BUOYANCY
ARCHIMEDES’ PRINCIPLE
Archimedes’ Principle is how we
put a number to the buoyant
force. It says that the buoyant
force on a submerged object is
equal to the weight of the liquid
displaced by that object.
By the principle of hydrostatic
𝐹1 = 𝑃1A
; 𝑃1 = ϒℎ1
= ϒℎ1(2)(2)
𝐹2 = 𝑃2A
; 𝑃2 = ϒℎ2
= ϒℎ2(2)(2)
If 𝐹2 > 𝐹1, Force is Unbalance
Therefore,
BF = ϒ V
where, weight (w) = Unit weight (ϒ) x Volume (V)
So,
BF = 𝑤𝑏𝑜𝑑𝑦
BF = ϒ𝐿𝑉𝐷
Where: BF – Buoyant Force
ϒ𝐿 - Unit weight of liquid
𝑉𝐷 – Volume displaced
For homogeneous solid of volume, V, “Floating” in a
homogeneous fluid at rest:
For a body of a height, H, with a constant horizontal
cross-sectional area:
𝑠𝑝. 𝑔𝑟. 𝑜𝑓 𝑏𝑜𝑑𝑦
𝐷=
(𝐻)
𝑠𝑝. 𝑔𝑟. 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑
𝛾𝑏𝑜𝑑𝑦
𝐷=
(𝐻)
𝛾𝑙𝑖𝑞𝑢𝑖𝑑
For a body of uniform cross-sectional area, the
area submerged, 𝐴𝑠 , is:
Where:
𝐴𝑠 is the submerged cross-sectional area of the
body
A is the area of the body
Procedure for solving buoyancy problem
1. Determine the objective of the problem solution. What
needs to be calculated- force, weight, volume, specific
weight.
2. Draw the free-body diagram of the object in the fluid.
Show all forces that act on the free body in the vertical
direction, including the weight of the body, the buoyant
force, and all external forces. If the direction of some is
unknown, assume the most probable and show it in the
free body.
3. Write the equation of static equilibrium in the vertical
directions, assuming the positive direction to be upward.
4. Solve for the desired force, weight, volume,
specific weight. Remember the following concept:
The buoyant force is calculated from BF = ϒL VD
The weight of a solid object is the product of its total
volume and its specific weight, W = ϒ V
An object with an average specific gravity less than that of
the fluid will tend to float because W < BF with the object
submerged.
An object with an average specific weight greater than that
of the fluid will tend to sink because W > BF with the object
submerged.
Neutral buoyancy occurs when a body stays in a given
position wherever it is submerged in a fluid. An object whose
average specific weight is equal to that of the fluid is
neutrally buoyant.
Example:
1. A piece of metal weighs 350 N in air and when it
is submerged completely in water, it weighs 240 N.
a. Find the volume of the metal.
b. Find the specific weight of the metal.
c. Find the specific gravity of the metal.
2. An iceberg having a specific gravity of 0.92 is
floating on salt water of sp. gr. 1.03. If the volume
of iceberg above the water surface is 1000 m³,
3. A container holds two layers of different liquids, one
fluid having a specific gravity of 1.2 is 200 mm deep and
the other fluid having a specific gravity of 1.50 is 250 mm
deep. A solid spherical metal having a diameter of 225
mm and a sp.gr. of 7.4 is submerged such a manner that
half of the sphere is on the top layer and the other half in
the bottom layer of fluids.
a. Compute the weight of the spherical metal.
b.Compute the buoyant force acting on the object.
c. Compute the tension in the wire holding the sphere to
maintain its position.
4. Two spheres each 1.20 m in diameter are
connected by means of a short rope. One weighs 4
KN and the other weighs 12 KN. When placed in
water
a.Compute the tension in the rope
b.Compute the depth of floatation of the 4 KN
sphere
c.Compute the volume of sphere exposed above the
water surface
5. A sphere of radius of 400 mm is immersed in
seawater (sp.gr. 1.026) by anchoring it to the
bottom of the seabed. The mooring line was
observed to have a tension of 800 N. Evaluate
the specific weight of the sphere in KN/m3.
6. A cube of wood 600 mm side floats in fresh
water with two faces horizontal other four faces
vertical. If its specific gravity is 0.80, obtain the
submerged depth of the wood in meters.
7. A prismatic object 200 mm thick by 200 mm wide by 400 mm
long is weighed in water at a depth of 500 mm and found to be 50
N.
a) Find its weight in air.
b) Find its specific gravity.
c) Find its specific weight.
8. A wooden buoy of sp. gr. 0.75 floats in a liquid with sp. gr. of
0.85.
a) What is the percentage of the volume above the
b) If the volume above the liquid surface is 0.0145 m^3, what is the
weight of the wooden buoy?
c) What load that will cause the buoy to be fully submerged?