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Hanging Sign problem

  1. Sep 21, 2007 #1
    I'm working on this hanging sign problem:

    A sign hangs precariously from your prof's office door. Calculate the magnitude of the tension in string 1, if θ1 = 27.42°, θ2 = 67.35°, and the mass of the sign is 9.9 kg.

    I was wondering if I could put 9.9*9.8 as a leg in the middle of the triangle to calculate the tension on the two strings. The only given hint is to calculate components.
     
  2. jcsd
  3. Sep 21, 2007 #2

    stewartcs

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    Draw a FBD then use Newton's second law, Fnet = Ma

    T1 + T2 + Fg = Ma

    The sign is in static equilibrium so Ma = 0, yielding,

    T1 + T2 + Fg = 0

    Find the x components and the y components using the given angles (noting their referenced axis) and then solve the two resulting simultaneous equations.

    CS
     
  4. Sep 21, 2007 #3
    could i use the force the sign is exerting as a leg in order to aid with component calculations?
     
  5. Sep 21, 2007 #4
    I guess what I'm trying to say, am I already give a Y component 97.02 N, and can I use that to calculate the remaining ones
     
  6. Sep 21, 2007 #5

    stewartcs

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    No. With only the Y component of force you do not have enough information to solve the problem. You must use Newton's second law and then break the problem down into componets in order to find the tension vector.

    The technique I posted is the proper way to solve the problem so you will conceptually understand what is going on with all of the forces.
     
  7. Sep 21, 2007 #6
    can i just make sure I have this right
    t
    T1+T2-Mg=Ma

    T1+T2-97.02=0

    That's what I got from my relatively simple diagram. I'm just not sure how to calculate components without a length.
     
  8. Sep 21, 2007 #7

    stewartcs

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    T1 + T2 - Mg = 0 is correct. This is your equilibrium equation.

    The components depend on the referenced axis and the direction of the angle. Is θ1 = 27.42° from the positive x axis or negative x axis (look at your drawing). What about θ2 = 67.35° referenced position (they should be opposite given your problem but maybe not)?

    If θ1 is a negative angle from the negative x axis and θ2 from the positive you have:

    x components: -T1cos(θ1) + T2cos(θ2) - 0 = 0

    y components: T1sin(θ1) + T2sin(θ2) - Mg = 0

    Solve those two equations simultaneously and then substitute back into the equilibrium equation.
     
  9. Sep 22, 2007 #8
    I know this is probably going to yield me a blaring duh, but i just solve the equation for T1.
     
  10. Sep 22, 2007 #9
    okay, so i forgot to mention that T2 is to the left of the Y

    my two equations are:

    X=-T2*cos(theta)+T1*cos(theta)=0
    Y= T1*sin(theta)+T2*sin(theta)-97.02=0

    I tried solving for X
    T2*cos(theta)=T1*cos(theta)
    T2=T1*cos(theta)/cos(theta)
    T2=T1

    It worries, me, I don't think that should be right.
     
  11. Sep 23, 2007 #10
    I got 66.4289 N for the tension in string 1

    is this right?
     
  12. Sep 23, 2007 #11
  13. Sep 23, 2007 #12

    stewartcs

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    No. The angles (theta 1 and theta 2 are different) so T1 is not equal to T2. If they were the same then you would be correct.
     
  14. Sep 23, 2007 #13
    do the cos not cancel out
     
  15. Sep 23, 2007 #14

    stewartcs

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    No. The angles are different so the cosines do not cancel. If the angles were the same they would, but they are not.
     
  16. Sep 23, 2007 #15
    Is the answer 187.011 N? I just got that by turning the whole thing into a triangle.
     
  17. Sep 23, 2007 #16

    stewartcs

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    No.

    Did you solve these two equations simultaneously?

    x components: -T1cos(θ1) + T2cos(θ2) - 0 = 0

    y components: T1sin(θ1) + T2sin(θ2) - Mg = 0


    Solve the x component equation for T2:

    -T1cos(θ1) + T2cos(θ2) - 0 = 0

    T2cos(θ2) = T1cos(θ1)

    T2 = T1(cos(θ1)/cos(θ2))


    Now substitue T2 into the y component equation:

    T1sin(θ1) + T2sin(θ2) - Mg = 0

    T1sin(θ1) + T1(cos(θ1)/cos(θ2))*sin(θ2) = Mg

    T1(sin(θ1) + (cos(θ1)/cos(θ2))*sin(θ2)) = Mg

    T1 = Mg/(sin(θ1) + (cos(θ1)/cos(θ2))*sin(θ2))

    Now you have the two equations (assuming the referenced angles were the same as I posted earlier). Solve this one for a numeric value of T1 since you have all the missing information for it. Then plug that answer into the T2 = T1(cos(θ1)/cos(θ2)) to find the other tension.
     
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