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Need help Newtons laws two diemnsions (vertical plane)

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A gardener pushes down on the handle of a lawnmover, applying a force of 250N. The handle is inclined at an angle 45* to the horizontal. If the coefficient of the kinetic friction between the wheels of the lawnmower and the ground is 0.40, what is the acceleration of the 20-kg lawnmower?


    2. Relevant equations
    Fnet=0=Fn+Fsin45-Fg
    Fn=mg-FSin45
    a=(Fx-Ff)/m
    Fnet=ma
    Ff=(UkFn) Uk=coefficient of friction miew? or meiu

    3. The attempt at a solution

    Answer is 1.38m/s^2
     
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2

    PhanthomJay

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    Your first equation applies on the y direction, but you have a signage error. The weight acts down and so does the vertical component of the applied force. The normal force acts up ; solve for it. Also, be sure to identify the horizontal component of the applied force, Fx.
     
  4. Sep 27, 2012 #3
    so are you saying my expression should be Fn= -mg+Fsin45 ?
     
  5. Sep 27, 2012 #4

    PhanthomJay

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    You must be very careful with plus and minus signs, or else they will get the better of you. . Draw a free body diagram of the mower. The weight acts down on it, the vertical component of the applied force on the handle acts down on it, and the normal force of the ground acts up on it (normal contact forces are generally 'pushing' forces). Choose the up direction as positive and the down direction as negative. The algebraic sum of these three forces in the y direction must equal zero. Watch plus and minus signs, and solve for N.
     
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