Fluids
Buoyant Force

The pressure force on the
sides of the volume balances
the weight of fluid in the
volume.
Fb = mg

The force remains even
without the original fluid.

This is the buoyant force.
• Equals the fluid weight
• Directed upward
• Acts on the volume
Fbuoy  Vg
Sinking and Rising


Fb = Vg
W = mg
Fb = Vg

An object in a fluid displaces
a volume that had some
mass.
If the object is heavier than
the fluid it sinks.
If the object is lighter it rises.
W = mg
Fnet  Vg  mg
a  ( Vg  mg ) / m
Archimedes’ Principle

An object suspended in a fluid has less apparent
weight due to buoyancy.
FT = mg - Vg
Fb = Vg
W = mg
Iceberg

An iceberg has an average
density of 86% of seawater.

What fraction of the iceberg
is underwater?

The buoyant force is the
weight of water displaced by
the iceberg: Fb = waterVsubg.

The weight is the total weight
of the ice: Wi = iceViceg.

Find the ratio of Vsub/ Vice
 We know the ratio ice/water
 waterVsubg = iceViceg
 Vsub/ Vice = ice/water = 0.86
Flow Rate

Streamlines in a fluid
represent the path of a
particle in the fluid.
•
•

Groups for fluid flow
Cross sectional area
The flow rate measures fluid
movement.
• mass per time
• density times area times
velocity
m
 Av
t
Conservation of Mass

The mass into a tube must
flow out at the same rate.

This is called the continuity
equation.

For constant density it only
requires the area and
velocity.
1 A1v1  2 A2v2
Canyon



A river flows in a channel
that is 40. m wide and 2.2 m
deep with a speed of 4.5
m/s.
The river enters a gorge that
is 3.7 m wide with a speed of
6.0 m/s.
How deep is the water in the
gorge?

The area is width times
depth.
 A1 = w1d1

Use the continuity equation.
 v1A1 = v2A2
 v1w1d1 = v2w2d2

Solve for the unknown d2.
 d2 = v1w1d1 / v2w2
 (4.5 m/s)(40. m)(2.2 m) /
(3.7m)(6.0 m/s) = 18 m
Fluid Energy



The kinetic energy in a fluid is the same as for any
other mass: K = ½ mv2.
The change in potential energy is: U = mgh.
The work done on a fluid is due to pressure.
• Pressure acting on a volume: W = PAx = PV.

From the work energy principle:
Win  Wout  K  U
P1V  P2V  12 m(v22  v12 )  mg ( y2  y1 )
Bernoulli’s Equation

The volume element is somewhat arbitrary in a
moving fluid.
• Mass divided by volume is density
• Divide by volume and separate states on each end
P1  P2  12  (v22  v12 )  g ( y2  y1 )
P1  gy1  12 v12  P2  gy2  12 v22

Bernoulli’s equation is equivalent to conservation of
energy for fluids.
Lift

If the height doesn’t change
much, Bernoulli becomes:
P1  12 v12  P2  12 v22
FL

Where speed is higher,
pressure is lower.

Speed is higher on the long
surface of the wing –
creating a net force of lift.