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MECHANICS FOR PP4

M1 TOPIC 1 : Forces and equilibrium

CIE A-Level Mechanics 1: Forces and Equilibrium

CIE A-Level Mechanics 1

Forces and Equilibrium

This topic covers the analysis of forces acting on particles and the conditions for equilibrium. Understanding forces is fundamental to solving problems in mechanics.

Note: Calculations are always required, not approximate solutions by scale drawing. Solutions by resolving are usually expected.

Identifying Forces

Common types of forces encountered in mechanics problems:

Weight (W)
Acts vertically downward through center of mass

Normal Reaction (R or N)
Perpendicular to contact surface

Friction (F)
Parallel to contact surface, opposes motion

Tension (T)
Pulling force through strings, ropes, or cables

Thrust/Compression
Pushing force through rods or struts

Typical Force Diagram

[Diagram showing weight, normal reaction, and friction forces on an object on an inclined plane]

Always draw clear force diagrams showing all forces acting on the particle.

Vector Nature of Force

Force is a vector quantity - it has both magnitude and direction.

Components of Forces

Forces can be resolved into perpendicular components, typically horizontal and vertical:

F_x = F cosθ F_y = F sinθ

Example: A force of 10 N at 30° to the horizontal:

F_x = 10 cos30° = 8.66 N F_y = 10 sin30° = 5 N

Resultant Force

The resultant of multiple forces is found by vector addition:

R = √(ΣF_x² + ΣF_y²)

θ = tan⁻¹(ΣF_y/ΣF_x)

Key Point: When finding resultants, always consider both magnitude and direction.

Equilibrium of a Particle

A particle is in equilibrium when the vector sum of all forces acting on it is zero.

ΣF = 0

This means:

ΣF_x = 0 and ΣF_y = 0

Problem-Solving Strategy for Equilibrium

Draw a clear diagram showing all forces
Resolve forces into perpendicular components
Apply equilibrium conditions: ΣF_x = 0 and ΣF_y = 0
Solve the resulting equations
Check your solution makes physical sense

Note: While solutions by resolving are usually expected, equivalent methods (triangle of forces, Lami's Theorem) are also acceptable but are not required knowledge.

Contact Forces

A contact force between two surfaces can be represented by two components: the normal component and the frictional component.

Components of Contact Force

[Diagram showing normal (R) and friction (F) components of contact force]

Smooth Contact Model

Smooth contact is a model that assumes no friction between surfaces.

Limitations: The smooth contact model is an idealization. In reality, most surfaces have some friction. This model is useful for simplifying problems where friction is negligible or not considered.

Friction

Limiting Friction and Limiting Equilibrium

Limiting friction (F_max) is the maximum value of friction that can prevent motion.

Limiting equilibrium occurs when a body is on the point of moving - friction is at its maximum value.

Note: Terminology such as 'about to slip' may be used to mean 'in limiting equilibrium' in questions.

Coefficient of Friction

Coefficient of friction (μ) is defined as the ratio of limiting friction to normal reaction.

F ≤ μR

F_max = μR

Where:

F = friction force
F_max = limiting friction
μ = coefficient of friction
R = normal reaction force

Key Point: The friction force only equals μR when motion is impending or occurring. Otherwise, F ≤ μR.

Example: A 5 kg box on a horizontal surface with μ = 0.4:

Weight = 5g = 49 N
Normal reaction R = 49 N
Maximum friction F_max = μR = 0.4 × 49 = 19.6 N

Newton's Laws of Motion

Newton's Third Law

Newton's Third Law: When two bodies interact, they exert equal and opposite forces on each other.

F_{A on B} = -F_{B on A}

Example: The force exerted by a particle on the ground is equal and opposite to the force exerted by the ground on the particle.

Important: Action and reaction forces act on different bodies, so they don't cancel each other when considering the motion of a single body.

All Three Laws Summary

First Law
An object remains at rest or in uniform motion unless acted upon by a resultant force

Second Law
F = ma (resultant force = mass × acceleration)

Third Law
Action and reaction are equal and opposite

Key Equations and Concepts Summary

Equilibrium condition
ΣF = 0

Component resolution
F_x = F cosθ
F_y = F sinθ

Limiting friction
F_max = μR

Friction inequality
F ≤ μR

Newton's Third Law
F_{A on B} = -F_{B on A}

Problem-Solving Approach:

Always start with a clear force diagram
Resolve forces into perpendicular components
Apply equilibrium conditions (ΣF_x = 0, ΣF_y = 0)
Use F = μR only when friction is limiting
Check if your answer is physically reasonable

Resultant Acting Along a Specific Direction

Quick reference: if the resultant acts horizontally, vertical components cancel; if it acts vertically, horizontal components cancel.

Resultant Direction	Meaning	Vertical Resolution	Horizontal Resolution	Key Condition
Positive x-direction	Resultant acts horizontally to the right.	Sum of vertical components = 0	Resultant = Sum of horizontal components (to the right)	Forces balance vertically
Negative x-direction	Resultant acts horizontally to the left.	Sum of vertical components = 0	Resultant = Sum of horizontal components (to the left)	Forces balance vertically
Positive y-direction	Resultant acts vertically upward.	Resultant = Sum of vertical components (upward)	Sum of horizontal components = 0	Forces balance horizontally
Negative y-direction	Resultant acts vertically downward.	Resultant = Sum of vertical components (downward)	Sum of horizontal components = 0	Forces balance horizontally

Abel Masitsa

CIE A-Level Mechanics: Triangle of Forces & Lami's Theorem

CIE A-Level Mechanics 1

Triangle of Forces & Lami's Theorem

Alternative Methods for Equilibrium Problems

While resolving forces into components is the primary method taught for solving equilibrium problems, there are two powerful geometric methods that can be more efficient for certain types of problems.

Syllabus Note: These methods are not required knowledge and will not be referred to in examination questions. However, they are acceptable alternative methods that can provide quicker solutions for specific scenarios.

When to use these methods: Both methods are particularly useful when dealing with three forces in equilibrium, especially when angles between forces are known or easily determined.

Triangle of Forces

Triangle of Forces Theorem: If three forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the three sides of a triangle taken in order.

Triangle of Forces Representation

[Diagram showing three forces F1, F2, F3 acting at a point, and their vector triangle representation]

The forces form a closed triangle when placed head-to-tail

Conditions for Using Triangle of Forces

Exactly three forces acting on the particle
All forces are coplanar (in the same plane)
The particle is in equilibrium
At least some information about force magnitudes or angles is known

Step-by-Step Procedure

Step 1: Draw a clear diagram showing all three forces acting on the particle

Step 2: Identify the angles between the forces

Step 3: Construct the force triangle by placing forces head-to-tail

Step 4: Use trigonometry (sine rule, cosine rule) to find unknown forces

Step 5: Verify the triangle closes (vectors return to starting point)

Important: The forces must be drawn in the correct order around the triangle. The direction around the triangle should be consistent.

Example Setup: A particle supported by two strings with tensions T₁ and T₂, with weight W acting downward.

Forces: T₁ (left), T₂ (right), W (downward)
Triangle: W (vertical), T₁ and T₂ completing the triangle

Lami's Theorem

Lami's Theorem: If three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces.

F₁/sinα = F₂/sinβ = F₃/sinγ

Lami's Theorem Angles

[Diagram showing three forces F1, F2, F3 with angles α, β, γ between them]

α = angle between F₂ and F₃, β = angle between F₁ and F₃, γ = angle between F₁ and F₂

Conditions for Using Lami's Theorem

Exactly three forces acting on the particle
All forces are coplanar
The particle is in equilibrium
Angles between the forces are known or can be determined

Step-by-Step Procedure

Step 1: Draw the three forces acting on the particle

Step 2: Identify the angles between the forces (α, β, γ)

Step 3: Apply Lami's theorem formula

Step 4: Solve for unknown forces using the ratios

Step 5: Check that α + β + γ = 360°

Angle Definition: The angle used for each force is the angle between the other two forces, not the angle the force makes with a reference direction.

Example: If F₁ = 50N, angle between F₂ and F₃ is 120°, then:

F₁/sinα = F₂/sinβ = F₃/sinγ
50/sinα = F₂/sinβ = F₃/sinγ

Method Comparison

Method	When to Use	Advantages	Limitations
Resolution of Forces	Any number of forces, general cases	Systematic approach, works for all scenarios	Can involve more algebra, simultaneous equations
Triangle of Forces	Exactly 3 forces in equilibrium	Geometric intuition, visual solution	Limited to 3 forces, requires scale drawing or trigonometry
Lami's Theorem	Exactly 3 forces with known angles	Direct calculation, efficient for angle-based problems	Limited to 3 forces, angles must be known

Examination Strategy: While resolution of forces is the expected method in exams, knowing these alternative methods can help you verify your answers or choose the most efficient approach for specific problems.

Worked Examples

Example 1: Triangle of Forces

Problem: A 100N weight is suspended by two strings making angles of 30° and 60° with the horizontal. Find the tensions in the strings.

Step 1: Forces: Weight W = 100N (down), T₁ (30° left of vertical), T₂ (60° right of vertical)

Step 2: Construct triangle: W vertical, T₁ and T₂ completing triangle

Step 3: Use sine rule in triangle with angles 30°, 60°, 90°

Step 4: T₁/sin60° = T₂/sin30° = 100/sin90°

Step 5: T₁ = 100 × sin60° = 86.6N, T₂ = 100 × sin30° = 50N

Example 2: Lami's Theorem

Problem: Three forces F₁, F₂, F₃ act at a point. F₁ = 80N, angle between F₂ and F₃ is 110°, angle between F₁ and F₃ is 130°. Find F₂ and F₃.

Step 1: Given: α = 110° (between F₂ and F₃), β = 130° (between F₁ and F₃)

Step 2: Find γ = 360° - 110° - 130° = 120° (between F₁ and F₂)

Step 3: Apply Lami's theorem: F₁/sinα = F₂/sinβ = F₃/sinγ

Step 4: 80/sin110° = F₂/sin130° = F₃/sin120°

Step 5: F₂ = 80 × sin130°/sin110° = 65.3N, F₃ = 80 × sin120°/sin110° = 73.6N

Verification Tip: Always check that your solution makes physical sense. Tensions should be positive, and the vector triangle should close properly.

Summary

Key Points to Remember

Both methods work only for three forces in equilibrium
Triangle of forces provides a geometric interpretation
Lami's theorem is algebraically efficient when angles are known
These are alternative methods to resolution of forces
In examinations, resolution method is expected but these alternatives are acceptable

Final Note: Mastering these methods gives you multiple approaches to solve equilibrium problems, allowing you to choose the most efficient method for each specific scenario.

Contact Force

Definition: A contact force is a force that acts only when two objects are touching. It arises from interactions at the surfaces or particles where the objects meet.

Common Examples

Contact Force	Description
Normal (reaction) force	The push a surface gives to an object in contact. Acts perpendicular to the surface.
Friction	The force that opposes motion (or the tendency to move) between two surfaces in contact. Acts along the surface.
Tension	The pull transmitted along a stretched string, rope, or cable. Acts along the string.
Air resistance / drag	The force of air (or fluid) on a moving object. Acts opposite to motion.
Upthrust (buoyant force)	The upward push a fluid exerts on a submerged object due to pressure differences.

Contact vs Non-contact Forces

Contact forces act only when objects touch (examples above).
Non-contact forces act even when objects are apart, e.g. gravity, electrostatic, and magnetic forces.

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Coplanar Forces - Mechanics Notes

Coplanar Forces in Mechanics

Definition

Coplanar forces are forces that act in the same plane. All the lines of action of these forces lie within a single two-dimensional plane.

🎯 Key Characteristics

All forces act in the same plane (2D)
Can be analyzed using 2D coordinate systems (x and y axes)
Simpler to analyze than non-coplanar (3D) forces
Common in many practical engineering and physics problems

Types of Coplanar Force Systems

Coplanar Concurrent Forces

Coplanar Parallel Forces

Type	Description	Example	Analysis Method
Coplanar Concurrent	All forces meet at a single point	Forces on a particle	Vector addition at point
Coplanar Parallel	All forces are parallel to each other	Weights on a beam	Algebraic sum, moments
Coplanar Non-concurrent	Forces don't meet at a point but are in same plane	Forces on a rigid body	Force resolution, moments

Analysis Methods for Coplanar Forces

1. Resolution of Forces

Break each force into x and y components:

Fₓ = F cosθ

Fᵧ = F sinθ

Force Resolution

2. Resultant Force Calculation

Find the vector sum of all forces:

Rₓ = ΣFₓ (sum of all x-components)

Rᵧ = ΣFᵧ (sum of all y-components)

R = √(Rₓ² + Rᵧ²) (magnitude of resultant)

θ = tan⁻¹(Rᵧ/Rₓ) (direction of resultant)

Example: Three Coplanar Forces

Given:

F₁ = 10 N at 0° (to the right)
F₂ = 15 N at 90° (upward)
F₃ = 8 N at 180° (to the left)

Solution:

Rₓ = 10 cos0° + 15 cos90° + 8 cos180° = 10 + 0 - 8 = 2 N

Rᵧ = 10 sin0° + 15 sin90° + 8 sin180° = 0 + 15 + 0 = 15 N

R = √(2² + 15²) = √(4 + 225) = √229 ≈ 15.13 N

θ = tan⁻¹(15/2) ≈ 82.4° above horizontal

Coplanar vs Non-Coplanar Forces

Aspect	Coplanar Forces	Non-Coplanar Forces
Dimension	2D (x-y plane)	3D (x-y-z space)
Analysis	Simpler - 2 components	Complex - 3 components
Equilibrium	ΣFₓ = 0, ΣFᵧ = 0	ΣFₓ = 0, ΣFᵧ = 0, ΣF𝓏 = 0
Moments	About z-axis only	About x, y, and z axes
Examples	Forces on a flat surface, bridge design	Space structures, aircraft in flight

Equilibrium of Coplanar Forces

Conditions for Equilibrium

For a body to be in equilibrium under coplanar forces:

ΣFₓ = 0 (sum of horizontal forces = 0)

ΣFᵧ = 0 (sum of vertical forces = 0)

ΣM = 0 (sum of moments about any point = 0)

Example: Beam in Equilibrium

Scenario: A 4m beam supported at both ends with loads:

Weight 200 N at center (2m from each end)
Find support reactions R₁ and R₂

Solution:

ΣFᵧ = 0: R₁ + R₂ - 200 = 0 → R₁ + R₂ = 200

ΣM about left end = 0: R₂ × 4 - 200 × 2 = 0 → R₂ = 100 N

Therefore: R₁ = 200 - 100 = 100 N

Problem-Solving Strategy

Identify all forces acting on the body
Draw a free-body diagram
Choose coordinate system (x-y axes)
Resolve forces into x and y components
Apply equilibrium equations: ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0
Solve the resulting equations

Real-World Applications

Engineering Applications

Bridge design - analyzing forces on trusses and beams
Building structures - calculating loads on floors and walls
Machine design - analyzing forces in mechanical components
Vehicle dynamics - studying forces on cars and aircraft

👨‍🏫 Teacher's Tip

Common student difficulties:

Forgetting to include all forces in the free-body diagram
Incorrect sign conventions for force components
Confusion between concurrent and non-concurrent systems
Errors in calculating moments

Teaching strategy: Use plenty of worked examples with clear diagrams. Emphasize the importance of consistent sign conventions.

🎯 Exam Tips

Always draw a clear free-body diagram
Use consistent sign conventions (e.g., right/up = positive)
Check your answer makes physical sense
Remember all three equilibrium conditions for rigid bodies
Practice with both concurrent and non-concurrent systems