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Homework Help: Accuracy of experimental values

  1. Oct 5, 2006 #1
    We did an experiment with abysmal accuracy. I'm now writing up my report and would like to meaningfully express the error between calculated and measured values.


    calculated T=0.284
    measured T = 0.201

    I want to express the difference between them as follows:

    % difference between the two = (1 - 0.201/0.284)(100%) = 29%

    But as ridiculous as the errors are throughout the experiment, they're not that bad!

    Take the difference between the two values and then divide by one of the two values? But which one?

    I know how to take percent error for something like [itex]107.45\pm\0.08[/itex] but this seems to be a different cup of tea.
    Last edited: Oct 5, 2006
  2. jcsd
  3. Oct 6, 2006 #2


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    Your 29% error value is correct. I assume that the calculation was also done with the measurements? It is impossible to comment on experiments, even if you tell us exactly what you did. Experiments are a very personal thing. We can only offer suggestions but even then may be missing something very obvious. So it is best to fool around with an experiment and try and push it in all directions in order to gain confidence in the result.
  4. Oct 6, 2006 #3
    The experiment was a basic one about Newton's second law:

    a dynamics cart was placed on a track, a mass was attatched to a string that ran from the cart over a pulley toward the floor.

    T is the tension in the string connecting the cart and the suspended mass.

    The calculation was [itex] T = \frac{(m_1)(m_2)}{m_1+m_2}g[/itex]
    and the measured value of T was obtained using a force sensor hooked up to a Pasco interface.

    So assuming the mass was decently accurate (and this balance hasn't created crazy abberations in other experiments), the calculated T should be OK.

    In the brief pre-lab, the professor asked us if we all believed that F=ma, and when we said yes, he replied that we wouldn't so much after this experiment. But 30%?!?! Yow.
  5. Oct 6, 2006 #4


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    Are you sure that the measured tension is the smaller one and not the other way round?
    Last edited: Oct 6, 2006
  6. Oct 9, 2006 #5
    In 6 out of 7 trials, the measured tension was indeed smaller than the calculated tension.

    Not only that, but measurments say that if there's a heavier cart and a lighter cart with the same suspended mass, the string attatched to the heavier cart will have a considerably smaller tension than the lighter cart--the exact opposite of what the calculations predict!

    I'm hoping for a really good post-lab to explain why we spent two hours on an experiment that the professor knew would give such screwy results.
  7. Oct 9, 2006 #6


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    Well according to my analysis the tension in the string should increase from a case where no rolling friction, f, is acting on the cart:

    [tex]T_o = \frac{m_1 m_2 g}{m_1 + m_2}[/tex]

    to a case where the cart do experience rolling friction:

    [tex]T_f = f(1 - \frac{m_1}{m_1 + m_2}) + T_o[/tex]

    which is not what you are reporting.

    The heavier cart will experience a larger rolling frictional force than the lighter one.

    Did you perhaps notice if the maximum force that the force sensor could measure was 50 N?
  8. Oct 9, 2006 #7


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    Your tensions look very small to me. They are approximately the weight of only 20 grams!
  9. Oct 9, 2006 #8
    :confused: Could you explain when you're talking about what a little more?

    49 N, actually :wink: . Why would that make a difference, if the largest measured tension was 0.267 N and the largest calculated tension was 0.4729 N?

    And you're on the money: 0.201 N was the measured tension when the suspended mass was 30g.

    Sorry, I really should draw up a table and attatch it... Do you want me to do that? At this point, I don't think I need enough help to make it worth it, but if you're interested in a screwy experiment, I'll be happy to share.
    Last edited: Oct 9, 2006
  10. Oct 9, 2006 #9
    And andrevdh, please pardon my manners (or lack thereof).
    Thank you very much for your help.
  11. Oct 11, 2006 #10


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    Both unexpected observations can be explained if the pulley is generating a bit of friction (taken that you did the experiment with quite small masses).

    To come back to the force sensor. If its maximum capability is 50 N (5 kg) one can expect that it will not be that accurate at 20 gram (0.020 kg). So any measurements at such small tensions may have quite significant errors (uncertainties). This means that the error in the measured value may be so large that it includes the calculated value as a possible outcome for the experiment.

    As far as friction on the pulley goes consider this: Friction in the pulley will increase the tension on the "input side" (masshanger pulling on the pulley) and decrease it on the "output side" (force sensor) of the pulley. This can account for the difference between the measured and calculated tension.

    A decrease in tension with a heavier cart can be explained as follows. The pulley is most likely to generate larger friction as lower rotational speeds - that is a heavier cart will experience a lower acceleration and therefore measure a lower tension than that of a lighter cart.

    You can test this hypothesis by using two force sensors simultaneously. Suspend another force sensor from the string and run the cart (without the mass hanger since the mass of the force sensor will be sufficient). Then compare the two tensions to see whether the input tension is more than the output tension.
    Last edited: Oct 11, 2006
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