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Minimum force to overcome static friction

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    http://postimage.org/image/530mgjivr/

    How come the solution states: (20)(9.8)(0.7cos30 - sin 30) isn't sin30 going in the positive direction? shouldn't the solution be Sin30 - .7cos30? Also, when solving these types of problems should I leave the mg part outside of the cos and sin area? Cause I got

    "F = 196sin30° - (.7 * 196cos30)"

    2. Relevant equations

    Newtons second law
    Fnet = ma

    3. The attempt at a solution

    N = 193cos30
    F = 196sin30° - (.7 * 196cos30)
     
  2. jcsd
  3. Mar 4, 2013 #2

    PhanthomJay

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    If you solve the equation for
    F and assume down the plane is positive, the solution is correct, although the conversion factor from newtons to pounds was left out in error. You get the same result in your attempt, except for the sign error.
     
  4. Mar 4, 2013 #3

    rude man

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    The static friction force is μ mg cosθ which acts up-ramp. The total force acting down-ramp is F + mg sinθ. Equate the two net forces to each other.
     
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