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

[Femme_physics]

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?

 

[pongo38]

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.

 

[bp_psy]

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.

 

[espen180]

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.

 

[Femme_physics]

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

 

[Studiot]

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.

 

[Femme_physics]

Ah, so in this case it's 0.45 + 0.6? Or just 0.45?

 

[Studiot]

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

 

[Dmytry]

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?

 

[espen180]

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.

 

[Femme_physics]

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.

 

[AlephZero]

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.

 

[Femme_physics]

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

 

[I like Serena]

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?

 

[Femme_physics]

Good point. Then it should be the first one, I reckon, yep. I agree, first one :)

 

[I like Serena]

Good point. Then it should be the first one, I reckon, yep. I agree, first one :)

I reckon I agree!

 

[Femme_physics]

If I look at the xz projection of this structure

http://img846.imageshack.us/img846/9219/thistructure.jpg

Don't I have to draw Tcd as a vector, even if it applies no moment on point O?

This is what I did:

http://img823.imageshack.us/img823/7761/tbax.jpg

In the solution manual they appeared to have ignored it! i.e. not even include it as a vector! This is false, right?

This is what the solution manual did (sorry for quality) ->>

http://img37.imageshack.us/img37/7189/solman00.jpg

 

[I like Serena]

This looks intimidating!
So you were already looking at the solution manual?

Did you try to solve the problem yourself?
What did you find?
Did you run into any problems?

 

[Femme_physics]

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?

 

[I like Serena]

Perhaps you have found an error in the solution manual.

Doesn't that thought excite you?

Show them wrong gurl!

(Quickly before someone else here proves me wrong! )

 

[Femme_physics]

I guess if someone didn't fix you after so long I must be correct

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?

 

[I like Serena]

Good morning Fp!

Getting back to the original diagram

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

Left!

I guess if someone didn't fix you after so long I must be correct

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?

 

[Femme_physics]

Good morning Fp!



Left!



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 T_AB

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.

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

Much of course thanks to you!

Anyway, off with the verbosity and on to physics

Yes, to repeat, it would be pulled over by T_AB

 

[I like Serena]

Yes, to repeat, it would be pulled over by T_AB

If T_AB pulls it over, what would happen to the cord AD?

 

[Femme_physics]

It pulls Tab preventing its pull

 

[I like Serena]

It pulls Tab preventing its pull

Is the system then still in equilibrium?

And what if we snip AD instead of CD?

 

[Femme_physics]

[/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.

 

[I like Serena]

It pulls Tab preventing its pull

Is the system then still in equilibrium?

Definitely not,

First you said Tad prevented the pull of Tab, and then you said it is not in equilibrium?

 

[Femme_physics]

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?

 

[I like Serena]

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.

 

[Femme_physics]

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!

 

[Femme_physics]

All the numbered lines are beams. Is this a 3D truss?

Do I use the same ol' method of joints and method of sections to solve it?

http://img708.imageshack.us/img708/9438/3dtruss2004summer7.jpg

 

[meldraft]

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)

 

[Studiot]

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.

 

[I like Serena]

Rotating counterclockwise

That's what I think also!

What will it be? Will it move or not?