ$ (measured from the horizontal axis).
Rectangular (component) form: Components along the $x$ - and $y$ -axes:

$V = V_{x} \hat{i} + V_{y} \hat{j} .$

Converting from polar to rectangular form

If a vector $V$ has magnitude $V$ and angle $θ$ , its components come from right-triangle trigonometry:

Horizontal component:

$V_{x} = V cos θ$
Vertical component:

$V_{y} = V sin θ$

Vector resolved into horizontal and vertical components

Converting from rectangular to polar form

If you’re given components,

$V = V_{x} \hat{i} + V_{y} \hat{j},$

you can recover the magnitude and direction:

Magnitude:

$V = V_{x}^{2} + V_{y}^{2}$
Direction:

$θ = tan^{- 1} (\frac{V _{y}}{V _{x}})$

Conversion from rectangular components to polar form

Decomposition along standard axes

When your axes are the usual horizontal ( $x$ ) and vertical ( $y$ ) directions, resolving a vector is straightforward. For a vector $V$ making an angle $θ$ with the horizontal:

Horizontal component:
$V_{x} = V cos θ$
Vertical component:
$V_{y} = V sin θ$

Example problem 1

A vector has a magnitude of 12 units and makes an angle of $4 0^{\circ}$ with the horizontal. Find its components.

Solution:

(spoiler)

$V_{x} = 12 cos 4 0^{\circ} \approx 12 \times 0.766 = 9.19$

$V_{y} = 12 sin 4 0^{\circ} \approx 12 \times 0.643 = 7.72$

Thus,

$V \approx (9.19, 7.72) .$

Resolving vectors in rotated or non-standard axes

In many physics problems, the most useful axes aren’t horizontal and vertical. For example, on an inclined plane it’s often easiest to choose one axis along the plane and the other perpendicular to it.

Decomposition in a rotated coordinate system

Assume the coordinate system is rotated by an angle $α$ relative to the horizontal. If a vector $V$ makes an angle $θ$ relative to the horizontal, then the angle relative to the rotated $x^{'}$ -axis is:

$ϕ = θ - α$

Using that relative angle, the components in the rotated system are:

Component along $x^{'}$ :

$V_{x^{'}} = V cos (θ - α)$
Component along $y^{'}$ :

$V_{y^{'}} = V sin (θ - α)$

Example problem 2

A force of 30 N acts at an angle of $7 0^{\circ}$ from the horizontal. If the coordinate system is rotated by $2 0^{\circ}$ clockwise, find the components of the force in the rotated system.

Solution:

(spoiler)

Compute the relative angle:

$ϕ = 7 0^{\circ} - 2 0^{\circ} = 5 0^{\circ}$
Resolve into components:

$F_{x^{'}} = 30 cos 5 0^{\circ} \approx 30 \times 0.643 = 19.29 N$

$F_{y^{'}} = 30 sin 5 0^{\circ} \approx 30 \times 0.766 = 22.98 N$

Projection of vectors

Why use projections?

A projection tells you how much of one vector points in the direction of another. This matters in physics because:

Simplification of problems: Often, only the component of a vector along a particular direction affects the situation you’re analyzing.
Understanding work and energy: In work calculations, only the component of force along the displacement contributes to work.
Vector decomposition: Projections give a systematic way to break vectors into parts aligned with specific directions, especially when the axes aren’t standard.

Scalar projection

The scalar projection (or component) of vector $A$ onto vector $B$ is:

$comp_{B} A = \frac{A \cdot B}{∣ B ∣} = ∣ A ∣ cos θ,$

where $θ$ is the angle between $A$ and $B$ .

Vector projection

The vector projection of $A$ onto $B$ is:

$proj_{B} A = (\frac{A \cdot B}{∣ B ∣ ^{2}}) B .$

Example problem 3

Find the scalar and vector projections of

$A = (5, 3)$

onto

$B = (4, 0) .$

Solution:

(spoiler)

Scalar projection:

$comp_{B} A = \frac{A \cdot B}{∣ B ∣} = \frac{5 \times 4 + 3 \times 0}{4} = \frac{20}{4} = 5.$
Vector projection:

$proj_{B} A = (\frac{20}{4 ^{2}}) B = (\frac{20}{16}) (4, 0) = (\frac{5}{4}) (4, 0) = (5, 0) .$

Additional questions

Now let’s apply these techniques in more integrated problems. Example problem 4

A block on an inclined plane experiences two forces. The first force is 40 N acting at an angle of $6 0^{\circ}$ from the plane upwards along the incline, and the second is 30 N acting at an angle of $2 0^{\circ}$ from the plane downwards along the incline. The plane is inclined at $3 0^{\circ}$ from the horizontal. Find the net force acting along the plane.

Solution:

(spoiler)

Set up the rotated axes: Align the $x^{'}$ -axis along the plane and the $y^{'}$ -axis perpendicular to the plane.
Resolve each force into $x^{'}$ components:
- For the 40 N force (acting upward along the plane):
$F_{1, x^{'}} = 40 cos 6 0^{\circ} = 40 \times 0.5 = 20 N .$
- For the 30 N force (acting downward along the plane):
$F_{2, x^{'}} = 30 cos 2 0^{\circ} \approx 30 \times 0.94 = 28.2 N .$
Determine the net force: Assuming the 28.2 N force acts in the opposite direction to the 20 N force, the net force along the plane is:

$F_{net, x^{'}} = 28.2 N - 20 N = 8.2 N .$

Forces and their resolution along the plane

Example problem 5 Problem: Vectors $A$ and $B$ have magnitudes of 6 N and 8 N respectively, and the angle between them is $12 0^{\circ}$ . Find the magnitude and direction of the resultant vector

$R = A + B .$

Solution:

(spoiler)

Magnitude (using the cosine rule):

$R^{2} = 6^{2} + 8^{2} - 2 \times 6 \times 8 cos 12 0^{\circ} .$

Since

$cos 12 0^{\circ} = - 0.5,$

$R^{2} = 36 + 64 + 96 \times 0.5 = 100 + 48 = 148,$

$R \approx 148 \approx 12.17 N .$

Direction (using the sine rule): Let $θ$ be the angle between $A$ and $R$ . Then:

$\frac{sin θ}{8} = \frac{sin 12 0 ^{\circ}}{12.17} .$

Calculate

$sin 12 0^{\circ} \approx 0.866$

and solve for $θ$ :

$sin θ \approx \frac{8 \times 0.866}{12.17} \approx 0.569,$

$θ \approx 34. 7^{\circ} .$

This is the angle between $A$ and the resultant $R$ .

Review and transition\

Vectors: quantities with magnitude and direction
Magnitude formula: $∣ v ∣ = v_{x}^{2} + v_{y}^{2}$
Key skills: breaking vectors into components, converting between polar and rectangular forms

Conversion between polar and rectangular forms\

Polar form: magnitude $V$ and angle $θ$
Rectangular form: components $V_{x} = V cos θ$ , $V_{y} = V sin θ$
From components to polar:
- Magnitude: $V = V_{x}^{2} + V_{y}^{2}$
- Direction: $θ = tan^{- 1} (V_{y} / V_{x})$

Decomposition along standard axes\

Standard axes: horizontal ( $x$ ) and vertical ( $y$ )
Components:
- $V_{x} = V cos θ$
- $V_{y} = V sin θ$
Used when axes are not rotated

Resolving vectors in rotated or non-standard axes\

Rotated axes: axes at angle $α$ from horizontal
Relative angle: $ϕ = θ - α$
Components in rotated system:
- $V_{x^{'}} = V cos (θ - α)$
- $V_{y^{'}} = V sin (θ - α)$

Projection of vectors\

Scalar projection: $comp_{B} A = \frac{A \cdot B}{∣ B ∣} = ∣ A ∣ cos θ$
Vector projection: $proj_{B} A = (\frac{A \cdot B}{∣ B ∣ ^{2}}) B$
Projections simplify problems, especially with non-standard axes

Applications and example problems\

Use component methods to resolve forces and motion along chosen axes
For multiple vectors, use cosine and sine rules to find resultant magnitude and direction
Always align axes with the problem’s geometry for simplification

Next | 3.1 Introduction and basic concepts

Components analysis

In this sub-chapter, we’ll cover:

Review and transition
Conversion between polar and rectangular forms
Decomposition along standard axes
Resolving vectors in rotated or non-standard axes
Projection of vectors
Applications and example problems

Review and transition

Here’s a quick recap of the key ideas we’ll use throughout this sub-chapter:

Vector basics: A vector is a quantity with both magnitude and direction. You can represent vectors graphically (arrows) and algebraically (components), and you can add or subtract them.
Magnitude of a vector: For a vector expressed as

$v = v_{x} \hat{i} + v_{y} \hat{j},$

the magnitude is

$∣ v ∣ = v_{x}^{2} + v_{y}^{2} .$

With those fundamentals in place, we can focus on two core tasks:

breaking a vector into components along chosen axes
converting between polar and rectangular forms

Conversion between polar and rectangular forms

Vectors are commonly written in two equivalent ways:

Polar form: A magnitude $V$ and an angle $θ$ (measured from the horizontal axis).
Rectangular (component) form: Components along the $x$ - and $y$ -axes:

$V = V_{x} \hat{i} + V_{y} \hat{j} .$

Converting from polar to rectangular form

If a vector $V$ has magnitude $V$ and angle $θ$ , its components come from right-triangle trigonometry:

Horizontal component:

$V_{x} = V cos θ$
Vertical component:

$V_{y} = V sin θ$

Converting from rectangular to polar form

If you’re given components,

$V = V_{x} \hat{i} + V_{y} \hat{j},$

you can recover the magnitude and direction:

Magnitude:

$V = V_{x}^{2} + V_{y}^{2}$
Direction:

$θ = tan^{- 1} (\frac{V _{y}}{V _{x}})$

Decomposition along standard axes

When your axes are the usual horizontal ( $x$ ) and vertical ( $y$ ) directions, resolving a vector is straightforward. For a vector $V$ making an angle $θ$ with the horizontal:

Horizontal component:
$V_{x} = V cos θ$
Vertical component:
$V_{y} = V sin θ$

Example problem 1

A vector has a magnitude of 12 units and makes an angle of $4 0^{\circ}$ with the horizontal. Find its components.

Solution:

(spoiler)

$V_{x} = 12 cos 4 0^{\circ} \approx 12 \times 0.766 = 9.19$

$V_{y} = 12 sin 4 0^{\circ} \approx 12 \times 0.643 = 7.72$

Thus,

$V \approx (9.19, 7.72) .$

Resolving vectors in rotated or non-standard axes

Decomposition in a rotated coordinate system

$ϕ = θ - α$

Using that relative angle, the components in the rotated system are:

Component along $x^{'}$ :

$V_{x^{'}} = V cos (θ - α)$
Component along $y^{'}$ :

$V_{y^{'}} = V sin (θ - α)$

Example problem 2

A force of 30 N acts at an angle of $7 0^{\circ}$ from the horizontal. If the coordinate system is rotated by $2 0^{\circ}$ clockwise, find the components of the force in the rotated system.

Solution:

(spoiler)

Compute the relative angle:

$ϕ = 7 0^{\circ} - 2 0^{\circ} = 5 0^{\circ}$
Resolve into components:

$F_{x^{'}} = 30 cos 5 0^{\circ} \approx 30 \times 0.643 = 19.29 N$

$F_{y^{'}} = 30 sin 5 0^{\circ} \approx 30 \times 0.766 = 22.98 N$

Projection of vectors

Why use projections?

A projection tells you how much of one vector points in the direction of another. This matters in physics because:

Simplification of problems: Often, only the component of a vector along a particular direction affects the situation you’re analyzing.
Understanding work and energy: In work calculations, only the component of force along the displacement contributes to work.
Vector decomposition: Projections give a systematic way to break vectors into parts aligned with specific directions, especially when the axes aren’t standard.

Scalar projection

The scalar projection (or component) of vector $A$ onto vector $B$ is:

$comp_{B} A = \frac{A \cdot B}{∣ B ∣} = ∣ A ∣ cos θ,$

where $θ$ is the angle between $A$ and $B$ .

Vector projection

The vector projection of $A$ onto $B$ is:

$proj_{B} A = (\frac{A \cdot B}{∣ B ∣ ^{2}}) B .$

Example problem 3

Find the scalar and vector projections of

$A = (5, 3)$

onto

$B = (4, 0) .$

Solution:

(spoiler)

Scalar projection:

$comp_{B} A = \frac{A \cdot B}{∣ B ∣} = \frac{5 \times 4 + 3 \times 0}{4} = \frac{20}{4} = 5.$
Vector projection:

$proj_{B} A = (\frac{20}{4 ^{2}}) B = (\frac{20}{16}) (4, 0) = (\frac{5}{4}) (4, 0) = (5, 0) .$

Additional questions

Now let’s apply these techniques in more integrated problems. Example problem 4

Solution:

(spoiler)

Set up the rotated axes: Align the $x^{'}$ -axis along the plane and the $y^{'}$ -axis perpendicular to the plane.
Resolve each force into $x^{'}$ components:
- For the 40 N force (acting upward along the plane):
$F_{1, x^{'}} = 40 cos 6 0^{\circ} = 40 \times 0.5 = 20 N .$
- For the 30 N force (acting downward along the plane):
$F_{2, x^{'}} = 30 cos 2 0^{\circ} \approx 30 \times 0.94 = 28.2 N .$
Determine the net force: Assuming the 28.2 N force acts in the opposite direction to the 20 N force, the net force along the plane is:

$F_{net, x^{'}} = 28.2 N - 20 N = 8.2 N .$

$R = A + B .$

Solution:

(spoiler)

Magnitude (using the cosine rule):

$R^{2} = 6^{2} + 8^{2} - 2 \times 6 \times 8 cos 12 0^{\circ} .$

Since

$cos 12 0^{\circ} = - 0.5,$

$R^{2} = 36 + 64 + 96 \times 0.5 = 100 + 48 = 148,$

$R \approx 148 \approx 12.17 N .$

Direction (using the sine rule): Let $θ$ be the angle between $A$ and $R$ . Then:

$\frac{sin θ}{8} = \frac{sin 12 0 ^{\circ}}{12.17} .$

Calculate

$sin 12 0^{\circ} \approx 0.866$

and solve for $θ$ :

$sin θ \approx \frac{8 \times 0.866}{12.17} \approx 0.569,$

$θ \approx 34. 7^{\circ} .$

This is the angle between $A$ and the resultant $R$ .

Achievable AP Physics 1

2. Vectors and their analysis

Our AP Physics 1 course is currently in development and is a work-in-progress.