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Fricional Force Problem

  1. Mar 6, 2007 #1
    1. The problem statement, all variables and given/known data

    A person pushes a 14kg lawn mover at constant sped with a force of 88N directed along the handle, which is at an angle of 45 degress to the horizontal. (a) draw the Free-body diagram showing all forces acting on the mower. (b) Calculate 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 accelerate it from rest to 1.5m/s in 2.5 seconds, assuming the same friction force?


    2. Relevant equations
    F=ma
    Fgy= (14)(9.8)cos45
    Fgx= Fg sin Theta
    Are those correct??
    3. The attempt at a solution

    The best I can figure is that each equals the same but im not sure what the frictional force is. Is it Fgy or Fgx im just not sure on much.
     
  2. jcsd
  3. Mar 6, 2007 #2

    PhanthomJay

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    You've got to first draw the free body diagram and identify all the forces acting and their direction. There's the applied force through the handle, the friction force at the ground opposing the motion, the normal force from the ground, and the mower's weight. Resolve into x and y components and use newton's laws in each direction.
     
  4. Mar 6, 2007 #3
    Last edited: Mar 6, 2007
  5. Mar 6, 2007 #4
    Well nothing any one has said seems to help in any . I have the free body diagram drawn and yall try to lead my somwhere i have no prior experience to do.
     
  6. Mar 6, 2007 #5
    Than complain on the original thread. Like help still don't get it! People do this all the time and I will call for help if I get stuck.

    We are here to do that. Plus there are some good tutorials on this site that supply the concepts needed. So here we go:

    think about this hard: if a body is at rest or moving at constant velocity the net forces acting upon it are zero. Now this applies to each dimension. If your lawnmower isnt sinking below your feet, that implies that the forces on it vertically are zero. How can this be? The mower has weight, and on top of that you are applying force downwards thru the handle. Yet is has no movement downward.

    So you have m*g (where gee is the earth-surface constant of 9.81m/s^2) and m is th mass expressed in Kilograms. You sit the mower on a scale and it weighs 14*9.8 Newtons. Now the push downwards is partly downward and partly horizontal. You add the vertical component of the component by using trig. Hint Sin (45). The sin is the Y (up/down direction).

    Now knowing that you have these two seemingly unopposed forces, but faced with the evidence that the mower just sits on the ground, there must then be a third force, referred to as the Normal force cuz it acts "normally" ie perpendicular to its own oriebtation. usually earth, so 90 degrees upward.

    Oddly enuf, friction does not care about the size or speed (aerodynamic heating etc being an exception) of the object, just the normal force. So you take your results from the above, Normal force, and multiply by the frictional cooeficient.

    So now that you have a normal force times the frictional cooefficient that resists movement--yes unless you have a self propelled unit--gotta push. This is the frictional force one has to overcome= N*u.

    Now to accelerate you need to provide more push than the resistance. The resistance does not go bigger the harder you push! It stays the same. Be careful here as the harder one pushes at 45 degrees the greater the friction. Does this help?
     
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