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Work from a constant force

  1. Feb 16, 2008 #1
    [SOLVED] work from a constant force

    1. The problem statement, all variables and given/known data

    [​IMG]


    i cant figure out the answer to hint 2


    2. Relevant equations

    w = f deltar

    w = F cos theta (deltar)


    is this img right?

    [​IMG]
     
    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2

    Doc Al

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    Your diagram looks good to me. Now how will you make use of it to find the dot product [itex]\vec{F}\cdot\vec{L}[/itex]?
     
  4. Feb 16, 2008 #3
    thats what i have been trying to figure out


    i have tried to relate angles and got

    sin (theta*PI/180)* tan (theta*PI/180)
     
  5. Feb 16, 2008 #4

    Doc Al

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  6. Feb 16, 2008 #5
    well the only reason i didnt say it was

    cos ( theta * PI /180)

    nor it is like cos ( (theta - 180) * PI /180)

    because that is the wrong answer so i dont know wth they want..

    and yes thats in my text i have been looking through it the whole time, not examples seem to be like this nor does my book explain this well at all
     
  7. Feb 16, 2008 #6

    Doc Al

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    [tex]W = F L \cos \phi[/tex]

    Express this in terms of [itex]\theta[/itex].
     
  8. Feb 16, 2008 #7
    I know how to set it up thats is not what i am trying to figure out,

    im trying to express PHI in terms of THETA, u see my IMG where it says FIND THIS i cant find that angle, i know how my FINAL answer is gonna be expressed.

    i cant figure out the how to answer the HINT 2: question


    it will be cos ( of some stuff in here * pi /180)
     
  9. Feb 16, 2008 #8

    Doc Al

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    Hint: [itex]\phi + \theta =[/itex] ?

    (Note: The angles are already in radians.)
     
  10. Feb 16, 2008 #9
    180..

    i have tried


    cos ( (180 - theta) * (PI /180))


    still wrong
     
  11. Feb 16, 2008 #10

    Doc Al

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    The angles are in radians, so no need to convert to radians:

    [tex]\phi + \theta = \pi[/tex]

    Now find [itex]\cos\phi = \cos(\pi - \theta)[/itex]

    (Review your trig identities if you need to.)
     
  12. Feb 16, 2008 #11
    in rads it would = PI..
     
  13. Feb 16, 2008 #12
    phi = pi - theta


    i have never seen that before in my life..
     
  14. Feb 16, 2008 #13
    thanks for your help
     
  15. Feb 16, 2008 #14

    Doc Al

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    So.... what's your final answer?
     
  16. Feb 16, 2008 #15
    L*F*cos(PI-THETA)

    what threw me off is how they switch between rads and degrees and they are liek dont forget to convert even whe nyou dont need to convert... and whe nyou haev to convert they dotn tell you to convert and when u submit the wrong answer they say, oh ya dont forget to convert.. so i assumed it was in degrees and well it was in rads go figure, threw off my answer big time, and when i went to submit my answer in the hint they told me to "not forget to switch to radians" screwey system
     
    Last edited: Feb 16, 2008
  17. Feb 16, 2008 #16

    Doc Al

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    OK, but you can simplify it a bit further using a trig identity. (And get rid of that [itex]\pi[/itex].)
     
  18. Feb 16, 2008 #17
    -cos(theta)
     
  19. Feb 16, 2008 #18

    Doc Al

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    Exactly! :cool:
     
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