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Finding Tension is three cables attached to a particular mass?

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    An object of mass m is suspended from a cable tied to two other cables, each fastened to a supporting beam, as shown below. The tensions in the three supporting cables are T1, T2, and T3. The diagram of the system is available via the link below:

    http://i40.tinypic.com/15goxg3.jpg

    The tension T1 = 1800 N.

    I worked out the first three parts, but could not work out parts d) and e)
    (a) Determine the horizontal component of T1.
    (b) Determine the horizontal component of T2.
    (c) Determine the tension T2.
    (d) Determine the tension T3.
    (e) Determine the mass of the suspended object.


    2. Relevant equations



    3. The attempt at a solution
    (a) Determine the horizontal component of T1.
    Th = 1800cos(60) = 900N
    (b) Determine the horizontal component of T2.
    Th = 1800cos(60) = 900N (The same component for T2) - The only thing is I do not understand why this is the same - I only knew to calculate it the same because I have the answers given.
    (c) Determine the tension T2.
    T2 = 900/cos(40) = 1175N

    Any help would be appreciated greatly
     
  2. jcsd
  3. Nov 12, 2011 #2
    T1/sin40=T2/sin60=T3/sin 80
    you can work out the T3 tension from this equality.and it is easily seen from figure mass m equals T3.
     
  4. Nov 12, 2011 #3

    PhanthomJay

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    Try applying Newton's 1st Law in the horizontal direction to solve for the horizontal component of T2
     
  5. Nov 13, 2011 #4
    using that method, I can't seem to get the correct answer:

    1800/sin40 = 1175/sin60 = T3/sin80

    ((1800/sin40) - (1175/sin60)) x sin80 = 1421.6N

    The answer for part d) T3 = 2313N

    What method do I apply to get this if it is not the one I currently used?
     
  6. Nov 13, 2011 #5

    PhanthomJay

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    Try applying Newton's 1st Law in the horizontal direction to solve for the horizontal component of T2. Then apply his 1st law in the vertical direction to solve for T3.

    Newton's 1st Law:

    [itex]\Sigma F_x = 0[/itex]
    [itex]\Sigma F_y = 0[/itex]
     
  7. Nov 14, 2011 #6
    Ok, i've got it now, thanks for your help:

    Information given as well as calculations done previously can help to solve part d and subsequently e.

    Part d)

    T3 = Tv1 + Tv2

    Tv1 = 1800sin(60) = 1560N
    Tv2 = 900tan(40) = 755N

    Therefore, T3 = 1560 + 755 = 2315N

    for part e)

    m = T3/g = 2315/9.8 = 236kg
     
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