Physics
Double Award
Double Award
Cambridge IGCSE
TOPIC 1A: FORCES and MOTION
Have a careful look at this YouTube video - watch the central span of the bridge as the wind blows across the Golden Gate Bridge in San Francisco:
Youtube: wtkphoto 2011
It's easy to miss - but the central section is moving! The whole road is moving up and down with the varying load driving across, and also the force produced by the wind.
The reason this happens is that this bridge is a suspension bridge. The road surface is supported by long steel cables. However these steel cables stretch under any load. The engineers who designed the bridge need to know in advance how the cables will behave, and this is where Hooke's Law comes in.
Sometimes even the best engineers make mistakes - it's really important to get it right! Here is the famous Tacoma Narrows bridge caught in a strong wind:
Youtube: Tacoma Narrows bridge collapse 1940
Robert Hooke was a scientist who first recorded the results he deduced for how things stretch. This basic principle, or 'law' on how cables and springs stretch is best understood by carrying out a practical investigation. In fact, this is a standard practical that you should know how to perform:
Method:
Improvements:
This experiment is quite simple, but results can be improved by repeat readings, using a clamp on a ruler to ensure it does not move, attaching a light horizontal marker to the weights like a long pin (to make it easy to measure the extension), avoiding parallax etc.
Goggles are an essential safety feature here as the springs can damage your eyes if released quickly.
Results
Here is a table showing typical (but simplified) values:
| Weight or Force , F (N) |
Total Length of Spring, L (cm) |
Extension, x (cm) |
|---|---|---|
| 0 | 6 | 0 |
| 1 | 8 | 2 |
| 2 | 10 | 4 |
| 3 | 12 | 6 |
Table 1: Typical Hooke's Law results
If you look at the results table, it is clear that the extension increases in uniform steps as the load is increased, and this can also be seen by the illustration in figure 1.
Conclusion
Hooke realised that the extension (NOT the length) was proportional to the weight or the force applied. This is true for springs and wires, but not for rubber bands.
(See below).
For the results for the spring above, the numerical value for the force applied (N) is half the extension (cm). We say that this spring has a spring constant k of 0.5 N/cm. We could convert this into the force required to stretch 1m, which according to the maths would be 50 N/m.
The spring constant k is defined as the force per unit extension, and we can use this Hooke's Law formula to calculate k:
| Spring constant (N/m or N/cm)= | Force applied (N) |
| extension (m or cm) |
| k = | F |
| x |
This law and formula means that engineers can predict the behaviour of wires, springs and many other materials, before building a structure.
Hooke's Law is only true for certain materials, and also as long as the force is not too high. As the force increases, springs and wires 'over-stretch' and after this point the extension is no longer proportional to the force. This value is known as the 'limit of proportionality' as shown on the graph below, figure 2.
If a material obeys Hooke's law, then the line on the force -extension graph will be straight. This linear region is the easiest way to identify where a substance is obeying Hooke's Law.
Figure 2: Hooke's Law graphs for springs, metal wires and rubber bands.
As can be seen from the above graph for the rubber band, there is no linear region. Therefore rubber bands do not obey Hooke's law. However a rubber band still demonstrates what is known as 'elastic behaviour' - shown by the ability to return to its original shape after the force has been removed. Try these questions to check you understand the law:
Take care: Some books and websites show this graph with the axes reversed. However a linear region still shows that Hooke's Law is obeyed, whichever way round the axes are given.Questions:
1. A wire of length 50 cm stretches to 70 cm under a load of 40 N.
a) We know that k = F/x.
The extension in this case is 70 - 50 = 20cm.
The force is 40N, therefore:
k= F/x = 40/20 = 2 N/cm
b) We need to find the extension before we find the total length.
Rearranging k = F/x gives x = F/k.
For 60 N force, the extension will be 60/2 = 30 cm.
However the original length of the spring was 50 cm, so the total length will be 50 + 30 = 80 cm.
2. The spring used to obtain the results as shown in table 1 is tested again. The spring constant k = 0.5 N /cm. Calculate the following values:
a) Rearranging k = F/x gives x = F/k.
Therefore 5 N will produces an extension x = 5/0.5 = 10cm extension
b) Using x = F/k and F = 6 N gives x = 6/0.5 = 12 cm extension.
However the original length of the spring was 6 cm, so the total length will be 6+12 = 18 cm.
3. How is Hooke's Law identified on a graph of the load on a spring against the spring's extension?
4. What do we mean by elastic behaviour?
Additional questions are available in these two end of unit tests - these cover sections 1.1 to 1.7.
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