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Homework Help: Tension; 2 cables holding weight

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A small sphere of weight W is held as shown by two wires AB and CD. If wire AB is cut, determine the tension in the other wire (a) before AB is cut, (b) immediately after AB has been cut. (Ans: (a) TCD = 0.742 W; (b) TCD = 0.940 W).

    http://img4.imageshack.us/img4/3763/picture11td.png [Broken]

    2. Relevant equations

    [tex]\Sigma[/tex]Fy=may
    Ty=Tsin[tex]\theta[/tex]

    3. The attempt at a solution

    TCDy+TABy-W=may ; ay=0 (not moving)

    TCDSin(70[tex]\circ[/tex])+TABSin(50[tex]\circ[/tex])-W=0 ; solve fot TAB and plug in

    TABSin(50[tex]\circ[/tex])=W-TCDSin(70[tex]\circ[/tex]) ; divide Sin(50[tex]\circ[/tex]) from both sides

    TAB=[W/Sin(50[tex]\circ[/tex])]-[TCDSin(70[tex]\circ[/tex])/Sin(50[tex]\circ[/tex])]

    TCDSin(70[tex]\circ[/tex])+[W/Sin(50[tex]\circ[/tex])]-[TCDSin(70[tex]\circ[/tex])/Sin(50[tex]\circ[/tex])]-W=0 ; Plugging in TAB from above

    TCD(Sin(70[tex]\circ[/tex])-(Sin(70[tex]\circ[/tex])/(Sin(50[tex]\circ[/tex])=W(1-1/(Sin(50[tex]\circ[/tex])

    Solving for TCD I get 1.066W when its supposed to be 0.742 W

    What am i doing wrong?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 17, 2010 #2

    rl.bhat

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    Homework Helper

    T(CD)*sin70 + T(AB)*sin50 - W = 0.
    After this write
    T(CD)*cos70 + T(AB)*cos50 = 0
    Now solve.
     
  4. Feb 17, 2010 #3
    so do the sum of the forces in the x direction?
    Wouldn't it then be T(CD)*cos70 - T(AB)*cos50 = 0 ?

    ---
    Worked out the sum of the forces in x and y direction
    solved for TAB Set them equal to each other and still wrong answer :(
     
    Last edited: Feb 17, 2010
  5. Feb 17, 2010 #4

    rl.bhat

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    Homework Helper

    In my equation I have given the vector sum. Your equation is correct.
    T(CD)*sin70 + T(AB)*sin50 = W ....(1)
    After this write
    T(CD)*cos70 = T(AB)*cos50
    T(AB) = T(CD)*cos70/cos50
    Put it in the eq, 1, you get
    T(CD)*sin70 + [T(CD)*cos70/cos50]*sin50 = W
    Now simplify.
    For part (b) T(CD) = component of W along the wire.
     
  6. Feb 17, 2010 #5
    thanks, turned out i just wasn't inputting in the calculator right.

    funny how this one of the simplest problems out of the assignment and i was having trouble with it and easily solved the complex ones. Always the easy stuff that gets me, i guess i think about it too hard, make it more complex then it is.

    been working on so much homework (thermo, dynamics, eaII, and solids) that my mind is fried!
     
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