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Slightly harder - pushing a lawnmower against friction

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    The handle of a 22-kg lawnmower makes a 35 degree angle with the horizontal. If the coefficient of friction between lawnmower and ground is .68, what magnitude of force is required to push the mower at constant velocity? Assume the force is applied in the direction of the handle.


    2. Relevant equations

    Fstatic = (mu)*(Forcenormal)

    3. The attempt at a solution

    Okay, so this problem comes with a solution at the back of the book. Says 342 N. Well, I thought about this logically, and said that the Force to get this thing going will have to be equal to the mu times the Forcenormal, that's what will have to be overcome. So I did that and got 146.6 Newtons. Well, it makes a 35 degree angle to the ground, so that is only the horizontal force we need. If we're pushing on the handle, I have to find the hypotenuse of that force triangle, and I got 178.9 Newtons, roughly half of the right answer. What am I doing wrong?
     
  2. jcsd
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