Buoyant Force
Contents:
•How to calculate
•Whiteboards
Buoyant force is the weight of displaced water:
B   fVf g
ρf = Density of fluid in kg m-3
Vf = Volume of displace fluid in m3
g = 9.81 N kg-1
Derive
Demo: Float not float/Cartesian Diver/hot air balloons/hydrometer
Buoyant force is the weight of displaced water:
B   fVf g
ρf = Density of fluid in kg m-3
Vf = Volume of displace fluid in m3
g = 9.81 N kg-1
What is the buoyant force on a 3.0 cm diameter air bubble
under water? ρH2O = 1.0E3 kg m-3
0.14 N
B   fVf g
ρf = Density of fluid in kg m-3
Vf = Volume of displace fluid in m3
g = 9.81 N kg-1
What is the buoyant force on a 5.45 kg iron shot submerged in
water? What is the weight of the shot in air, and what is its
apparent weight submerged?
ρFe = 7.8E3 kg m-3, ρH2O = 1.0E3 kg m-3, ρ = m/V so V = m/ρ
6.85 N, 53.5 N, 46.6 N
B   fVf g
ρf = Density of fluid in kg m-3
Vf = Volume of displace fluid in m3
g = 9.81 N kg-1
The King’s crown has a mass of 14.7 kg, but appears to have a
mass of only 13.4 kg when weighed when it is submerged in
water. What is the density of the crown? Is it gold?
ρAu = 19.3E3 kg m-3, ρH2O = 1.0E3 kg m-3, ρ = m/V so V = m/ρ
11.3x103 kg m-3, so no
B   fVf g
ρf = Density of fluid in kg m-3
Vf = Volume of displace fluid in m3
g = 9.81 N kg-1
The King’s crown has a mass of 14.7 kg, but appears to have a
mass of only 13.4 kg when weighed when it is submerged in
water. What is the density of the crown? Is it gold?
ρAu = 19.3E3 kg m-3, ρH2O = 1.0E3 kg m-3, ρ = m/V so V = m/ρ
???????
11.3x103 kg m-3, so no
Buoyant Forces
1|2|3
What is the buoyant force on a rectangular block of wood that
measures 12x23x15 cm if it is submerged in the Dead Sea
where the density of the water is 1240 kg m-3?
(convert cm to m first)
1240*.12*.23*.15*9.81 = 50.36 N
50. N
A 15x15x5.0 cm piece of wood floats in water (1000. kg m-3)
face down in the water with the waterline 3.1 cm up the 5.0
cm side:
What is its mass?
What is its density?
Volume of water displaced: .15*.15*.031 = 6.975E-4 m3
mass of water displaced: 6.975E-4*1000 = 0.6975 kg
mass of block = 0.6975 kg
density of block = 0.6975/(.15*.15*.05) = 620 kg m-3
btw: (3.1/5)*1000 = 620
amount submerged = % of fluid’s density object is
620 kg m-3
A 5.0x4.0x4.0 cm piece of wood with a density of 530 kg m-3
is tied to the bottom of a pail of water (1000. kg m-3) with a
string and held completely submerged. What is the tension on
the string?
mass of the block: .04*.04*.05*530 = 0.0424 kg
weight of block: 0.0424 * 0.415944 N
Buoyant force on the block: .04*.04*.05*1000*9.81 = 0.7848 N
y Equilibrium of block: -0.415944+.7848+F = 0, F = -0.368856 N
0.37 N
A 25x25x10 cm block of iron (7.80x103 kg m-3) floats
on mercury (13.6x103 kg m-3) If one of the 25x25 cm
faces is down into the mercury, how far into the
mercury does the block sink before coming to
equilibrium?
mass of iron block: .25*.25*.10*7800 = 48.75 kg
Volume of displaced mercury: 48.75/13,600 = 3.584E-3 m-3
depth into mercury: .25*.25*D = 3.584E-3 m-3
D = 0.057 35 ≈ 5.7 cm. (BTW 7.8E3/13.6E3 = .5735…..)
5.7 cm