Newton’s laws of motion
Study Newton’s laws of motion for Year 12 Specialist Mathematics in Queensland (QCAA). Forces acting on a particle are combined into a single resultant, and Newton’s second law \(\mathbf{F}=m\mathbf{a}\) links that resultant to the acceleration, taking g as \(9.8\ \text{m/s}^2\).
You will learn the first law (equilibrium and constant velocity), the second law for concurrent forces, and the third law of action and reaction, then apply them to weight, the normal reaction and apparent weight in a lift — core modelling skills in Unit 4 motion.
Theory
Newton’s laws of motion govern how forces change the motion of a particle in Year 12 Specialist Mathematics (QCAA, Queensland). The first law gives equilibrium (zero resultant means constant velocity), the second law \(\mathbf{F}=m\mathbf{a}\) links the resultant force to the acceleration, and the third law pairs equal and opposite action–reaction forces. This page adds the resultant of concurrent forces, weight \(mg\), the normal reaction and apparent weight, with \(g=9.8\ \text{m/s}^2\).
A force is a push or pull, measured in newtons (N). The single force that has the same effect as all the forces acting on a particle is the resultant (or net) force, found by adding the forces as vectors. Newton’s three laws describe how that resultant governs the motion.
Newton’s first law: a particle stays at rest or moves with constant velocity unless a non-zero resultant force acts. So a body in equilibrium (or moving at constant velocity) has resultant force zero.
Newton’s second law: the resultant force equals mass times acceleration, \(\mathbf{F}=m\mathbf{a}\). The acceleration is in the direction of the resultant force, and its size is \(a=\dfrac{F}{m}\). This is the law used to find an unknown force, mass or acceleration under concurrent forces.
Newton’s third law: every action has an equal and opposite reaction. The two forces are the same size and opposite in direction but act on two different bodies — so they never cancel on one body. The weight of a particle is \(W=mg\) (down); the surface it rests on pushes back with the normal reaction \(N\). In a lift, the floor’s reaction \(R\) is the apparent weight.
Newton’s second law, in vector and in scalar (one-direction) form:
The resultant of concurrent forces is their vector sum; the acceleration follows from it:
Weight is the force of gravity; the equilibrium (first-law) condition is a zero resultant:
For a particle in a lift (taking up as positive), Newton’s second law gives the apparent weight \(R\):
Applying Newton’s second law to a particle
- Draw the free-body diagram: mark every force on the particle — weight \(mg\), normal reaction \(N\), applied forces, friction or resistance.
- Choose a positive direction (along the motion), then find the resultant by adding the forces, taking those against the positive direction as negative.
- Apply \(\mathbf{F}=m\mathbf{a}\): set the resultant equal to \(ma\) (or work component by component for \(\mathbf{i},\mathbf{j}\) forces).
- Solve for the unknown — acceleration, mass, an applied force or the reaction \(R\) — and state the units.
Apply \(F=ma\) and make \(a\) the subject:
| \(F\) | \(=\) | \(ma\) |
| \(36\) | \(=\) | \(8a\) |
| \(a\) | \(=\) | \(\dfrac{36}{8}\) |
| \(=\) | \(4.5\) |
The acceleration is \(4.5\ \text{m/s}^2\), in the direction of the force.
Find the resultant along the floor (take right as positive):
| \(F\) | \(=\) | \(40-15\) |
| \(=\) | \(25\) |
Now apply Newton’s second law:
| \(a\) | \(=\) | \(\dfrac{F}{m}\) |
| \(=\) | \(\dfrac{25}{5}\) | |
| \(=\) | \(5\) |
The acceleration is \(5\ \text{m/s}^2\) to the right.
Take up as positive; the resultant is \(R-mg\), so \(R-mg=ma\):
| \(R-mg\) | \(=\) | \(ma\) |
| \(R\) | \(=\) | \(m(g+a)\) |
| \(=\) | \(65\times(9.8+2)\) | |
| \(=\) | \(65\times 11.8\) | |
| \(=\) | \(767\) |
The apparent weight is \(R=767\ \text{N}\) (greater than the true weight \(637\ \text{N}\)).
Add the forces to get the resultant:
| \(\mathbf{F}\) | \(=\) | \((1+5)\mathbf{i}+(5+3)\mathbf{j}\) |
| \(=\) | \(6\mathbf{i}+8\mathbf{j}\) |
Divide by the mass, \(\mathbf{a}=\dfrac{\mathbf{F}}{m}\):
| \(\mathbf{a}\) | \(=\) | \(\tfrac{1}{2}(6\mathbf{i}+8\mathbf{j})\) |
| \(=\) | \(3\mathbf{i}+4\mathbf{j}\) |
Take the magnitude:
| \(|\mathbf{a}|\) | \(=\) | \(\sqrt{3^2+4^2}\) |
| \(=\) | \(\sqrt{25}\) | |
| \(=\) | \(5\) |
The acceleration is \((3\mathbf{i}+4\mathbf{j})\ \text{m/s}^2\), of magnitude \(5\ \text{m/s}^2\).
Common pitfalls
Frequently asked questions
What are Newton's three laws of motion?
The first law: a particle keeps constant velocity (or stays at rest) unless a resultant force acts. The second law: \(\mathbf{F}=m\mathbf{a}\). The third law: every action has an equal and opposite reaction on another body.
How do you find acceleration from forces?
Add all the forces to get the resultant \(\mathbf{F}\), then divide by the mass: \(\mathbf{a}=\dfrac{\mathbf{F}}{m}\). The acceleration points in the direction of the resultant force.
What is the difference between mass and weight?
Mass \(m\) (in kilograms) measures how much matter a body has; weight is the gravitational force on it, \(W=mg\) (in newtons), with \(g=9.8\ \text{m/s}^2\).
Why do action and reaction forces not cancel out?
They act on different bodies. Newton’s second law is applied to one body at a time, so only the forces acting on that body appear — its reaction partner acts on the other body.
Why do you feel heavier in a lift that accelerates upwards?
The floor must both support your weight and accelerate you up, so its reaction is \(R=m(g+a)>mg\). That larger reaction is your apparent weight; accelerating down gives \(R=m(g-a)
What is the resultant force when a car travels at constant velocity?
Zero. Constant velocity means zero acceleration, so by \(F=ma\) the resultant is zero (Newton’s first law) — the driving force exactly balances the resistance.