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Pseudo Force ~

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Prism 1 with bar 2 of mass m placed on it gets a horizontal acceleration [tex]\omega[/tex] directed to the left. At what maximum value of this acceleration [tex]\omega[/tex] will the bar be still stationary relative to the prism, if the coefficient of friction between them [tex]k < cot\alpha[/tex]

    th_untitled3.jpg

    2. Relevant equations


    [tex]mg sin \alpha - m \omega cos \alpha[/tex] = Force trying to move the body along the prism.
    k ([tex]m\omega sin \alpha + mg cos \alpha[/tex]) = Frictional force.


    3. The attempt at a solution
    Since the prism is accelerating to the left, the bar experiences a pseudo force [tex]m\omega [/tex] directed to the right. Resolving it has components
    [tex]m \omega cos \alpha [/tex] opposed to [tex]mg sin \alpha [/tex] and the component [tex]m\omega sin \alpha [/tex] acting along the direction of [tex]mg cos \alpha[/tex]

    At equilibrium, the frictional force equals the force which acts along the prism.
    Equating, [tex]mg sin \alpha - m\omega cos\alpha[/tex] = k ([tex]m\omega sin \alpha + mg cos \alpha[/tex])
    Further simplifying,

    [tex]\omega = \frac {g (sin \alpha - k cos \alpha)}{cos \alpha + k sin \alpha}[/tex]

    This is not the correct answer. Can you please point out where I've made a mistake?
     
  2. jcsd
  3. Mar 19, 2009 #2

    Doc Al

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    Staff: Mentor

    Looks like you made a sign error. Which way does the friction force act? Which direction are you calling positive?

    (I'd use ΣF = 0 parallel to the wedge, coupled with a consistent sign convention.)
     
  4. Mar 19, 2009 #3
    the maximum value of k is cot(alpha). So have you tried putting that in?
     
  5. Mar 19, 2009 #4
    Oh. Thanks for that correction. I've got the right answer now. :biggrin:
    I reckon I can use the sign system in any direction as long as I put the signs correctly?
     
  6. Mar 19, 2009 #5
    Its given [tex]k < cot\alpha[/tex]. What is the importance of this data in this question? I got the correct answer although I didn't have this point in mind. So I'm thinking why its given..
     
    Last edited by a moderator: Mar 19, 2009
  7. Mar 19, 2009 #6

    Doc Al

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    Absolutely.
    Consider the sign of your expression for the acceleration if that condition were not met. :wink:
     
  8. Mar 19, 2009 #7
    Sign of the expression for acceleration? I don't notice any particular change when [tex]k < cot\alpha[/tex]!! :frown: Where am I going wrong?
     
  9. Mar 19, 2009 #8

    Doc Al

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    Staff: Mentor

    Take a look at your (corrected) expression. What happens to the sign of the denominator when k > cot α ?
     
    Last edited: Mar 19, 2009
  10. Mar 19, 2009 #9
    The denominator turns out to be zero when [tex]k= cot\alpha[/tex]. :tongue:
    Thanks for the help.
     
    Last edited: Mar 19, 2009
  11. Mar 19, 2009 #10
    I see that the tendency of the body 2 is to move up. Why is it like that, usually it always tends to move down, right?
     
  12. Mar 20, 2009 #11

    Doc Al

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    Staff: Mentor

    The block would tend to move down the prism, if the prism weren't accelerating towards it. That changes things by introducing additional forces.
     
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