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Right angle equilibrium

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data

    the right angle ruler shown in the figure hangs from a peg . it is made of unifrom sheet metal. let lambda equal the leiner dinsity. one arm is Lcm long and the other is 2L cm long. find the angle theta at which it will hang in equilibrium

    2. Relevant equations



    3. The attempt at a solution

    I belive ij ust sum the forces, that is the forces in the y direction equal zero as well as the torqs. i dont think i need the x right?

    F(y) = 0 = P - .5mg - mg = P - .75mg

    F(torque) = 0 = .5Amg - Bmg hwere A = .5LcosQ and B = LsinQ

    F = 0 = .5(.5LcosQ)mg - (LsinQ)mg = .25cosQ - sinQ = 0

    is this correct, how do i solve this
     

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    Last edited: Dec 8, 2009
  2. jcsd
  3. Dec 8, 2009 #2

    PhanthomJay

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    you are on track, but you have the force P equal to .75mg, where instead, it must equal mg, where mg is the total weight of the right angled metal. It doesn't make much difference, though....note that sin theta/cos theta =tan theta, when you solve your equation for theta.
     
  4. Dec 9, 2009 #3
    something is not right

    .25cosQ = sinQ
    cosQ = 4sinQ
    0 = 4tanQ

    Q = 0

    what did i do wrong
     
  5. Dec 9, 2009 #4

    PhanthomJay

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    Your algebra.

    if .25cosQ = sinQ, then divide both sides of the equation by cosQ, which yields

    .25cosQ/cosQ = sinQ/cosQ, or
    .25 = tanQ

    Solve for the angle, Q.
     
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