Fluid - any substance that flows, typically a liquid or a gas
Hydrostatics - the study of fluids at rest, such as the pressure of a fluid at a particular depth, or the buoyant force acting on an object in a fluid
Hydrodynamics - the study of fluids in motion
Density is the degree of compactness of a substance. It is the mass per unit divided by the volume of the unit. The symbol for density is (rho).
It is most commonly measured in kilograms per cubic meter or kg/m3.
The equation for density is: 𝑚 𝜌 =
𝑉
Where ρ is the density, m is the mass of the object, and V is the volume of the object.
Practice: Page 278 of College Physics. Example 9.4; Part (a)
Equation Help: http://quizlet.com/43876040/density-flash-cards/
Pressure is the force per unit of area. Pressure is scalar which means it will always be positive. It is measured in Pascals (Pa). One pascal is equal to one newton per meter (1 Pa = 1
N/m 2 ).
The equation for pressure is:
𝑃 =
𝐹
𝐴
Where P is the pressure, F is the force, and A is the area of the object.
There is more than one type of pressure, though. There is atmospheric pressure, gauge pressure, and absolute pressure.
The equation that relates them is:
𝑃 𝑎𝑏𝑠
= 𝑃 𝑔𝑎𝑢𝑔𝑒
+ 𝑃 𝑎𝑡𝑚
Practice:
A boy swims a lake and initially dives 0.5 m beneath the surface. When he dives 1 m beneath the surface, how does the absolute pressure change?
(A) It doubles (B) It quadruples (C) It cut to a half (D) It slightly increases (E) It slightly decreases
Solution: D
More Practice:
Page 278 of College Physics. Example 9.4; Part (b)
Page 279-280 of College Physics. Example 9.5
Page 280 of College Physics. Example 9.6

Even More Practice: http://www.learningpod.com/pod/density-and-pressure-practice/e3a9ddd5-
3368-42e0-9d88-2a7658d65b28
Equation Help: http://quizlet.com/13787803/pressure-flash-cards/
Videos to Help:
Ms. Twu Fluid Mechanics Videos 1-4 https://sites.google.com/site/twuphysicslessons/home/fluids
Level 3: Barometers
Barometers are devices used to measure air pressure. A barometer consists of a large dish and a long tube that is sealed at one end. The tube is inverted into the dish so the open end is in the liquid. When using the formula
P=gh, P is the atmospheric pressure, g is gravity, h is the height of the fluid in the tube, and is density of the fluid.
The equation used is:
𝑃 𝑎𝑡𝑚
= 𝜌𝑔ℎ
Practice:
What is the difference between the pressure on the bottom of a pool and the pressure on the water surface?
(A) 𝜌𝑔ℎ (B) 𝜌𝑔 ℎ 𝜌
(C) 𝑔ℎ
(D) 𝑔ℎ 𝜌
(E) 0
Solution: A
Practice:
An open bottle is filled with a liquid which is flowing out through a spigot located at the distance h below the surface of the liquid. What is the velocity of the liquid leaving the bottle?
(A) √𝑔ℎ (B) 2𝑔ℎ (C) 4𝑔ℎ (D) 𝜌𝑔ℎ (E) √2𝑔ℎ
Solution: E
More Practice:
Page 283 of College Physics. Quick Quiz 9.3. Answer on A.61 (#3).
Page 284 of College Physics. Quick Quiz 9.4. Answer on A.61 (#4).
Free Response:

A glass U-tube with a uniform diameter of 0.850 cm is used to determine the density of an oil.
As shown in the figure above, a 24.5 cm column of water balances a 27.2 cm column of the oil so that interfaces A and B of the mercury with the other liquids are at the same height. The density of water is 1.00 x 10 3 kg/m 3 .
(a) Calculate the density of the oil.
(b) Calculate the absolute pressure at B, the interface between the water and the mercury.
A new tube, identical to the U-tube except for a cone shape on the left, as shown above, is filled with the same volume of mercury that was in the U-tube. The mercury is at the same height on both sides of the new tube as it was in the U-tube, as shown by the dashed line. The same volumes of oil and water that were in the U-tube are now poured into the new tube, on the left and right respectively.
(c) Indicate the new position of B relative to A.
____ Above A ____ Below A ____ At the same height as A
Justify your answer.
(d) A small piece of wood with density less than that of the oil is placed so that it floats in the left side of the tube. Indicate whether the pressure at the bottom of the tube increases, decreases, or remains the same.
____ Increases ____ Decreases ____ Remains the same
Solution: http://apcentral.collegeboard.com/apc/public/repository/ap12_physics_b_scoring_guidelines.pdf
Page 8

Any object that is in a fluid experiences a buoyant force that partially supports the object.
An anchor at the bottom of the ocean, a submarine traveling on the water’s surface, and a balloon filled with helium all feel a force acting on them. The buoyant force is a net upward force that is done on the object. It is felt from all sides, perpendicular to the object. The horizontal forces cancel, but the vertical forces do not. As depth increases, the pressure increases, so the forces pushing up on the object are greater than the forces pushing down on the object. This is because the bottom of the object is at a greater depth than the top of the object (P=gh). The different forces on the top and bottom of the object cause the buoyant force.
Archimedes’ Principle states that the force of buoyancy is equal to the weight of the fluid that is displaced by the object.
𝐹 = 𝜌𝑔𝑉 where is the density of the fluid, g is the acceleration due to gravity, and V is volume of the displaced fluid
If the force of buoyancy is equal to or greater than the weight of the object, the object will float, but if the force of buoyancy is less than the weight of the object, the object will sink.
Because the force of buoyancy and the density of the fluid are directly proportional, if the density of the fluid increases, the force increases. Thus, if the density is greater than the object, the object will float.
To what extent to which the object is floating depends on the object’s density in relation to the fluid’s density. The fraction submerged is the ratio of the volume submerged to the volume of the object.
Fraction submerged =
𝑉 𝑓𝑙𝑢𝑖𝑑
𝑉 𝑜𝑏𝑗𝑒𝑐𝑡 𝑔𝜌 = 𝑚
𝑉
, so 𝑉 = 𝑚𝜌
Thus, put the equation for Volume into the fraction submerged equation to get 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
⁄ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
Fraction submerged = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡
⁄ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡
Because the object is floating, its mass and that of the displaced object are equal, so the cancel from the equation leaving
Fraction submerged = 𝜌 𝑜𝑏𝑗𝑒𝑐𝑡 𝜌 𝑓𝑙𝑢𝑖𝑑

The fraction submerged is also the specific gravity . If an object floats, its specific gravity is less than one, if it sinks; its specific gravity is less than one. If the specific gravity is exactly one, the object stay suspended, neither floating nor sinking. Scuba divers try to obtain this specific gravity so they can hover in the water.
Archimedes’ principle states that the buoyant force on the object equals the weight of the fluid displaced. This means that the object appears to weigh less when submerged; we call this measurement the object’s apparent weight .
Apparent weight (F
A
) = weight of the object - force of buoyancy
The object suffers an apparent weight loss equal to the weight of the fluid displaced.
Alternatively, on balances that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced.
Apparent weight loss = weight of fluid displaced or
Apparent mass loss = mass of fluid displaced
Practice:
A raft is constructed of wood having a density of 600 kg/m
3
. Its surface area is 5.7 m
2
, and its volume is 0.60 m
3
. When the raft is placed in fresh water of density 1000 kg/m
3
, to what depth does the raft sink in the water?
Solution:
𝐹
𝐵
= 𝑊𝑒𝑖𝑔ℎ𝑡 = 𝜌𝑔𝑉 = 𝜌𝑔𝐴ℎ 𝜌𝑔𝑉 = 𝜌𝑔𝐴ℎ 𝜌𝑉 = 𝜌𝐴ℎ ℎ = 𝜌𝑉 𝜌𝐴 ℎ =
(600)(0.6)
(1000)(5.7) ℎ = 0.06 𝑚
*Note: ρ are not the same in this equation, so they do not cancel. The left is the density of the raft, while the right is the density of the water.
Practice:
A cube of steel that measures 5.0 cm on each side is immersed in water. The density of steel is 9.0 x 10
3
kg/m3. The density of water is 1.0 x 10
3
kg/m3. What is the (a) buoyant force acting on the cube and what is (b) its apparent weight?
Solution:
(a)
3
−4
3
(b) 𝐹
𝐴
= 𝐹
𝑊
− 𝐹
𝐵
𝐹
𝑊
= 𝜌𝑔𝑉
3
−4
𝐹
𝑊
𝐹
𝑊
= (9 𝑥 10 3 )(9.8)(1.25 𝑥 10 −4 )
= 11 𝑁

𝐹
𝐴
𝐹
𝐴
= 𝐹
𝑊
− 𝐹
𝐵
= 11 − 1.23
𝐹
𝐴
= 9.77 𝑁
Fluids can undergo two types of flow: laminar flow and turbulent flow .
Laminar flow, also known as streamline flow , is when the fluid moves in a smooth line, one after each other. This is represented by the drawing of streamlines, which are example paths of particles. Turbulent flow occurs at high flow rates when the fluid is moving past irregular shapes. In turbulent flow, the motion of the fluid becomes chaotic, and it forms eddies and whirlpools. Turbulent flow absorbs energy and increases the frictional drag throughout the fluid.
Turbulent flow is very complicated and the actual motion of the fluid cannot be precisely calculated; the mathematics is not precise and it is often very hard to describe the behavior using models
In dealing with flow through pipes and tubes, we will assume that the fluid is incompressible and you can assume that they encounter no internal friction as they flow through the pipe.
The rate of flow is the volume of fluid that passes a certain cross-section area in a given time.
where Q is the rate of flow, v is the velocity, and A is the area of a cross-section or the volume of the fluid passing a point at any given time
where V is the volume of the fluid and t is time
The flow rate is constant throughout the entire pipe, because friction is not put into account.

Because the flow rate is continuous,
1
1
1
2
2
2
A horizontal pipe has a constriction in it as shown. At point 1 the diameter is 0.06 m, while at point 2 it is only 0.02 m. At point 1, v1 is 2 m/s and P1 is 180 kPa. Calculate v
2
(speed in the smaller pipe).
Solution:
1
2
1
2
2
2
2
1
2
2
1
1
2
1
2
2
1
2
2
2

When a fluid flows through a narrow opening, a nozzle, its velocity will increase.
That is the main idea of the nozzle effect. As you should remember from the last problem:
𝐴
1 𝑣
1
= 𝐴
2 𝑣
2
, so because the area is smaller, the velocity must increase. This is why a fire truck hose has the attachment at the end. Because of the nozzle effect, the fire truck is able to spray the fire and end it. This also allows you to spray your friends in the summer with water guns.
Practice:
Water flows through a rubber hose 3.0 cm in diameter at a velocity of 5.0 m/s. If the hose is coupled into a nozzle that has a diameter of 0.50 cm, what is the new speed of the fluid?
Solution:
𝐴
1 𝑣
1 𝑣 𝑣 𝑣 𝑣
2
2
2
2
= 𝐴
2 𝑣
2
=
=
=
𝐴
1 𝑣
1
𝐴 𝜋𝑟
2
2
1 𝑣
1
2 𝜋𝑟
2
0.015
2
0.005
(5)
2
= 180 𝑚/𝑠
More Practice: Page 317 of College Physics. Number 46.
Page 317 of College Physics. Number 47.
Video to Help:
Ms. Twu Fluid Mechanics Video 17 https://sites.google.com/site/twuphysicslessons/home/fluids/fluids-page-3
Developed in 738 by Swiss physicist Daniel Bernoulli, Bernoulli's Equation compares the pressure at two different locations in a moving fluid. Bernoulli’s is an equation that is extremely important in fluids. It allows calculations for the differences in two different locations in a moving fluid. Your book describes Bernoulli’s as a consequence of energy conservation as applied to an ideal fluid. Bernoulli’s is beautiful. 𝜌𝑣
1
2 𝜌𝑣
2
2
𝑃
1
+ 𝜌𝑔ℎ
1
+
2
= 𝑃
2
+ 𝜌𝑔ℎ
2
+
2
When P is pressure, ρ is density of the fluid, h is the vertical height at the point, and v is the velocity of the fluid at the point

While the equation looks big and scary it really is not. For most of the problems involving this things will cancel out making it really easy to use.
Practice:
You have a couple of garden hoses connected together to water your backyard. Because you're a physics nerd, you put a gauge where the two hoses connect to each other. The hose ends with your sprinkler. When you put a crimp in the hose to stop water flow, you noticed that the pressure registered at 5.00 x 10 3 Pa. When the sprinkler is running, what is the velocity of the water flowing out?
Solution:
𝑃
𝑃
1
1
+ 𝜌𝑔ℎ
1
= 𝑃
2
+ 𝜌𝑣
1
2
+ 𝜌𝑣
2
2
2
2
= 𝑃
2
+ 𝜌𝑔ℎ
2
+ 𝜌𝑣
2
2
2 𝑣
2
= √
2(𝑃
1
− 𝑃
2
) 𝜌 𝑣
2
= √
2((5.00 𝑥 10 3 + 1.0 𝑥 10 5 ) − 1 𝑥 10 5 )
1000 𝑣
2
= 3.16 𝑚/𝑠
More Practice: Page 293 of College Physics.
Page 295 of College Physics.
Equation Help: http://www.cram.com/flashcards/ap-physics-b-siegel-unit-6-2001121
Torricelli’s is an application of Bernoulli’s. It states that if you have a tank of fluid and there is a small hole in the bottom of the tank, the fluid will leave through that hole with the same velocity as it would experience if you dropped it from the same height to the level of the hole.
Torricelli’s can only be used if the top of the tank is open to the atmosphere. 𝑣
1
2 = 𝑣
2
2 + 2𝑔ℎ

Practice:
V
2
is assumed to be zero because it is the level of the water receding 𝑣 = √2𝑔(ℎ
2
− ℎ
1
)
The water tower in the drawing above is drained by a pipe that extends to the ground. The amount of water in the top spherical portion of the tank is significantly greater than the amount of water in the supporting column. The density of water is 1000 kg/m 3 . The valve is open. What is the speed of the water past the valve?
Solution:
Videos to Help:
Ms. Twu Fluid Mechanics Videos 20-21 https://sites.google.com/site/twuphysicslessons/home/fluids/fluids-page-4
Wings
Birds fly due to Bernoulli’s Theorem. As the wing passes through the air, the flow rate in front of the wing on the top and bottom has to equal the flow rate at the end of the wing on the top and bottom. The air passing over the top of the wing has to travel a longer distance than the air passing under the wing. This is because of the curve on the top of the wing (the camber).
Since it is covering a longer distance in the same time, the velocity of the air on the top of the wing is greater than the velocity of the air on the bottom of the wing. From Bernoulli's principle, we know that the pressure decreases as the velocity increases. So the pressure above the wing is lower than the pressure under the wing. Air pressure pushes the wing up to try and equalize the pressures. This upward force is called lift.

For something to fly, of course, the lift must be greater than the weight. Birds are engineered to be extremely light in weight. They have hollow bones, no teeth, no bony tail, etc.
This helps them fly as the amount of lift their wings must develop is very small.
Entrainment
Imagine a perfume atomizer bottle. Low pressure is at the top of the tube between the pump and spritzer. When a person starts to pump, it forces air into a tube. The flow is forced through a nozzle which accelerates the flow rate, but drops the pressure. Liquid is drawn into the low pressure region and pushed out the nozzle due to a high velocity.