Find the force acting on the roller in the direction of the spring

A torque meter with a triangular slab extension is inserted into a corresponding triangular slot. The C-shaped arm features a V-shaped dent on which a roller is seated. This roller is held in compression by a spring. The roller's positions are labeled '0' for the initial state and '1' for the final state. The torque meter reads 40 in-lb when transitioning from state 0 to state 1. My goal is to determine the force exerted on the spring during this transition.

My current thought process:

1. Find the lever arm distance 'r', this includes the extension length. assume r= 7.5''
2. M = r X F, F = 5 lbs, this is the force applied at the torque meter.
3. Then, using the pressure angles, I think there is a way to determine the force component acting in the direction of the spring. This is the step that I am really confused at.

Is this even a correct approach?
Concerning the torque reading, if the extension length from the torque wrench socket to pivot A is 1.5 inches, does this mean that the actual torque applied at point A is 46 in-lb? The highlighted length represents the extension/adaptor.

JackLee said:

Then, using the pressure angles, I think there is a way to determine the force component acting in the direction of the spring. This is the step that I am really confused at.

The 'V' notch moves along an arc about point A.
The roller rides on the contact angle of the moving notch.
You show two parallel positions for the roller and the spring. What freedom does the roller have, and in what direction is the spring oriented?

JackLee said:

TL;DR Summary: Torque adjustment due to custom extension. Force Transmitted by torque at an angle.

Concerning the torque reading, if the extension length from the torque wrench socket to pivot A is 1.5 inches, does this mean that the actual torque applied at point A is 46 in-lb? The highlighted length represents the extension/adaptor.

From what I can make of your diagram the torque meter tells you the torque everyhere on that U shape. So it tells you the torque about the pivot. (That's how torque works). I can't see that 'r' is relevant. When you push down on the lever, the relevant force on the contact point will be only on the left hand slope, once the roller starts to move. The force on the notch will be normal to its surface (if the roller is free) and that makes things easier to work out.

Your diagram shows the left hand slope line of the Vee passes through the pivot. So you can draw in that line. You then have the normal force producing a torque that's equal and opposite to the torque you measure on your meter (when it just starts to lift). The moment of the force on the roller will be the Normal force times the distance from the roller contact point to the pivot.

The force on the spring will be the normal force, modified by the slope θ.
Try modifying your diagram a see if it makes any more sense and I'll try to help with that modification if it gives you a problem.

Please, anyone, chip in if you can.

The "force exerted on the spring" (difficult) is different from the "force exerted on the spring/roller system" (easier). The connection between the arm and the spring is the roller; the force applied by the roller is radial. The force on the spring depends on the angle formed by that radial and the spring axis.

Imagine that the roller contact point is at 9:00 (on the roller) - no force is transferred to the spring because there is no component of the radial force in the direction of the roller axis.

Imagine that the roller contact point is at 6:00 (on the roller) - the radial force is the spring force.

Your geometry is somewhere between these 2 extremes. The dimensions of the roller/well are required.

Baluncore said:

The 'V' notch moves along an arc about point A.
The roller rides on the contact angle of the moving notch.
You show two parallel positions for the roller and the spring. What freedom does the roller have, and in what direction is the spring oriented?

My bad, the diagram is misleading, the roller and the spring are constrained to only move up and down as it's pushed by the U-shape cam. The two positions are relative to the cam only.

sophiecentaur said:

From what I can make of your diagram the torque meter tells you the torque everyhere on that U shape. So it tells you the torque about the pivot. (That's how torque works). I can't see that 'r' is relevant. When you push down on the lever, the relevant force on the contact point will be only on the left hand slope, once the roller starts to move. The force on the notch will be normal to its surface (if the roller is free) and that makes things easier to work out.

Your diagram shows the left hand slope line of the Vee passes through the pivot. So you can draw in that line. You then have the normal force producing a torque that's equal and opposite to the torque you measure on your meter (when it just starts to lift). The moment of the force on the roller will be the Normal force times the distance from the roller contact point to the pivot.

The force on the spring will be the normal force, modified by the slope θ.
Try modifying your diagram a see if it makes any more sense and I'll try to help with that modification if it gives you a problem.

Please, anyone, chip in if you can.

Yes, the roller is free to roll, and is constrained to move only along spring axis. I think I now understand why that 'r' is irrelevant, even though the force is applied at the handle of the torque meter, it is then transmitted through the whole torque meter + extension + cam system, treating them as one linkage and the location where the force is transmitted to the roller is at the left slope of the Vee. This new R is about .77'', this makes a lot more sense about the force. 40/.77=51.9 lbs.

My diagram was terrible, above is a better representation of the cam; It looks like the left slope does not go through the pivot or I might be confused on that.

I guess 51.9 lbs is the force of the cam (normal to the left slope), to find the component acting in the spring direction (y-axis), it's 51.9 lbs * Cos (90-theta), theta is ~ 46, I get force =37.3 lbs. The spring compressed from .81'' to .70'', a deflection of .11'' (free length is 1.25''), there are total of two springs connected in parallel, so for a single spring, the spring constant is about 170 which is about 5 times higher than old spring data. I am expecting a spring rate of 30 - 50. I am thinking 37.3 - force of friction in the y component, which would be 37.3 - 16.1 = 21.2 lbs. (Static friction force = .6*37.3 = 22.38, static friction force in the y-axis = 22.38*cos (44) = 16.1.)
I still get a crazy high spring rate: (21.2/2)/.11 = 96. The rolling resistance is not accounted, will have to read a little more into it.

JackLee said:

I think I now understand why that 'r' is irrelevant,

A bent handle behaves the same as a wheel of the same radius when pulled tangential to the motion. Gearing systems only consider radii and not where the various gear wheels are placed. The guy who first spotted that deserves a medal.

JackLee said:

I still get a crazy high spring rate: (21.2/2)/.11 = 96.

Why do you need to know the spring rate? Is it all stiffer than you expect or perhaps sloppy?
Do you have the old spring? You could see what load is needed to compress it. Alternatively, if you don't actually have a spring yet, you could use a force measurer to see what force is produced by a given torque. Hold the U bracket in a vise on a pin and measure the force on bathroom scales. Is that spring spring rate just going against your intuition? Imperial makes no sense to my intuition at all. Is it a matter of units?

I'm just thinking of your " 51.9 lbs * Cos (90-theta)".
The force triangle should, I think, indicate

Fspring = Fcam/sin(θ)

That would reduce the spring rate needed. Waddya think?

Dullard said:

The "force exerted on the spring" (difficult) is different from the "force exerted on the spring/roller system" (easier). The connection between the arm and the spring is the roller; the force applied by the roller is radial. The force on the spring depends on the angle formed by that radial and the spring axis.

Imagine that the roller contact point is at 9:00 (on the roller) - no force is transferred to the spring because there is no component of the radial force in the direction of the roller axis.

Imagine that the roller contact point is at 6:00 (on the roller) - the radial force is the spring force.

Your geometry is somewhere between these 2 extremes. The dimensions of the roller/well are required.

I am a little lost on that radial force. Here is a better view of the spring - roller system, both springs and roller are connected by a pin, the spring sits on a guide. The roller moves the pin, and the pin carries the guide which compresses the springs.

JackLee said:

above is a better representation of the cam

I actually managed to miss that improved diagram - sorry!!

But now you have a better idea of the angles, you should be able to get a better force estimate.
It makes sense to have that extra slope on the face of the cam to give an increasing force needed as the roller moves upwards. That makes things stable.

You can do the force calculations at the bottom and at the top of the roller movement and check the force magnitude goes the right way.

sophiecentaur said:

A bent handle behaves the same as a wheel of the same radius when pulled tangential to the motion. Gearing systems only consider radii and not where the various gear wheels are placed. The guy who first spotted that deserves a medal.

Why do you need to know the spring rate? Is it all stiffer than you expect or perhaps sloppy?
Do you have the old spring? You could see what load is needed to compress it. Alternatively, if you don't actually have a spring yet, you could use a force measurer to see what force is produced by a given torque. Hold the U bracket in a vise on a pin and measure the force on bathroom scales. Is that spring spring rate just going against your intuition? Imperial makes no sense to my intuition at all. Is it a matter of units?

I'm just thinking of your " 51.9 lbs * Cos (90-theta)".
The force triangle should, I think, indicate

Fspring = Fcam/sin(θ)

That would reduce the spring rate needed. Waddya think?

The goal is to properly re-define the spring rate; they are currently incorrect and failing the test. I suppose I wanted to brush up on my rusty physics too, and to have an analytical method for verification.
COS 44 = F_s / 51.9, no?

On a side note, is this a similar problem to a wheel going over a bump/step? (The difference here is that the roller is not applying the force, it is the cam that is applying the force.) Would this be equivalent to the wheel being stationary and the bump runs into the wheel?

sophiecentaur said:

I actually managed to miss that improved diagram - sorry!!

But now you have a better idea of the angles, you should be able to get a better force estimate.
It makes sense to have that extra slope on the face of the cam to give an increasing force needed as the roller moves upwards. That makes things stable.

You can do the force calculations at the bottom and at the top of the roller movement and check the force magnitude goes the right way.

No worries! What extra slope were you referring to?The radius where F_cam peak is?

JackLee said:

No worries! What extra slope were you referring to?The radius where F_cam peak is?

The original sketch had the face of the cam in line with the pivot. Your improved version shows the face leaning away from that line.

JackLee said:

(The difference here is that the roller is not applying the force, it is the cam that is applying the force.)

There are equal and opposite forces between cam and roller in all positions. If the forces were not equal then there would be a resulting motion but the operation is slow and we can ignore any acceleration. This is another potential intuition problem in mechanics.

JackLee said:

I suppose I wanted to brush up on my rusty physics too, and to have an analytical method for verification.

Remember that there is another force involved and that is the lateral force on the rod and slot (the guide) that hold the spring. The compression force and this lateral force add vectorially to produce the force from the cam. In your diagram you only have two forces when there must be a third force. The two forces act together 'against' the force from the cam.

Are you trying to do all this with no extra help from a text book? Referring to your improved diagram, you need to use the fact that the cam force is no longer at right angles to the line to the pivot. The torque (moment) is strictly force times perpendicular distance from the pivot and on your diagram, the angle of the face is appreciable. The normal force on the roller needs modifying. (Sorry if this is obvs.) If you stick to symbols in your equations then it gets the message across better than "44" and "51.9". (which I call "magic numbers") The numbers should all go in at the last minute

"I am a little lost on that radial force."

Assuming that's still the case:

For a static analysis of the roller, you may legitimately assume that the applied force is along a line between the roller contact point and the roller axle. This force may be broken into analytically useful components: along the spring axis, and orthogonal to the spring axis - Sophiecentaur explains it pretty well in post #12.

JackLee said:

Here is a better view of the spring - roller system, both springs and roller are connected by a pin, the spring sits on a guide. The roller moves the pin, and the pin carries the guide which compresses the springs.

Could you explain the number parts corresponding to the roller and the pin?
Sorry, I fail to see anything resembling a symmetrical wheel in that drawing.

JackLee said:

I am a little lost on that radial force. Here is a better view of the spring - roller system, both springs and roller are connected by a pin, the spring sits on a guide. The roller moves the pin, and the pin carries the guide which compresses the springs.

That diagram of the roller carriage is from the wrong angle for your calculations. The diagram needed would be pictured from a direction of the roller axis and the pivot.

Without friction or distortion, the force on the circumference of the roller will always be radial (through the axis of the roller. That will be the equilibrium position - any other angle and the roller will rotate and sit on a different position on the cam face.

@JackLee You seem not to have drawn a diagram with all the relevant angles labelled (symbols not numbers). The angle between the axis of the spring carrier and the the line from the pivot and AND the angle between the cam contact point and the pivot. If you do the calculation for two positions, you will get the force on the spring (for the applied torque) and its displacement. That will give you the spring rate.
Your problem may be harder than you thought when you started on it. It's doable but a bit 'fiddly' as a starter problem.