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Solution to Buoyancy II

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  • Solution to Buoyancy II

    I want to thank everyone for their input over the past few weeks
    into the discussion on buoyancy. I would like to summarize the responses
    for the latest posting. These responses all pointed to a measurement error
    as a prime suspect for the strange results. One message I received from Hans
    Mattes noticed that a systematic addition to R1 and R2 seemed to solve the
    problem. After pulling a few hairs out looking at the old data I decided to
    re-run the validation. The numbers I produced agreed quite well (within 5
    mm) with the predicted CB regardless of the positioning of the 2 supporting
    forces. Thus there was an error in the first set of data. Upon further
    inspection of the original data I noticed that the force data were almost
    identical to the new validation. However, there were some errors in the
    recording of the moment arms in the first set of data. With the moment arms
    corrected these data provided a validation similar to the new set of data
    (i.e., within 5 mm). I appreciate your time (and patience) with this matter.
    A summary of the responses is appended.

    Scott Mclean



    Dr. Hinrichs and I were pleased by the response to our previous
    posting concerning the center of buoyancy. We now have a problem which may
    be more interesting. The method of locating the center of buoyancy was
    validated using a 180 cm long piece of PVC pipe of approximately 2 inch
    diameter. One end was filled with concrete. The ends were sealed so the
    other end was filled with air. The CM of the pipe was measured to be 65 cm
    from the heavy end. The CV was at 90 cm or half way along the length of the
    pipe due to the symmetry of the pipe.

    The measurement of the CB was validated by supporting this pipe with
    two straps (one on each side of the CB) while the pipe was fully submerged.
    The fully submerged CV would be coincident with the CB. A 4 kg mass was
    added to the strap on the light end and a 2 kg mass was added on the heavy
    end to keep the pipe submerged. The free body diagram is pictured below.

    /|\ R1 | /|\ R2
    |--dR1--| | |
    heavy | CM | | light
    ||================================================ ===========||
    end | CB end

    R1 = supporting force at the heavy end
    R2 = supporting force at the light end
    B = buoyant force
    W = weight
    dR1 = distance from heavy end to R1
    dR2 = distance from heavy end to R2
    dB = distance from heavy end to CB
    dW = distance from heavy end to CM

    The translational and rotational equations of statis equilibrium for this
    system are

    sum of forces = 0

    eqn (1) R1 + B + R2 - W = 0.

    sum of moments = 0

    eqn (2) (R1*dR1) + (B*dB) + (R2*dR2) - (W*dW) = 0.

    Solving these equations for the location of the CB in terms of measured
    forces and distances gives

    eqn (3) (W*dW) - (R1*dR1) - (R2*dR2)
    dB = ------------------------------.
    W - R1 - R2

    The net forces (i.e., with the added weight due to the added masses
    subtracted out), R1 and R2 were measured using calibrated load cells. R2
    was found to be negative (implying that the application of a downward force
    was required to keep this end of the pipe submerged).

    The measurement of dB was approximately 90 cm (within 3 or 4 mm) when R1 and
    R2 were positioned symmetrically about the CB regardless of the distance
    between the load cells. However, when the heavy end support remained
    stationary and the light end support was moved towards the middle, the
    calculated dB grew linearly from 89.5 cm to 109.8 cm (over 9 locations).
    The opposite was found when the light end supporting force remained
    stationary and the heavy end supporting force was moved. dB decreased
    linearly from 89.5 cm to 69.1 cm over 9 locations.

    We suspected that there were errors in the measurements and perhaps caused
    the strange results. But upon inspection, the buoyant force was predicted to
    be approxmately 3380 g (* 9.81/1000 N). This calculation held for every
    condition. This compared well with the calculated volume of the pipe
    (3309 cm~3). The systematic (linear) nature of the deviation suggests that
    the results were not due to poor data.

    The system contains two unknowns, the magnitude of the buoyant force (B) and
    the point of application of the buoyant force (dB). If the data we have
    collected are real then moving the load cells asymmetrically changes either
    the magnitude of the buoyant force (B) or the point of application of this
    force. The fact that the measured buoyant force agrees so well with the
    theoretical buoyant force would indicate that dB changes. This makes no
    sense based on the definitions of the buoyant force and the CB.

    Can anyone see an error in the derivations we have made or the logic we are
    using? Any help would be greatly appreciated. We will post a summary of
    all responses we receive.

    Thanks again,

    Scott Mclean
    ================================================== =========================


    I don't have the solution for your problem, but I have some
    suggestions that might help you to solve it.

    I think we can take it for certain that the CB is exactly in the
    middle of the bar and cannot move. In my opinion, we can always
    rest sure of that. No way it can be or move somewhere else.

    I suspect that the CM could have moved, or the Weight could have
    changed. Did you try to weigh the bar at the end of the experiments?
    Some water might have leaked into the cilinder. Assuming that the light end
    of the bar will always tend to be slightly higher (closer to the surface)
    than the heavy end, the water inside the cilinder would both move
    the CM closer to the heavy end and make W larger.

    Are you sure that the scale you used to weigh the cilinder was
    calibrated correctly? The best thing would be to use the load cells to
    sure W. Did you?

    Your method was correct, so the error must be searched in the
    collected data. Since you wrote that the load cells were calibrated,
    the only data that can be wrong seem to be W and dW.

    Something could be said also about possible couples of forces
    generated by the straps, which would change your equation for
    static rotational equilibrium.

    Also, the point of application of the forces exerted by
    the straps might not coincide with the center of the straps. How wide
    were the straps? Did the masses suspended below the straps deform them?

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    Panta rei :-)
    ================================================== ==========================

    Hi Scott:

    When you say "... the added weight due to the added masses was subtracted
    out", do you mean that the upthrusts on the two masses have also been

    Bryan Finlay, PhD
    Director, Orthopaedic Research Laboratory
    University Hospital 519-663-3063
    P.O. Box 5339 519-663-3904 FAX
    London, Ontario, CANADA, N6A 5A5
    ================================================== ==========================

    Here are my wild guesses:

    Maybe you have not taken into account the weight forces of the 2 Kg
    and 4 Kg masses? They don't show up in your equations, but maybe you
    included them already as part of R1 and R2 --I don't know.

    Another (although unlikely) possibility is that you need to take
    into account the buoyancy forces exerted on each of the two masses. But I
    suppose you used metal weights, so the volume would be small enough to make
    the buoyancy forces negligible.

    Jesus Dapena
    Department of Kinesiology
    Indiana University
    Bloomington, IN 47405, USA
    1-812-855-8407 (office phone) (email)
    ================================================== ==========================

    Hi Scott

    Just a thought--have you checked the accuracy of your load cells over a
    range of loads. When they are positioned at an equal distance to the left
    and to the right of the center, you presumably get the same readings. Now,
    if you moved one towards the center the readings at both locations would
    change--one would go up and the other would go down. (I have not checked
    this, but I think this is the case.) What would happen if one reading
    increased too much and the other dropped too much? Your net buoyancy force
    could stay the same, but the calculated CB would change.

    Regards, Brian Davis
    ================================================== ==========================

    Hi, Scott,

    I believe this is the same question as that you proposed before. If you feel
    no problem with that simple one, then this one dose not exsist. Please see

    > The translational and rotational equations of statis equilibrium for this
    > system are
    > sum of forces = 0
    > eqn (1) R1 + B + R2 - W = 0.
    > sum of moments = 0
    > eqn (2) (R1*dR1) + (B*dB) + (R2*dR2) - (W*dW) = 0.

    Now denoting

    eqn(4) R=R1+R2
    eqn(5) dR=(R1*dR1+R2*dR2)/(R1+R2)

    then eqn(1) and (2) become

    eqn(6) R+B-W=0
    eqn(7) R*dR+B*dB-WdW=0

    The eqn(6) and (7) are same as yours in the previous posting. That means no
    matter how many supporting forces (eg. R1, R2) you would apply, their sum
    (R) should balance the difference of the body weight and buoyancy, and the
    position (dR) at which the sum of all supporting forces is applied is
    determined by the net moment due to the weight and buoyant. For a given body,
    R, W, dB, dW are all fixed, so you can only find one value for R and one
    value for dR to reach a static equilibrium. You can not change dR, moving the
    support will destroy the rotational balance.

    Back to your purpose, if you want to measure the buoyancy and CB for a given
    body, I do not think your approach is very good particularly for the CB. You
    will have to move the support to locate a point at which the rotational
    balance is realized, this point gives dR (or dR1 when dR2 is fixed, or dR2
    when dR1 is fixed). Then the equations can be used to find out CB. The
    problem is, as I understand, this kind static equilibrium may be unstable.
    This increases the difficult of the measurement. If you have any rough idea
    about CB in advance, your method might work well.

    Jianyu Cheng
    ================================================== ==========================


    I enjoyed our conversation the other day, but I haven't gotten back to
    you because, until now, I haven't had time to look at the problem. Now
    that I do look at the problem it isn't clear to me that there is any
    structural issue with the model that's being used for analysis. What
    does appear, however, is a clear skewing of the numerical results whenever
    the weights are moved assymetrically. Could this be an experimental
    measurement error rather than a conceptual one? It's interesting to look
    at the data from the "CB Calculations" page of the Excel spreadsheet you
    gave me. I plotted "d" versus "R Position." The first third is flat
    (consistent COB with symmetrical weights) while the second and third
    thirds trend up and down (with assymetrical loadings). If, however, the
    "CB tare" values for "LC1" and "LC2" are increased by 870 (grams?) and
    890 grams respectively (to 5023 and 2864) the values which compute for
    "d" and virually constant for ALL positions of the weights!!!

    Is it possible that cables or brackets or clamps or something else
    were not accounted for? Might bouyancy of some of the apparatus be
    accountable? Not having seen the equipment, it's hard for me to make any
    assessment, but its a REMARKABLE situation that corrections of about
    880 in both tare weights totally remove the anomolous results.

    I seems worth thinking about.

    Hans Mattes
    (201) 386-3266
    ================================================== ==========================

    Scott, Hi...
    Greetings to you all. Due to e-mail problems I missed your first problem and
    the responses to it, so I might be repeating what someone else has already
    thought. But, here it goes:
    a) If the pipe rotates when submerged, then you need to apply a couple
    (torque), that is R1 and R2 must be equal.
    b) Now, the pipe will not rotate, but it may sink or float. In those cases
    you should apply a force, let's say RSTAR, through the point around which you
    perform your torque equilibrium (ie. CM). This RSTAR should always be used
    (even when you add the couple mentioned above for to prevent rotation!) and
    is the one that gives you the force equilibrium (B=W+RSTAR). Note R1 and
    R2 always cancel at the force equilibrium.
    For simplicity, you may want to apply R1 also at the CM. But, If you vary R2
    without varying R1 you do not have force equilibrium. Your system is either
    coming out of the water (thus B has changed) or is unbalanced (thus moving
    up or down).
    Well that is what I could come up with from here! I hope it helps. Good luck
    and thanks for the problem...