Transcript Chapter 9

Physics
Chapter 8
Fluid Mechanics
Density
ρ=M
V
Therefore:
ρV = M
Chapter 8
Density
Fig. T9.3, p. 262
Slide 13
Archimedes’s Principle
Any object completely or partially submerged
in a fluid is buoyed up by a force whose
magnitude is equal to the weight of the fluid
displaced by the object.
Buoyant Force
• The upward force is
called the buoyant force
• The physical cause of
the buoyant force is the
pressure difference
between the top and
the bottom of the
object
Buoyant Force, cont.
• The magnitude of the buoyant force always equals
the weight of the displaced fluid
FB   fluidVg  w fluid
• The buoyant force is dependent on the volume of the
object and the density of the fluid it is submerged in.
• For a floating object, the buoyancy force must equal
the weight of the object.
B=W
• Some objects may float high or low
Example: A balloon having a volume of 1.5 cubic meters is filled with
ethyl alcohol and is tethered to the bottom of a swimming pool.
Calculate the tension in the cord tethering it to the bottom of the
swimming pool.
 Fy = B – T – W = 0
Therefore, T = B – W
B
FBD
T = waterVg – alcoholVg
T = (water – alcohol )Vg
T
W
T = (1000kg/m3 – 806kg/m3) (1.5m3)(9.8m/s2)
T = 2851.8 N
11.6 Archimedes’ Principle
Example 9 A Swimming Raft
The raft is made of solid square
pinewood. Determine whether
the raft floats in water and if
so, how much of the raft is beneath
the surface.
11.6 Archimedes’ Principle
Vraft  4.0 m4.0 m0.30 m  4.8 m
FBmax   Vg   waterVwater g



 1000kg m 3 4.8m 3 9.80 m s 2
 47000N

11.6 Archimedes’ Principle
Wraft  mraft g   pineVraft g



 550kg m 3 4.8m 3 9.80 m s 2
 26000N  47000N
The raft floats!

11.6 Archimedes’ Principle
If the raft is floating:
Wraft  FB
26000N  waterVwater g



26000N  1000kg m3 4.0 m4.0 mh 9.80m s2
h

26000N
 0.17 m
3
2
1000kg m 4.0 m4.0 m 9.80 m s




Pressure
P=F
A
N
m2
Pascal’s Principal
Pressure applied to a fluid in a closed
container is transmitted equally to every point
of the fluid and to the walls of the container.
Hydraulic Lift
P1 =P2
Therefore:
F1 = F 2
A1
A2
F2, A2
F1, A1
Pressure at a Depth
P = F = mg = ρVg = ρAhg = ρhg
A A
A
A
Find the pressure on the bottom
of the submarine due to the water
above it.
330 m
Find the pressure on the bottom
of the submarine due to the water
And air above it.
Equation of Continuity
• A1v1 = A2v2
• The product of the crosssectional area of a pipe
and the fluid speed is a
constant
– Speed is high where the
pipe is narrow and speed
is low where the pipe has
a large diameter
• Av is called the flow rate
How Airplanes Wings Work