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Finding angle of inclination. Hanging object.

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I have to find the angles from horizontal of AB=30cm and BC=40cm.
    http://peecee.dk/upload/view/356001
    I don't know how to tackle this one.

    I was told that i could see them as lever arms, and find out at which angle they cancel eachother out. I'm not sure how to do that.

    Thanks you in advance.
     
  2. jcsd
  3. Mar 11, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi EVriderDK! Welcome to PF! :wink:

    Draw the forces (the two weights), and take moments about B …

    what do you get? :smile:
     
  4. Mar 11, 2012 #3
    Thank you :)

    I could do that, if it was a straight line, but how to do it when it is bend in 90 degrees?
     
  5. Mar 11, 2012 #4

    tiny-tim

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    Have you done moments of forces?
     
  6. Mar 11, 2012 #5
  7. Mar 11, 2012 #6

    tiny-tim

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    But do you know how to calculate the moment of a general force about a general point? :confused:
     
  8. Mar 11, 2012 #7
    Yes. It is just Force/length ?
     
  9. Mar 11, 2012 #8

    tiny-tim

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    Noooo :redface:

    You need to look it up in your book.

    (and i'm off to bed :zzz:)
     
  10. Mar 11, 2012 #9
    Woops meant F*r :D
     
  11. Mar 12, 2012 #10

    tiny-tim

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    what is r ?
     
  12. Mar 12, 2012 #11
    Radius or length
     
  13. Mar 12, 2012 #12

    tiny-tim

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    of what?

    from what to what?
     
  14. Mar 12, 2012 #13
    From the "rotation point" B to the end of an arm/lever.
     
  15. Mar 12, 2012 #14

    tiny-tim

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    ah, no :redface:

    it's the perpendicular distance from B to the line of the force

    (in this case, the force is the weight, so the line of the force is the vertical line through the centre of the rod)

    another formula is F.d (or Fdsinθ)

    where d is the vector from B to the centre of mass, and θ is the angle between F and d

    see http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html :wink:
     
  16. Mar 12, 2012 #15
    Hmm.

    B is a point. How can something be perpendicular to a point?

    Centre of the mass? So i have to make a triangle out of it, and find the center of gravity?
     
  17. Mar 12, 2012 #16

    tiny-tim

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    i don't understand

    either follow the diagram, http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html, or follow my instructions
     
  18. Mar 12, 2012 #17
    I'm looking at the site, but it doesn't make sense. Maybe i'm just not that good at english :) It's confusing.

    I don't know what you mean by F.d. What does "." represent ?

    If the Force is the force of gravity, and the force of gravity is a vertical force through the point B, how can i make a line perpendicular to the force(Fg) and B?
     
    Last edited: Mar 12, 2012
  19. Mar 12, 2012 #18

    tiny-tim

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    That's the dot product (or scalar product), see http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html

    If you don't like the idea of dot product, jsut use the Fdsinθ formula :smile:
    No, you make it through B (or from B), and perpendicular to that vertical line.

    In other words: if d is the distance from B to the centre of the rod,

    then dsinθ is the horizontal distance from B to the vertical line through the centre of the rod.​
     
  20. Mar 12, 2012 #19
    So d is a line from the middle of AB, perpendicular to AB, which goes to F_g, which is the vertical force through B ?

    Dot product and vector calculations is long gone. Cannot remember how it works, so another method is preferrable :)
     
  21. Mar 12, 2012 #20

    tiny-tim

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    No!!

    Read the instructions

    it's perpendicular to the force

    and it's through B (the pivot point)​
     
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