I Question about force from a torque arm extension

[vtsteam]

I have a question that I've gone back and forth on several times, trying to visualize it's driving me nutz! It relates to a Prony brake I've built for a model hot air engine, but I'm going to generalize it here as best I can as just a mechanics problem. See attached sketch.

My question is, with a given stable torque, and a horizontal arm of length (A), and another vertical arm (B) normal to the first which is on a scale (top drawing):

Does the scale read differently if you increase the length of the vertical arm? Like say increasing to 2B? (bottom drawing)

In other words, is the effective arm in both cases, the diagonal (C) the distance between the center of pivot and the contact point on the scale. Or is it simply A?

Thanks for any clarification! It really is bugging me.

ps. if you want any details about the actual engine or brake I can give a link, but I didn't want to complicate the question.

thanks again

 

[vtsteam]

Still thinking of how to think!

Maybe if I imagine a circle around the pivot with radius A. Then the vertical arm B is tangent to it. No matter how long it is, it's still on the tangent.

So any force on B is in line with it. So the scale should read the same no matter how long it is. It's just straight compression.....?

 

[A.T.]

In other words, is the effective arm in both cases, the diagonal (C) the distance between the center of pivot and the contact point on the scale. Or is it simply A?

If the scale surface is friction-less, so it only transmits vertical forces, then the lever arm is simply A, so it will read the same in both cases.

Otherwise it is a statically indeterminate problem, because you can have many different force vectors from the scale to balance that torque.

 

[berkeman]

Are you familiar with basic vector math? That's the easiest way to visualize what is going on, IMO. Also, are you familiar with Free Body Diagrams (FBDs) already?

 

[vtsteam]

Thanks @A.T. for practical engine measurements purposes it is low enough - a brass rounded stylus on a stainless steel scale surface, and this whole thing is small -- measuring single digit Watts.

Thanks @berkeman I'm familiar with only pretty basic graphic vector solutions.... too long since college, I'm 76.

Here's a pic of the first setup, until I started to doubt the wisdom of the brass stylus. Then went back and forth on it, and finally arrived here...

 

[berkeman]

I'm familiar with only pretty basic graphic vector solutions....

I'm not completely understanding the setup and the question, but after my first reading of your OP, it seemed like you were asking if the same torque would produce different forces through different lever arm lengths (which of course they do). There are some angles involved in your diagrams, but the basic vector equation is just $\vec \tau = \vec r \times \vec F$

 

[vtsteam]

hi Berkeman, no it was the variable length of the vertical extension B not the lever arm A. I was wondering if varying that made any difference in the scale reading. And so far the conclusion is that it doesn't (in a friction free connection to the scale)

 

[Baluncore]

Here's a pic of the first setup, until I started to doubt the wisdom of the brass stylus.

That should work OK, so long as the arm remains parallel to the scale surface, and you press the tare button to cancel the unbalanced weight of the brass bolt and torque arm.

 

[berkeman]

no it was the variable length of the vertical extension B not the lever arm A.

Which changes your length C and some angles. Can you draw the FBDs with angles shown and do the torque equation calculations?

 

[vtsteam]

Thanks @Baluncore, I agree.

Also, re. taring procedure: I lift the arm before running the engine. It is loose with no brake pressure at that point. I turn on the scale and then lower the arm to get the static weight, and record that. Then after running and taking readings, I transfer the weight and readings to a spreadsheet I wrote which makes the tare adjustment from the static weight, as well as calculating the Newton-Meters and Watts (using the recorded RPM reading at each data point).

I found that way was the best way to go because the scale sometimes times out during a run, and shuts off. And re-taring the static weight of the arm on the fly is impossible without stopping the engine. The arm weight doesn't vary, anyway, so just taking it once is enough.

@berkeman thanks. I believe (now) that distance C (an imaginary diagonal) is irrelevant. I can't do the math you are asking for, but I do feel I have an adequate mental picture now, once I considered that B is a tangent to a circle with radius A around the pivot point.

The force transferred in a tangential direction is always perpendicular to the radius at that point. No matter how long the tangent line is, the momentary force is along that tangent line. That is the vector direction, if you will.

Since the scale is perpendicular to that tangent (and force) it will always read the same, whether the tangent is short or long.

 

[jack action]

I believe (now) that distance C (an imaginary diagonal) is irrelevant.

In your setup, it is relevant. The force $F$ at the tip of the bolt on the scale is with respect to the length $C$ ($F=TC$), and that force is perpendicular to the arm $C$.

Your scale can only read the vertical component of the force $F$. Technically, you should do some math to find the true value of $F$:

So the relationship between the torque $T$ and the vertical force $F_v$ would be:
$$T = FC = \frac{F_v}{\sin \theta}C$$
With your setup, the difference might be negligible, but it is the correct way to look at it.

To eliminate that problem, you need to change the way you fix your vertical link. It should be fixed with respect to the scale, and arm $A$ should only rest on top of the vertical link $B$. That way, the torque $T$ is $FA$ where $F$ is truly perpendicular to the scale, thus no correction needed on the reading. You are basically extending the scale surface instead of extending the arm.

 

[A.T.]

Thanks @A.T. for practical engine measurements purposes it is low enough - a brass rounded stylus on a stainless steel scale surface,

Brass on steel can have a static friction coefficient of up to u = 0.51 (dry) or u = 0.19 (lubricated):
https://www.schneider-company.com/coefficient-of-friction-reference-chart/

Your max. relative error for the torque will be roughly u * B / A.

To minimize the error you can:
- lubricate the scale surface
- attach a bearing / roller to the stylus tip
- make B as small as possible compared to A (lift the scale instead of extending the stylus)
- make sure A is parallel to the scale surface

 

[vtsteam]

In your setup, it is relevant. The force $F$ at the tip of the bolt on the scale is with respect to the length $C$ ($F=TC$), and that force is perpendicular to the arm $C$.

Your scale can only read the vertical component of the force $F$. Technically, you should do some math to find the true value of $F$:

View attachment 361811​

So the relationship between the torque $T$ and the vertical force $F_v$ would be:
$$T = FC = \frac{F_v}{\sin \theta}C$$
With your setup, the difference might be negligible, but it is the correct way to look at it.

To eliminate that problem, you need to change the way you fix your vertical link. It should be fixed with respect to the scale, and arm $A$ should only rest on top of the vertical link $B$. That way, the torque $T$ is $FA$ where $F$ is truly perpendicular to the scale, thus no correction needed on the reading. You are basically extending the scale surface instead of extending the arm.

Yes, and I believe in this case theta = 90 degrees. I may be wrong, but this is exactly the crux of why I've gone back and forth on this. And looking at the posts above by others, why they don't all seem to agree.

(btw in practical reality I did shorten the length of B as much as I could, and raised the scale a few days ago, hedging my bets. But I am still interested in this as a theoretical problem. I do want to understand, for sure.)

 

[A.T.]

And looking at the posts above by others, why they don't all seem to agree.

BTW: I do agree with this suggestion:

To eliminate that problem, you need to change the way you fix your vertical link. It should be fixed with respect to the scale, and arm $A$ should only rest on top of the vertical link $B$.

This is, in principle, equivalent to lifting the scale, so B = 0. The horizontal friction between the vertical and horizontal link will not generate any torque around the motor axis, if they are on the same height.

However, due to their internal workings, scale readings of the vertical force might be skewed, if you additionally apply a torque to the scale surface. And the setup with the vertical link fixed to the scale would enable this. So, you it might be safer to stick with your setup, and just minimize the friction the stylus has on the scale surface.

 

[vtsteam]

I believe I am not applying torque to the scale surface, assuming everything is square, flat, and level, and the scale uses a solid pressure transducer, not a spring action. I doubt whether greasing the top of the scale or adding a strip of Teflon, would make a measurable difference. But I certainly could do that and check the difference. But that is a practical matter anyway, and I really, trying to clarify the theoretical situation -- the ideal case.

As I see it now, this situation can be mimicked by a weight on a pin ended beam in statics. No matter what length support is put under the end to the fixed surface, it still sees the same vertical force (100grams). As long as everything is, again, square, straight, level, and unmoving. Here's a pic of how I'm imagining this as a static situation:

And in that situation there is no horizontal component, so no friction. Again this is the ideal case I'm trying to picture.

 

[A.T.]

As I see it now, this situation can be mimicked by a weight on a pin ended beam in statics. No matter what length support is put under the end to the fixed surface, it still sees the same vertical force (100grams). As long as everything is, again, square, straight, level, and unmoving. Here's a pic of how I'm imagining this as a static situation:

And in that situation there is no horizontal component, so no friction. Again this is the ideal case I'm trying to picture.

'No friction' is an assumption or specification you have to make, in order to conclude, that the vertical force on the fixed surface is 100g * g.

It doesn't follow from the whole thing being static alone, because it is staticaly indeterminate, if friction is not assumed to be zero.

 

[vtsteam]

It sounds like you are talking about the real life brake, and not the idealized case drawn above, but maybe I'm wrong about that?

 

[A.T.]

It sounds like you are talking about the real life brake, and not the idealized case drawn above,

I am talking about what I quoted. It's not clear what your assumptions and what your derived conclusions are there.

 

[vtsteam]

Okay in that case I was talking about an Idealized state of equilibrium as a means of visualization for myself. If it's unclear to you or others, I apologize.

In the case of the real life brake, I believe that friction could have a small effect because of vibration of the stylus on the scale. How much of an effect, I don't know. My guess is not much. That makes my last statement a theory. It could be totally wrong. An experiment could either back that up or disprove it. It would be fairly simple to get a scale reading dry while running, and then add a tiny amount of lubricant and see if the scale reading reduces.

 

[A.T.]

Okay in that case I was talking about an Idealized state of equilibrium as a means of visualization for myself.

And my comment was, that equilibrium alone doesn't imply that there is no friction, because you can have equilibrium with or without friction here.

In the case of the real life brake, I believe that friction could have a small effect because of vibration of the stylus on the scale.

If your point is, that you can construct this setup carefully, such that there is indeed no relevant friction, then that's correct. You just cannot assume it from equilibrium alone.

It would be fairly simple to get a scale reading dry while running, and then add a tiny amount of lubricant and see if the scale reading reduces.

That would be the simplest possible test for your practical purpose: to check how much friction affects your specific measurement.

However, if you are interested in the general case and want to see that the setup is statically indeterminate, you would have to compare best case to the worst case:
- minimal friction: with lots of lubricant or ideally a bearing roller
- maximal friction: with hi-friction-rubber or double-sided-tape, scale shifted horizontally towards / away from the axis until the traction almost breaks, while the motor applies the torque.

 

[256bits]

@vtsteam
Is this the friction part of the Prony break?

If so, it seems to have contacts at the left and right, rather than just top and bottom.
That can skew your lever arm length A.

 

[erobz]

Sum torques about $O$:

$$ T -\mu N b - aN = 0 $$

$$ \implies N = \frac{T}{\mu b + a } $$

I imagine if you take proper care to set it up, and that deflections are not large it should be in the ballpark.
At any rate, even if a small bit of friction is present at the tip, how sensitive is the scale to lateral loads?

 

[vtsteam]

A.T. In the case of the drawing, it is ideal perfect construction, and the line representing the beam has no mass, is absolutely rigid, measurements are perfect, etc, etc. It's a an ideal representation which I was using to picture (to myself at least) why the direction of the force on the scale in an idealized case was not normal to a diagonal force from the center of the shaft. That would have added a horizontal component to the force on the scale. In other words an angled vector, rather than a strictly vertical one.

In real life beams flex, measurements aren't perfect, construction isn't, parts vibrate, etc. etc. We all know this, including me. Which is why I was trying to establish whether you were referring to the idealized case of the drawing, or the real life case of the wooden brake.

In the idealized case there is no movement. It is static because all parts are rigid have no thickness or mass , etc. and the force is vertical on the surface which is perpendicular to it. There are no horizontal components of force.

My understanding of the word friction requires an oppositional force between two contacting surfaces. There is none in my idealized drawing. If your definition of friction is different, then that could explain the difference in our imaginary scenarios. It will suit the individual definition.

Switching consideration to the real life brake with all its real life aches and pains, and my own imperfections, etc. Yes it is carefully built within the simple expectations I have for the level of readings I need. That is about +- .1 gram on up to about 20 grams expected experimental range, including weight of arm, etc. The scale reads in tenths, so it's probably even cruder than that, but adequate for the range I need for testing different displacers in the engine, etc. And of course relative comparisons are easier to achieve than absolute measurements. It's a certainty that my readings will contain some flaws, as a result of all these real world complications, and I'm willing to minimize those through any suggested improvements, if they make a reasonable difference. If lubricating the point of the stylus makes a difference in the above scenario, lubricate I will!

Thanks @256 bits, you're seeing the nuts loose and the block raised a lot before tightening to get a reading. Also there is a metal flange on the drum in front of a relieved area on the face of the wood block. The real contact area is inside this. The blocks were bored on a lathe to a .001" interference fit, and that area is out of sight. In use, the contact pattern on the actual wear area shows it is even on the top of the block.

The photo shows the thumbscrews not tightened, and no springs under them, because it is an early demo running pic with no pressure, and nothing permanent. All components are now bolted down on a permanent base, etc.

Sorry, I really hadn't meant to get into an engineering discussion of the real brake here, but just wanted a clarification in my mind about whether measuring the arm center to center distance was reasonable for calculating the torque with a scale, or whether I had to measure that arm on a diagonal to the point of cvontact on the scale, as a result of using (originally) a longish stylus.

 

[vtsteam]

Hi @erobz, just saw your post after I wrote mine. Thanks for the great diagram and formulas! The scale uses a pressure cell rather than a spring and has a kilogram range, while my actual measurements are in the sub 20 gram range, so deflection is nil. I would say that the only horizontal forces would be the result of vibration from the engine. It is a small Stirling engine, and everything is solidly bolted down to a single base plate now. I mean, I think the setup is reasonably good for output measurements for a hobbyist experimenting with mechanical engine changes. I just wanted to make sure I wasn't making a serious blunder by thinking the moment arm was length A (in my original post) instead of length C.

 

[A.T.]

There are no horizontal components of force.

If that is ensured, then my first post already answerd the question:

If the scale surface is friction-less, so it only transmits vertical forces, then the lever arm is simply A

 

[256bits]

Thanks @256 bits, you're seeing the nuts loose and the block raised a lot before tightening to get a reading.

Another question.
Is the mating surface for the brass pin exactly flush and even?
A mis-alignment could have the force downwards being transmitted by error of the radius of the brass pin lengthening or shortening lever arm A, rather down the centre of the pin, with then some friction then also being possible at the scale, since the downward force is not exactly vertical ( assuming the pin has a rounded tip where it meets the scale ).

 

[vtsteam]

I'm satisfied with an understanding of the mechanics now, which contributors to this thread helped me to come to. It's a great resource, and I'm glad you all took the time to think about it and explain. As for the actual physical brake. I think it's going to work well for me, as presently constructed, within a realistic range of precision. I'll keep an eye on this thread in case something comes up to substantially challenge my understanding. I'm going to switch over to actually doing something with it now -- taking a series of measurements with different displacer cylinders and seeing what happens. Thank you all!

 

[Patxitxi]

I have a question that I've gone back and forth on several times, trying to visualize it's driving me nutz! It relates to a Prony brake I've built for a model hot air engine, but I'm going to generalize it here as best I can as just a mechanics problem. See attached sketch.

My question is, with a given stable torque, and a horizontal arm of length (A), and another vertical arm (B) normal to the first which is on a scale (top drawing):

Does the scale read differently if you increase the length of the vertical arm? Like say increasing to 2B? (bottom drawing)

In other words, is the effective arm in both cases, the diagonal (C) the distance between the center of pivot and the contact point on the scale. Or is it simply A?

Thanks for any clarification! It really is bugging me.

ps. if you want any details about the actual engine or brake I can give a link, but I didn't want to complicate the question.

thanks again

View attachment 361798

You are mixing up torque (then bigger C produces more torque) with pressure (compression).
See sketch.

You need the following (or SIMILAR) if you want to apply more VERTICAL pressure on the SCALE. Then the more length you give to C the more pressure you apply. Note the blue dot must be fixed and rotating.

 

[vtsteam]

Hi, thanks, Please read the whole thread carefully.

 

[vtsteam]

Gotta be AI.

 

[Patxitxi]

Gotta be AI.

Maybe my brain got an update without me knowing it?

 

[vtsteam]

LOL!