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.


    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

    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))

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

  9. Shamanics

    Shamanics Banned

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