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IGCSE Physics Notes Forces

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Syllabus Objectives

IGCSE Physics Revision Notes Effects of forces

CORE:

  • Know that forces may produce changes in the size and shape of an object
  • Sketch, plot and interpret load–extension graphs for an elastic solid and describe the associated experimental procedures
  •  Determine the resultant of two or more forces acting along the same straight line
  • Know that an object either remains at rest or continues in a straight line at constant speed unless acted on by a resultant force
  • State that a resultant force may change the velocity of an object by changing its direction of motion or its speed
  • Describe solid friction as the force between two surfaces that may impede motion and produce heating
  •  Know that friction (drag) acts on an object moving through a liquid
  •  Know that friction (drag) acts on an object moving through a gas (e.g. air resistance)

EXTENDED:/ SUPPLEMENT:

  • Define the spring constant as force per unit extension; recall and use the equation k = F /x
  • Define and use the term ‘limit of proportionality’ for a load–extension graph and identify this point on the graph (an understanding of the elastic limit is not required)
  •  Recall and use the equation F = ma and know that the force and the acceleration are in the same direction
  •  Describe, qualitatively, motion in a circular path due to a force perpendicular to the motion as:
  • (a) speed increases if force increases, with mass and radius constant
  • (b) radius decreases if force increases, with mass and speed constant
  • (c) an increased mass requires an increased force to keep speed and radius constant (F = mv2 /r is not required)

1.5.2 Turning effect of forces

CORE OBJECTIVES:

  • Describe the moment of a force as a measure of its turning effect and give everyday examples
  •  Define the moment of a force as moment = force × perpendicular distance from the pivot; recall and use this equation
  •  Apply the principle of moments to situations with one force each side of the pivot, including balancing of a beam 
  • State that, when there is no resultant force and no resultant moment, an object is in equilibrium

EXTENDED/SUPPLEMENT OBJECTIVES:

  • Apply the principle of moments to other situations, including those with more than one force each side of the pivot
  •  Describe an experiment to demonstrate that there is no resultant moment on an object in equilibrium

1.5.3 Centre of gravity

CORE OBJECTIVES:

  • State what is meant by centre of gravity
  • Describe an experiment to determine the position of the centre of gravity of an irregularly shaped plane lamina
  •  Describe, qualitatively, the effect of the position of the centre of gravity on the stability of simple objects

SUPPLEMENT  OBJECTIVES:

[NONE]

 

Revision Notes

Forces Revision Notes IGCSE Physics

Video  preview of the revision notes for IGCSE Physics , Topic Force, strictly by Cambridge trained experts

IGCSE Physics / Centre of Gravity, Moments and Equilibrium /  Practice

 
How to use this page: Read the short notes first, then practise the structured questions. All worked solutions are built around the mark scheme suggestions only, with clear method steps and exam wording. You can insert your own images in the dashed boxes wherever needed.

1Key Ideas You Must Know

Centre of gravity
The point through which the whole weight of an object may be considered to act.
Equilibrium
An object is in equilibrium when there is no resultant force and no resultant moment.
Moment of a force
Moment = force × perpendicular distance from the pivot.
Uniform object
If a shape is uniform and symmetrical, its centre of gravity lies at its geometrical centre.
Important Formula
moment = force × perpendicular distance
Unit of moment: newton metre, written as N m

2Equilibrium Conditions

Condition Meaning
No resultant force The forces balance, so the object does not accelerate in a straight line.
No resultant moment Clockwise moment = anticlockwise moment, so the object does not turn.
Object in equilibrium Both of the above must be true at the same time.
Examiner tip: In a theory question on equilibrium, the safest complete answer is: “There is no resultant force and no resultant moment.”

3MCQ – Centre of Gravity of a Hanging Tile

MCQ Practice

A hole is made in a square tile of uniform thickness. The tile hangs loosely on a nail. Where is the centre of gravity of the tile?

Correct option: D
Reveal Answer and Explanation
The tile is a uniform square, so its centre of gravity is at the centre of the square.
When the tile hangs freely, the centre of gravity must lie vertically below the point of support.
Among the labelled points, only D is at the central position of the tile and is the correct location for the centre of gravity.
Exam wording to remember: For a freely hanging object, the centre of gravity lies on the vertical line below the point of suspension.

4Worked Example – Rock Climber in Equilibrium

A rock climber, of total mass 62 kg, holds herself in horizontal equilibrium against a vertical cliff.

(a) Calculate her weight. [1]

Reveal Answer
Step 1: Use weight = mass × gravitational field strength.
Step 2: Weight = 62 × 10
Answer: 620 N
Built from mark scheme: accepted answer is 620 N.

(b) State the two conditions for the climber to be in equilibrium. [2]

Reveal Answer
There is no resultant force.
There is no resultant moment.
Built from mark scheme: exactly these two points are required for equilibrium.

(c)(i) Calculate the moment of the climber’s weight about the point where her feet touch the cliff. The perpendicular distance is 0.90 m. [2]

Reveal Answer
Step 1: Use moment = force × perpendicular distance.
Step 2: Moment = 620 × 0.90
Step 3: Moment = 558 N m
Step 4: To 2 significant figures, this is 560 N m.
Built from mark scheme: accepted answer is 560 N m using 620 × 0.90 or F × x?.

(c)(ii) The rope makes an angle of 60° with the climber’s body. The distance from the feet to the hands is 1.2 m. Calculate the tension T in the rope. [3]

Reveal Answer
Step 1: Since the climber is in equilibrium, clockwise moment = anticlockwise moment.
Step 2: Use the moment from part (c)(i): 560 N m.
Step 3: The perpendicular distance for the tension is 1.2 sin 60°.
Step 4: Set up the moment equation:
T × 1.2 sin 60° = 560
Step 5: Rearranging gives:
T = 560 ÷ (1.2 × sin 60°)
Step 6: T ? 538.9 N
Answer: 540 N
Built from mark scheme: use any moment method and apply T × 1.2 sin 60° = 560, giving 540 N.

5Short Notes – How to Answer These Questions

Centre of gravity question
Look for the point where the object’s whole weight can be considered to act. For a hanging object, this point must be vertically below the support.
Moment question
Always use the perpendicular distance from pivot to line of action of the force.
Equilibrium question
State no resultant force and no resultant moment.
Tension from moments
Balance the turning effects about the pivot, then solve for the unknown force.

6Extra Exam-Style Practice

Question 2

State what is meant by the centre of gravity of an object. [1]

Reveal Answer
The centre of gravity is the point through which the whole weight of the object may be considered to act.
Question 3

Why does a freely suspended object come to rest with its centre of gravity vertically below the point of suspension? [2]

Reveal Answer
If the centre of gravity is not vertically below the point of suspension, the weight produces a moment.
The object turns until there is no resultant moment, which happens when the centre of gravity is vertically below the support.
Question 4

A force of 25 N acts perpendicularly at a distance of 0.40 m from a pivot. Calculate the moment. [2]

Reveal Answer
Step 1: Moment = force × perpendicular distance
Step 2: Moment = 25 × 0.40
Answer: 10 N m
Question 5

State the unit of moment. [1]

Reveal Answer
N m

7Common Mistakes

Mistake 1:
Using the ordinary distance instead of the perpendicular distance in a moments calculation.
Mistake 2:
Giving only one condition for equilibrium. You need both.
Mistake 3:
Choosing a corner of a uniform square as the centre of gravity. It is at the centre.
Mistake 4:
Forgetting to round the final answer suitably when the mark scheme uses a rounded value.

8Rapid Revision Summary

Centre of gravity
Point where whole weight acts.
Moment
Force × perpendicular distance.
Equilibrium
No resultant force and no resultant moment.
Hanging object
Centre of gravity lies vertically below support.
Unit of moment
N m
MCQ tile answer
D

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