physics chapter 8 fluid mechanics. density ρ = m v therefore:ρv = m

PhysicsChapter 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

• 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

fluidfluidB wVgF

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

T = waterVg – alcoholVg

T = (water – alcohol )Vg

T = (1000kg/m3 – 806kg/m3) (1.5m3)(9.8m/s2)

T = 2851.8 N

B

T W

FBD

11.6 Archimedes’ Principle

Example 9 A Swimming Raft

The raft is made of solid squarepinewood. Determine whetherthe raft floats in water and ifso, how much of the raft is beneaththe surface.

11.6 Archimedes’ Principle

N 47000

sm80.9m8.4mkg1000 233

max

gVVgF waterwaterB

m 8.4m 30.0m 0.4m 0.4 raftV

11.6 Archimedes’ Principle

N 47000N 26000

sm80.9m8.4mkg550 233

gVgmW raftpineraftraft

The raft floats!

11.6 Archimedes’ Principle

gVwaterwaterN 26000

Braft FW

If the raft is floating:

23 sm80.9m 0.4m 0.4mkg1000N 26000 h

m 17.0sm80.9m 0.4m 0.4mkg1000

N 2600023

h

Pressure

P = F N A m2

Pascal’s PrincipalPressure 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

F2, A2F1, A1

P1 =P2

Therefore:F1 = F2A1 A2

Pressure at a Depth P = F = mg = ρVg = ρAhg = ρhg

A A A A

330 m

Find the pressure on the bottom of the submarine due to the waterabove it.

Find the pressure on the bottom of the submarine due to the waterAnd air above it.

Equation of Continuity• A1v1 = A2v2

• The product of the cross-sectional 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