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Static Equilibrium of a

  1. Jul 26, 2015 #1
    1. The problem statement, all variables and given/known data



    A uniform sphere of radius R is supported by a rope attached to a vertical wall,as shown in the figure. The point where the rope is attached to the sphere is located so a continuation of the tope would intersect a horizontal line through the sphere’s centre a distance R/2 beyond the centre, as shown. What is the smallest possible value for the coefficient of friction between wall and sphere.

    wtf2q.png

    2. Relevant equations



    ##T_x=F_N\\\\F_f+T_y+Mg\\\\\Sigma\tau=0=\frac{1}{2}RMg-RF_f\\\\tan30^o=\frac{T_x}{T_y}\\\\F_f=\mu F_N##




    3. The attempt at a solution

    okr3sz.png

    ##\frac { 1 }{ 2 } RMg=RF_{ f }\\ \\ \frac { 1 }{ 2 } Mg=\mu F_{ N }\\ \\ \\ \frac { 1 }{ 2 } Mg=\mu T_x\\ \\ ##

    equation 1



    ##F_f+\frac{T_x}{tan30^o}=Mg\\\\\mu F_N+\frac{T_x}{tan30^o}=Mg\\\\\mu T_x+\frac{T_x}{tan30^o}=Mg\\\\##

    equation 2


    Solve equation 1 for Tx and substitute into equation 2.


    ##\mu=\frac{1}{tan30^o}=1.73…##


    Back of the book says 0.87
     
  2. jcsd
  3. Jul 26, 2015 #2

    haruspex

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    How do you get your first equation, ##\frac 12 RMg=RF_f##?
     
  4. Jul 27, 2015 #3
    I used the end of the rope where it connects to the sphere to be the pivot position and set all the torques to 0. This equated the Frictional Force with the Center of Mass but now that I see it, that would be that the Frictional force is 1.5 R away not R. I will try and recalculate it based on this, I think I did this before though....checking..
     
  5. Jul 27, 2015 #4
    yup, that was it.

    Thanks again.
     
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