Dr. Mohamed Tarek Shamaa

BUOYANCY AND FLOTATION

The Archimedes' principle states, "Whenever a bodyis immersed wholly or partially in a fluid, it isbuoyed up (i.e., lifted up) by a force equal to theweight of fluid displaced by the body (the force ofbuoyancy or simply buoyancy). Or in other words,whenever a body is immersed wholly or partially ina fluid, the resultant force acting on it, is equal to thedifference between the upward pressure of the fluidon its bottom, and the downward force due togravity.

A stone weigh 105Ib in air. When submerged in water, it weight 67.0 Ib. Find the volume and specific gravity of the stone.

Solution

Buoyant force Fb = weight of water displaced by stone = Win air - Win

water

Fb = 105 – 67 = 38 Ib

weight of water displaced = 38 = water V = 62.4 V

V = V stone = 0.609 ft3

W in air = stone Vstone = (SG) water Vstone = (SG) Fb ,

SG = weight of stone in air / weight of equal volume of water

SG = 105 / [(0.609)(62.4)] = 2.76

Archimedes, when asked by King Hiero if the new crown was pure gold (SG = 19.3), found the crown weight in air to be 11.8 N and in water to be 10.9 N. Was it gold?

Solution: The buoyancy is the difference between air weight and underwater weight:

Fb = Wc air-Wc water = 11.8 - 10.9 = 0.9 N = water Vcrown

But also Wair = crown Vcrown = (SG) water Vcrown = (SG) Fb ,

Wc water = Wc air- Fb = (SG) Fb - Fb

so Wc in water= Fb(SG-1)

Solve for SGcrown = 1+ Wc in water /Fb = 1 + 10.9/0.9 = 13.1

(not pure gold)

The centre of buoyancy is the point, through

which the force of buoyancy is supposed to act. It

is always the centre of gravity of the volume of

the liquid displaced. In other words, the centre of

buoyancy is the centre of area of the immersed

section.

A completely submerged body, such as a submarine, will remain in

equilibrium at rest only, if (i) the buoyant force is equal to the

weight W and (ii) the centre of buoyancy (B) lies directly above

the center of gravity(G). Submarines are required to use

balancing tanks, to makes FB = W and trimming tanks to brink the

centre of buoyancy above the centre of gravity.

if the center of gravity of the completely submerged body is above

the center of buoyancy, the resulting couple formed by the weight and

the buoyant force will cause the body to overturn and move to a new

equilibrium position. Thus, a completely submerged body with its

center of gravity above its center of buoyancy is in an unstable

equilibrium position.

Whenever a body, floating in a liquid, is given a small angulardisplacement, it starts oscillating about some point. This point,about which the body starts oscillating, is called metacentre.

the distance between the centre of

gravity G and the metacentre is

known as the metacentric height. This

height will be positive when M lies

above G, and represents the condition

for stable equilibrium of the vessel.

If, however, the metacentre M lies

below G (in between B and G), the

metacentric height is negative

signifying unstable equilibrium.

The metacentric height will be positive when M lies above G, and

represents the condition for stable equilibrium of the vessel. If,

however, the metacenter M lies below G (in between B and G), the

metacentric height is negative signifying unstable equilibrium.

DETERMINATION OF METACENTRIC HEIGHT

The buoyant force acting through B'

F'b = W + df - df = W = Fb

By moments about B,

F'b × BB' = df × gg

BB' = df × gg / F'b = df × gg/W

BB' = df × gg / ρ g V (1)

Where V is the volume of the displaced

fluid.

df = ρg×(1/2) AA'×(1/2)b×L

AA' = (1/2) b θ and gg =( 2/3) (1/2)b + 2/3(1/2)b = 2/3b

Substitute in (1)

BB' = MB θ = (ρg ×(1/4)b θ × (1/2)b ×L ×(2/3)b)/ ρgV

Or MB = (1/12) Lb3/V = I/V

Where I is the second moment of the plan area of the body at water

level about its longitudinal axis. Hence, the metacentric height:

MG = MB – GB

MG = I/V – GB

MB = I/V = MG + GB

If the centre of gravity G is below B, then the metacentricheight is given by:

MG = I/V + GB

MB = I/V = MG - GB

The distance MB = I/V is known as the metacentric radius

A rectangular barge is 20 m long, 7 m wide and 3 m deep. it has a

draft of 2 m when fully loaded. The C.G. of barge is on the axis of

symmetry at the water surface. Determine the stability condition of

the barge and the metacentric height.

Sol. A floating body remains in stable equilibrium so long as it has a

positive metacentric height i.e.,

MG = MB – GB = I/V - GB

I = L × d3/12 = 20 × (7)3/12

V = L × d × depth of draft = 20 × 7 × 2

BG = 2 – 1 = 1 m

Making substitutions, the metacentric height is determined as:

MG = [20 × (7)3/12] / (20 × 7 × 2 ) - 1 =1.0417 m

Under the given conditions, the barge is stable and its metacentric

height is 1.0417

Example

A rectangular pontoon is 7m long and 3m wide the weight of the

pontoon is 30 ton. Determine the position of center of gravity

above the base of the pontoon such that it does not overturn in

seawater take the specific weight of water as 0.98 ton/ m3.

Example

Sol. Wtotal= Wdis = Fb = water V dis

30 = 7*3*h*0.98

h=1.45m

I = 7× 33 /12 = 15.75m4

V= 7 × 3 × 1.45 =

MB = I/V = 15.75/30.45 = 0.517m

OM = OB + MB

OB = h/2 = 1.45/2 = 0.725m

OM = 0.725 + 0.517 = 1.242m

OG = OM = 1.242m (Neutral case)

0.725 OG 1.242

O

Determine the volume and density of an object that weight 4 N in water and 5 N in an alcohol of SG 0.8.

Example

Solution:

Fb = Weight of displaced fluid

[(9790)(V object) = W object - 4

[(0.8)(9790)(V object) = W object - 5

subtracting the second equation from the first gives

9790 V object - 7832 V object = 1 , 1958 V object =1

V object = 0.0005107m3

[(9.79)×(1000) ×(0.0005107) = W object - 4 ,

W object = 9.0 N

= W object / V object = 9.0/0.0005107 = 17623 N/m3

= / g = 17623/9.81= 1796 kg/m3 ,

Solution:

Fb rod = 9790(π/4)(0.04)2(8) = 98.42 N,

W rod = rod V rod = (SG) water V rod = (SG) Fb

W rod - Fb rod = (SG) Fb rod - Fb rod = (SG -1 ) Fb rod

W lead =2(9.81) = 19.62 N,

F b lead = water V lead = (water / lead ) ×w lead = W lead / SGlead =

F b lead = 19.62/11.4 = 1.72 N.

Sum moments about B:

MB = 0 = (SG-1)(98.42)(4 cos30o ) - (19.62- 1.72)(8cos30o ) = 0

Solve for SGrod = 0.636

ExampleThe uniform rod in the figure is hinged at B

and in static equilibrium when 2 kg of lead (SG

= 11.4) are attached at its end. What is the

specific gravity of the rod material?

Example

A cylinder made of wood (SG = 0.6) and circular in cross section is

required to float in oil (SG = 0.9), if d = the diameter of the cylinder

and L = height. Prove that L cannot exceed 0.75d for the cylinder to

floating with its longitudinal axis vertical.

Sol. OG = 0.5 L

OB = 0.5h'

W = Fb = Wdis

(d2/4) ×L×0.6 × water = (d2/4) ×h'×0.9 × water

h'=0.67×L OB=0.5h'=0.34×L

BG = OG – OB = 0.5L - 0.34L = 0.16 L

MB = I/V = [(π/4)×(d/2)4 ]/ [(π/4)×(d)2 ×h'] = 0.0933 × (d2/L)

BG = MB

0.16 L = 0.0933 × (d2/L)

d = 1.3L

L = 0.75d