Download Velocity and Acceleration

Velocity and
Acceleration
motion (MO-shlm) n.: the displacemerlt of an object
in relntion to objects that are considered to be
stationary.
VELOCITY
OBJECTIVES
. Define displacement, speed.
velocity, and acceleration.
. Solve problems dealing with
.
accelerated motion.
Discuss and apply Newton's
three laws of motion.
. Discuss and apply Newton's
of universal gravitation.
411
law
3.1 The Nature of Motion
If you see a car in front of
your house and later see it farther along the street, you are
correct in assuming that the car has moved. To reach this
conclusion, you observed two positions of the car and you
also noted the passage of time. You might not know how
the car got from one position to the other. It might have
moved at a steady rate or it might have speeded up and
then slowed down before it got to its second position. It
might also have been moving when you first noticed it and
might still have been moving at its second location. The
car may have moved from one location to the other in a
straight line, or it may not have. But none of these possibilities changes the truth of the statement that the car has
moved.
Usually, when we see something move, we do not just
make two observations.
Unless we are only concerned
with the beginning and end of an object's motion (for example, in determining average speed), we usually observe
the object moving continuously. However, even a case of
continuous observations can be broken down into a series
of shorter observations for any two points along an object's trajectory.
Figure 3-1 shows several important things about the nature of motion. First, it shows the changing positions of a
moving object. Such a change of position in a particular direction is called a displacement. A displacement has magni-
,
,i
f'
VELOCITY AND ACCELERATION
41
~
.
~
p
~
,
~
~
~
r
tude and direction and is, therefore, a vector quantity. A
displacement is always a straight line segment from one
point to another point, even though the actual path of the
moving object between the two points is curved. Displacement vectors can be combined like other vector quantities.
For example, the statement that an airplane flew 500
kilometers south describes the airplane's displacement,
since the statement specifies the magnitude and direction
of the airplane's motion. Whether the airpJane flew in a
straight line or not, the airplane's displacement vector is a
straight arrow directed south and representing a magnitude of 500 kilometers.
Figure 3-1 also shows that relative motion can be detected by comparing the displacement of one object with
respect to another. The reference object is usually considered to be stationary. Thus in the case of the moving car,
the street and houses are the stationary references. So
motion may be defined as the displacement of an object in relation to objects that are considered to be stationary.
Figure 3-1. A study of motion. The
disk was photographed every 0.1 s
as it moved from left to right
along the meterstick. Note that
the displacement of the disk is
equal in equal time periods.
Displacement is a change of position.
Motion is relative displacement.
Speed is the time rate of motion.
10
9
in Chapter
5.
3.2 Speed The time rate of motion is called speed. For example, if you are walking at a speed of 3.0 mIs, in 1.0 second
you travel 3.0 meters. In 2.0 seconds, you travel 6.0 meters, and in 3.0 seconds you travel 9.0 meters. A graph of
this motion, and of several other motions with constant
speeds, is shown in Figure 3-2. Notice that the lines that
represent greater speeds are more steeply inclined. The
slope ofa line indicates how steep the line is. The slope is
the ratio of the vertical component of a line to the corresponding horizontal component. For example, the blue
segment in line A has a vertical component of 3.0 units
and a horizontal component of 1.0 unit. The slope of the
.
.
ImelS-=
3.0
1.0
3 .0.
This definition of the slope ofa line on a graph may be
used to represent
a physical quantity such as speed. If the
8
8
This chapter deals with the simplest type of motionthe motion of an object along a straight line. Motion in a
curved line will be discussed
A
71-
,
.'05 '
o
~
o
4
3
c
2.D
-~~
o
1
I
2
I
3
I
4
3.0
5
,
Elapsedtime.s
Figure 3-2. Three graphs of motion with different constant
speeds. A is for 3.0 mis, B is for
1.0 mis, and C is for 0.33 m/s. In
each case, speed is represented
by the slope of the line.
I
7
I
8
,
.
CHAPTER
42
vertical axis represents
distance
and the horizontal
3
axis
represents elapsed time, then the slope of the line represents speed. For each of the speeds graphed in Figure 3-2
the data for computing the slope are given. A graph such
as this and a single computation of the slope will deter~
mine the speed over the entire motion only if the speed is
constant. If the determination of speed over any randomly chosen intervalgives the same value, the speed is constant. The
graph of constant speed is a straight line.
Average speed is found by dividing the total distance by the
elapsed time. For example, if you run 112 m in 20.0 5, you
might not run at constant speed. You might speed up and
slow down. But as long as you move a total distance of
112m
112 m in 20.0 s, your average speed is
= 5.60 m/s.
20.0 s
Suppose you ride a bicycle 120 m in 2'0 s. Your average
120 m
speed is
_
= 6.0 m/s. A graph of your speed might
20 s
look like Figure 3-3. You start at a speed of 0.0 mls and
increase your speed. Then from about 3 s to 15 s you
travel at constant speed. Then at about 16 s you begin to
tire and slow down until at 2'0 s you stop. Instantaneous
"0
100
90
,
80
. '"
o
B 60
is
50'40
speed is the slope of the line that is tangent to the curve at a given
point. Without a speedometer it is difficult to determine
30
20
10
20
Elapsed1ime,'
Figure 3-3. A graph of motion
variable speed.
with
your instantaneous
speed. It is the speed that you would
have if your speed did not change from that point on.
Instantaneous speed can be measured approximately by
timing the motion over a short distance. In Figure 3-3 instantaneous speed is found approximately by dividing a
small amount of time, measured on the graph, into the
corresponding
distance for that time. The region of the
curve marked A represents your change in speed as you
move faster. An interval of time of 2.3 s is indicated on the
horizontal axis. The corresponding
distance for this time
interval, approximated to the nearest 0.1 m, is 9.0 m. The
slope of the blue line changes over the interval A, but the
9.0 m
average is
= 3.9 m/s. This method gives only an
2.3 s
approximation
of the instantaneous
speed because we
found the average speed for the time interval indicated. This
value for the average speed is the same as the instantaneous speed (the slope of the tangent line) at some point
in region A, but we cannot precisely determine that point.
Since the graph of your motion is a straight line in the
region B, it represents constant speed. The elapsed time in
region B is 9.0 s - 7.0 s = 2.0 s. The corresponding dis-
tance is 56 m
-
41 m = 15 m. The instantaneous speed in
region B can be determined
precisely
because
it is the
VELOCITY AND ACCELERATION
15 ill
= 7.5 m/s.
2.0 s
Region C on the graph is curved. The elapsed time is
18.0 s - 16.0 5 = 2.0 s. The corresponding
distance is
118.0 ill - 109.0 ill = 9.0 m. The average speed over the
interval C (as an approximation
of the instantaneous
9.0 ill
speed) is
= 4.5 m/s.
2.0 s
Both instantaneous
speed and average speed are completely described in terms of magnitude alone. Hence
speed is a scalar quantity.
same
,
,
f
,
I
I
I
I
Ir
I
I
43
,-,
as the average speed. It is
3.3 Velocity
When both speed and direction are specified for the motion, the term velocity is used. In other
words, velocity is speed in a particular direction. A person
walking eastward does not have the same velocity as a
person walking northward, even though their speeds are
the same. Two persons also have different velocities if
they walk in the same direction at different speeds. As
stated in Section 2.11, velocities are vector quantities. Mag+
nitude and direction are necessary to describe velocities.
For the problems in this chapter, in which the motion is
along a straight line, the word "velocity" and the symbol
for velocity will be used without stating the direction
angle. The values given for velocity will indicate only the
magnitude and not the direction. In these cases the velocity may be either a positive or a negative number to indicate in which direction the object is moving. A negative
sign as part of velocity, therefore, does not mean that the
speed is negative; speed always represents the magnitude
of the velocity. Hence, when the direction angle of velocity is not given, the equations for velocity and speed appear the same.
Average velocity is defilled as the total displacementdivided
by the total elapsed time.
df
-
dj
tf
-
t;
Velocity is speed in a given direction.
where Vavis the average velocity, di is the initial position at
time ti, and df is the final position at time tf. This relationship may also be written
Vav
I1d
= -----:1t
where ~d represents the change of position (displacement)
during the time intervall1t. The Greek letter 11(delta) designates a change in the quantity of variables d and t.
The Greek letter 6. in an equation
means "change of."
CHAPTER 3
44
7.0
6.0
5.0
E 4.01
B
oo"
.~
A
3.0
2.0
1.0
Ol
o
2.0
B.O
6.0
'.0
10.0
ElaP'ledtime,s
70
,
I
6.0
V
,
5.0
/
,
D
c
o
.
I
-~ 3.0J
C
2.0J
o
IP
l/
1.00
0.50 m/s. The same value for 'l\w is obtained by choosing
any other two points on the graph, such as A and B. In
this case, ilt is 2.0 5 and ~d is 1.0 m and v,w again equals
0.50 mis. The velocity, represented
by the slope of the
straight line, is constant. The motion of the object is uniform.
In Figure 3-5, the velocity is variable. The average velocity
5.0 m
for the entire range of the gra p h v,'" = 10
_ s ,or
velocity
be0.50 mis, as before. Ho\,-,'ever, the average
2.0 m
tween points C and D is
, or 1.0 mis.
2.0 s
To find the instantaneous
velocity at any point on the
graph in Figure 3-5, we use the same method that was
used to find instantaneous
speed in Section 3.2. For example, the instantaneous
velocity at point C is found by
drawing a tangent to the curve at C and measuring the
slope of the tangent. The slope is 0.5 mis, the instantaneous velocity at C.
The reason that the slope method can be used to measure instantaneous velocity at a specific point on a curved
line is that a very small segment of a curve is almost identical to a straight line segment. This becomes apparent
when you look closely at the area around point C in Figure
3-5 and imagine it to be magnified several times. The blue
and black line segments near C would then virtually coincide.
Instantaneous velocity is the velocity that n moving body
would have if it did Hot speed liP, slow down, or change direction
from that point Oil. The speedometer of a car shows instantaneous speed. But since a speedometer does not indicate
direction, it is not a velocity meter. To determine velocity, a
speedometer must be used in conjunction with a compass.
.
Figure 3-4. A graph of motion with
constant velocity.
o
-4.0
I
o
2.0
.
Graphic representations
of the velocity of motion along
a straight line are shown in Figure 3-4 and Figure 3-5.
These graphs indicate that the direction of motion does
not change in either case because the magnitude of displacement is continuously increasing. Furthermore, since
the graphs do not show the direction of motion, they have
the same form as the speed graphs in Figure 3-2 and Figure 3-3. In Figure 3-4, ~t for the entire range of the graph
5.0 ill
_
is 10 sand ild is 5.0 m. Consequentlv, - v~v = 10
_ s ,or
4.0
6.0
8.0
Elapsed time,s
10.0
Figure 3-5. A graph of motion with
variable velocity.
3.4 Solving Velocity Vector Problems
As vector quantities, velocities can be combined by the parallelogram
method discussed in Section 2.11. The following examples
show how to solve velocity problems of this kind.
~.
-/~
VELOCITY AND ACCELERATION
45
EX;UIPU~
A pitcher can throw a ball with a velocity of 125 km/h
relative to the ground in still air. If he throws the ball with that velocity
when a crosswind of 28 km/h is blowing from the left, what is the velocity
of the ball with respect to the ground?
-
,
,
Solution
Since this is a vector problem, the parallelogram method described in
Section 2.11 can be used to find the answer. Both the graphic and trigonometric solutions are shown below.
Graphic solution. Refer to Figure 3-6. Using an appropriate scale (such as
10 mm = 1.0 km/h), draw the velocity vector VBto represent the velocity
of the ball in still air. Then draw the velocity vector Vw to the same scale
at a right angle to VBto represent the velocity of the crosswind. Complete
the parallelogram formed by these two vectors, and draw the vector Vl{'
This represents the resultant velocity.
Measure the length of VRand convert it to km/h. The answer should be
130 km/h. Measure () with a protractor. The answer should be 12° clockwise from VB'
Trigonometric solution. Use the Basic equations above as follows:
(a) Working equation:
() = arctan
=
arctan
(~:)
28 km/h
(125
kmlh )
~ 12°
.
.
(b ) wvvor k mg equatwn:
Vw
VI{= ----=--
smH
~
28 km/h
sin 12°
130 kmlh
28 km/h
0.21
I~XA~IPI.E
A pilot wants to fly a plane at a velocity of 500.0 km/h
eastward with respect to the ground. A wind is blowing southward at a
velocity of 75.0 km/h. What velocity must the pilot maintain with respect
to the air to achieve the desired ground velocity?
CHAPTER 3
46
Solution
o
"p",S07kmlh
_
If:I",8.53"
,,~=500.0kmlh
_________
---------------
The vector diagram for this problem is shown in Figure 3-7. The desired velocity, VR, is the diagonal of the parallelogram formed by the
vectors Vw (the velocity of the wind) and Up (the velocity needed to keep
the plane at 500.0 km/h). The trigonometric solution is shown below:
(a) Working equation:
8 = arctan
75.0 km/h
(vwVR ) = arctan (500.0 km/h
-
= 8.53° north
.
(b) Workmg
.
equatIOn:
Vp
=-
Vw
=
sin ()
\I
of east
75.0 km/h
sin 8.53°
~ 507 km/h
I~X.nIPI.1\
The velocity of an airplane with respect to the air is
425 km/h in a direction 45.0° north of east. A wind is blowing northward
at 75.0 km/h. What is the resultant velocity of the airplane?
Solution
The vector diagram for this problem is shown in Figure 3-8. Since there
arc no right angles in the problem, the laws of sines and cosines must be
used for the trigonometric solution, as follows:
r
VELOCITY AND ACCELERATION
,
[(360.0' - 2(45.0')]
(a ) (j =
2
47
135.0'
-
Working equation:
,
tlf(
= VVw-~ +
Dp 2 -
2 DI/I,.'Vpcas
f:!
'
~V(75.0 km/h)' + (425 km/h)'
-
2(75.0 km/h)(425 km/h)(cos 135.0')
~481 km/h
(b) Working equation:
.
v,v sin (j'
sm(j=Vjl.
.
{J = arCS1l1
[
.
(75.0 km/h)(sin 135.0')
]
(481 km/h)
<
E
.o
~6.33°
Hence, the resultant velocitv is 481 km/h in
a direction 45.0° + 6.33°, or 51.3°, north of east.
QUESTIONS:
GROUP A
1. Explain the statement" All motion is
relative."
2. (a) What is a reference object? (b)
When we describe motion, what is a
common reference object?
3. On a distance-versus-time
graph,
what represents the speed?
4. What is the shape of a distanceversus-time graph for an object moving at constant speed?
5. What is the difference between average speed and instantaneous
speed?
6. How are instantaneous
speed and
average speed related for an object
with constant speed?
GROUP B
7. What is the relationship between
speed and velocity?
8. Answer the following by referring to
Figure 3-5. (a) Over which interval
did the object have its greatest velocity? (b) Over which interval was the
.,
"~ 0
object's velocity decreasing? (c) Over
which interval was its speed increasing?
9. Figures 3-4 and 3-5 show two objects
in motion. Compare the distance each
traveled. Compare their displacements.
10. Sketch a graph of motion (distance as
a function of time) that would correspond to the following bicycle trip.
You start your bicycle from rest and
increase your speed until you reach a
comfortable traveling speed. You continue at this speed for a few minutes
and complete your trip by slowing
down uniformly to a stop.
PROBLEMS:
GROUP A
1. If you live 10.0 km from your school
and it takes the school bus 0.53 h to
reach the school, what is the average
speed of the bus?
2. Determine the average speed of the
disk in Figure 3-1.
CHAPTER 3
48
3. A small plane is traveling toward the
east at 155 kill/h. It encounters a wind
blowing toward the east (a tail wind)
of 35 km/h. What is the velocity of the
plane relative to an observer on the
ground?
4. The pilot of a plane measures an air
speed of 165 km/h north. An observer
on the ground sees the plane pass
overhead at a velocity of 145 km/h
toward the north. What is the velocity
of the wind that is affecting the plane?
GROUP B
5. A motorboat heads due east at 12 mJs
across a river that flows toward the
south with a current of 3.5 m/s.
(a) What is the resultant velocity relative to an observer on the shore? (b) If
the river is 1360 ill wide, how long
docs it take the boat to cross? (c) How
far downstream is the boat when it
reaches the other side?
6. Find the air velocity of a plane that
must have a relative ground velocity of
250.0 kmlh north if it encounters a
wind pushing it toward the northeast at
75.0 km/h.
7. A swimmer can swim in still water at
a speed of 9.50 m/s. He intends to
swim directly across a river that has a
downstream current of 3.75 m/s.
(a) What must his heading be?
(b) What is his velocity relative to the
bank?
ACCELERATION
Acce/emtion is a chl/llge ill velocity over time. It can be positive or
negative.
3.5 The Nature of Acceleration
The gas pedal of a car
is sometimes called the accelerator. When you drive a car
along a level highway and increase the pressure on the
accelerator, the speed of the car increases. We say that the
car is accelerating. When you apply the brakes of the car
that you are driving, the speed of the car decreases. The
speedometer indicates that the car's speed is decreasing,
and we say that the car is decelerating.
Acceleration is defined as the time rate of changeof velocity.
An automobile that goes from O.Ukm/h to 60 km/h in 10
seconds has a greater acceleration than a car that goes
from 0.0 km/h to 60 kmlh in 15 seconds. The first car has
6U km/h
an average acceleration
of
= 6.0 (km/h)/s. The
IOs
second car has an average acceleration of 4.0 (kmJh)/s.
Acceleration, like velocity, is a vector quantity. In the
case of a car moving along a straight road, acceleration can
be represented by a vector in the direction the car is moving. Deceleration occurs in the opposite direction. For the
cases in which the motion is known to be along a straight
line, the direction of the acceleration is described as positive or negative. In Section 3.3 we saw that the equations
for speed and velocity appear the same, since the equa~
tions do not specify the direction of displacement. For the
same reason, the equations for rate of change of speed and
acceleration appear the same. These equations involve
only the magnitude and not the direction of the motion.
.,
,
,
f
f
I
r
,
I
VELOCITY AND ACCELERATION
Average acceleration
49
is found by the relationship
-
Vf
a av
7.0
Vi
6.0
~
tf - t;
where aav is the average acceleration, Viis the initial velocity at time ti, and Vf is the final velocity at time tf. This
relationship may also be stated
a av
4v
--
,.
",
..
-g
E
4t
>0
where /1L' is the change in velocity for time interval ill.
Figures 3-9 and 3-10 are velocity-time graphs for accelerated motion. In Figure 3-9 the velocity curve is a straight
line. Hence the acceleration does not change. This is an
illustration of constant acceleration.
The motion illustrated in Figure 3-10 is an example of
variable acceleration. As we mentioned in the case of finding
instantaneous speed or instantaneous velocity, finding the
average acceleration between two points on the curve only
approximates
the instantaneous
acceleration.
Instanta+
neous acceleration is equal to the slope of the line that is
tangent to the curve at the given point.
Notice that the slope of the curve in Figure 3-10 is not
always positive. That is, the velocity does not increase
during all time intervals. For the interval between points
A and 8, ilt is 3.0 sand /1L' is -2.0 m/s. Consequently,
oo
~
5.0
4.0
3.0
2.0
LO
2.0
6.0
4.0
8.0
Figure 3-9. A graph of velocity
versus time for constant acceleration
"
-2.0 m/s
3.0 s
~
-0.67-
m/s
s
This reduction in speed is called deceleration, or negative
acceleration.
The dimensions of acceleration should be carefully ex-
7.0
6.0
amined. Acceleration is the rate of change of the rate of
~ 5.0
displacement, or (/1dl/1t)//1t. If /1d is in meters and /1t is in
seconds, acceleration is given in mlsls, or m/s2.
~.
,~ 4.0
3.6 Solving
Acceleration
Problems
Galileo Galilei
(1564-1642) was the first scientist to understand
clearly
the concept of acceleration. He found that the acceleration
of a ball rolling down an inclined plane and the acceleration of a falling object were both examples of the same
natural phenomenon
and were described by the same
mathematical rules.
Let us consider one of Galileo's experiments. One end
of a board is raised so that it is just steep enough to allow a
ball to roll from rest to the 0.5-m position in the first second. The apparatus is shown in Figure 3-11. As the ball
rolls down the board, its position is shown at one-second
A
1/
o
E
,o
i='3.0
~
2.0
\
/
/
B
1.0)
-~
2.0
.~
4.0
-6.0~
8.010.0
Elapsed time. s
Figure 3-10. A graph of velocity
versus time for motion with variable acceleration.
10.0
CHAPTER
50
3
Endof1stsecond
Start
l
.
End of 2nd second
11.$
End of 3td second
End of 4th second
!I
22.533.$
4
Figure 3-11. The ball rolls down
the plane with constant accelera-
tion.
End of 5th second
4.555.566.577.5
1
2
3
4
5
Table 3-1
ACCELERATION
Time at TIme, t, Position at Position at Distance, J.d,
traveled
start of
end of
start of at end of
In interval
Interval
Interval
interval
Interval
1'1
0.0
10
2.0
3.0
4.0
1'1
10
2.0
3.0
4.0
5.0
Galileo's
experiment verified
property
of inertia.
Iml
0.0
05
2.0
4.5
8.0
the
Iml
0.5
2.0
4.5
80
12.5
Iml
05
15
2.5
3.5
45
Average
Average
InstanInstal'acceleration,
taneous
taneous
velocity. va.,
velocity, VI. velocity, V" aa.. during
during
Interval
at start
at end of
interval
interval
of Interval
(m/s2)
(m/s)
(m/s)
(m/s)
0.5
15
2.5
3.5
4.5
0.0
10
2.0
3.0
4.0
10
2.0
3.0
4.0
50
1.0
10
10
10
10
The initial velocity of the ball is zero, Its acceleration, or
rate of gain in velocity, is 1.0 m/52, Therefore its velocity at
the end of the first second is 1.0 m/s. (If the ball at that
instant were no longer accelerated, it would continue to
move with a constant velocity of 1.0 m/s.) At the end of
the next second, its velocity is 2.0 m/s. (2.0 s x 1.0 m/s2 =
2.0 m/s.) Similarly the ball's velocity at the end of the fifth
second is 5.0 m/s. These data show that the final velocity of
an object starting from rest and accelerating at a constant rate
equals the product of the acceleration and the elapsed time. If an
object does not start from rest, its final velocity will equal the sum
of its initial
Galileo was among the first scien~
tists to use experiments
in testing
hypotheses.
11.51212.51
intervals. The data obtained are shown in the first six columns of Table 3-1.
The data in the last four columns of Table 3-1 show that
the velocity at the end of each second is 1.0 m/s greater
than at the end of the previous second. Thus the acceleration is 1.0 (m/s)/s, or 1.0 m/52. The motion of a balJ rolling
down an inclined plane is an example of constant acceleration.
CONSTANT
Interval
number
88.599.51010.511
velocity and the increase in velocity produced by the
acceleration.
We may arrive at this same conclusion by mathematical
reasoning. The equation defining acceleration is
Vf - Vi
tf - ti
---"
VELOCITY AND ACCELERATION
51
For constant acceleration aay becomes a. Then assuming
that Vi is the velocity in the direction of the acceleration
when t; = a and tf is elapsed time ilt,
-
Vf
Vi
lI.t
Multiplying
both sides of the equation
by £It,
adt=Vf-Vi
Isolating
the term Vf yields
Vf
= Vi
+ adt
If the object starts from rest, Vi =
and
=
Vf
1-
(Equation
1)
(Equation
2)
a
al1t
In Section 3.3 we saw that
l
Vav
or
lI.d
=
I1d=
A.t
dt
Vay
f
I
For constant acceleration, the average velocity for any
given interval of time can be found by taking one-half the
sum of the initial and final velocities:
Vi
+
Vf
2
By substituting
Equation 3,
(Equation 3)
the value of Vf (obtained in Equation 1) in
Vi
+
(Vi
+ aA.t)
2
(Equation 4)
Now, substituting the value of Vay(obtained in Equation 4)
in Equation 2 gives
lI.d~
Vi lI.t + !all.t'
(Equation
5)
If the object starts from rest,
V;dt= 0
and
From this equation
dd=!a de
we see that if an object starts from rest
CHAPTER 3
52
and travels with constant acceleration, the displacement of
the object during the first second is numerically equal to
one-half the acceleration.
Now let us solve Equation 1 for ilt, and substitute this
value of ill in Equation 5. From Equation I,
Vf
At~
The Mathematics
Refresher may
Substituting
in Equation
come in handv herc_
Vf
i1d = Vi(
Multiplying
:
-
Vi
a
5,
Vi)
,
+ "2a
(
VI -
by 2a and expanding
a
)'
terms,
2a i1d =..2-v,vf - Zv/ + v/
Combining
V,
-
..2Y-rVi
+
Vi2
terms,
2a i1d = v/ -v/
Solving
for Uf,
Vf =
~+
2ai1d
(Equation
6)
With constant acceleration, the final velocity equals the
square foot of the sum of the square of the initial velocity
and twice the acceleration times the displacement. If the
object starts from rest,
Table 3-2
ACCELERATION EQUATIONS
With Initial
velocity
Starting from rest
u,= n "'-I
"'-d
I',
= 1', cl/ -;n
= v'V,' +
211
cl/2
3d
"'-d= ia
M~
and
Vf
=
Y2a,1d
The equations just given are summarized in Table 3-2.
They apply whether the constant acceleration is positive
or negative; hence it is important to note the sign of the
displacement. When it is negative, the equations may be
used to calculate the distance an object will travel when its
motion is constantly decelerated and the time that elapses
in bringing the object to a stop. The following example
problems show how the equations are used in problems
dealing with both positively and negatively accelerated
motion.
I':X"UWU:
Starting from rest, a ball rolls down an indine at a constant
acceleration of 2.00 m/s2. (a) What is the velocity of the ball after 8.5 s? (b)
How far does the ball roll in 10.0 s7
VELOCITY
-
AND
ACCELERATION
~
Given
.....
'1'..."'.
5.1
-
. .ill... ..........
Unknown
Vi= 0.0 m/s
.Inl
BasiCequations
Vf == Vi
a == 2.00 m/s2
fltr:!:: 8.5 s
flt2 = 10.0 s
~
+ aflt
i1d
Solution
(a) Working equation: Vf = Vi + aM1 = 0.0 m/s + (2.00 m/s2)(8.5 s)
= 17m/s
(b) Working equation: fld = vjM2 + ~aM/ = 0.0 + H2.00 m/s2) (10.0 sf
~
IDa rn
EXA.\'lPLE
A car traveling at 88 km/h undergoes a constant deceleration of 8.0 m/s2. (a) How long does it take the car to come to a stop? (b)
How far does the car move after the brakes are appJied?
,,;,. /
. .... '-.~
~~,.<...I
.- .~t~;~~t~Sri.~!,
Given
Unknown
i:
i1t
Pi = 88 kmlh
i"f - Vi
,
.
..
Vf==0.0 km/h
a =:c-8.0 mls2
fld= viM + iaM2
I
~
I
.
Solution
(a) Working equation: M =
Vf
-
Vi
a
(0.0 km/h - 88 krn/h)(IOOO rn/krn)
(-8.0
m/s')(3600
s/h)
= 3.15
(b) Working equation: tJ.d = i'\M + }atJ.t2
(88 km/h)(3.1 s)(IOOOm/krn)
~
3600 s'"
+
,(
' _ 8.0
.2
rnls 2 )(3.1 s)
= 37 m
PR\CTICE PHOBLEM
An object starts from rest and moves with constant acceleration for a distance of 150 m in 5.0 s. What is the acceleration
of the object?
Ails. 12 m/s2
CHAPTER 3
54
3.7 Freely Falling Bodies
In New York City if a body
falls freely from rest, it reaches a velocity of 9.803 m/s in
one second of time. Since a = vf/8t, the acceleration
equals 9.803 m/s2 at this location. We must consider the
location because the value of the acceleration depends on
the earth's gravity at that location. Since the gravity on the
earth may vary from place to place, the value of the acceleration may also vary. (The nature of gravity will be discussed more fully in Sections 3.11-3.13.) A rounded value
of 9.80 m/s2 is approximately valid for most locations. (Air
friction is not considered in this example.)
The equations for accelerated motion apply to freely fa11ing bodies. Since the acceleration due to gravity, g, is the
same for all objects at a given location, we may substitute g
for a in these equations. Thus
Vf=V;+gi1t
i1d= vii1t+
!giit2
Vf = Vv/ + 2giid
Figure3.12. Objects
fall with the
same acceleration in a vacuum
(left) but with different accelerations through air (right).
When freely falling bodies start from rest, Vi = 0, and
the motion is described by the following equations:
Vf
Vf
Figure 3-13. Skydivers in free fall.
Air resistance allows skilled divers
to control position and direction of
their bodies even before the parachutes are opened.
= gi1t
i1d=!gi1t2
~
Y2gfJ.d
In the case of free-falling bodies, ild is always a vertical
distance. The vector quantities v and l1d in these equations
are customarily assigned plus signs if they are directed
downward and minus signs if they are directed upward.
(The description of the motion is simplified by considering
the downward
direction of gravity as positive.) These
equations apply only to objects that are faUing freely in a
vacuum because they do not take into consideration the
resistance of air, as illustrated in Figure 3-12. These equations do not apply to objects whose air resistance is great
compared to their weight, such as leaves faUing in air.
An object thrown upward is constantly decelerated by
the force of gravity until it fina]]y stops rising. Then as the
object falls, it is constantly accelerated by the force of gravity. If the effect of the atmosphere is neglected, then the
time required for the object to fall is the same as the time
required for it to rise. Study the following example.
EXAI\WI.E
A ball is thrown vertically upward with a velocity of
115 m/s. (a) To what height will it rise? (b) How long will it take for the
ball to fall back to the earth?
,
VELOCITY
AND
~
",,;"&,"'J:',o
Given
-115
...
0lil Vf
0lil
m/s
0.0 m/s
9.80
Basic equations'
Unknown
---=
Vf =
iiiII g =
Vj
-
ACCELERATION
..
m/52
LIt
Vf
~
=
'Iv,' +
'-\
2gLldllill
+ gt.!.t
Solution
(a) Working equation: t1d
=
=
(b) Working equation: f1t =
2g
0- (-115 m/s)'
(2)(9.80 mls')
-675
Vf -
g
(The value for Vi is negative
because the velocity is
upward.)
(The negative value obtained for ild
indicates an upward displacement.)
ill
v-1
=
0-
= 11.7 s
Neglecting the effect of the atmosphere,
amount shown above, or 23.4 s.
(-115 m/s)
9.80 m/s2
the total time will be twice the
I)R.U:TI{~E pn()8U~I\IS
1. A ball drops from rest and attains a velocity
of 53 m/s. How much time has elapsed?
Ans. 5.4 s
2. How far did the ball in the previous
second?
Ans. 25 m
QUESTIONS:
GROUP
II
1. (a) Define acceleration verbally and
mathematically.
(b) What type of
quantity is acceleration? (c) What is
deceleration?
2. A certain car is accelerating at
5.0 km/h/s. Explain what is happening
to the motion of tne car.
3. Give an example of a moving object
that has a velocity vector and an acceleration vector in the same direction
and one that has velocity and acceleration vectors in opposite directions.
GROUP B
4.
Can an object be moving at a constant
problem
fall during
the third
speed and accelerating?
5. What would be the shape of a speedversus-time graph for an object in free
fall?
6. What is the algebraic sign (positive or
negative) for designating the displacement and veJocity vectors of objects in
free fall?
PROBLEl'vIS: GROUP A
Note: Use g = 9.80 m/s2. Disregard the
effect of the atmosphere.
1. A car moving along a level road increases its speed uniformly from
16 m/s to 32 mls in 10.0 s. (a) What is
the car's acceleration? (b) What is its
~-
CHAPTER 3
56
average speed? (c) How far did it
move while accelerating?
2. A ball initially at rest rolls down a hill
with an acceleration of 3.3 m/s2, (a) If
it accelerates for 7.5 Sf how far will it
move? (b) How far would it have
moved if it had started at 4.0 rnls
rather than from rest?
3. A penny dropped from the roof of a
building takes 6.3 s to strike the
ground. How high is the roof?
4. A flower pot falls from a windowsill
21).0 m above the sidewalk. (a) How
fast is the flower pot moving when it
strikes the ground? (b) How much
time does a passerby on the sidewalk
below have to move out of the way
before the flower pot hits the grou'nd?
of a runway, undergoes a constant acceleration of 1.6 m/s2 for a distance of
1600 m before takeoff. (a) What is its
speed upon takeoff? (b) What is the
time required for takeoff?
7. A physics student throws a softball
straight up into the air. Her friend has
a stopwatch and determines that the
ball was in the air for a total of 3.56 s
(a) With what speed was it thrown?
(b) How high did it rise?
8. A water rocket starts from rest and
rises straight up with an acceleration
of 5.00 m/s2 until it runs out of water
2.50 s later. If it continues to rise until
its speed is zero, what will be its maximum height?
PHYSICS ACTIVITY
CROUP B
5. A boy sliding down a hill accelerates
at 1.40 m/52. If he started from rest, in
what distance would he reach a speed
of 7.00 m/s?
6. A plane, starting from rest at one end
Determine your reaction time by having a
friend unexpectedly drop a vertically held
meterstick between your hands. Catch the
meterstick and calcuhte the time from the
distance the meterstick has fallen through
your grasp.
NEWTON'S
3.8
Figure 3-14. Sir Isaac Newton formulated the law of gravitation
and
three laws of motion that describe
how forces act on matter,
Law
of Inertia
LAWS
OF MOTTON
Thus far in Chapter
3, we have dis-
cussed motion apart from its causes. Now we shall study
the effects that forces have on motion. A force is a physical
quantity that can affect the state of motion of an object. A push
or a pull is a force. Since a force has magnitude and direction, and can be combined with other forces the way displacements are combined, it is a vector quantity.
The relationships among force, mass, and motion were
described clearly for the first time by Sir Isaac Newton
(1642-1727) in three laws of motion that bear his name.
While some aspects of Newton's laws of motion can be
tested only under carefully controlled conditions,
repeated experiments and observations show that the laws
are universally true. That is, the laws apply to objects
throughout the universe.
Newton's first law of motion deals with the motion of a
body on which no net force is acting. That is, either there
is no force at all acting on the body or the vector sum of all
forces acting on the body is zero. The word "net" refers to
--
r
VELOCITY AND ACCELERATION
~..
the second situation. Even though a body may have many
forces acting on it, these forces may act against each other.
They may balance each other in such a way that the body
does not change its state of motion. If such a body is at
rest, it will remain at rest. If it is in motion, it will continue
moving ill a straixht line with constant speed. Newton's first
law of motion may therefore be stated as follows: If there is
no net force actirlg 011II body, it will continue in its state of rest or
will continue moving along 11straight line with constant speed.
As we saw in Section 1.8, this property of a body that
opposes any change in its state of motion is called inertia.
Hence Newton's first law of motion is known as the law of
inertia.
At first glance, this law seems to contradict our everyday experiences. If a car is to be kept moving with a constant velocity, the car's engine must apply a constant force
to it. If the engine stops applying this force, the car comes
to a stop. Only then does the car seem to obey the part of
Newton's law that stahi>s that objects at rest will remain at
rest unless acted upon by an unbalanced force.
Close shtdy of a moving car shows, however, that it is
the force of friction that brings the car to a stop and not the
absence of the force provided by the engine. If it were
possible to remove this friction, the car would keep rolling
without applying a constant force. That is what happens
to a space vehicle beyond the earth's atmosphere.
The
vehicle keeps moving even when its rocket engine is shut
off, and the ship's velocity is changed only by the force of
gravity of other objects in space.
The study of the motion of a car in the absence of friction is an example of a thought experiment since the study
cannot be performed under actual conditions. It was a
thought experiment of this kind that led Galileo to an understanding of inertia even before Newton described it.
Galileo noticed that if a ball rol1s down one incline and up
a second one, as shown in Figure 3-15, the ball will reach
57
The law of inertia states that unless a net force acts on an object,
the motion
of an object (or lack of
it) does not change.
....
Figure3-15. A diagram of Galileo's thought experiment. In the
upper and middle drawings, the
ball reaches almost the height at
which it started. In the lower drawing, the ball continues to move
indefinitely at constant velocity. In
each drawing, the successive images show the positions of the
ball after equal time intervals.
CHAPTER 3
58
almost the same height on the second incline as the height
from which it started on the first incline. Galileo concluded that the difference in height is caused by friction
and that if friction could be eliminated the heights would
be exactly alike. Then he reasoned that the ball would
reach the same height no matter how shallow the slope of
the second incline. Finally, if the second slope were eliminated altogether, the ball would keep rolling indefinitely
with constant velocity. This is the same idea as the one
expressed in Newton's first law of motion. lnertia keeps a
stationary object stationary and a moving object moving.
The greater the mass (and inertia) of an object, the
greater is the force required to produce a given acceleration. A pencil lying on the floor has relatively little mass
and therefore little inertia. You can produce acceleration
easily by kicking it with your foot. A brick has much more
mass and more inertia. You can easily tell its difference
from the pencil if you try to kick it.
Newton's first law specifies which forms of motion have
a cause and which forms do not. Constant motion in a
straight line is the only motion possible for an object far
removed from other objects. Variable motion is always
caused by the presence of some other object.
3.9 Law of Acceleration
Newton's first law of motion
tells us how a body acts when there is no net applied
force. Let us consider what happens when there is either a
single applied force or two or more applied forces whose
vector sum is not zero. (Remember that forces are vector
quantities.) In the following discussion, we shall use "applied force" or just "force" to mean the vector sum of all
forces applied to the body.
Newton's second law of motion states: The effect of an
applied force is to cause the body to accelerate in the direction of
the force. The acceleration is in direct proportion to the force and
in inverse proportion to the mass of the body.
If the body is at rest when the force is applied, it will
begin to move in the direction of the force and will move
faster and faster as long as the force continues.
If the body is moving in a straight line and a force is
Figure 3-16. Ademonstration of
,.,applied
in the direction of its motion, it will increase in
inertia. In the bottom photo, the
J-speed
and
continue to do so as long as the force continues.
table has been quickly pulled
If the force is applied in the direction opposite to the moaway. and the chicken dinner is
tion, the acceleration will again be in the direction of the
momentarily suspended in midair.
force, causing the body to slow down. If such a force continues long enough, the body will slow down to a stop and
then begin to move with increasing speed in the opposite
direction.
59
VELOCITY AND ACCELERATION
1-
I
~
I
~
If the force applied to a moving object is not along the
line of its motion, the acceleration will still be in the direction of the force. The effect in general will be to change
both the direction and the speed of the motion. We shall
postpone detailed consideration of this more complex case
until Chapter 5. For the present we shall consider only the
speed of an object initially at rest or one that is moving in
a straight line that is changed only by a force acting along
that line.
It should be emphasized again that a body to which no
force is applied has zero acceleration. If such a body is in
motion, its velocity does not change. If it is at rest, its
velocity remains unchanged; its velocity stays at zero. And
when there is no change in velocity, the acceleration is
zero. When a force is applied to a body, the acceleration is
not zero, the velocity changes, and the acceleration is in
the direction of the applied force.
If different forces are applied to the same body, the
magnitude of the acceleration is directly proportional to
the amount of the force. That is, doubling the force causes
the acceleration to double, and so on.
We can express such a direct proportion as an equation
by saying that one of the quantities equals the other one
times a constant, or
F= ka
in which F is the applied force, a is the resulting acceleration, and k is the constant of proportionality.
Since F and a
are both vectors, this equation also says that the applied
force and the resulting acceleration are in the same direction (which is a restatement of the first part of Newton's
second law of motion).
What happens when forces are applied to different objects? This question brings us to the third part of Newton's
second
law. It turns
out that the equation
F
=
ka is always
true but that the value of k is different for different objects.
In fact, we find that if we properly define the units of force
and mass, the value of k for any object is identical with the
mass, m, of that object. Thus the equation is written
F= ma
In the metric system, the unit of mass is the kilogram
and acceleration has the units of meters per second
squared. (See Section 3.5.) The force required to accelerate
1 kilogram of mass at 1 meter per second squared is 1 kg .
mfs2. This relationship defines the newton in terms of fundamental units. That is,
1 N = 1 kg.mJS2
Because of friction and gravity,
stepping on tfle accelerator of a
car does not necessarily accelerate the car.
CHAPTER 3
60
Since the acceleration acquired by a particular object is directly proportional to the amount of force applied, this law
of acceleration can be expressed as the proportion
F
a
F'
a'
--
where F and F' are two different
the corresponding
accelerations.
forces and a and a' are
In the case of a freely falling object, one of the forces is
known. It equals the weight, FWfof the object. The acceleration, g, is also known. It is the acceleration due to gravitation, 9.80 m/52. Making these substitutions,
the proportion becomes
a
g
F
and
FINn
F~-
Therefore, we can
desired acceleration
Since F = ma, we
above equation and
g
calculate the force needed to give any
to an object of known weight.
can substitute this value for F in the
get
Notice the similarity between
lI1a=-
f;" = mg and F = ma. What is
g
the reason for this similarity?
Dividing both sides of the equation
Fw
m=-
and
Fw
by a, we have
g
= mg
Provided the value of g is known, these two equations are
used to convert from newtons to kilograms, and vice
versa. The following examples show how Newton's second law is used in solving acceleration problems.
EXAJ\WU:
A car weighs 1.50 x to'! N. What force is required to accelerate the car at 4 m/s2? (Neglect the effect of friction and air resistance in
this problem.)
Given
III
Fw ~ 1.50 X 104N
Ea =4m/s2
;17''T"''''",~
11_1" .Iwn
u."
F
It'":
"'.,",,_,..,..11.1..,1,
.,1,.,.
Basic equations
-i\~::-~--'~~g
F'=
mn
1
1
I
s
.
,
r
61
VELOCITY AND ACCELERATION
Solution
Substituting
the first Basic equation
Working equi7tion:
F=
into the second,
we get the
(Ew)(")
g
(1.50 x 10' N)(4 m/s')
9.80 m/52
=6x103N
l'lucnt:l~
I'RORl.rl\lS
at sea level is 72 kg?
1. W:hat is the weig:ht of a person whose mass
AilS. 710 N
2. A car weighs 1.15 x 104 N. A force of 895 N is used to accelerate the
car from rest to 25 m/s. How much time is required for the acceleration?
Ails. 33 s
3. What force is required to accelerate the car in Question
to 50 mls in 5 s?
Ans. 6 x lO' N
2 from 25 mls
3.10 Law of Interaction
Newton's third law of motion For every action there is an equal
may be stated as follows: When one body exerts a force on and opposite reaction.
another, the second body exerts on thefirst a force of equalmagnitude in the oppositedirection. To illustrate this law, consider
some of the forces that are exerted when a book is resting
on the top of a level table. The book's weight acts downward against the table. The table top exerts an upward
force on the book. These forces are equal in magnitude
and opposite in direction. When you walk forward on a
level floor your feet exert a horizontal force against the
floor, and the floor pushes against your feet with a force of
equal magnitude, but in the opposite direction.
In each of these situations we have two objects. In the
first instance, the objects are the book and the table. In the
second instance, they are the foot and the floor. Two
forces are involved in each situation. In the first, they are
the weight of the book on the table and the force of the
table against the book. In the second, they are the force of
the foot against the floor and the force of the floor against
the foot. In Newton's 0\\'11 words: "To every action there is
always opposed an equal reaction."
Unaccompanied
forces do not exist in nature.
Newton's third law holds true for a11objects at all times,
whether they are stationary or moving. Every force is accompanied by an equal and opposite force, independent
62
CHAPTER
,
k.
Figure 3-17. A demonstration of
action-reaction. The kilogram mass
exerts a downward force of 9.8
newtons, while the persons hand
exerts an equal upward force of
9.8 newtons. (The weights of the
balances are not considered in
this illustration.)
QUESTIONS:
3
of the motion of the objects involved. Let us consider another example. A boy rows a boat toward the shore of a
lake and, when he is a meter or so from shore, he attempts
to leap ashore. He exerts a force against the boat and the
boat exerts an equal but opposite force against him. The
force exerted by the boat against the boy accelerates him.
The direction of his acceleration is opposite to the direction in which his force caused the boat to be accelerated. If
we assume the resistance to the motion of the boat by the
water to be negligible, the amount of acceleration of each
object is inversely proportional to its mass. The boy might
judge the force he must exert to reach shore on the basis of
his experience in jumping the same distance from an object fixed on the earth. If so, when he jumps from the boat,
he might not reach shore, but instead fall into the water.
Suppose a kilogram mass is suspended from two spring
balances that are connected, as shown in Figure 3-17. As
we learned in Section 2.4, a kilogram weighs about 9.8
newtons near the earth's surface. Then why do both balances have a reading of 9.8 newtons? The answer is Newton's third law. The downward force exerted by the mass
is accompanied by an equal upward force exerted by the
person's hand. Neither force can exist without the other.
If additional balances are added in this experiment, each
balance will still read 9.8 newtons (provided the weights
of the balances are neglected). Can you explain why?
You may think that motion cannot occur if opposing
forces are equal. If two people pull on a lightweight wagon
with equal force in opposite directions, the wagon will not
move. But this is not an example of action and reaction.
There are two forces, it is true, and they have the same
magnitude and opposite directions, but they areboth exerted
on the same object. Action and reaction apply when forces
are exerted on different objects. For example, action and
reaction are involved in the force that each person's feet
exert against the ground and the equal but opposite force
the ground exerts against the feet.
GROUP A
1. (a) What is a force? (b) What type of
quantity is a force?
2. Explain the term net force.
3. State the law of inertia.
4. Draw a sketch of your physics book
lying at rest on the desk. Draw arrows representing all the forces acting
on the book. (This sketch is called a
free-body diagram.) What is the net
force acting on the book?
5. Push your physics book across the
desk at a constant speed. Draw a
free-body diagram and indicate the
net force acting on the book.
6. In your diagram for Question 5, is
there a force of which people were
unaware before Galileo's time?
~
,-
f
,
c
,
I
7. Give your physics book a sudden
push so that it moves across the
desk. What happens to its motion?
(c) Compare the forces that the astronaut would have to exert to push the
llO-kg crate across a level, frictionless
Draw a free-body diagram. Is there
an unbalanced force acting?
8. What physical quantity is a measure
of the amount of inertia an object
has?
surface on the moon and on the
earth.
~
GROUP A
Note: Use g = 9.80 m/52, All forces are net
forces.
1. What force is needed to give a mass of
25 kg an accelemtion
of 20.0 m/52?
Hon change when tvvo people push
compared to when only one person
pushes?
2. A force of 30.0 N acting on an object
gives it an acceleration of 5.00 m/s2.
(a) What is the mass of the object?
(b) What is its weight?
3. What acceleration will you give to a
24.3-kg box if you push it with a force
of 90.0 N?
4. A car weighs 19 600 N. (a) What is the
mass of the car? (b) If a braking force
of 1250 N is needed to stop the car,
what is its acceleration?
GROUP B
r--
PROBLEMS:
9. You are having trouble starting your
car. A friend offers to give a push. A
second friend comes along and also
offers to help. How does the accelera-
f
r
6.1
VELOCITY AND ACCELERATION
10. A 70-kg defensive back and a 110-kg
nose tackle are chasing you down the
football field. If you do some brokenfield running (change direction continuaJly), which one will have a more
difficult time changing his direction to
follow you? Explain.
11. An astronaut performed an experiment on the moon in which he
dropped a feather and a hammer
from the same height. He found they
had the same acceleration and landed
on the surface at the same time. Why
did both objects have the same acceleration?
12. Two children are pulling on a lightweight wagon to move it across the
floor. If the force of the children on
the wagon equals the force of the
wagon on the children, why does it
move? Draw a free-body diagram to
show all the forces on the children
and wagon.
13. An astronaut on the moon has a 110kg crate and a 220-kg crate. (a) Which
crate weighs more on the earth? On
the moon? (b) How do the forces
needed to lift the crates straight up
on the moon compare to the forces
needed to lift them on the earth?
GROUP B
Note:
Use g
=
9.80 m/s2. All forces
are net
forces.
5. An 1800-kg car starting from rest accelcrates constantly during the first 9.50 s
of its motion. If the force acting on the
car is 2150 N, what is the car's speed
at the end of this time?
6. What force must be exerted on an electron to move it from a state of rest to
a speed of 3.5 x 107 mls in a distance
of 0.75 m?
7. During a throw, a baseball pitcher exerts a force on the ball through a distance of 3.20 m before releasing it. If
the D.12-kg ball leaves the pitcher's
hand with a speed of 35.0 mis, how
much of a force did the pitcher exert
on the ball?
8. A catcher stops the baJJ in Problem 7
by pulling back 25.0 em on the mitt
during the catch. How much force did
the catcher exert on the ball?
r.
l
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64
CHAPTER 3
GRAVITATION
Newton developed his laws of
motion and gravitation when he
was only twenty~three years old
and while he was home from college on an extended vacation during the Great Plague in 1665.
Support
3.11 Newton's
Law at Universal Gravitation
In addition to formulating his three laws of motion, Newton
described the force that makes falling bodies accelerate
toward the earth. In doing so, he made use of the laws of
planetary motion that were developed by Johannes Kepler
almost a century before Newton's time. See Section 1.4. In
Newton's account of his study of falling bodies, he states
that he wondered whether the force that makes an apple
fall to the ground was related to the force that keeps the
planets in their orbits. If so, a single law could be used to
describe the attraction between objects in the entire universe.
Newton explained Kepler's laws by stating that the force
of attraction between two objects is directly proportional to the
product of the masses of the objects and inversely proportional to
the square of the distance between their centers of mass. (The
center of mass of an object is that point at which all its
mass can be considered to be concentrated. This concept is
similar to that of "center of gravity," which will be discussed in Chapter 4.) Newton called this attractive force
the force of gravitation. Hence the statement is known as
Newton's law of universal gravitation. In equation form
the law is
Cm1m2
F=
d'
where F is the force of gravitation, mj and m2 are the
masses of the attracting objects, d is the distance between
their centers of mass, and G is a proportionality constant
Figure 3-18. The Cavendish exper- called the universal gravitational constant.
iment. The force of gravitation, F,
Newton was only able to confirm his theory with astrobetween the movable and stationnomical observations. He knew the approximate value of
ary spheres can be measured by
G, but the first precise measurement was made in 1797 by
observing, with the help of a mirror
the English scientist Henry Cavendish (1731-1810). A
and light beam, the amount of
schematic
diagram of this experiment is shown in Figure
twist in the suspending fiber.
3-18. A small lead sphere is attached to each end of a
lightweight rod. The rod is suspended by a thin quartz
fiber that has a mirror attached to it. Two large lead spheres
are placed in fixed positions near the small spheres. The
gravitational force between the fixed and movable spheres
draws the movable spheres toward the fixed ones. This
motion causes the suspending fiber to twist. The fiber offers a slight resistance to twisting. The resistance increases
as the twisting increases, and the angle of twist is proportional to the force of gravitation between the fixed and the
movable spheres. The angle of twist could be measured
directly from the rod's movement, but the sensitivity of
,
VELOCITY
65
AND ACCELERATION
the instrument can be increased by shining a light into the
mirror. The light is reflected onto a distant scale, and small
movements of the mirror result in large movements of the
reflected light across the scale. Such an arrangement
is
called an optical lever. Since G has a very small value, extreme care must be used to isolate the Cavendish apparatus from outside forces such as those produced by air currents and electric charges. The present value of G is
6.67 X 10-11 N. m2/kg2. (Some scientists think that this
value will decrease as the universe keeps expanding.)
3.12 The Mass of the Earth
Once G has been measured, the equation for Newton's law of universal gravitation can be used to find the force of gravitation between
any two objects of known mass with a known distance
between their centers of mass. Or if the force of gravitation, the distance, and one of the masses are known, the
equation can be used to find the value of the other mass.
For example, the equation can be used to find the mass of
the earth. In the equation, use m,,, the mass of the earth,
for mI. For 1n2' use mp' the mass of a particle on the earth's
surface. Newton showed that, for a spherical mass, we
can consider all of the mass to be located at the center. So
in this case d will be the radius of the earth. The force of
gravitation, F, will be the weight of the particle, so we
shall replace F by Fw.
.
Fw
memp
_- C-
Dividing both sides of the equation
Fw
-
(Equation
d'
mp
by I1lp'
Gme
-
(Equation 2)
d'
(Equation 3)
But
Fw
-~g
mp
Or
Substituting
1)
Equation
(Equation 4)
4 in Equation
g
Gme
~
----;j'
2,
(Equation 5)
Since both G and m" are constants, Equation 5 shows
that the acceleration due to gravity in a given location depends only on the square of the distance from the center of
the earth.
.-~
66
CHAPTER 3
Equation 5 may be used to determine
earth. Solving this equation for mel
How does this value compare with
the one given in Section 2.9?
the mass of the
gd'
mc=C
Substituting
9.80 m/s2 for g, 6.37 x 106 ill for d, and
6.67 x lQ-11 N. m1/kg1 for C, me = 5.96 X 1024 kg, the
mass of the earth.
Figure 3-19. Saturn's rings. Each
of the millions of particles circling
the planet is held in a specific
orbit by the force of gravitation
between the planet and the particle.
Table 3-3
SURFACE GRAVITIES OF THE
MOON AND PLANETS
Body
Jupiter
Neptune
Saturn
Uranus
Earth
Venus
Mercury
Mars
Pluto
Moon
--
Relative gravIty
(Earth = 1.00)
2.53
1.18
1.07
0.92
1.00
0.91
0.38
0.38
0.09
0.16
3.13 Relation Between Gravity and Weight
The term
gravity is used to describe the force of gravitation on an objecton
or near the surface of a celestial body, such as the earth. Thus we
say that the moon has less gravity than the earth because
the force of gravitation near the moon's surface is less than
the force of gravitation near the earth's surface. A calculation of the gravity of a celestial body near the surface must
take into consideration the size of the body since the force
of gravitation varies inversely with the square of the distance to the center of mass of the body. See Table 3-3 for
the surface gravities of the moon and planets.
The value of the mass of the earth can be used to determine the value of the gravity at any given location on or
above the earth's surface. Such additional factors as local
variations in the composition of the earth's crust and the
effect of the earth's rotation must also be taken into consideration, however.
Equation 1, Section 3.12 shows that the weight of an object is the measure of the force of attraction between the
object and the earth. When we say that a person weighs
900 N, we mean that the force exerted on that person by
the earth is 900 N. The person also exerts a force of 900 N
on the earth. Equation 3 tells us that the weight of an object equals the product of its mass and the acceleration due
to gravity. The mass of an object is constant. But Equation
5 indicates that the acceleration due to gravity is inversely
proportional to the square of the distance from the center
of the body to the center of the earth. All parts of the
earth's surface are not the same distance from its center.
The variations range from 393 m below sea level at the
shore of the Dead Sea to 8848 m above sea level at the top
of Mount Everest. An object at the top of a mountain,
where gravity is less, will weight less there than it does at
sea level. Also, since the earth is slightly flattened at the
poles and since the earth's rotation counteracts the force of
gravitation, an object weighs a little more at the North
Pole than it does at the equator.
3.14 Gravitational Fields The value of g, the accelera] tion due to gravity, at a particular point is sometimes
,
,-
(-
VELOCITY AND ACCELERATION
67
called the gravitational field strength at that point. The effect
of rotation, if any, is usually included in determining g. A
regionof space in whicheach point is associated with the value of
g at that point is called a gravitational field. Since g is an
acceleration, it is a vector. Hence, a gravitational field is a
region of space in which each point has associated with it
a vector equal to the value of g at that point and which is
called the gravitational field strength.
If an object of mass m is located at any point in a gravitational field, the force of gravitation on the object can be
calculated by the equation in Section 3.9
Fw = mg
If the mass of the object is 1 kg, then Fw is numerically
equal to g. Another way to think of the gravitational field
strength at a point is that it equals the force of gravitation
that would be exerted on a mass of one kilogram if the
mass were located at that point, or
g~-
\
I
,
/
Fw
m
The value of g for the earth varies with the distance from
the center of the earth and with certain other aspects of the
earth's motion and composition, In Figure 3-20, vectors
are used to represent the gravitational field strength at
various altitudes, Table 3-4 lists the values of gravitational
field strength for various locations on the earth's surface,
as determined by careful measurements,
Table 3-4
GRAVITATIONALFIELD STRENGTH OF THE EARTH
Location
North Pole
Greenland
Stockholm
Brussels
Banff
Chicago
New York
Denver
San Francisco
Canal Zone
Java
New Zealand
Latitude
90° North
70° North
590 North
51° North
51" North
42° North
41" North
40Q North
38° North
9° North
6° South
37° South
1m)
FIeld
strength
(N/kg)
0
20
45
102
1376
182
38
1638
114
6
7
3
9.832
9.825
9.818
9.811
9.808
9.803
9.803
9.796
9.800
9.782
9.782
9.800
Altitude
I
\
Figure 3-20, Simplified diagram
of the earth's gravitational field.
The inner circle of vectors represents the magnitude and direction
of the field at various places on
the earth's surface. The outer vectors represent the field at different altitudes.
-,
68
CHAPTER
The concept of force fields
ful in describing
various
is usenatural
phenomena.
QUESTlONS:
3
The concept of a field is more complex than the idea of
forces pushing on objects. Yet the field concept can explain more physical phenomena (such as electromagnetic
waves) than can the simpler concept of forces. In the field
concept, objects are surrounded
by regions that exert
forces on masses in those regions. Thus a ball falls to the
earth because of interaction between the earth's gravitational field and that of the ball.
Another way to describe a gravitational field is to say
that an object changes the properties of the space in its
vicinity. However, the field concept does not explain the
cause of gravitation any more than the idea of acceleration
does. Even though scientists are able to express certain
aspects of gravitation in terms of acceleration and force
fields, the real nature and origin of the force of gravitation
are still unknown.
CROUP A
1. (a) What force pulled the apple to
earth in the story of Sir Isaac Newton
under the apple tree? (b) Was there
another force acting here? Explain.
2. (a) If you constructed a graph of the
force of gravity as a function of distance from the earth, what would it
look like? (b) Would the curve ever
reach the abscissa?
3. Comment on this statement, "There is
no gravity in outer space."
4. Where would you live if you wanted
to have the smallest possible weight
and still remain on the surface of the
earth? (Refer to Table 3-4.)
5. If all objects have a gravitational field,
why do you not find yourself attracted
to the lockers in the school's hallway
as you walk to class?
6. Your textbook weighs about 13 N. If
you carry it from the kitchen table
(1 m from the earth's surface) to the
second floor (3 ill from the earth's surface), is its weight now one ninth of
13 N? Explain.
7. The legend is that Galileo dropped two
objects from the top of the Tower of
Pisa to show that objects in free fall
have the same acceleration. (a) What
are the action/reaction forces involved
here? (b) What type of forces are they?
8. (a) What is a gravitational field?
(b) How does field theory explain why
a falling object drops to earth?
PROBLEMS:
CROUP
11
1. Approximate the gravitational force of
attraction between a 50.0-kg girl and a
60.0-kg boy if they are sitting 2.50 m
apart in a physics class?
2. What is the weight of a 50.0-kg girl if
the acceleration due to gravity is
9.80 m/s2?
3. What would a 70.0-kg boy weigh on
the moon, where g = 1.62 m/52?
CROUP B
4. What is the force of gravitation between the proton (mass = 1.67 X
10-27 kg) and the electron in a hydrogen atom if they are 1.00 x 1O-1t m
apart?
5. Calculate g at a point 6.37 x 106 ill
above the surface of the earth.
6. Two satellites of equal mass are orbit-
VELOCITY
AND
ACCELERATION
69
iog the earth 50.0 ill apart. If the gravitational force between them is 1.50 x
10-7 N, find the mass of each.
7. The moon orbits the earth at a dis-
tance of 3.90 x lO!!m. If the gravitational force of attraction between them
is 1.90 x 102u N, what is the mass of
the moon?
Displacement is a change of position of
an object in a particular direction. Motion
is the displacement of an object in relation to stationary objects. Speed is the
time rate of change of position (displacement). Velocity is speed in a particular
direction. Speed is a scalar quantity, but
velocity is a vector quantity. Graphs can
be used to determine average and instantaneous speed and velocity. Velocity
problems can be worked either graphically or trigonometrically
by use of
the parallelogram method of vector
addition.
Acceleration is the rate of change of
velocity. The relationships among displacement, time, initial and final velocity,
and acceleration can be expressed as
equations. In the case of freely falling
bodies, the acceleration caused by the
earth's gravity is used.
A force is a quantity that can affect the
state of motion of an object. Newton's
laws of motion are (1) a body will con-
onue in its state of rest of uniform motion in a straight line unless a net force
acts on it; (2) the acceleration of a body is
directly proportional to the net force exerted on it, inversely proportional to its
mass, and in the same direction as the
force; (3) when one body exerts a force
on another body, the second body exerts
an equal force on the first body but in the
opposite direction.
The force of gravitation is the mutual
force of attraction between bodies. Newton's law of universal gravitation states
that the force of gravitation between two
objects is directly proportional to the
product of their masses and inversely
proportional to the square of the distance
between their centers of mass. The mass
of the earth can be calculated by means
of this law. The force of gravitation near
the surface of a celestial body is called
gravity. The values of gravity at various
distances from a celestial body make up
the gravitational field of the body.
f
1-
,-
I
~
~
~
r
~
~..
VOCABULARY
I
acceleration
average acceleration
average speed
average velocity
constant speed
constant velocity
displacement
gravitational field
gravity
instantaneous
speed
instantaneous
velocity
law of acceleration
law of inertia
law of interaction
law of universal
gravitation
motion
slope
speed
thought experiment
variable acceleration
variable velocity
velocity