Newton's laws of motion, the resultant of forces acting on a particle, the resolution of forces into components, the relationship $\mathbf{F} = m\mathbf{a}$, and the analysis of equilibrium and of motion under constant forces including weight, normal reaction and friction

What this dot point is asking

VCAA wants you to apply Newton's laws to a particle, find the resultant of several forces, resolve forces into perpendicular components, use $\mathbf{F} = m\mathbf{a}$ to relate net force and acceleration, and analyse both equilibrium (zero acceleration) and motion under constant forces such as weight, normal reaction and friction. This connects the vector and calculus work to physical situations.

Newton's laws

First law. A particle continues at rest or in uniform straight-line motion unless a non-zero resultant force acts. This defines the natural state and identifies acceleration as the signature of a net force.

Second law. The resultant force on a particle equals its mass times its acceleration:

This is a vector equation, so it holds component by component.

Third law. When body A exerts a force on body B, body B exerts an equal and opposite force on A. These act on different bodies, so they do not cancel on a single particle.

Resolving forces

To use the second law, find the resultant of all forces. The most reliable method is to resolve each force into perpendicular components, sum the components in each direction, and apply $\mathbf{F} = m\mathbf{a}$ separately along each axis. Choose axes to simplify the problem: horizontal and vertical for free motion, or parallel and perpendicular to a surface for motion on an incline.

Common forces:

• Weight: magnitude $mg$, directed vertically downward.
• Normal reaction $N$: the push of a surface, perpendicular to it.
• Friction: parallel to the surface, opposing relative motion, with magnitude up to $\mu N$ where $\mu$ is the coefficient of friction. When sliding, friction equals $\mu N$.

Equilibrium and motion

In equilibrium the acceleration is zero, so the resultant force is zero and each component balances. For motion under constant forces the acceleration is constant, and you can find the resultant, divide by the mass to get $\mathbf{a}$, and then use constant-acceleration kinematics for velocity and displacement.

Examples in context

Example 1. A $2\ \text{kg}$ mass hanging in equilibrium on a string has string tension $T = mg = 2\times 9.8 = 19.6\ \text{N}$.

Example 2. A horizontal force of $10\ \text{N}$ on a $4\ \text{kg}$ block on a smooth surface gives $a = \frac{10}{4} = 2.5\ \text{m/s}^2$.

Try this

Q1. State Newton's second law as an equation. [1 mark]

• Cue. $\mathbf{F}_{\text{net}} = m\mathbf{a}$.

Q2. A $3\ \text{kg}$ object has a resultant force of $12\ \text{N}$. Find its acceleration. [1 mark]

• Cue. $a = \frac{12}{3} = 4\ \text{m/s}^2$.

Q3. Resolve the weight of a $10\ \text{kg}$ block on a $20^\circ$ incline into components along the slope. [2 marks]

• Cue. Down-slope $mg\sin 20^\circ$, into-slope $mg\cos 20^\circ$, with $mg = 98\ \text{N}$.