Skip to content
Cambridge IGCSE0625

Forces

Physics 0625 Chapter Notes

What this chapter covers

Forces - Effects of forcesForces - Turning effect of forcesForces - Centre of gravity
ShareWhatsAppPost
Forces notes

Read online free. Download a watermarked copy with a free account.

Read the notes

The full Forces notes as text: skim, search, and jump between subtopics.

~22 min read

In this chapter

  1. 1Stretching a spring: load and extension
  2. 2The spring experiment
  3. 3Resultant of forces along a line
  4. 4Resultant force and motion
  5. 5Friction and drag
  6. 6Force, mass and acceleration
  7. 7Spring constant and the limit of proportionality
  8. 8Motion in a circular path
  9. 9Moment of a force
  10. 10Principle of moments
  11. 11Equilibrium
  12. 12More than one force each side
  13. 13Experiment: balancing a metre ruler
  14. 14What centre of gravity means
  15. 15Finding the centre of gravity of a lamina
  16. 16Stability and the position of the centre of gravity
  17. 17Comparing the stability of objects

1. Stretching a spring: load and extension

Forces may produce changes in the size and shape of an object. A load hung from a spring makes it longer.

The extension is the new length of the spring minus its length with no load. It is not the length of the spring.

On a load–extension graph, a straight line through origin shows that load and extension are directly proportional. A line that is not a straight line, or not through origin, does not show this.

x = l - l₀

Examiner insight

“Most candidates did not take the weight of the can into account when reading the graph.” “Many candidates referred to the points being close to the line or the line being straight. Reference to the line passing through the origin was not included.”

Worked example 11 mark

The diagram shows an extension–load graph for a spring. An empty can of weight 3.0 N is suspended from the spring. Liquid is poured into the can until the extension is 8.0 cm. What is the weight of the liquid?

A 4.0 N · B 10.0 N · C 13.0 N · D 16.0 N

  1. 1

    at an extension of 8.0 cm the graph gives a total load of 16.0 N

  2. 2

    weight of liquid = 16.0 - 3.0 = 13.0 N

  3. 3

    C [1]

2. The spring experiment

Measure the length of the spring with no load. Hang a load from the spring, record the new length and calculate the extension. Repeat with larger loads.

For an accurate length, keep the ruler close to spring and the ruler parallel to spring, and read with your eye level with reading. A marker or set square is used to align spring with ruler.

Plot the graph with each axis labelled with its quantity and unit, a scale filling more than half the grid, and a single thin best-fit line with a balance of points. Do not overload the spring.

Examiner insight

“Referred to the use of a set square, without stating that it was used to align the spring with the ruler.” “Lines of best fit were forced through the origin, required rotation or did not have ‘balanced’ points.”

Worked example 12 marks

A student investigates the behaviour of a spring. Describe two techniques the student uses when measuring the length of the spring to ensure an accurate reading.

  1. 1

    ruler close to the spring, or ruler parallel to the spring [1]

  2. 2

    eye level with the reading [1]

3. Resultant of forces along a line

For forces acting along the same straight line, add the forces that act in the same direction and subtract the forces that act in the opposite direction.

A resultant has a size and a direction. Give the direction in words the diagram supports, such as to the right or in direction of the larger force.

Examiner insight

“Some candidates added the two forces.” “Some candidates stated that the direction of the force was East/West. The diagram did not show which direction corresponded to East or West.”

Worked example 12 marks

Figure 2.3 shows the horizontal forces on the pendulum bob after the student releases it. Calculate the resultant horizontal force on the load.

  1. 1

    resultant force = 0.5 - 0.1 = 0.4 N [1]

  2. 2

    direction is to the right [1]

4. Resultant force and motion

An object either remains at rest or continues in a straight line at constant speed unless acted on by a resultant force. So when an object moves in a straight line at constant speed there is no resultant force on it.

A resultant force may change the velocity of an object by changing its direction of motion or its speed.

Examiner insight

“The majority of candidates had the misconception that there was a resultant force on the car because it was going downhill, even though it was travelling at constant speed.” “Only more able candidates recognised they should also include a change in direction.”

5. Friction and drag

Solid friction is the force between two surfaces that may impede motion and produce heating.

Friction also acts on an object moving through a liquid or through a gas. This friction is called drag. In air it is air resistance.

Drag is a resistive force: it opposes the motion. A bird flying to the right has air resistance to the left. A ball falling through oil has an upward drag force.

Examiner insight

“Better prepared candidates identified friction as the force stopping the bicycle but many candidates were then unable to explain that friction takes place when two surfaces are rubbing together.” “Some candidates stated that a larger force was needed to get the ship to move or to accelerate the ship, without making any reference to resistive forces.”

Worked example 12 marks

Figure 3.2 shows a metal ball. The gravitational force on the ball is 1.2 N. The metal ball falls into a container of oil. After a short time, the metal ball is falling through the oil at a constant speed. Complete the labels on Figure 3.2 to show the forces on the metal ball as it falls at a constant speed.

  1. 1

    the upward arrow is labelled drag force [1]

  2. 2

    constant speed, so there is no resultant force: the upward force is 1.2 N [1]

6. Force, mass and acceleration

In F = ma, F is the resultant force, m is the mass and a is the acceleration. The force and the acceleration are in the same direction.

When more than one force acts, find the resultant first. For a rocket moving upwards, resultant force = force due to rocket – mg.

F = ma

Examiner insight

“A few candidates were unable to recall the equation relating Force and acceleration. These responses usually went on to divide mass by acceleration.” “Candidates needed to recognise that as the car was decelerating, the resultant force would act in a backwards direction.”

Worked example 12 marks

The mass of the ship is 3.5 × 10⁷ kg. The ship accelerates at 0.75 m / s². Calculate the force which causes this acceleration.

  1. 1

    F = ma = 3.5 × 10⁷ × 0.75 [1]

  2. 2

    = 2.6 × 10⁷ N [1]

7. Spring constant and the limit of proportionality

The spring constant k is force per unit extension. Its unit matches the units used for F and x: N / m or N / cm.

Up to the limit of proportionality, the extension is proportional to the load and the graph is a straight line. The limit of proportionality is the point at the end of the straight section. Beyond it the line curves.

k = F/x

Examiner insight

“Weaker candidates confused the limit of proportionality with the elastic limit and referred to the spring not returning to its original length.” “Common errors included omitting the unit from the answer or the use of an incorrect unit, often Nm.”

Worked example 11 mark

An 800 g mass is suspended from a light spring with a spring constant of 20 N / m. The mass is pulled down a further 10.0 cm from the equilibrium position without the spring going past the limit of proportionality. The mass is then released. Which row about the resultant force on the mass and its acceleration immediately after its release is correct?

A 2.0 N upwards, 2.5 m / s² upwards · B 7.8 N upwards, 9.8 m / s² upwards · C 9.8 N downwards, 9.8 m / s² downwards · D 200 N upwards, 0.25 m / s² upwards

  1. 1

    further extension = 10.0 cm = 0.10 m

  2. 2

    extra force from the spring = kx = 20 × 0.10 = 2.0 N upwards, which is the resultant force

  3. 3

    a = F/m = 2.0/0.80 = 2.5 m/s² upwards

  4. 4

    A [1]

8. Motion in a circular path

An object moves in a circular path when a force acts perpendicular to motion. The force is towards the centre of the circle.

Speed increases if force increases, with mass and radius constant. Radius decreases if force increases, with mass and speed constant. An increased mass requires an increased force to keep speed and radius constant.

Examiner insight

“An answer of centripetal force did not answer the question.” “Many drew arrows pointing outwards which suggested a misunderstanding of the action of a resultant force.”

Worked example 11 mark

A satellite is in a circular orbit around the Earth. State the direction of the force acting on the satellite.

  1. 1

    perpendicular to the motion, towards the centre of the orbit [1]

Recap

  • Forces may change the size and shape of an object.
  • Extension = new length − length with no load.
  • Directly proportional = straight line through the origin.
  • Resultant along a line: add forces in the same direction, subtract opposite ones, and state the direction.
  • No resultant force: the object stays at rest or keeps moving in a straight line at constant speed.
  • Friction acts between two surfaces, impedes motion and produces heating. Drag is friction in a liquid or a gas.
  • Extended: k = F ÷ x, force per unit extension. The limit of proportionality is the end of the straight section.
  • Extended: F = ma with F the resultant force. Force and acceleration are in the same direction.
  • Extended: circular path = force perpendicular to the motion, towards the centre.

9. Moment of a force

The moment of a force is a measure of its turning effect. Loosening a nut with a spanner and lifting the handle of a trolley are both turning effects about a pivot.

Moment = force × perpendicular distance from the pivot. The unit is N m, or N cm for a distance in cm.

moment = F × d

F = moment/d

Examiner insight

“There was some confusion between the command words ‘define’ and ‘describe’, with weaker candidates often stating that the moment of a force was its turning effect.” “The most common error was to divide the force by the distance to give 0.83.”

Worked example 13 marks

Figure 1.2 shows the empty trolley lying on the floor. The student lifts the handle by applying a force of 12 N, as shown in Figure 1.2. Calculate the moment of the 12 N force about the pivot.

  1. 1

    moment = force × perpendicular distance from pivot [1]

  2. 2

    = 12 × 0.90 [1]

  3. 3

    = 11 N m (10.8 to 3 significant figures) [1]

10. Principle of moments

A beam balances when the clockwise moment equals the anticlockwise moment. With one force each side of the pivot, write one moment equal to the other and solve for the unknown. State that they are equal. Each distance is measured from the pivot.

clockwise moment = anticlockwise moment

F₁ d₁ = F₂ d₂

Examiner insight

“Some candidates separately wrote out the clockwise and anticlockwise moments but made mistakes in equating them, and did not state that they were equal, limiting the credit that could be awarded.” “Weaker candidates failed to use the principle of moments and unsuccessfully attempted to balance vertical forces.”

Worked example 12 marks

Figure 2.1 shows a diagram of a car brake pedal. The driver applies a force of 14 N to the end of the brake pedal. The piston applies a force of 72 N to the brake pedal.

The brake pedal is stationary. The piston exerts a force of 72 N on the brake pedal. Calculate the distance d.

  1. 1

    The pedal is stationary, so the moments about the pivot balance: 72 × d = 14 × 19.6

  2. 2

    d = (14 × 19.6)/72 [1]

  3. 3

    = 3.8 cm [1]

11. Equilibrium

An object is in equilibrium when there is no resultant force and no resultant moment.

Both conditions are needed. ‘No resultant force’ on its own is not a complete description.

resultant force = 0 and resultant moment = 0

Examiner insight

“A smaller number of candidates only considered the equilibrium of the moments of the forces.” “For an object to be in equilibrium, both the resultant force on the object and the resultant moment about a point must be zero.”

Worked example 11 mark

Which statement gives a complete description of any object that is in equilibrium?

A There are no forces acting. · B There is no resultant force. · C There is no resultant force and no resultant turning effect. · D There is no resultant turning effect.

  1. 1

    C: there is no resultant force and no resultant turning effect. [1]

12. More than one force each side

With several forces, add up the moments on each side of the pivot. The sum of the clockwise moments equals the sum of the anticlockwise moments.

The weight of a uniform beam acts at its centre: include its moment. Use the total distance from the pivot for each force. Work line by line and label every distance on the diagram.

Σ (clockwise moments) = Σ (anticlockwise moments)

Examiner insight

“Stronger answers were produced by those candidates who worked logically, line by line, with a clear method, and who correctly labelled all distances on the diagram.” “Weaker candidates often tried to work backwards from the given value of the mass.”

Worked example 11 mark

A student is designing a toy. She attaches a model plane to one end of a uniform metal rod. The weight of the metal rod is 0.50 N. The weight of the model plane is 3.5 N.

She uses a counterbalance to keep the system in equilibrium. What is the weight of the counterbalance?

A 2.5 N · B 2.8 N · C 3.5 N · D 4.7 N

  1. 1

    clockwise moment = 3.5 × 15 = 52.5 N cm

  2. 2

    anticlockwise moment = (W × 20) + (0.50 × 5.0) = 20W + 2.5

  3. 3

    20W + 2.5 = 52.5

  4. 4

    W = 50/20 = 2.5 N: answer A [1]

13. Experiment: balancing a metre ruler

To show that there is no resultant moment when an object is in equilibrium, balance a metre ruler on a pivot. Place the ruler with the 50.0 cm mark exactly above the pivot. Mark the centre of P and align it with the 10.0 cm mark. The distance x is from the centre of P to the 50.0 cm mark.

Place object Q and adjust it until the ruler is as close as possible to being balanced: the position between just tilting one way and then just tilting the other way. Record y, the distance from the centre of Q to the 50.0 cm mark. Repeat for other values of x.

At balance the moments are equal, so a graph of y against x is a straight line. Its gradient G is the ratio of the weights.

W_P x = W_Q y

G = W_P/W_Q

Examiner insight

“Some described moving the ruler backwards and forwards on the pivot but only a small number recognised the technique of achieving a position between the ruler just tilting one way and then just tilting the other way.” “Occasionally, candidates gave the readings of the positions of the masses on the metre rule at balance, and not the distances of the masses from the pivot.”

Worked example 12 marks

A student investigates the balancing of a metre ruler. The student places an object P with its centre on the metre ruler at the 10.0 cm mark.

Explain how to place the centre of object P on the 10.0 cm mark as accurately as possible. Draw a diagram to illustrate your answer.

  1. 1

    mark the centre of P and align it with the 10.0 cm mark on the ruler [1]

  2. 2

    or use the scale readings either side of P and average them to give 10.0 cm

  3. 3

    a diagram correctly illustrating the method [1]

Recap

  • The moment of a force is a measure of its turning effect.
  • Moment = force × perpendicular distance from the pivot.
  • Balanced beam: clockwise moment = anticlockwise moment.
  • Equilibrium: no resultant force and no resultant moment.
  • Extended: add all the moments on each side, including the weight of a uniform beam at its centre.
  • Extended: in the ruler experiment use distances from the pivot, and the gradient of y against x is W_P / W_Q.

14. What centre of gravity means

The centre of gravity is the point through which the weight of an object appears to act. Say weight, not gravity. Say centre of gravity, not centre of mass. On a uniform metre ruler it is at 50 cm.

Examiner insight

“Many referred to gravity rather than weight or the force of gravity.” “Weaker candidates referred to the centre of gravity as being the point where most of the weight acts or the point at which the object balances.”

Worked example 11 mark

What is the name of the point through which the weight of an object appears to act?

A the centre of gravity · B the moment · C the pivot · D the turning point

  1. 1

    A: the centre of gravity. [1]

15. Finding the centre of gravity of a lamina

A plane lamina is a flat sheet. Suspend it from one point so that it hangs freely. The centre of gravity is vertically below the point of suspension, so draw a vertical line down from that point.

Suspending the lamina from just one point is not sufficient. It gives a line, not a position. Suspend it from a second point and draw the vertical line again. The centre of gravity is where the two lines cross.

Examiner insight

“Many candidates selected option D as the position of the centre of mass, not realising that it must be vertically below the point of suspension.”

16. Stability and the position of the centre of gravity

The lower its centre of gravity, the greater the stability of an object. A bigger area of base also makes an object more stable.

An object topples when the vertical line through its centre of gravity falls outside the area of the base. With a lower centre of gravity or a bigger base, a larger tilt is needed before this happens.

When the centre of gravity is high and the base area is small, a small tilt is enough.

Examiner insight

“Only stronger candidates gave the expected answer relating the idea that a small tilt or rotation makes G no longer vertically above the base”

Worked example 12 marks

Figure 3.1 shows two containers, A and B, partly filled with water. Explain why container A is more stable than container B.

  1. 1

    the centre of gravity of A is lower [1]

  2. 2

    the area of the base of A is bigger [1]

  3. 3

    also credited: a larger tilt is needed before the vertical line through the centre of gravity falls outside the area of the base (any two of the three points)

17. Comparing the stability of objects

To compare objects, look at two things: the height of the centre of gravity and the area of the base.

The most stable object has the lowest centre of gravity and the biggest area of base.

When you explain a change, link it to the centre of gravity and say more stable or less stable.

Examiner insight

“Stronger candidates clearly stated that the centre of gravity was lowered and that made the ship more stable. The question asked for the effect on stability so the comparative ‘more stable’ was needed.” “Many candidates gained partial credit for “making the base heavier”, but this was rarely linked to lowering the centre of gravity as the reason for more stability.”

Worked example 11 mark

Four blocks are made of the same metal. They have the same volume and weight. They are placed as shown in the diagrams. The centre of gravity of each block is labelled G. Which block is the most stable?

  1. 1

    C: its centre of gravity is the lowest and the area of its base is the biggest [1]

Recap

  • Centre of gravity: the point through which the weight of an object appears to act.
  • Lamina: draw a vertical line down from two points of suspension. G is where the lines cross.
  • One point of suspension is not sufficient: it gives only a line.
  • The lower its centre of gravity, the greater the stability of an object.
  • A bigger area of base means a larger tilt is needed before the object topples.
  • Say ‘more stable’: the comparison is needed.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. (a) State what a force may change about an object, other than its motion. (b) State the two ways in which a resultant force may change the velocity of an object.3 marks
  2. A box slides across a floor and comes to rest. (a) Name the force that stops the box and state where it acts. (b) State one other effect of this force.3 marks
  3. A spring is 42 mm long with no load. With a load of 3.0 N its length is 66 mm. Calculate the extension of the spring.2 marks
  4. A sledge is pulled forwards along level snow with a force of 85 N. Friction of 30 N and air resistance of 15 N act backwards on the sledge. Calculate the resultant force on the sledge and state its direction.3 marks
  5. A student plots load against extension for a spring. The best-fit line is straight and crosses the load axis at 0.4 N. State and explain whether load and extension are directly proportional.2 marks
  6. A steel ball falls through oil in a straight line at a constant speed. The gravitational force on the ball is 0.60 N. State the name, the size and the direction of the other force on the ball.3 marks
  7. A spring is 8.0 cm long with no load. With a load of 6.0 N its length is 12.0 cm. The limit of proportionality is not reached. Calculate the spring constant in N / m.3 marks
  8. The graph shows an extension–load graph for a spring. (a) State the load at the limit of proportionality. (b) Calculate the spring constant in N / cm.3 marks
  9. A drone has a mass of 2.5 kg and a weight of 24.5 N. Its rotors produce an upward force of 32 N. Calculate the initial acceleration of the drone and state its direction.4 marks
  10. An ice skater of mass 50 kg holds a rope tied to a post. She moves around the post in a circle at constant speed. The rope pulls on her with a force of 120 N, perpendicular to her motion. Friction is negligible. (a) State the direction of her acceleration. (b) Calculate her acceleration. (c) The force from the rope increases. Her speed stays the same. State what happens to the radius of her path. (d) A skater of mass 70 kg replaces her, at the same speed and radius. State and explain how the force from the rope changes.6 marks
  11. State what is meant by the moment of a force.1 mark
  12. A mechanic pushes with a force of 40 N at right angles to the end of a spanner of length 25 cm. Calculate the moment of the force about the nut. Give your answer in N m.3 marks
  13. A moment of 18 N m is needed to turn a nut. A spanner of length 45 cm is used, with the force at right angles to it. Calculate the minimum force needed.3 marks
  14. A light beam is balanced on a pivot at its centre. A weight of 300 N hangs 1.2 m from the pivot on one side. A weight of 450 N hangs on the other side. Calculate the distance of the 450 N weight from the pivot.3 marks
  15. State the two conditions that must be true for an object to be in equilibrium.2 marks
  16. A light metre rule is pivoted at its 50.0 cm mark. A weight of 2.0 N hangs from the 20.0 cm mark. A weight of 3.0 N is hung on the other side so that the rule balances horizontally. At which mark on the rule is the 3.0 N weight hung?3 marks
  17. A light beam is pivoted at its centre. Two downward forces of 6.0 N and 4.0 N act on the left of the pivot, at 0.50 m and 1.0 m from it. A single downward force of 5.0 N acts on the right. Calculate its distance from the pivot when the beam balances.3 marks
  18. A uniform metre rule of weight 1.2 N is pivoted at its 30.0 cm mark. A weight W hangs from the 0 cm end. The rule balances horizontally. Calculate W.4 marks
  19. A car park barrier is a uniform pole of length 4.0 m and weight 60 N. It is pivoted 0.50 m from end A. A counterweight is fixed to the pole with its centre 0.20 m from end A. A sign of weight 15 N hangs from the pole 1.0 m from the other end, B. The pole balances horizontally. (a) Calculate the mass of the counterweight. Use g = 9.8 N/kg. (b) The sign falls off. State which end of the pole moves down, and explain why.6 marks
  20. A student balances a uniform metre ruler of weight 1.0 N on a pivot. The weight of the ruler acts at the 50.0 cm mark. The student puts the pivot under the 40.0 cm mark by mistake. Object P, of weight 2.0 N, has its centre at the 10.0 cm mark. Object Q, of weight 2.5 N, is moved along the other side until the ruler is as close as possible to being balanced. (a) Calculate the scale reading at the centre of Q. (b) The student calculates only the moments of P and Q about the pivot, finds they are not equal, and says there is a resultant moment on the balanced ruler. Use numbers to explain why this is wrong.6 marks
  21. State what is meant by the centre of gravity of an object.1 mark
  22. A uniform metre ruler is balanced horizontally on a pivot at the 20 cm mark. A weight hangs from the 0 cm end. At which mark is the centre of gravity of the ruler?1 mark
  23. A low coffee table and a tall bookcase have the same weight and the same area of base. Which is more stable? Give one reason.2 marks
  24. Describe an experiment to find the position of the centre of gravity of an irregularly shaped lamina.4 marks
  25. A student suspends a lamina from point A and draws a vertical line down from A. She suspends it from point B and draws a second line. The lines cross at X. She now suspends the lamina from a third point C. Where does the vertical line down from C pass? Explain.2 marks
  26. Three vases have the same weight. Each has a circular base. Which vase is the most stable? Explain your answer. | vase | height of centre of gravity / cm | diameter of base / cm | | --- | --- | --- | | P | 12 | 8 | | Q | 8 | 8 | | R | 8 | 5 |2 marks
  27. A tall block stands on a table. A student tilts it a little about one bottom edge and lets go. It does not topple. She tilts it further and it topples. Explain both results.3 marks
  28. A roadside sign stands on a flat base. It falls over in a strong wind. Suggest two changes to the sign that make it more stable, and explain how each change works.4 marks
  29. A student suspends an irregularly shaped lamina from point P and draws a vertical line down from P. She then suspends it from point Q and draws a vertical line down from Q. Point Q is on the first line, near the opposite edge of the lamina. (a) Explain why her second line lies along her first line. (b) Explain why she cannot yet find the centre of gravity, and state what she should do next.4 marks
  30. A van stands on a platform that tilts sideways. The platform is tilted slowly until the van starts to topple, and the angle of tilt is recorded. The van carries the same boxes in each test, so its weight is the same. The three angles recorded are 31°, 38° and 44°. State which angle belongs to each test. Explain your answer. | test | heavy boxes | left and right wheels | | --- | --- | --- | | 1 | on the floor, under the light boxes | normal distance apart | | 2 | on top of the light boxes | normal distance apart | | 3 | on the floor, under the light boxes | further apart |5 marks

Go deeper

Practise and revise with member-only material for this chapter.

Workbook

Exam-style practice questions with full worked solutions.

Unlock with a plan

Chapter at a Glance

A single-page summary of every key idea for fast revision.

Unlock with a plan

Free notes are just the start.

Unlock every Workbook and Chapter at a Glance, and generate your own worksheets and predicted papers.

Explore plans

Related chapters

Physical quantities and measurement techniquesMotionMass and weightDensityMomentumEnergy, work and power