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Homework Help: Pulling at an angle

  1. Jul 3, 2011 #1
    1. The problem statement, all variables and given/known data
    A box of mass 30kg is being pulled along rough horizontal ground at a constant speed using a rope. The rope makes an angle of 20 degrees with the ground. The coefficient of friction between the box and the ground is 0.4. The box is modeled as a particle and the rope as a light, inextensible string. The tension in the rope is P newtons.

    2. Relevant equations

    3. The attempt at a solution

    The normal force is
    [tex]30g-P\sin 20^\circ.[/tex]
    The force of friction is
    [tex](30g-P\sin 20^\circ)0.4,[/tex]
    which should be equal to the pulling force,
    [tex]P\cos 20^\circ.[/tex]
    Hence
    [tex](30g-P\sin 20^\circ)0.4=P\cos20^\circ.[/tex]
    Taking g as 9.8 and solving the equation yields
    [tex]P=109,[/tex]
    corrected to 3 significant figures.

    However, the answer says that it should be 125 instead of 109. What did I do wrong? Also, the book seems to prefer to have 3 significant figures but 9.8, the numerical value of g, only has 2, which seems weird to me. Any justifications for it?

    (Sorry, I don't know how to type the degree sign in LaTeX. Can anyone teach me how to?

    EDIT: added the degree signs
     
    Last edited: Jul 4, 2011
  2. jcsd
  3. Jul 3, 2011 #2

    rock.freak667

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    I would say that you are correct. Even putting g = 10 N/kg would not get it to 125 N.
     
  4. Jul 3, 2011 #3

    SammyS

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    20^\circ

    [tex](30g-P\sin 20^\circ)0.4=P\cos20^\circ[/tex]
     
  5. Jul 3, 2011 #4
    I am also getting 190 N.

    You can also do degrees by making a lower case o into a superscript with the x2 button at the top of the message composing window: cos20o
     
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