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Derivation of equation for mass on pulley and displacement

  1. Nov 14, 2006 #1
    [​IMG]

    Derivation of equation for mass on pulley and displacement


    Sorry that the picture stinks, but its all I got. The system is in equilibrium.The counter mass on the left is mass A and the mass on the right is mass B, both of mass m. The center mass of mass M is denoted as B. The length of the system is denoted as L. h stands for the vertical displacement of the center mass. The equation is..

    h= ML / sqrt(16m^2 - 4M^2)

    I wrote the equations for the sum of the forces and my teacher told me I could derive it from those but I can't get any further than what I have.
    Forces in x direction = T(sub c)cos(theta) - T(sub a)cos(theta) =0 and
    Forces in y direction = T(sub c)sin(theta) + T(sub a)sin(theta) - T(sub b) = 0
     
  2. jcsd
  3. Nov 14, 2006 #2
    You don't have the tensions T described in the drawing.

    Also -- can you give us expressions for the tensions in terms of m, M, L, h?
     
  4. Nov 16, 2006 #3
    [​IMG]
    T1 = T2 = T3 = T4 = mg
    T5=Mg

    I don't know how to bring L or h into this.
     
    Last edited: Nov 16, 2006
  5. Nov 17, 2006 #4

    radou

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    In what relation is T5 with T2 and T3?
     
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