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Static equilibrium question

  1. Jan 25, 2008 #1
    This question was grabbed out from the textbook Physics for Scientists and Engineers with Modern Physics, 7th edition by Serway and Jewett.

    A uniform pole is propped between the floor and the ceiling of a room. The height of the room is 7.80 ft, and the coefficient of static friction between the ceiling and the pole is 0.576. The coefficient of static friction between the pole and the floor is greater than that. What is the length of the longest pole that can be propped between the floor and the ceiling?

    Here's my diagrams:
    [​IMG] general picture
    [​IMG] free body diagram

    where N_1 and N_2 are the normal forces from the floor and ceiling respectively and f_1 and f_2 are the static frictional forces due to the floor and ceiling respectively.

    The relevation equations are:

    [tex]\Sigma[/tex] [tex]T[/tex] = 0
    [tex]\Sigma[/tex] F = 0
    L = [tex]\frac{d}{sin\theta}[/tex]

    So for the x-component of force:
    f_2 - f_1 = 0
    then: f_2 = f_1
    however, f_2 < f_1 and thus N_2 < N_1

    for the y-component of force:
    N_1 - N_2 - mg = 0

    and for the torque, taking the axis of rotation at the very bottom of the pole:

    [tex]\frac{1}{2}[/tex]Lmgcos[tex]\theta[/tex] + LN_2cos[tex]\theta[/tex] - Lf_2sin[tex]\theta[/tex] = 0

    (taking counterclockwise as positive rotation)

    And im kinda much stuck here.

    Well L looks like it will be canceled out in the torque equation, but I was wondering maybe I could find theta and plug it into the last equation to find L. but im having trouble find theta
     
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 26, 2008 #2
    Since you have the coefficient of static friction between the pole and the ceiling you can correlate f_2 and N_2.
     
    Last edited: Jan 26, 2008
  4. Jan 26, 2008 #3
    oh ok. so i replaced f_2 with 0.576N_2. then rearranging the equation, i got [tex]\frac{1}{2}[/tex]mg + N_2(1-t0.576tan[tex]\theta[/tex]) = 0

    then i got theta as about 60 degrees. and then i plugged it into the last equation and got the length as 9.00 ft...

    i dunno if what i did was right. all i did was set 1-0.576tantan[tex]\theta[/tex] = 0 and solved for theta...
     
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