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Forces and Newton-Identifying Acceleration

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A man pulls a rope, with a force of 200N, attached to a box, with a mass of 35kg, 30° from the horizontal. The [tex]\mu[/tex]k is 0.3. Find the acceleration of the box.


    2. Relevant equations
    [tex]\Sigma[/tex]F = ma
    Ffr = [tex]\mu[/tex]FN

    3. The attempt at a solution
    FBD
    http://img255.imageshack.us/img255/2181/forceandnewtonaccelerat.jpg [Broken]

    Net Force Equations
    [tex]\Sigma[/tex]Fy = ma
    FN - FG = ma
    FN - mg = ma
    FN = mg + ma
    Acceleration is 0 because the box is not moving up or down so
    FN = mg


    [tex]\Sigma[/tex]Fx = ma
    Fappcos[tex]\theta[/tex] - Ffr = ma
    Fappcos[tex]\theta[/tex] - [tex]\mu[/tex]FN = ma
    Fappcos[tex]\theta[/tex] - [tex]\mu[/tex]mg = ma
    (Fappcos[tex]\theta[/tex] - [tex]\mu[/tex]mg) / m = a
    ((200N)(cos30°) - (0.3)(35kg)(9.8m/s²)) / 35kg = a
    2.0087m/s² [tex]\approx[/tex] a

    I am told this is wrong and that the correct answer is around 2.87m/s².

    What did I do wrong so that I did not arrive at the correct answer?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 23, 2009 #2
    You are missing a force in the y-dir.
     
  4. Feb 23, 2009 #3
    It's in there with:

    Net Force Equations
    [tex]\Sigma[/tex] Fy = ma
    FN - FG = ma
    FN - mg = ma
    FN = mg + ma
    Acceleration is 0 because the box is not moving up or down so
    FN = mg
     
  5. Feb 23, 2009 #4

    Delphi51

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    Homework Helper

    All your work looks correct to me.
    Don't forget to check the question - very easy to copy something incorrectly.
     
  6. Feb 23, 2009 #5
    Since the Fapp is at a 30 degrees, isn't there a component of it working in the y-dir?
     
  7. Feb 23, 2009 #6
    Yes I now see that I forgot to include fapp in the y direction. The correct net equation is:

    FN = mg -Fappsin[tex]\theta[/tex]

    (Fappcos[tex]\theta[/tex] - [tex]\mu[/tex](mg -Fappsin[tex]\theta[/tex]) / m = a
    ((200N)(cos30°) - (0.3)((35kg)(9.8m/s²)-((200N)(sin30°))) / 35kg = a
    2.866m/s² = a

    This is correct. Thank you.
     
    Last edited: Feb 23, 2009
  8. Feb 23, 2009 #7

    Delphi51

    User Avatar
    Homework Helper

    Thanks for catching that, slmg! I forgot about the pull affecting the normal force!
     
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