Will there be a rotation around z axis?

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1. Oct 21, 2015

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 cant 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 ?

Last edited: Oct 21, 2015
2. Oct 21, 2015

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 ?

3. Oct 21, 2015

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.

:)

4. Oct 21, 2015

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 ?

5. Oct 21, 2015

peter010

ok,

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

Last edited: Oct 21, 2015
6. Oct 21, 2015

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 ?

7. Oct 22, 2015

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>

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8. Oct 22, 2015

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 ...

9. Oct 22, 2015

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 ?

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Last edited: Oct 22, 2015
10. Oct 22, 2015

BvU

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 ?

11. Oct 22, 2015

peter010

sorry im 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 ?

Last edited: Oct 22, 2015
12. Oct 22, 2015

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.

13. Oct 22, 2015

peter010

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