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Tension Force - accelerometer

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data

    This is a solved problem. It is attached. Why is sin theta horizontal, and cosine theta vertical?
    2. Relevant equations

    All relevant equations are given.
    3. The attempt at a solution
    The solution is given.
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2015 #2

    billy_joule

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    You'll have to be more specific.
    The short answer is: Because trigonometry says so.
     
  4. Oct 13, 2015 #3
    Its been a while since I have seen trig. Can you explain how I should look at it? It says theta is relative to the vetical.
     
  5. Oct 13, 2015 #4

    billy_joule

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    Are you familiar with SOH CAH TOA?
    Can you draw a labelled right angled triangle with FT as the hypotenuse?
     
  6. Oct 13, 2015 #5
    Yes remember those 3 trig functions. I can draw a right triangle with the hypotenuse labeled Tension force. Where are you going with this?
     
  7. Oct 13, 2015 #6

    billy_joule

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    I'm trying to ascertain what exactly you are having trouble with. Do you understand vector decomposition? Do you understand why you need to do it in this case? Can you use trig to find lengths of a right triangle sides?

    'draw a labelled right angled triangle' = include all information you have. Orientation relative to x & y axis, theta, Label the other sides descriptively ie FT,X FT, Y. Using SOH CAH TOA, what are the lengths of the opposite and adjacent sides?
     
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