Chapter 15 compatibility mode by triphu95

Lecture Outline Chapter 15 Physics (101)

Chapter 15 Fluids

Units of Chapter 15 • Density • Pressure • Static Equilibrium in Fluids: Pressure and Depth • Archimedes’ Principle and Buoyancy • Applications of Archimedes’ Principle • Fluid Flow and Continuity • Bernoulli’s Equation

15-1 Density The density of a material is its mass per unit volume:

The water has the following density:

15-2 Pressure Pressure: is a measure of amount of force F per unit area A:

EXAMPLE

15-1

Find the pressure exerted on skin of a balloon if you press in a force of 2.10 N using (a) Your finger (b) A needle. Assume the area of your fingertip is 1.0 x 10-4 m2 and the area of the needle tip is 2.5 x 10-7 m2 (c ) Find the minimum force necessary to pop the balloon with the needle, given that the balloon pops with a pressure of 3.0 x 10-3 N∕m2 Copyright © 2010 Pearson Education, Inc.

solution Part (a)

F 2. 1 4 2 P   2 . 1  10 N m A 1.0 10  4 Part (b)

F 2.1 6 2 P   8 . 4  10 N m A 2.5 10 7 Part (c)

F P   F  PA A F  (3.0  105 )(2.5  10 7 )  0.075 N The same force applied over a smaller area results in greater pressure – think of poking a balloon with your finger and then with a needle. Copyright © 2010 Pearson Education, Inc.

15-2 Pressure Atmospheric pressure is due to the weight of the atmosphere above us.

The pascal (Pa) is 1 N/m2. Pressure is often measured in pascals. There are a number of different ways to describe atmospheric pressure. In pascals: In pounds per square inch: In bars: Copyright © 2010 Pearson Education, Inc.

EXSERCISE 15-1 Find the force exerted on the palm of your hand by atmospheric pressure. Assume your palm measure 0.080 m by 0.10 m.

solution

F Pat   F  Pat A A F  (1.01 105)(0.080)(0.10)  810 N Since atmospheric pressure acts uniformly in all directions, we don’t usually notice it. The pressure in a fluid acts equally in all direction, and acts at right angles to any surface.

15-2 Pressure

Therefore, if you want to, say, add air to your tires to the manufacturer’s specification, you are not interested in the total pressure. What you are interested in is the gauge pressure – how much more pressure is there in the tire than in the atmosphere? To deal with such situation, we introduce the gauge pressure, Pg, define as follows :

EXAMPLE

15-2

Estimate the gauge pressure in a basketball by pushing down on it and noting the area of contact it makes with the surface in which it rests. Suppose that we push down with a moderate force of 22 N. the circular area of contact will probably have a diameter of about 2 cm.

solution

15-3 Static Equilibrium in Fluids: Pressure and Depth The increased pressure as an object descends through a fluid is due to the increasing mass of the fluid above it.

cosider a cylindrical contianer filled to hieght h with a fluid of density  , The top surface of fluid is open to atmosphere, with a pressure Pat . If the cross - secional area of the conianer is A, the downward force exered on the op surface b he atmosphere is

At the bottom of the container, the downward force is

Fbottom  Ftop  W V  hA M  V W  Mg  Vg  hAg Ftop  Pat A Fbottom  Pat A   (hA) g The pressure at the bottom

Pbottom Pbottom

Fbottom Pat A   (hA) g    Pat  gh A A  Pat  gh

At a depth h below the surface of the fluid, then, the pressure P is given by

If the pressure at one point is P1, the pressure P2 at a depth h below that point is following

EXSERCISE 15-2 The Titanic was found in 1985 lying in the bottom of the North Atlantic at a depth of 2.5 miles. What is the pressure at this depth?

solution

15-3 Static Equilibrium in Fluids: Pressure and Depth A barometer compares the pressure due to the atmosphere to the pressure due to a column of fluid, typically mercury. The mercury column has a vacuum above it, so the only pressure is due to the mercury itself. A fluid that is often used in such a barometer is mercury (Hg), with the density of ρ=1.3595x104 kg∕m3. the corresponding height for a column of mercury is 5

15-3 Static Equilibrium in Fluids: Pressure and Depth This leads to the definition of atmospheric pressure in terms of millimeters of mercury:

In the barometer, the level of mercury is such that the pressure due to the column of mercury is equal to the atmospheric pressure.

Water Seeks Its Own Level :

15-3 Static Equilibrium in Fluids: Pressure and Depth This is true in any container where the fluid can flow freely – the pressure will be the same throughout.

If two different liquid, with different densities, are combined in the U tube, the levels in the arm are not the same. The pressure at the base of each arm must be equal

15-3 Static Equilibrium in Fluids: Pressure and Depth Pascal’s principle: An external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid. Suppose we push down in piston 1 with the force F1. this increases the pressure in that cylinder by the a mount

F1 P  A1

 The pressure in cylinder 2 increases by the same a mount

F2 P   F2  PA2 A2

If we move piston 1 down d1, piston 2 rises a distance d2 therefore

A1d1  A2 d 2 A1 d 2  d1 ( ) A2

EXSERCISE 15-3 To inspect a 14500 N car, it is raised with a hydraulic lift. If the radius of the small piston is 4.0 cm, and the radius of the large piston is 17 cm, find the force that must be exerted in the small piston to lift the car.

solution

15-4 Archimedes’ Principle and Buoyancy We assume that the cubical block is of length L on a side, and that the pressure on the top surface is P1. the downward force on the block, then, is F1 = P1A = P1L2 With a difference depth of h=L: P2 = P1 + ρgL The upward force exerted on the bottom force of the cub is F2 = P2A = (P1+ρgL)L2 =P1L2+ρgL3 = F1+ρgL3 The net vertical force exerted by the fluid on the block-that is, the buoyant force Fb is

Archimeds principle An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object

ď&#x192;&#x2DC;The volume V may be the total volume of the object, or any fraction of the total volume.

Fb= ρwatergV

V, mwater

Vs, mwater

VS Vs, mwater

V, mwater

w object= m object g w water = m water g

=m w0object g m0 =wρobject oV S w water = m water g

all the object is submerged

portion of the object is Submerged

15-5 Applications of Archimedes’ Principle An object floats when it displaces an amount of fluid equal to its weight.

EXAMPLE

15-5

A person who weight 720.0 N in air is lowered into a tank of water to about chin level. He sits in a harness of negligible mass suspend from a scale that read his apparent weight. He now exhales as much air as possible and dunk his head underwater, submerging his entire body. If his apparent weight while submerged is 34.3 N, find (a) His volume (b) His density

solution

Part (a)

Fb   waterV p g W  Wa Wa   waterV p g  W  0  V p   water g W  Wa 720.0  34.3 2 3 Vp    6 . 99  10 m  water g (1.00 103 )(9.81) Part (b)

W   pV p g W 720.0 3 p    1050 kg m V p g (6.99 10  2 )(9.81)

Siris formula

(4950 kg m 3 ) xf   4.50 p xf : the body fat fraction ρp : density of a persons body

If ρp=900 kg∕m3 (all fat) If ρp=1100 kg∕m3 (no fat)

xf = 1 xf = 0

ACTIVE EXAMPLE

15-1

A piece of wood with a density of 706 kg ∕ m is tied with a string to the bottom of a water-filled flask. The wood is completely immersed, and has volume of 8.00 x 10-6 m3. what is the tension of the string?

solution

F

 ma y

ay  0

equilibirum

Fb  T  mwood g  0 T  Fb  mwood g

 wood Wwood

mwood   mwood   woodV  (706)(8.00 10 6 )  5.65 10 3 kg V  mwood g  (5.65 10 3 )(9.81)  5.54  10  2 N

Fb   water gV  (1000)(8.00 10 6 )(9.81)  7.85 10  2 N T  7.85 10  2  5.54  10  2  2.3110  2 N

mwater  water   mwater   waterV V Fb  Wwater  mwater g   waterVg

Fb= ρwatergV

V, mwater

15-5 Applications of Archimedes’ Principle An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water. An object float when it displaces an amount of fluid whose weight is equal to the weight of the object.

15-5 Applications of Archimedes’ Principle The fraction of an object that is submerged when it is floating depends on the densities of the object and of the fluid.

15-6 Fluid Flow and Continuity Continuity tells us that whatever the volume of fluid in a pipe passing a particular point per second, the same volume must pass every other point in a second. The fluid is not accumulating or vanishing along the way. This means that where the pipe is narrower, the flow is faster, as everyone who has played with the spray from a drinking fountain well knows.

15-6 Fluid Flow and Continuity

ρ1 : density of a fluid in the pipe 1 ρ2 : density of a fluid in the pipe 2

: speed of a fluid in the pipe 1 v2 : speed of a fluid in the pipe 2

Most gases are easily compressible; most liquids are not. Therefore, the density of a liquid may be treated as constant, but not that of a gas.

15-7 Bernoulli’s Equation When a fluid moves from a wider area of a pipe to a narrower one, its speed increases; therefore, work has been done on it.

W1  F1x1  P1 A1x1 V1  A1x1 W1  P1V1

, W2   P2 V2

The kinetic energy of a fluid element is:

Wtotal  K 1 1 2 2 ( P1  P2 )V  ( v2  v1 )V 2 2

Equating the work done to the increase in kinetic energy gives:

1 2 P  v 2

15-7 Bernoulli’s Equation If a fluid flows in a pipe of constant diameter, but changes its height, there is also work done on it against the force of gravity. The total work equal to zero (since ∆K = 0) yields

Wtotal  W pressure  Wgravity 0  ( P1  P2 )V  g ( y2  y1 )V Equating the work done with the change in potential energy gives:

15-7 Bernoulli’s Equation The general case, where both height and speed may change, is described by Bernoulli’s equation:

This equation is essentially a statement of conservation of energy in a fluid. Thus, in general, the quantity within the a fluid

1 2 P  v  gy 2

is a constant