electric motor

[MARCOC]

Good morning
a three-phase motor has data injectors
2800 g iri/min
1.1 kw power
3.8 nm pair
diamer tree exit 19mm

if I want to know the peripheral force on the tree and just apply m=f*b
and then get 400n

Thank you very much

 

[Fulvio Romano]

I'm sorry.

 

[eih64]

I'm sorry.

forse mi perdo qualcosa ...

 

[PierArg]

forse mi perdo qualcosa ...

Bye. I would have said 400 n.
Where do you get lost?

 

[volaff]

performance!!!

 

[eih64]

Bye. I would have said 400 n.
Where do you get lost?

: b= 19 mm m=3.8 f= m/b= 3.8x1000/19 = 200
No?

 

[Focus]

: b= 19 mm m=3.8 f= m/b= 3.8x1000/19 = 200
No?

19 is the diameter, b is 19/2.

Hi.

 

[eih64]

19 is the diameter, b is 19/2.

Hi.

We're not

 

[Fulvio Romano]

performance!!!

What performance?
the engine torque is at the axis, you are calculating the force to the axis, so at the same point. . .

 

[volaff]

i had written performance because i thought that eih64 was referring to what "dead". i ask for venia!

 

[Focus]

We're not

please explain yourself.

 

[Fulvio Romano]

We're not

About what? Could you be a little more eloquent?

 

[eih64]

we are not on the 400 n... it seems clear

 

[eih64]

are waiting for answers and not only visits .... force ;

 

[volaff]

in practice sec eih64 the "arm" is not the radius bensi the diameter to what I understood!

 

[volaff]

I think!

 

[Fulvio Romano]

we are not on the 400 n... it seems clear

in practice sec eih64 the "arm" is not the radius bensi the diameter to what I understood!

You may be right, but it depends on how you interpret the initial question.

Imagine tightening the tree in a clamp. If you want to know the f1 force that has to exercise every single jaw, of course this will be f1 * d = c and then f1 = 200n. Thanks, the jaws are two...

If instead, as I seemed to understand from the question, we have only one force, think for example to a grinder, approach the piece and snap it with a certain force, the force (tangenziale) on the piece will be such that f2 *r = c, and then f2 = 400n as we said.

What's wrong with you?

 

[Focus]

You may be right, but it depends on how you interpret the initial question.

Imagine tightening the tree in a clamp. If you want to know the f1 force that has to exercise every single jaw, of course this will be f1 * d = c and then f1 = 200n. Thanks, the jaws are two...

If instead, as I seemed to understand from the question, we have only one force, think for example to a grinder, approach the piece and snap it with a certain force, the force (tangenziale) on the piece will be such that f2 *r = c, and then f2 = 400n as we said.

What's wrong with you?

however the f1 are 2 and agree on the axis, therefore total 400n

 

[eih64]

The initial conditions are:
3.8 nm pair
shaft diameter output 19mm.

then everyone sees how he wants

 

[Fulvio Romano]

The initial conditions are:
3.8 nm pair
shaft diameter output 19mm.

then everyone sees how he wants

Yes, like the sex of angels.
you are talking about the input data, not the question.

if I want to know the peripheral force on the tree and just apply m=f*b
and then get 400n

the question is "the peripheral force". peripheral force is 400n. point

"But if I attack two diametric forces what happens? "
Okay, in this case the forces are two, you have the possibility to increase the arm, and then they will be each 200


"But if I apply three forces and a tilted torque of an alpha angle to the axis of rotation, and all immersed in the molasses? "

the answer is 42 tongue