Holding a Cone Up: Minimum Force & θ Explained

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To hold an upside-down cone motionless with two fingers, the minimum normal force required from each finger can be derived from the balance of forces acting on the cone. The friction force, which is dependent on the static coefficient of friction (u) and the normal force (N), must counteract the gravitational forces acting on the cone. The equation u(F + mg sinθ/2) = (mg cosθ)/2 is established to relate these forces. The minimum finger force (F) can be expressed as F > (mg/2)(cosθ/u - sinθ), indicating the relationship between the angle θ and the forces involved. Understanding these dynamics is crucial for solving the problem effectively.
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Homework Statement



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?

Homework Equations



Friction force = uN
Gravity = mg
etc...

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?
 
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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.
 
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Shameless bump :(
 
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