This document presents information about buoyancy and Archimedes' principle. It discusses how a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the body. The buoyant force acts through the center of buoyancy, which is the centroid of the submerged volume. For an object floating in two immiscible fluids, there are two buoyant forces acting through separate centers of buoyancy. An example calculation is given to determine the volume of water displaced and center of buoyancy for a wooden block floating horizontally in water.

1. FLUID MECHANICS
PRESENTATION
ON
BUOYANCY
PREPARED BY:
HARSH BHANDARI(150450119051) s

2. BUOYANCY
When a body is immersed in a fluid, there is tendency of a fluid to exert force on the
submerged body. This upward force is same as weight of fluid displaced by the body is known
as force of buoyancy.
Archimedes principal :
“Whenever a body is immersed wholly or partially in a fluid it is lifted up by a force
equal to the weight of fluid displaced by the body.”

3. From above figure, The body subjected following two forces.
(1) Downward gravitational force(W)
(2) Upward force of liquid(FB )
The upward force of liquid is known as force of buoyancy(FB ) and is equal to weight of
water displaced by body.
FB = weight of water displaced by body Where, h= depth of body immersed in liquid
= ρg × volume of water displaced by body b= width of body
= ρg × (h × b × l) l= length of body
G= Centre of gravity of body
B= Centre of buoyancy

4. Centre of Buoyancy
• It is point on body where the force of buoyancy acts. It is always the centre of gravity of
displaced liquid.
• If a body is immersed such a way that part of its volume V1 is immersed in a fluid of density
ρ1 and the rest of its volume V2 in another immiscible fluid of mass density ρ2 as shown in
figure

5. The force of buoyancy on body due to fluid 1
FB1
= ρ1 g V1 , acting through B1 centroid of volume V1
The force of buoyancy on body due to fluid 2
FB2
= ρ2 g V2 , acting through B2 centroid of volume V2
Total upward force of buoyancy F = F1 + F2 = ρ1 g V1 + ρ2 g V2
For equilibrium
Upward force of buoyancy = Weight of body
∴ ρ1 g V1 + ρ2 g V2 = ρgV
∴ ρ1V1 + ρ2V2 = ρV

6. Example:
Calculate the volume of water displaced and position of centre of buoyancy for a
wooden block (density 680 kg/m3) of width 3m and depth 2m, when it floats horizontally in
water. The length of wooden block is 5m.
Given data:
ρbody = 680 kg/m3
ρw = 1000 kg/m3
H= 2m
b= 3m
l= 5m
Weight of water displaced = Weight of wooden block
ρw × g × volume of water displaced = ρbody × g × volume of wooden block

7. ∴ 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 =
𝜌 𝑏𝑜𝑑𝑦 × 𝑔 × (𝐻 × 𝑏 × 𝑙)
𝜌 𝑤 × 𝑔
=
𝜌 𝑏𝑜𝑑𝑦(𝐻×𝑏×𝑙)
𝜌 𝑤
=
680 2×3×5
1000
= 𝟐𝟎. 𝟒𝒎 𝟑
Now, volume of water displaced = volume of wooden block in water
∴ 20.4 m3 = (h × b × l)
∴ 20.4 = h × 3 × 5
∴ h = 1.36m (depth of block is in water)
∴Position of centre of buoyancy = h/2 = 1.36/2 = 0.68m