1. BUOYANCY AND ARCHIMEDES' PRINCIPLE
Archimedes’ Principle states that a body wholly or partly immersed in a fluid is buoyed up by a force equal to the weight of the fluid it displaces.
An object lowered into a fluid “appears” to lose weight. The force that causes this apparent loss of weight is referred to as the buoyant force. The buoyant force is considered to be acting upward through the center of gravity of the displaced fluid.
FB = mF g = ρF g VF

2. The pressure difference between the top and bottom of this object must support the weight of the fluid that would be there in equilibrium if the object were not there.
In general, objects floats on a fluid if its density is less than that of the fluid. At equilibrium (when floating) the buoyant force on an object has a magnitude equal to the weight of the object.

3. Buoyant Force Due to a Fluid

4. The fraction of the floating object that is submerged is equal to the ratio of the density of the object to that of the fluid.

5. 10. 5 An aluminum object has a mass of 27. 0-kg and a density of 2
10.5 An aluminum object has a mass of 27.0-kg and a density of 2.70x103 kg/m3. The object is attached to a string and immersed in a tank of water.
a. Determine the volume of the object
m = 27 kg
ρAl = 2700 kg/m3
ρwater = 1000 kg/m3
= 0.01 m3

6. FT Vwater = VAl Fgwater = ρwVwg FB = 1000(0.01)(9.8) = 98 N
b. Determine the tension in the string when it is completely immersed. FT
Vwater = VAl
Fgwater = ρwVwg
= 1000(0.01)(9.8)
= 98 N
FB = Fgwater FB Fg
ΣF = FT + FB - Fg = 0
FT = Fg - FB
= 27(9.8) - 98
= N

7. FB = Fg - Fgapparent = 0.086(9.8) - (0.073)(9.8) = 0.1274 N FB = ρwVwg
10.6 A piece of alloy has a mass of 86 g in air and 73 g when immersed in water. Find its volume and its density.
mair = kg
mwater = kg
ρwater = 1000 kg/m3
FB = Fg - Fgapparent
= 0.086(9.8) - (0.073)(9.8)
= N
FB = ρwVwg
= 1.3x10-5 m3
= 6.6x103 kg/m3

8. FB = Fg - Fgapparent = (0.067-0.045)(9.8) = 0.2156 N = 2.48x10-5 m3
10.7 A solid aluminum cylinder with density of 2700 kg/m3 has a measured mass of 67 g in air and 45 g when immersed in turpentine. Find the density of turpentine.
FB = Fg - Fgapparent
= ( )(9.8)
= N
ρAl = 2700 kg/m3
mair = kg
mwater = kg
= 2.48x10-5 m3
= kg/m3