---
title: "Stability of conical buoy with light"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:conical-buoy-stability"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/conical-buoy-stability
site: BCT Engineering
---

# Stability of conical buoy with light

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2078 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/conical-buoy-stability
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

A buoy, floating in sea-water of density 1025 kg/m³ is conical in shape with a diameter across the top of 1.2 m and a vertex angle of 60°, its mass is 300 kg and its centre of gravity is 750 mm from the vertex. A flashing guiding light is to be fitted to the top of the buoy. If this unit is of mass 55 kg, what is the maximum height of its centre of gravity above the top of the buoy if the whole assembly is not be unstable? (The centroid of a cone of height h is at 3h/4 from the vertex.)

## Short answer

Draft d =0.997 m gives KB =0.748 m, BM =0.249 m, KM =0.997 m. Combined KG from the vertex must not exceed KM, so the light unit's CG may be at most1.31 m above the buoy top.

## Answer

**Data:** sea-water $\rho=1025\ \text{kg/m}^3$; cone vertex down, top diameter 1.2 m, vertex angle $60^\circ$ (half-angle $30^\circ$); buoy mass 300 kg, $G_1$ at 0.75 m from the vertex; light mass 55 kg at height $y$ above the top.

```text
      light at y
      ____
     \    /   top dia 1.2 m
      \  /
 water \/  vertex (down)
```

**Floating depth $d$** (vertex to waterline). Total mass $M=355$ kg:

$$
V=\frac{355}{1025}=0.3463\ \text{m}^3=\frac13\pi\,(d\tan30^\circ)^{2}d
$$

$$
d^{3}=\frac{3V}{\pi\tan^{2}30^\circ}=0.9922\ \Rightarrow\ d=0.9974\ \text{m}
$$

(The buoy height is $0.6/\tan30^\circ=1.0392$ m, so it floats with the top just above water.)

**Waterline radius:** $r_w=d\tan30^\circ=0.5758$ m.

**Centre of buoyancy:** $KB=\tfrac34d=0.7480$ m from the vertex.

$$
I=\frac{\pi r_w^{4}}{4}=0.0863\ \text{m}^4,\qquad BM=\frac{I}{V}=0.2493\ \text{m}
$$

$$
KM=KB+BM=0.7480+0.2493=0.9974\ \text{m}
$$

**Limit of stability:** $KG\le KM$. Combined centre of gravity from the vertex:

$$
KG=\frac{300(0.75)+55(1.0392+y)}{355}\le0.9974
$$

$$
225+55(1.0392+y)\le354.07\ \Rightarrow\ 1.0392+y\le2.3468
$$

$$
y\le1.308\ \text{m}
$$

**Answer: the centre of gravity of the light unit may be at most about 1.31 m above the top of the buoy** (2.35 m from the vertex) for the assembly not to be unstable.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the floating depth d found?
   A. Displaced volume = total mass 355 kg / 1025 = 0.3463 m^3 equated to cone volume (1/3)π(d tan30°)^2 d
   B. Using only the buoy mass 300 kg
   C. Using the light's mass alone
   D. Volume equals the whole cone volume
   Answer: A. The buoy and light together float, so total mass 355 kg is balanced by buoyancy. (Key point: Floating depth)

2. What is the limit for the assembly not to be unstable?
   A. KG ≥ KM
   B. KG ≤ KM, where KM = KB + BM
   C. KG ≤ KB
   D. BM ≤ BG
   Answer: B. G must not rise above M; at the limit GM = 0. (Key point: Stability limit)

3. How is BM obtained?
   A. BM = V / I
   B. BM = I of the top diameter / V
   C. BM = I / V with I = π r_w^4 / 4 of the waterline circle
   D. BM = 3d/4
   Answer: C. I is the second moment of the waterline area; r_w = d tan30° = 0.5758 m gives BM = 0.249 m. (Key point: Metacentric radius)

4. What is the maximum height of the light's centre of gravity above the top of the buoy?
   A. About 1.31 m
   B. About 2.35 m
   C. About 0.75 m
   D. About 1.04 m
   Answer: A. Solving KG = [300(0.75) + 55(1.0392 + y)]/355 ≤ 0.9974 gives y ≤ 1.308 m (2.35 m from the vertex). (Key point: Result)

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