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Trig. identities method problem

  1. Oct 19, 2006 #1
    I've been struggling with the followings equations (I know that can be solved by Trig. identities method) and I tried almost everything but I can't get them right.
    the equations are the following:
    -Tab*cos(alpha)+Tbc*cos(beta) = 0 (eq.1)
    We know the angular values for alpha and beta, and also I reviewed the answers on my workbook and they are; Tab=9.67 and Tbc=12.334, I would like if someone can show me how to solve those identities step by a step.
    Thanks in advance
    Those equations are part of a problem in mechanics of materials, in which two wires are supporting a weigth (15) both in different known angles (that's why the equations of equilibrium yield those two precedent equations) but I can't get the correct results for each tension in the wires, I know they are trig. identities and something I am doing is not coming well enough in the algebraic procedure.
  2. jcsd
  3. Oct 19, 2006 #2
    According to the question
    for my convenience let alpha=A and beta=B
    there are two equations
    Tbc*cosB - Tab*cosA = 0 ------i
    From i
    Tbc*cosB = Tab*cosA
    => Tbc=Tab*cosA/cosB dividing by cosB both sides
    Placing this value in ii
    => Tab*sinA + (Tab*cosA/cosB)*sinB = 15
    => Tab*(sinA + cosAsinB/cosB) = 15
    => Tab*(sinAcosB + cosAsinB)/cosB = 15
    => Tab*sin(A + B) / cosB = 15 As sin(A+B)=(sinA*cosB)+(cosA*sinB)
    => Tab = 15*cosB/sin(A + B)
    Placing this value in
    Tbc = Tab * cosA / cosB
    now u can calculate the values
  4. Oct 19, 2006 #3


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    Naumaan gave a perfectly good answer but I would do it a slightly different way.
    Multiply the first equation by [itex]T_{ab}cos(\alpha)+T_{bc}cos(\beta)[/itex] to get [itex]T_{bc} cos^2(\alpha)-T_{ab}cos^2(\beta)= 0[/itex].
    Multiply the second equation by [itex]T_{ab} sin(\alpha)- T_{bc}sin(\beta)[/itex] to get [itex]T_{ab}^2 sin^2(\alpha)- T_{bc}^2 sin^2(\beta)= 15T_{ab}sin(\alpha)- 15T_{bc}sin(\beta[/itex] and add the two equations. since [itex]sin^2(x)+ cos^2(x)= 1[/itex], that gives [itex]T_{ab}^2 - T_{bc}^2= 15T_{ab}sin(\alpha)- 15T_{bc}sin(\beta[/itex] which involves only sines of [itex]\alpha[/itex] and [itex]\beta[/itex].

    Use [itex]T_{ab}sin(\alpha)+ T_{bc}sin(\beta)= 15[/itex] to replace either [itex]sin(\alpha)[/itex] or [itex]sin(\beta)[/itex] by the other and solve the resulting equation.
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