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Friction Forces

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Well I don't think my teacher explained this very well so I need some help.. Thank you in advance.

    26.) A person pushes a 14.0-kg lawn mower at constant speed with a force of F = 88.0N directed along the handle which is at an angle of 45.0(degrees) to the horizontal.
    a) Draw the free-body diagram showing all forces acting on the mower. Calculate b) the horizontal friction force on the mower, then c) the normal force exerted vertically upward on the mower by the ground. d) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force.

    2. Relevant equations
    Fx = F cos (theta)
    Fy = F sin (theta)
    F = ma
    a = (vf-vi)/t

    3. The attempt at a solution
    a) My answer was:

    b) Fx = 88 cos 45 = 62.2N

    c) I thought it would be the vertical component, but it's not, not sure where to go from that now..

    d) Not sure :(

  2. jcsd
  3. Sep 27, 2010 #2
    Im pretty sure your Fp should be pointing in the third quadrant, you don't push up on a lawn mower. To find the Normal just sum the forces in the Y direction and solve for Fn
    Last edited: Sep 28, 2010
  4. Sep 28, 2010 #3
    The answer is 199 though :|
  5. Sep 28, 2010 #4
    yes the answer is 199.43 N, you need to correct your diagram

    [PLAIN]http://img836.imageshack.us/img836/3172/blockl.png [Broken]

    Now sum the forces in the Y direction

    Forces in Y = Fn - Fg - Fpsin(t) = 0;

    solve for Fn and you will find your answer
    Last edited by a moderator: May 4, 2017
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