Using a hexagonal key on a square-headed screw

[Heirot]

Homework Statement

Suppose that you want to undo a square-headed screw of side $$a$$ using a key that has a regular hexagon hole of side $$b$$. What relations must hold between these two number in order for you to succed?

Homework Equations

Basic geometry and trigonometry.

The Attempt at a Solution

The obvious solution is that the square must be the largest one that can be inscribed in a given hexagon, e.g. see here (2nd. picture): http://www.yucs.org/~gnivasch/cube/index.html .

I have a problem visualizating the situation for a slightly smaller square - can it also be undone with the same key? Of course, now there wouldn't be four point of contact between the key and the screw, so my question is, could you undo a screw with three points of contact? If yes, what's the lower limit on a size of a screw?

Thanks!

 

[Dick]

I would say if the radius of the circumcircle of the square is larger than the radius of the inscribed circle of the hexagon it would work. That gives you two points of contact, doesn't it? It gives you nonzero torque.

 

[Heirot]

Yes, that seem very reasonable. The thing that's really bothering me - are two or three points of contact enough to eliminate slipping? Shouldn't the screw be loose in that case?

 

[Dick]

Yes, that seem very reasonable. The thing that's really bothering me - are two or three points of contact enough to eliminate slipping? Shouldn't the screw be loose in that case?

I don't think the question is about whether it makes a good screwdriver or not. Just whether it can make contact with the screw. Though I would think if it's enough larger than the inscribed circle (but not to large to fit in the hexagon), and you're not worried about stripping the metal if the screw is tight, it wouldn't be that bad.

 

[Heirot]

So, could we say it this way - in case of no friction, there is only one possibility - the largest square inscribed in a hexagon. On the other hand, for perfectly rough material, the smallest screw is given in your example?

 

[Dick]

I don't think the problem is really about friction either, just geometry. If the square is too large to turn freely in the hexagon, i.e. bigger than the minimum then it will jam in the hexagon regardless of friction. I'd just worry about the geometry part.

 

[Heirot]

Yes, you're absolutely right. I see it know. Thanks!