Questions: Floating Body Stability and Equilibrium

Question 1

A narrow cylindrical buoy floats vertically with its center of buoyancy (B) located below its center of gravity (G). What can be correctly concluded about its stability?

• A. It is definitely unstable — buoyancy must act above gravity for a floating body to be stable
• B. It may still be stable if the metacenter (M) is located above the center of gravity (G)
• C. It is definitely stable — all floating objects in equilibrium are stable by definition
• D. It is neutrally stable — B and G being on the same vertical line means the body is in balance

Answer: (B)

The most common misconception is that B must be above G for stability. This is only true for fully submerged bodies. For floating bodies, B can be below G and the body can still be stable — as long as the metacenter M lies above G. When the body tilts, B shifts toward the submerged side, and the line of action of the buoyant force rises to intersect the original centerline at M. If M is above G, this creates a restoring moment. Narrow cylinders often have B below G; their stability depends on whether BM (= I/V) is large enough to raise M above G.

Question 2

A ship designer wants to increase a vessel's metacentric height (GM) to improve stability. Which modification most directly achieves this?

• A. Raising heavy machinery from the keel to the upper deck to elevate the center of gravity
• B. Widening the hull at the waterline to increase the second moment of the waterplane area
• C. Replacing dense steel plating with lighter aluminum to reduce total displacement
• D. Increasing the vessel's draft by adding ballast, which lowers the center of buoyancy

Answer: (B)

GM = KB + BM − KG. BM = I/V, where I is the second moment of the waterplane area about the longitudinal axis and V is displaced volume. Widening the hull dramatically increases I (which scales as the cube of half-width), raising BM and therefore M. This is the dominant geometric effect in ship stability — flat-bottomed barges are stable precisely because their wide waterplane creates a large I. Option A raises G, which decreases GM. Options C and D have secondary effects but do not directly target the largest term in BM.

Question 3

For a floating body, achieving stable equilibrium requires that the metacenter (M) lie above the center of gravity (G), not merely that the center of buoyancy (B) lie above G.

• T. True
• F. False

Answer: True

This is the core principle. For fully submerged bodies (submarines, balloons), B must be above G — there is no shift of B when the body tilts, so no metacenter. For floating bodies, B shifts when the body tilts, and the metacenter M defines where the resulting buoyant force acts. Stability requires M above G (positive metacentric height GM > 0), regardless of the relative positions of B and G in equilibrium. A body with B below G can be perfectly stable if BM is large enough to raise M above G.

Question 4

A floating body that is in equilibrium — meaning the buoyant force exactly equals its weight — is necessarily in stable equilibrium.

• T. True
• F. False

Answer: False

Equilibrium and stability are distinct conditions. Equilibrium simply means net force and net moment are zero. Stability asks what happens when the body is perturbed. A pencil balanced vertically on its tip is in equilibrium but unstable. A floating body is in unstable equilibrium when the metacenter M is below G (negative GM): any tilt creates a capsizing moment that increases the tilt rather than restoring it. Equilibrium is necessary but not sufficient for stability.

Question 5

Explain why a wide flat-bottomed barge is more stable than a narrow upright log of the same weight, using the concept of how the center of buoyancy moves when each body is tilted.

Model answer: When the barge tilts, its wide bottom means the submerged volume shifts dramatically toward the lower side — the center of buoyancy (B) moves far in the direction of tilt. The line of action of the buoyant force then intersects the original vertical axis at a metacenter (M) high above the center of gravity (G), creating a strong restoring couple that rights the barge. The narrow log, when tilted, shifts very little submerged volume — B moves only slightly, placing M close to B and possibly below G. If G is also elevated (e.g., dense wood floating mostly above water), GM becomes negative and the log is unstable. The key is that wide waterplane area shifts B far on tilt, raising M.

This is captured in the formula BM = I/V, where I is the second moment of the waterplane area. For a rectangle of width w, I scales as w³. Doubling the width multiplies BM by 8, raising M substantially. Ship designers exploit this by preferring wide, low hull forms, and use ballast to lower G when wide hulls aren't possible (e.g., sailing yacht keels).