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00:14
Hi is it possible I can ask a small question about moments here? It's bugging me for a bit now, and I can't seem to find a solution for it.
Go for it
, @Axel
Thanks! So I'm trying to understand the concept of moments around an axis. I'm at a problem, and they ask to find a moment around the z-axis. There is a force in the J direction, and it says that that force pointing in the J direction has no contribution to the moment about the z-axis
Why is that? Is it because there is no Z component to the force?
@DanielSank
"J direction"?
@Axel: Please explain "J direction".
In a 3D plane, the y axis, there's a force going downards in the y axis
@DanielSank
@Axel: So "J direction" meant "y direction"?
00:24
Yes!
i61.tinypic.com/dp96dh.png, it's saying that Tab does not contribute to the moment about z axis
@DanielSank
@Axel: Ah. Indeed not.
Why is that though? @Daniel
Do you know the formal equation for a torque about an axis?
@DanielSank *
...for a given force.
$\tau = \vec{r} \times \vec{F}$
Turn on chatjax if that doesn't render for you.
00:27
Okay, I get that.
ok
So, if you look at the cross product in this case, what do you get?
It would be 0
Indeed.
That's your answer.
So in reality, the force must have some sort of component in the axis we are rotating
@Axel: I don't understand what you mean.
The force Tab in that picture points along the y axis only.
But that's not the important part.
The important part is that, starting from the z axis, $\vec{r} \times \vec{F} =0$.
This is basically because the force is long the line which joins the point of application of the force to the z axis.
00:34
omg lol im such an idiot
i get it now
@Axel: Great :) You're not an idiot.
Glad you understand now.
So any force along the line which joins the point of application would not contribute.
@Axel: Right.
:D Thank you! This stuff is pretty cool once you get it
Like, imagine pushing on a bike tire such that the force is directed straight at the center of the wheel. That won't make the wheel spin.
Right?
00:38
Yupp
You need to apply it else where
@Axel: In general, whenever you get some weird-ass formula like $\tau = \vec{r} \times \vec{F}$, try to find a simple case so that you can get intuition.
Well, with the bike wheel, you have to push so that the force is not directed straight at the center of the weel.
I understand it now :)
Thank you really, saved me hours.
@Axel: Groovy. That's why I like this site.
 
13 hours later…
13:36
@Qmechanic: With regards to this Meta thread, should we comment on why we're downvoting a particular answer?
 
3 hours later…
17:02
Is anyone else having trouble logging into Meta? I'm logged in to the main site, and my other SE accounts automatically recognize me and log me in. But none of the Meta sites recognize me at all, even when I manually click on "log in."
no problem here
17:32
Ah, I have discovered the problem: it's a browser plugin. I can log in to Meta now.
 
1 hour later…
19:01
I'd like to encourage everyone to read The NEW new “Be Nice” Policy “Code of Conduct” - Updated with your feedback (also linked in the featured sidebar on-site). Good stuff there.
3
 
3 hours later…
22:02
lol - had almost finished an answer.... then another member got there just before me (so, as my answer would have been an unnecesary duplicate, I discarded what I had done)... gotta get faster with the math jax...

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