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Tension Forces Question

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A bag of cement hangs from three wires as shown. Two of the wires make angles theta1 and theta2, respectively, with the horizontal.
    a)Show that, if the system is in equilibrium, then:
    T1=Fgcostheta2/sin(theta1+theta2)

    b) Given that Fg=325 N, theta1=10.0degrees, and theta2=25.0degrees, find the tensions T1, T2, and T3 in the wires.



    3. The attempt at a solution

    Part A:

    (Fnet)x=max (a=0)
    T2cosx2+T1cosx1=0
    T2=-T1cosx1/cosx2

    (Fnet)y=may (a=0)
    T1sinx1+T2sinx2-Fg=0
    I substituted in the equation I got for T2 which gave me:

    T1sinx1+T1cosx1sinx2/cosx2-Fg=0

    I rearranged the above equation to solve for T1 and got:

    Fgcostheta1/sinx1cosx2+sinx2

    I assume I need a cosx1 in the denominator so it can be changed to sin (x1+x2) but I'm not sure where I get the cosx1 from? Did I skip it in one of the equations?

    Now for Part B:

    Fg=325 N (cause it's the weight of the cement bag)
    x1=10.0 degrees
    x2=25.0 degrees

    I plugged them into the above equations that I had found:

    T1=Fgcosx2/sin(x1+x2)
    T1=325 N x cos 25.0 degrees/sin (10.0degrees + 25.0degrees)
    T1=514 N

    T2= -T1cosx1/cosx2
    T2= -514 x cos10.0degrees/cos25.0degrees
    T2=-558

    I don't understand why tension 2 would be negative though. I would asume they would all be upwards vertical forces. Maybe I solved one of the equations wrong?

    I feel like I should just drop the negative sign, but I don't know if that would be correct.
     
  2. jcsd
  3. Sep 9, 2007 #2
    Ok, I think I see what I've done wrong already. Perhaps T1cosx2 should be negative to begin with and then when I rearrange the equation it will become positive?

    I still don't see where the second cosx2 came from though.
     
  4. Sep 9, 2007 #3
    Maybe more details will help?

    I went back over my equation and got the same thing:

    T1sinx1+T2sinx2-Fg=0
    Substituting in the first equation:
    T1sinx1+(T1cosx1/cosx2)sinx2-Fg=0
    So now I bring over my numbers:

    T1sinx1+T1cosx1/cosx2)sinx2=Fg
    Then I divide the cosx2 in the denominator to bring it to the other side:

    T1sinx1+T1cosx1sinx2=Fgcosx2

    Then I divide over the sinx1, cosx1, and sinx2, which leaves me with:

    T1+T1=wcosx2/sinx1+cosx1sinx2

    How do I get rid of the second T1 and where doest he cosx2 in the denominator come from? Did I make a mistake in my calculations?
     
  5. Sep 9, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Can you give a diagram? I don't understand how the wires are arranged. What's the angle of the third wire?
     
  6. Sep 9, 2007 #5
  7. Sep 9, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    OK.
    OK.
    Error: You must multiply all terms by cosx2.
     
  8. Sep 9, 2007 #7
    So, should I have:

    T1sinx1cosx2+T1cosx1sinx2cosx2=Fgcosx2?

    I still don't understand what happens to the second T1
     
  9. Sep 9, 2007 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Almost, but you have an extra cosx2 factor in the second term.

    Not sure what you mean by "second" T1. Realize that: A*X + A*Y = A*(X + Y).
     
  10. Sep 9, 2007 #9
    Let me try to explain where I'm not getting it:

    I understand now where the second cosx2 came from, but I wind up with this equation:

    T1+T1=Fgcosx2/sinx1cosx1sinx2cosx2

    Doesn't that mean it's really
    2(T1)=Fgcosx2/sinx1cosx1sinx2cosx2

    I want to know how to get just:

    T1=Fgcosx2/sinx1cosx1sinx2cosx2
     
  11. Sep 9, 2007 #10

    learningphysics

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    Homework Helper

    This should be T1sinx1cosx2 + T1cosx1sinx2 = Fgcosx2

    So T1(sinx1cosx2 + cosx1sinx2) = Fgcosx2

    Divide both sides by (sinx1cosx2+cosx1sinx2)

    T1 = Fgcosx2/(sinx1cosx2+cosx1sinx2).

    there isn't a factor of 2.
     
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