Which force acts perpendicular to the side faces of an axe head?

[NathanRB]

Homework Statement

Which force acts perpendicular to the side faces of an axe head, if this axe head encloses an angle of 10 degrees (the sharpness of the axe, so to say), and a force of 1000N is acted on the back of the axe head as it cuts through a piece of wood.

Relevant Equations

Fk = F * sin (0.5 α)
Fx = Fk * cos (0.5 α)

Description:
F: Force acted on the back of the axe head
Fx: horizontal force
Fk: Force acting perpendicular to each of the side faces of the axe head
α: angle enclosed by the two side faces of the axe = 10 degrees

Say that these pictures are accurate.

Personally, I think I can solve this problem, but the issue is, that I had a debate on it with my Physics teacher.

This is how my teacher would solve it; my teacher says that each Fk on each side is constructed from half of F and Fx. So, F = 2 * Fk * sin (0.5 α), hence Fk = F / (2 * sin (0.5 α)) and Fx = Fk * cos (0.5 α).
The question is actually to find the force acting perpendicular to the faces, but for some reason, my teacher went and calculated Fx instead of just calculating Fk. That is not the issue, however.

My issue with this is, as the angle α approaches 0, the Force Fk and also Fx approaches infinity. I presented this fact and my teacher said that as the angle becomes closer, the equation doesn't apply anymore because if the angle is so small, it wouldn't be an axe anymore, and rather, a knife, and we would need another equation for that. I personally think that explanation makes absolutely no sense.

This is actually how I would solve it. I think that Fk and Fx are components of F. So, Fk = F * sin (0.5 α) but still, Fx = Fk * cos (0.5 α). I think it's the right way to go, but I'm here because I'm open to any explanation. If I am wrong, please do explain why or how. If my teacher's method is wrong, it would be great if more than one person, would be willing to give their opinions (so it would be enough to convince my teacher). And if both methods are wrong, please provide the correct method and explanation. Thank you!

 

[mfb]

Your teacher is correct - if we can neglect friction, an "axe" with zero opening angle and finite contact area with the material is just a frictionless knife and has no vertical forces (in the orientation shown here).

 

[NathanRB]

Your teacher is correct - if we can neglect friction, an "axe" with zero opening angle and finite contact area with the material is just a frictionless knife and has no vertical forces (in the orientation shown here).

But I still don't understand why it would be a valid answer if the horizontal Force Fx approaches infinity as the angle approaches 0.

 

[mfb]

Your axe is getting better, basically. In practice it is more useful to consider the opposite case: Keep Fx constant and consider what happens to the vertical force: It goes towards zero. If your axe is converted to a super sharp knife the force needed to push it into the material goes down.

 

[NathanRB]

Your axe is getting better, basically. In practice it is more useful to consider the opposite case: Keep Fx constant and consider what happens to the vertical force: It goes towards zero. If your axe is converted to a super sharp knife the force needed to push it into the material goes down.

I agree that the axe gets "better", as the angle becomes smaller; but I still can't comprehend that it's possible to give 1000N in and the axe produces 5737N out on each side, or that it would be possible to let go an axe head with the mass of 0.68 kg (I searched on Google) and a sharpness angle of 0.66 degrees, touching the surface of a piece of wood and see it chop through the wood because it produces about the same force of 5737N to both sides. Don't these examples violate Thermodynamics somehow?

And can you point out the errors in my method, so that I can understand better?

I'm sorry if I seem stubborn, but I just can't seem to understand it.

 

[mfb]

If you stop a door with a wedge you use the same concept. The small forward force from the door is enough to block the door - the vertical force is so big that the wedge doesn't slide across the floor despite being pushed by the door. Smaller wedge angles work better.

It is a bit like a lever. If your axe is very sharp you need to move it inwards a lot for the same gain in width.

 

[NathanRB]

If you stop a door with a wedge you use the same concept. The small forward force from the door is enough to block the door - the vertical force is so big that the wedge doesn't slide across the floor despite being pushed by the door. Smaller wedge angles work better.

It is a bit like a lever. If your axe is very sharp you need to move it inwards a lot for the same gain in width.

It's very hard to wrap my mind around, but every single source before I came to this forum also say the same thing, so I'll just have to accept it. Thanks anyway, for the replies, I really appreciate them.