Is the transfer from 2D mechanics to 3D mechanics intimidating?

In summary: AC to BC will give you the same answer.In summary, in 3D classical mechanics, using vector algebra, you can solve problems without resorting to 2D, with the added convenience of using conventional algebra alone. Additionally, you need to be familiar with parametric notation, and be aware of inertia tensors.
  • #36
I like Serena said:
What will it be? Will it move or not?

If any of the cords are snipped, it moves.
 
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  • #37
Femme_physics said:
If any of the cords are snipped, it moves.

But I thought the other cord and the stick itself would block the movement? :confused:

Btw, did you try to do the math and find all the internal forces?
Can you solve the system assuming it is in equilibrium?
 
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  • #38
But I thought the other cord and the stick itself would block the movement?
Well EVENTUALLY it would have to come to a halt. It will move until it assumes a new equilibrium position.

Btw, did you try to do the math and find all the internal forces?
Can you solve the system assuming it is in equilibrium?

Yes I did, I solved the problem I just don't have it here since I forgot my USB drive at home :( I'll post it later though :smile:

Much appreciated!
 
  • #39
What happened to post#32?
 
  • #40
Studiot said:
What happened to post#32?

Sorry. I was trying to solve it today but it appeared too long, but I did notice it just all boils down to the method of joints (after I peaked at the manual). Only treated in 3D. But same method applies :)

Your comment was spon on though, and exactly how I started solving it. Sigma Fy appears to give me my first result, and since the first step is the hardest, I think everything is just a matter of numbers thenceforth. :smile: thanks.

At any rate, It doesn't appear that hard anymore, just kinda long. I might get to it sooner or later.
 
  • #41
Femme_physics said:
Well EVENTUALLY it would have to come to a halt. It will move until it assumes a new equilibrium position.
Yes I did, I solved the problem I just don't have it here since I forgot my USB drive at home :( I'll post it later though :smile:

Much appreciated!
I don't remember if I had posted the solution in the other thread abut the problem but here it is!

http://img11.imageshack.us/img11/830/800sum7.jpg
 
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  • #42
This will take me a little more time than I have right now.
Are you here in the evening?
 
  • #43
Yes, but can we worry about the other threads first? See, I'm not worried about this since we agreed we got the right answers, I'm more worried about getting the right answers to the other stuff before the test tomorrow! :rolleyes:
 
  • #44
Femme_physics said:
I don't remember if I had posted the solution in the other thread abut the problem but here it is!

http://img11.imageshack.us/img11/830/800sum7.jpg

In your yx-view you have an equation for sum MO, which is not correct.
You forgot the moment of TADy.

You will find that if you add TADy, you can not solve the system (too many unknowns).

Effectively you snipped cord AD and solved the system!
(And so, with the cord snipped, it will not move, but it will be in equilibrium :wink:)

This is a statically indeterminate system, meaning there is more than 1 solution.
In your case, you have to effectively snip 1 of the 2 cords to be able to solve the system.
In practice the forces would divide themselves over the 2 cords depending on material properties.For reference, if you snip cord AD (what you did), the solution has:

Ox=0 N
Oy=866 N
Oz=900 N

TCDxy=1732 N

TADy=0 N (since it is snipped)
And if you snip cord CD, the solution has:

Ox=1500 N
Oy=866 N
Oz=900 N

TCDxy=0 N (since it is snipped)

TADy=866 N
 
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  • #45
You will find that if you add TADy, you can not solve the system (too many unknowns).

Tab doesn't act on the y plane
 
  • #46
Good morning Fp! :smile:

I write AD, you respond with AB.
Do you mean to say there is no cord between A and D?
 
  • #47
Well there is no "Tad", so if anything, you invented a wire! :P
 
  • #48
Now I understand why our discussion didn't go anywhere! :smile:

Apparently you were thinking I was talking about AB, when I was talking about the non-existing cord AD.
 
  • #49
Ah, so there is no cord AD? So I was right? :wink:
 
  • #50
That depends...
Did you sneak in and remove the cord while I wasn't watching? :-p
 
  • #51
What? No! There were always only two cords! Can't believe you're suspecting me, is this what our marriage based on?!? :-p

But, seriously, two cords. When did you go on inventing a cord?
 
  • #52
Playing tricks on each other that work out funny? Definitely! :approve:

The very first time I looked at the picture I immediately saw a cord between A and D, and I thought: Hey, that's a funny problem! :smile:

And when you had the question: "In the solution manual they appeared to have ignored it [Ed. CD]! i.e. not even include it as a vector! This is false, right?"
I drew the wrong conclusion they had removed CD to make the problem solvable.

Looking more carefully at the picture, I guess AD was not intended as a cord.
 
  • #53
And when you had the question: "In the solution manual they appeared to have ignored it [Ed. CD]! i.e. not even include it as a vector! This is false, right?"
I drew the wrong conclusion they had removed CD to make the problem solvable.

I was referring to Tcdx when I said that! That's what they completely seemed to ignore!

Looking more carefully at the picture, I guess AD was not intended as a cord.
Yea, it's not mentioned in the question, it's just a line to define the shape I guess, but I can see why you got confused! :smile:
 

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