Chapter 13 - Solutions
Description: Find the weight of a cylindrical iron rod given its area and length and the density of iron.
Part A
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800
92.1
rod,
Hint
A.1
and diameter 2.75
, length
from a storage room to a machinist. Calculate the weight of the
. The acceleration due to gravity,
= 9.81
.
General approach
To calculate the weight of a body you need to know its mass. If the mass is unknown, and instead
the problem gives you information on the volume of the body and the density of the material of
which it is made, you can calculate the mass of that body by applying the definition of density.
Hint
A.2
Weight of a body
A body of mass
has weight with magnitude
given by
,
where
= 9.81
is the acceleration due to gravity.
Hint
A.3
Find the mass of the rod
From the length and the cross-sectional area of the rod and the density of iron, find the mass
the rod.
Hint
A.3.1
Finding mass from volume and density
Consider a mass
of homogeneous material, whose volume is
material is defined as
. The density
.
Hint
A.3.2
Volume of a cylinder
The volume
of a cylinder of length
If the cross section is a circle with radius
and cross-sectional area
, then
Express your answer numerically in kilograms.
ANSWER:
=
is given by
of that
of
If the cross section is a circle with radius
, then
Express your answer numerically in kilograms.
ANSWER:
=
Express your answer numerically in newtons.
ANSWER:
=
Part B
Now that you know the weight of the rod, do you think that you will be able to carry the rod without a
cart?
Hint
B.1
Weight and mass
A weight measured in newtons may not be very familiar to you; therefore it is more convenient to
determine the corresponding mass measured in kilograms, or in pounds.
Hint
B.2
Unit conversion factors
The unit convertion factors for newtons, kilograms, and pounds are
,
.
ANSWER:
yes
no
± Volume of Copper
Description: ± Includes Math Remediation. Calculate the volume of a given number of moles of
copper.
Part A
What is the volume
of a sample of 3.70
of copper? The atomic mass of copper (Cu) is 63.5
, and the density of copper is
Hint
A.1
.
How to approach the problem
Since the number of moles of copper is known, calculate its mass, and then use the mass and the
density of copper to find the volume of the copper.
Hint
A.2
What is the mass of the copper?
Calculate the mass
of 3.70
Express your answer in grams.
of copper.
ANSWER:
=
Hint
A.3
Using the density in calculations
Recall that density is defined as the ratio of the mass to volume:
the units, since the volume should be expressed in cubic centimeters.
. Be careful with
Express your answer in cubic centimeters to 3 significant figures.
ANSWER:
=
Pressure in the Ocean
Description: Short conceptual problem on hydrostatic pressure at various depths in the ocean. This
problem is based on Young/Geller Conceptual Analysis 13.2
The pressure at 10
below the surface of the ocean is about 2.00×105
.
Part A
Which of the following statements is true?
Hint A.1
How to approach the problem
Since pressure is defined as force per unit area, the pressure at a given depth below the surface of
the ocean is the normal force on a small area at that depth divided by that area. Thus, given the
pressure at 10
you can determine the normal force on a unit area at that depth. Also recall that,
in general, the pressure in a fluid varies with height, but it also depends on the external pressure
applied to the fluid. In the case of the ocean, the external pressure applied at its surface is the
atmospheric pressure.
ANSWER:
The weight of a column of seawater 1
in cross section and 10
high
pressure at 10
you can determine the normal force on a unit area at that depth. Also recall that,
in general, the pressure in a fluid varies with height, but it also depends on the external pressure
applied to the fluid. In the case of the ocean, the external pressure applied at its surface is the
atmospheric pressure.
ANSWER:
The weight of a column of seawater 1
is about 2.00×105
in cross section and 10
high
.
The weight of a column of seawater 1
in cross section and 10
high
plus the weight of a column of air with the same cross section extending up
to the top of the atmosphere is about 2.00×105
The weight of 1
of seawater at 10
.
below the surface of the ocean
is about 2.00×105
.
The density of seawater is about 2.00×105 times the density of air at sea
level.
The pressure at a given level in a fluid is caused by the weight of the overlying fluid plus any external
pressure applied to the fluid. In this case, the external pressure is the atmospheric pressure. Tropical
storms are formed by low-pressure systems in the atmosphere. When such a storm forms over the
ocean, both the atmospheric pressure and the fluid pressure in the ocean decrease.
Part B
Now consider the pressure 20
is true?
Hint B.1
below the surface of the ocean. Which of the following statements
Variation of pressure with height
If the pressure at a point in a fluid is
the fluid is
, then, at a distance
below this point, the pressure
in
,
where
is the density of the fluid and
pressure at a depth
is the acceleration due to gravity. This means that the
below a reference height of pressure
. Note that the quantity
is greater than
is related to the weight of a column of fluid of height
in cross section.
ANSWER:
by an amount
The pressure is twice that at a depth of 10
.
The pressure is the same as that at a depth of 10
.
and 1
The pressure is equal to that at a depth of 10
plus the weight per 1
cross sectional area of a column of seawater 10
The pressure is equal to the weight per 1
column of seawater 20
high.
cross sectional area of a
high.
Tactics Box 13.1 Hydrostatics
Description: Knight/Jones/Field Tactics Box 13.1 Hydrostatics is illustrated.
Learning Goal: To practice Tactics Box 13.1 Hydrostatics.
In problems about liquids in hydrostatic equilibrium, you often need to find the pressure at some point
in the liquid. This Tactics Box outlines a set of rules for thinking about such hydrostatic problems.
Hydrostatics
Draw a picture. Show open surfaces, pistons, boundaries, and other features that affect
pressure. Include height and area measurements and fluid densities. Identify the points at which
you need to find the pressure. These objects make up the system; the environment is everything
else.
TACTICS BOX 13.1
1.
2.
Determine the pressure
at surfaces.

Surface open to the air:

Surface in contact with a gas:

Closed surface:
, usually 1
.
.
, where
is the force the surface, such as a piston,
3.
exerts on the fluid and
is the area of the surface.
Use horizontal lines. The pressure in a connected fluid (of one kind) is the same at any point
along a horizontal line.
4.
Allow for gauge pressure. Pressure gauges read
5.
Use the hydrostatic pressure equation:
.
, where
is the density of the
fluid, is the acceleration due to gravity, and is the height of the fluid.
Use these rules to work out the following problem: A U-shaped tube is connected to a box at one end
and open to the air at the other end. The box is full of gas at pressure
mercury of density 1.36×104
mercury column is
= 16.0
, and the tube is filled with
. When the liquid in the tube reaches static equilibrium, the
high in the left arm and
= 6.00
high in the right arm, as
shown in the figure.
What is the gas pressure
inside the box?
Part A
Find the pressures
and
atmospheric pressure.
at surfaces A and B in the tube, respectively. Use
Express your answers, separated by a comma, in terms of one or both of the variables
to denote
and
.
ANSWER:
,
=
Part B
As stated in rule 3 in the Tactics Box, it is always convenient to use horizontal lines in hydrostatic
problems. In each one of the following sketches, a different horizontal line is considered. Which
sketch would be more useful in solving the problem of finding the gas pressure?
ANSWER:
ANSWER:
It is useful to draw a horizontal line as in this sketch because it allows you to relate pressure in the
mercury at critical points in the tube. For example, even though you are not given the pressure at
point C, you know that
, because point C is at the same height as any point on
surface B. At the same time, you can also use the hydrostatic pressure equation to relate the pressure
at point C to that at surface A, where pressure is determined by the gas pressure in the box.
Part C
Assume
Hint
C.1
. What is the gas pressure
?
Conversion factor between atmospheres and pascals
Recall that
.
Hint
C.2
Find the height of the liquid column above point C
What is the height
of the mercury column above point C?
Express your answer in meters.
ANSWER:
=
Express your answer in pascals to three significant figures.
ANSWER:
=
ANSWER:
=
A Submerged Ball
Description: Quantitative question involving the application of Archimedes' principle.
A ball of mass
and volume
is lowered on a string into a fluid of density
.
Assume that the object
would sink to the bottom if it were not supported by the string.
Part A
What is the tension
shown in the figure?
Hint
A.1
in the string when the ball is fully submerged but not touching the bottom, as
Equilibrium condition
Although the fact may be obscured by the presence of a liquid, the basic condition for equilibrium
still holds: The net force on the ball must be zero. Draw a free-body diagram and proceed from
there.
Hint
A.2
Find
Hint
A.2.1
Find the magnitude of the buoyant force
, the magnitude of the buoyant force.
Archimedes' principle
Quantitatively, the buoyant force can be found as
,
Hint
A.2.1
Archimedes' principle
Quantitatively, the buoyant force can be found as
,
where
is the force,
acceleration due to gravity, and
Hint
A.2.2
is the density of the fluid,
is the magnitude of the
is the volume of the displaced fluid.
Find the mass of the displaced fluid
Compute the mass
of the fluid displaced by the object when it is entirely submerged.
Express your answer in terms of any or all of the variables
,
,
, and
.
ANSWER:
=
Express your answer in terms of any or all of the variables
,
,
, and
.
ANSWER:
=
Express your answer in terms of any or all of the given quantities and
acceleration due to gravity.
, the magnitude of the
ANSWER:
=
± Buoyant Force Conceptual Question
Description: ± Includes Math Remediation. Conceptual question on the properties of a floating wooden
block.
A rectangular wooden block of weight
floats with exactly one-half of its volume below the
waterline.
Part A
What is the buoyant force acting on the block?
Hint
A.1
Archimedes' principle
The upward buoyant force on a floating (or submerged) object is equal to the weight of the liquid
displaced by the object.
Mathematically, the buoyant force
on a floating (or submerged) object is
,
where is the density of the fluid,
acceleration due to gravity.
Hint
A.2
is the submerged volume of the object, and
What happens at equilibrium
The block is in equilibrium, so the net force acting on it is equal to zero.
ANSWER:
The buoyant force cannot be determined.
is the
Part B
The density of water is 1.00
Hint
B.1
. What is the density of the block?
Applying Archimedes' principle
In Part A, you determined that the buoyant force, and hence the weight of the water displaced, is
equal to the weight of the block. Notice, however, that the volume of the water displaced is onehalf of the volume of the block.
Hint
B.2
Density
The density
of a material of mass
and volume
is
.
ANSWER:
2.00
between 1.00 and 2.00
1.00
between 0.50 and 1.00
0.50
The density cannot be determined.
Part C
Masses are stacked on top of the block until the top of the block is level with the waterline. This
requires 20
Hint
C.1
of mass. What is the mass of the wooden block?
How to approach the problem
When you add the extra mass on top of the block, the buoyant force must change. Relating the
mass to the new buoyant force will allow you to write an equation to solve for
Hint
C.2
.
Find the new buoyant force
With the 20- mass on the block, twice as much of the block is underwater. Therefore, what
happens to the buoyant force on the block?
ANSWER:
The buoyant force doubles.
The buoyant force is halved.
The buoyant force doesn't change.
Hint
ANSWER:
The buoyant force doubles.
The buoyant force is halved.
The buoyant force doesn't change.
Hint
C.3
System mass
Before the 20-
mass is placed on the block, the mass of the system is just the mass of the block (
), and this mass is supported by the buoyant force. With the 20of the system is now
Hint
C.4
mass on top, the total mass
.
The mass equation
The original buoyant force was equal to the weight
is twice the old buoyant force, then
of the block. If the new buoyant force
,
where
is the mass of the block.
The new buoyant force must support the weight of the block and the mass, so according to
Newton's 2nd law
.
By setting these two expressions for
ANSWER:
equal to each other, you can solve for
.
40
20
10
Part D
The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid,
only one-third of the wooden block is submerged. Is the unknown liquid more or less dense than
water?
ANSWER:
more dense
less dense
Part E
What is the density of the unknown liquid
Hint
E.1
?
Comparing densities
From Part C, the mass of the block is 20
. In water, one-half of the block is submerged, so one-
half of the volume of the block times the density of the water must be equivalent to 20 . In the
unknown substance, one-third of the block is submerged, so one-third of the volume of the block
times the density of the unknown substance must be equivalent to 20 . You can use these
observations to write an equation for the density of the unknown substance.
Hint
E.2
Setting up the equation
The buoyant force on the block in water
unknown liquid
is the same as the buoyant force in the
,
,
so
,
where
is the density of water,
volume of the entire block, and
is the density of the unknown liquid,
is the
is the acceleration due to gravity.
Express your answer numerically in grams per cubic centimeter.
ANSWER:
=
Understanding Bernoulli's Equation
Description: Conceptual questions on Bernoulli's equation applied first to a horizontal converging flow
tube and then to an elevated one.
Bernoulli's equation is a simple relation that can give useful insight into the balance among fluid
pressure, flow speed, and elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids
with a constant density and no internal friction forces, whose flow patterns do not change with time.
Despite its limitations, however, Bernoulli's equation is an essential tool in understanding the behavior
of fluids in many practical applications, from plumbing systems to the flight of airplanes.
For a fluid element of density
that flows along a streamline, Bernoulli's equation states that
,
where
is the pressure,
is the flow speed,
is the height,
is the acceleration due to gravity, and
subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli's
equation becomes clearer if we rearrange the terms of the equation as follows:
.
The term
on the left-hand side represents the total work done on a unit volume of fluid by
the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The
two terms on the right-hand side represent, respectively, the change in potential energy,
, and the change in kinetic energy,
, of the unit volume during its
flow from point 1 to point 2. In other words, Bernoulli's equation states that the work done on a unit
volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic
energy per unit volume that occurs during the flow. This is nothing more than the statement of
conservation of mechanical energy for an ideal fluid flowing along a streamline.
Part A
Consider the portion of a flow tube shown in the figure.
Point 1 and point 2
are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point
2. If the cross section of the flow tube at point 1 is greater than that at point 2, what can you say about
the pressure at point 2?
Hint
A.1
How to approach the problem
Apply Bernoulli's equation to point 1 and to point 2. Since the points are both at the same height,
their elevations cancel out in the equation and you are left with a relation between pressure and
flow speeds. Even though the problem does not give direct information on the flow speed along the
flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward
point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at
point 2 is greater than or smaller than the flow speed at point 1. With that information and
Bernoulli's equation, you will be able to determine the pressure at point 2 with respect to the
pressure at point 1.
Hint
A.2
Apply Bernoulli's equation
Apply Bernoulli's equation to point 1 and to point 2 to complete the expression below. Here
and
point 2 is greater than or smaller than the flow speed at point 1. With that information and
Bernoulli's equation, you will be able to determine the pressure at point 2 with respect to the
pressure at point 1.
Hint
A.2
Apply Bernoulli's equation
Apply Bernoulli's equation to point 1 and to point 2 to complete the expression below. Here
and
are the pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2.
Also, use
Hint
A.2.1
for elevation with the appropriate subscript, and use
for the density of the fluid.
Flow along a horizontal streamline
Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In
other words, when applying Bernoulli's equation to any two points of the streamline,
and they cancel out.
Express your answer in terms of some or all of the variables
.
,
,
,
,
,
, and
ANSWER:
=
Along a horizontal streamline, Bernoulli's equation reduces to
,
or
.
Since the density of the fluid is always a positive quantity, the sign of
of
Hint A.3
, allowing you to determine whether
Determine
gives you the sign
is greater than or smaller than
.
with respect to
By applying the continuity equation, determine which of the following is true.
Hint A.3.1
The continuity equation
The continuity equation expresses conservation of mass for incompressible fluids flowing in a
tube. It says that the amount of fluid
time interval
flowing through a cross section
of the tube in a
must be the same for all cross sections, or
.
Therefore, the flow speed must increase when the cross section of the flow tube decreases, and
vice versa.
time interval
must be the same for all cross sections, or
.
Therefore, the flow speed must increase when the cross section of the flow tube decreases, and
vice versa.
ANSWER:
ANSWER:
lower than the pressure at point 1.
The pressure at point 2 is
equal to the pressure at point 1.
higher than the pressure at point 1.
Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of
an ideal fluid.When the cross section of the flow tube decreases, the flow speed increases, and
therefore the pressure decreases. In other words, if
, then
and
.
Part B
As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows
through a flow tube with decreasing cross section moves toward a region of lower pressure.
Physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the
fluid element as a net force that causes the fluid to __________.
Hint B.1
Effects from conservation of mass
Recall that, if the cross section
of the flow tube varies, the flow speed must change to
conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid
to increase or decrease speed depending on whether the fluid is flowing through a portion of the
tube with a smaller or larger cross section.
ANSWER:
decrease in speed
increase in speed
remain in equilibrium
Part C
Now assume that point 2 is at height
with respect to point 1, as shown in the figure.
The ends of the flow
tube have the same areas as the ends of the horizontal flow tube shown in Part A. Since the cross
section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward
point 2 from point 1 moves toward a region of lower pressure. In this case, what is the pressure drop
experienced by the fluid element?
Hint C.1
How to approach the problem
Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you
must take into account the difference in elevation between points 1 and 2. Do you need to add this
additional term to the other term representing the pressure drop between the two ends of the flow
tube or do you subtract it?
ANSWER:
The pressure drop
is
smaller than the pressure drop occurring in a purely
horizontal flow.
equal to the pressure drop occurring in a purely
horizontal flow.
larger than the pressure drop occurring in a purely
horizontal flow.
Part D
From a physical point of view, how do you explain the fact that the pressure drop at the ends of the
elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely
horizontal flow from Part A?
Hint D.1
Physical meaning of the pressure drop in a tube
As explained in the introduction, the difference in pressure
between the ends of a flow
tube represents the total work done on a unit volume of fluid by the pressure forces of the
surrounding fluid to move that volume of fluid from one end to the other end of the flow tube.
ANSWER:
A greater amount of work is needed
to balance the
increase in potential energy from the
elevation change.
decrease in potential energy from the
elevation change.
larger increase in kinetic energy.
surrounding fluid to move that volume of fluid from one end to the other end of the flow tube.
ANSWER:
A greater amount of work is needed
to balance the
increase in potential energy from the
elevation change.
decrease in potential energy from the
elevation change.
larger increase in kinetic energy.
larger decrease in kinetic energy.
In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube
had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an
elevated flow tube, the difference in pressure must also balance the increase in potential energy of the
fluid; therefore a higher pressure is needed for the flow to occur.
Flow Velocity of Blood Conceptual Question
Description: Conceptual question on plaque changing the effective cross sectional area and blood
pressure of an artery.
Arteriosclerotic plaques forming on the inner walls of arteries can decrease the effective cross-sectional
area of an artery. Even small changes in the effective area of an artery can lead to very large changes in
the blood pressure in the artery and possibly to the collapse of the blood vessel.
Imagine a healthy artery, with blood flow velocity of
and mass per unit volume of
. The kinetic energy per unit volume of blood is given by
Imagine that plaque has narrowed an artery to one-fifth of its normal cross-sectional area (an 80%
blockage).
Part A
Compared to normal blood flow velocity,
blockage?
Hint A.1
, what is the velocity of blood as it passes through this
Continuitity equation and reduced cross-sectional area
By the equation of continuity, as the cross-sectional area of an artery decreases because of plaque
formation, the velocity of blood through that region of the artery will increase. The new flow speed
can be calculated by rearranging the equation of continuity,
;
so
where
and
are the initial and final cross-sectional areas, and
and final velocities of the blood, respectively.
ANSWER:
and
are the initial
ANSWER:
Part B
By what factor does the kinetic energy per unit of blood volume change as the blood passes through
this blockage?
ANSWER:
25
5
1
Part C
As the blood passes through this blockage, what happens to the blood pressure?
Hint
C.1
Blood pressure and blood velocity
Bernoulli's equation states that the sum of the pressure, the kinetic energy per volume, and the
gravitational energy per volume of a fluid is constant. For initial and final pressures
initial and final velocities
and
, and mass per unit volume of blood,
of changes in gravitational energy leads to
and
,
, ignoring the effects
.
Basically, the sum of kinetic energy and pressure must remain constant in an artery. This leads to a
very serious health risk. As blood velocity increases, blood pressure in a section of artery can drop
to a dangerously low level, and the blood vessel can collapse, completely cutting off blood flow,
owing to lack of sufficient internal pressure.
Hint
C.2
Calculating the change in blood pressure
From Bernoulli's equation, the change in pressure is the negative of the change in kinetic energy
per unit volume. For initial and final kinetic energies per unit volume of the blood,
respectively,
,
or
,
where
. Rearranging this equation yields
and
,
,
or
,
where
. Rearranging this equation yields
or
.
ANSWER:
It increases by about 250
It increases by about 41
.
.
It stays the same.
It decreases by about 41
It decreases by about 250
.
.
Since the kinetic energy increases by a factor of 25,
.
Bernoulli's equation tells you that
.
As the blood velocity increases through a blockage, the blood pressure in that section of the artery can
drop to a dangerously low level. In extreme cases, the blood vessel can collapse, completely cutting
off blood flow, owing to lack of sufficient internal pressure. In the next three parts, you will see how
a small increase in blockage can cause a much larger pressure change.
For parts D - F imagine that plaque has grown to a 90% blockage.
Part D
Relative to its initial, healthy state, by what factor does the velocity of blood increase as the blood
passes through this blockage?
Express your answer numerically.
ANSWER:
Part E
By what factor does the kinetic energy per unit of blood volume increase as the blood passes through
this blockage?
Express your answer numerically.
ANSWER:
Express your answer numerically.
ANSWER:
Part F
What is the magnitude of the drop in blood pressure,
, as the blood passes through this blockage?
Use
as the normal (i.e., unblocked) kinetic energy per unit volume of the blood.
Express your answer in pascals using two significant figures.
ANSWER:
=
Video Tutor: Air Jet Blows between Bowling Balls
Description: An air jet is directed between a pair of bowling balls hanging close to each other. How do
the balls respond?
First, launch the video below. You will be asked to use your knowledge of physics to predict the
outcome of an experiment. Then, close the video window and answer the question on the right. You can
watch the video again at any point.
Part A
A strong wind blows over the house shown in the figure. The wind is much stronger over the house's
roof than lower down, and the house has an open chimney. A window on the ground floor is open,
and so are the doors inside the house. Which way will air flow through the house?
Hint A.1
How to approach the problem
This problem asks you to apply what you've learned about pressure and fluid flow to a new
situation. Is the pressure above the chimney lower or higher than that at the window? Remember
that the wind is much faster at the level of the chimney than at the level of the open window.
Wind will flow from a region of higher pressure to a region of lower pressure. Use this idea to
determine the direction of air flow.
ANSWER:
In the window and out the chimney
In no direction; there is no air flow through the house.
In the chimney and out the window
Air pressure is lower at the chimney and higher outside the window, so air will flow in through the
window and out through the chimney.
Video Tutor: Pressure in Water and Alcohol
Description: Pressure is measured at the bottom of two equal-height columns of liquid, one consisting
of water and the other of alcohol. How do the pressures compare?
First, launch the video below. You will be asked to use your knowledge of physics to predict the
outcome of an experiment. Then, close the video window and answer the question on the right. You can
watch the video again at any point.
Part A
Rank, from smallest to largest, the pressures in the tank of motionless fluid shown in the figure.
Hint
A.1
How to approach the problem
Recall that the pressure at any point in an open container of fluid is due to the weight of the
overlying fluid plus the weight of the overlying air.
Also, think about the fact that this fluid is motionless. What does that tell you about the pressure at
any two points on the same horizontal level, such as points B and C?
To rank items as equivalent, overlap them.
ANSWER:
Pressure depends on vertical distance from the fluid surface. As we descend deeper in the fluid, the
pressure increases.
View
Pressure depends on vertical distance from the fluid surface. As we descend deeper in the fluid, the
pressure increases.
Video Tutor: Weighing Weights in Water
Description: Steel and aluminum cylinders of identical size and shape are attached to spring scales and
lowered into water. Which, if either, shows a greater reduction in weight?
First, launch the video below. You will be asked to use your knowledge of physics to predict the
outcome of an experiment. Then, close the video window and answer the question on the right. You can
watch the video again at any point.
Part A
Suppose that we repeat the experiment shown in the video, but we replace one of the cylinders with a
cylinder that has twice the radius (and use larger containers of water). If the height of the original
cylinder is
the video?
Hint A.1
, how deeply must we submerge the new cylinder to get the same weight reduction as in
How to approach the problem
How does the reduction in the cylinder’s weight relate to the amount of water the cylinder
displaces?
How does the volume of a cylinder depend on the cylinder’s radius and height?
ANSWER:
2
To get the same reduction in weight, the same volume of cylinder must be submerged. The volume of
a cylinder is
=
, so if we double the radius, we must reduce the depth of immersion by a
4
To get the same reduction in weight, the same volume of cylinder must be submerged. The volume of
a cylinder is
factor of four.
=
, so if we double the radius, we must reduce the depth of immersion by a
Q13.33. Reason: Since the question talks about the “extra pressure” we will ignore the air pressure above the
water; you have an equal amount of air pressure on the inside. The 7 N quoted is the “increased pressure.”
Therefore the equation we need is Equation 13.5 without the po term,
where we want to solve for
We will also use
and round g to
for one significant figure accuracy.
The correct answer is D.
Assess: This one significant figure calculation can easily be done in the head without a calculator. The answer
seems plausible, and the other choices seem too small. The units cancel appropriately to leave the answer in m.
Q13.35. Reason: We’ll assume the ball is not being forcibly held under the water (because then the net force
would be zero). A free-body diagram shows only two forces on the ball: the downward force of gravity
and the upward buoyant force, whose magnitude equals the weight of the displaced fluid (the weight
of
of water).
The downward gravitational force has a magnitude of
The upward buoyant force has a magnitude of
Subtracting gives the magnitude of the net force,
The correct answer is B.
Assess: This is enough force to cause the basketball to accelerate upward fairly impressively, as you have
probably seen in the swimming pool (if you haven’t, have your teacher assign you a field trip to the pool to try
this out).
Q13.37. Reason: We’ll use the equation of continuity, Equation 13.12, and make all the assumptions inherent
in that equation (no leakage, etc.). We’ll also use
for v1.
Call the plunger point 1 and the nozzle point 2, and solve
The correct answer is B.
Assess: The answer seems to be a reasonable slow and steady pace. The ratio of the speeds is the square of the
ratio of the radii. The last step (the computation) can easily be done without a calculator.
Q13.38. Reason: The question says to ignore viscosity, so we do not need Poiseuille’s equation; Bernoulli’s
equation should suffice. And because the pipe is horizontal we can drop the
same on both sides).
terms (because they will be the
We are given
and
We will also use the equation of continuity to solve for v2.
Putting it all together
The magnitude of this is 12,700 Pa, so the correct answer is D.
Assess: Since Pa = N/m2 the units work out.
Problems
P13.7. Prepare: Vf is the volume of the fish, Va is the volume of air, and Vw is the volume of water displaced
by the fish + air system. The mass of the fish is mf, the mass of the air is ma, and the mass of the water displaced
by the fish is mw.
Solve: Neutral buoyancy implies static equilibrium, which will happen when
Assess: Increase in volume of 8.0% seems very reasonable.
P13.8. Prepare: The density of sea water is 1030 kg/m3. Also,
Solve: The pressure below sea level can be found from Equation 13.5 as follows:
Assess: The pressure deep in the ocean is very large.
This means
P13.14. Prepare: The pressure required will be the atmospheric pressure exerted by the air on the water in the
container subtracted by the pressure exerted by the column of water.
Solve:
Assess: The principle here is similar to the principle in a barometer.
P13.30. Prepare: The pipe is a flow tube, so the equation of continuity, Equation 13.12, applies. To work with
SI units, we need the conversion 1 L = 10–3 m3.
Solve: The volume flow rate is
Using the definition Q = vA, we get
Assess: This is a reasonable speed for water flowing through a 2.0-cm pipe.
P13.55. Prepare: Knowing the volume flow rate, the number of vessels and the diameter of each vessel, we
can use Equation 13.13 to solve the problem. We can either say that each vessel handles 1/2000 of the volume
flow rate or we can say the total area for this flow rate is 2000 times the area of each vessel. Either way we will
get the same result.
Solve: Starting with the definition of volume flow rate and the knowledge that each vessel handles 1/2000 of
this volume, we may write
Solving for v
Assess: At this rate, water could travel to the top of an 18 m tree in a day. That seems reasonable.
P13.61. Prepare: Please refer to Figure P13.61. Assume the gas in the manometer is an ideal gas. Assume that
a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the
gas cell. This means the volume of the chamber can be considered constant.
Solve: In the ice-water mixture the pressure is
p1 = patmos + ρHgg (0.120 m) = 1.013 × 105 Pa + (13,600 kg/m3)(9.8 m/s2)(0.120 m) = 1.173 × 105 Pa
In the freezer the pressure is
p2 = patm + ρHgg (0.030 m) = 1.013 × 105 Pa + (13,600 kg/m3)(9.8 m/s2)(0.030 m) = 1.053 × 105 Pa
For constant volume conditions, the gas equation is
Assess: This is a reasonable temperature for an industrial freezer.