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Finding projected force along line

  1. Dec 21, 2007 #1
    [​IMG]

    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?
     
    Last edited: Dec 21, 2007
  2. jcsd
  3. Dec 21, 2007 #2
    Would it be the dot product [itex]\vec{F}\cdot\vec{r}_{ab}[/itex]?
     
  4. Dec 21, 2007 #3
    Nobody?...
     
  5. Dec 22, 2007 #4

    Dick

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    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.
     
  6. Dec 22, 2007 #5

    rl.bhat

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    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
     
  7. Dec 22, 2007 #6
    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
     
  8. Dec 22, 2007 #7
    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,
     
  9. Dec 22, 2007 #8

    Dick

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    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.
     
  10. Dec 22, 2007 #9

    TMM

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    It seems to me that components have to be vectors, since they must add up to the vector itself.
     
  11. Dec 22, 2007 #10

    Dick

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    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".
     
  12. Dec 23, 2007 #11
    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
    {-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.
     
    Last edited: Dec 23, 2007
  13. Dec 23, 2007 #12
    arrrgghh...I'll go through it again and scan in my work. Maybe someone can spot it...
     
  14. Dec 23, 2007 #13

    Dick

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    Your u_AB looks correct. What do you get for u_DA? Because then you just want 80*u_AB.u_DA.
     
  15. Dec 23, 2007 #14

    Dick

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    Ah, I see you've already posted that. So just do the calc. What do you get for u_AB.u_DA?
     
  16. Dec 23, 2007 #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!
     
  17. Dec 23, 2007 #16

    Dick

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    You did a bad round off on the last component of u_AB. You're welcome!
     
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