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Calculating angle with coefficient of static friction value!

  1. Feb 27, 2012 #1
    There's a box sitting on a plank. The coefficient of static friction is 0.33. I want to figure out at which angle the box will begin to slide if the plank is tilted.


    I know the factors that come into play are FN, Fg or Fgx, and Fs.



    So far I have

    Fnet= FN-Fs-Fg
    =FN- μs-sin∅
    =cos-0.33-sin∅

    This doesn't look right... can anyone help me out?
     
  2. jcsd
  3. Feb 27, 2012 #2

    tiny-tim

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    Hi Fireant! :smile:
    I don't understand that at all. :confused:

    First, use components in the normal direction to find N.

    Then you know the friction is 0.33*N, and you can use components in the slope direction to find θ.

    Show us what you get. :smile:
     
  4. Feb 27, 2012 #3
    Hi thank you for responding!

    The mass of the box is 10kg. So the normal force here would have to be mg...?

    normal force=98N

    Ff=μs(FN)
    =0.33(98)
    =32.34 N


    and then
    tanθ= 97.5/98
    θ=44°

    I got 97.5 (for the x component) from pythagorean theory...

    I'm pretty lost!
     
  5. Feb 27, 2012 #4

    tiny-tim

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    Hi Fireant! :smile:
    No, the weight, W, is mg (98 N).

    The normal force, N, is the normal component of the reaction force.

    You find N by taking components in the normal direction (of N and W), and equating the sum to zero …

    what do you get?
     
  6. Feb 27, 2012 #5
    I think i'm getting confused when you say 'the components in the normal direction'... do you mean on the y axis? Isn't that just FN? Which would be -98N ?
     
  7. Feb 27, 2012 #6

    tiny-tim

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    Some people call that the y axis, some people call the vertical the y axis :confused:

    That's why I say "the normal direction", so there's no confusion.
    What is the normal component of W ?
     
  8. Feb 27, 2012 #7
    Sorry I'm at work, that's why it's taking so long to reply.

    the normal component of W would be mgcostheta?
     
  9. Feb 27, 2012 #8

    tiny-tim

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    Yes, so N = mgcosθ.

    ok, now find the friction force, and then write the equation for components along the plank.
     
  10. Feb 27, 2012 #9
    so Ff=μsFN
    =0.33(mgcosθ)
    =0.33(98)(cosθ)
    =32.34cosθ
     
  11. Feb 27, 2012 #10

    tiny-tim

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    ok

    now the equation along the plank
     
  12. Feb 27, 2012 #11
    would be...

    Fnet=Fn-Ff-Fg
    0=32.34cosθ-98

    ?
     
  13. Feb 27, 2012 #12
    but i'm missing

    sintheta

    right?
     
  14. Feb 27, 2012 #13

    tiny-tim

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  15. Feb 27, 2012 #14
    0=32.34costheta-sintheta-98
     
  16. Feb 27, 2012 #15
    im not sure how i would isolate theta...
     
  17. Feb 27, 2012 #16

    tiny-tim

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    put = in the middle and divide :smile:
     
  18. Feb 27, 2012 #17
    lol i'm hopeless...! thank you so much for bearing with me

    ok gonna try this out...

    32.34costheta=sintheta-98

    sintheta=32.34costheta+98
    sintheta/32.34=costheta+98

    Sorry, what am i dividing by and how?
     
  19. Feb 27, 2012 #18
    or would it be easier if:

    costheta=(sintheta-98)/32.34
     
  20. Feb 27, 2012 #19

    tiny-tim

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    sin/cos = tan ? :wink:
     
  21. Feb 27, 2012 #20
    .......oh wow...lol..forgot about that.

    so

    32.34costheta=sintheta-98
    tantheta=130.34
    theta=89.6 or 90

    that seems wrong..
     
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