1. Buoyancy The relationship is → B = ρVg
Any time an object is immersed in a fluid, it will experience an upward force (buoyant force) from the fluid. (you feel lighter in a swimming pool.)
The relationship is → B = ρVg
Where:
B = buoyant force (N)
ρ = density of the fluid (kg/m3)
V = volume of displaced fluid (m3)
g = 9.8 m/s2

2. Archimedes’ Principle
the volume of the displaced fluid is equal to the volume of the submerged portion of the displacing object.
Problem solving applications
Use B as you would any other force in ΣF calculations. (B always is directed up)

3. Examples
Calculate the weight of a rock of mass 32 kg and density 2800 kg/m3 when fully submerged in water of density 1000 kg/m3 ?
GIVEN
mrock = 32 kg (Fg = N)
ρrock = 2800 kg/m3
g = 9.8 m/s2
ρfluid = 1000 kg/m3

4. The weight of the rock in the water is
First the volume of the rock can be determined.
ρ = m / V therefore V = m/ρ
V = (32 kg)  ( 2800 kg/m3) = 1.14 x 10-2 m3
this is also the volume of the displaced fluid (since the rock is fully submerged)
the buoyant force calculated as follows:
B = ρVg
therefore
B = (1000) ( 1.14 x 10-2) (9.8)
B = 112 N
The weight of the rock in the water is
therefore 112 N less than it is in air or N

5. the buoyant force calculated as follows:
A helium balloon of volume 12 x 10-3 m3 has a total mass of 12 grams. Calculate the initial acceleration of the balloon when released in air (density = 1.3 kg/m3).
GIVEN
mballoon = kg (Fg = N)
Vballoon = 12 x 10-3 m3 since the balloon is fully submerged in air, this is also the volume of the displaced air
g = 9.8 m/s2
ρfluid = 1.3 kg/m in this case the fluid the balloon is fully submerged in is air
the buoyant force calculated as follows:
B = ρVg
therefore
B = (1.3) ( 12 x 10-3) (9.8)
B = N

6. A free body diagram for the balloon is shown to the right.
Σ F = ma Fg
B - Fg = ma a
Σ F = B - Fg
= .012 (a)
a = 2.95 m/s2

7. A wooden raft of density 600 kg/m3 is placed in
sea water ( = 1030 kg/m3). If the mass of the
raft is 300 kg, what additional mass can it
hold and remain afloat?
First the volume of the raft can be determined.
ρ = m / V therefore V = m/ρ
V = (300 kg)  ( 600 kg/m3) = 0.50 m3
this is also the volume of the displaced fluid (since the raft can remain afloat when fully submerged)

8. The weight the raft itself = (300)(9.8) = 2940 N
the buoyant force calculated as follows:
B = ρVg
therefore
B = (1030) ( 0.5) (9.8)
B = 5047 N
The weight the raft itself = (300)(9.8) = 2940 N
The additional weight the fully submerged raft can hold is therefore – 2940 = 2107N
The additional mass is ÷ 9.8 = 215 kg

9. A gold ( = 19,300 kg/m3) bar is suspected to have a hollow
center. It has a mass of g in air and an apparent mass of
36.22 g in water. How large is the hole in the center?
First the volume of the bar, if it were solid gold, can be determined.
ρ = m / V therefore V = m/ρ
V = ( kg)  ( 19,300 kg/m3) = 1.98 x 10-6 m3
The buoyant force can be determined as the difference in weight of the bar in air and in water.
B = ( )(9.8) – ( )(9.8) = N
B = ρVg
therefore
0.02 = (1000) (V) (9.8)
V= 2.03 x 10-6 m3

10. Therefore the hole is 5 x 10-8 m3
The calculated volume is bigger than the volume of a solid gold bar by
2.03 x 10-6 m3 – 1.98 x 10-6 m3
Therefore the hole is 5 x 10-8 m3