Force
Study force for Year 12 Specialist Mathematics in Queensland (QCAA). A force is a vector measured in newtons, so several forces on a body combine by vector addition into a single resultant that governs how the body moves.
You will learn to work with momentum, the resultant of concurrent forces, weight and the normal reaction, and to apply Newton’s laws — the foundation for modelling motion under forces later in the course.
Theory
Force in Year 12 Specialist Mathematics (QCAA, Queensland) treats a force as a vector measured in newtons. This page covers momentum \(p=mv\), the resultant of concurrent forces, weight \(W=mg\), the normal reaction and applied force, Newton’s second law \(F=ma\), and Newton’s third law of action and reaction. Take \(g=9.8\) m/s\(^2\).
A force is a push or pull measured in newtons (N). Because it has a magnitude and a direction, a force is a vector, written in component form as \(\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}\). Several forces acting at the same point are called concurrent forces.
The momentum of a body of mass \(m\) moving with velocity \(v\) is \(p=mv\) (units kg\(\,\)m/s). Along a line, take one direction as positive so that a velocity in the opposite direction is negative. The impulse delivered by a force equals the change in momentum, \(\Delta p=m(v_2-v_1)\).
The resultant force is the vector sum of all forces acting on a body, \(\mathbf{R}=\sum\mathbf{F}\). For two perpendicular forces the magnitude follows from Pythagoras, \(|\mathbf{R}|=\sqrt{F_x^{\,2}+F_y^{\,2}}\). A body is in equilibrium exactly when the resultant is the zero vector, \(\sum\mathbf{F}=\mathbf{0}\); otherwise it accelerates.
The weight of a body is the force of gravity on it, \(W=mg\), acting vertically down. The normal reaction \(N\) is the support force a surface pushes back with, perpendicular to the surface. Newton’s second law links the resultant force to acceleration, \(\mathbf{F}=m\mathbf{a}\), while Newton’s third law says every action has an equal and opposite reaction that acts on the other body.
Momentum of a body, and the impulse as its change in momentum:
The resultant of concurrent forces, and its magnitude when the components are perpendicular:
Weight, and Newton’s second law relating the resultant force to acceleration:
Working with forces
- Identify every force on the body — weight \(W=mg\) down, the normal reaction \(N\), any tension or applied force — and draw a free-body diagram.
- Add the forces as vectors to get the resultant \(\mathbf{R}=\sum\mathbf{F}\); for perpendicular forces use \(|\mathbf{R}|=\sqrt{F_x^{\,2}+F_y^{\,2}}\).
- Apply the law you need: \(p=mv\) for momentum, \(\Delta p=m(v_2-v_1)\) for impulse, or \(\mathbf{F}=m\mathbf{a}\) to link the resultant to the acceleration.
- Check equilibrium: a zero resultant means no acceleration; a non-zero resultant gives acceleration \(a=\dfrac{|\mathbf{R}|}{m}\) in the direction of \(\mathbf{R}\).
Momentum is mass times velocity:
| \(p\) | \(=\) | \(mv\) |
| \(=\) | \(2000 \times 8\) | |
| \(=\) | \(16\,000\) |
The momentum is \(16\,000\) kg m/s.
Add the forces as vectors (east is \(\mathbf{i}\), north is \(\mathbf{j}\)):
| \(\mathbf{R}\) | \(=\) | \(5\mathbf{i}+12\mathbf{j}\) |
Perpendicular components combine by Pythagoras:
| \(|\mathbf{R}|\) | \(=\) | \(\sqrt{5^2+12^2}\) |
| \(=\) | \(\sqrt{25+144}\) | |
| \(=\) | \(\sqrt{169}\) | |
| \(=\) | \(13\) |
The resultant is \(5\mathbf{i}+12\mathbf{j}\) N, of magnitude \(13\) N.
(i) Weight is the force of gravity, \(W=mg\):
| \(W\) | \(=\) | \(mg\) |
| \(=\) | \(6 \times 9.8\) | |
| \(=\) | \(58.8\) |
(ii) Rearrange \(F=ma\) to make the acceleration the subject:
| \(a\) | \(=\) | \(\dfrac{F}{m}\) |
| \(=\) | \(\dfrac{18}{6}\) | |
| \(=\) | \(3\) |
(i) The weight is \(58.8\) N. \;\; (ii) The acceleration is \(3\) m/s\(^2\).
Change in momentum \(\Delta p=m(v_2-v_1)\); take the initial direction positive, so the rebound velocity is negative:
| \(\Delta p\) | \(=\) | \(0.4\,(-3-5)\) |
| \(=\) | \(0.4 \times (-8)\) | |
| \(=\) | \(-3.2\) | |
| \(|\Delta p|\) | \(=\) | \(3.2\) |
The change in momentum has magnitude \(3.2\) kg m/s.
Common pitfalls
Frequently asked questions
What is the formula for momentum?
Momentum is \(p=mv\), the mass times the velocity, measured in kilogram metres per second. Along a line, choose a positive direction so a velocity the other way counts as negative.
How do you find the resultant of two perpendicular forces?
Add them as vectors, then take the magnitude with Pythagoras: \(|\mathbf{R}|=\sqrt{F_x^{\,2}+F_y^{\,2}}\). For \(6\) N and \(8\) N at right angles this is \(\sqrt{36+64}=10\) N.
What is the difference between mass and weight?
Mass is the amount of matter in kilograms; weight is the gravitational force on that mass, \(W=mg\), in newtons. With \(g=9.8\) m/s\(^2\), a \(5\) kg mass weighs \(49\) N.
What does Newton’s third law say?
Every action has an equal and opposite reaction, and the two forces act on different bodies. If the Earth pulls a book down, the book pulls the Earth up with an equal and opposite force.
How is impulse related to momentum?
The impulse of a force equals the change in momentum it produces, \(\Delta p=m(v_2-v_1)\). For a rebound, the initial and final velocities have opposite signs.
When is a body in equilibrium?
A body is in equilibrium when the resultant force is the zero vector, \(\sum\mathbf{F}=\mathbf{0}\); it then stays at rest or moves at constant velocity. A non-zero resultant produces acceleration \(a=\dfrac{|\mathbf{R}|}{m}\).