Can we use a spring to accurately measure force?
18/10/2011 § 2 Comments
- Independent variable: The independent variable is the weight (N) of the masses.
- Dependent variable: The dependent variable is the length (cm) that the spring stretches to.
- Controlled variables: used the same spring & the way we measured with the ruler (top of ruler against the scale and as perpendicular as possible)
Summary of Method
Alisa and I would accomplish each step quite quickly and the only trouble we would have was deciding which centimeter on the ruler lined up with the arrow that hung from the spring. I steadied the ruler against the scale and the desk and tried to make it stand as perpendicular to the floor and as parallel to the scale as possible.Alisa was in charge of adding and removing the weights for each stage of the experiment. We would both switch occasionally to stop the weight and spring from swinging too much. Both Alisa and I would work together to see the closest centimeter on the ruler that the weight would fall next to. I would record the results on the computer. Everything was simple and very straightforward.
I noticed that with every weight we added, the spring would stretch about 3 to 4 cm longer. From weights 0N to 2.5N, the average length the spring would stretch was about 3 cm but not 4 cm. It isn’t till 3.0N that the spring stretches an extra 4 centimeters. Our results also showed that between the three trials, the extension lengths weren’t exactly the same at each point but with a range of 0.1 – 0.3 centimeters (a range of 0.5cm for 0.5N).
The graph above is a visual representation of the results from our experiment. One can see that the line graph that was produced from our experiment’s results is not constant but follows a general pattern.
The trend is seen here; as I explained earlier, the change in length with every 0.5N added was a constant change, between 3cm to 4cm in change. Because the changes in length between each added 0.5N weight were pretty much the same and didn’t have that big of a difference, the slope of the line is still quite straight.
Firstly, the error bars on the graph are there but they are so small that they simply cannot be seen. If you take a look at the graph and the results table image again, the title of the graph states that the error bars are ½ the range of the lengths between the three trials. The ½ range column of the results table show that the highest ½ range is 0.25cm, which is still less than 1cm, less than even half a centimeter, therefore it would not be seen on a graph where the y-axis (the cm axis) has a scale of 2 cm.
Our results are valid. There aren’t many reasons or causes in the designed experiment to allow for many mistakes or possible errors to occur. It was too simple for that. However, my partner and I were still able to do the experiment and receive the most accurate results possible. The information seen in the visual graph and the results table are quite reliable because my partner and I did the exact same things for each of the fifteen measurements. We used the same spring, the same side of the ruler and the same method of measuring where spring with the ruler. We hardly had any reason to change our method, which makes our measurements consistent and therefore makes our data reliable.
The relationship seen in the graph is that the extension of the spring depended on how much the mass weighed. We can see that the more the mass weighed, the longer the spring would extend. If the mass weighed less (like 0.5N or 1.0N), the spring wouldn’t stretch as far as if the mass were 2.5N or 3.0N, where it stretched to an average of 34.4cm.
Conclusion & Mystery Weight
My group’s mystery weight was #3 and our calculated weight was 1.25. The true weight was actually 1.27. We found this by using the line graph that we already created. The red line forming a box at the bottom left corner of the graph represents our estimated guess of how much the mystery mass weighed. The mystery mass stretched the spring to 22.0cm. By drawing the red line, we could estimate that the mass’s weight was about 1.25N (the line in between 1.0N and 1.5N is 1.25N). The true weight was 1.27 but we weren’t far away, only 0.02N off. We found the %error through a few calculations, shown below.
- The equation: (calculated weight – true weight) / true weight x 100%
- (1.25 – 1.27) = -0.02
- -0.02 / 1.27 = -0.015748
- -0.0157… x 100% = 1.57%
This makes sense because if we recall the error bars discussed earlier, they couldn’t even be seen because they were so small. We were only 0.02N away from the actual weight of the #3, and also, if we look at the numbers, 0.02 looks to be about one to two percent of 1.57 (notice position of the decimals, 0.02 and 1.57 could look like .002 and .157, but that’s just a little observation).
To answer the original question, yes, we can measure the force of an object using a scale. When force is applied to scales, the scales respond with elastic force – they want to return to their original shape. The amount of change in its shape will be equal to the force applied to it, and since we measured the change in the spring’s shape (the centimeters of extension), we were able to calculate the force (How a spring scale is used, 2011).
For future investigations similar to this one, perhaps using scales that are in better shape would make measuring easier. Our scale was a little wobbly which made it hard for my group to align the ruler correctly. Also, if instead of a straight ruler, we could use something flexible, like measuring tape.
Self-Assessment, the MYP Way
I took quite some time to write this blog up and make sure everything was clear and sound. I included everything that was required but also tried to maintain a format that was straightforward and precise. Nothing is confusing, in my opinion. Everything follows a normal lab write-up format, even with the References at the bottom. My graphs are fine, my results table is quite colorful, and I tried to analyze the data as best as I could using scientific terminology such as ‘elastic force’, ‘force’, ‘respond’, ‘apply’, etc.