Forces in 3-D: Homework Solutions

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SUMMARY

The forum discussion revolves around solving a 3-D statics problem involving forces and moments, specifically focusing on the relationship between points C and D and the unknown distance L. Participants analyze the components of forces BE and CD, with BE represented as (-4i + 8j - 8k) and CD as (6j - 8k). The discussion emphasizes the need to break down forces into components and set moments around point A to find the value of L that maintains static equilibrium.

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  • #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)
 
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  • #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)
 
  • #36
haruspex said:
I have another few minutes...
Write out the cross product of the tension in CD and the vector AC (or AD, whichever is easier).
(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)
 
  • #37
VectorA said:
(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)
Yes, except that you need to divide by the magnitude of vector CD and multiply by the unknown tension in CD, of course.
Next step is to write that the sum of the three torque vectors is zero.
 
  • #38
haruspex said:
Yes, except that you need to divide by the magnitude of vector CD and multiply by the unknown tension in CD

Is it true that i cannot calculate the magnitude of the vector? because L is a unknown quantity.
 
  • #39
VectorA said:
Is it true that i cannot calculate the magnitude of the vector? because L is a unknown quantity.
L will feature in the expression for the length of CD, but we will eventually be able to solve for it.
 
  • #40
haruspex said:
L will feature in the expression for the length of CD, but we will eventually be able to solve for it.

Ok i will try to solve it now. I hope i get the right answer
 
  • #41
I tried and still having some troubles. :(
 
  • #42
VectorA said:
I tried and still having some troubles. :(
I can only help if you take the trouble to post your working as far as you can get.
To make it easier, use the method I suggested in post 21: invent symbols to represent awkward expressions that occur repeatedly, like ##\sqrt{L^2+100}##.
Or start using LaTeX. To see how others have used it, right click on a line of algebra (like mine above), select Show Math As, etc., copy the text and paste it into your own post. After copying it you will need to insert a pair of consecutive hash characters (#) at each end of it. Use Preview to see if it comes out right.
 

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