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Holding a cone up!

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    With two fingers, you hold an cone motionless, upside down. The mass of the cone is m, and the static coefficient u. The angle of the tip, when viewed from the side, is 2θ. What is the minimum normal force required to hold the cone up (with each finger)? And, in terms of u, what is the minimum value of θ that allows you to hold up the cone?

    2. Relevant equations

    Friction force = uN
    Gravity = mg
    etc...

    3. The attempt at a solution

    I included a terrible paint drawing of my progress so far:

    cone.jpg

    The small "f" denotes the friction force, and i have split my mg force into mgsinθ and mgcosθ. Considering just one side, for the cone to remain motionless, I would assume that the friction force f must = (mgcosθ)/2. and f itself = uN. Also, N = F+mgsinθ/2, where F is the applied force from the one finger.

    So now i have u(F+mgsinθ/2) = (mgcosθ)/2. Does this make sense? And if so, how do I find the minimum normal force that I need to apply with each finger?
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 8, 2009 #2
    I expanded, with uN > f, that the finger force F must be at a minimum:

    F > (mg/2)(cosθ/u - sinθ)

    I would absolutely be indebted to anyone who would take the time to verify this.
     
    Last edited: Sep 8, 2009
  4. Sep 9, 2009 #3
    Shameless bump :(
     
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