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Homework Help: Irregular beam physics homework

  1. Jun 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A 1750-N irregular beam is haning horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.25m long and weighing 2.50N. The center of gravity of this beam is one third of the way along the beam from the end where wire A is attached. A) If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? B) Which pulse arrives first?


    2. Relevant equations

    T_a + T_b = 1750
    Sum of all forces = 0.
    wave speed = sqrt(force/linear mass density)
    C_g is L/3 from A and 2L/3 from B

    3. The attempt at a solution

    I really just have hit a logical block and have no idea what to do.
    ------
    L = 1.25
    m = .255
    mu = mass/length = .204
    To find the force I tried: T = F * L/3
    with F = 1750, a guess, really
    1.25*1750/3 = 729.1666

    Now I can't decide what to do with that number or even if it is a correct answer.
     
    Last edited: Jun 7, 2010
  2. jcsd
  3. Jun 7, 2010 #2

    Doc Al

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    Staff: Mentor

    Re: Beam

    OK.
    Don't guess. Set the net torque on the beam to zero and solve for the two tensions.
    Not sure what you're doing here. Why did you multiply by 1.25?

    Once you find the force, use it and mu to find the speed of the pulse. How long does each pulse take to travel to the ceiling?
     
  4. Jun 7, 2010 #3
    Re: Beam

    Such that:
    F1L1-F2L2=0

    L1 = 1.25/3 = .417
    L2 = 2*1.25/3 = .833

    ?
     
  5. Jun 7, 2010 #4

    Doc Al

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    Staff: Mentor

    Re: Beam

    Yes.

    No, the 1.25 is the length of the strings, not the length of the beam. Call the length of the beam "L". Then what are L1 and L2 in terms of L? You don't need the actual length of the beam--which isn't given--to solve for the forces.
     
  6. Jun 7, 2010 #5
    Re: Beam

    This made me feel particularly... obtuse.

    L1 = L/3
    L2 = 2L/3

    Then:
    F1L1 - F2L2 = 0
    (F1*L)/3 - (F2*2L)/3 = 0

    Should F1+F2 = -1750?
     
  7. Jun 7, 2010 #6

    Doc Al

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    Staff: Mentor

    Re: Beam

    Good. Keep simplifying this. How does F1 relate to F2?

    Yes, but get rid of the minus sign. Then you can solve for the forces by combining the two equations.
     
  8. Jun 7, 2010 #7
    Re: Beam

    F1 = 2F2
    I got that from isolating F1 in the earlier equation.

    By plugging that into the lower equation

    F1=1166.67
    F2=583.33

    v = sqrt(f/mu)

    v1=75.62m/s
    v2=53.46m/s

    t=d/v;d=1.25m

    t1=.017s
    t2=.023s

    Delta_t = .006s
    The wave on String A arrives first.
     
  9. Jun 7, 2010 #8

    Doc Al

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    Re: Beam

    Excellent!

    My only suggestion would be to not round off to two significant figures until the end. (Use 3 sig figs when you calculate the time--that will give you a slightly different Delta t.)
     
  10. Jun 7, 2010 #9
    Re: Beam

    Thank you very much for your help, good Sir.

    I'll also modify my answer as you advised.

    Thanks again!
     
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