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Mechanics and a bit of fluids (mech).

  1. Nov 24, 2014 #1
    Hi,
    This problem has been driving me mad!
    Can anyone simplify the physics of this problem? Because I can't solve due to to many unknowns.
    Question ref: Engineering Materials, Benham, Crawford & Armstrong.
    Please refer to attachments.
    1.9.jpg 1.9 fb.jpg
     
  2. jcsd
  3. Nov 25, 2014 #2

    SteamKing

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    First of all, you have described way too many forces acting on this beam.
    You've accounted for the weight of the beam and the forces in the pin at A, both of which are good.

    It's where this beam goes into the water that your analysis goes off the rails.

    If the water is pressing down on the beam from above, as denoted by Fwa, then the water will also be pressing up on the beam from below, which I presume is represented by FB. The location where these two forces act seems a bit sketchy.

    But a shrewd man known as Archimedes already encountered something like this long ago. It seems that you have overlooked how to apply his principle to this problem.
     
  4. Nov 26, 2014 #3
    Thanks for your post.

    Yes, buoyancy force is equal to weight of displaced fluid, thanks to Archimedes. But, taking a second to think about introducing a Fwa term. This term is necessary because the weight of the water (ignoring air pressure) above the beam will add to the moment acting on the beam, while the force due to buoyancy pushes up (the beam) not the water under the beam. Yes, I did some revision and the direction and position of the buoyancy force was incorrect.

    A way around this (refer to attached), could we employ a moment of area method to calculate the centroid distance? (Ignoring effects due Fwa).

    1.9 FB.jpg
     
    Last edited: Nov 26, 2014
  5. Nov 26, 2014 #4

    SteamKing

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    You're getting warmer, but I think you can solve the original problem using the equations of static equilibrium alone.

    IOW, you don't need to know anything about metacenters and whatnot. You just need to identify the forces acting on the plank and where their centers of action are located w.r.t. the pin at A.
     
  6. Nov 26, 2014 #5
    I'd like to note. 1.9 Fwa.jpg
     
  7. Nov 26, 2014 #6

    SteamKing

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    Remember, you are furnished the dimensions of the plank and its specific gravity. Fiddling around with pressures and whatnot is peripheral to solving this problem. After all, if you apply Archimedes principle to the submerged portion of the plank, you can calculate the buoyant force in a more direct manner, as well as determine its center of application.
     
  8. Nov 26, 2014 #7
    p { margin-bottom: 0.25cm; line-height: 120%; }
    I'm trying to simplify the problem, otherwise I have to take into account separate vertical & horizontal forces.


    So what your saying is, simply let the buoyancy force act through centre gravity of the volume which is occupied by the portion wood which is submerged?
     
    Last edited: Nov 26, 2014
  9. Nov 26, 2014 #8

    SteamKing

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    Exactly. The buoyant force acting on the submerged portion of the plank must equal the net of the hydrostatic pressure acting over the surface of the plank.
     
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