Force,Density & Pressure

A-Level Physics: Turning Effects of Forces

A-Level Physics Notes

4.1 Turning Effects of Forces

1. Centre of Gravity

Definition

The Centre of Gravity of an object is the single point through which its entire weight can be considered to act. It is the average location of the weight distribution.

Key Points

For a uniform, regular object (like a ruler or a sphere), the centre of gravity is at its geometric centre.
For an irregular object, it can be found experimentally by suspending it from different points and tracing the vertical plumb lines; the centre of gravity is where these lines intersect.
An object is in stable equilibrium when the line of action of its weight (the vertical line downwards from its centre of gravity) falls within its base.

Why it's important: The concept of a centre of gravity allows us to simplify problems by treating the distributed force of gravity as a single force acting at a specific point.

2. Moment of a Force

Definition

The Moment of a Force (or Torque) is the measure of its turning effect. It is defined as the product of the force and the perpendicular distance from the pivot (or fulcrum) to the line of action of the force.

Moment = Force × Perpendicular Distance from the Pivot
M = F × d

M = Moment of the force (Newton-metres, Nm)
F = Force (Newtons, N)
d = Perpendicular distance from the pivot to the line of action of the force (metres, m)

PIVOT ● │ │ d │ ↓ F ┌─────────┐ │ Spanner │ └─────────┘

The perpendicular distance `d` is from the pivot to the line of action of the force `F`. The moment is `F × d`.

Key Principles

Principle of Moments: For a body to be in rotational equilibrium (i.e., not turning), the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about the same point.

Sum of Clockwise Moments = Sum of Anticlockwise Moments
The moment is a vector quantity—it has both magnitude and direction (clockwise or anticlockwise).
To maximise the moment, you can either increase the force or increase the distance from the pivot.

Worked Example

A seesaw is balanced. A person weighing 600 N sits 2 m from the pivot. Where must a person weighing 400 N sit to balance it?

Clockwise Moment = Anticlockwise Moment
(400 N) × d = (600 N) × (2 m)
400d = 1200
d = 3 m

The 400 N person must sit 3 m from the pivot.

3. Couple

Definition

A Couple is a pair of forces that are:

Equal in magnitude.
Opposite in direction.
Parallel to each other but not collinear (they do not act in the same straight line).

Effect of a Couple: A couple produces a turning effect only (rotation). It results in no net linear force (i.e., no resultant force), so the object does not accelerate linearly—it only rotates.

Real-World Examples

Turning a steering wheel.
Opening a tap.
Using a screwdriver.
The forces on the handles of a bicycle's handlebars when turning.

4. Torque of a Couple

Definition

The Torque of a Couple (often just called the "moment of a couple") is the total turning effect produced by the couple. It is defined as the product of one of the forces and the perpendicular distance between the lines of action of the two forces.

Torque of a Couple = One Force × Perpendicular Distance between the Forces
τ = F × s

τ (tau) = Torque of the couple (Newton-metres, Nm)
F = Magnitude of one of the forces (N)
s = Perpendicular distance between the lines of action of the two forces (m)

← F ┌─────┐-------------┐ │ Object │ └─────┘-------------┘ → F <--- s --->

The torque is calculated using the perpendicular distance `s` between the two parallel forces `F`.

Alternative Calculation

The torque of a couple can also be found by calculating the sum of the moments of the two forces about any point.

← F → F ┌───┴───┐ ┌───┴───┐ │ a │ P │ b │ └───────┴───┴───────┘ <--- s = a + b --->

Taking moments about point P between the two forces:
Moment due to left force F is F × a (clockwise).
Moment due to right force F is F × b (clockwise).
Total Moment (Torque) = F × a + F × b = F(a + b)
Since a + b = s, Torque = F × s

This confirms that the torque of a couple is the same about any point, as it depends only on the force and the separation `s`.

Summary & Key Differences

Feature	Moment of a Single Force	Torque of a Couple
Forces Involved	One single force.	Two equal, opposite, and parallel forces.
Resultant Force	Can have a non-zero resultant force, causing linear acceleration.	Zero resultant force. Causes rotation only, no linear acceleration.
Calculation	M = F × d (distance from pivot to force)	τ = F × s (distance between the two forces)

Exam Tip

When solving problems on moments:

Identify the pivot point clearly.
Always use the perpendicular distance from the pivot to the line of action of the force. If the force is at an angle, you must resolve it to find the perpendicular component.
Label clockwise and anticlockwise moments to avoid sign errors when applying the Principle of Moments.
Remember that a couple provides a pure rotation with no net force.

A-Level Physics: Equilibrium of Forces

A-Level Physics Notes

4.2 Equilibrium of Forces

1. Principle of Moments

Principle of Moments

For a body to be in rotational equilibrium (i.e., not turning), the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about the same point.

Σ Clockwise Moments = Σ Anticlockwise Moments

┌─────────────────┐ │ │ │ Object │ │ │ └───┬─────────┬───┘ │ │ ↓ F₁ ↓ F₂ d₁ d₂ CW ACW

Key Points

The principle applies to any rigid body in equilibrium
You can take moments about any point - the pivot doesn't have to be the actual axis of rotation
Moments are calculated as force × perpendicular distance from the pivot
Clockwise and anticlockwise directions must be clearly identified

Worked Example

A uniform beam of length 4m and weight 100N is supported at its centre. A 200N weight is placed 1m from the left end. What force must be applied at the right end to balance the beam?

200N 100N F=? ↓ ↓ ↓ │ │ │ ┌────┴───┴───┴───┴───┴───┴───┐ │ │ └────┬───┬───┬───┬───┬───┬───┘ │ │ │ │ │ │ 1m 2m 3m 4m 5m 6m ↑ ↑ ↑ ACW Pivot CW

Solution:

Take moments about the pivot (centre at 2m):
Clockwise moment = F × 2m
Anticlockwise moment = (200N × 1m) + (100N × 0m) = 200Nm
Apply principle: F × 2 = 200
Therefore: F = 100N

2. Conditions for Equilibrium

Equilibrium Definition

A system is in equilibrium when there is no resultant force and no resultant torque acting on it.

Two Conditions for Equilibrium

No Resultant Force (Translational Equilibrium)
ΣF = 0 (Sum of all forces = 0)

This means the forces in any direction balance out:
- ΣFₓ = 0 (sum of horizontal forces = 0)
- ΣFᵧ = 0 (sum of vertical forces = 0)
No Resultant Torque (Rotational Equilibrium)
Στ = 0 (Sum of all moments = 0)

This means the clockwise and anticlockwise moments balance out.

Types of Equilibrium

Static Equilibrium: Body is at rest (not moving or rotating)
Dynamic Equilibrium: Body moving with constant velocity (no acceleration)
In both cases, both conditions (ΣF = 0 and Στ = 0) must be satisfied

Worked Example

A ladder rests against a smooth wall. The ladder has weight W and a person of weight 2W stands at its midpoint. Show that both equilibrium conditions are satisfied.

Wall (smooth) │ │ Normal R₂ ←─────┐ │ │ │ │ Ladder │ │ │ │ Floor└─────┴─────→ Normal R₁ ↑ ↑ Friction Weight 3W total

Solution:

Vertical forces: R₁ = 3W (upwards) balances total weight 3W (downwards)
Horizontal forces: Friction = R₂ (left) balances R₂ (right)
Moments: Take moments about base - clockwise and anticlockwise moments balance
Therefore: ΣFₓ = 0, ΣFᵧ = 0, and Στ = 0

3. Vector Triangle for Coplanar Forces

Coplanar Forces

Coplanar forces are forces that act in the same plane. When three coplanar forces act on a body in equilibrium, they can be represented by a closed vector triangle.

Vector Triangle Principle

If three forces acting at a point are in equilibrium, then:

They can be represented in magnitude and direction by the three sides of a triangle
The triangle must be closed (no gaps)
The vectors follow each other in tip-to-tail fashion

Force Diagram Vector Triangle F₃ ↖ ↗ F₂ F₁ ● P ┌───┐ ↙ │ │ F₁ └───┘ ↗ ↖ F₂ F₃

Steps to Construct Vector Triangle

Draw the first force to scale in the correct direction
From the tip of the first vector, draw the second force to scale
From the tip of the second vector, draw the third force to scale
If the forces are in equilibrium, the triangle will close perfectly

Worked Example

A 100N weight is suspended by two strings making angles of 30° and 60° with the horizontal. Find the tensions in the strings using a vector triangle.

T₁ T₂ ↗ ↖ / 30° 60° \ ●─────────────● │ │ │ 100N │ ↓ (Weight)

Vector Triangle Solution:

T₂ (60° from horizontal) ┌─────────────────┐ │ │ │ │ │ │ └─────────────────┘ ↗ ↖ T₁ (30°) 100N

Using trigonometry or scale drawing:
T₁ = 100N × cos60° = 50N
T₂ = 100N × cos30° = 86.6N

Applications of Vector Triangles

Suspended objects with multiple supports
Forces in frameworks and trusses
Resolution of forces in equilibrium
Checking if forces can be in equilibrium

Summary of Key Concepts

Concept	Description	Mathematical Expression
Principle of Moments	For rotational equilibrium, sum of clockwise moments equals sum of anticlockwise moments	ΣM_CW = ΣM_ACW
Translational Equilibrium	No resultant force in any direction	ΣFₓ = 0, ΣFᵧ = 0
Rotational Equilibrium	No resultant moment about any point	Στ = 0
Vector Triangle	Three coplanar forces in equilibrium form a closed triangle when drawn tip-to-tail	F₁ + F₂ + F₃ = 0

Exam Tips

Always state both conditions for equilibrium (ΣF = 0 AND Στ = 0)
When taking moments, choose your pivot wisely - often at a point where an unknown force acts
For vector triangles, ensure the directions are consistent and the triangle closes perfectly
Remember that weight always acts through the centre of gravity
For objects in contact, consider Newton's Third Law pairs

A-Level Physics: Density and Pressure

A-Level Physics

4.3 Density and Pressure

Density

Definition: Density (ρ) is defined as mass per unit volume.

ρ = m/V

ρ (rho) = density (kg m⁻³)

m = mass (kg)

V = volume (m³)

Note: 1 g cm⁻³ = 1000 kg m⁻³. Water has a density of approximately 1000 kg m⁻³.

Pressure

Definition: Pressure (p) is defined as the normal force per unit area.

p = F/A

p = pressure (pascals, Pa)

F = normal force (newtons, N)

A = cross-sectional area (m²)

Note: 1 Pa = 1 N m⁻². Pressure is a scalar quantity.

Hydrostatic Pressure

The pressure at a point in a fluid at rest due to the weight of the fluid above it.

Derivation of Δp = ρgΔh

Consider a fluid column of height Δh, uniform cross-sectional area A, and density ρ:

Volume of fluid column: V = AΔh
Mass of fluid column: m = ρV = ρAΔh
Weight of fluid column: W = mg = ρAΔhg
This weight is the force exerted on the base, so the pressure is:
p = F/A = (ρAΔhg)/A = ρgΔh

Thus, the pressure difference between two points separated by a vertical height Δh is:

Δp = ρgΔh

Total Hydrostatic Pressure

The total pressure at a depth h in a liquid is given by:

p = p₀ + ρgh

p₀ = atmospheric pressure at the liquid surface (Pa)

ρ = density of the fluid (kg m⁻³)

g = gravitational field strength (N kg⁻¹)

h = depth below the surface (m)

Upthrust and Archimedes' Principle

Origin of Upthrust

An object submerged in a fluid experiences a resultant upward force (upthrust) because the pressure at the bottom of the object is greater than the pressure at the top, creating a pressure difference.The upthrust acting on an object in a fluid is due to a difference in hydrostatic pressure.

Archimedes' Principle

Statement: The upthrust on a body immersed in a fluid is equal to the weight of the fluid displaced by the body.

F = ρ_fluid g V_displaced

F = upthrust (N)

ρ_fluid = density of the fluid (kg m⁻³)

V_displaced = volume of fluid displaced (m³)

Conditions for Flotation

An object will float if the upthrust equals its weight (weight of fluid displaced equals object's weight)
An object will sink if its weight is greater than the maximum possible upthrust

Summary of Equations

ρ = m/V

p = F/A

Δp = ρgΔh

p = p₀ + ρgh

F_upthrust = ρ_fluid g V_displaced

Mastering Inclined Planes in Physics

Mastering Inclined Planes in CIE A-Level Physics

Note: While inclined planes aren't explicitly listed in the CIE Physics syllabus, they frequently appear in exam questions as applications of core principles like force resolution, moments, and energy.

Why Inclined Planes Matter

Inclined planes test your ability to apply physics principles to realistic situations. They're not a new topic - just a new context for applying what you already know about forces, moments, and energy.

The Core Concept: Force Resolution

The key to solving inclined plane problems is understanding how to resolve forces into components.

Visual representation of an inclined plane

F_parallel = mg sinθ (down the slope)
F_{perpendicular} = mg cosθ (into the slope)

Memory Aid: "Sine for Slope" - the sine component acts down the slope.

Step-by-Step Problem Solving

Step 1: Draw and Label

Draw the inclined plane with angle θ. Mark all forces: weight (mg), normal reaction (R), and friction (if present).

Step 2: Resolve Forces

Resolve the weight vector into components parallel and perpendicular to the plane.

Step 3: Apply Physics Principles

Use equilibrium conditions, F=ma, or moments equations as needed for the specific problem.

Common Scenarios

1. Simple Sliding Block

For a block on an inclined plane:

Force down slope: mg sinθ
Friction force: μR = μmg cosθ
Net force: mg sinθ - μmg cosθ

2. Ladder Problems (Moments)

A common exam question combining inclined planes with moments:

Ladder Example

A uniform ladder rests against a smooth wall with its base on rough ground. When taking moments about the base:

Clockwise moment = mg ×

cosθ
Anticlockwise moment = R_wall × L sinθ

3. Energy Problems

For objects moving on inclined planes:

Gravitational PE: mgh, where h = d sinθ
Work done: Force × distance along slope

Worked Examples

Example 1: Basic Force Resolution

Problem: A 5 kg block rests on a 30° frictionless incline. Find the acceleration.

Solution:

F_parallel = mg sinθ = 5 × 9.8 × sin30° = 5 × 9.8 ×
12
= 24.5 N
F = ma ⇒ 24.5 = 5a
a = 4.9 m/s²

Example 2: Ladder with Moments

Problem: A uniform ladder (length 4m, mass 20kg) leans at 60° to a smooth wall with rough ground.

Solution:

Weight moment: 20g × 2 × cos60° = 20g × 2 ×
12
= 20g
Wall reaction moment: R_wall × 4 × sin60° = R_wall × 4 ×
√32
Equation: R_wall × 2√3 = 20g
R_wall =
20g2√3
=
10g√3
≈ 56.6 N

Example 3: Fraction Application

Problem: A block slides down a 45° incline. What fraction of its weight acts parallel to the slope?

Solution:

F_parallel = mg sin45° = mg ×
√22
Fraction of weight =
F_parallelmg
=
√22
Approximately
√22
≈ 0.707 or about 71% of the weight

Common Angles Reference

Angle θ	sinθ	cosθ	tanθ
0°	0	1	0
30°	12	√32	1√3
45°	√22	√22	1
60°	√32	12	√3
90°	1	0	∞

Common Pitfalls & Solutions

Pitfall 1: Wrong Angle Identification

Solution: Remember θ is always between the slope and horizontal. Test with extremes: if θ=0°, plane is flat; if θ=90°, plane is vertical.

Pitfall 2: Confusing sinθ and cosθ

Solution: Use the memory aid "Sine for Slope." Verify with extreme cases where the answer should be obvious.

Pitfall 3: Forgetting Force Components in Moments

Solution: When taking moments, always use the perpendicular distance from the pivot to the line of action of the force.

Pitfall 4: Incorrect Fraction Application

Solution: Remember that sin30° =

, not sin30° =

√32

. Use the table above for reference.

Practice Strategy

Week 1: Master basic force resolution on different angles
Week 2: Add friction to your practice problems
Week 3-4: Focus on ladder problems and moments applications
Ongoing: Solve mixed problems from past papers

Exam Tips

            Don't panic when you see an inclined plane - it's just force resolution
Draw immediately - a clear diagram is half the solution
Resolve weight first - this should always be your first step
Check with extremes - verify your equations make sense for θ=0° and θ=90°
Allocate time wisely - diagram (1-2 min), resolution (2-3 min), equations (2-3 min)
Use exact fractions when possible - 12
 is better than 0.5 in many cases

        

Final Thought

Inclined planes don't make physics more difficult - they make it more interesting. They're an opportunity to demonstrate your deep understanding of fundamental principles.

You already know the physics - the slope just gives you a new way to apply it.

A-Level Physics Practical: Equilibrium of a Wooden Strip

Paper 9702/31 - October/November 2024

Practical Overview

This experiment investigates the equilibrium position of a wooden strip under varying loads. You'll measure how the extension of a spring relates to the angle of the wooden strip, and analyze the relationship between these variables.

Key Objectives

Set up apparatus to investigate equilibrium of a pivoted strip
Make accurate measurements of length and angle
Record, present and analyze data in a table and graph
Determine unknown constants in a physical relationship

Required Apparatus

Stand with boss and clamp
Metre rule & 30cm ruler (mm scale)
String (~1mm diameter, ~50cm)
Expendable spring (~25 N/m)
Wooden strip with screw

10g mass hanger + five 10g slotted masses
Adhesive putty (~6g)
Wooden block
180° protractor (1° divisions)
Ten paper clips
Balance (shared, 0.1g precision)

Methodology

Initial Setup

Set up the apparatus as shown in the diagrams. Ensure the clamp rod is approximately 15cm above the bench. The wooden strip should rest against the base of the stand.

Use adhesive putty to fix the string centrally on the wooden strip in line with the spring. Wrap the string around the screw.

Arrange the block and protractor so the initial angle θ₀ is between 75° and 85°. Measure and record L₀ (length of coiled spring section) and θ₀.

Mass Measurement

Make a hook from one paper clip and hang nine paper clips from it. Measure and record the total mass m of all ten paper clips (should be between 3.0g and 6.0g).

First Data Point

Hang a 40g mass (mass hanger + slotted masses) from the string loop. Also hang the paper clips from the same loop.

Calculate and record the total mass M. Measure and record the new angle θ and length L of the coiled spring section.

Calculate e = L - L₀ (the extension).

Data Collection

Vary M using different combinations of slotted masses and paper clips. For each value of M, measure and record M, θ and L.

Collect at least six sets of readings. Ensure the range M_max - M_min > 30g.

As M increases, θ should decrease - check your results show this trend.

Data Table Requirements

Column headings must follow format: "Quantity / Unit"
All L values to nearest mm
sin θ values to 3-4 significant figures
Include columns for: M, θ, L, e, sin θ

Graph Plotting

Y-axis: e (extension)
X-axis: sin θ
Points must occupy >50% of graph area
Draw line of best fit
Calculate gradient and y-intercept

Analysis

The relationship is:

e = P sin θ + Q

Where:

P = gradient of your graph
Q = y-intercept of your graph

Don't forget units for P and Q!

Common Mistakes

θ₀ not in range 75°-85°
Incorrect calculation of total mass M
Poor table formatting
Small graph scale
Forgetting units for P and Q

Marking Scheme Overview

Initial measurements (θ₀, m)

2 marks

Data collection (table with 6+ readings)

4 marks

Graph plotting

3 marks

Line of best fit

1 mark

Gradient and intercept

2 marks

Determining P and Q

2 marks

TOTAL

20 marks