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Help: Newton's Laws - cement bag at rest.

  1. Oct 1, 2004 #1
    Q: A bag of cement of weight 339 N hangs from three wires as suggested in the figure below. Two of the wires make angles and with the horizontal. If the system is in equilibrium, find the tensions in the wires?


    The answer requires T1, T2, T3.

    I know T3 = Fgravity = 339N.

    I made a free-body diagram at the knot:


    Now, I can resolve the forces into their x, y components:

    T1: x: -T1(cos62) y: T1(sin62)
    T2: x: T2(cos25) y: T2(sin25)
    T3: x: 0 y: -339N

    I can make the two required equations:

    (1) Fnetx = T2(cos25) - T1(cos62) = 0

    (2) Fnety = T1(sin62) + T2(sin25) - 339 = 0

    I can solve (1) for T2 in terms of T1:

    T2 = T1(cos62/cos25)
    T2 = 0.961 T1

    My problem occurs here, substituting T2 into (2) to get T1:

    T1(sin62) + (0.961 T1)(sin25) - 339 = 0

    I do not know how to solve for T1. Could someone be of assistance, and also double check that I got everything up until the point I was stuck, correct?

    Thank you!

    PS. This is first-year University Physics.
  2. jcsd
  3. Oct 1, 2004 #2


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    Now, I haven't bothered to check your numerical values; your approach has no flaws as far as I can see.
    "T1(sin62) + (0.961 T1)(sin25) - 339 = 0"
    Rewrite this as:

    T1(sin62 + 0.961sin25) = 339
    You should be able to solve it now.
  4. Oct 1, 2004 #3
    That helps tremendously. Thank you very much!
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