Mathematics | Vectors in mechanics

Understanding Vectors in Mechanics: Your Guide to Motion in 2D

Welcome to the exciting world of Vectors in Mechanics! Until now, many of your mechanics problems focused on movement in a straight line (one dimension). Now, we are stepping into two dimensions (2D), where objects can move forwards, backwards, left, right, and everything in between!

This chapter is absolutely vital because it teaches you the mathematical tools needed to handle forces and motion that aren't perfectly horizontal or vertical. Mastering vectors makes the whole rest of the Mechanics curriculum manageable!

1. Scalars vs. Vectors: The Fundamental Difference

Before we start adding things up, we need to know what we are measuring.

What is a Scalar?

A Scalar quantity only has magnitude (size or amount). It doesn't tell you the direction.

• Examples: Distance, Speed, Mass, Time, Energy.
• Analogy: If you say your car is traveling at 60 km/h, that's speed (a scalar).

What is a Vector?

A Vector quantity has both magnitude (size) and direction.

• Examples: Displacement, Velocity, Acceleration, Force, Momentum.
• Analogy: If you say your car is traveling at 60 km/h North-East, that's velocity (a vector).

2. Representing Vectors in Two Dimensions

In 2D mechanics (which is what M1 focuses on), we define direction using two perpendicular axes: horizontal (\(\mathbf{i}\)) and vertical (\(\mathbf{j}\)).

Unit Vectors \(\mathbf{i}\) and \(\mathbf{j}\)

• \(\mathbf{i}\) is the unit vector (a vector of magnitude 1) in the positive horizontal direction (usually East or right).
• \(\mathbf{j}\) is the unit vector in the positive vertical direction (usually North or up).

Any vector \(\mathbf{a}\) can be written as a combination of these two components:

$$\mathbf{a} = x\mathbf{i} + y\mathbf{j}$$

Column Vector Notation

Sometimes, it’s cleaner to write vectors using column notation:

$$\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}$$

Example: A velocity vector of \(\mathbf{v} = 3\mathbf{i} - 5\mathbf{j}\) means the object is moving 3 units right and 5 units down. In column form: \(\mathbf{v} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}\).

3. Magnitude and Direction (Finding the Size and Angle)

When you are given a vector in \(\mathbf{i}\) and \(\mathbf{j}\) form, you often need to find its total size (magnitude) and its precise direction (angle).

Step 1: Calculating the Magnitude

The magnitude is simply the length of the vector. Since the horizontal (\(x\)) and vertical (\(y\)) components form a right-angled triangle, we use Pythagoras' Theorem.

If \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\), the magnitude (written as \(\lvert \mathbf{a} \rvert\) or \(|\mathbf{a}|\) or sometimes \(a\)) is:

$$\lvert \mathbf{a} \rvert = \sqrt{x^2 + y^2}$$

Example: If \(\mathbf{v} = 4\mathbf{i} - 3\mathbf{j}\), the magnitude is:

$$\lvert \mathbf{v} \rvert = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Step 2: Calculating the Direction

The direction is usually given as the angle (\(\theta\)) the vector makes with the positive \(\mathbf{i}\) axis (the positive x-axis).

We use basic trigonometry (SOH CAH TOA) based on the components:

$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{y}{x}$$

Important Tip for Direction! (The Quadrant Check)

The calculation \(\theta = \arctan \left( \frac{y}{x} \right)\) only gives you the angle in the first quadrant (between 0° and 90°). You must draw a sketch to ensure you give the correct angle relative to the positive x-axis (\(\mathbf{i}\)).

Common Mistake to Avoid: If your vector is \(-4\mathbf{i} + 3\mathbf{j}\) (second quadrant), calculating \(\arctan(3/-4)\) will give a negative angle. You must add 180° to find the correct angle from the positive x-axis.

Key Takeaway: Magnitude uses Pythagoras; Direction uses Trigonometry, always check your quadrant!

4. Vector Arithmetic (Adding and Subtracting)

Working with vectors is much simpler than working with numbers that have angles attached. We simply combine the components separately.

Addition and Subtraction

To add or subtract two vectors, combine the \(\mathbf{i}\) components and combine the \(\mathbf{j}\) components.

Let \(\mathbf{a} = x_1\mathbf{i} + y_1\mathbf{j}\) and \(\mathbf{b} = x_2\mathbf{i} + y_2\mathbf{j}\).

Addition (Finding the Resultant Vector)

The Resultant Vector (\(\mathbf{R}\)) is the single vector that represents the combined effect of all the vectors acting together.

$$\mathbf{R} = \mathbf{a} + \mathbf{b} = (x_1 + x_2)\mathbf{i} + (y_1 + y_2)\mathbf{j}$$

Analogy: If you walk 3m East, then 4m North, your total displacement is found by adding the vectors. Your total movement East is \(3+0 = 3\). Your total movement North is \(0+4 = 4\).

Subtraction

$$\mathbf{a} - \mathbf{b} = (x_1 - x_2)\mathbf{i} + (y_1 - y_2)\mathbf{j}$$

Multiplication by a Scalar

If you multiply a vector by a scalar \(k\), you multiply both the \(\mathbf{i}\) and the \(\mathbf{j}\) components by \(k\).

$$k\mathbf{a} = k(x_1\mathbf{i} + y_1\mathbf{j}) = (kx_1)\mathbf{i} + (ky_1)\mathbf{j}$$

Example: If \(\mathbf{a} = 2\mathbf{i} - 5\mathbf{j}\), then \(3\mathbf{a} = 6\mathbf{i} - 15\mathbf{j}\). The vector is now three times as long, but still points in the same direction.

Did you know? Vector addition is commutative: \(\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}\).

Key Takeaway: Treat the \(\mathbf{i}\) and \(\mathbf{j}\) components as completely separate equations until the very end.

5. Vectors in Kinematics (Motion)

The concepts of displacement, velocity, and acceleration that you used in linear motion (SUVAT) still apply, but now all those quantities are vectors!

Position and Displacement Vectors

Position Vector (\(\mathbf{r}\))

The Position Vector (\(\mathbf{r}\)) defines the location of a point or particle relative to a fixed origin (O).

If a particle P is at coordinates (3, 7), its position vector is \(\mathbf{r} = 3\mathbf{i} + 7\mathbf{j}\).

Displacement Vector (\(\mathbf{s}\))

Displacement is the change in position. If a particle moves from position \(\mathbf{r}_1\) to position \(\mathbf{r}_2\), the displacement vector \(\mathbf{s}\) is:

$$\mathbf{s} = \mathbf{r}_2 - \mathbf{r}_1$$

Vector Velocity and Acceleration

When working with constant acceleration in 2D, our familiar SUVAT equations become vector equations:

The most commonly used vector kinematics formula is:

$$\mathbf{v} = \mathbf{u} + \mathbf{a}t$$

Where:

• \(\mathbf{v}\) is the final velocity vector.
• \(\mathbf{u}\) is the initial velocity vector.
• \(\mathbf{a}\) is the constant acceleration vector.
• \(t\) is the scalar time elapsed.

Step-by-Step for Kinematics Problems

1. Separate the components: Write down separate equations for the \(\mathbf{i}\) components (horizontal) and the \(\mathbf{j}\) components (vertical).
2. Solve each dimension independently: Use standard scalar kinematics formulas (like \(v = u + at\)) for the \(\mathbf{i}\) column and the \(\mathbf{j}\) column.
3. Recombine (if needed): If the question asks for the final speed or resultant force, recombine the components found in step 2 using the magnitude formula (Pythagoras).

Example: A particle starts with velocity \(\mathbf{u} = 5\mathbf{i}\) and accelerates at \(\mathbf{a} = 2\mathbf{j}\) for \(t = 4\) seconds.

$$\mathbf{v} = \mathbf{u} + \mathbf{a}t$$

$$\mathbf{v} = (5\mathbf{i}) + (2\mathbf{j}) \times 4$$

$$\mathbf{v} = 5\mathbf{i} + 8\mathbf{j}$$

The final speed (magnitude of \(\mathbf{v}\)) is \(\sqrt{5^2 + 8^2} = \sqrt{89} \approx 9.43 \text{ units/s}\).

Key Takeaway: Vector kinematics problems are just two simultaneous 1D kinematics problems happening at once. Do not mix the \(\mathbf{i}\) and \(\mathbf{j}\) components!

6. Resolving Vectors (Forces at Angles)

One of the most powerful applications of vectors in M1 is dealing with forces that act at an angle. To use Newton's Laws (like \(\mathbf{F} = m\mathbf{a}\) or equilibrium conditions), we must break diagonal forces into their horizontal and vertical components.

This process is called resolving the vector.

The Resolution Rule

If a force \(F\) acts at an angle \(\theta\) to the horizontal (the \(\mathbf{i}\) axis):

1. The component adjacent (next) to the angle uses Cosine.
2. The component opposite the angle uses Sine.

If \(\theta\) is measured from the horizontal:

• Horizontal Component (\(\mathbf{i}\)): \(F_x = F \cos \theta\)
• Vertical Component (\(\mathbf{j}\)): \(F_y = F \sin \theta\)

Step-by-Step Resolution Process

Imagine a force \(F = 10\) N acting at 30° above the horizontal:

1. Draw a Right-Angled Triangle: Sketch the force vector and draw its horizontal and vertical components forming a triangle.
2. Calculate Horizontal Component (x-axis): This component is adjacent to the 30° angle.
   \(F_x = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} \approx 8.66\) N.
3. Calculate Vertical Component (y-axis): This component is opposite the 30° angle.
   \(F_y = 10 \sin(30^\circ) = 10 \times 0.5 = 5\) N.
4. Write as a Vector: The force vector is \(\mathbf{F} = 8.66\mathbf{i} + 5\mathbf{j}\) N.

Equilibrium Condition (The Sum of Vectors is Zero)

When a particle is in equilibrium (it is stationary or moving at a constant velocity), the resultant force acting on it is zero. This means:

$$\sum \mathbf{F} = \mathbf{0}$$

In component form, this means:

• The sum of all horizontal components must be zero. (\(\sum F_x = 0\))
• The sum of all vertical components must be zero. (\(\sum F_y = 0\))

Don't worry if this seems tricky at first! Resolving vectors is a skill that improves rapidly with practice. Always sketch your forces clearly and check which component is adjacent and which is opposite the given angle.

Key Takeaway: We resolve forces to simplify complex diagonal problems into two manageable, separate, perpendicular problems.