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

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Transitioning from 2D to 3D mechanics is generally manageable, as many principles remain consistent, with the added complexity of more variables. Familiarity with vector algebra, particularly dot and cross products, is beneficial, as is understanding parametric notation. While some problems can be simplified by identifying symmetries, not all scenarios allow for reduction to 2D. The inertia tensor may appear intimidating, but it is more straightforward than it seems. Overall, the discussion emphasizes that with practice and understanding, the shift to 3D mechanics can be navigated effectively.
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How much more than is to learn when you go from 2D to 3D? I'm curious to know whether all these problems I solve in 2D are less helpful for when we move to 3D, where there are more elements?
 
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3D can be managed using conventional algebra alone, but it is more convenient to use vector algebra using dot and cross products. So if you have done 2D using vectors, then 3D is a reasonably simple extension of that. You also need to be familiar with parametric notation.
 
Femme_physics said:
How much more than is to learn when you go from 2D to 3D? I'm curious to know whether all these problems I solve in 2D are less helpful for when we move to 3D, where there are more elements?

There should be absolutely no difficulty.Actually there isn't really a transition from 2D to 3D all classical mechanics is in 3D you sometimes ignore one or 2 dimensions.
 
Yes, many problems you will encounter will allow you to reduce the number of dimensions by some symmetry of the problem. You just have to identify the symmetry.

Otherwise it just like 2D, just with potentially more variables.
 
Thanks for the reassurance. So I can always "reduce" the problem to 2D, is what you're saying?


That reminds me, I've been staring at a certain problem since my first semester trying to think which distances to take when I calculate the arm of the force in the middle, the 600 [N], to point A. Do I take the diagonal distance here from 600 to A, or do I just use the horizontal one to the wall (that being 0.45)?

http://img860.imageshack.us/img860/6118/40893341.jpg
 
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You can do either but you must be consistent when forming the equation.

That means that all other moments in the equation must be calculated about the same point or axis.

In this problem it is probably best to take moments about the line through AB and use the perpendicular distance to that line ie the perp dist to the wall.
 
Ah, so in this case it's 0.45 + 0.6? Or just 0.45?
 
The force on the plate has a moment of

600 x 0.45 about the z axis (line through AB)

600 x 0.6 about the x-axis (line through BD)

It has zero moment about the y-axis since it is parallel to it.


and so on
 
The one slightly intimidating thing about 3D is the inertia tensor but it is actually far easier than it sounds (and its dynamics).

For your problem, yep, what studiot says, you can pick any line as axis of rotation, the sum of moments around that axis will be zero. I'd use diagonal BC as axis.
Also, don't forget that force on A is pulling that hinge out of the wall. That may be what you need to calculate?
 
  • #10
Femme_physics said:
Thanks for the reassurance. So I can always "reduce" the problem to 2D, is what you're saying?

No, this is not always possible, but most simple systems will allow this. The simplest example I can think of where you cannot reduce the number of dimensions is when modelling the motion of a charged particle in a magnetic field, in which the force is always acting perpendicular to the direction of motion, and even then you can sometimes (for example if the field is homogenous) make a coordinate transformation which allows you to reduce the problem to 2 dimensions.
 
  • #11
Dmytry said:
The one slightly intimidating thing about 3D is the inertia tensor but it is actually far easier than it sounds (and its dynamics).

For your problem, yep, what studiot says, you can pick any line as axis of rotation, the sum of moments around that axis will be zero. I'd use diagonal BC as axis.
Also, don't forget that force on A is pulling that hinge out of the wall. That may be what you need to calculate?

That's okay, I was just giving an example I wasn't trying to solve the problem. Your replies cleared it :)

Thanks.
 
  • #12
Femme_physics said:
That reminds me, I've been staring at a certain problem since my first semester trying to think which distances to take when I calculate the arm of the force in the middle, the 600 [N], to point A. Do I take the diagonal distance here from 600 to A, or do I just use the horizontal one to the wall (that being 0.45)?

For a 3D situation like this, you are not taking moments about a point, you are taking moments about a line. In a 2D problem, the line is always perpendicular to the 2D plane, so the there isn't an obvious difference between the line and the point where it intersects the plane.

If you take monents about the line AB to get an equation for the forces at C, you use the perpendicular distance from AB. If yuu take moments about AC (which would be a good way to find the vertical reaction at B) you use the perpendicular distance from AC.
 
  • #13
AlephZero said:
For a 3D situation like this, you are not taking moments about a point, you are taking moments about a line. In a 2D problem, the line is always perpendicular to the 2D plane, so the there isn't an obvious difference between the line and the point where it intersects the plane.

If you take monents about the line AB to get an equation for the forces at C, you use the perpendicular distance from AB. If yuu take moments about AC (which would be a good way to find the vertical reaction at B) you use the perpendicular distance from AC.

We recently started 3D so I finally understand this comment.

May I ask though, taking the yx view for the diagram I posted above, is CE drawn like that?

http://img862.imageshack.us/img862/7495/ce1.jpg
Or like this?

http://img535.imageshack.us/img535/2858/ce2v.jpg IMO it's the latter that's correct
 
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  • #14
Femme_physics said:
taking the yx view for the diagram I posted above, is CE drawn like that?

Perhaps you should include the stick and the wall in your drawing, and then treat it as a regular 2D problem.
If you would do that, would you get a visual cue from it? :wink:
 
  • #15
Good point. Then it should be the first one, I reckon, yep. I agree, first one :)
 
  • #16
Femme_physics said:
Good point. Then it should be the first one, I reckon, yep. I agree, first one :)

I reckon I agree! :smile:
 
  • #17
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  • #18
This looks intimidating!
So you were already looking at the solution manual? :wink:

Did you try to solve the problem yourself?
What did you find?
Did you run into any problems?
 
  • #19
Well, I guess just lingering curiosity since I got a bit confused as to how to solve it, but I was being kinda stupid because if I'd thought a bit more I'd find out it's really equation. Just take the yx view and do sum of all moments on O.

But I decided to look at the manual and that's when the question came up... because if I do sum of all forces on X it DOES exist, whereas in their diagram it doesn't. What's the deal?
 
  • #20
Perhaps you have found an error in the solution manual.

Doesn't that thought excite you? :smile:

Show them wrong gurl!

(Quickly before someone else here proves me wrong! :wink:)
 
  • #21
I guess if someone didn't fix you after so long I must be correct :smile:

Getting back to the original diagram

http://img860.imageshack.us/img860/6118/40893341.jpg [/QUOTE]

In the YZ view. I think the left drawing is the correct one. Correct?

attachment.php?attachmentid=36775&stc=1&d=1309245965.jpg
 

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  • #22
Good morning Fp! :smile:

Femme_physics said:
Getting back to the original diagram

In the YZ view. I think the left drawing is the correct one. Correct?

Left! :wink:

Femme_physics said:
I guess if someone didn't fix you after so long I must be correct :smile:

There is a little more to it, but some things are best experienced rather than be told.
(The others in this thread seem to agree, because I'm sure they know what's going on.)

However, since I love you so much and you are apparently very reluctant to do the work, I'll give you a hint. ;)

Suppose you snip the cord CD, what would happen with the construction? Would it still be in equilibrium or not?
 
  • #23
I like Serena said:
Good morning Fp! :smile:



Left! :wink:



There is a little more to it, but some things are best experienced rather than be told.
(The others in this thread seem to agree, because I'm sure they know what's going on.)

However, since I love you so much and you are apparently very reluctant to do the work, I'll give you a hint. ;)

Suppose you snip the cord CD, what would happen with the construction? Would it still be in equilibrium or not?


Of course not, it would be pulled over by TAB

And I'm not "very reluctant to do the work"! And I learn tons of things the hard way, and I like it. Sometimes I try it the easy way because sometimes the easy way works too and enlightens me too. :smile:

And I'm actually trying to solve it right now, after solving 7 questions on my own yesterday (statics+dynamics including 3D) from a test that was last year. So am always practicing to keep myself in shipshape :biggrin:

Much of course thanks to you!

Anyway, off with the verbosity and on to physics :smile:

Yes, to repeat, it would be pulled over by TAB :wink:
 
  • #24
Femme_physics said:
Yes, to repeat, it would be pulled over by TAB :wink:

If TAB pulls it over, what would happen to the cord AD?
 
  • #25
It pulls Tab preventing its pull
 
  • #26
Femme_physics said:
It pulls Tab preventing its pull

Is the system then still in equilibrium?

And what if we snip AD instead of CD?
 
  • #27
[/quote]Is the system then still in equilibrium?
Is the system then still in equilibrium?[/quote]

Definitely not,
And what if we snip AD instead of CD?
There would be nothing to prevent its torque. And after it begins its torque then CD will pull it to its way.
 
  • #28
Femme_physics said:
It pulls Tab preventing its pull

Femme_physics said:
Is the system then still in equilibrium?

Definitely not,

:confused:
First you said Tad prevented the pull of Tab, and then you said it is not in equilibrium?
 
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  • #29
yes, if we snip it then it won't be in equilibrium anymore

TCD does indeed prevent the pull.

I don't see how my affirmations are conflicting?
 
  • #30
Femme_physics said:
yes, if we snip it then it won't be in equilibrium anymore

TCD does indeed prevent the pull.

I don't see how my affirmations are conflicting?

So we snip CD.

If the system is not in equilibrium anymore, that would mean that point A starts moving.
In which direction will it move then?

If this movement is prevented by Tad, then that would mean the system is still in equilibrium.
Edit: the movement is prevented by a combination of Tad and Foa.
 
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  • #31
In which direction will it move then?
Rotating counterclockwise
Edit: the movement is prevented by a combination of Tad and Foa.

That's what I think also!
 
  • #33
Just a pointer, when you have a hinge, it can only produce reactions co-linear to the resultant force you are applying, which is probably why your book's solution is like it is :) (although the way it's drawn it doesn't look like it can actually move towards the negative y axis, since the rotation is blocked by the wall) Also, your solution manual should also have a moment at C, unless it assumes that there are joints in either E or C (which it probably does)
 
  • #34
Since you are not given any other information, I assume triangle ABC is horizontal.

Therefore there is only one force in the frame that can balance the moment of W about BC, that is the compression in strut AD. (can you see why AD must be a strut?)

This should get you started.
 
  • #35
Femme_physics said:
Rotating counterclockwise

That's what I think also!

What will it be? Will it move or not?
 
  • #36
I like Serena said:
What will it be? Will it move or not?

If any of the cords are snipped, it moves.
 
  • #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
 

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