Solving 3D Statics Problem: Calculating Tension and Reactions

[Femme_physics]

I've kicked it up a notch and tried tackling 3D statics

Homework Statement

http://img849.imageshack.us/img849/4640/attempt01drawing.jpg

A rectangular board is held in the condition shown in the drawing by clutching devices A and B (I couldn't find the exact translation in Hebrew for what they wrote :( by I think "clutching devices is the next best thing), and cable EF. The weight of the board is 150 [N].

It's given that the reaction at the X axis in clutching device A is zero.

Calculate:

1) Tension in cable EF
2) The reaction at the clutching devices

The Attempt at the Solution

I clearly did not pay attention to the 40mm that the cable is slanted at, but I'm not really sure how to pay attention to it. It made sense to me to just look at it as though it was a 2D problem, since that's what I know!

http://img41.imageshack.us/img41/876/attemptedtodo.jpg

 

[tiny-tim]

Hi Femme_physics!

… clutching devices A and B (I couldn't find the exact translation in Hebrew for what they wrote :( by I think "clutching devices is the next best thing), …

They're clamps!

Sorry, but you can't treat this as 2D!

You have the known W and unknowns T A_y A_z B_x B_y and B_z.

Start with a moments equation.

(And don't forget that in 3D, moments are taken about an axis not a point! )

(in 2D, it's also an axis, but it always comes vertically out of the page, so it looks like a point!)

 

[I like Serena]

Hi Fp!

I clearly did not pay attention to the 40mm that the cable is slanted at, but I'm not really sure how to pay attention to it. It made sense to me to just look at it as though it was a 2D problem, since that's what I know!

For starters, you forgot the forces that clamps A and B exert.

And you seem to have calculated W, but that can't be right, because W is given as 150 N.
In 2D problems you usually look at the x components of each force and the y components of each force.
Their sums have to be zero.

In 3D problems, you get an extra equation, because each force also has a z component.
Their sum has to be zero too.

What you would typically do, is make a side view, as you have done, looking only at the y and z components, and balancing them.
What you missed however, is that you're not looking at T now, but only at T_y and T_z.
(And you forget the force at A and at B.)After that you would make a second side view, looking for instance at the x and z components, and balancing them.
You will have to do that yet.After that you can calculate the total tension T from T_x, T_y, and T_z.

 

[Femme_physics]

Thanks-- I'll reply tomorrow morning :)

And clamps! *smacks forehead* lol. Thanks tiny-tim ;)

 

[SteamKing]

I don't think A and B are clamps so much as they are hinges, which would allow the board to rotate if it were not supported by the cable EF.

 

[WatermelonPig]

Yes I would agree with that just from the picture. And when you are taking moments I'd just recommend taking the determinants of matrices rather than trying to visiualize it.

 

[Ouabache]

And clamps! *smacks forehead* lol.

Taking a wild guess, are they אוחז במכשיר?
I would also describe them as http://www.fitmaxhardware.com/hinges"
(like a door swings on, but mounted horizontally).

You are doing some nice exercises, with multi-dimensional vector analysis.

 

[Femme_physics]

Sorry, but you can't treat this as 2D!

Damn it! Would've made it a lot easier ;)

For starters, you forgot the forces that clamps A and B exert.

Hmm, come to think of it, they do exert forces in the Z axis, my bad.

And you seem to have calculated W, but that can't be right, because W is given as 150 N.

Ah, supposed to have been T.
Boy I've messed things up in here.

What you would typically do, is make a side view, as you have done, looking only at the y and z components, and balancing them.
What you missed however, is that you're not looking at T now, but only at Ty and Tz.
(And you forget the force at A and at B.)

I see. I'm not used to 3D I got confused!

Taking a wild guess, are they אוחז במכשיר?

תופסנים

to be more exact :)

I would also describe them as hinges
(like a door swings on, but mounted horizontally)

Yes, that's what I thought they are!

You are doing some nice exercises, with multi-dimensional vector analysis.

Thanks :D

Yes I would agree with that just from the picture. And when you are taking moments I'd just recommend taking the determinants of matrices rather than trying to visiualize it.

The determinants of what the what what?

----------

Anyway, for my latest attempt I just decided to do free body diagrams without any calculations:


http://img577.imageshack.us/img577/2875/zaxisyaxis.jpg

http://img199.imageshack.us/img199/3015/yaxis.jpg

 

[I like Serena]

Good morning Fp!

Boy I've messed things up in here.

Anyway, for my latest attempt I just decided to do free body diagrams without any calculations:

I'm glad to see that you nicely unmessed things this time!


Perhaps you could name the various vectors as with Bx, By, Bz, etcetera?
Btw, Ax would be zero, as per the problem statement.

And then you could try and find a few useful equilibrium equations...

(According to tiny-tim, I shouldn't use the same smiley 3 times in a row, so the third became a wink! )

 

[Femme_physics]

Good morning

http://img43.imageshack.us/img43/2682/myattemptl.jpg

Perhaps you could name the various vectors as with Bx, By, Bz, etcetera?
Btw, Ax would be zero, as per the problem statement.

Ok, I'll name in my next attempts

(According to tiny-tim, I shouldn't use the same smiley 3 times in a row, so the third became a wink! )

Ah yes, tiny-tim's guide to emoticocns must be adhered when homeworking helping!

 

[I like Serena]

Ah yes, tiny-tim's guide to emoticocns must be adhered when homeworking helping!

Indeed! Hail mighty tiny-tim!
(Actually, I'm abusing one of his guidelines here, but I like a bit of artistic freedom! )

http://img43.imageshack.us/img43/2682/myattemptl.jpg

The T you're showing is not the real T.
You only have the Ty and Tz components in this view.
Those are the 2 that you should calculate here...

Btw, the T you have calculated now is actually $\sqrt {T_y^2 + T_z^2}$.

 

[Femme_physics]

Ohhhhhhhhhhhhhhhh!

Wait, then how do I find the combined T vector for all the axes?

Do I do something weird like

http://img62.imageshack.us/img62/1295/tzzzzz.jpg ?

Gosh, I miss 2D :((and lol at "artistic freedom"!)

 

[I like Serena]

Ohhhhhhhhhhhhhhhh!

Wait, then how do I find the combined T vector for all the axes?

Do I do something weird like
?

That looks weird! :uhh:

But, I think...

Yes! That is entirely correct!

Gosh, I miss 2D :(

I think you'll grow to like it! I did! :D

 

[Femme_physics]

1 more question,

In this view

http://img171.imageshack.us/img171/9382/thisdrawing.jpg

Uploaded with ImageShack.us

Do I not treat the other hinge, hinge A? I mean, it's "behind" hinge B from that view, but like... I can't draw it...so I leave it out?

Yes! That is entirely correct!

Really? Wow! Awesome. Then, I think I got what I need now! :) I'll work it out.

I think you'll grow to like it! I did! :D

If you did, I will! :)

 

[I like Serena]

1 more question,

In this view

Do I not treat the other hinge, hinge A? I mean, it's "behind" hinge B from that view, but like... I can't draw it...so I leave it out?

No, you treat A and B as if they are in the same place. :)
The forces Ay and By add up.
And the forces Az and Bz add up too.

Really? Wow! Awesome. Then, I think I got what I need now! :) I'll work it out.

If you did, I will! :)

:D

 

[Femme_physics]

No, you treat A and B as if they are in the same place. :)

So I'm in fact calculating Ay+By in that view when I look at the hinge at the y axis?


BTW, I did score the correct result for T (=99 [N])

But I was looking at the answers and Bx does not equal zero there! Even though the question clearly states it should. I reckon I found another mistake!


=http://imageshack.us/photo/my-images/843/theanswers.jpg/]http://img843.imageshack.us/img843/9730/theanswers.jpg [/URL]

 

[I like Serena]

So I'm in fact calculating Ay+By in that view when I look at the hinge at the y axis?

Yep!

BTW, I did score the correct result for T (=99 [N])

Nice!

But I was looking at the answers and Bx does not equal zero there! Even though the question clearly states it should. I reckon I found another mistake!

I didn't check the numbers, but your answers can't be right.
By would not be zero.

And you wouldn't be able to solve both Ax and Bx. The system can only be solved by assuming a fixed value for one of the two. That's why the problem statement says that Ax is zero.

 

[Quinzio]

Ok for T.
$A_z + B_z$ should equal the force from the string.

I think also the forces towards y and x should be revised.

 

[Femme_physics]

I didn't check the numbers, but your answers can't be right.
By would not be zero.

These are not MY answers, these are the solution manual answers, it must be wrong then!


Sorry, really wanted to make it clear :) I'll always admit a mistake, but don't tag me with someone else's!

And you wouldn't be able to solve both Ax and Bx. The system can only be solved by assuming a fixed value for one of the two. That's why the problem statement says that Ax is zero.

No, wait, the problem says Bx is zero, I believe I wrote it as well

Yep

Hmm, so I have the "combined" value for By and Ay, but I can't really tell how the value splits between them, right?

 

[Quinzio]

You can determine all the forces there.
I don't know why they say that Ax is zero ... it seems it's not true.
The rules are the same as in 2D. Net sum of forces =0, net sum of torques =0.
I think you may need help on how to calculate torques in 3D.

 

[I like Serena]

No, wait, the problem says Bx is zero, I believe I wrote it as well

But I was looking at the answers and Bx does not equal zero there! Even though the question clearly states it should. I reckon I found another mistake!

It's given that the reaction at the X axis in clutching device A is zero.

There! Aren't you saying here that Ax = 0?

So I'm in fact calculating Ay+By in that view when I look at the hinge at the y axis?

Yes! :)

These are not MY answers, these are the solution manual answers, it must be wrong then!

Sorry, really wanted to make it clear :) I'll always admit a mistake, but don't tag me with someone else's!

[PLAIN]http://img843.imageshack.us/img843/9730/theanswers.jpg[/QUOTE]

All right, here are my answers:

The plus and minus signs are not so relevant, since a difference in sign would only mean a different choice for the chosen direction.

Ax + Bx = -24

Ay = -10
Az = -8

By = 85
Bz = 68

Tx = 24
Ty = 75
Tz = 60
T = 99

It appears, as I have done it, that the axes are different from your solutions manual.
I tend to believe that your solution manual is wrong here!

Hmm, so I have the "combined" value for By and Ay, but I can't really tell how the value splits between them, right?

Leaving aside the matter of axes for now, yes, there is no way to distinguish them.
Ultimately it is solvable of course, but that requires material properties that are outside of the scope of your current class material.

You can determine all the forces there.
I don't know why they say that Ax is zero ... it seems it's not true.
The rules are the same as in 2D. Net sum of forces =0, net sum of torques =0.
I think you may need help on how to calculate torques in 3D.

Sorry, but no, you cannot determine all the forces.
The 2 forces involved act on the same line.
This means you cannot distinguish them in their force sum.
And you can not distinguish them in any moment sum.
Cheers!

 

[Quinzio]

Sorry, but no, you cannot determine all the forces.
The 2 forces involved act on the same line.
This means you cannot distinguish them in their sum.
And you can not distinguish them in any moment sum.

Fine. Excuse me.

 

[Femme_physics]

There! Aren't you saying here that Ax = 0?

Oops, my bad. My indiscretion has been working extra hours lately! Should be Bx = 0

Oops :(

Leaving aside the matter of axes for now, yes, there is no way to distinguish them.
Ultimately it is solvable of course, but that requires material properties that are outside of the scope of your current class material.

But if I look at it from a different axis point then it becomes solvable with the scope of my current class material, right?

I'll work on that now, but I'd hate to work on a problem which has a mistake! Then again, that means I'll get to submit the author the right solutions! So, I just might :)

Thank you, ILS! I got to help a student in mechanics now ;) So I'll probably only reply tomorrow, as usual :)

 

[I like Serena]

Oops, my bad. My indiscretion has been working extra hours lately! Should be Bx = 0

Oops :(

I'm glad you're willing to admit to a mistake.
It takes a strong person to admit he/she is wrong!

But if I look at it from a different axis point then it becomes solvable with the scope of my current class material, right?

The problem as given is solvable within the scope of your current class material.
Think of the hinge or clamp at B as one that allows sideways movement, and you're set.

I'll work on that now, but I'd hate to work on a problem which has a mistake! Then again, that means I'll get to submit the author the right solutions! So, I just might :)

The problem does not have a mistake!
It's only the solution manual.

Actually, I just realized what they have done in the solution manual.
The solution manual took another coordinate system in which z is up, y is to the right, and x is to the front, which is a regular choice.
Tbh, the choice of coordinate system in the problem is unusual with z not being up or down.
Still, there seems to be yet another mistake...

Thank you, ILS! I got to help a student in mechanics now ;) So I'll probably only reply tomorrow, as usual :)

See you tomorrow then! I'll look forward to it! Now scooch!

 

[WatermelonPig]

I just mean that instead of trying to make 2D views of this you could calculate the cross products instead of the moment arms of the forces which would give you the same equations but you can also draw the 2D views if you want to.

 

[Femme_physics]

I'm glad you're willing to admit to a mistake.
It takes a strong person to admit he/she is wrong!

I agree!

Actually, I just realized what they have done in the solution manual.
The solution manual took another coordinate system in which z is up, y is to the right, and x is to the front, which is a regular choice.
Tbh, the choice of coordinate system in the problem is unusual with z not being up or down.
Still, there seems to be yet another mistake...

I just mean that instead of trying to make 2D views of this you could calculate the cross products instead of the moment arms of the forces which would give you the same equations but you can also draw the 2D views if you want to.

Thanks, I prefer the 2D views :) lots easier to me!

Now if Bx = 0 does this mean your answers are true, or did you consider Ax = 0 when you wrote the answers?

 

[I like Serena]

Now if Bx = 0 does this mean your answers are true, or did you consider Ax = 0 when you wrote the answers?

The answer I gave is true regardless.

I wrote Ax + Bx = -24

You can set Ax = 0, or Bx = 0, or anything else, and the solution will still hold!

 

[Femme_physics]

Awesome, then I'll try to solve it with you being the answer manual this time! :)

At worst, if you made mistakes, I'll send you the mistakes/corrections to your email after ILS confirms that these are indeed mistakes and confirms my new corrections.;)

 

[I like Serena]

Awesome, then I'll try to solve it with you being the answer manual this time! :)

At worst, if you made mistakes, I'll send you the mistakes/corrections to your email after ILS confirms that these are indeed mistakes and confirms my new corrections.


;)

Who is this ILS! Do you think he (she?) knows better?

 

[Femme_physics]

:tongue:


Okay my Bz is off

http://img851.imageshack.us/img851/8830/23444837.jpg


Can you tell me why?

 

[I like Serena]

Okay my Bz is off

Can you tell me why?

You're using T, but what you should use, is Tx and Tz.

 

[Femme_physics]

Oh, right! Hah, see it's tricky! ;)Thanks, working on it!

 

[I like Serena]

Oh, right! Hah, see it's tricky! ;)


Thanks, working on it!

Yeah, I know.
It's like bending your mind into a new direction it's not accustomed to!

Welcome to the world of 3D!

And perhaps you'd also like "Flatland" (see for instance http://en.wikipedia.org/wiki/Flatland" ), where you need your mind to get rid of a dimension!

 

[Femme_physics]

Woah, I've been living in a 2D world this whole time, there's a whole new world open in front of me! Unbe*******livable!

I'm scared. Mom?

 

[I like Serena]

Woah, I've been living in a 2D world this whole time, there's a whole new world open in front of me! Unbe*******livable!

I'm scared. Mom?

And?
Did it help?
Could your mom protect you from the third dimension (wait till you get to the fifth!)
Or could she help you to understand it?

Or are you on your own exploring the world!