Forces

Welcome to Forces (Unit 2: Mechanics)

Welcome to one of the most exciting and practical areas of GCSE Further Mathematics! Mechanics is the study of how objects move and how the forces acting upon them shape that motion. Whether you are launching a rocket, pulling a sledge up a snowy hill, or riding in an elevator, the rules governing your motion are the exact same mechanics principles you will master right here.

Don't worry if mechanics seems a bit unfamiliar at first. We will break every concept down into bite-sized, step-by-step pieces. By the end of this chapter, you will have all the tools needed to tackle any forces question on your CCEA Unit 2 exam with confidence!

Key Takeaway: Mechanics connects the mathematics of algebra and trigonometry to the physical world through the study of forces and motion.

1. Understanding Forces: The Fundamentals

What is a Force?

A force is simply a push or a pull acting on an object. In mathematics and physics, force is a vector quantity, meaning it always has two vital properties:
• Magnitude: How strong the force is (size).
• Direction: The line and heading along which the force pushes or pulls.

The standard SI unit of force is the Newton (\(\text{N}\)). One Newton is defined as the force required to accelerate a mass of \(1\text{ kg}\) at a rate of \(1\text{ m s}^{-2}\):
\(1\text{ N} = 1\text{ kg m s}^{-2}\)

Common Forces in Mechanics Problems

When drawing a free-body diagram (a sketch showing all the forces acting on a single particle), you will encounter these standard forces:

1. Weight (\(W\) or \(mg\)): The downward pull of gravity on an object. Weight always acts vertically downwards toward the centre of the Earth.
\(W = mg\)
where \(m\) is the mass in kilograms (\(\text{kg}\)) and \(g\) is the acceleration due to gravity (taken as \(g = 9.8\text{ m s}^{-2}\) unless stated otherwise on your CCEA exam paper).

Did you know? Mass and weight are not the same thing! Mass (in \(\text{kg}\)) is the amount of matter in an object and never changes, whereas weight (in \(\text{N}\)) depends on gravity.

2. Normal Reaction Force (\(R\) or \(N\)): The supportive push from a surface in contact with the object. The word normal in mathematics means perpendicular, so \(R\) always acts at \(90^\circ\) to the contact surface.

3. Tension (\(T\)) and Thrust:
• Tension (\(T\)): The pulling force transmitted through a taut string, rope, or cable. Tension always pulls away from the object along the line of the string.
• Thrust (or Compression): The pushing force transmitted through a rigid rod. Thrust pushes towards the object.

4. Frictional Force (\(F\) or \(F_r\)): A resistive force that acts along the contact surface opposing the direction of motion or intended motion.

5. Driving Force / Applied Force (\(P\) or \(F\)): An active pulling or pushing force applied to an object (such as the driving force of an engine or a person pulling a rope).

Key Takeaway: Every force has magnitude and direction. Always start a mechanics problem by identifying all active forces and drawing a clear diagram.

2. Modelling Assumptions: Simplifying the Real World

To make calculations solvable at GCSE level, mathematicians use simplified models. In your CCEA Further Maths exam, you may be asked to explain what these terms mean:

• Particle: The object's entire mass is treated as acting at a single point. This means we can ignore the dimensions (size/shape) of the object, air resistance, and any rotational effects (spinning).
• Light string / rod: The string or rod has zero mass (\(m = 0\)). As a result, the tension is uniform (the exact same value) throughout its entire length.
• Inextensible string: The string does not stretch. Therefore, all connected objects move together with the same acceleration and the same velocity.
• Smooth surface / Smooth pulley: There is no friction at the contact surface or in the pulley wheel bearings. For a smooth pulley, the tension is identical on both sides of the string passing over it.

Key Takeaway: Memorise these four definitions—they are frequent 1-mark and 2-mark theory questions in Unit 2!

3. Resolving Forces into Perpendicular Components

When forces act at an angle, we split (or resolve) them into two perpendicular directions: horizontally and vertically.

The Golden Rule of Resolving

For any force \(F\) at an angle \(\theta\) to a reference line:
• Adjacent to the angle: Component \(= F \cos\theta\)
• Opposite to the angle: Component \(= F \sin\theta\)

Memory Trick: "Close to the angle = Cosine" (both start with Co). If you must swing through the angle \(\theta\) to reach your axis, use \(\cos\theta\). If you go away from the angle, use \(\sin\theta\).

Equilibrium of a Particle

An object is in equilibrium when all the forces acting on it cancel out completely, meaning the resultant force is zero (\(\sum \mathbf{F} = \mathbf{0}\)). An object in equilibrium is either completely stationary or moving at a constant velocity.

For coplanar forces (forces in a 2D plane), equilibrium gives us two separate equations:
\(\sum F_x = 0\) (Total forces pulling Right = Total forces pulling Left)
\(\sum F_y = 0\) (Total forces pulling Up = Total forces pulling Down)

Step-by-Step Example: Resolving on a Horizontal Surface

A crate of mass \(10\text{ kg}\) rests on a smooth horizontal floor. A rope pulls the crate with a force of \(50\text{ N}\) at an angle of \(30^\circ\) above the horizontal. Find the normal reaction \(R\) and the horizontal acceleration.

Step 1: Resolve vertically (\(\sum F_y = 0\))
The upward forces are the normal reaction \(R\) and the vertical component of the pull, \(50 \sin 30^\circ\). The downward force is the weight \(W = mg = 10 \times 9.8 = 98\text{ N}\).
\(R + 50 \sin 30^\circ = 98\)
\(R + 50(0.5) = 98\)
\(R + 25 = 98 \implies R = 73\text{ N}\)

Notice: Because the rope pulls slightly upwards, it takes some weight off the floor, so \(R\) is less than \(mg\)!

Step 2: Find the resultant horizontal force
The only horizontal force is the horizontal component of the pull:
\(F_{\text{net}} = 50 \cos 30^\circ = 50 \times 0.8660 = 43.30\text{ N}\)

Key Takeaway: Always resolve forces into two perpendicular directions. In equilibrium, the sum of components in opposite directions must balance to zero.

4. Forces on an Inclined Plane (Slopes)

Inclined plane problems are a staple of CCEA Mechanics. When an object sits on a plane tilted at an angle \(\alpha\) to the horizontal, it is always best to resolve forces parallel and perpendicular to the slope.

Splitting the Weight on a Slope

The weight \(mg\) acts straight down. By geometry, the angle between the downward vertical and the line perpendicular to the slope is equal to the slope angle \(\alpha\):
• Component of weight pulling down the slope (parallel): \(mg \sin\alpha\)
• Component of weight pressing into the slope (perpendicular): \(mg \cos\alpha\)

Since the object does not lift off or sink into the slope, forces perpendicular to the plane must balance:
\(R = mg \cos\alpha\)

Memory Trick: "Slide down = Sine" (both start with S). The component that makes things slide down the slope is always \(mg \sin\alpha\).

Key Takeaway: For inclined planes at angle \(\alpha\), weight components are always \(mg \sin\alpha\) down the slope and \(mg \cos\alpha\) into the slope.

5. Newton's Laws of Motion

Newton's First Law

An object will remain at rest or continue moving at a constant velocity in a straight line unless acted upon by a non-zero resultant force.

Newton's Second Law: \(F_{\text{net}} = ma\)

When an unbalanced (resultant) force acts on an object of mass \(m\), it produces an acceleration \(a\) in the direction of the force:
\(F_{\text{net}} = ma\)

In practice, we apply this along the line of motion using the equation:
\((\text{Forces in direction of motion}) - (\text{Forces opposing motion}) = ma\)

Newton's Third Law

If object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

Step-by-Step Example: Newton's Second Law

A toy car of mass \(2\text{ kg}\) is pushed along a rough horizontal table with a forward force of \(14\text{ N}\). The frictional resistance is \(4\text{ N}\). Find the acceleration of the toy car.

Step 1: Write down Newton's Second Law along the line of motion
\(F_{\text{net}} = ma\)
\((\text{Forward Force}) - (\text{Friction}) = ma\)

Step 2: Substitute known values and solve
\(14 - 4 = 2a\)
\(10 = 2a \implies a = 5\text{ m s}^{-2}\)

Key Takeaway: Newton's Second Law links the net force directly to acceleration: always subtract opposing forces from driving forces before equating to \(ma\).

6. Friction and Rough Surfaces

When two rough surfaces slide or attempt to slide past one another, friction opposes the motion.

Smooth vs. Rough Surfaces

• Smooth surface: Friction is zero (\(F_r = 0\)).
• Rough surface: Friction resists motion up to a maximum threshold.

Limiting Friction and the Coefficient of Friction (\(\mu\))

Friction adjusts to match the applied force up to a maximum limit. When an object is on the point of slipping (limiting equilibrium) or is already moving, friction reaches its maximum possible value, \(F_{\text{max}}\):
\(F_{\text{max}} = \mu R\)

where:
• \(\mu\) (the Greek letter mu) is the coefficient of friction (a dimensionless number, usually between \(0\) and \(1\)).
• \(R\) is the normal reaction force in Newtons (\(\text{N}\)).

When the object is stationary and not on the verge of moving, static friction is simply equal to whatever force is trying to move it:
\(F_r \le \mu R\)

Step-by-Step Example: Object on a Rough Slope

A block of mass \(4\text{ kg}\) is held on a rough plane inclined at \(30^\circ\) to the horizontal. The coefficient of friction between the block and the plane is \(\mu = 0.2\). When released, the block slides down the plane. Find its acceleration (taking \(g = 9.8\text{ m s}^{-2}\)).

Step 1: Find the normal reaction \(R\)
Resolving perpendicular to the slope:
\(R = mg \cos 30^\circ = (4)(9.8) \cos 30^\circ = 39.2 \times 0.8660 = 33.95\text{ N}\)

Step 2: Calculate the frictional force \(F_r\)
Since the block is moving, friction is at its maximum:
\(F_r = \mu R = 0.2 \times 33.95 = 6.79\text{ N}\)

Step 3: Apply \(F_{\text{net}} = ma\) down the slope
\(mg \sin 30^\circ - F_r = ma\)
\((4)(9.8)\sin 30^\circ - 6.79 = 4a\)
\(39.2(0.5) - 6.79 = 4a\)
\(19.60 - 6.79 = 4a\)
\(12.81 = 4a \implies a = 3.20\text{ m s}^{-2}\text{ (to 3 s.f.)}\)

Key Takeaway: Maximum friction is \(F_{\text{max}} = \mu R\). Always calculate the normal reaction \(R\) first before finding friction.

7. Connected Bodies: Pulleys, Strings & Lifts

Connected particle problems look challenging because there is more than one object, but the method is always identical: write an \(F = ma\) equation for each object separately, then solve the simultaneous equations.

System 1: Smooth Pulley with Two Hanging Masses

Consider two masses \(m_1\) and \(m_2\) connected by a light inextensible string passing over a fixed smooth pulley, with \(m_1 > m_2\):

• Mass \(m_1\) accelerates downwards: \(m_1 g - T = m_1 a\)
• Mass \(m_2\) accelerates upwards: \(T - m_2 g = m_2 a\)

Adding the two equations cancels the internal tension \(T\), allowing you to solve for acceleration \(a\) immediately!

System 2: Mass on Table Connected to Hanging Mass

Mass \(A\) (\(m_A\)) rests on a table and is connected by a string over a smooth pulley at the edge to hanging mass \(B\) (\(m_B\)):

• For hanging mass \(B\) (moving downwards): \(m_B g - T = m_B a\)
• For mass \(A\) on the table (moving horizontally): \(T - F_r = m_A a\)

System 3: Lifts and Apparent Weight

When a person of mass \(m\) stands on a scale inside a lift of mass \(M\), the scale reads the normal reaction \(R\) exerted on the person:

• Lift accelerating upwards: The upward force must beat gravity:
\(R - mg = ma \implies R = m(g + a)\) (You feel heavier!)

• Lift accelerating downwards: Gravity beats the upward reaction:
\(mg - R = ma \implies R = m(g - a)\) (You feel lighter!)

• Lift moving at constant velocity (\(a = 0\)):
\(R = mg\) (Normal weight)

Key Takeaway: For connected particles, set up individual equations of motion along each object's direction of acceleration, then solve them simultaneously.

8. Quick Summary & Common Pitfalls to Avoid

Common Exam Mistakes

• Mistake 1: Confusing Mass and Weight. Remember: Mass is \(m\) (in \(\text{kg}\)). Weight is \(mg\) (in \(\text{N}\)). Never multiply by \(g\) twice or forget to multiply mass by \(g\) when finding downward gravity force!
• Mistake 2: Assuming \(R = mg\) always. If there is an angled pull, or if the particle is on an incline, \(R\) is not equal to \(mg\). Always resolve perpendicular to the plane to find \(R\).
• Mistake 3: Swapping Sine and Cosine on Inclines. Remember: Parallel to slope is always \(mg \sin\alpha\), perpendicular is always \(mg \cos\alpha\).
• Mistake 4: Inconsistent Sign Conventions. In connected particle problems, make sure the direction of acceleration is treated as positive for each separate body.

Mechanics Quick Review Checklist

• Weight: \(W = mg\) downwards.
• Resolving at angle \(\theta\): Adjacent \(= F \cos\theta\), Opposite \(= F \sin\theta\).
• Inclined slope at angle \(\alpha\): Parallel \(= mg \sin\alpha\), Perpendicular \(= mg \cos\alpha\).
• Newton's Second Law: \(F_{\text{net}} = ma\).
• Limiting Friction: \(F_{\text{max}} = \mu R\).