Welcome to Vectors in Mechanics!
Hello future mathematician! This chapter is where we take the fundamental ideas of motion you learned previously (like speed, distance, and acceleration) and make them much more powerful by introducing vectors. Instead of only moving in a straight line, we can now analyze motion in two dimensions (like a bird flying or a boat sailing).
Don't worry if this seems tricky at first. Vectors are simply tools that help us keep track of both magnitude (how much) and direction (where to). By the end of these notes, you’ll be a pro at applying vector calculus to forces and motion!
I. Position and Displacement Vectors
In mechanics, we usually set up a fixed point, called the Origin (\(O\)), from which all measurements are taken. Think of this as the "Home" button on your GPS.
1. Position Vectors (\(\mathbf{r}\))
A Position Vector, denoted by \(\mathbf{r}\), tells you exactly where a particle \(P\) is relative to the origin \(O\). We often express vectors using the unit vectors \(\mathbf{i}\) (in the x-direction) and \(\mathbf{j}\) (in the y-direction).
If a particle \(P\) is 3 units across (right) and 4 units up, its position vector is:
\(\mathbf{r} = 3\mathbf{i} + 4\mathbf{j}\)
We can also write this in column vector form, which is often easier for calculations:
$$\mathbf{r} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$
2. Displacement Vectors
Displacement is the change in position. It tells you the shortest straight-line distance and direction from point \(A\) to point \(B\).
If you start at point \(A\) (position \(\mathbf{r}_A\)) and move to point \(B\) (position \(\mathbf{r}_B\)), the displacement vector from \(A\) to \(B\) is:
$$ \vec{AB} = \mathbf{r}_B - \mathbf{r}_A $$
Example: If \(A\) is at \(\mathbf{r}_A = 2\mathbf{i} + 5\mathbf{j}\) and \(B\) is at \(\mathbf{r}_B = 6\mathbf{i} - \mathbf{j}\), the displacement \(\vec{AB}\) is:
$$\vec{AB} = (6\mathbf{i} - \mathbf{j}) - (2\mathbf{i} + 5\mathbf{j}) = (6-2)\mathbf{i} + (-1-5)\mathbf{j} = 4\mathbf{i} - 6\mathbf{j}$$
Quick Review: Magnitude (Distance)
If a vector is \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\), its magnitude (length), written as \(|\mathbf{a}|\), is found using Pythagoras’ theorem:
$$ |\mathbf{a}| = \sqrt{x^2 + y^2} $$
This magnitude represents the distance moved or the speed (if \(\mathbf{a}\) is a velocity vector).
II. Velocity, Speed, and the Constant Velocity Model
In mechanics, vectors are essential for describing how fast and in what direction something is moving.
1. Velocity (\(\mathbf{v}\))
Velocity is the rate of change of position. It is a vector quantity.
- If a particle moves displacement \(\mathbf{s}\) over time \(t\), the average velocity is \(\mathbf{v} = \frac{\mathbf{s}}{t}\).
- Velocity vectors tell us the components of speed in the \(x\) and \(y\) directions. E.g., \(\mathbf{v} = 2\mathbf{i} + 5\mathbf{j}\) means the particle is moving 2 units/second horizontally and 5 units/second vertically.
2. Speed (Scalar)
Speed is the magnitude of the velocity vector. It is a scalar quantity (it only has size, no direction).
If \(\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\), then:
$$ \text{Speed} = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} $$
Memory Aid: Velocity is a Vector. Speed is a Scalar.
3. The Constant Velocity Equation
When an object is moving at a constant velocity (\(\mathbf{v}\)), its position changes predictably over time \(t\).
If \(\mathbf{r}_0\) is the initial position (at \(t=0\)), its position \(\mathbf{r}\) at time \(t\) is given by:
$$ \mathbf{r} = \mathbf{r}_0 + \mathbf{v}t $$
Analogy: This is the vector equivalent of the simple formula: Final Position = Start Position + (Speed × Time).
How to Solve Problems Using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\)
You must separate the components:
- Write out the full vector equation.
- Separate the equation into two scalar equations (one for \(\mathbf{i}\) and one for \(\mathbf{j}\)).
i-component: \(x = x_0 + v_x t\) - j-component: \(y = y_0 + v_y t\)
- Solve the resulting simultaneous equations or substitute values to find the required time or position.
✎ Common Mistake Alert
Do not mix up displacement from the origin (\(\mathbf{r}\)) and displacement travelled (\(\mathbf{s}\)). If a question uses the term "displacement vector," it usually means the vector \(\mathbf{s}\), which is often \(\mathbf{r} - \mathbf{r}_0\).
III. Constant Acceleration and Vector SUVAT
Just like in straight-line motion, we can use the SUVAT equations when the acceleration is constant, but now everything (except time \(t\)) is a vector!
1. Acceleration (\(\mathbf{a}\))
Acceleration is the rate of change of velocity. It is also a vector.
$$ \mathbf{a} = \frac{\mathbf{v} - \mathbf{u}}{t} $$
Where \(\mathbf{u}\) is the initial velocity and \(\mathbf{v}\) is the final velocity.
2. Vector SUVAT Equations
The standard SUVAT variables become vectors:
- \(\mathbf{u}\): Initial Velocity (vector)
- \(\mathbf{v}\): Final Velocity (vector)
- \(\mathbf{a}\): Constant Acceleration (vector)
- \(\mathbf{r}\): Displacement from the initial position (vector)
- \(t\): Time (scalar)
The two key SUVAT equations you will use in vector form are:
1. Velocity-Time Relationship:
$$ \mathbf{v} = \mathbf{u} + \mathbf{a}t $$
2. Displacement-Time Relationship:
$$ \mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 $$
(Note: Although the squared form of SUVAT, \(v^2=u^2+2as\), exists, it is rarely used in vector mechanics because it involves the dot product, which is often outside the scope of M1.)
Solving Problems with Vector SUVAT
The secret weapon for solving vector mechanics problems is decomposition:
Step 1: Decompose
Rewrite the vector equation as two separate scalar equations: one for the horizontal (\(\mathbf{i}\)) components and one for the vertical (\(\mathbf{j}\)) components.
Example: If \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) becomes:
i-direction: \(x = u_x t + \frac{1}{2}a_x t^2\)
j-direction: \(y = u_y t + \frac{1}{2}a_y t^2\)
Step 2: Solve Components
Use the scalar equations to find unknown values like time (\(t\)). Time is the quantity that links the two dimensions together.
Step 3: Recombine (if needed)
Once you have found the components of a required vector (like \(\mathbf{v}\)), recombine them to state the final vector, or find its magnitude (speed/distance).
🔍 Quick Review: Key Takeaways
- Constant Velocity: Use \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\).
- Constant Acceleration: Use \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) or \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\).
- Always separate into \(\mathbf{i}\) and \(\mathbf{j}\) components to solve!
IV. Forces and Newton’s Laws in Vector Form
The power of vectors truly shines when dealing with forces, as forces always have a direction.
1. Forces and Resultant Force
A Force (\(\mathbf{F}\)) is a vector. If multiple forces act on a body, their combined effect is called the Resultant Force (\(\mathbf{R}\)).
The resultant force is simply the vector sum of all the individual forces acting on the particle:
$$ \mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + \dots $$
Example: If \(\mathbf{F}_1 = 5\mathbf{i} + 2\mathbf{j}\) N and \(\mathbf{F}_2 = -3\mathbf{i} + 6\mathbf{j}\) N, then:
\(\mathbf{R} = (5-3)\mathbf{i} + (2+6)\mathbf{j} = 2\mathbf{i} + 8\mathbf{j}\) N
2. Equilibrium
If a particle is in Equilibrium (meaning it is stationary or moving with constant velocity), the resultant force acting on it must be zero.
$$ \mathbf{R} = \mathbf{0} \quad \text{or} \quad \sum \mathbf{F} = \mathbf{0} $$
This means the components must separately equal zero:
- Sum of \(\mathbf{i}\) components = 0
- Sum of \(\mathbf{j}\) components = 0
3. Newton’s Second Law (\(\mathbf{F} = m\mathbf{a}\))
This is the fundamental principle of mechanics. When expressed using vectors, it states that the resultant force (\(\mathbf{F}\)) acting on an object of mass \(m\) produces an acceleration (\(\mathbf{a}\)) in the same direction.
$$ \mathbf{F} = m\mathbf{a} $$
Important Note: Mass (\(m\)) is a scalar, but Force (\(\mathbf{F}\)) and Acceleration (\(\mathbf{a}\)) are vectors.
How to Use \(\mathbf{F} = m\mathbf{a}\)
Scenario 1: Finding Acceleration
- Calculate the Resultant Force \(\mathbf{F}\) (sum all individual force vectors).
- Divide the vector \(\mathbf{F}\) by the scalar mass \(m\).
$$\mathbf{a} = \frac{\mathbf{F}}{m}$$
(Remember: Dividing a vector by a scalar means dividing both the \(\mathbf{i}\) and \(\mathbf{j}\) components.)
Scenario 2: Finding a Missing Force or Unknown Value
- Calculate the required acceleration vector \(\mathbf{a}\) using vector SUVAT (if necessary).
- Calculate the required Resultant Force \(\mathbf{F}_{required} = m\mathbf{a}\).
- Set up the equation: $$\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_{missing} = \mathbf{F}_{required}$$
- Solve for the missing force vector or unknown component by equating \(\mathbf{i}\) and \(\mathbf{j}\) components.
Did you know? In 3D space, we would use a third unit vector, \(\mathbf{k}\), representing the z-axis, but for M1, we usually stick to the 2D plane (\(\mathbf{i}\) and \(\mathbf{j}\)).
V. A Final Review of Vector Components
To succeed in this chapter, you must be comfortable manipulating components.
Let \(\mathbf{a} = a_x\mathbf{i} + a_y\mathbf{j}\) and \(\mathbf{b} = b_x\mathbf{i} + b_y\mathbf{j}\). Let \(k\) be a scalar.
1. Vector Addition/Subtraction
$$ \mathbf{a} + \mathbf{b} = (a_x + b_x)\mathbf{i} + (a_y + b_y)\mathbf{j} $$
You add/subtract components separately.
2. Scalar Multiplication (e.g., in \(m\mathbf{a}\) or \(\mathbf{v}t\))
$$ k\mathbf{a} = (ka_x)\mathbf{i} + (ka_y)\mathbf{j} $$
The scalar multiplies both components.
3. Finding the Unit Vector (\(\hat{\mathbf{a}}\))
A Unit Vector has a magnitude of 1 and points in the same direction as \(\mathbf{a}\).
$$ \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} $$
We hope these notes give you the confidence to tackle vector problems! Remember, vectors just help us stay organized. Keep your \(\mathbf{i}\)'s separate from your \(\mathbf{j}\)'s, and you will do great!