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Practical - Finding the value of gravity

  1. Feb 28, 2014 #1
    1. We had to set up an experiment that looked like the attached diagram. By adding different masses (independent variable) we would measure angle theta (dependent variable). Our task was to draw an appropriate graph and from the gradient, see what our gravity value came to.




    So far i have had no luck and what I am stuck on is the variables being plotted on the graph, as my first graph attempt didn't produce a value close to 9.81m/s^2 or 10N/kg. All I'm asking is if I can have a bit of direction with the variables I am supposed to display on my graph to find the correct value of gravity.
     

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  3. Mar 1, 2014 #2

    ChrisVer

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    how much did it produce?
     
  4. Mar 1, 2014 #3
    The angle does not depend on ##g##.
     
  5. Mar 2, 2014 #4
    I've tried various ways to produce the value of gravity (guess and check method - not ideal but it may help me realize), therefore I have achieved numerous results. The closest I got was a magnitude of 7.
     
  6. Mar 2, 2014 #5

    Redbelly98

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    After looking at the problem more carefully, I agree with voko. There is no way to get the value of g from measuring the angle in this setup.

    It might have been possible to determine g if the blocks were moving (say the middle mass falling, and the 200 g masses [STRIKE]falling[/STRIKE] rising), and then the time taken for one of the blocks to fall a certain distance were measured. But this setup does not seem to allow for that sort of measurement, and you would probably have wanted the angle to be zero -- i.e., all the strings vertical.

    (EDIT: made correction as indicated above.)
     
    Last edited: Mar 3, 2014
  7. Mar 2, 2014 #6

    PhanthomJay

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    Since g cannot be determined from the given setup and information, how did you arrive at any number for g?? For a given hanging mass, theta will be the same whether the experiment is conducted on Earth , Jupiter,, or the Moon.
     
  8. Mar 3, 2014 #7
    Yeah I realised the problem. The reason why I still attempted was because I was intrigued why it was so difficult. Anyway, long story short, we were not told until today that a mistake had been made, and we were meant to assume we already knew the tension force of 2N for both strings. From that we can graph cos(theta/2) against mass.
     
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