Force exerted on a cone

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?

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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.

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Just want to add that you'd have to subtract the gravitational acceleration from your $a$ to get the cone's actual acceleration if you hold it upright as in the OP's question. And the speed is then calculated as $v(t) = (a-g)t$. – Wouter Feb 10 '13 at 13:50
I'm sorry but I don't find this correct. Forces perpendicular to the cone can't make it move in the horizontal direction. The only answer to this question is when the forces are acting perpendicular to the surface, to which you do get the same result that you got. (That of course, as you can see is because the work calculation is same for both the cases). – Alraxite Feb 10 '13 at 15:53
The cone not moving horizontally may seem contrary to intuition but that's because your fingers end up applying a force perpendicular to the surface of the cone instead. It's hard to apply a force directly perpendicular (along $y$-axis) to the cone, but if you do manage it, it will not move. I think you may say that the actual effective force is not (of course) going to be along the $y$-axis, but then there would be no point of drawing the forces in that direction in the first place if they are going to act perpendicular to the surface instead. – Alraxite Feb 10 '13 at 15:54
@Alraxite: it's a fair cop. That's why I used a work calculation. – John Rennie Feb 10 '13 at 18:17
My apologies for the late response, but thank you for that answer. I used the icy cone as a simple way of describing the action. In reality I'm considering the cone in space with no interaction with gravity as Wouter pointed out. This is all world-building research for a fictional story. I was investigating to see if (with added elements) this could be used as a method of propulsion. Liberties are taken with HOW the force is generated, but I wanted to remain as realistic as possible. I'm going to dust off the calculator now to see if I can make sense/use of this. Thanks again! – Wes Feb 11 '13 at 3:37