Force exerted on a cone

Question

This is probably a pretty elemental question, but I can't figure out what the concept is called or how to calculate it. Say you have a solid cone shaped piece of ice, you hold it in your hand and squeeze the cone and it slips up as you continue the same force until it slips all the way up and out of your grip. What is this called and how do you calculate the speed of the cones movement with a given amount of force?

Answer 1

I don't think there is a particular name for what you describe, however you can work out how fast the cone would move.

The diagram shows you holding the cone with finger and thumb so you're applying a force $F$ to either side. We'll assume the ice is very slippery so we can neglect friction. The angle $\theta$ is half the cone angle. The cone experiences a force $F'$ in the direction shown so it moves in that direction as you squeeze. The acceleration of the cone is just the force $F'$ divided by the mass of the cone:

$$ a = \frac{F'}{m} $$

So you just need to calculate the force $F'$. There are various ways to do this: I'll do it by calculating the work done as you squeeze the cone.

Suppose the cone moves a distance $dx$ in the direction of $F'$, then the work done is force times distance:

$$ W = F'dx $$

When the cone moves a distance $dx$ your fingers each move inwards by a distance $sin(\theta)dx$, so the work done by your fingers is:

$$ W = 2Fsin(\theta)dx $$

In the absence of friction the work done by your fingers equals the work done on the cone so:

$$ F'dx = 2Fsin(\theta)dx $$

$$ F' = 2F \space sin(\theta) $$

So the cone accelerates as you squeeze it, with the acceleration given by:

$$ a = \frac{2F \space sin(\theta)}{m} $$

Footnote:

Alraxite's comment is quite correct. In an effort to keep the explanation simple I have made it a bit misleading, and as I've drawn the forces they cannot move the cone. What would actually happen is that as you squeeze the cone your fingers will deform and end up applying the force in a different direction. However, because I've used energy balance in my calculation the answer comes out the same, so we don't need to worry about exactly how the force gets applied to the cone. If you're interested in pursuing this (and I'm guessing not!) perhaps ask a another more detailed question and we can address the problem there.