Mechanical Advantage of a Compound Wedge

[goodboybaddog]

Homework Statement

I have a knife with compound bevel edge. It is going through a block of cheese. What is the mechanical advantage at point 1 and point 2. See my solution below. I think it would be the same as a single wedge which is length of wedge over width of edge. Please correct me if I am wrong.

Relevant Equations

Length of wedge/width of wedge = mechanical advantage

Length of wedge/width of wedge = mechanical advantage

 

[phinds]

So, you are saying that the mechanical advantage of a wedge 3 feet long and 1 inch wide is enormously greater than that of a 2 inch long wedge 1 inch wide?

 

[hmmm27]

Friction :

- a thinner blade means an easier cut since the sides of the knife aren't pushing the material out of the way as much, but it also means the knife is prone to breakage

- differing angles on the side of the blade means different coefficients of static friction (relative to the force applied to the knife), which means the blade won't get stuck as much, with a compound edge.

 

[haruspex]

Length of wedge/width of wedge = mechanical advantage

Please clarify what you mean by length and width in this context.

 

[haruspex]

differing angles on the side of the blade means different coefficients of static friction

Perhaps you mean different normal forces. The coefficient should be taken to be the same. And kinetic friction would be more relevant?

 

[hmmm27]

^ yes, thanks : stuck for words, why I put "relative to...".

 

[goodboybaddog]

Please clarify what you mean by length and width in this context.

please view photo attached to post. I am referring to that the length from tip to second wedge end of the compound divided by the width or horizontal displacement equals the mechanical advantage of the compound blade.

 

[haruspex]

please view photo attached to post. I am referring to that the length from tip to second wedge end of the compound divided by the width or horizontal displacement equals the mechanical advantage of the compound blade.

Sorry, still not clear. Is "second wedge end" where the two wedges meet or the top of the upper wedge? If the former, how are you defining m.a. for the upper wedge?

Btw, this is perhaps an odd use of the concept of mechanical advantage. It would be perfectly appropriate if the objective were to achieve a given separation laterally. E.g. we could use a wedge to drive apart two blocks so as to insert something between. A narrow wedge makes the force required less, but it has to be exerted over a greater distance for the same separation.
But to cut a cheese block we are only concerned with achieving some separation, no matter how small. The force has to be exerted over the same distance regardless of the wedge angle.

 

[goodboybaddog]

So, you are saying that the mechanical advantage of a wedge 3 feet long and 1 inch wide is enormously greater than that of a 2 inch long wedge 1 inch wide?

its not so much the length as it is the angle or slope. the steeper the slope the higher the MA. i assume being that the total displacement is covered once the blade is engaged it would be the total length divided by the total width of the blade. At least i think this would be true for a compound blade.

 

[goodboybaddog]

also to be clear this is an ideal example where no friction or energy is lost beside for cutting.

 

[haruspex]

Incidentally, it is easy to disprove the claim in the image. Consider the third section of blade (parallel sides) as a third wedge. The analysis says that the m.a. will continue to increase as the cut proceeds, which is obviously nonsense.
Unfortunately a full analysis would require a model for the behaviour of the material being cut.

 

[goodboybaddog]

Incidentally, it is easy to disprove the claim in the image. Consider the third section of blade (parallel sides) as a third wedge. The analysis says that the m.a. will continue to increase as the cut proceeds, which is obviously nonsense.
Unfortunately a full analysis would require a model for the behaviour of the material being cut.

that is not part of the wedge system. it does not contribute to displacement. I would not think that would be a way to disprove this. I would think the standard input displace over output displacement average would suffice for MA for this ideal system.

 

[goodboybaddog]

also to be clear mechanical advantage at the 2 points is referring to the blade engagement. so when only the tip is engaged verse the full length of the blade.

 

[haruspex]

that is not part of the wedge system.

That does not invalidate my argument. E.g. take it as as an extremely shallow wedge, an angle of 0.1°, say.
At the same time, take the first wedge to be extremely blunt. The claimed solution is obviously wrong.

You could try a model in which the surrounding material exerts a constant and uniform horizontal pressure on the blade, ignoring friction. See how much work is done in pushing the blade down dx when (i) only the wide wedge is engaged (ii) after the second wedge is engaged.

 

[Tom.G]

I get the M.A. of the second wedge (the one away from the cutting edge) as 6.5.
They wrongly computed the slope from the cutting edge instead of from where the taper starts, 0.78mm from the edge at a thickness of 0.7mm.

9.2-0.78= 8.42 length of run
2-0.7= 1.3 thickness
8.42/1.3≅ 6.5https://www.physicsforums.com/attachments/reddit-question-jpg.253227/

 

[haruspex]

They wrongly computed the slope from the cutting edge instead of from where the taper starts

But why is that right? That would give the same result as if the blade were much deeper and sharper, using the shallower taper all the way down. Wouidn't you expect that to make cutting easier than with the actual blunter tip?

 

[goodboybaddog]

I get the M.A. of the second wedge (the one away from the cutting edge) as 6.5.
They wrongly computed the slope from the cutting edge instead of from where the taper starts, 0.78mm from the edge at a thickness of 0.7mm.

9.2-0.78= 8.42 length of run
2-0.7= 1.3 thickness
8.42/1.3≅ 6.5https://www.physicsforums.com/attachments/reddit-question-jpg.253227/

I would think in this case being it is soft cheese that both wedges would be engaged. Being that a formula to find mechanical advantage is displacement input over displacement output i think this would hold up. Even if you were to extend the edge beyond the 9.2 mm the width would increase which would bring you closer to the steep slopes MA but you would never quite reach it due to the blunt point on the tip.

 

[haruspex]

I would think in this case being it is soft cheese that both wedges would be engaged.

So did you try my model in post #14?

 

[haruspex]

Thinking about this some more, the nature of the material being cut is critical.

With a block of cheese, and a knife that extends across the block (as opposed to, say, pushing a screwdriver into the block) the cheese will exert little pressure on the upper part. The blunter tip will push the cheese apart at such an angle that there might not be any contact higher up. Thus, the M.A. might never be much more than that afforded by the lower wedge.

In the screwdriver example things are different because the surrounding material will help to keep lateral pressure on all the way up.

With a wedge driven into a block of wood, the split in the wood may precede the tip of the wedge. As a result, the principal lateral force could be at the top of the block (in which case the taper of the wedge at that point determines the M.A.) or at the shoulder between the two wedge tapers.

All of the above is still ignoring friction.