linear force calculation by a couple

[teseo]

hello to all:
I have a problem of simple physics (why didn't I study?? ? ? )

I have a motor that develops me a pair of 0.062nm to which is connected a lever of 2.5cm and to which through a reference is connected a linear guide I would like to know with what force the linear guide is moved allego schema thanks.

 

[Focus]

the force you have depends on the angle of the arm itself. therefore during the motion the force will be variable.

 

[teseo]

Thank you, I imagined that I was interested in 90° and when the three elements will be aligned.
I came to calculate the force at the tip of the first arm that is 2.48n but then I miss the data of the other carriers to get the result. I think it works out like three levers were carriers, right?

 

[Focus]

Thank you, I imagined that I was interested in 90° and when the three elements will be aligned.
I came to calculate the force at the tip of the first arm that is 2.48n but then I miss the data of the other carriers to get the result. I think it works out like three levers were carriers, right?

the force, along the guide, when the arm is at 90° (vertical) is the same as you have at the tip of the arm (they have the same direction), when the whole will be aligned the force will be obligatory 0.

 

[teseo]

thanks to the answer,
I have to use this system to get a deep pressure on the guide, so when I am aligned I have pressure 0?
Oh, do you think I made a mistake?
I'm sorry, isn't he very similar to a kneecap? when I am aligned I have the maximum pressure because in theoretical way I can no longer go back pushing the linear guide... I'm not there?

 

[Focus]

you will have a high force just before the dead point, then playing with the elasticity of the system you can keep it at the dead point itself, but it will not be a force generated by the drive couple.

 

[PierArg]

If I understand correctly, you have pressure.
when you have everything aligned horizontally you have reverse motion.
the only maximum component you have is the inertia force of the alternate part (your last member).

 

[MBT]

seeks the laws of the motion of a biella-manovella mechanism.
Now I don't fit, but if I can tonight I'll find you some formula on the jacosa

 

[teseo]

Thanks mbt I try too

 

[MBT]

I would say by p. 227 and following
http://books.google.it/books?id=kbw...a=x&ei=ncxbuctlkao04atuwohiaq&ved=0cdwq6aewaa

 

[teseo]

Okay, guys, I got it.