Pressure and buoyancy in fluids
•  FCQ’s for lecture and
tutorials will be next week.
•  Buoyancy force today
•  Fluid dynamics on Monday
(along with the loudest
demonstration of the
semester).
•  Review on Wednesday and
Friday next week.
•  Final exam at 7:30am on
Tuesday, May 5.
Pressure in a liquid
p0 A
Consider the bottom of a column
of water with depth d and cross
sectional area A inside a container
open to the atmosphere.
Atmospheric pressure p0
pushes down with force of p0A.
pA
mg
p0 A
d
pA
The weight of the column pushes down with force
mg. For a liquid with density ρ, m=ρV=ρAd.
Because the liquid is in static equilibrium, the upward force from
pressure, pA, must equal the downward forces
pA = p0 A + ρ Adg
so p = p0 + ρ gd (hydrostatic pressure)
When you descend in a liquid, the weight of the liquid above you
2
causes the pressure to increase.
Clicker question 1
Set frequency to BA
Three vessels are full of the same liquid and open to the same
atmosphere. The pressure is measured in each at a distance of 3
m below the surface. What can we say about the pressures?
A. only two are the same
B. all three are different
C. all three are the same 3 m
The hydrostatic pressure is p = p0 + ρ gd
It only depends on the pressure on top and the amount of water
in a column directly overhead so it is the same for all 3.
3
Pressure in a liquid
In the problem, the liquids
had the same height because
they were filled that way.
If they were all connected
(as shown), the liquid levels
would have to be the same.
A
B
Why? Well, assume the first container had a higher level.
Then, since p = p0 + ρ gd, pressure at A would be greater than at B.
Assume a gradual decrease from A to B. Then at any point
between them, pressure from the left is greater than from the right
resulting in a net force to the right (not equilibrium). Therefore,
fluid will flow to the right until equilibrium is reached.
4
Some rules for pressure
Anywhere in a connected, static, uniform density fluid, the
pressure at a given height is the same.
Pascal’s law: A pressure change at one point in an incompressible
fluid appears undiminished at all points in the fluid.
A force F is applied to a
piston of area A, increasing
pp00
the pressure by F/A.
F
p = p0 +
A
dd1
F
p = p0=+p0 + ρ gd1
A
F
d2
p = p0 + ρ gd2
5
Clicker question 2
Set frequency to BA
Uma Thurman (mass of 60 kg) is standing on a piston, connected as
shown to another piston on which a 6000 kg stretch Hummer is
resting. How much bigger in area is the piston under the Hummer
compared to the one under Uma? Try writing down two expressions
for the pressure at the dotted line.
A. same
On the left side we have
B. 10 times
FL
FL = mU g
pL = p0 +
C. 100 times
AL
D. 1000 times
E. 10000 times Right side: p = p + FR
FR = mH g
R
0
AR
Both pressures at same height so must be the same. Setting equal:
FL
FR
p0 +
= p0 +
AL
AR
so FL AR = FR AL
FR
so AR = AL = 100 ⋅ AL
FL
Buoyancy
p0 A
Take a volume of water with
density ρf and area A extending
from a depth of d2 to d1.
Summing the forces of the
free body diagram gives
∑F
y
p0 A + ρf gAd1
mf g
p0 A + ρf gAd2
d2
d1
pA
= p0 A + ρf gAd1 − p0 A − ρf gAd2 − mf g
= ρf gA(d1 − d2 ) − mf g = 0
If we replace the fluid with an object, the only difference is mass.
∑F
y
= ρf gA(d1 − d2 ) − mo g = 0
Note that A(d1-d2) is the volume Vf of fluid displaced
Upward force ρ f gA(d1 − d2 ) = ρ f Vf g equals the weight of displaced fluid
Archimedes principle
A body partially or fully immersed in a fluid feels an
upward force equal to the weight of the displaced fluid.
This force is called the buoyant force: FB = ρfVf g
As shown, it is due to the increase of pressure with depth in a fluid.
If the object is fully immersed then the volume of the displaced
fluid is equal to the volume of the object: Vf = Vo
Note that volume is related to mass and density: mo = ρoVo
If an object is only partially submerged, the volume of the
displaced fluid is less than the volume of the object: Vf < Vo
Buoyancy example
A 2 cm by 2 cm by 2 cm cube of iron (ρ=8 g/cm3) is weighed
with the iron outside, half in and fully in the water, as shown
in the diagram. What is the measured weight in each case?
Iron mass: mo = ρoVo = 8 g/cm 3 ⋅ 8 cm 3 = 64 g = 0.064 kg
T − mo g = 0 so
Out of the T
water:
mo g T = mo g = 0.064 kg ⋅10 m/s2 = 0.64 N
FB T
In the
water:
mo g
FB + T − mo g = 0 so T = mo g − FB
FB = ρfVf g = 1 g/cm3 ⋅ 8 cm3 ⋅10 m/s2 = 0.08 N
so T = mo g − FB = 0.64 N − 0.08 N = 0.56 N
FB T
½ in the
water:
mo g
FB + T − mo g = 0 so T = mo g − FB
FB = ρfVf g = 1 g/cm 3 ⋅ 4 cm 3 ⋅10 m/s2 = 0.04 N
so T = mo g − FB = 0.64 N − 0.04 N = 0.60 N
Archimedes crown
Archimedes lived from 287 BC to 212 BC. King Hiero II gave a goldsmith
gold to make a crown. When the crown was made, it was found to weigh
the same as the gold given but the king suspected that silver had been
mixed in so he asked Archimedes to find out (without damaging the crown).
One day while getting into his tub, Archimedes noticed water was displaced
in an amount equal to his volume and figure out a way to determine the
density of the crown. He ran naked through the streets of Syracuse
shouting ‘Eureka’, Greek for “I found it”. What had he found?
Balance the weight of the
crown with pure gold in air.
Submerge both in water. If the
crown is less dense than the gold,
it must have a larger volume (in
order to have the same weight).
In that case, the crown will displace more water than the gold, leading to
a larger buoyant force on the crown, and a smaller apparent weight.
Clicker question 3
Set frequency to BA
An ice cube is floating in a glass of water. As the
ice cube melts, the level of the water…
A. rises.
The buoyant force is always
B. falls.
C. stays the same. equal to the weight of the liquid
displaced by the object.
Since the ice cube is in equilibrium, the weight of
water displaced is equal to the weight of the ice cube.
When the ice cube melts it becomes water with the same
weight as the ice cube so real water takes the place of the
displaced water.