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Torque and friction

  1. Nov 6, 2007 #1
    A rod of mass M rests vertically on the floor, held in place by static friction. IF the coefficient of static friction is [tex]\mu_{s}[/tex], find the maximum force F that can be applied to the rod at its midpoint before it slips.

    I'm not exactly sure what i am suppose to be looking for.. Obviously there will be a FBD of the rod and friction will oppose the force, i need to find the maximum force of static friction and than anything greater than that will push the rod over. Where do i start with this?
     
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  3. Nov 6, 2007 #2

    PhanthomJay

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    The rod is going to rotate but not necessarily slip if F is small enough. You must look at the torque and resulting acceleration of the rod, then apply Newton 2 for the translational motion of the rod's center of mass. You need to find I of the rod also before you begin.
     
  4. Nov 6, 2007 #3
    so are you saying i would use:

    [tex]\alpha = LF/I = LF/(1/3)ML^{2}[/tex]
    than
    [tex]a= \alpha L[/tex]
    than
    [tex]F = m[ \alpha L + \mu_{s}g][/tex]
     
  5. Nov 6, 2007 #4

    PhanthomJay

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    good, you are almost there; however, the load is applied at midpoint...adjust your torque and the tangential aceleration of the center of mass accordingly.
     
  6. Nov 6, 2007 #5
    Ok i did that and i got as an answer

    [tex]F = (3F/2) + \mu_{s}Mg[/tex]

    is that right?

    (I'm trying to think through this verbally here:)
    I can't see how its possible for a force to depend on the force itself?... or can i re-arrange the formula and than i would get:

    [tex]F = (-2 \mu_{s}Mg)[/tex]

    But why the negative?
     
    Last edited: Nov 6, 2007
  7. Nov 7, 2007 #6

    PhanthomJay

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    I don't know what you did here...your original approach was good, its just that you forgot thet the torque was FL/2 and not FL; and the acceleration was (alpha)L/2 and not (alpha)L. Redo and resubmit, please.
     
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