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Homework Help: Help w/ this force problem please

  1. Oct 19, 2004 #1
    hi, i've approach the first question, but i don't know how to start w/ this question. I already have my body diagram and everything. Could someone tell me how ?
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    Consider a lawnmower of weight which can slide across a horizontal surface with a coefficient of friction . In this problem the lawnmower is pushed using a massless handle, which makes an angle with the horizontal. Assume that , the force exerted by the handle, is parallel to the handle.

    Take the positive x direction to be to the right and the postive y direction to be upward.
    http://session.masteringphysics.com/probhtml/MFS.cf.8_a.jpg [Broken]
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    The first 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.
    >>> For this I got
    [tex]Fh=(-(\mu*w))/(-cos(\theta)+\mu*sin(\theta))[/tex]
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    This question I've problem with
    The solution for Fh has a singularity (that is, becomes infinitely large) at a certain angle theta critical . For any angle [tex]\theta>\theta critical [/tex] , the expression for Fh will be negative. However, a negative applied force Fh would reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawnmower at all.

    Find an expression for [tex]tan(\theta) critical[/tex]
     
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Oct 19, 2004 #2
    The force will become infinitely large whenever the denominator of your expression for the force goes to zero. Thus, the singularity occurs at

    [tex]\mu \sin \theta - \cos \theta = 0[/tex]
     
  4. Oct 19, 2004 #3
    I'm not quite understand how [tex]\mu \sin \theta - \cos \theta = 0[/tex]
    relate w/ [tex] tan(\theta critical) [/tex] ?
     
  5. Oct 20, 2004 #4
    hi, i think i got it . it's 1/mu after all :)
     
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