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Dynmics, motion along a line

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A machine has an 850g steel shuttle that is pulled along a square steel rail by an elastic cord. The shuttle is released when the elastic cord has 23.0N tension at a 45 degree angle.
    What is the initial acceleration of the shuttle?

    2. Relevant equations

    F = ma
    v1^2 = v0^2 + at
    v1^2 = v0^2 +2a(delta x)

    3. The attempt at a solution

    F = ma
    23 = .85a
    a= 27.06 m/s^2
    This is the acceleration when the cord is let go, but I have no idea how to find the inital acceleration. I know how to split the force into the x/y components but I'm not sure how that will help. Thanks

    Attached Files:

    Last edited: Sep 30, 2008
  2. jcsd
  3. Oct 1, 2008 #2
    perhaps you should use newton's second law to break the force into components:

    sum of forces in x-direction = Tcos(theta) = ma_x
    sum of forces in y-direction = N + Tsin(theta) - mg = ma_y = 0

    Which a are you solving for? (for you to figure out)

    You are given enough information from these equations and values to easily solve this I think.
  4. Oct 1, 2008 #3
    so Fx = 23cos45 = .85a
    so a_x = 19.13m/s^2

    and Fy = 0
    but how do these equations give you the initial acceleration? Don't they only solve for the acceleration when F = 23N

    I am confused by the initial acceleration part and the part describing the string as elastic. I don't know how elasticity affects the problem. And also, should I include friction of steel on steel?
  5. Oct 1, 2008 #4


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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi bluebear19! Welcome to PF! :smile:

    (welcome also to chansong :wink:)

    "initial" means immediately the cord is released …

    in other words, while it's at its full length, and it is still at 23N tension. :smile:

    So you're there! :biggrin:

    (and no, you can ignore friction in any exam question that doesn't give you a coefficient of friction :wink:)
  6. Oct 1, 2008 #5
    Is the problem really this simple?

    I tried solving it in this way, by finding the Fx component of the tension (23*cos45), and then dividing by the mass (.850 kg), using F=ma. However, this does not yield the correct answer. What is missing?
  7. Oct 1, 2008 #6
    Do we need to account for steel on steel friction for this problem? I have been having such a headache with this, it is driving me crazy!
  8. Oct 1, 2008 #7
    ok I got it, my friends helped me. so you have to find the normal force which is Fy - mg
    then to find the friction force you multiply the normal force by the kinetic coefficient of friction for steel on steel. then u take that number and substract it from the Fx. then F = ma and ur done
  9. Oct 1, 2008 #8
    Cool, thanks!

    So acceleration is negative?

    EDIT: Ah nvm, I've been at this way too long. Thanks again, everyone!
  10. Oct 1, 2008 #9
    i was assuming there was no friction in your problem (as there was none in the one i had to solve myself)

    but it looks like you've solved it so congrats
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