In Specialist Mathematics, forces are treated as vector quantities because they possess both magnitude (measured in Newtons, $N$) and direction. The study of forces acting on a body at rest or moving with constant velocity is known as Statics, while the study of forces causing motion is Dynamics.
A force $\mathbf{F}$ acting in a two-dimensional plane can be expressed in terms of its components using unit vectors $\mathbf{i}$ and $\mathbf{j}$.
To solve complex problems, we “resolve” a force into two perpendicular components, usually horizontal and vertical:
* $F_x = |\mathbf{F}| \cos(\theta)$
* $F_y = |\mathbf{F}| \sin(\theta)$
EXAM TIP: When resolving forces, always draw a large, clear force diagram (Free Body Diagram). VCAA examiners often award method marks for correctly showing the direction and labeling of forces, even if the final calculation is incorrect.
The resultant force ($\mathbf{R}$ or $\sum \mathbf{F}$) is the vector sum of all individual forces acting on a body.
If forces $\mathbf{F}_1, \mathbf{F}_2, \dots, \mathbf{F}_n$ act on a particle:
$$\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \dots + \mathbf{F}_n = \left(\sum x_i\right)\mathbf{i} + \left(\sum y_i\right)\mathbf{j}$$
The relationship between the resultant force, mass ($m$), and acceleration ($\mathbf{a}$) is:
$$\mathbf{R} = m\mathbf{a}$$
Where:
* $\mathbf{R}$ is the resultant force in Newtons ($N$).
* $m$ is the mass in kilograms ($kg$).
* $\mathbf{a}$ is the acceleration vector in $m/s^2$.
KEY TAKEAWAY: If the resultant force $\mathbf{R} = \mathbf{0}$, the acceleration $\mathbf{a}$ must be $\mathbf{0}$. This means the object is either at rest or moving with a constant velocity.
A particle is in equilibrium if the vector sum of all forces acting on it is the zero vector.
For a system of forces in a plane to be in equilibrium:
1. The sum of the horizontal components must be zero: $\sum F_x = 0$
2. The sum of the vertical components must be zero: $\sum F_y = 0$
Mathematically, if $\mathbf{R} = x\mathbf{i} + y\mathbf{j}$, then $x=0$ and $y=0$.
If three forces act on a particle in equilibrium, they can be represented in magnitude and direction by the sides of a closed triangle taken in order.
For three coplanar forces $\mathbf{F}_1, \mathbf{F}_2, \mathbf{F}_3$ in equilibrium, where $\alpha, \beta, \gamma$ are the angles opposite to the respective forces:
$$\frac{|\mathbf{F}_1|}{\sin(\alpha)} = \frac{|\mathbf{F}_2|}{\sin(\beta)} = \frac{|\mathbf{F}_3|}{\sin(\gamma)}$$
COMMON MISTAKE: Students often forget that “moving at a constant velocity” implies equilibrium ($\sum \mathbf{F} = 0$). Do not automatically assume a moving object has a non-zero resultant force.
When modeling physical systems, several specific forces frequently appear:
| Force | Description | Direction |
|---|---|---|
| Weight ($\mathbf{W}$) | Due to gravity: $W = mg$ ($g \approx 9.8$) | Always vertically downwards |
| Normal Reaction ($\mathbf{N}$) | Force from a surface | Perpendicular to the surface |
| Tension ($\mathbf{T}$) | Force in a string or rod | Along the string, away from the body |
| Friction ($\mathbf{F}_f$) | Resistance to motion | Parallel to surface, opposing motion |
REMEMBER: Weight is $mg$, not just $m$. Mass is a scalar (kg), while weight is a vector (N). In VCE Specialist Math, use $g = 9.8 \, m/s^2$ unless stated otherwise.
Problems involving inclined planes are a staple of VCE exams. It is usually most efficient to resolve forces parallel and perpendicular to the plane rather than horizontally and vertically.
Let $\theta$ be the angle of the incline to the horizontal:
* Component of Weight parallel to the plane: $mg \sin(\theta)$ (acting down the plane)
* Component of Weight perpendicular to the plane: $mg \cos(\theta)$ (acting into the plane)
VCAA FOCUS: VCAA frequently asks questions where a force is applied at an angle to an inclined plane. In these cases, you must resolve the applied force into components parallel and perpendicular to the plane before setting up your equilibrium equations.
STUDY HINT: Practice converting between “magnitude-direction” form and “$\mathbf{i}, \mathbf{j}$” component form quickly. Speed in resolving vectors is essential for finishing the Specialist Mathematics Exam 2.
