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Resolving Forces

  1. Mar 3, 2009 #1
    Hi,

    Attached is a link to an image:

    http://i423.photobucket.com/albums/pp315/skaboy607/Image.jpg [Broken]

    Probably a very easy question but I cant work out how they get the vertical and horizontal reaction forces to be F/cos(fi) and Ftan(fi). Any help would be most appreciated.

    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 4, 2009 #2

    tiny-tim

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    Hi skaboy607! :smile:

    (have a phi: φ :wink:)

    I assume there's no friction at the top, so taking vertical components (assuming there's no acceleration) should give you an almost-vertical force of F/cosφ :wink:
     
  4. Mar 4, 2009 #3
    Hi,

    Thanks for the phi φ!

    When I take vertical forces I get Fcosφ. Not F/cosφ . And I have no idea where the Ftanφ comes from.

    Thanks for your help.
     
  5. Mar 4, 2009 #4

    tiny-tim

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    No … if the unknown force is G, vertical components give you F = Gcosφ :wink:
     
  6. Mar 4, 2009 #5
    oooooooooohhhhhhhhhhhh. Thanks! Just confused me because they used the same F. Any ideas on the Ftanφ.

    Thanks
     
  7. Mar 4, 2009 #6

    tiny-tim

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    erm … I was going to ask you that! :wink:

    (probably got something to do with resolving horizontally :rolleyes:)

    (btw, is the bottom fixed?)
     
  8. Mar 4, 2009 #7
    oh I don't know, it is the horizontal force but how they got to that i dont know. i'm thinking along the lines sin/cos=tan?

    Yea point Q doesnt move.
     
  9. Mar 4, 2009 #8

    tiny-tim

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    Hint: what is the horizontal component of the F/cosφ force?
     
  10. Mar 4, 2009 #9
    Sorted-think ive got it. Horizontal component of F/cosφ force (F/cosφ)(sinφ) which is equal to Ftanφ!
     
  11. Mar 4, 2009 #10

    tiny-tim

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    :biggrin: Woohoo! :biggrin:
     
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