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[meccanicamg]

Here we are with a little puzzle.
I have three wheels: one in the bottom big and two small in contact. on the big wheel there is a force upwards of f.
to determine reactions to small wheels, how do I break forces? as a model on the left or as a model on the right?
I feel like I've become no longer able to find the way.
then how do I compose to calculate the result on the bearing? r3 and r1 are already radial. r2 is perpendicular to r1 and therefore opposes rotation. r4 should be decomposed along the perpendicular of the radial and parallel line to be added with r3. On the other side should I do pitagora between r1 and r2?
Or isn't it okay?Can you give me your opinion, please? I'm losing myself in a glass of water.
I think the right solution is the right one.

 

[meccanicamg]

so if I use the right model I have that r4 breaks into r5 and r6 that are the tangential and radial.then the right roller bearing will have as radial equivalent |r3-r5|. but r6 how does it react to the bearing? transport to the center and do pitagora with |r3-r5|?

but therefore do not get the same result that I calculated on the left directly?

 

[meccanicamg]

or with such simplicity will we have to break down along the two radial directions to get r7 and r8?

 

[volaff]

for how the system is schematized is symmetrical compared to the f action line: I imagine therefore that the reaction will also be symmetric.

the force between small and large wheel is exchanged at the contact point: then you have f/2 to the left and f/2 to the right.

such reaizone will be decomposed in radial and tangential direction.
for me it is left solution then mirrored symmetrically to the right.

I'm waiting for someone to say his.

 

[meccanicamg]

for how the system is schematized is symmetrical compared to the f action line: I imagine therefore that the reaction will also be symmetric.

the force between small and large wheel is exchanged at the contact point: then you have f/2 to the left and f/2 to the right.

such reaizone will be decomposed in radial and tangential direction.
for me it is left solution then mirrored symmetrically to the right.

I'm waiting for someone to say his.

definitely left and right must be equal.
but....to the left is done following a different way of decomposing than the right way.

 

[volaff]

Let's talk together then.
the right roller system - left has to balance the f force of the big roller.
where do they exchange this force (equal to f/2)?
Do you agree with me that it is in the contact point? like toothed wheels to mean us.

 

[Marco F inox]

so if I use the right model I have that r4 breaks into r5 and r6 that are the tangential and radial.View attachment 68506then the right roller bearing will have as radial equivalent |r3-r5|. but r6 how does it react to the bearing? transport to the center and do pitagora with |r3-r5|?

but therefore do not get the same result that I calculated on the left directly?

here pitagora does not center, there are no straight corners.
brings up the force f until its origin coincides with the center of the big wheel.
from the tip of this f, tracks the two parallels to the two interaxes formed by the big wheel with the small wheels.
merge these points with the center of the big wheel, these two segments are the breakdown of force f.
Don't be surprised if these two forces look great, they could be even larger than f. force all depends on the angle that the small wheels form compared to the center of the big wheel.

 

[volaff]

here pitagora does not center, there are no straight corners.
brings up the force f until its origin coincides with the center of the big wheel.
from the tip of this f, tracks the two parallels to the two interaxes formed by the big wheel with the small wheels.
merge these points with the center of the big wheel, these two segments are the breakdown of force f.
Don't be surprised if these two forces look great, they could be even larger than f. force all depends on the angle that the small wheels form compared to the center of the big wheel.

You mean that?
to well think about the action lines of the piccla-great wheel are precisely the lines of action of the forces.

I think you're right

 

[meccanicamg]

here pitagora does not center, there are no straight corners.
brings up the force f until its origin coincides with the center of the big wheel.
from the tip of this f, tracks the two parallels to the two interaxes formed by the big wheel with the small wheels.
merge these points with the center of the big wheel, these two segments are the breakdown of force f.
Don't be surprised if these two forces look great, they could be even larger than f. force all depends on the angle that the small wheels form compared to the center of the big wheel.

So you confirm that it is like post number 3 image?

 

[meccanicamg]

You mean that?
View attachment 68509to well think about the action lines of the piccla-great wheel are precisely the lines of action of the forces.

I think you're right

I also got there after the pip of the first post.
But then... why can't I have two vertical reactions?

 

[volaff]

I also got there after the pip of the first post.
But then... why can't I have two vertical reactions?

Before dismantling the forces, you have to understand what the action rules are.
the small-big wheel action line what are you thinking?

 

[meccanicamg]

Before dismantling the forces, you have to understand what the action rules are.
the small-big wheel action line what are you thinking?

the action and reaction lines are the joints between small wheel and large wheel..... that is post 3.
means that I replace f with a system of equivalent forces r7 and r8.
analyzing the contact point between each small and large wheel I will have an equal and opposite reaction of sign (verse) to r7 and r8.

 

[volaff]

the action and reaction lines are the joints between small wheel and large wheel..... that is post 3.
means that I replace f with a system of equivalent forces r7 and r8.
analyzing the contact point between each small and large wheel I will have an equal and opposite reaction of sign (verse) to r7 and r8.

exactly.
Why do you want to avert a vertical reaction if the right of azone of forces is what you described little?

 

[meccanicamg]

exactly.
Why do you want to avert a vertical reaction if the right of azone of forces is what you described little?

because at first it seemed logical, as it was for you: a force up and two reaction down.....
in reality there is but it is necessary to disassemble r7 to find a vertical and a horizontal.

 

[volaff]

because at first it seemed logical, as it was for you: a force up and two reaction down.....
in reality there is but it is necessary to disassemble r7 to find a vertical and a horizontal.View attachment 68510

Yeah.
Sometimes we forget that, before breaking a force, we need to understand what are the rules of action of binding reactions, obviously when it is possible to determine them graphically in this case.

 

[cacciatorino]

Are these wheels also exchanged torque moment or are they only discharges of forces, like rolling cage?

 

[Stan9411]

because at first it seemed logical, as it was for you: a force up and two reaction down.....
in reality there is but it is necessary to disassemble r7 to find a vertical and a horizontal.View attachment 68510

I agree with this schematization.

 

[Marco F inox]

You mean that?
View attachment 68509to well think about the action lines of the piccla-great wheel are precisely the lines of action of the forces.

I think you're right

This is the exact interpretation of what I said, the r7 and r8 forces can be decomposed, as has already been done, obtaining a modest uphill force, but a remarkable lateral force outwards.
basically this big wheel acts on the small as a divaricator wedge.

 

[biz]

because at first it seemed logical, as it was for you: a force up and two reaction down.....
in reality there is but it is necessary to disassemble r7 to find a vertical and a horizontal.View attachment 68510

I agree

 

[meccanicamg]

Are these wheels also exchanged torque moment or are they only discharges of forces, like rolling cage?

the wheel in the bottom process the material and therefore all three turn but the motion of the material is made by another machine that is there after.