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Homework Help: Inclined plane problem

  1. Nov 2, 2004 #1
    empty post

    empty post
     
    Last edited: Nov 2, 2004
  2. jcsd
  3. Nov 2, 2004 #2
    i chose d

    marlon
     
  4. Nov 2, 2004 #3
    You have four forces. Suppose the x-acis is along the ramp and the force F for pulling the object is aligned along this x-axis. Y-axis perpendicular to x-axis.

    gravity
    [tex]-mg\sin(\theta)*e_{x} - mg\cos(\theta)*e_{y}[/tex]

    normal force
    [tex] N * e_{y} [/tex]

    friction
    [tex]-{\mu}N*e_{x} [/tex]

    pull-force
    [tex]F*e_{x} [/tex]

    The sum of y-componets yield :

    [tex] N = mg\cos(\theta) [/tex]

    The sum of x-componets yield :

    [tex] -{\mu}N -mg\sin(\theta) + F = 0 [/tex]

    So we have that [tex] F = {\mu}mg\cos(\theta) + mg\sin(\theta) [/tex]

    or [tex] F = mg(\mu\cos(\theta) + \sin(\theta)) [/tex]

    let's take the derivative with respect to the angle theta and set this equal to 0. Thus we get :

    [tex] 0 = mg(-{\mu}\sin(\theta) + \cos(\theta)) [/tex]
    or
    [tex] 0 =-{\mu}\sin(\theta) + \cos(\theta) [/tex]

    or
    [tex] \mu = \frac {\cos(\theta)}{\sin(\theta)} = \cot(\theta)[/tex]
    Thus [tex]\tan(\theta) = \frac {1}{\mu}[/tex]

    marlon.
     
  5. Nov 2, 2004 #4
    Answer d yields a maximum value of Work. We are looking for a minimum value.
     
  6. Nov 3, 2004 #5
    please prove your statement...

    marlon
     
  7. Nov 3, 2004 #6
    Besides i disagree because the first derivative is zero in [tex]\mu = \cot(\theta)[/tex]

    Yet if [tex]\mu > \cot(\theta)[/tex] then thefirst derivative is negative
    Yet if [tex]\mu < \cot(\theta)[/tex] then the first derivative is positive

    So the mu corrsponds to a minimal value here. You are mixing extrema with maxima. An extremum is a general name for both local max and min values of a function
    marlon
     
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