1. Physics 203 College Physics I Fall 2012
S. A. Yost
Chapter 10
Part 2
Pascal’s Principle, Archimedes’ Principle, Buoyancy

2. Announcements
Thursday: exam on chapters 7 – 9 (only the sections covered in the homework) Today: Problem set 10A was due. (You remembered, right?) Problem set 10B is scheduled for the Tuesday after Thanksgiving, but this depends on what we discuss in today’s class. Watch the due date for possible changes.

3. Atmospheric Pressure
The weight of the air above us produces atmospheric pressure at sea level equal to
1 atm = × 105 N/m2.
Pressure is also measured in Pascals:
1 Pa = 1 N/m2.

4. Gauge Pressure
Pressure gauges are normally set to zero when only atmospheric pressure is present.
Gauge pressure is then the additional pressure beyond that due to the atmosphere.
The total pressure including atmospheric pressure is the absolute pressure.

5. Suction
Pump
A negative gauge pressure corresponds to suction. If we produce a negative gauge pressure on a straw, water will be “sucked” up the straw.
Why does the water go up the straw?
P < Patm h

6. Suction What is the highest a suction pump can draw water up a tube?
The absolute pressure can’t be less than zero.
Assume the pressure in the tube is reduced to P = 0.
P < Patm h

7. Suction
Pump
The water must be supported by atmospheric pressure.
rgh = 1 atm = × 105 N/m2
P = 0 h
1.013 × 105 N/m2
(1000 kg/m3)(9.8 m/s2)
h =
= 10.3 m.
P = rgh

8. Hydraulic Lift must push out the same volume into a
A hydraulic lift is a simple machine which uses the fact that any fluid pushed into a cylinder on one end V
must push out the
same volume into a
cylinder at the other
end. L2 V L1

9. Hydraulic Lift
Since the volumes are the same, the distances the cylinders move is related to their areas:
A1 L1 = V = A2 L2.
→ L2/L1 = A1/A2. V A2 L2 V A1 L1

10. Hydraulic Lift
The work done by a force on one cylinder will equal the work done by the other cylinder.
F1 L1 = W = F2 L2. F2 V L2 V F1 L1

11. Hydraulic Lift Mechanical advantage F2 / F1 = L1 / L2 = A2 / A1. F2 A2

12. Pascal’s Principle Know: F2 / F1 = A2 / A1 Rewrite: F2/A2 = F1/A1
→ DP2 = DP1. F2 V A2 V A1 F1

13. Pascal’s Principle DP = F1/A1 = F2/A2
When an force is applied to a closed vessel, the pressure increases by the same amount throughout the vessel. F2 A2 DP A1 F1
DP = F1/A1 = F2/A2

14. Pascal’s Demonstration
Pascal demonstrated his principle by inserting a thin 12 m long tube of diameter 6 mm into a wine barrel of diameter 40 cm.
Filling the tube with water caused the barrel to burst!
What was the weight of water in the tube?
What was the force on the lid? h D

15. Pascal’s Demonstration
d = m
What was the weight of the water in the tube?
m = r (pr2) h
= 1000 kg/m3
× (2.8 × 10-5 m2) × 12 m
= 340 g.
h = 12 m
r = m D
mg = kg × 9.8 m/s2
= 3.33 N (0.75 lb)

16. Pascal’s Demonstrationh
What was the force on the lid of the barrel?
The force on the bottom of the tube is
F1 = mg = 0.75 lb on area
A1 = pd2/4 (= 2.8 x 10-5 m2) D

17. Pascal’s Demonstrationh
The force on the lid of the barrel is
F2 = mg (A2/A1) with
A2 = pD2/4
Then F2 = mg (D/d)2 D
Area is proportional to the length squared.

18. Pascal’s Demonstration
d = m h
The force on the lid of the barrel is
F2 = mg (D/d)2
= 0.75 lb (0.40 m / m)2
= 0.75 lb × 4400
= 3300 lb
D = 0.4 m
big number!

19. Pressure in water
Balancing the net force of pressure and the weight for an imaginary box of water gives
F2 – F1 = mw g
where F1 = P1 A and F2 = P2 A. F1 V F2
submerged box

20. Buoyancy
Suppose I replace the imaginary box of water by an actual box.
What is the net force due to pressure on the box?
F2 – F1 = mw g , just as before!
This is the buoyant force.
Archimedes Principle: The buoyant force equals the weight of the water displaced. F1 V F2
submerged box

21. Floating Example
A plastic block with specific gravity 0.3 floats in water. What fraction of the block’s volume is under water?
The weight of the block equals the buoyant force.

22. Floating Example 30% of the block is under water.V Vw
If the volume of the block is V and the volume under water is Vw, we need to find Vw/V.
Use FB = rw Vwg
= mg = rVg.
Then rw Vw = rV, and
Vw/V = r/rw = SG = 0.3
30% of the
block is
under water.

23. Buoyancy Puzzle
Two identical cups have water in them, to the same level. But one also has a plastic block floating in it.
Which is heavier? A B
C they are the same A: B:

24. Buoyancy Puzzle The total mass in case A is MA = rV
What is the mass in case B?
Don’t guess – use physics. v1 v2

25. Buoyancy Puzzle Remember Archimedes’ Principle:
The weight of the plastic block equals the weight of the water that would have occupied the submerged volume of the block.

26. Buoyancy Puzzle V A: B:
This means the mass of the whole block equals the mass of water that would occupy volume v2.
Therefore the two cups have the same total mass. v1 v2

27. Buoyancy Puzzle Another way to look at it:
The force on the bottom of the cup is the weight of the contents.
The force on the bottom of the cup is also the pressure times the area.
Both have the same depth, so they have the same pressure. V A: B: v1 v2

28. Volume Rate of Flow
The volume rate of flow (also called flux) is defined to be the volume of fluid that flows past a point in a pipe per unit time.
Q = DV / Dt = Av
where A is the area, v the flow velocity.
DV A v

29. Volume Rate of Flow The volume rate of flow is measured in m3/s.
Another common unit is liters/second (L/s)
1 liter = 1000 mL = 1000 cm3
1 m3 = 1000 L. DV

30. Volume Rate of Flow
The volume rate of flow of an incompressible fluid is the same throughout a pipe. Q DV

31. Volume Rate of Flow
Q = Av is fixed: the fluid’s velocity is inversely proportional to the cross- sectional area. v3 v1 v2 DV DV
DV A

32. Bernoulli Principle
Suppose a volume V of incompressible fluid is pushed into a pipe with a pressure difference and height difference between the ends.
The same volume of fluid will be pushed out the other end. P1 P2 v1 v2 h

33. Bernoulli Principle
Pushing a volume V of fluid into a pipe under pressure P does work on it:
W = F L = F V/A = PV. F L
V A P

34. Bernoulli Principle
Bernoulli’s Principle is an expression of energy conservation:
PV + ½ m v2 + mgh = constant.
Work + kinetic energy
+ potential energy =
constant. P1 P2 v1 v2 h

35. Bernoulli Principle Dividing by volume, P + ½ r v2 + rgh = constant.
Pressure + kinetic energy density
+ potential energy density =
constant. P1 P2 v1 v2 h

36. Question
1. A fluid flows through the pipe shown. In which section is the flow velocity the greatest?
Selections: A B C D The same C A B

37. Answer
The volume rate of flow Q = vA is constant for an incompressible fluid. The fluid moves fastest where the pipe is narrowest, section B. (It moves slowest in section C.) C A B

38. Question
2. In which section is the pressure of the fluid the greatest?
Selections: A B C D The same C A B

39. Answer
This is Bernoulli’s principle: The pressure in a fluid decreases when the flow velocity increases. The fluid moves most slowly at C, so the pressure is highest there (and lowest at B). C A B

40. Water Tower and Fountain
A water tower feeds a fountain, which shoots water straight up in the air.
How fast does the water leave the fountain?
Assume the top of the water is a height h = 55 m above the fountain. h v

41. Water Tower and Fountain1
We’ll assume the tank is big, so the top of the water stays fixed:
h1 = 55 m,
v1 = 0,
P1 = 0 h v
gauge
pressure

42. Water Tower and Fountain1
At the fountain,
h2 = 0
v2 = v
What is P2? h
unknown v 2

43. Water Tower and Fountain1
Careful!
This is not hydrostatics.
If the fountain were turned off, the pressure would be
P2 = rgh = 1000 kg/m3
x 9.8 m/s2 x 55 m
= 5.4 x 105 N/m2. h
no flow! 2

44. Water Tower and Fountain1
When the fountain is flowing, this changes!
The pressure just outside the pipe is P2 = 0,
normal atmospheric pressure. h P2 v 2

45. Water Tower and Fountain1
The velocity is given by Bernoulli’s equation with
P1 = P2 = 0
h1 = h, h2 = 0
v2 = 0, v1 = v h P2 v 2

46. Water Tower and Fountain1
The only terms remaining are
½ r v2 = rgh
The result is the same as if the water had fallen from the top of the tower:
v = √ 2gh = 33 m/s. h v 2

47. Water Tower and Fountain1
What is the volume rate of flow if the pipe has diameter 1 cm?
Q = Av
A = p (0.5 cm)2
= cm2 h v 2

48. Water Tower and Fountain1
Q = Av
A = cm2
v = 33 m/s = 3300 cm/s
Q = 2600 cm3 /s
= 2600 mL /s = 2.6 L/s h v 2

49. Water Tower and Fountain1
How high does the water rise from the fountain?
Bernoulli’s equation between points 1 and 2:
P1 = P2 = 0,
v1 = v2 = 0
implies rgh1 = rgh2. 2 h v

50. Water Tower and Fountain1
The water rises to the height of the tower.
This assumes energy conservation:
No friction (viscosity or air resistance) or turbulence is considered. 2 h v

52. Water Tower and Fountain
What is the pressure inside the pipe feeding the fountain?
If there is no nozzle constricting the flow, then v is the same in both places, so the pressure must be zero both inside and outside the pipe. v 2 1 v

53. Water Tower and Fountain
Putting a nozzle on the hose would change this.
Then v is bigger outside the hose, and the pressure is higher inside the hose. If the height is about the same, then
P1 + ½ rv12 = P2 + ½ rv22. v2 2 1 v1

54. Water Tower and Fountain
The gauge pressure in the hose is
P1 - P2 = ½ r(v22 – v12).
Continuity: A1 v1 = A2 v2.
P1 – P2 = ½ rv22 (1 - A22/A12) v2 2 A2 A1 1 v1

55. Water Tower and Fountain
What is the gauge pressure in the pipe, assuming it’s diameter is still 1 cm at point 1, but the nozzle has diameter 1/8 cm at point 2. v2 A2 2
P1 – P2 = ½ rv22 (1 - A22/A12)
A2/A1 = (1/8)2 = 1/64 A1 1 v1

56. Water Tower and Fountain
P1 – P2 = ½ rv22 (1 - A22/A12)
= 0.5 (1000 kg/m3)(33 m/s)2 (1 – 1/4096)
= 5.4 x 105 Pa
Note that P1 – P2 ≈ rgh = 5.4 x 105 Pa
The pressure inside the pipe is nearly the same as when the pipe is shut off entirely! = v2 2 1 v1

57. Water Tower and Fountain
If v2 is the same with or without the nozzle, the water shoots as high either way.
What good is the nozzle? h v2

58. Water Tower and Fountain
This was an idealized problem: we assumed the water loses no energy in the pipe, not matter how fast it goes.
In reality, there is viscosity, a fluid friction that increases with speed. And 33 m/s is really fast! Bernoulli’s equation would be a poor approximation without the nozzle. h v2 v1

59. Water Tower and Fountain
The nozzle allows the water to move more slowly through the pipe:
v1 = A2v2/A = v2/64 = 0.52 m/s
just inside the nozzle, decreasing the energy lost to viscosity. h v2 v1