What force opens a folding knife?

AI Thread Summary
The discussion centers on the mechanics of opening a folding knife with a flick of the wrist. It clarifies that the force involved is not centrifugal or centripetal but rather the inertia of the blade. When the handle is flicked quickly, the blade lags behind due to insufficient force, eventually snapping into the open position as the handle accelerates. The term "gravity knife" is deemed inaccurate, as gravity does not play a role in this action. Ultimately, the key force at play is the inertia of the blade in relation to the rapid motion of the handle.
bklynrob
Messages
2
Reaction score
0
If I flick open a folding knife with a flick of my wrist, what force is that?

Is it centrifugal force, centripetal force, or some other force?
 
Physics news on Phys.org
The way you phrased it, it could be a whole lot of forces. Let's stay in 2D.

Suppose you pull linearly towards your knuckles and keep going at constant velocity. The blade's C of G will want to stay put, causing a moment about the pivot, which can open it.

Alternatively, you could spin your little finger knucklewards keeping the pivot in one place. That can also open it. Or give a push the other way to get the blade moving and then reverse the handle.

If you were nifty with your kit, I reckon you'd also take advantage of the ballet dancer's trick of pulling their arms into increase angular velocity. I think that would mean moving the pivot towards a place just behind the backbone of the blade once it was already moving.

But by this time you'd already have a bottle in your face, which is what I would have used in the first place.

Happy drinking,
Adrian.
 
Adrian, thanks. I'm sure I worded my question poorly. Here's another try: there are certain folding knives (knives with blades that fold into the handle via a pivot on one end) which have a blade that is loose enough that, when you hold the handle and flick your wrist, the blade will snap out and lock into place.

Some call these "gravity knives" but I'm certain the term is wrong. Gravity plays no role in the blade opening, right? Some laws say these knives open with centrifugal force, but I'm not sure that's right either. It seems to me, as a layman, that the force of flicking your wrist coupled with inertia or centripetal force is what causes the blade to snap into place.
 
bklynrob said:
If I flick open a folding knife with a flick of my wrist, what force is that?

Is it centrifugal force, centripetal force, or some other force?

Due to there being no such force, you can be certain that it is not centrifugal force - though many a non-physics person will claim it is (even a knife manufacture/advertiser may be tempted).

I think it is most likely the inertia of the blade that results in it to opening. You force the handle to accelerate very quickly, and since there is insufficient force transmitted to the blade, it is "left behind" until it snaps into the "open" position,at which point the handle supplies sufficient force to let the blade then keep up with the handle.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top