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Inclined slope with acceleration

  1. Nov 28, 2008 #1
    in the following question, a box with a mass of m is tied to a pole at the top of a slope with an incline of (alpha), the whole slope accelerated at "a", find the tension "T" in the rope.

    http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273613370813875458

    what i did was:

    F=T-mg[sin(alpha)]=ma[cos(alpha)]
    T=m[(gsin(alpha))+(a(cos(alpha))]
    T=m(cos(alpha))[tg(alpha)-a]

    which according to the answer sheet is only a part of the correct answer,
    it says this is correct when-- a<g/tg(alpha)

    but when a>g/tg(alpha)
    then
    T=m*sqrt(a^2+g^2)

    why specifically a>g/tg(alpha) or a<g/tg(alpha), what happens physically at that point? is that the critical acceleration where N=0??
    how do i see this from the equations?, the second T, seems to me like the sum of the 2 vectors ma and mg, what is the logic in doing that
     
    Last edited by a moderator: Nov 28, 2008
  2. jcsd
  3. Nov 28, 2008 #2

    tiny-tim

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    Hi devanlevin! :smile:
    No, you've left out N. :frown:
     
  4. Nov 28, 2008 #3
    but N is 90 degrees to the x axis im talking about, the axis parallel to T, look at the picture,
     
  5. Nov 28, 2008 #4

    tiny-tim

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    Hi devanlevin! :smile:

    (have an alpha: α and a square-root: √ :wink:)
    Sorry, I should have been more specific. :redface:

    You didn't consider what hapens when N is zero …

    then the box is hanging on for dear life (at an angle smaller than α). :biggrin:
     
  6. Nov 28, 2008 #5
    thats the second case, the one i noted i dont know how to build the equation for,
    when a>g/tg(alpha)
     
  7. Nov 28, 2008 #6
    how do i find this
     
  8. Nov 28, 2008 #7

    tiny-tim

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    The rope is now at an unknown angle.

    Just write the horizontal and vertical equations, and solve by eliminating the angle. :smile:

    (or just use a vector triangle :wink:)
     
  9. Nov 28, 2008 #8
    from what i understand,
    i say, when a>g/tg(alpha) N=0, and so the total force is sqrt((T)^2+(mg)^2) and that is the size of the force on the box, now that is equal to ma???

    so T=sqrt((ma)-(mg)^2)

    but that isnt right, where am i going wrong,,, can you please show me how to solve this both ways, both with equation and with vector triangle
     
  10. Nov 28, 2008 #9

    tiny-tim

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    No, that is not the total force …

    T and g are not perpendicular.
     
  11. Nov 29, 2008 #10
    right, so how would i divide my axises, parallel to a or parallel to t??
    if parallel to a
    F=Tcos(alpha)=ma

    if parallel to a
    F=F=T-mg[sin(alpha)]=ma[cos(alpha)]
    which is what i did in the 1st place
     
  12. Nov 29, 2008 #11

    tiny-tim

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    waaah! :cry:

    it's not alpha!!

    and you need two axes!
     
  13. Nov 29, 2008 #12
    so when a<g/tg(alpha) i divide the vector components parallel to T
    and when a>g/tg(alpha) i divide them parallel to a

    shouldnt i use the same x,y axises in both cases??? here once i look at it on the slopes axis once on the grounds axis?? how do i know how to divide and when
     
  14. Nov 29, 2008 #13

    tiny-tim

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    No, you can use any pair of axes in both cases …

    the only difference is that if a ≤ g/tanα, then the rope is at angle α, but if a > g/tanα, the rope is at an angle less than α.
     
  15. Nov 29, 2008 #14
    i can only see the 1st case with an axis parallel to t

    F=T-mg[sin(alpha)]=ma[cos(alpha)]
    T=m[(gsin(alpha))+(a(cos(alpha))]
    T=m(cos(alpha))[tg(alpha)+a]

    and with a regular axis, parallel to a for the second case

    Ma(y)=0=Mg-T(y)
    Ma(x)=ma=T(x)

    T=sqrt[(mg)^2 + (ma)^2]

    if i try the other way what are my equations? i get

    for the 1st case
    Ma(x)=ma=Tcos(alpha)-Nsin(alpha)
    Tx=Nsin(alpha)+ma
    Ma(y)=0=Tsin(alpha)+Ncos(aplha)-mg
    Ty=mg-Ncos(alpha)

    then to find T==>Tx^2+Ty^2
    but i get an equation with N in it



    could you please show me the equations you get for each axis
     
  16. Nov 30, 2008 #15

    tiny-tim

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    Hi devanlevin! :smile:

    Well that's pretty good :approve:

    you've tried both methods (x-y axes and T-N axes), and you've got the right equations for them (except there should be minus signs in the Tx equation) …

    you just haven't spotted the trick for solving the x-y one :wink:

    you simply need to eliminate N …

    so multiply the first equation by cosα, and the second by sinα, and add! :smile:
     
  17. Dec 1, 2008 #16


    where should there be a minus?? dont see it
     
  18. Dec 1, 2008 #17

    tiny-tim

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    look at your own diagram … a is to the left, and n is to the right
     
  19. Dec 1, 2008 #18
    X+ in the direction of a

    Ma(x)=ma=Tcos(alpha)-Nsin(alpha) (here there is a minus)
    if i wan Tcos(alpha) on one side of the =, i move -Nsin(alpha) to the side of ma,
    becomes + Nsin(alpha)

    Tx=Nsin(alpha)+ma

    still dont see it
     
  20. Dec 1, 2008 #19
    did what you said using my equations, without the - and it works, you sure there is a minus i havent written
     
  21. Dec 1, 2008 #20
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