Homework Help: Forces in 3-D

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1. Aug 5, 2015

VectorA

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I am kind of confused as to why CD will be a function of L.

For BE i think it is ( -4i + 8j - 8k)
For CD i think it is (6j - 8k) ---> I don't think there is any i value because they seem to be lined up together. I think i am not understanding the question properly. Is it that D is not lined up with C and so there will be an i value?

2. Aug 5, 2015

SteamKing

Staff Emeritus
Well, the distance L is unknown, so how can you be certain the points C and D are aligned in the x-direction? What if L can move back and forth parallel to the x-axis?

3. Aug 5, 2015

VectorA

Ok that makes sense so L is unknown and it can be anything, which is later confirmed in part(b). So Can you please help me in part(b) ?

4. Aug 5, 2015

haruspex

You want the value of L for which the system is static. What equations can you write down for static systems?

5. Aug 5, 2015

VectorA

First i broke BE into components and i got 0.33BE(i) + 0.66BE(j) - 0.66BE(k) ----------(1)
Then there is a ball and socket joint at A so it is broken into components Ax, Ay, Az ------------ (2)
Force at P = 5kn so it is simply 5 Kn (j) ----------- (3)

Problem arises at CD. This is what i think looks like when you break CD into components given that L is unknown

-------------- (4)

Now after that I would take x component from each of the above equations and combine them together, same with y and same with z. This way i would have x components of all the equations separated in one equations and then same with y and z and i can set the coefficient equal to zero and try solve for it but there are too many unknowns.

Next step is to take moment around A and set it to zero but again the CD is the problem as there are too many unknowns :(

6. Aug 5, 2015

VectorA

above are the equations that i can think of.

7. Aug 6, 2015

haruspex

Which axis through A?
Other than L and the tension in CD, what unknowns do you have in that equation?

8. Aug 6, 2015

VectorA

Well, i also don't have BE to begin with. My main problem that i am stuck at is that neither do i have CD nor L. I am sorry but this problem is really confusing me :(
Can you please give me more hint in order to solve it?

9. Aug 6, 2015

haruspex

Right, you have three unknowns, if you avoid involving the forces at A. To keep those forces out of it, you are taking moments about A. But you haven't answered my question about axis. Which axis through A are you taking moments about? What are the choices?

10. Aug 6, 2015

VectorA

I am sorry i don't quite understand the question here. I take the distance from A to the B and then do a cross product with the force BE. Take distance from A to C and then do a cross product of that distance with the force CD. Take distance from A to P and do a cross product of it with 5Kn.

11. Aug 6, 2015

haruspex

Ok, that's fine. So you get a 3D vector equation. That's equivalent to 3 scalar equations, and you have three scalar unknowns.

12. Aug 6, 2015

VectorA

I can take moment about BE and P but i cant around CD. Even if i do take moment around CD there are more than one unknowns.

13. Aug 6, 2015

haruspex

I agreed that what you did in taking those cross products was fine. (It's equivalent to what I was thinking of in terms of taking moments about three axes through A.)
That one vector statics equation should give you all you need.

14. Aug 6, 2015

VectorA

First i am posting the components of each force, that is BE, CD and P

15. Aug 6, 2015

VectorA

Moment around P = -40(i) + 15(k)

Moment around BE = -5.336(i) + 2.64(j)

Now moment around CD is a bit complicated.

16. Aug 6, 2015

haruspex

Posting your working as images is frowned upon. It makes it hard for those making comments to point out which equations or steps they are commenting on. Please learn to use LaTeX.

17. Aug 6, 2015

VectorA

I am really sorry. This is my first day here at the forums.

18. Aug 6, 2015

haruspex

You mean the moment of the force at P about A.
Again, you mean the moment of the tension in BE about A, yes? And you need the magnitude of that tension as a factor.
Why is it any more complicated? Just take the cross product of vector CD with either AC or AD.

19. Aug 6, 2015

VectorA

well you see the components of CD are a little different than BE or P because CD has L. As you can see in the picture above the components involves fraction and square root. I took the moment and this is what i get.

Again i am really sorry for this picture. I will learn LATEX with time.

20. Aug 6, 2015

VectorA

So now i am confused on how to solve this further.

21. Aug 6, 2015

haruspex

First, we need to get the correct expression for the vector CD. What you had in post #5 is wrong. How do the x coordinates of C and D compare?
Next, to make the algebra simpler, invent a symbol for 1/(length of CD).
Now you should be able to type your working in without having to use LaTeX.
(I also advise against plugging in any actual lengths, but instead invent symbols for all the given distances. It makes it much easier to follow the working and minimises errors.)

22. Aug 6, 2015

VectorA

They are the same?

23. Aug 6, 2015

haruspex

No, look again. Please post what you think the vectors AC, AD look like.

24. Aug 6, 2015