1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Center of buoyancy location

  1. Dec 2, 2017 #1
    Hello everyone,

    According to Archimedes' principle, a solid body immersed in fluid (gas or liquid) always experiences a net upward buoyant force due to pressure variations occurring within he hosting fluid. Pressure calculations show that the buoyant force magnitude is equal to the magnitude of the weight of the fluid displaced by the solid body regardless of the overall shape of the body itself. I believe Archimedes empirically discovered that impressive and not intuitive result. The body immersed in the host fluid does not have to be a solid and can be a liquid or a gas.

    If the body has uniform density, the buoyancy force acts exactly at the body's center of mass CM. But if the body is NOT uniform in density, the net buoyancy force acts at a different point called center of buoyancy CB. The CB is the center of mass of the displaced fluid (say water). We can calculate the location of CB by knowing the weight and shape of the volume of water displaced by the body.

    Does anyone know where I can find a mathematical derivation of this interesting result, i.e. the fact that the net buoyant force acts at the CM of the displaced fluid? That does not seem to be an intuitive result. The magnitude of the buoyant force only depends on the weight of the displaced water but the point of application of the force depends both on the shape of the displaced volume of water and its weight...

    thanks!
     
  2. jcsd
  3. Dec 2, 2017 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The position of the center of buoyancy is given by a position average similar to the center of mass,
    $$\vec{R}_{BF}=\frac{\int{\vec{r}d(BF)}}{BF}$$
    Here ##BF## stands for "buoyant force". Now ##BF=\rho g V##, where ##\rho## is the density of the fluid and V the displaced volume. The element of buoyant force is
    ##d(BF)=\rho ~g~dV=g dm## where ##dm## is a mass element of the displaced fluid. Thus,
    $$\vec{R}_{BF}=\frac{\int{\vec{r}g~dm}}{\rho ~g~V}=\frac{\int{\vec{r}dm}}{m}$$
    The ratio on the far right is the position of the center of mass of the displaced fluid.
     
  4. Dec 2, 2017 #3
    Thanks kuruman! I think I get it.

    To paraphrase, on the surface of the immersed object acts an infinitesimal pressure which produces an infinitesimal buoyancy force at each infinitesimal surface element of the object. the integral of the infinitesimal buoyancy forces give the net upward buoyancy force. The various infinitesimal forces act on the surface on the immersed body. However, The location of the center of buoyancy CB is calculated using the infinitesimal buoyancy forces produced by each elemental fluid volume dV and not just the fluid touching the surface of the immersed body. I guess that is Archimedes principle at work which tells us that the surface integral of the infinitesimal forces over the surface area of the immersed object gives the same volume integral of the infinitesimal forces over the entire displaced volume...
     
  5. Dec 4, 2017 #4

    tech99

    User Avatar
    Gold Member

    My understanding is that the magnitude of buoyancy and position of CB do not depend on the material inside the object, which is considered as zero density. Having found the CB we then find the CG. The total force acting on the body, both uplift and rotation, can then be found from the position and magnitude of these two. I am sure that is what is done with ship design.
     
  6. Dec 4, 2017 #5

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your understanding is correct. Here is a simple way to think about it. Start with a floating irregular object (figure left). Remove the object and consider the submerged volume (figure right) enclosed by the dotted line. The mass of the water enclosed by this volume floats on water (profound thought) and its center of mass (CM) coincides with its center of buoyancy (CB). The CB in both pictures is at the same position and depends on the shape of the displaced volume of fluid not on the specifics of what's inside this volume. What's inside the volume determines the position of the CM not the CB.

    CenterOfBuoyancy.png
     
  7. Dec 4, 2017 #6
    Thank you kuruman and tech99.

    I see how, in the picture on the right, the water inside the volume enclosed by the dotted lines is floating and supported. The center of mass of that enclosed volume of water is the CB and depends a) on the shape of the volume and b) on the mass contained in that volume. The formula for the CB depends on the density of the material (water in this case). The equation for CB seems to be the average mass position of the water within that displaced volume.

    Could you then please clarify the meaning of "... the magnitude of buoyancy and position of CB do not depend on the material inside the object, which is considered as zero density. ..."?
     
  8. Dec 4, 2017 #7

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I cannot speak for @tech99 and what is meant by "zero density". For the "material inside the object" consider this.
    1. Put a thin massless shell around the dotted line (figure on right) to keep water from flowing back in.
    2. Dump the water that's inside and get a number of masses the sum of which is equal to the water that you took out.
    3. Place the masses such that their CM is at the CB (or CM) of the water you took out.
    You do all this and you will see that, as far as the water outside the volume is concerned, nothing has changed and the new object will float. Yet the density inside the volume is not uniform, but I wouldn't call it zero. Also note that if the CM of the masses is placed vertically below the BF, the object will float in stable equilibrium. Small angular displacements will produce a restoring torque and small vertical displacements will produce a restoring force.
     
  9. Dec 4, 2017 #8

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    "Water" is not the "material inside the object".
     
  10. Dec 6, 2017 at 3:18 PM #9
    Thanks, I am making some progress:

    The volume of water that previously existed where we now have the submerged volume of the object had a certain weight that was exactly supported by the pressure forces of the water around it. That implies that the net upward buoyancy force (surface integral of the pressure distribution on the submerged volume surface) has to be equal to the weight of the water displaced. The displaced volume of water, before the object was submerged, was in equilibrium (no net force or net torque): its weight and the buoyancy force must act at points along the same vertical line. If the immersed object is homogeneous, the center of mass of the object and the CB are the same point. But, why, conceptually, does an object having non-homogeneous density have a CB that is not coincident wit the object's center of mass CM?

    Is it correct to consider the center of buoyancy CB as a point that pertains to the immersed object and not to the displaced fluid?

    Thanks!
     
  11. Dec 8, 2017 at 3:20 AM #10

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    There is no need for the water to have ever existed. You can float an object in far less water than the submerged volume of the object:

    battleship.jpg


    Because the CM does depend on the density distribution of the object, while the CB doesn't.
     
  12. Dec 9, 2017 at 7:37 AM #11
    Thanks A.T.

    1) That is really cool. I follow your diagram and it makes sense. I am still amazed though. Is that really possible for a ship to truly float on that little water or is there some real life approximation that does not allow that? This situation doesn't seem to fully connect with Archimedes principle teachings. According to Archimedes, the buoyancy force (which must be equal to the ship's weight for floating to occur) is equal to the weight of the displaced fluid. But in this situation (gallon of water problem) we are not really displacing a volume of water having weight equal to the weight of the ship....Somehow, the surface integral of the pressure exerted by those few gallons of water produces a force equal to the weight of the ship and this happens only because that little water is distributed into this deep shaped hole with the same shape as the ship...

    2) I see how the CM is a fixed point of the object (only determined by the mass distribution of the object) while the CB is the CM of displaced volume of fluid. The fluid is incompressible but can change shape hence the location of the CB changes too;
     
  13. Dec 9, 2017 at 8:11 AM #12

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Real water is not a continuum, but consists of molecules. You cannot make the water film arbitrarily thin. But the point here is not whether it's 1 gallon or 1 million gallons. Both is less than the displacement of a battle ship.

    displaced fluid = volume of the object below the fluid surface

    No actual fluid needs to have been in that volume.
     
    Last edited: Dec 9, 2017 at 8:37 AM
  14. Dec 9, 2017 at 10:51 AM #13
    I see how "No actual fluid needs to have been in that volume".

    But how is the pressure distribution able to provide sufficient upward net force to support the ship? In the more common case of a ship in the ocean, the ship moves (displaces) a volume of fluid so that its surface touches water at a certain depth that has a larger pressure than the water at a shallower depth. The water at a higher depth has a larger hydrostatic pressure because of the amount of water above it.

    In the problem we are discussing, things are slightly different. But I think it works this way from the picture below: regardless of the amount of water, the pressure at the bottom of that thin and curved layer of water around the ships' hull has a sufficient pressure for Stevin principle. The pressure only depends on depth and not on the shape and size of the container.

    upload_2017-12-9_11-50-37.png


     
  15. Dec 9, 2017 at 11:07 AM #14

    jbriggs444

    User Avatar
    Science Advisor

    Yes. The pressure on the ship's surface is the same regardless of whether the water layer outside is thin or thick.
     
  16. Dec 9, 2017 at 9:52 PM #15

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Or put it another way: The pressure in that thin layer at a wall doesn't depend on whether the rest of the volume is filled with water or a ship. The thin layer exerts the same pressures and thus total buoyancy on both.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Center of buoyancy location
  1. Fluids - buoyancy (Replies: 4)

  2. Buoyancy and density (Replies: 1)

  3. Buoyancy conservation (Replies: 3)

Loading...