1. Static Fluids

2. Objectives: you should be able to:
Define and apply the concepts of density and fluid pressure to solve physical problems.
Define and apply concepts of absolute, gauge, and atmospheric pressures.
State and apply Archimedes’ Principle to solve physical problems.

3. Pressure
Pressure is the ratio of a force F to the area A over which it is applied:
A = 2 cm2
1.5 kg
= 73,500 N/m2
Pressure is a scalar; the SI units of pressure are pascals:
1 Pa = 1 N/m2

4. Fluid Pressure
Water flow shows  F
The force F exerted by a fluid on the walls of its container always acts perpendicular to the walls. F
Fluids exert pressure in all directions.
Fluids at rest have no component of force parallel to any solid surface (otherwise fluid would flow)

5. Pressure vs. Depth in Fluid
Pressure = force/area h mg
Area
P = rgh
Fluid Pressure:
Pressure at any point in a fluid is directly proportional to the density of the fluid and to the depth in the fluid. (applies only to incompressible fluids)

6. Independence of Shape and Area.
Water seeks its own level, indicating that fluid pressure is independent of area and shape of its container.
At any depth h below the surface of the water in any column, the pressure P is the same. The shape and area are not factors.

7. A diver is located 20 m below the surface of a lake (r = 1000 kg/m3)
A diver is located 20 m below the surface of a lake (r = 1000 kg/m3). What is the pressure due to the water?
The difference in pressure from the top of the lake to the diver is: h
r = 1000 kg/m3
DP = rgh
h = 20 m; g = 9.8 m/s2
DP = 196 kPa

8. Gravity and Pressure A) P1 > P2 B) P1 = P2 C) P1 < P2
Two identical “light” containers are filled with water. The first is completely full of water, the second container is filled only ½ way. Compare the pressure each container exerts on the table.
A) P1 > P2 B) P1 = P2 C) P1 < P2 1 2
Example w/ container of water? Double water double weight, calculate using density, see how A disappears.
P = F/A
= mg / A
Cup 1 has greater mass, but same area 31

9. Dam Question A) FA > FB B) FA=FB C) FA< FB
Two dams of equal height prevent water from entering the basin. Compare the net force due to the water on the two dams.
A) FA > FB B) FA=FB C) FA< FB
F = P A, and pressure is rgh. Same pressure, same Area same force even though more water in B! 35

10. Patm = 101,300 Pa (also called 1 atmosphere)
Atmospheric Pressure
atm h
Mercury
P = 0
One way to measure atmospheric pressure is to fill a test tube with mercury, then invert it into a bowl of mercury.
Density of Hg = 13,600 kg/m3
Patm = rgh
h = m
Patm = (13,600 kg/m3)(9.8 m/s2)(0.760 m)
Patm = 101,300 Pa (also called 1 atmosphere)

11. Absolute Pressure = Gauge Pressure + Atmospheric Pressure
DP = 196 kPa
1 atm = kPa
Absolute Pressure: The sum of the pressure due to a fluid and the pressure due to atmosphere.
Gauge Pressure: The difference between the absolute pressure and the pressure due to the atmosphere:
Absolute Pressure = Gauge Pressure + Atmospheric Pressure
Pabs = 196 kPa kPa
DP = 196 kPa
1 atm = kPa
Pabs = 297 kPa

12. Archimedes’ Principle
An object that is completely or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.
2 lb
The buoyant force is due to the displaced fluid. The block material doesn’t matter.

13. Calculating Buoyant Force
The buoyant force FB is due to the difference of pressure DP between the top and bottom surfaces of the submerged block.
FB = F2 – F1
= P2 A – P1 A
FB = rf gVf
Buoyant Force:
Vf is volume of fluid displaced.

14. Buoyancy and Archimedes’ Principle
The net force on the object is then the difference between the buoyant force and the gravitational force.

15. Fraction of object submerged:
Floating objects:
When an object floats, partially submerged, the buoyant force exactly balances the weight of the object. FB mg
FB = rf gVf
mx g = rxVx g
rf gVf = rxVx g
Floating Objects:
rf Vf = rxVx
Fraction of object submerged:
If Vf is volume of displaced water Vwd, then the fraction of object x submerged is given by:

16. Problem Solving Strategy
1. Draw a figure. Identify givens and what is to be found. Use consistent units for P, V, A, and r.
2. Use absolute pressure Pabs unless problem involves a difference of pressure DP.
3. The difference in pressure DP is determined by the density and depth of the fluid:

17. Problem Strategy (Cont.)
4. Archimedes’ Principle: A submerged or floating object experiences an buoyant force equal to the weight of the displaced fluid:
5. Remember: m, r and V refer to the displaced fluid. The buoyant force has nothing to do with the mass or density of the object in the fluid. (If the object is completely submerged, then its volume is equal to that of the fluid displaced.)

18. Problem Strategy (Cont.)
6. For a floating object, FB is equal to the weight of that object; i.e., the weight of the object is equal to the weight of the displaced fluid: FB mg

19. All forces are balanced:
A 2-kg brass block is attached to a string and submerged underwater. Find the buoyant force and the tension in the rope (rb= 8700 kg/m3).
All forces are balanced:
FB + T = mg
FB = rwgVw mg
FB = rgV T
Force diagram
Vb = Vw = 2.30 x 10-4 m3
Fb = (1000 kg/m3)(9.8 m/s2)(2.3 x 10-4 m3)
FB = 2.25 N

20. **This force is sometimes referred to as the apparent weight.
(cont.) A 2-kg brass block is attached to a string and submerged underwater. Now find the the tension in the rope.
FB = 2.25 N
FB + T = mg
T = mg - FB
T = (2 kg)(9.8 m/s2) N
T = 19.6 N N
T = 17.3 N mg
FB = rgV T
Force diagram
**This force is sometimes referred to as the apparent weight.

21. A student floats in a salt lake with 33
A student floats in a salt lake with 33.3% of his body above the surface. If the density of his body is 970 kg/m3, what is the density of the lake water?
Fraction submerged
rw Vw = rsVs
33.3%
66.7%
rw = 1454 kg/m3

22. Buoyancy and Archimedes’ Principle
This principle also works in the air; this is why hot-air and helium balloons rise.

23. Summary of Properties of Fluid Pressure
The forces exerted by a fluid on the walls of its container are always perpendicular.
The fluid pressure is directly proportional to the depth of the fluid and to its density.
At any particular depth, the fluid pressure is the same in all directions.
Fluid pressure is independent of the shape or area of its container.

24. Archimedes’ Principle:
Summary
P = rgh
Fluid Pressure:
Pascal:
Archimedes’ Principle:
Buoyant Force:
FB = rf gVf