Welcome to Forces and Newton's Laws
Have you ever wondered why you feel pushed back into your seat when a car accelerates, or why a book resting on a table doesn't fall straight through to the floor? The answers lie in the study of forces and Newton's laws of motion.
In this unit of CCEA AS 2 Applied Mathematics, we explore the rules that govern how and why objects move. Don't worry if mechanics feels a bit abstract at first! Once you learn how to draw clear diagrams and apply a few simple rules systematically, solving these problems becomes a repeatable, step-by-step process.
1. Understanding Forces: The Basics
A force is simply a push or a pull acting upon an object. Forces are vectors, which means they have both a magnitude (size) and a direction. In the SI system, force is measured in Newtons (\(\text{N}\)).
Common Types of Forces You Must Know
• Weight (\(W\)): The gravitational pull of the Earth on an object. It always acts vertically downwards towards the centre of the Earth. We calculate weight using the formula \(W = mg\), where \(m\) is mass in kilograms (\(\text{kg}\)) and \(g\) is acceleration due to gravity (for CCEA, take \(g = 9.8\text{ m s}^{-2}\)).
• Normal Reaction (\(R\) or \(N\)): The supporting contact force exerted by a surface on an object. The word normal means perpendicular (\(90^\circ\)) to the surface of contact.
• Tension (\(T\)): The pulling force transmitted through a taut string, rope, cable, or tow-bar. Tension always pulls away from the object along the line of the string.
• Thrust / Compression: A pushing force exerted by a rigid rod when compressed.
• Friction (\(F\)): A contact force that opposes the relative motion (or tendency of motion) between two rough surfaces.
• Driving Force (\(D\) or \(P\)): The forward force generated by an engine or motor.
• Resistance / Drag: Opposing forces such as air resistance or water resistance that act in the direction opposite to motion.
Did you know? Mass and weight are not the same thing! Your mass is the amount of matter in your body (measured in \(\text{kg}\)) and never changes whether you are on Earth or the Moon. Your weight is a force (in \(\text{N}\)) and depends on the local gravitational field strength \(g\).
Free-Body Force Diagrams
Before doing any calculation in mechanics, your first step must always be to sketch a free-body diagram. This is a simple diagram showing the object as a single particle with every single external force acting on it represented as an arrow pointing in the correct direction.
Top Tip: Always label the direction of acceleration with a double arrow (\(\implies\)) next to the diagram so you don't confuse acceleration with a force!
Key Takeaway: Always draw every force acting on the particle, label them clearly with their symbol, and specify the direction of expected acceleration before writing any equations.
2. Resolving Forces into Perpendicular Components
Forces often act at an angle rather than neatly horizontal or vertical. To solve problems, we split (or resolve) an angled force into two perpendicular components: one along the direction of motion and one perpendicular to it.
The Golden Rule for Resolving
If a force of magnitude \(F\) makes an angle \(\theta\) with a given direction:
• The component adjacent (next) to the angle is \(F \cos\theta\).
• The component opposite to the angle is \(F \sin\theta\).
Memory Trick: Remember "Cos for Close" — if you have to turn the force across the angle \(\theta\) to reach the line, use \(\cos\theta\). If you turn away from the angle, use \(\sin\theta\).
Forces on an Inclined Plane
When an object sits on a slope inclined at an angle \(\alpha\) to the horizontal, it is usually easiest to resolve forces parallel to the slope and perpendicular to the slope.
• The component of weight pulling the object down the slope is \(mg \sin\alpha\).
• The component of weight pressing into the slope is \(mg \cos\alpha\).
• The normal reaction \(R\) acts perpendicular to and away from the slope.
Key Takeaway: Resolving breaks complex 2D vector problems into two simple, independent 1D problems.
3. Newton's First Law of Motion and Equilibrium
Newton's First Law states: An object will remain at rest or continue to move with constant velocity in a straight line unless acted upon by a non-zero resultant force.
What is Equilibrium?
When the vector sum of all forces acting on a particle is zero, the forces are balanced. The particle is said to be in equilibrium. This means:
• Acceleration is zero (\(a = 0\)).
• The resultant force is zero (\(\sum F = 0\)).
To solve equilibrium problems in 2D, we resolve in two perpendicular directions:
• \(\sum F_x = 0\) (Forces to the right = Forces to the left)
• \(\sum F_y = 0\) (Forces upwards = Forces downwards)
Example: A block of mass \(4\text{ kg}\) rests in equilibrium on a smooth horizontal table, pulled by a horizontal rope with tension \(T = 20\text{ N}\) and opposed by a horizontal force \(P\). Since it is in equilibrium horizontally: \(P = T = 20\text{ N}\). Vertically: \(R = mg = 4 \times 9.8 = 39.2\text{ N}\).
Key Takeaway: If an object is "at rest", "stationary", or moving at "constant speed / constant velocity", acceleration is \(0\), so forces balance in every direction.
4. Newton's Second Law of Motion: \(F = ma\)
Newton's Second Law states: The rate of change of momentum of an object is directly proportional to the resultant force acting on it and takes place in the direction of the force.
For an object with constant mass \(m\), this gives the most famous formula in mechanics:
\(F_{\text{net}} = ma\)
where:
• \(F_{\text{net}}\) (or \(\sum F\)) is the resultant force acting in the direction of acceleration (in \(\text{N}\)).
• \(m\) is the mass of the body (in \(\text{kg}\)).
• \(a\) is the acceleration (in \(\text{ m s}^{-2}\)).
Step-by-Step Method for Applying \(F = ma\)
Step 1: Draw a clear force diagram showing all forces and mark the direction of acceleration with a double arrow (\(\implies a\)).
Step 2: Choose your positive direction to be the direction of acceleration.
Step 3: Calculate the resultant force in that direction: \(\text{Resultant Force} = (\text{Forces in direction of } a) - (\text{Forces opposing } a)\).
Step 4: Set up the equation \(F_{\text{net}} = ma\) and solve for the unknown quantity.
Connecting Forces with Kinematics (SUVAT)
In many CCEA exam questions, you will combine \(F = ma\) with the constant acceleration equations (SUVAT):
• \(v = u + at\)
• \(s = ut + \frac{1}{2}at^2\)
• \(v^2 = u^2 + 2as\)
• \(s = \frac{(u + v)}{2}t\)
Strategy: Use forces (\(F = ma\)) to find acceleration \(a\), then plug \(a\) into SUVAT to find displacement, time, or velocity (or vice versa!).
Key Takeaway: Always subtract opposing forces from forward forces to get the resultant force: \(F_{\text{forward}} - F_{\text{opposing}} = ma\).
5. Newton's Third Law of Motion
Newton's Third Law states: When body A exerts a force on body B, body B exerts an equal and opposite force on body A.
Every "action" has an equal and opposite "reaction". However, remember that the two forces in a Newton's third law pair act on different bodies, which is why they do not simply cancel each other out!
6. Connected Particles and Common Applications
1. Motion in a Lift (Elevator)
Consider a person of mass \(m\) standing on the floor of a lift. The forces acting on the person are their weight \(mg\) downwards and the normal reaction from the floor \(R\) upwards (which represents the reading on a weighing scale).
• Lift accelerating upwards (\(a\) upwards):
\(R - mg = ma \implies R = m(g + a)\)
You feel heavier because the floor must push harder to accelerate you upwards!
• Lift accelerating downwards (\(a\) downwards):
\(mg - R = ma \implies R = m(g - a)\)
You feel lighter!
• Lift moving at constant velocity (\(a = 0\)):
\(R - mg = 0 \implies R = mg\)
2. Connected Particles: Cars, Caravans, and Tow-bars
When two bodies are connected by a light, inextensible tow-bar or rope, they share the same acceleration (\(a\)).
You can approach connected particles in two ways:
1. The Whole System: Treat both objects as one large combined mass \((m_1 + m_2)\). Internal forces (like the tension in the tow-bar) cancel out.
\(F_{\text{engine}} - \text{Total Resistances} = (m_1 + m_2)a\)
2. Individual Particles: Apply \(F = ma\) to just the car or just the trailer to find internal tension \(T\).
3. Pulleys and Pegs
Two masses \(m_1\) and \(m_2\) (\(m_1 > m_2\)) connected by a light inextensible string passing over a smooth fixed pulley.
Key modeling assumptions and their mathematical meaning:
• "Light string / Light pulley": Mass is negligible, so tension \(T\) is constant throughout the string, and the pulley adds no mass to the system.
• "Inextensible string": The string cannot stretch, meaning both masses have the same magnitude of acceleration.
• "Smooth pulley": There is no friction between the string and the pulley, meaning tension \(T\) is the same on both sides of the pulley.
Equations of motion for the system:
• For heavier mass \(m_1\) (moving downwards): \(m_1 g - T = m_1 a\)
• For lighter mass \(m_2\) (moving upwards): \(T - m_2 g = m_2 a\)
Adding the two equations eliminates \(T\), allowing you to solve for \(a\) easily!
Key Takeaway: For connected particles, find the common acceleration first by looking at the whole system or adding individual equations to eliminate tension.
Common Mistakes to Avoid
• Mixing up Mass and Weight: Never use mass \(m\) as a force. Always multiply by \(g\) (\(9.8\text{ m s}^{-2}\)) to get the weight \(mg\) in Newtons.
• Assuming \(R = mg\) always: The normal reaction \(R\) is only equal to \(mg\) for a stationary object on a horizontal plane with no other vertical forces. On a slope, \(R = mg \cos\alpha\). In a lift, \(R = m(g \pm a)\).
• Sign errors in \(F = ma\): Always define your positive direction in the direction of acceleration so that \(a\) remains positive.
• Forgetting to resolve weight on a slope: Weight acts straight down vertically, NOT perpendicular to the slope.
Quick Review: Essential Formulas
• Weight: \(W = mg\) (where \(g = 9.8\text{ m s}^{-2}\))
• Newton's Second Law: \(\sum F = ma\)
• Equilibrium Condition: \(\sum F_x = 0\) and \(\sum F_y = 0\)
• Component parallel to angle \(\theta\): \(F \cos\theta\)
• Component perpendicular to angle \(\theta\): \(F \sin\theta\)
• Weight components on an incline of angle \(\alpha\): Parallel = \(mg \sin\alpha\), Perpendicular = \(mg \cos\alpha\)