helpful pantograph cylinder?? where do i cry?

[s.melotto]

Hello everyone,
I should calculate the effort required to a simple cylinder effect controlled by a hand pump, to lift a pantograph frame with a total of 300kg weight force, centered on the upper floor (for now we leave any friction or other forces in play).
I'm inflating in the balance of forces, I'm not sure I took the right arm for the cylinder. I must say that I have to use all the force generated by the cylinder without breaking it (but I have serious doubts) multiplied by the perpendicular distance to its axis until the straight parallel to the axis and passing through the zipper of the pantograph.
I have two doubts:
Do I have to take all my strength or just a component?? ? I can think of taking it all, but I can think of it, xe I think that, at least in this position, the vertical component does not have effect (it has open pantograph), the horizontal one seems to open the scissor but that has no effect once opened....that is, from wherever I look I am doubted.... I hope to have explained myself sufficiently well.
- the arm is correct taken perpendicular to the axis of the cylinder, or should I take the quota vertically from the ground? ?
igo imageI hope you can help me!! ! !
thanks for the courteous attention

 

[meccanicamg]

That design is really bad.
but it is really a start with a very high force because the arm is very small.
I hope it's not really like this and give a bar pattern.
Try to look at the patterns of other pantographs.

 

[Archimed&]

Hello, even I mostly from cell I struggle to understand. For example, but maybe my limit, that arm of 46,039 from where it is taken... It would be better if I could, to make a static scheme. simple lines + constraints + external forces.
the second thing I would do, very important, is to fix me a system of x,y coordinates, and then to break all the forces in those two directions. then choice of the pole and finally writing of the cardinal equations. in this way you immediately notice if the components make a contribution at the moment. This method is very "academic" but it is fundamental, if diagrams and schematics are not drawn (well) always leaves something, even if you are calledfeynman, and the accounts get high.

 

[s.melotto]

Hi.
I know I did something very quick.
attached a more exhaustive sketch. . .
I have to understand what strength the cylinder needs (vinculated to the beam through df) to lift the q load
in yellow the cylinder according to its axis, must raise the strength represented in green; I have indicated the hinges and carts, my doubt is how I have to break down and where to apply the driving force; I'm messing up. .
- Where should I apply strength?
- which components contrast the load and how do they calculate them?
- What is the expression/s of balancing?
please help me
thank you so much for the future help
Good weekend

 

[Archimed&]

Hello, my friend, I'm sorry I can ask you more because I still have a little difficulty understanding your scheme

1. I guess the trolleys are two, right? One above and one below. if you had only one isostatic yield all, since you would have 3 mobile bodies in total (the two x rods plus the platform that rises, then 3x3 nine degrees of freedom) but 4 hinges and a cart that take away 2*4+1*1=9 degrees of freedom, then all. from your scheme it seems that at the top right you have a zipper;

2. the yellow scum that changes in your three solutions what would it be? the cylinder component?

3. If yes, solutions 1 and 3 are equal because you are moving a force applied to a rigid body along its act of action, and this is influential in the writing of balance equations (transmissibility principle)

Thank you!

 

[Archimed&]

I hope you can put external files, in addition to the numerous posts already present on the forum regarding the topic, I suggest you take a look at this document, from page 55 onwards, where the calculation you are looking for is explained

https://webthesis.biblio.polito.it/9433/1/tesi.pdf

 

[Marco F inox]

before making so many calculations, if it represents the cylinder and its stem, the push arm is given by the distance of this axis from point b, center of rotation.
This arm, already put as it has already been said, tends to cancel when the stem reaches point b, the platform stops and no longer rises.
it is necessary to change the anchor points of the cylinder.