Hooke's Law, Moments and Centre of Mass
Hooke's law states that the extension of a spring is directly proportional to the force applied, up to its limit of proportionality, given by force = spring constant x extension (F = k x e). A moment is the turning effect of a force about a pivot, calculated as moment = force x perpendicular distance from the pivot, and the principle of moments states that for an object in equilibrium the sum of clockwise moments equals the sum of anticlockwise moments. The centre of mass is the single point at which the entire weight of an object can be considered to act, and an object is stable when a vertical line from its centre of mass falls within its base. The extended IGCSE course requires quantitative use of the principle of moments, including finding unknown forces or distances, and analysis of the elastic potential energy stored in a stretched spring.
Before you start
Make sure you're comfortable with these topics first:
Method
- For spring problems, use F = k x e, where extension e is the increase in length of the spring beyond its natural (unstretched) length, keeping length in metres or a consistent unit throughout.
- Check whether the spring is still within its limit of proportionality; beyond this limit, extension is no longer proportional to force, and F = k x e no longer applies.
- Calculate elastic potential energy stored in a stretched spring using elastic potential energy = 0.5 x k x e^2, provided the limit of proportionality has not been exceeded.
- For moments problems, identify the pivot and every force acting on the object, and calculate each moment using moment = force x perpendicular distance from the pivot.
- Apply the principle of moments: for an object in equilibrium, sum of clockwise moments = sum of anticlockwise moments; set up this equation and solve for the unknown force or distance.
- For stability questions, find the centre of mass (for a symmetrical, uniform object this is at its geometric centre) and consider the size of its base: a low centre of mass and a wide base give a more stable object, since the vertical line from the centre of mass is less likely to fall outside the base when the object is tilted.
Worked example
A uniform beam of length 4.0 m is balanced on a pivot at its centre. A force of 30 N acts downward at a point 1.5 m to the left of the pivot. Calculate the force that must act 2.0 m to the right of the pivot to keep the beam in equilibrium.
- Calculate the anticlockwise moment produced by the 30 N force: moment = force x distance = 30 x 1.5.
- 30 x 1.5 = 45 N m.
- By the principle of moments, for equilibrium the clockwise moment must equal this anticlockwise moment, so the clockwise moment = 45 N m.
- Write the equation for the unknown clockwise moment: 45 = F x 2.0, where F is the unknown force at 2.0 m from the pivot.
- Rearrange to find F: F = 45 / 2.0.
- State the final answer with its unit: F = 22.5 N.
Practice questions
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Q1State Hooke's law.Show answer
Answer: The extension of a spring is directly proportional to the force applied, up to the limit of proportionality (force = spring constant x extension).
Q2A spring has a spring constant of 40 N/m. Calculate the force needed to extend it by 0.050 m.Show answer
Answer: 2.0 N (40 x 0.050).
Q3Calculate the moment of a force of 12 N acting 0.40 m from a pivot.Show answer
Answer: 4.8 N m (12 x 0.40).
Q4State the principle of moments.Show answer
Answer: For an object in equilibrium, the sum of the clockwise moments about a pivot equals the sum of the anticlockwise moments.
Q5State what is meant by the centre of mass of an object.Show answer
Answer: The single point at which the entire weight of the object can be considered to act.
Q6A spring with spring constant 25 N/m is extended by 0.20 m within its limit of proportionality. Calculate the elastic potential energy stored. Use elastic potential energy = 0.5 x k x e^2.Show answer
Answer: 0.50 J (0.5 x 25 x 0.20^2 = 0.5 x 25 x 0.04).
Q7State why a Formula 1 car is designed with a low centre of mass and a wide wheelbase.Show answer
Answer: To make it more stable, since the vertical line from its centre of mass is less likely to fall outside its base when it corners, reducing the risk of toppling.
Exam-style questions
Written in the style of a IGCSE Science exam paper, with a full mark scheme.
A spring has an unstretched length of 0.12 m. When a force of 6.0 N is applied, its length becomes 0.16 m. (a) Calculate the extension of the spring. (1 mark) (b) Calculate the spring constant of the spring. Use F = k x e. (3 marks)
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A uniform see-saw of length 5.0 m is pivoted at its centre. A child of weight 300 N sits 2.0 m from the pivot on the left side. A second child sits on the right side, 2.5 m from the pivot, and the see-saw is balanced. (a) Calculate the weight of the second child. (4 marks) (b) The first child then moves to sit only 1.0 m from the pivot. Explain, using the idea of moments, why the see-saw is no longer balanced, and calculate the size of the extra downward force needed at 2.5 m from the pivot on the left side to rebalance it. (2 marks)
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Free printable worksheet
Want more practice on paper? Download the hooke's law, moments and centre of mass worksheet pack - 6 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
This topic is chapter 24 of IGCSE Science Workbook, the whole course as one free printable PDF.
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