A Bar Suspended by Two Wires

Discussion in 'Homework' started by Ingolf, Dec 7, 2010.

  1. Ingolf

    Ingolf Well-Known Member

    A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle phi_1 with the horizontal, and the right wire makes an angle phi_2. The bar has length L.

    [​IMG]

    Find the position of the center of mass of the bar, x, measured from the bar's left end.
    Express the center of mass in terms of L, phi_1, and phi_2.
     
    Last edited: Dec 7, 2010
  2. 3.14159

    3.14159 Well-Known Member

    zz
    Edit: I wrote ZZ coz i have no idea how to solve this question, not because i think that this problem is lacking data. I wouldnt know if it is.
     
    Last edited: Dec 9, 2010
  3. Shamanics

    Shamanics Banned

    The problem is lacking data.
     
  4. Ingolf

    Ingolf Well-Known Member

    No it wasn't. I finally solved it using equilibrium.
     
    Last edited: Dec 13, 2010
  5. Shamanics

    Shamanics Banned

    Show me your solution, I insist the problem lacks data.
     
  6. Kriegskanzler

    Kriegskanzler Well-Known Member

    How exactly were you able to solve this? And what formula? Torque?
     
  7. Mr.DarC

    Mr.DarC Well-Known Member

    How did you use equilibrium?

    The strings are massless so mg=0
     
  8. Ingolf

    Ingolf Well-Known Member

    My solution:

    The bar is in equilibrum so the net torque and forces acting on it must be zero. I take the torque of the left end of the bar.

    The formula for the net torque looks like this:

    (1) L*T_2*sin(phi_2)-w*x=0

    where T_2 is the tension in wire #2 and w is the weight of the bar (m*g).

    The net forces in x- and y-directions must also be 0, so we have

    (2) F_x: T_2*cos(phi_2)-T_1*cos(phi_1)=0
    (3) F_y: w-T_1*sin(phi_1)-T_2*sin(phi_2)

    where T_1 is the tension in wire #1. I have taken the positive y-direction to be downward so that w (weight) is positive.

    By rearranging (2) I get that

    (4) T_1/T_2=cos(phi_2)/cos(phi_1)

    I solve for x in (1) and get

    (5) x=L*T_2*sin(phi_2)/w

    Solving for w in (3) and inserting in (5) gives

    (6) x=L*T_2*sin(phi_2)/(T_1*sin(phi_1)+T_2*sin(phi_2))
    x=L/((T_1/T_2)*(sin(phi_1)/sin(phi_2))+1)

    By inserting (4) in (6) I get that

    x=L/(cos(phi_2)*sin(phi_1)/(cos(phi_1)*sin(phi_2))+1)
    x=L/(tan(phi_1)*tan^(-1)(phi_2)+1)=L/(tan(phi_1)/tan(phi_2)+1)
     
  9. Shamanics

    Shamanics Banned

    I guess you dont use flexibility at all.... But meh, that is ok.
     
    Last edited: Dec 12, 2010