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Hanging chandelier

  1. Feb 5, 2006 #1
    **** nvm**** i accidentally gave up. im actually emailing my prof right now to see if i can undo it

    why must they make this so difficult?

    here is the problem:


    my answer is:

    t1= mg/[cos(90-theta1)+ sin(90- theta1)]

    this is definately right. the only problem is that i have to include theta 2 in my formula. this has stumped me
    Last edited: Feb 5, 2006
  2. jcsd
  3. Feb 5, 2006 #2
    is it a right triangle, because if it's a right triangle, you know that 90 -theta1 is just theta2
  4. Feb 5, 2006 #3
    I got this when I did it.
    [tex] T_1 = \frac{mg - T_2sin\theta_2}{sin\theta_1} [/tex]
    Surely [tex] cos(90 - \theta_1) [/tex] is the same as [tex] sin(\theta_1) [/tex]?

    Edit: Dam it, beaten by a minute
  5. Feb 5, 2006 #4
    o. well i thought it was right
  6. Feb 5, 2006 #5
    i do not believe that it is a right triangle, but it could be i guess

    this is what led me to believe that was right:


    look at problem number 4 on page 3
  7. Feb 5, 2006 #6
    i get what you guys are staying, but that wouldnt include theta2 without using T2
  8. Feb 5, 2006 #7
    The expression you gave in your original post was re-arranged wrong, but the correct one is the same as the one I posted earlier. To get rid of theata2, resolve vertically and substitute in.
  9. Feb 5, 2006 #8

    but doesnt it say not to include t2 in it?
  10. Feb 5, 2006 #9
    i did it your way and it said:

    The correct answer involves the variable theta_2 which was not part of your answer.
  11. Feb 5, 2006 #10
    in the link that you gave, theta2 was 45. The answer includes a tangent of theta2, but since tangent of 45 is just 1, it does not appear in the answer there.
  12. Feb 5, 2006 #11
    guys forget it, i accidentally gave up on this question

    thanks for your help, and sorry to waste your time
  13. Oct 9, 2008 #12
    hey, i have the same question for my assignment. the asnwer is:

    mg/ sin(theta1)+tan(theta2)*cos(theta1)
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