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Physics 151: Lecture 29 Today’s Agenda. Today’s topics Fluids under static conditions, Ch. 14.1 through 14.4 Pressure Pascal’s Principle (hydraulic lifts etc.) Archimedes’ Principle (floatation). See text: 14.1. Fluids. At ordinary temperature, matter exists in one of three states

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physics 151 lecture 29 today s agenda
Physics 151: Lecture 29 Today’s Agenda
  • Today’s topics
    • Fluids under static conditions, Ch. 14.1 through 14.4
    • Pressure
    • Pascal’s Principle (hydraulic lifts etc.)
    • Archimedes’ Principle (floatation)
fluids

See text: 14.1

Fluids
  • At ordinary temperature, matter exists in one of three states
    • Solid - has a shape and forms a surface
    • Liquid - has no shape but forms a surface
    • Gas - has no shape and forms no surface
  • What do we mean by “fluids”?
    • Fluids are “substances that flow”…. “substances that take the shape of the container”
    • Atoms and molecules are free to move.
    • No long range correlation between positions.
fluids3

See text: 14.1

Fluids
  • What parameters do we use to describe fluids?
    • Density

units :

kg/m3 = 10-3 g/cm3

r(water) = 1.000 x103 kg/m3 = 1.000 g/cm3

r(ice) = 0.917 x103 kg/m3 = 0.917 g/cm3

r(air) = 1.29 kg/m3 = 1.29 x10-3 g/cm3

r(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3

fluids4

n

A

Fluids
  • What parameters do we use to describe fluids?
    • Pressure

units :

1 N/m2 = 1 Pa (Pascal)

1 bar = 105 Pa

1 mbar = 102 Pa

1 torr = 133.3 Pa

1atm = 1.013 x105 Pa

= 1013 mbar

= 760 Torr

= 14.7 lb/m2 (=PSI)

  • Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface.
    • Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as:
pressure vs depth incompressible fluids liquids

p

0

F

1

y

1

y

2

p

1

A

p

2

mg

F

2

See text: 14.2

Pressure vs. DepthIncompressible Fluids (liquids)
  • When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid
  • For an incompressible fluid, the density is the same everywhere, but the pressure is NOT!
  • Consider an imaginary fluid volume (a cube, face area A)
    • The sum of all the forces on this volume must be ZERO as it is in equilibrium: F2 - F1 - mg = 0
pressure vs depth

For a fluid in an open container pressure same at a given depth independent of the container

  • Fluid level is the same everywhere in a connected container, assuming no surface forces
  • Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium?
  • Imagine a tube that would connect two regions at the same depth.

See text: 14.2

Pressure vs. Depth
  • If the pressures were different, fluid would flow in thetube!
  • However, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium.
lecture 29 act 1 pressure

A) r1 < r2

B) r1 = r2

C) r1 > r2

Lecture 29, ACT 1Pressure
  • What happens with two fluids?? Consider a U tube containing liquids of density r1 and r2 as shown:
    • Compare the densities of the liquids:

dI

r2

r1

example
Example
  • A U-tube of uniform cross-sectional area, open to the atmosphere, is partially filled with mercury. Water is then poured into both arms. If the equilibrium configuration of the tube is as shown in Figure on the right, with h2 = 1.00 cm.
  • Determine the value of h1.
example9
Example
  • Figure on the right shows Superman attempting to drink water through a very long straw. With his great strength he achieves maximum possible suction. The walls of the tubular straw do not collapse.
  • (a) Find the maximum height through which he can lift the water.
pascal s principle

Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel.

See text: 14.2

Pascal’s Principle
  • So far we have discovered (using Newton’s Laws):
    • Pressure depends on depth: Dp = rgDy
  • Pascal’s Principle addresses how a change in pressure is transmitted through a fluid.
  • Pascal’s Principle explains the working of hydraulic lifts
    • i.e. the application of a small force at one place can result in the creation of a large force in another.
    • Does this “hydraulic lever” violate conservation of energy?
      • Certainly hope not.. Let’s calculate.
pascal s principle11

See text: 14.2

Pascal’s Principle
  • Consider the system shown:
    • A downward force F1 is applied to the piston of area A1.
    • This force is transmitted through the liquid to create an upward force F2.
    • Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid.
  • F2 > F1 : Have we violated conservation of energy??
pascal s principle12

See text: 14.2

Pascal’s Principle
  • Consider F1 moving through a distance d1.
    • How large is the volume of the liquid displaced?
  • This volume determines the displacement of the large piston.
  • Therefore the work done by F1 equals the work done by F2 We have NOT obtained “something for nothing”.
lecture 29 act 2a hydraulics

M

dA

A) dA=(1/2)dB

B) dA = dB

C) dA = 2dB

A1

A10

M

dB

A2

A10

Lecture 29, ACT 2aHydraulics
  • Consider the systems shown to the right.
    • In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference dI in the liquid levels.
    • If A2 = 2´A1, compare dA and dB.
lecture 29 act 2b hydraulics

M

dA

A1

A10

A) dA = (1/2)dC

B) dA = dC

C) dA = 2dC

dC

M

A1

A20

Lecture 29, ACT 2bHydraulics
  • Consider the systems shown to the right.
    • In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference dI in the liquid levels.
    • If A10 = 2´A20, compare dA and dC.
archimedes principle

W2?

W1

W1 > W2

W1 < W2

W1 = W2

See text: 14.4

Archimedes’ Principle
  • Suppose we weigh an object in air (1) and in water (2).
    • How do these weights compare?
  • Why?
    • Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.
archimedes principle16

W2?

W1

Archimedes:

The buoyant force is equal to the weight of the liquid displaced.

See text: 14.4

Archimedes’ Principle
  • The buoyant force is equal to the difference in the pressures times the area.
  • The buoyant force determines whether an object will sink or float. How does this work?
sink or float

y

F

mg

B

See text: 14.4

Sink or Float?
  • The buoyant force is equal to the weight of the liquid that is displaced.
  • If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink.
  • We can calculate how much of a floating object will be submerged in the liquid:
  • Object is in equilibrium

Animation

the tip of the iceberg

y

F

mg

B

See text: 14.4

The Tip of the Iceberg
  • What fraction of an iceberg is submerged?
lecture 29 act 3 buoyancy

Pb

styrofoam

A) It sinks

B)

C)

D)

styrofoam

Pb

Lecture 29, ACT 3Buoyancy
  • A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown.
    • If you turn the styrofoam+Pb upside down, what happens?
act 3 a more fun with buoyancy

Cup II

Cup I

(D) can’t tell

(C) the same

(A) Cup I

(B) Cup II

See text: 14.4

ACT 3-AMore Fun With Buoyancy
  • Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it.
    • Which cup weighs more?
act 3 b even more fun with buoyancy

oil

water

(B) move down

(C) stay in same place

(A) move up

See text: 14.4

ACT 3-BEven More Fun With Buoyancy
  • A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (roil < rball < rwater) is slowly added to the container until it just covers the ball.
    • Relative to the water level, the ball will:
recap of today s lecture
Recap of today’s lecture
  • Chapter 14.1-4
    • Pressure
    • Pascal’s Principle
    • Archimedes Principle