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FORCES SPRINGS This section of work is also known as Hookes Law.

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Presentation on theme: "FORCES SPRINGS This section of work is also known as Hookes Law."— Presentation transcript:

1 FORCES SPRINGS This section of work is also known as Hookes Law.

2 When a force is exerted on a spring it will either compress (push the spring together) or stretch the spring if the weight is hung on it.

3 Some objects like bridges will also behave like springs. When a weight is placed on a bridge parts will be stretched and under tension, other parts will be be squashed together or compressed

4 In compression the particles making up the material are pushed closer together When it is under tension the particles making the material are pulled further apart

5 In this experiment the extension of the spring is measured with increasing weights being added 1N, 2N, 3N to 5N. The extension is a measure of how much the spring stretched. (Not the total length of the spring)

6 The results of the Extension are plotted against the Force added. The spring stretches in proportion to the mass added but at 5N the spring is stretched beyond its elastic limit and becomes deformed.

7 Because springs stretch proportionally we can use them as a spring balance to measure a force.

8 The extension will double if the force doubles. It will treble if the force trebles. If the load exceeds the elastic limit of the spring then the law no longer applies and the spring will not return back to its original length.

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10 When the weight and the upward, restoring force are equal the spring is said to be in equilibrium

11 Read the questions carefully and answer these questions on paper. When you have finished answering them then go to the next slides to check if you got the answers correct.

12 Answer to Question A & B In the first experiment the student tested three different diameter springs made with the same thickness of wire. To make it a fair test the springs would have to be the same length. 3N, 5N, 7N, 9N & 11N were added to each spring and the extension was recorded for each weight. Conclusion = the greater the diameter of the spring the greater the extension. For every 0.5cm increase in diameter the spring extends an extra 1cm.

13 In the Second experiment the pupil kept the diameter and the length of the spring constant but changed the thickness of the wire used for each spring. Conclusions – the thicker the wire used the smaller the extension. If you double the thickness of the wire you halve the extension. This is because the restoring force from the thicker wire will be stronger so it will be more difficult to stretch it.

14 Answer to C To make the springs less soft you need to make the spring with thicker wire and have as small a diameter as possible for the spring.

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17 Results of Experiments Mass in gFinal length of spring cm Original length cm Extension of spring in cm. 100 200 300 400

18 Graph showing extension in series

19 SPRINGS IN SERIES Each spring extends as a result of the force. If there are identical springs the extension will be double that of a single spring. 5cm Total 10 cm extension

20 SPRINGS IN PARALLEL In parallel the load (weight) is shared by the springs. So if you have 2 identical springs the extension will halve. 5cm 2.5cm

21 Question 8 calculate the length for each if 1kg stretches the spring 5cm

22 Answers to Q9 The force acting on the ceiling is 10N ( It may be a little bit more if you include the weight of the spring as well.) Downwards The spring would stretch less because the force of gravity on the moon is 1/6 th that of Earth so the weight acting would be less.

23 Answers question 8 A = 5cm B = 10 cm C = 2.5 cm D = 5cm So B stretches the most.

24 SUMMARY When weights are added to a spring it stretches proportionally. When 2 springs are connected in series each spring will stretch proportionally so the extension will be double that of a single spring with the same load. When two springs are in parallel the load is shared so the extension will be the same as half the weight on a single spring. When the load is too great the spring will be stretched beyond its elastic limit and will not go back to its original length. We find the extension by subtracting the original length of the spring from the total length with the load on.

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