1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  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.

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

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

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

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

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

  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...

  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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Inclined plane problem
  1. Inclined plane problem (Replies: 2)