Finding projected force along line

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Homework Help Overview

The discussion revolves around finding the projected component of a force along a specified line axis in a three-dimensional context, specifically related to a pipe. The original poster has identified the unit vector along the line but is uncertain about the next steps in the calculation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of the dot product to find the projection of the force vector along the line. There are questions about the interpretation of components as vectors and the mathematical procedures involved in the projection calculation.

Discussion Status

Participants are actively engaging with the problem, offering hints and clarifying concepts. Some have verified their vectors and are attempting to reconcile their calculations with expected results. There is a recognition of potential rounding issues affecting the final answer.

Contextual Notes

There are references to a diagram that is causing confusion, and the original poster is working under constraints of homework rules that may limit the information they can share. The discussion includes verification of vector components and the need for clarity in mathematical procedures.

Saladsamurai
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So I have found the unit vector [itex]u_{ab}[/itex] now I am stuck as to what to do.

I need to find the projected component of the 80 N force along the line axis AB of the pipe.

I have the coordinates of B and A as well. But I am a little lost now.

Any hints are appreciated.

Thanks,
Casey

Also I am having trouble reading the diagram...I have A at [itex]r_A=6i+7j-2k[/itex] meters...does that look right to you?
 
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Would it be the dot product [itex]\vec{F}\cdot\vec{r}_{ab}[/itex]?
 
Nobody?...
 
Ok, your diagram is really confusing, because it looks like the force is pointed along one of the isometric lines. The beginning point of the force is A, and the final point is 0i+0j-12k. Yes, dot product.
 
If AD is the vector F, then CD = 0i + 0j - 12k and CA = 6i + 7j - 2k. the AD = -6i -7j -10k
Find unit vector along AD and force F in vector form. Then projection = F.rA
 
Dick said:
Ok, your diagram is really confusing, because it looks like the force is pointed along one of the isometric lines. The beginning point of the force is A, and the final point is 0i+0j-12k. Yes, dot product.

Dick! Thank you! I did not even see that! It DOES LOOK like it is parellel to the x-axis, but you are right, the tip lies at -12...damn. I'd been staring at that all night.

Thanks again,
Casey
 
So now if [itex]\vec{r}_{AD}=-6i-7j-10k[/itex] then [itex]\vec{u}_{AD}=\frac{\vec{r}_{AD}}{|r|_{AD}}=-.441i-.515j-.735k[/itex] so [itex]\vec{F}=-35.3i-41.2j-58.8k[/itex].

Now the component of [itex]\vec{F}[/itex] projected on [itex]\vec{r}_{AB}[/itex] should be equal to [itex](\vec{F}\cdot\vec{u}_{AB})*\vec{u}_{AB}[/itex]=

(a scalar)* a unit vector= a Vector?

I am just confused because I thought that components were not vectors.

Someone smack me,
 
Is this just a semantic question? If a vector is say, i+2j+3k, I don't see anything terribly wrong with saying either "the y component is 2" or "the y component is 2j". They both seem to convey the same information to me.
 
It seems to me that components have to be vectors, since they must add up to the vector itself.
 
  • #10
TMM said:
It seems to me that components have to be vectors, since they must add up to the vector itself.

Sure. The given question could also be answered by just giving the magnitude of the "projection of the force". That's why I said "semantics".
 
  • #11
Dick said:
Is this just a semantic question? If a vector is say, i+2j+3k, I don't see anything terribly wrong with saying either "the y component is 2" or "the y component is 2j". They both seem to convey the same information to me.
Right I realized this after some sleep.

But, my problem is mathematical. The answer says 31.1 N and I cannot for he life of me get that out of the numbers I have for F (above post) and [itex]\vec{u}_{AB}=\frac<br /> {-6i-3j+2k}{7}=-.857i-.429j+.288k[/itex]

Well, my Professor verified my vectors are correct via e-mail...so I must be doing something wrong in the procedure.
 
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  • #12
arrrgghh...I'll go through it again and scan in my work. Maybe someone can spot it...
 
  • #13
Saladsamurai said:
arrrgghh...I'll go through it again and scan in my work. Maybe someone can spot it...

Your u_AB looks correct. What do you get for u_DA? Because then you just want 80*u_AB.u_DA.
 
  • #14
Ah, I see you've already posted that. So just do the calc. What do you get for u_AB.u_DA?
 
  • #15
I think I got it. I did [itex]\vec{F}_{AD}\cdot\vec{u}_{AB}=30.97 N[/itex] which is very close to the 31.1 N I am looking for. I am sure that is from round-off.

Dick, thank you. It's always a pleasure...at least for me!
 
  • #16
You did a bad round off on the last component of u_AB. You're welcome!
 

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