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Forces-Newton's Laws and Vectors

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stoplight to 28 m/s in 6.0 s, what angle [tex]\vartheta[/tex] does the string make with the vertical?


    2. Relevant equations
    [tex]\sum[/tex]F=ma


    3. The attempt at a solution
    I found the acceleration to be about 4.667, but I honestly have no clue where to go from here. I've looked for other examples in the book that are similar but I haven't found any. The angle part is really throwing me off. I'm completely lost and I just want to be pushed in the right direction.
     
  2. jcsd
  3. Feb 15, 2009 #2

    jgens

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    Your value for the acceleration seems alright. Have you drawn a free-body diagram yet and carefully labeled all the forces acting on the dice? If you have, write it out here.
     
  4. Feb 15, 2009 #3

    djeitnstine

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    Try drawing a free body diagram

    edit: jgens said it before me
     
  5. Feb 15, 2009 #4
    Ok. Would it be something like this?



    [tex]\uparrow[/tex] FT
    [ ][tex]\leftarrow[/tex] Acceleration?
    [tex]\downarrow[/tex] FW
     
  6. Feb 15, 2009 #5

    jgens

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    Not quite. Since, the dice are going to make an angle, Ft should not be positioned directly above mg; hence, your tension has two components, one in the x direction and one in the y. Additionally, your diagram should not show the acceleration as a force.

    If it's easier you could just write out the results of your free-body diagram as:
    ∑Fx = ...
    ∑Fx = ...
     
  7. Feb 15, 2009 #6
    FTcos[tex]\theta[/tex]+FTsin[tex]\theta[/tex]+FW.

    Is that correct? Would i have to break it into two equations because of the x and y components?
     
  8. Feb 15, 2009 #7

    jgens

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    Yes, it should be broken into two x and y components.
     
  9. Feb 15, 2009 #8
    [tex]\sum[/tex]Fx=FTsin[tex]\theta[/tex]
    [tex]\sum[/tex]Fy=FTcos[tex]\theta[/tex]+FW

    Is that correct? I can substitute FW for mass times gravity right?
     
  10. Feb 15, 2009 #9

    jgens

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    Your equations look correct and Fw can be replaced with mg (just make sure you use g < 0). However, you can extrapolate more still. Are the dice accelerating in the vertical component and what does this suggest about ∑Fy? Eventually, the dice must be accelerating at the same rate as the car, what does this suggest about ∑Fx (think about Newton's second law)?
     
  11. Feb 15, 2009 #10
    Oh! So,

    [tex]\sum[/tex]Fy=0

    because it is not accelerating, and

    FTcos[tex]\theta[/tex]+mg=ma

    Would gravity be negative?
     
  12. Feb 15, 2009 #11

    jgens

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    Well, it seems you've gotten your forces in the x and y directions mixed up. Your equations should read:

    ∑Fx = Ftsin(theta) = ma
    ∑Fy = Ftcos(theta) - abs(g)m = 0.
     
  13. Feb 15, 2009 #12
    Oh yeah, sorry about that. But that still gives me three variables in both equations. I still only know that acceleration is 4.667m/s2 and gravity is 9.8m/s2.
     
  14. Feb 15, 2009 #13

    jgens

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    True, but you can easily reduce it. Divide ∑Fy by ∑Fx and solve, or using ∑Fy come up with an expression for Ft and substitute it into ∑Fx and solve. Either method should work.
     
  15. Feb 15, 2009 #14
    Ok, I got answer, 25 degrees, but I just want to make sure my math is correct.

    I substituted Ftinto [tex]\sum[/tex]Fx.

    (mg/cos[tex]\theta[/tex])sin[tex]\theta[/tex]=ma

    mg*tan[tex]\theta[/tex]=ma

    tan[tex]\theta[/tex]=a/g

    tan-1(a/g)=[tex]\theta[/tex]
     
  16. Feb 15, 2009 #15

    jgens

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    Your math it correct. I can't check your angle measure because I don't have a calculator or table of values at this point in time but I would presume it is correct. Good job!
     
  17. Feb 15, 2009 #16
    Thanks so much for your help! I've been having trouble understanding force and just going through that problem with you helped me a lot. Thanks so much for your time. :)
     
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