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Tension of two wires at 40 and 45 degrees

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data
    An advertising sign of mass 40kg is suspended by two wires as shown in the diagram. What is the tension in each wire?

    http://img132.imageshack.us/img132/9681/untitledwx0.png [Broken]

    I have no idea how to solve this, or what formulas a re relavent to it. Any suggestions are greatly appreciated.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 17, 2007 #2
    The 1st step, you should do is draw a free body diagram for the hanging sign, that shows the forces acting on it. Include in your free body diagram, the angles of the rope.

    2nd step should be listing the relevant Force equations in the x and y direction.
    Hint: (Fnet)x = 0 = ....
    (Fnet)y = 0 = ...

    then try to find a relationship amonst them to solve for the unknown quantities. (Hint: i think you will be better off starting with the (fnet)y equation).
  4. Mar 17, 2007 #3
    okay thanks

    so the (Fnet)y=-400 + 400=0
    therefore the combined (F)y of the two wire is equal to 400 N?
  5. Mar 17, 2007 #4
    uhh no... how did u get -400+400??
  6. Mar 18, 2007 #5
    i assumed that gravity acceleration is 10 m s-1
    then used w=mg to find 400
    since the sing pulls down with 400N the two wire must pull up with 400N for the system to be in equilibrium so there 400N acting in both direction for it to cancel each other out
  7. Mar 18, 2007 #6


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    You are correct, so you can write;

    [tex]T_1\cos40 + T_2\cos45 = 400[/tex]

    Now, do the same for the x components.
  8. Mar 18, 2007 #7
    okay thanks Hootenanny
    so if i do the following and solve simultaneously, will it yield me the correct result?
    T_1 \cos 40 + T_2 \cos 45 = 400 \\
    T_1 \sin 40 + T_2 \sin 45 = 0 \\
    \left[ {\begin{array}{*{20}c}
    {\cos 40} & {\cos 45} & {400} \\
    {\sin 40} & {\sin 45} & 0 \\
    \end{array}} \right] \\
    \left| {T_1 } \right| = 354.9408\,N \\
    T_2 = 310.8120\,N \\

  9. Mar 18, 2007 #8
    oh i c.. ye thats correct... you wrote (Fnet)y=-400 + 400=0, but it's
    Fnet =-400 + 400=0.
  10. Mar 19, 2007 #9
    okay thanks for the help, greatly appreciated
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