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Why is this rigidity constant decomposed using a squared cosine?

  1. Mar 11, 2017 #1
    Good night guys. I've been studying for a test on mechanical vibrations and I came across this problem of a telescopic crane (or whatever its name is). It's necessary to obtain a equivalent constant kp and then it has to be decomposed into its vertical component and the author uses a squared cosine of 45 degrees to do so. I would have used a sine of 45 degrees instead without squaring it. I know the sine and the cosine have both the same result, I'm just using a sine because the vertical component is needed.
    Could you please tell me why the author is using a squared cosine? Also, why is the force pulling and not pushing when the weight of the guy is obviously going downwards?
    Thank you
    IMG_1119.JPG IMG_1120.JPG
     
  2. jcsd
  3. Mar 12, 2017 #2
    Off the top of my head, I don't have a good answer to your question. That said, let me note that the axial vibration is rarely ever the major concern for a system such as this. In reality, it needs to be treated as a cantilever in bending, with non-uniform bending stiffness. Bending is almost always the dominant deformation mode when both axial and bending deformations are present. I'm going to have to think a while longer about your original question.
     
  4. Mar 12, 2017 #3
    Yes please Dr.D. thanks
     
  5. Mar 12, 2017 #4
    By the way. I'm sorry if this was posted in the wrong forum. I'll be posting this kind of threads in the homework section in the future. Thanks for your understanding.
     
  6. Mar 12, 2017 #5
    The problem statement appears to be in Spanish (or is it Portuguese?), which just about wipes me out. I can guess at what the problem is asking, but based on that, the answer given makes no sense at all. To do much better, I would suggest that you translate the problem statement into English. After that, I and probably others as well, will be able to do more to help you.
     
  7. Mar 13, 2017 #6
    Yes, I will translate it. Thank You. :)
     
  8. Mar 14, 2017 #7
    The attached PDF gives an explanation of this problem that I hope will make clear the very limited role that the boom angle plays.
     

    Attached Files:

  9. Mar 14, 2017 #8

    JBA

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    As I translate it, the first section of the problem is simply a calculation of the composite kb (stiffness) for the combined three boom segments' k1, k2, k3 along the axis of the boom, ignoring the bending of the beam; with a following conversion of that linear beam kb to vertical component k. As a result, I cannot see any reason for the use of cos^2 45° rather than simply sin 45°.
     
  10. Mar 15, 2017 #9
    Request to Guidestone:
    May we have a citation for the book where this problem is found (author, title, publisher, date, page number), please? Also, is the language Spanish?
     
  11. Mar 17, 2017 #10
    Hi, thank you for the answers and the material you have posted. I've been really busy with exams so I don't have the chance to be around. The book's name is mechanical vibrations and it was written by Singiresu S.Rao. I will make my best to get the English version of it. I don't remember the number of the page but I'll make it available as soon as I finish with school stuff. The language is Spanish indeed. :)
    Thanks again
     
  12. Mar 20, 2017 #11
    Guidestone, I look forward to your translation and the page number (don't forget the publisher, date, and edition also). Thanks,
     
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