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Dimensional Analysis: Inverse Cosine

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data

    For the following dimensional equation, find the base dimensions of the parameter f:

    M M-3 = a cos( f L )


    2. Relevant equations

    M represents mass, a represents acceleration due to gravity, in terms of mass * length over seconds squared [[M * L]/[t2]] where L represents length and t represents time.

    For example, solving for k in the equation:

    ML2 = k L t M2

    results in k = L M-1 t-1


    3. The attempt at a solution

    The answer, which is given, is L-1,

    So I got

    cos-1(M-2 a-1) = f L

    the output of the cos-1 function results in a dimensionless unit of measure (radian/degree).

    Therefore,

    c = f L

    where c is a constant/dimensionless quantity,

    therefore f = L -1

    The problem is that I cannot understand how the inverse cosine function works in this sense. Can its input be "dimensionally inequivalent", I cant find the word for it.
     
    Last edited: Aug 27, 2014
  2. jcsd
  3. Aug 27, 2014 #2

    berkeman

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    Why are you taking the inverse cosine? It looks like you are given a simple cos() equation to analyze...

    BTW, it looks like there is a typo in your latex for this line: "results in k = L M-1 t" -- the t appears in the exponent in your equation instead of in a denominator...
     
  4. Aug 27, 2014 #3
    Correct, I will edit that. And I didnt use latex btw, I used what and is called.

    I am using inverse cosine so I can "remove" f and L from inside the parenthesis. I didnt think of another way.
     
  5. Aug 27, 2014 #4

    berkeman

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    No need to remove or simplify anything in this case. What are the units for the argument for any trig function?
     
  6. Aug 27, 2014 #5
    Degrees/radians?
     
  7. Aug 27, 2014 #6

    berkeman

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    Well, good thought, but both of those are dimensionless. There are no degrees or radians in the MKS or cgs systems of units, I believe.

    Think about what a trig function is defined as in terms of triangles. cos() = what over what?
     
  8. Aug 27, 2014 #7
    A length over a length--in essence, a ratio.
     
  9. Aug 27, 2014 #8

    berkeman

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    correct-a-mundo! Which is why the argument to trig functions is dimensionless. Makes sense?
     
  10. Aug 27, 2014 #9
    Yes absolutely. BUT (and there's a big but lol)

    In this question, cos() = M-2 a-1.

    The right side of the equation is not a ratio... it has units.. ):

    Btw, M-2 a-1 = M-2 M-1 L-1 t2 = M-3 L-1 t2
     
    Last edited: Aug 27, 2014
  11. Aug 27, 2014 #10

    berkeman

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    No the right side cannot have units, because cos() does not have units. So what do the units of a have to be?

    And the problem had asked you for the units of f, which only depend on what else is in the argument for the cos() funstion.
     
  12. Aug 27, 2014 #11

    berkeman

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    Actually now I see they are saying a is acceleration, so the original equation does not look dimensionally correct to me... Is that the actual question?
     
  13. Aug 27, 2014 #12
    AHHH THIS IS IT! After looking at my question, variables with unknown units are italicized.

    I thought a was acceleration! Thanks very much!


    I guess we are stuck again. Ill try to take a picture and post it. Is that allowed?
     
  14. Aug 27, 2014 #13

    berkeman

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    Ah, sweet! :smile:
     
  15. Aug 27, 2014 #14
    I think we are stuck again... this was my original problem. I will take a picture and host it on this site.
     
  16. Aug 27, 2014 #15
    Theres the equation see attached
     

    Attached Files:

  17. Aug 27, 2014 #16

    berkeman

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    No attachment yet...
     
  18. Aug 27, 2014 #17
    Sorry I forgot to press upload, its there now
     
  19. Aug 27, 2014 #18
    I just want to add, I want to ask my prof. but I'm nervous, I'm a freshman, and I dont know when would be appropriate, since class time = lecture time, not homework help time... Is that what office hours are for?
     
  20. Aug 27, 2014 #19

    berkeman

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    Yes, office hours, or if there is a TA for the class, you can ask them during their office hours.

    Can you zoom out on the pic to show the whole problem statement? I need to leave work now, but will check in later tonight from home to see if I can help any more with the question.
     
  21. Aug 27, 2014 #20
    updated attachment
     

    Attached Files:

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