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Kinetic friction of crate being pulled

  1. Aug 4, 2005 #1
    I've worked this problem an dget an answer, but when i enter it into our homework site, it says my answer of 229 N is wrong, do any of you see what im doing wrong.
    A 1.06×102 kg crate is being pulled across a horizontal floor by a force P that makes an angle of 28.9° above the horizontal. The coefficient of kinetic friction is 0.193. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

    I found the force applied by friction to be 200.5 N (F= (0.193)(106kg)(9.8m/s2)

    so to find the force of P i said the force in the opposite x direction needs to be equal to the frictional force

    cos28.9= 200.49/P

    =229 N


    If you see anything I'm doing wrong or another way to approach it tell me.

    I don't think this is suppose to be a difficult problem.

    Thanks in advance!
     
  2. jcsd
  3. Aug 4, 2005 #2

    OlderDan

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    Science Advisor
    Homework Helper

    In your calculation of the frictional force you have neglected the vertical component of P. The normal force is the weight of the crate, less the vertical component of P.

    Draw a free body diagram with ALL of the forces acting on the crate. Then add separately all the vertical components and all the horizontal components.
     
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