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Pushing a lawn mower

  1. Feb 5, 2006 #1
    this question didnt give me any figures

    this is the question:

    Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.Express the required force in terms of given quantities.

    so im thinking, x and y look like this:

    however, i was marked wrong when i put the answer as:

    Fh= y/sin(theta)

    can someone please help me out? thanks

    EDIT: after i answered this question, it marked me wrong and said ,"the correct answer does not depend on the variable: y"
    Last edited: Feb 5, 2006
  2. jcsd
  3. Feb 5, 2006 #2
    did you try drawing a free body diagram? what forces are acting on the lawnmower?
  4. Feb 5, 2006 #3
    thanks for your response
    Last edited: Feb 5, 2006
  5. Feb 5, 2006 #4

    list what forces are acting on the lawnmower
  6. Feb 5, 2006 #5
  7. Feb 5, 2006 #6
    don't forget friction
  8. Feb 5, 2006 #7

    i dont really see how this helps though
  9. Feb 5, 2006 #8
    now, you know that if it's moving at a constant velocity, the horizontal and vertical net vectors must be zero. So, write the equations for the vectors

    the [itex] \Sigma F [/itex] equations
  10. Feb 5, 2006 #9
    thanks for your help

    would i just be finding the x-component of Fpush since it is independent form the y-component?
  11. Oct 1, 2007 #10
    I am working on this exact same problem right now, and I am really stumped. I'm pretty sure that if this problem included some actual numbers, then I could knock it out pretty quick, but notationally, this think is kicking my butt :(

    I have a few formulas already worked out from the hints given in the homework.

    One of the formulas that I have worked out is the sum of the forces in the y direction, [​IMG]. If I solve this for [​IMG], I get, [​IMG]. The other formula is the sum of the forces in the x, [​IMG]. Solving this one for [​IMG] yields, [​IMG]. Now, the problem states that I should solve the equations simultaneously for [​IMG].

    I guess this is about where I'm stumped. Am I supposed to do a substitution? When I hear solve simultaneously, I think "system of equations", but I don't see how that would work here. Sigh, I really don't see how this is supposed to work.
  12. Oct 1, 2007 #11


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    Gold Member

    I was thinking about about the way I mow my lawn. I don't exert force parallel to the handle, at least not after I get the thing moving. I exert force parallel to the ground, so I don't think the angle of the handle comes into play. But since they want you to assume that the force exerted is parallel to the handle, you have to do it that way.

    In your example, what is n and w? Why are you summing your forces by subtracting? The y-component of the force of your push adds to the weight of the lawn mower.

    Your only force repelling the motion of the lawn mower is friction. F=mu * force normal.
    Force normal is the weight of the lawn mower (mg) + the y-component of the force you are exerting (F*sin(theta)). You must balance this force with the x-component of your push (F*cos(theta)).
  13. Oct 1, 2007 #12
    Ok, I finally got it. Here it is. (uw)/(cos(theta)-usin(theta))

    Thanks for the help

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