Forces in 3-D: Homework Solutions

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In summary: So i am going to write the components of the distance first.D = (L-4)i + (4)j + (4)kMoment around CD = (1.25)(L-4) - (0.66)(4) - (0.66)(4)In summary, the conversation is about a physics problem involving a static system with multiple unknown variables. The main focus is on finding the value of L for which the system is static. The conversation involves discussing different equations and components of forces, as well as taking moments around different axes to find a solution.
  • #1
VectorA
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Homework Statement



djWgK96.png

Homework Equations

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?
 
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  • #2
VectorA said:

Homework Statement



djWgK96.png

Homework Equations

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?

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
SteamKing said:
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?

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
VectorA said:
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) ?
You want the value of L for which the system is static. What equations can you write down for static systems?
 
  • #5
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

vpWukY8.png
-------------- (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
haruspex said:
You want the value of L for which the system is static. What equations can you write down for static systems?

above are the equations that i can think of.
 
  • #7
VectorA said:
take moment around A and set it to zero
Which axis through A?
Other than L and the tension in CD, what unknowns do you have in that equation?
 
  • #8
haruspex said:
Which axis through A?
Other than L and the tension in CD, what unknowns do you have in that equation?

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
VectorA said:
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.
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
haruspex said:
Which axis through A are you taking moments about? What are the choices?

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

I can take moment about BE and P but i can't around CD. Even if i do take moment around CD there are more than one unknowns.
 
  • #13
VectorA said:
I can take moment about BE and P but i can't around CD. Even if i do take moment around CD there are more than one unknowns.
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.
Please post it.
 
  • #14
First i am posting the components of each force, that is BE, CD and P

SfgqiVA.png
 
  • #15
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
VectorA said:
First i am posting the components of each force, that is BE, CD and P

SfgqiVA.png
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
haruspex said:
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.

I am really sorry. This is my first day here at the forums.
 
  • #18
VectorA said:
Moment around P = -40(i) + 15(k)
You mean the moment of the force at P about A.
VectorA said:
Moment around BE = -5.336(i) + 2.64(j)
Again, you mean the moment of the tension in BE about A, yes? And you need the magnitude of that tension as a factor.
VectorA said:
Now moment around CD is a bit complicated.
Why is it any more complicated? Just take the cross product of vector CD with either AC or AD.
 
  • #19
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.
Nh4U7Qs.png
 
  • #20
haruspex said:
Why is it any more complicated? Just take the cross product of vector CD with either AC or AD.

So now i am confused on how to solve this further.
 
  • #21
VectorA said:
So now i am confused on how to solve this further.
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
haruspex said:
How do the x coordinates of C and D compare?

They are the same?
 
  • #23
VectorA said:
They are the same?
No, look again. Please post what you think the vectors AC, AD look like.
 
  • #24
haruspex said:
No, look again. Please post what you think the vectors AC, AD look like.

AD: L(i) + 6(j)

AC: 6(i) + 8(k)
 
  • #25
VectorA said:
AD: L(i) + 6(j)

AC: 6(i) + 8(k)
Right, so take the difference to find the vector CD.
 
  • #26
haruspex said:
Right, so take the difference to find the vector CD.

I am so sorry for asking stupid questions but does the difference look like this?

6-L(i) + 6(j) + 8(k)
 
  • #27
VectorA said:
I am so sorry for asking stupid questions but does the difference look like this?

6-L(i) + 6(j) + 8(k)
No, you have some signs wrong.
To calculate CD from AC and AD, which do you subtract from which?
 
  • #28
AC - AD

6-L (i) + 8(k) - 6(j)
 
  • #29
haruspex said:
No, you have some signs wrong.
To calculate CD from AC and AD, which do you subtract from which?
is my answer correct?
 
  • #30
VectorA said:
AC - AD
Ummm, no. We're agreed that "CD" means "the vector from C to D", yes? If the above were correct, we'd have CD+AD=AC. Does that look right?
 
  • #31
haruspex said:
Ummm, no. We're agreed that "CD" means "the vector from C to D", yes? If the above were correct, we'd have CD+AD=AC. Does that look right?
yes

I guess then it is AD-AC

L-6(i) +6(j) -8(k)
 
  • #32
VectorA said:
yes

I guess then it is AD-AC

L-6(i) +6(j) -8(k)
Yes.
Sorry, but I have to sign off now for the day.
 
  • #33
haruspex said:
Sorry, but I have to sign off now for the day.

Can you please just quickly tell me what the next step should be? It would be really appreciated.
 
  • #34
VectorA said:
Can you please just quickly tell me what the next step should be? It would be really appreciated.
I have another few minutes...
Write out the cross product of the tension in CD and the vector AC (or AD, whichever is easier).
 
  • #35
(8k +6i) X (L-6i +6j -8k) ----> i replaced L-6 with x

So the cross product that i got is = 8x+48(j) - 48(i) +36(k)
 

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