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momentum weight roller

  • Thread starter Thread starter asblo
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asblo

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Good morning to all

I apologize in advance for ignorance, I usually do not make particular calculations, but in this case I would like to come to head. I would like to know what the weight force a roller applies on a roll in support during rotation.

I have a 50 kg mass bound with hinge, I spotted its mass center and tracked a vertical straight of 500 (n). I draw a straight from the hinge to the mass center and the project, until the orthogonal straight fp does not meet the vertical straight. this will be its weight strength 375, I think it is right here. If the weight force passed through the center of the roller I could consider this, but in this case it is out of center.

How should I proceed now?1666076599749.webp
 
You have to break down the weight force in the only two directions that make sense, and that prevent the mass from turning around the world:)

one is the straight that you have already drawn and that passes by baricentro and hinge, the other is the straight that passes by baricentro and tangency point between roller and coil.

nb: By weight I mean 50kg, not the one you indicated with fp (i.e., by weight).

the weight force acts vertically always (in a shed not subject to accelerations :) ), the reactions that prevent him from falling are precisely the reaction (inclined) given by the coil that generates a traction, contrasted by the arm bound to the hinge.
 
if you have doubts or problems making calculations (but you also do it graphically, similar to how you started) ask, up to these things I can help you.
 
for me it is yes! the thrust on the reel is 37,8kg.

If, like me, you like to estimate before (when feasible) a value "acceptable", spannometric, I expected a result similar to yours.

the baricentro/cerniera axis resembles a 45°, in that case the value would have been 50/√2, which makes 35.4...
 
so the correct composition would be like this? f1 and f2?
hi, in my opinion this decomposition is right only if the diameter of the coil does not change and if the position of the pin is on the straight that connects the center of the system with the center of the coil.
otherwise I would do like this:
calculation of the torque generated by the system according to the alpha angle (vertical arm )
this pair opposes the moment generated on the roller pin.
calculation of the force exerting on the pin knowing that the direction is normal to the pivot/center of rotation of the system.
at this point I break the force along the direction of the straight passing through the center of the pin and the center of the coil and the direction of the arm pivot center rotation.
from the decomposing I get the force that presses on the reel the other is to load the thrust system.
by making an excel you can set everything according to the alpha angle and see how the roller force changes on the reel.

Hi.
 
Last edited by a moderator:
hi, in my opinion this decomposition is right only if the diameter of the coil does not change and if the position of the pin is on the straight that connects the center of the system with the center of the coil.
otherwise I would do like this:
calculation of the torque generated by the system according to the alpha angle (vertical arm )
this pair opposes the moment generated on the roller pin.
calculation of the force exerting on the pin knowing that the direction is normal to the pivot/center of rotation of the system.
at this point I break the force along the direction of the straight passing through the center of the pin and the center of the coil and the direction of the arm pivot center rotation.
from the decomposing I get the force that presses on the reel the other is to load the thrust system.
by making an excel you can set everything according to the alpha angle and see how the roller force changes on the reel.

Hi.
the coil changes, but found the system calculation with ømax and ømin coil.
Wanting to do as you say, the first step would be what I started doing, right? with a result from 375n. If I could put down a sketch, I'd be grateful, to understand better.
 
I train you a sketch, I forced the situation to show you that the couple given by weight moves to the pin and from them perpendicular to the coil.
there is definitely a condition where baricentro /centro perno roller /centro reel are on the same axis but in most cases six under these conditions and the thrust of the reel varies because it varies its diameter.
with
- precise measures of the hinge position
- the weight center
- the position of the roller pin compared to the baricentre
-the minimum and maximum diameters of the coil

you can do the exact construction


Hi.
 

Attachments

I train you a sketch, I forced the situation to show you that the couple given by weight moves to the pin and from them perpendicular to the coil.
there is definitely a condition where baricentro /centro perno roller /centro reel are on the same axis but in most cases six under these conditions and the thrust of the reel varies because it varies its diameter.
with
- precise measures of the hinge position
- the weight center
- the position of the roller pin compared to the baricentre
-the minimum and maximum diameters of the coil

you can do the exact construction


Hi.
I'm sorry, but I realized there's a bit of reasoning missing as soon as I can add it to you.
 
I attach the image I think you can explain what I meant
Hi.
 

Attachments

  • modifica test.webp
    modifica test.webp
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simply:
_You have the weight force applied to the mass center of your object
_You have the same and contrary force applied to the contact point
_the balance of moments compared to pole to (circum) comes from fp*b=fp1*b1 (we are working with orthogonal references in a Cartesian reference system "standard")
_You make it considering the contact point.
I have no way of schematizing what I said, but it should be clear.
 

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