I don't understand a solution to a homework question. Specifically, I don't understand the logical sequence of each bolded statement in the answer.

(b) To ensure that a ship is in stable equilibrium, would it be better if its center of buoyancy was above, below, or at the same point as, its center of gravity? Explain.


(b) From the diagram, if the center of buoyancy (the point where the buoyancy force acts) is above the center of gravity (the point where gravity acts) of the entire ship, when the ship tilts, the net torque about the center of mass will tend to reduce the tilt. If the center of buoyancy is below the center of gravity of the entire ship, when the ship tilts, the net torque about the center of mass will tend to increase the tilt. Stability is achieved when the center of buoyancy is above the center of gravity.

Although I don't understand the effects of torque on a 3D structure, I understand that torque gets created when forces act at different points (with the exception of at the center of mass) along any structure. I also understand that the setup contributing to the least torque is the more stable one. I don't understand why one net torque would be stronger than the other.

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Is it that $Fb$ and $Fw$ are, for the most part, close together when $Fb$ acts above $Fw$?

  • $\begingroup$ Look at the radial distance of the forces from the HINGE point(not just ANY point) in either case, i.e. the point about which toppling is to occur. (Torque is directly proportional to radial distance) That should answer your question. $\endgroup$ – GRrocks Mar 20 '16 at 12:11

Although "center of gravity below center of buoyancy" may be the right answer for this question it's not the right answer in practical naval architecture!

In practice, naval architects will purposely design a surface vessel such that the center of buoyancy is actually below the center of gravity. This indeed will cause the ship to roll in one direction or another, but the roll angle is limited by another opposing force. As the ship rolls, the "water plane" that projects through the vessel increases and this shifts the center of buoyancy such that it is raised above the center of gravity - which never changes position. So the center of gravity, and what's known as the 'righting arm' from a changing center of buoyancy lead to a very slow harmonic motion in the ship's roll motions. Drag forces between the ship's surface and water tend to dampen this motion.

Why do the designers do this? By having the center of gravity above the center of buoyancy the vessel becomes easier to maneuver in terms of changing the ship's heading. If the ship cannot roll easy it tends to resist changes in heading. A ship with its center of gravity below the center of buoyancy is called 'bottom heavy', and some ships are designed in this way such as sail boats. Sail boats typically have a long keel that lowers the center of gravity well below the center of buoyancy to balance out forces from the wind onto the sail's and masts.

Stability is usually not measured by the center of buoyancy and center of gravity directly, but rather indirectly by a measure derived from the two and also taking into account the effects of water plane: the Metacentric Height . Metacentric height can be numerically derived from a ship's data, but more accurately determined empirically using the inclining experiment .

But getting back to your specific, original question. To understand how the center of gravity behaves relative to center of buoyancy imagine a pendulum constructed from a rotating joint, and a stick connecting the rotating joint to a weight at the other end of the stick - the pendulum 'bob'. The weight is the ship's center of gravity, and the pivot point is the ship's center of buoyancy.

If you start with the bob directly over the center of rotation (center of gravity over center of buoyancy) and just give it a small nudge, the pendulum will move away from the nudge and go through a large swing so that the bob eventually comes to rest below the center of rotation.

If however you start with the bob directly below the center of rotation (center of gravity below center of buoyancy) and then give it a small nudge you will feel the pendulum push back towards the nudge, attempting to maintain it's lower position.

The first case - the 'inverted' pendulum is 'unstable'. The latter case, the pendulum as we normally see it, is 'stable'.

That's ship stability in a nutshell.

  • $\begingroup$ @user21945 Note I've used 'force' in my description, but as Gummy bears pointed out in describing the roll motion of ships we are actually dealing with torques which involve forces and moment arms. So you can generally replace 'force' with 'torque' throughout my answer. $\endgroup$ – docscience Jan 17 '15 at 18:53

Well to understand this solution, I think it is important to know the basics about torque. Torque is the analogue of force in rotation. Force causes motion in linear mechanics, while torque causes rotation in rotational mechanics.

As we are considering rotation, it is important to understand that we take everything about a certain point. We need to consider one point as the reference point. For our convenience, that is usually the center of mass/gravity of the body.

Now coming to your question: It is not that one net torque is stronger than the other. Rather, if the distance of center of buoyancy from the center of mass is the same, the net torque would be the same as well. Instead, it is that the torques are acting in different directions.

The torque will always act towards the reference point. We take torque to be either clockwise or counterclockwise.

If the center of buoyancy is below the center of mass, when the object tilts (as your diagram shows) the center of buoyancy is less displaced than the center of mass. [Consider a circular ride, the closer to the center you stand, the less dizzy you feel. This is because if you are closer to the center, your speed is lower, as is your angular displacement.] Thus, as said above, the torque will act in the direction of the center of mass, which happens to be in the direction of the tilt in case of the center of buoyancy being lower. Thus this causes the body to tilt further, decreasing stability.

When the center of buoyancy is above the center of mass, it's the opposite. Center of buoyancy gets displaced more than center of mass, and the direction in which the torque acts happens to be opposite to that of the tilt, thus increasing stability.

  • $\begingroup$ "If the center of buoyancy is below the center of mass, when the object tilts (as your diagram shows) the center of buoyancy is less displaced than the center of mass." How can that be? If a system rotates, the components of the system (in this case, points), rotate to the same extent. If they rotate, with respect to each other, to same extent, how can displacement take place? If I've misunderstood your reasoning, could you give me another example or refine the merry-go round one: 1) closer you stand to mgoround center, less dizzy = I understand. 2) ang displacement varies? how? $\endgroup$ – Muno Jan 18 '15 at 14:30

In the left hand case the twisting effect is clockwise, tending to make the ship more upright: in the right hand case it is counter-clockwise tending to tip it over.

  • $\begingroup$ That's just coincidental, isn't it? The diagram for $Fb$ above $Fw$ could easily have been redrawn with $Fb$ acting towards the right of the center of mass. $\endgroup$ – Muno Jan 18 '15 at 14:29
  • $\begingroup$ In that situation, user21945, the twisting effects would also be reversed, still tending to make the ship more upright on the left, and capsizing on the right. $\endgroup$ – Harry Weston Jan 19 '15 at 15:08

The diagrams at the beginning of this post show how a fully submerged vessel maintains stability, by keeping the CG below the CB. As alluded to in a previous comment, this is impractical for a vessel which is floating.


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