Applications of vectors in a plane - Mathematics A - H240 - Cambridge OCR A Level

Welcome to Applications of Vectors in a Plane!

In this chapter, we’re going to take the vector skills you’ve learned in Pure Maths and put them to work in the real world. We are moving into the Mechanics section of your course, where we look at how forces make things move (or stay still!).

Think of a vector as a set of instructions: "Go this far in that direction." In Mechanics, vectors allow us to describe Forces, Velocity, and Acceleration in a 2D space (a plane) without getting tangled up in messy drawings. By the end of these notes, you’ll be able to calculate exactly where a particle is going and how fast it’s getting there using just a few vector rules.

Don't worry if this seems tricky at first! We will break every process down into small, manageable steps.

1. Prerequisite Check: Vector Basics

Before we jump into the mechanics, let’s quickly refresh the "tools" we need from the syllabus (Ref 1.10). In 2D, we usually write vectors in two ways:

1. Component Form: \( x\mathbf{i} + y\mathbf{j} \), where i is one unit to the right and j is one unit up.
2. Column Form: \( \begin{pmatrix} x \\ y \end{pmatrix} \)

Quick Review: Magnitude and Direction
To find the Magnitude (the size or length) of a vector \( \mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix} \), we use Pythagoras:
\( |\mathbf{a}| = \sqrt{x^2 + y^2} \)

To find the Direction (the angle \( \theta \) it makes with the positive x-axis), we use trigonometry:
\( \tan \theta = \frac{y}{x} \)

Key Takeaway: Vectors have both size and direction. In Mechanics, the "size" of a force vector is how hard you are pushing, and the "direction" is where you are pushing.

2. Resultant Forces (Ref 3.03p)

In the real world, objects rarely have just one force acting on them. Imagine a boat being pulled by two different tugboats. The Resultant Force is the single "super-force" that has the same effect as all the individual forces added together.

How to find the Resultant:

To find the resultant of two or more forces, you simply add the vectors together. If you have force \( \mathbf{F_1} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \) and force \( \mathbf{F_2} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} \), the resultant \( \mathbf{R} \) is:
\( \mathbf{R} = \mathbf{F_1} + \mathbf{F_2} = \begin{pmatrix} 3+1 \\ 4-2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} \)

Did you know? If the resultant force is zero (\( \begin{pmatrix} 0 \\ 0 \end{pmatrix} \)), the object is in equilibrium. This means it’s either perfectly still or moving at a constant speed in a straight line!

Direction of Motion vs. Direction of Force:

This is a common "trick" in exams! Here is the rule to remember:
- The Velocity vector tells you the direction of motion (where it is going right now).
- The Acceleration vector (and the Resultant Force) tells you the direction of the force (where it is being pushed).

Example analogy: Imagine you are sliding on ice going North (Velocity), but a strong wind starts blowing you East (Force/Acceleration). You are moving North, but you are being pushed East!

Key Takeaway: To find the total effect, just add the vectors. The Resultant Force always points in the same direction as the acceleration.

3. Resolving Forces into Components (Ref 3.03p)

Sometimes a force is given as a magnitude and an angle (e.g., "A force of 10N at 30 degrees to the horizontal"). To do any math with it, we need to "break it down" into horizontal and vertical parts. This is called Resolving.

If a force \( F \) acts at an angle \( \theta \) to the positive x-axis:
- Horizontal Component (\( \mathbf{i} \)): \( F \cos \theta \)
- Vertical Component (\( \mathbf{j} \)): \( F \sin \theta \)

Memory Aid: "Cos is Cross"
Think: \( \cos \) is the component going cross (horizontal) the angle, and \( \sin \) is the other one!

Step-by-Step Explanation:
1. Identify the magnitude (\( F \)) and the angle (\( \theta \)).
2. Calculate the horizontal part: \( x = F \cos \theta \).
3. Calculate the vertical part: \( y = F \sin \theta \).
4. Write it as a vector: \( \mathbf{F} = \begin{pmatrix} F \cos \theta \\ F \sin \theta \end{pmatrix} \).

Key Takeaway: Resolving turns "diagonal" forces into easy-to-use horizontal and vertical vectors.

4. Dynamics in a Plane: \( \mathbf{F} = m\mathbf{a} \) (Ref 3.03q)

Newton’s Second Law (\( F=ma \)) works perfectly with vectors! The formula becomes:
\( \mathbf{F} = m\mathbf{a} \)
Where \( \mathbf{F} \) is the Resultant Force vector, \( m \) is the mass (a scalar/number), and \( \mathbf{a} \) is the Acceleration vector.

Wait, what if the force changes with time?
If the force depends on time (\( t \)), we use Calculus. Remember these links:
- Displacement (\( \mathbf{r} \)) \( \xrightarrow{\text{differentiate}} \) Velocity (\( \mathbf{v} \)) \( \xrightarrow{\text{differentiate}} \) Acceleration (\( \mathbf{a} \))
- Acceleration (\( \mathbf{a} \)) \( \xrightarrow{\text{integrate}} \) Velocity (\( \mathbf{v} \)) \( \xrightarrow{\text{integrate}} \) Displacement (\( \mathbf{r} \))

Example Problem:
A particle of mass 2kg is acted on by a force \( \mathbf{F} = \begin{pmatrix} 6t \\ 4 \end{pmatrix} \). Find the acceleration at \( t=3 \).
1. Use \( \mathbf{a} = \frac{\mathbf{F}}{m} \)
2. \( \mathbf{a} = \frac{1}{2} \begin{pmatrix} 6t \\ 4 \end{pmatrix} = \begin{pmatrix} 3t \\ 2 \end{pmatrix} \)
3. At \( t=3 \), \( \mathbf{a} = \begin{pmatrix} 3(3) \\ 2 \end{pmatrix} = \begin{pmatrix} 9 \\ 2 \end{pmatrix} \text{ ms}^{-2} \).

Common Mistake to Avoid: Don't forget the Constant of Integration (\( +\mathbf{c} \)) when integrating! In vectors, \( \mathbf{c} \) is also a vector, like \( \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} \). You usually find this using the "initial conditions" (e.g., "at \( t=0 \), the particle is at the origin").

Key Takeaway: \( \mathbf{F} = m\mathbf{a} \) allows you to switch between force and motion. If the force isn't constant, use differentiation or integration on each component separately.

5. Quick Review Box: Tips for Success

- Check your units: Forces in Newtons (N), Mass in kg, Acceleration in \( \text{ms}^{-2} \).
- Draw a sketch: Even if the syllabus says "no scale drawings," a quick 2-second sketch helps you see if your angle should be positive or negative.
- Bold vs. Underline: In textbooks, vectors are bold. When you write them by hand in your exam, you should underline them (e.g., \( \underline{a} \)) to show they aren't just regular numbers.
- Magnitudes are positive: The magnitude (size) of a force can never be negative. If your Pythagoras gives a negative, check your squaring!

Summary of Applications

What we covered:

- Resultants: Adding force vectors to find the total push.
- Components: Using \( \sin \) and \( \cos \) to break diagonal forces apart.
- Direction: Velocity shows where it's going; Force/Acceleration shows where it's being pushed.
- F=ma: Connecting the total force to the acceleration.
- Calculus: Using integration and differentiation when forces change over time.

Keep practicing! Vectors in Mechanics are just about being organized with your horizontal and vertical numbers. You've got this!

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