1. Steve Aycox Kurtis Massey Damilola Koleowo January 29, 1999
Group 6 Buoyancy Force
Steve Aycox
Kurtis Massey
Damilola Koleowo
January 29, 1999

2. Buoyancy is an upward force exerted by any fluid upon
a body placed in it. It is caused by the density difference
between the submerged body and the displaced fluid. d
This is a cube submerged
in water with side lengths
equal to h and a volume of V.
The distance between the
surface of the water and top
of the cube is d. There are a
total of 5 forces acting on
the box.
note: the positive y-axis
points downward.
H2O x h y

3. The first force is the Hydrostatic Pressure -
acting on the top surface: H d H
H2O A x h y
The pressure exerted on the top surface of the box is equal to the
atmospheric pressure plus the pressure of the water above it times
the surface area:
(Po +rgd)A
Where A is the surface area which can also be written as V/h:
(Po +rgd)V/h

4. The second and third forces are the Hydrostatic
Pressure - acting on the side surfaces:
H2O x y
Note that these forces are equal in magnitude at the same heights. They
act on both sides of the box making them opposite in direction. Thus the
hydrostatic pressure on the sides of the box cancels out.

5. The fourth force is the Hydrostatic Pressure -
acting on the bottom surface: H d
H2O h x y H
The pressure exerted on the bottom surface of the box is equal to the
summation of the atmospheric pressure and the pressure of the water
at the bottom level times the surface area:
(Po +rg(d+h))A
Where A is the surface area which can also be written as V/h:
[(Po +rg(d+h))V/h]

6. The fifth force is the Weight of the box:d
H2O h x W y
Weight is the force due to gravity acting on the mass of the box.
W=mg
Note: when density (m/v) is given, weight can be rewritten as:
W= rBgV

7. The Buoyant force is calculated by the summation of the
forces acting on the box.
S F= (rBgV) + ((Po +rWgd)V/h) - [(Po +rWg(d+h))V/h]
Use the distributive property:
S F= rBgV + PoV/h + rWgdV/h - PoV/h - rWgDV/h - rWghV/h
Canceling out like terms:
S F= rBgV - rWghV/h
Which simplifies to:
S F=(rB - rW)gV

8. During WWII, the Japanese launched the first intercontinental
Example Problem:
During WWII, the Japanese launched the first intercontinental
ballistic attack. Hydrogen-filled balloons carrying explosives,
ballast and control devices were released near Tokyo to ride on
the newly discovered jet stream to the United States. Technical
Major Teiji Takeda helped to design these balloons which needed
a lifting capacity of 450 kg at sea level and 60 F. Estimate
the diameter of the spherical balloon.
rair(1 atm, 289 K)= kg/m3
rH2(1 atm, 289 K)= kg/m3 *
Note: 60ºF
289 K *
*Wetty, J.R., Wicks, C.E., and R.E. Wilson. Fundamentals of Momentum Heat and Mass Transfer.
3rd edition: Wiley, pg 766.

9. Answer:
The buoyant force must exceed the weight of the hydrogen
gas sufficiently that the net force upward will be able to lift
the accompanying weight. Mathematically, this says:
mballon(g) = (rair - rH2)gV
Which simplifies to:
V=m/(rair - rH2)
Substituting in for Volume and rearranging to get r:
r=[(3m)/(4p(rair - rH2))]1/3
V=(4/3) pr3

10. Numerically:
r=[(3(450kg))/(4p( ))]1/3
r=18.20m
D = r/2 = 9.10m