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Three Pulling Strings

  1. Oct 22, 2004 #1
    Q.
    Three strings, in the horizontal plane, meet in a knot and are pulled with three forces such that the knot is held stationary. The tension in string 1 is T1 = 3 N. The angle between strings 1 and 2 is q12 = 130° and the angle between strings 1 and 3 is q13 = 120° with string 3 below string 1 as shown.

    a) Find the tension in string 2.

    Ans. Decomposed the tensions into there x and y components and found the tension T2 as. 2.75 N, which is correct.


    b) Find the tension in string 3.

    Did the same as a) and found T3 to be 2.44 N which is correct again.

    A mass of 2 kg is now placed on the knot and supported by a frictionless table in the plane of the strings.

    c) Find the acceleration of the mass if all the forces remain the same as above.

    Ans. The forces remain the same, so the acceleration must be zero, which is correct.

    d) If the sizes and directions of T2 and T3 remain the same, but T1 is increased by 1.4 N, what is the acceleration of the mass?

    Ans. Stuck on this part ! Help me on this one. , what does the question mean when it says T2 and T3 the sizes and directions remain the same, does it mean T2 and T3 have same magnitude, I know T1 is increased in part d, which means new T1 would be orignal T1 + 1.4N which is 3.4 N , now.

    How to setup the eqns here. PLEASE HELP ME :cry:
     
  2. jcsd
  3. Oct 22, 2004 #2

    NateTG

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    Science Advisor
    Homework Helper

    The way I read it, it's like you had the equilibrium, and an additional 1.4N were added along the first string - as if someone's pulling a little harder on rope 1.

    You can either decompose all of the vectors and directions and figure things out, or you can take advantage of the fact that you already know the previous net force was zero.
     
  4. Oct 22, 2004 #3
    I went with the decomposition and did as follows:

    Original T1 + New T1 = 3N + 1.4N = 4.4N ; T2 = 2.75 N and T3 = 2.44 N

    so,

    4.4N + T2cos (130 ) + T3 cos (120) = m * a ( along the x - direction )

    4.4N + 2.75N * cos (130) + 2.44N cos (120 ) = 2 * a ( along the x - direction )

    Tried to solve, for a and I got 0.71 m/s2, which in incorrect. Can somebody tell me if I am missing something.

    or , if there is an alternate and a much shorter method, please let me know. the fact that you already know the previous net force was zero.[/QUOTE]

    Thanks,

    Naeem
     
  5. Oct 22, 2004 #4
    Yeah, I got it folks , problem with sig figs.... answer is 0.70 m/s2..
    Thanks anyway
    !!!!!!!!!!!
     
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