Will there be a rotation around z axis?

[peter010]

Hello people,

I know it seems like a silly question. In case 1, its clear that no rotation will occur around z axis. Anyhow, case number 2 is confusing me; I can't decide clearly if a rotation will occur since a moment should be generated around z axis, but in the same time, the line of the acting force is perpendicular on the pivot !

what do you think please, will it rotate in case 2 ?

 

[BvU]

Hello Peter, welcome to PF !

So what are these ? flexible water hoses or stiff metal rods ?
A bit of description would really help clear this up !
What is the criterion to decide whether or not somethign will rotate arountd the z-axis ?

 

[peter010]

Hello,
Cheers :)

--------------
any of the two structures consists of beams.

The structure looks like the English letter "L", where in the second case, the beam in y-axis has a bucking,

in the second case, rotation means that the beam in Y axis, which is in XY plan, will rotate to the x axis.

:)

 

[BvU]

So what is the full problem statement ?
Why are those pivot points marked ?
Is there any support point somewhere ?
The F indicate loads along the y-axis ?
What is the criterion to decide whether or not something will rotate around the z-axis ?

 

[peter010]

ok,

I will upload another drawing for case 2> to be clearer :)

 

[BvU]

no spports ? not even or the pivot points ?

And in case 1 everything is in the yz plane ?
And in case 2 everything is in the yz plane ?

the full problem statement ?

 

[peter010]

Hello,

Kindly, find this sketch.

the statement is about to answer whether the curved beam will rotate around z axis or not, at the current position, and from the steady state status, due to the applied load>

 

[BvU]

I would call this case 3, to me it looks different from the others

I take it the left-handed coordinate system is accidental -- the axis that's pointing to the right is the z axis ( -- although 'z' ?)

And I assume the 'triangle' beam - beam - curved beam can only rotate around this z-axis, and that the arc with two arrows is in the xy plane.

The perpendicular distance from the line of action is zero: the line of action goes through the candidate rotation axis.
So the torque is zero and no rotation will occur.

But it's an unstable equilbrium ...

 

[peter010]

Hi,

yes, its corresponding assumptions.

The thing which is confusing me, that this load will be analysed like in the photo, and thus a torque should be generated..

So, is there any physical way that can help to overcome this anomaly position ?

 

[BvU]

this load will be analysed

by whim ?

If $\vec F$ is in the yz-plane, there can't be a net component pointing outside that plane. Your sketch suggests $F_z$ in the yz-plane and $F_1$ along the curvature or something ?

 

[peter010]

sorry I am not good in drawing :)

Is there is any way to to overcome this anomaly situation, so we can make the curved beam rotates. I said that in the meaning of modifying the curved beam design somehow in order to divert the direction of the applied force and then to create a torque that able to create rotation ?

 

[BvU]

• So far we've ignored the weight of the curved beam. That alone will make it rotate anti-clockwise.
• Even if the curved beam has weight 0, it still is an unstable equilibrium, so in reality it will rotate, either clockwise or anti-clockwise.

 

[peter010]

So, this may happen while the exerted force is existing in this position. right ? :)