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Physics For Scientists And Engineers 6E - part 107

S E C T I O N 1 4 . 2 • Variation of Pressure with Depth

425

Quick Quiz 14.2

The pressure at the bottom of a filled glass of water ($ !

1 000 kg/m

) is P. The water is poured out and the glass is filled with ethyl alcohol

($ ! 806 kg/m

). The pressure at the bottom of the glass is (a) smaller than P

(b) equal to P (c) larger than P (d) indeterminate.

Example 14.2 The Car Lift

The air pressure that produces this force is

This pressure is approximately twice atmospheric pressure.

1.88 " 10

P !

1.48 " 10

(5.00 " 10

1.48 " 10

In a car lift used in a service station, compressed air exerts a
force on a small piston that has a circular cross section and a
radius of 5.00 cm. This pressure is transmitted by a liquid to
a piston that has a radius of 15.0 cm. What force must the
compressed air exert to lift a car weighing 13 300 N? What
air pressure produces this force?

Solution Because the pressure exerted by the compressed air
is transmitted undiminished throughout the liquid, we have

(5.00 " 10

(15.0 " 10

(1.33 " 10

Example 14.3 A Pain in Your Ear

We estimate the surface area of the eardrum to be approxi-
mately 1 cm

1 " 10

. This means that the force on

it is F ! (P

bot

% 5 N. Because a force on the eardrum

of this magnitude is extremely uncomfortable, swimmers of-
ten “pop their ears” while under water, an action that pushes
air from the lungs into the middle ear. Using this technique
equalizes the pressure on the two sides of the eardrum and
relieves the discomfort.

! 4.9 " 10

! (1.00 " 10

kg/m

)(9.80 m/s

)(5.0 m)

bot

Estimate the force exerted on your eardrum due to the wa-
ter above when you are swimming at the bottom of a pool
that is 5.0 m deep.

Solution First, we must find the unbalanced pressure
on the eardrum; then, after estimating the eardrum’s sur-
face area, we can determine the force that the water exerts
on it.

The air inside the middle ear is normally at atmospheric

pressure P

. Therefore, to find the net force on the eardrum,

we must consider the difference between the total pressure
at the bottom of the pool and atmospheric pressure:

Example 14.4 The Force on a Dam

Water is filled to a height H behind a dam of width w (Fig.
14.5). Determine the resultant force exerted by the water on
the dam.

Solution Because  pressure  varies  with  depth,  we  cannot
calculate  the  force  simply  by  multiplying  the  area  by  the
pressure.  We  can  solve  the  problem  by  using  Equation
14.2  to  find  the  force  dF exerted  on  a  narrow  horizontal
strip  at  depth  h and  then  integrating  the  expression  to
find  the  total  force.  Let  us  imagine  a  vertical  y axis,  with
y ! 0 at the bottom of the dam and our strip a distance y
above the bottom.

We can use Equation 14.4 to calculate the pressure at

the depth h; we omit atmospheric pressure because it acts
on both sides of the dam:

P ! $gh ! $g(H & y)

Figure 14.5 (Example 14.4)

Interactive

You can adjust the weight of the truck in Figure 14.4a at the Interactive Worked Example link at http://www.pse6.com.

14.3 Pressure Measurements

During  the  weather  report  on  a  television  news  program,  the  barometric  pressure is
often  provided.  This  is  the  current  pressure  of  the  atmosphere,  which  varies  over
a small  range  from  the  standard  value  provided  earlier.  How  is  this  pressure
measured?

One instrument used to measure atmospheric pressure is the common barometer,

invented by Evangelista Torricelli (1608–1647). A long tube closed at one end is filled
with mercury and then inverted into a dish of mercury (Fig. 14.6a). The closed end of
the tube is nearly a vacuum, so the pressure at the top of the mercury column can be
taken as zero. In Figure 14.6a, the pressure at point A, due to the column of mercury,
must equal the pressure at point B, due to the atmosphere. If this were not the case,
there would be a net force that would move mercury from one point to the other until
equilibrium is established. Therefore, it follows that P

gh, where $

is the den-

sity of the mercury and h is the height of the mercury column. As atmospheric pressure
varies, the height of the mercury column varies, so the height can be calibrated to mea-
sure atmospheric pressure. Let us determine the height of a mercury column for one
atmosphere of pressure, P

1 atm ! 1.013 " 10

Pa:

Based on a calculation such as this, one atmosphere of pressure is defined to be the
pressure equivalent of a column of mercury that is exactly 0.760 0 m in height
at 0°C.

A device for measuring the pressure of a gas contained in a vessel is the open-tube

manometer illustrated in Figure 14.6b. One end of a U-shaped tube containing a liq-
uid is open to the atmosphere, and the other end is connected to a system of unknown
pressure P. The pressures at points A and B must be the same (otherwise, the curved
portion of the liquid would experience a net force and would accelerate), and the
pressure at A is the unknown pressure of the gas. Therefore, equating the unknown
pressure P to the pressure at point B, we see that P ! P

gh. The difference in pres-

sure P & P

is equal to $gh. The pressure P is called the

absolute pressure, while the

difference P & P

is called the

gauge pressure. For example, the pressure you mea-

sure in your bicycle tire is gauge pressure.

9: h !

1.013 " 10

(13.6 " 10

kg/m

)(9.80 m/s

)

0.760 m

426

C H A P T E R 1 4 • Fluid Mechanics

P = 0

(a)

(b)

A B

Figure 14.6 Two devices for mea-

suring pressure: (a) a mercury

barometer and (b) an open-tube

manometer.

Answer We know from Equation 14.4 that the pressure
varies linearly with depth. Thus, the average pressure due to
the water over the face of the dam is the average of the pres-
sure at the top and the pressure at the bottom:

Now, the total force is equal to the average pressure times
the area of the face of the dam:

which is the same result we obtained using calculus.

F ! P

A ! (

gH )(Hw) !

gwH

top

bottom

0 % $gH

Using Equation 14.2, we find that the force exerted on the
shaded strip of area dA ! w dy is

Therefore, the total force on the dam is

Note that the thickness of the dam shown in Figure 14.5 in-
creases with depth. This design accounts for the greater and
greater pressure that the water exerts on the dam at greater
depths.

What If?

What if you were asked to find this force without

using calculus? How could you determine its value?

gwH

F !

P dA !

g(H & y)w dy !

dF ! P dA ! $g(H & y)w dy

14.4 Buoyant Forces and Archimedes’s Principle

Have you ever tried to push a beach ball under water (Fig. 14.7a)? This is extremely
difficult to do because of the large upward force exerted by the water on the ball. The
upward force exerted by a fluid on any immersed object is called a

buoyant force. We

can determine the magnitude of a buoyant force by applying some logic. Imagine a
beach ball&sized parcel of water beneath the water surface, as in Figure 14.7b. Because
this parcel is in equilibrium, there must be an upward force that balances the down-
ward gravitational force on the parcel. This upward force is the buoyant force, and its
magnitude is equal to the weight of the water in the parcel. The buoyant force is the
resultant force due to all forces applied by the fluid surrounding the parcel.

Now imagine replacing the beach ball&sized parcel of water with a beach ball of the

same size. The resultant force applied by the fluid surrounding the beach ball is the same,
regardless of whether it is applied to a beach ball or to a parcel of water. Consequently, we
can claim that

the magnitude of the buoyant force always equals the weight of the

fluid displaced by the object. This statement is known as Archimedes’s principle.

With the beach ball under water, the buoyant force, equal to the weight of a beach

ball-sized parcel of water, is much larger than the weight of the beach ball. Thus, there
is a net upward force of large magnitude—this is why it is so hard to hold the beach ball
under the water. Note that Archimedes’s principle does not refer to the makeup of the
object experiencing the buoyant force. The object’s composition is not a factor in the
buoyant force because the buoyant force is exerted by the fluid.

S E C T I O N 1 4 . 4 • Buoyant Forces and Archimedes’s Principle

427

Quick Quiz 14.3

Several common barometers are built, with a variety of

fluids. For which of the following fluids will the column of fluid in the barometer be
the highest? (a) mercury (b) water (c) ethyl alcohol (d) benzene

Quick Quiz 14.4

You have invented a spacesuit with a straw passing through

the faceplate so that you can drink from a glass while on the surface of a planet. Out
on the surface of the Moon, you attempt to drink through the straw from an open glass
of water. The value of g on the Moon is about one sixth of that on Earth. Compared to
the difficulty in drinking through a straw on Earth, you find drinking through a straw
on the Moon to be (a) easier (b) equally difficult (c) harder (d) impossible.

(a)

(b)

Figure 14.7. (a) A swimmer attempts to push a beach ball underwater. (b) The forces

on a beach ball–sized parcel of water. The buoyant force

B on a beach ball that replaces

this parcel is exactly the same as the buoyant force on the parcel.

Archimedes

Greek Mathematician,
Physicist, and Engineer
(287–212

Archimedes was perhaps the

greatest scientist of antiquity. He

was the first to compute

accurately the ratio of a circle’s

circumference to its diameter,

and he also showed how to

calculate the volume and surface

area of spheres, cylinders, and

other geometric shapes. He is

well known for discovering the

nature of the buoyant force and

was also a gifted inventor. One

of his practical inventions, still in

use today, is Archimedes’s

screw, an inclined, rotating,

coiled tube used originally to lift

water from the holds of ships. He

also invented the catapult and

devised systems of levers,

pulleys, and weights for raising

heavy loads. Such inventions

were successfully used to

defend his native city, Syracuse,

during a two-year siege by

Romans.

To understand the origin of the buoyant force, consider a cube immersed in a liquid

as in Figure 14.8. The pressure P

at the bottom of the cube is greater than the pressure

at the top by an amount $

fluid

gh, where h is the height of the cube and $

fluid

is the den-

sity of the fluid. The pressure at the bottom of the cube causes an upward force equal to
P

A, where A is the area of the bottom face. The pressure at the top of the cube causes a

downward force equal to P

A. The resultant of these two forces is the buoyant force

(14.5)

where V is the volume of the fluid displaced by the cube. Because the product $

fluid

V is

equal to the mass of fluid displaced by the object, we see that

where Mg is the weight of the fluid displaced by the cube. This is consistent with our
initial statement about Archimedes’s principle above, based on the discussion of the
beach ball.

Before we proceed with a few examples, it is instructive for us to discuss two com-

mon situations—a totally submerged object and a floating (partly submerged) object.

Case 1: Totally Submerged Object

When an object is totally submerged in a fluid of den-

sity $

fluid

, the magnitude of the upward buoyant force is B ! $

fluid

gV ! $

fluid

obj

where V

obj

is the volume of the object. If the object has a mass M and density $

obj

its weight is equal to F

Mg ! $

obj

, and the net force on it is B & F

fluid

obj

)gV

obj

. Hence, if the density of the object is less than the density of the

fluid, then the downward gravitational force is less than the buoyant force, and the un-
supported object accelerates upward (Fig. 14.9a). If the density of the object is greater
than the density of the fluid, then the upward buoyant force is less than the downward
gravitational force, and the unsupported object sinks (Fig. 14.9b). If the density of the
submerged object equals the density of the fluid, the net force on the object is zero and
it remains in equilibrium. Thus,

the direction of motion of an object submerged in

a fluid is determined only by the densities of the object and the fluid.

Case 2: Floating Object

Now consider an object of volume V

obj

and density $

obj

(

fluid

static equilibrium floating on the surface of a fluid—that is, an object that is only partially
submerged (Fig. 14.10). In this case, the upward buoyant force is balanced by the down-
ward gravitational force acting on the object. If V

fluid

is the volume of the fluid displaced

by the object (this volume is the same as the volume of that part of the object that is
beneath the surface of the fluid), the buoyant force has a magnitude B ! $

fluid

Because the weight of the object is F

Mg ! $

obj

, and because F

B, we see that

fluid

obj

, or

(14.6)

fluid

obj

fluid

B ! Mg

B ! (P

)A ! ($

fluid

gh)A ! $

fluid

428

C H A P T E R 1 4 • Fluid Mechanics

(a)

(b)

Active Figure 14.9 (a) A totally submerged object that is less dense than the fluid in

which it is submerged experiences a net upward force. (b) A totally submerged object

that is denser than the fluid experiences a net downward force and sinks.

▲

PITFALL PREVENTION

14.2 Buoyant Force is

Exerted by the Fluid

Remember that

the buoyant

force is exerted by the fluid. It
is  not  determined  by  properties
of  the  object,  except  for  the
amount of fluid displaced by the
object. Thus, if several objects of
different  densities  but  the  same
volume  are  immersed  in  a  fluid,
they will all experience the same
buoyant force. Whether they sink
or float will be determined by the
relationship between the buoyant
force and the weight.

Figure 14.8 The external forces

acting on the cube of liquid are the

gravitational force

and the buoy-

ant force

B. Under equilibrium

conditions, B ! F

At the Active Figures link

at http://www.pse6.com, you

can move the object to new po-

sitions as well as change the

density of the object to see the

results.

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