Holding a cone

With two fingers, you hold an ice cream cone motionless upside down. The mass of the cone is m, and the coefficient of static friction between your fingers and the cone is \mu. When viewed from the side, the angle of the tip is 2\theta.

  1. What is the minimum normal force you must apply with each finger in order to hold up the cone?
  2. In terms of \theta, what is the minimum value of \mu that allows you to hold up the cone?

Assume that you can supply as large a normal force as needed.

Screenshot 2015-01-18 at 13.37.02

Problem Source : Introduction to Classical Mechanics

We can see that the horizontal partial forces acting on the cone cancel each other out. So, we only need to concern the vertical partial forces.

\begin{aligned}  2f \cos \theta &\geq mg + 2 F \sin \theta \\ 2\mu F \cos \theta &\geq mg + 2F\sin \theta \\ F &\geq \frac{mg}{2(\mu\cos\theta-\sin\theta)}  \end{aligned}

The normal force needed depends on \mu\cos\theta-\sin\theta. It is safe to say that it must be larger than 0.

\begin{aligned}  2(\mu\cos\theta-\sin\theta) &\geq 0 \\ \mu &\geq \tan\theta  \end{aligned}

Holding a cone

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