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Equilibrium and a log

  1. Jun 13, 2006 #1
    A log weighs 800N and has its center of gravity at the midpoint. Find the lowest coefficient of static friction necessary to keep the log from slipping.

    Answer (1.01)

    [​IMG]

    Ok, so, i started out by finding my torque and force equations.
    Torque = (800)*(3.5)*(cos(25)) = (Tv)*(6.5)*(cos(25)) + (Th)*(6.5)*(sin(25))

    Normal Force = (800) - (Tv)

    Force of Friction = (Normal Force) (Coefficient of Friction)

    Force of Friction = (Th)

    tan(25) = (Th)/(Tv)

    So, solving for (Th) I got...
    (Th) = tan(25)/Tv

    So, Torque = (800)*(3.5)*cos(25) = (Tv)*(6.5)*cos(25) + (tan(25)/(Tv))*(sin(25))

    Where (Tv) is the vertical component of the tension and (Th) is the horizontal component of tension.

    Solving for (Tv) I get (Tv) = 353.83
    Solving for (Th) (Th) = tan(25)(353.83) = 164.99

    If Normal Force = Weight - (Tv) then N = 800 - 353.83 = 446.17

    so (Normal Force)(Coefficient of Friction) = (Th)
    Then (446.17)(Coe. of Friction) = 164.17
    The coe. of friction is not equal to 1.01

    Any help as to where I went wrong would be much appreciated! Thanks so much in advance for any help!

    Dan
     
  2. jcsd
  3. Jun 14, 2006 #2
    hmm, i don't claim to be a physic pro but the coeff i calculated is the same as yours. Maybe someone wanna come in and help?
     
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