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Blocks on an Incline

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the following two scenarios with the same box sliding up two differently inclined ramps. The coefficient of static friction is 0.7, the coefficient of kinetic friction 0.5. Eventually, the box will stop. Up to what maximum angle of the incline (in degrees) would it not slide back down?


    2. Relevant equations
    sin = o/h
    cos= a/h
    tan= o/a

    3. The attempt at a solution

    cos-1 (.7/.5) = error
    cos-1 (.5/.7)= 44.41 = wrong
    cos-1 (.4) = 66.42= wrong
    cos-1 (1) = 0 so I thought it could also equal 360 so I added 360+66.42 = 426.42 = wrong

    sin-1 (.7/.5)= error
    sin-1 (.5/.7) =45.58 (I think its wrong)

    tan-1 (.7/.5) = 54.46 = wrong
    tan-1 (.5/.7)= 35.53 = scared to try to see if its wrong

    HELP PLEASE !!!!!!!
     

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    Last edited: Nov 3, 2013
  2. jcsd
  3. Nov 3, 2013 #2

    tiny-tim

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    hi lollikey! :smile:
    this is silly

    you're not even trying to work out the correct answer, you're just trying every answer you can think of :redface:

    that's no way to learn physics, and it's no way to pass your exams!

    start again …

    what equation(s) do you think relevant? :smile:
     
  4. Nov 3, 2013 #3
    I KNOW!!!! but I'm on the last question and this unit has been confusing and torturous I just want it to end :(

    I first though that the reverent equations would be the kinetic friction and the static friction equation but then that doesn't really matter because there are no number
     
  5. Nov 3, 2013 #4

    tiny-tim

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    but the numbers are .7 and .5 :confused:
     
  6. Nov 3, 2013 #5
    yeah but that's the coefficient so i thought if i could find the normal force and all the other forces i could use them but the i saw that that was wrong so i just tried doing trig functions as you can see i have no idea what i'm doing at this point
     
    Last edited: Nov 3, 2013
  7. Nov 3, 2013 #6

    tiny-tim

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    yes … so call the mass "m", and carry on from there :smile:

    (m will cancel out in the end)
     
  8. Nov 3, 2013 #7
    Fg = 9.8m

    Fn= (9.8m)cos(theta)

    Fk= (9.8m)cos(theta) * .5

    Fst= (9.8m)cos(theta) * .7

    would these be right?
     
    Last edited: Nov 3, 2013
  9. Nov 3, 2013 #8

    tiny-tim

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    no, Fst (9.8m)cos(theta) * .7 :wink:

    ok, so what is the maximum angle at which there can be static equilibrium?
     
  10. Nov 3, 2013 #9
    would it be

    (9.8m)cos(theta)*.5 ≤ (9.8m)cos(theta)*.7

    then solve for theta?
     
  11. Nov 3, 2013 #10

    tiny-tim

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    but that's obviously true for any theta, isn't it? :redface:

    do a static equilibrium equation​
     
  12. Nov 3, 2013 #11
    i don't know what that is .....
     
  13. Nov 3, 2013 #12

    tiny-tim

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    an equation that tells you why nothing is moving! :rolleyes:

    (and i'm off to bed :zzz:)
     
  14. Nov 3, 2013 #13
    the only equations i know that he taught us for this unit is newtons second law, static friction and kinetic friction!!


    (sooo regretting taking ap physics now!!!!)
     
  15. Nov 3, 2013 #14
    You don't need anything more than those equations...
     
  16. Nov 3, 2013 #15
    then why did tiny-tim say i did?
     
  17. Nov 3, 2013 #16
    so the force of static friction is only on an object while it is not in motion, the force acted upon an object needs to be greater than this static friction in order for it to move. What does newtons second law equation say?
     
  18. Nov 3, 2013 #17
    mass*acceleration = sum of forces
     
  19. Nov 3, 2013 #18
    good. when the object is not moving what is the force?
     
  20. Nov 3, 2013 #19
  21. Nov 3, 2013 #20
    Yes that is good, so sum up all the forces in the x direction, and keep in mind that m*a = 0 when the object is not in motion.
     
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