Download Further Applications of Newton`s Laws

Newton’s laws are fundamental in
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—
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his chapter continues our study of Newton’s laws and emphasizes their fun
damental importance in physics. We cover some important applications of
Newton’s laws, including friction and the very important subject of the
dynamics of circular motion (we studied its kinematics in Chapter 3). Although
some material in this chapter may seem to repeat topics covered in Chapter 4. in
fact, new elements are involved.
T
An object moving to
the right (in a table or floor. The two
surfaces in contact are rough, at least
on a microscopic scale.
/
\
106
/
/
/IkFN.
relation is not a fundamental law; it is a useful relation between the magni
of the friction force Ftr which acts parallel to the two surfaces. and the magni
of the normal force FN which acts perpendicular to the surfaces, It is not a
or equation since the two forces are perpendicular to one another. The term
• called the coefficient of kinetic friction. and its value depends on the nature of
two surfaces. Measured values for a variety of surfaces are given in ThbIe 5—1.
‘,e are only approximate. however. since ,iik depends on whether the surfaces
wet or dry. Ofl hO’W much they have been sanded or rubbed. if any burrs
ain. and other such factors. But ,a is roughly independent of the sliding speed
dl as the area of contact.
What we have been discussing up to now is kinetic friction, when one object
s over another. There is also static friction, which refers to a force parallel to
two surfaces that can arise even when they are not sliding. Suppose an object
as a desk is resting on a horizontal floor. If no horizontal force is exerted on
desk, there also is no friction force. But now, suppose you try to push the desk,
t doesn’t move. You are exerting a horizontal force, but the desk isn’t moving.
here must he another force on the desk keeping it from moving (the net force
•
ro on an object that doesn’t move). This is the force of .sta’ic friction exerted
• he floor on the desk. If you push with a greater force without moving the desk.
force of static friction also has increased. If you push hard enough. the desk
• eventually start to move, and kinetic friction takes over. At this point.
have exceeded the maximum force of static friction, which is given by
/ max)
/I.,F\. where ,u is the coefficient of static friction (Table 5—1). Since
force of static friction can vary from zero to this maximum value, we rite
I
•
Further Applications of
Newton’s Laws
FIGURE 5—1
experiment shows that the friction force is approximately proportional to the
ia/force between the two surfaces, which is the force that either object exerts
ic other, perpendicular to their common surface of contact (see Fig. 5—2). The
•
of friction between hard surfaces typically depends very little on the total
ce area of contact: that is, the friction force on this book is roughly the same
her it is being slid on its wide face or on its spine. assuming the surfaces have
ame smoothness. We thus consider a reasonable model of friction in which we
me the friction force is independent of area. Then we can write the propor
ility between the friction force F. and the normal force F as an equation by
ting a constant of proportionality. ,u
:
1
•
physics, and this chapter covers
Some different applications. These
photos show two situations of using
Newton’s laws which involve some
new elements in addition to those
discussed in the previous chapter.
The downhill skier illustrates
friction on an incline, although at
this moment she is not touching the
snow, and SO is retarded only by air
resistance which is a velocitydependent force (an optional topic
in this chapter). The people on the
rotating amusement park ride on the
right illustrate the dynamics of
circular motion.
Applications of Newton’s Laws Involving
Friction
Until now we have ignored friction, but it must he taken into account in niost prac
tical situations. Friction exists between two solid surfaces because even the
smoothest looking surface is quite rough on a microscopic scale, Fig. 5—1. When we
try to slide an object across another surface, these microscopic bumps impede the
motion. Exactly what is happening at the atomic level is not yet fully understood.
It is thought that the atoms on a bump of one surface may come so close to the
atoms of the other surface that attractive electric forces between the atoms could
“hond as a tiny weld between the two surfaces. Sliding an object across a surface
is often jerk. perhaps due to the making and breaking of these bonds. Even when
a round object rolls across a surface, there is still some friction, called rolling
fric
[loll, although it is generally much less than when objects slide across a
surface. We
focus our attention now on sliding friction, which is usually called kinetic friction
(kinetic is from the Greek for “moving”).
When an object slides along a rough surface, the force of kinetic friction acts
opposite to the direction of the objecfs velocity. The magnitude of the force of
kinetic friction depends on the nature of the two sliding surfaces. For given sur
FA
ing
FIGURE 5—2 When an object is
pulled by an applied force (F)
along a surface, the force of friction
Fir opposes the motion. The
magnitude of Fir is proportional to
the magnitude of the normal force F.,.
•
r
1
F
ABLE 5—1
<
Coefficients of Friction
urfaces
Coefficient of
Static Friction,
iod on wood
Coefficient of
Kinetic Friction, I’k
0,4
0.2
0.1
tal on metal (lubricated)
0.15
0.03
0.07
eel on steel (unlubricated)
0.7
0.6
on ice
uhher on dry concrete
1.0
ubber on wet concrete
0.7
uhber on other solid surfaces
JIon® on Teflon in air
eflon on steel in air
1—4
•uhricated ball hearings
<
vnovial joints (in human limbs)
0.04
0.01
0.01
0.5
0.04
<0.01
0.01
‘values are approximate and are intended only as a guide.
SECTION 5—1
AppI ications of Newton’s Laws Involving Friction
107
You may have noticed that it is often easier to keep a heavy object moving, such
as pushing a table, than it is to start it moving in the first place. Indeed, we can sec
from Table 5--i that
is generally greater than bk
FN
y
A
F
0
—————
FA
Friction: static and kinetic. Our 10.0-kg mystery box rests
= Q4O and the coeffi
on a horizontal floor. The coefficient of static friction is
cient of kinetic friction is k
0.30. Determine the force of friction, Ffr acting on
the box if a horizontal external applied force F is exerted on it of magnitude:
(a) 0, (b) iON. (c) 20N, (d) 38Nand (e) 40N.
t L
‘
—
IL
F
(b)
(a)
SOLUTION
The free-body diagram of the box is shown in Fig. 5—2. Examine it
carefully. In the vertical direction there is no motion, so
= ma
0 yields
mg = 0. Hence the normal force for all cases is
F
mg
FIGURE 5—2 Repeated for
I ‘ample 5—1.
FIGURE 5—3 Magnitude of the force
ol Ii iction as a function of the external
trOCc applied to a body initially at rest.
\s the applied force is increased in
inaenitude, the force of static friction
increases linearly to just match it, until
the applied force equals
If the
applied force increases further, the
hod will begin to move, and the friction
loice drops to a roughly constant value
diaracteristic of kinetic friction,
5()
40
fig
FN
=
)
2
(lOMkg)(9.80m/s
=
98ON.
Fir
(a) Since no force is applied in this first case, the box doesn’t move, and Ffr = 0.
(h) The force of static friction will oppose any applied force up to a maximum of
,‘ISFN
(0.40)(98.ON)
/
30
=
The applied force is FA = iON. Thus the box will not move; since
10 N.
F
0 then Fir
‘fr
(c) An applied force of 20 N is also not sufficient to move the box. Thus
2() N to balance the applied force.
Fir
(d) The applied force of 38 N is still not quite large enough to move the box: so
the friction force has now increased to 38 N to keep the box at rest.
( e) A force of 40 N vill start the box moving since it exceeds the maximum force
of static friction, ,uF\
(0.40)(98.0 N) = 39 N. Instead of static friction, we now
have kinetic friction, and its magnitude is
20
Fir
50 60
0 Applied force, f
PN
no
motion
4
slidino
I-’k’N
=
(0.30)(98.ON)
=
29N.
There is now a net (horizontal) force on the box of magnitude F
11 N, so the box will accelerate at a rate
Kinetic
friction
Static
friction
10
10 20 30
=
0
ar
F/m
11 N/10 kg
40 N
29 N
=
as long as the applied force is 40 N. Figure 5—3 shows a graph that summarizes
this Example.
CONCEPTUAL EXAMPLE 5—2
To push or to pull a sled? Your little sister
wants a ride on her sled. If you are on flat ground. will you exert less force if you
push her or pull her? See Figs. 5—4a and h. Assume the same angle 0 in each case.
RESPONSE Free-body diagrams are shown in Figs. 5—4c and d. If you push her,
and 0 > 0, there is a vertically downward component to your force. Hence the
normal force exerted upward by the ground will be larger than mg (where in is
the mass of sister plus sled). If you pull her, your force has a vertically upward
component, so the normal force FN can be less than mg. Because the friction
force is proportional to the normal force, it will be less if you pull her. So you
exert less force if you pull her.
CHAPTER 5
Further Applications of Newton’s Laws
tag
(c)
(d)
A box against a walL You can hold a box
ainst a rough wall (Fig. 5—5) and prevent it from slipping down by pressing
ird horizontally. How does the application of a horizontal force keep an object
)m moving vertically?
ONCEPTUAL EXAMPLE 5—3
This won’t work well if the wall is slippery. You need friction. Even
en, if you don’t press hard enough. the box will slip. The horizontal force you
ESPONSE
F
iply produces a normal force on the box exerted by the all (why?). The force
gravity, tag. acting downward on the box, can now be balanced by an upward
Pulling against friction. A 10.0-kg box is pulled along a
orizontal surface by a force F of 40.0 N which is applied at a 30.0 angle. This is
ke Example 4—li in the preceding chapter except now there is friction, and we
ssume a coefficient of kinetic friction of 0.30. The free-body diagram is shown in
ig. 5—6. calculate the acceleration,
#
•
rng
iction force whose magnitude is proportional to the normal force. The harder
u push, the greater J is and the greater fr can be. If you don’t press hard
nough, then mg
SFN and the box begins to slide down.
1.1 rn/s
2
Now we look at some Examples involving kinetic friction in a variety of situ
ations. Note that both the normal force and the friction force are forces exerted by
one surface on another; one is perpendicular to the contact surfaces (the normal
force), and the other is parallel (the friction force).
108
Fir
F
mg
39N.
—
FfrUsFN
Example 5 2,
e
.
—
FIGURE 5—4
F
F,,
—
FIGURE 5—5
Example 5 3.
FIGURE 5—6
Example 5 4.
1
F
SOLUTION
The force of kinetic friction on the box opposes the direction of
notion and is parallel to the surfaces of contact. The calculation for the vertical
y) direction is just the same as before (Example 4—11): there is no motion in the
20.0 N is less
sin 30
(40.0 N)(sin 30.0 )
ertical direction because f
(10.0 kg)(9.80 m/s
)
2
98.0 N. ‘lking y as posi
han the weight of the box nig
30.0
—
Fr
ve upward, we have
.
1
F
N
ina,
rng +
98.0 N
+
20.0 N
=
ing
0
and the normal force is F\ = 78.0 N. Next we apply Newton’s second law for the
rnrizontal (x) direction (positive to the right):
=
,na
SECTION 5—1
Applications of Newton’s Laws Involving Friction
109
The friction force is kinetic as long as Ffr
‘fr
(0.30)(78.ON)
/‘kFN
—
ILk
=
=
[‘
is less than
which it is since
234 N
and
Fcos30.0
=
(40.ON)(0.866)
=
=
34.6N
[If fLk FN were greater than F
, what would you conclude?] Hence the box does
1
accelerate:
Fp
34.6 N
23.4 N
10.() kg
‘r
x
-—
in
=
l.lm/s
In the absence of friction, as we saw in Example 4 11, the acceleration ould be
much greater. Note that our final answer should have only to significant figures
because our least significant input value has two
0.30)
5.0 kg
II
2.0 kg
(a)
Two boxes and a pulley. In Fig. 5—7a, two boxes are con
nected by a cord running over a pulley. The coefficient of kinetic friction between
box I and the table is 0.20. We ignore the mass of the cord and pulley and any fric
tion in the pulley, which means we can assume that a force applied to one end of
the cord will have the same magnitude at the other end. We wish to find the accel
eration, a, of the system. which will have the same magnitude for both boxes
assuming the cord doesn’t stretch. As box II moves down, box I moves to the right.
—
g
1
m
)
2
(5.Okg)(9.8m/s
-
—
49N.
In the horizontal direction, there are two forces on box I (Fig. 5—7b): F
, the ten
1
sion in the cord (whose value we don’t know), and the force of friction
FT
Ffr
11
F
(0.20)(49N)
9.8 N.
The horizontal acceleration is what we wish to find; we use Newton’s second law
in the x direction,
a. which becomes (taking the positive direction to
1
rn
the right and setting a
1
a):
1
F
FT
The skier. The skier in Fig. 5—8a has just begun descending
30° slope. Assuming the coefficient of kinetic friction is OAO, calculate (a) her
eleration and (b) the speed she will reach after 4.0 s.
OLUTION First we draw the free-body diagram showing all the forces acting
the skier. Fig. 5—8b: her weight (F(, = mg) downward, and the two forces
rted on her skis by the snow—the normal force perpendicular to the snow’s
face, and the friction force parallel to the surface. These three forces are
wn acting at one point in Fig. 5—8b, for convenience. Also for convenience, we
ose the x axis parallel to the snow surface with positive direction downhill,
d the y axis perpendicular to the surface. With this choice, we have to resolve
ly one vector into components, the weight, and its components are shown as
shed lines in Fig. 5—Sc. They are given by
Fux
Fr
—
ma.
[box I]
Next consider box II. The force of gravity F(,
g
11
in
19.6 N pulls downward;
and the cord pulls upward with a force F
1 So we can write Newton’s second law
for box II (taking the downward direction as positive):
.
1
F
11
m
g
—
I’
11
m
a,
1 is not equal to g.J
0, then F
11 We have two unknowns, a
m
and F
1 and e also have to equations. We solve the box I equation for F
:
1
(a)
sin 0,
mere we have stayed general, using (9 rather than 30 for now.
To calculate her acceleration down the hill, ar, we apply Newton’s second law
the x direction:
F
mgsinO
—
mar
mar
/..LkFN
ere the two forces are the component of the gravity force (-x direction) and
friction force ( x direction). We want to find the value of ar, but we don’t yet
ow FN in the last equation. Let’s see if we can get F\ from the y component of
wton’s second law:
—
mg
cos 0
ma,
(h)
FӍl u
0
ma
()because there is no motion in the y direction (perpendicular
ere we set av
the slope). Thus we can solve for F:
..
1
F
—
rng
cos (9
(c)
d we can substitute this into our equation above for mar:
mg
[box IIj
[Note here that if a
mg
mgcos0,
SOLUTION Free-body diagrams are shown for each box in Fig. 5—7h and c. Box
I does not move vertically, so the normal force just balances the weight.
Fv
(h)
Next we take an example of an object moving down an incline, as in Example
1 of Chapter 4; but now we include friction,
sin 0
,u(mg cos 0)
Example 5 6:
A skier descending a slope.
FIGURE 5—8
mar
iere is an in in each term hich can he canceled out. Thus (setting 0
=
30 and
0.10):
,
(c)
11
m
g
FIGURE 5—7 Example 5-5.
1
F
Fir
+
a
1fli(1.
“fr
a
1
n
=
a.
11
1n
ml
—
1
F
19.6 N
5.0kg
+ m
1
—
9.8N
=1.4m/s
2.0kg
-
--
which is the acceleration of box I to the right, and of box II down. If we wish, we
can calculate FT using the first equation:
1
F
110
CHAPTER 5
—
F
+
Further Applications of Newton’s Laws
a
1
rn
=
9.$N
+
cos 30
050 g
(0.l0)(0.866)g
0.41 g.
)
2
(5.Okg)(1.4m/s
PROBLEM
SOLVING
pend on the mass. That such a cancellation sometimes occurs, and thus may
ye a useful conclusion as well as saving calculation, is a big advantage of work
g with the algebraic equations and putting in the numbers only at the end,
The speed after 4.0 s is found. since the acceleration is constant, by using
Now we solve for a and put in numerical values:
g
11
rn
g
e skier’s acceleration is 0.41 times the acceleration of gravity, which in num
4.0 m/s°. It is interesting that the mass canceled
)
2
(0.41 )(9.8 m/s
rs is a
useful conclusion that the acceleration doesn ‘t
have
the
we
and
so
ut here,
and substitute this into the box II equation:
g
1
In
g sin 30
=
17N.
q. 2—12a:
v
Co + at
—
0
+
)(4.0 s)
2
(4.0 m/s
—
16 rn/s.
ere we assumed a start from rest.
SECTION 5—1
Applications of Newton’s Laws Involving Friction
111
Measuring I’k Suppose in Example 5—6 that the snow is
slushy and the skier moves down the 30 slope at constant speed. What can you
say about the coefficient of friction, ph’?
V
FT
SOLUTION Now the skier moves down the slope at constant speed. and w
want to find ik. The free-body diagram and the F
ma equations for the
and y components will be the same as above, except that now we are give i
(l = 0. Thus
=
=
—
nig
nig cosO
sin 0
—
1
ma
’N
k T
1
/
=
nla
=
mg
=
0.
From the first equation. we have F...,
ond equation:
sin 0
tug
Now ‘,ye solve for
—
,uk(mg cos 0)
k=
1
which for 0
=
=
11
F
0.
=—
mg c05
0
COS
0
tan 0
tan 3()
=
(case
1
a
1 since F is related to
0 and we solve for F
ing (see Fig. 5—9c). Thus
lngsin0
11
F
-
(whose value we are
,ii
rn
g
2
.
Fl
pmgcos0, the minimum value that
F...
5
Ffr can be at most t
0) is. after dividing through by g.
to prevent motion (a
Jib
1111 sin
—
0
—
1
mgsin0 + F
0.40
2.8kg.
)
cos37
1
F
ill2g
iii,
1n
mgsin0 ±
sin(9
+ eiil
without causing acceleration is given by
/il1l
C
1
OS0
cos0
0.10 cos 37
(10.0 kg)(sin 37
=
0.
rna
Fr
2 can have
the maximum value m
1
in
SOLUTION (a) Figure 5—9b shows two free-body diagrams for box n1
1 The
tension force exerted by the cord is labeled Fr. The force of friction can be either
up or down the slope, and we show both possibilities in Fig. 5—9h: (i) if iii
0 or
is sufficiently small. 1
t would tend to slide down the incline, so Fr would be
directed up the incline; (ii) if in
2 is large enough,
will tend to be pulled up the
plane, so Fir would point down the plane. For both cases, Newton’s second law for
the v direction (perpendicular to the plane) is the same:
can
if 1112 < 2.8 kg. then box I will slide down the incline.
for case (ii). Newton’s second law is
fl
A plane, a pulley, and two boxes. A box of mass
10.0 kg rests on a surface inclined at (
37 to the horizontal. It is con
nected by a lightweight cord, which passes over a massless and frictionless pulle
to a second box of mass ni, which hangs freely as shown in Fig. 5-9a. (a) If the
coefficient of static friction is
0.40, determine what range of values for
mass m will keep the system at rest. (b) If the coefficient of kinetic friction is
0.30. and m
10.0 kg. determine the acceleration of the system.
2
m
0
/L, lilt C05
(10.0 kg)(sin 37
tan0
to determine ,u under a variety of conditions. All we need to do is observe at
what slope angle the skier descends at constant speed. Here is another reason
why it is often useful to plug in numbers only at the end: we obtained a general
result useful for other situations as well.
(c)
Example 5-$.
siRE 5—9
ing) by F
1
0.58.
ii)
(h)
(a)
11
=tan0
lIltg
(case i)
cos 0; we substitute this into the sec
Notice that we could use the equation
k
1
/
g
1
ni
O=37/
ant
sin0
Ill,
0
19
30 is
I’k
rn =
10.0 kg
0
Pk:
imiç’sinO
FT
)
=
9.2 kg.
s to prevent motion. ‘e have the condition
2.8 kg
9.2kg.
1112
0.30. then
If 1112 = 10.0 kg and /
k
1
li an acceleration a given by
iii
will fall and
nit
will rise up the plane
.
1fl a
El
—
g sin 0
1n
1
2 also accelerates. F
e in
—
k I\.
t
Im,a
g
2
in
(see Fig. 5-9c) and we substitute this
the equation above:
1
rn
—
lilt
g cos 0
alt a
0
111 a
m2g
g sin (3
m2a
-
Rk
“N
solve for the acceleration a and substitute F..,
1172
since there is no v motion. So
—
=
=
COS (9.
and then
1 0.() kg. to find
Jib g
—
hhul i,
sin 0
—
I’k fl1t
g
COS
0
a=----—
—
lflt + Jib
mgcos0.
Now for the x motion. We consider case (i) first for which F
ni g
112
CHAPTER 5
sin 0
—
Further Applications of Newton’s Laws
1
F
—
Fir
in a,
—
ma gives
(9.8m/s
)
2
(10.Okg)(1
—
sin37
0,30cos37
20.0 kg
SECTION 5—1
0.079g
-
.
2
0.7$rn/s
Applications of Newton’s Laws Involving Friction
113
Friction can be a hindrance. It slows down moving objects and causes heating
and binding of moving parts in machinery. Friction can be reduced by using lubri
cants such as oil. More effective in reducing friction between two surfaces is t
maintain a layer of air or other gas between them. Devices using this concept
which is not practical for most situations, include air tracks and air tables (o
games) in which the layer of air is maintained by forcing air through many tin
holes. Another technique to maintain the air layer is to suspend objects in air using
magnetic fields (“magnetic levitation”). On the other hand, friction can he helpful
Our ability to walk depends on friction between the soles of our shoes (or feet
and the ground. (Does walking involve static friction or kinetic friction?) Thc
movement of a car, and also its stability, depend on friction, When friction is low
such as on ice, safe walking or driving become difficult.
——
/
/
/
/
A force is required
an object moving in a circle.
FIGURE 5—11 Swinging a ball on
the end of a string.
E 5—10
peed is constant, the force
ted toward the center of
Ic.
I
here is a common misconception that an object moving in a circle has an
rd force acting Ofl it, a so-called centrifugal (“center-fleeing”) force. Con-
br example, a person swinging a ball on the end of a string around her head
11). If you have ever done this yourself, you know that you feel a force
outward on your hand. The misconception arises when this pull is interas an outward “centrifugal” force pulling on the ball that is transmitted
the string to your hand. This is not what is happening at all. To keep the ball
g in a circle, you pull inwardly on the string, which in turn exerts the force on
II. The ball exerts an equal and opposite force on your hand (Newton’s third
md this is the force your hand feels (see Fig. 5—Il). The force on the hal/is
me exerted inward/v on it by the string.
For even more convincing evidence that a “centrifugal force” does not act Ofl
ill, consider what happens when you let go of the string. If a centrifugal force
acting, the ball would fly outward, as shown in Fig. 5—12a. But it doesn’t: the
lies off tangentially (Fig. 5—12b), in the direction of the velocity it had at the
i rnt it was released, because the inward force no longer acts. Try it and see!
r
This acceleration, aR, is called radial or centripetal acceleration because it is
directed toward the center of the circle. For uniform circular motion (v
com
stant), the magnitude of the acceleration is constant, but its direction is continually
changing. Hence the acceleration vector an is a variable acceleration. The direction
of an is always perpendicular to the velocity v. [The equation an
/r is valid
2
v
even if v is not constant, and we will treat that case in Section 5—4.j
An object moving uniformly in a circle, such as a ball swung horizontally on
the end of a string, must have a net force applied to it to keep it moving in that
circle, instead of in a straight line. That is. a net force is necessary to give it a
centripetal acceleration. The magnitude of the required net force can be calculated
using Newton’s second law for the radial component,
man, where
/r is the centripetal acceleration, and
2
is the total (net) force in the
an = v
radial direction:
—
man
=
m
‘
[circular motionl
(5—1)
Since an is directed along the radius toward the center of the circle at any moment,
the net force too must he directed toward the center of the circle for uniform circu
lar motion (v
constant). A net force is clearly necessary because otherwise, if no
net force were exerted on the object, it would not move in a circle hut in a straight
line, as Newton’s first law tells us. To pull an object out of its “natural” straight-line
path, a net force to the side is necessary. For uniform circular motion, this sideways
net force must act toward the circle’s center (see Fig. 5—10). The direction of the
force is thus continually changing so that it is always directed toward the center of
the circle, This force is sometimes called a centripetal (“aiming toward the cen
ter”) force. But be aware that “centripetal force” does not indicate some new kind
of force. The term merely describes the direction of the force: that the net force is
directed toward the circle’s center. The force must he applied by other objects. For
example. when a person swings a ball in a circle on the end of a string, the person
pulls on the string and the string exerts the force on the ball,
CHAPTER 5
Force on hand
/
In Section 3—9 we saw that a particle revolving in a circle of radius r with uniforn
speed v undergoes a radial (or centripetal) acceleration which at any moment is
given by
114
“5
exerted by
string
Dynamics of Uniform Circular Motion
Fn
.\
/
F
Ii
an
.—‘‘
Force on ball
exerted by
string
Further Applications of Newton’s Laws
I
If centrifugal force existed, the ball would fI off as in (a) when released.
it flies off tangentially as in (h). Similarly, sparks fly in straight lines tangentially from
Ige of a rotating grinding wheel (c).
RE 5—12
t,
\
/
I
/
(a)
\
(c)
(b)
SECTION 5—2
Dynamics of Uniform Circular Motion
115
FT
-
FIGURE 5—13
Example 5 9.
—
__
0
r
—
—
)
We solve the two equations above for
Lsin0):
.igr
/
mg
sin 0
i
1
ill
ill
: ESTIMATE
:
mate the force a person
Force on revolving ball (horizontal). Est
must exert on a string attached to a 0.150-kg ball to maL
the hail revolve in a horizontal circle of radius 0.60(3 m. The ball makes 2.00 revo
lutions per second (T = 0.500 s). as in our earlier Example 3—11.
SOLUTION First we draw the free-body diagram for the ball. Fig. 5—13, whic
shows the two forces acting on the hail: the force of gravity, rng; and the tensio
force F
1 that the string exerts (which occurs because the person exerts that sam
force on the string). The hall’s weight complicates matters and makes it impossi
hie to revolve a ball with the cord horizontal. But if the weight is small enough. w
can ignore it. Then F will act nearly horizontally (b
0 in Fig. 5—13) and provid
the force necessary to give the ball its eentripetal acceleration. We apply Newton
second law to the radial direction, which now is horizontal, so we call it .v:
i11(i
where a
v
/
2
r and v
2r/7
2ir(0.60() m )/(0.50() s)
-
,
1
F
in
(0.h0k)
I
(7.54 rn/s)
2
(0.b00 m)
-
7.54 rn/s. Thus
vertical component equal to nig
something like Example 5 l() below.]
Lg
mg
COS
sin 1)
cos 0
01
e period T is the time required to make one revolution, a distance of
c = 2irL sin 0/T: then
= 2L sin 0. The speed r can thus he
written
T
2rrLsin0
ijLSji2f
20
[Lg sin
I,
V
2
A
/Lcos0
g
ing
Cos 0
Revolving ball (vertical circle). A 0.150-kg ball on the end
1.10-rn-long cord (negligible mass) is swung in a vertical circle. (a) Determine
minimum speed the ball must have at the top of its arc so that it continues
ing in a circle. (h) Calculate the tension in the cord at the bottom of the arc
e ball is moving at twice the speed of part (a).
i
SJLUTION The free-hod diagram is shown in Fig. 5—15 for both situations.
.
1
At the top (point A). two forces can act on the ball: ing. its weight: and F
lension force the cord exerts at point A. Both act downward, and their vector
a acts to give the ball its centripetal acceleration aR. We apply Newton’s second
for the vertical direction. choosing downward as positive (toward the center):
14N.
where we have rounded off because our estimate ignores the hall’s eight
We keep only two significant figures here in the answer because mg
)
2
(0.lsokg)(9.80m/s
1.5 N. being about l/1() of our result, is small. hut not s’
small as to justify stating a more precise answer since we ignored the effect o
mg. [Note: if you want to include the effect of ing here, resolve F
1 in Fig. 5—1
into components. and set the horizontal component of F
1 equal to ,nv
/r and it
2
/
1 between them (and
by eliminating F
‘
1
F
FIA + fig
=
—
nmR
/
\
/
1
FTn
/
\
/
/
....__.o
nig
Example 5—Il. with
lree—bod diagrams at the l o positions.
FIGURE 5—15
m
5
, at A will get larger if v
1
this equation we can see that the tension force F
s speed at top of circle) is made larger. as expected. But we are asked for the
mum speed to keep the hail moving in a circle. The cord will remain taut as
as there is tension in it: hut if the tension disappears (because v is too
dl) the cord can go limp, and the ball will fall out of its circular path. Thus, the
in i/r. Then:
0. for which we have ing
1 \
mum speed will occur if F
—
Example 5 10.
Conical pendulum.
FIGURE 5—14
—
the string makes with the vertical (Fig. 5—14). (a) In what direction is the acceie
atioii of the ball. and what causes the acceleration? (b ) Calculate the speed anc
period (time required for one revolutiotl) of the hall in terms of L. 0. g. and iii.
—
—
frn0
Fu
__
SOLUTION (a) Tile acceleration points horizontally toward the center of thc
ball’s circular path (not along the cord). The force responsible for the acceleratior
is the net force which here is the vector sum of the forces acting on the mass m
its weight
(of magnitude F
mg) and the force exerted by the tension in thc.
. The latter has horizontal and vertical components of magnitude F
1
cord, F
1 sin (
and F
1 eos 0, respectively. We apply Newton’s second law to the horizontal and
vertical directions, In the vertical direction, there is 110 motion, so the acceleratior
is zero and the net force in tile vertical direction is zero:
1 eos 0
F
nig
--
116
CHAPTER 5
a circular path.
cord exerts its tension force
ipward whereas tile force of gravity. iiig. acts dosnward. So Newton’s second
I his time choosing upward as positive (toward the center). gives
nig
111
F
—
In
speed v
0 is given as twice what e found in (a), namely 6.56 m/s. [Note that
peed changes here because gravity acts on the ball all along the path, hut
/r.j We solve for F
2
mi’
111 in the last equation:
I still remains valid. FR
0.
111
F
+ nig
=
-
=
Further Applications of Newton’s Laws
3.28rn/s.
\t the bottom of the circle (see Fig. 5—15) the
second law tells us:
in
=
is the minimum speed at the top of the circle if the ball is to continue mo—
0
In the horizontal direction there is o1ll’ one force, of magnitude F sin 0. that acb
toward tile center of the circle aild gives rise to the acceleration i’
/r. Newton’s
2
sinO
1
F
)( 1.10111)
2
\ (9.SOm/s
\ gr
Conical pendulum. A small ball of mass in. suspended b’
a cord of length L. revolves in a circle of radius r = L sin 0. where 0 is the angh
-
i
(0.bokg)
(6.56
m/s)
+ (0.b0kg)(9.SOm/si
=
7.34N.
(1.10111)
that the cords tension not only provides the centripetal acceleration, but
niau to compensate for tile downward force of gravity.
he even larger than
SECTION 5—2
Dynamics of Uniform Circular Motion
117
Skidding on a curve.
A 1000-kg car rounds a curve on a
oad of radius 50 m at a speed of 50 km/h (14 m/s). Will the car make the
I, Draw a free-body diagram, showing all the forces act
ing on each object under consideration. Be sure you
can identify the source of each force (tension in a
cord, Earth’s gravity, friction, normal force, and so
on), so you don’t put in something that doesn’t
belong (like a centrifugal force).
2. Determine which of these forces, or which of their
components, act to provide the centripetal accelera
tion—that is, all the forces or components that act
radially, toward or away from the center of the circu
lar path. The sum of these forces (or components)
provides the centripetal acceleration, aR
/r.
2
v
3. Choose a coordinate system. and positive and nega
tive directions, and apply Newton’s second law to the
radial component:
FR
_
maR
= in
;:
[radial directionj
Include only radial components of force.
or will it skid, if: (a) the pavement is dry and the coefficient of static friction
0.60; (b) the pavement is icy and ,u = 0.25?
S
UTION Figure 5—18 shows the free-body diagram for the car. The normal
N. on the car is equal to the weight since the road is flat and there is no
al acceleration:
FN
=
mg
=
)
2
(l000kg)(9.8m/s
=
9800N.
c horizontal direction the only force is friction, and we must compare it to
orce needed to produce the centripetal acceleration to see if it is sufficient,
net horizontal force required to keep the car moving in a circle around the
is
FR
Highway Curves, Banked and Unbanked
An example of centripetal acceleration occurs when an automobile rounds a curve
In such a situation, you may feel that you are thrust outward. But there is not somc
mysterious centrifugal force pulling on you. What is happening is that you tend t
move in a straight line, whereas the car has begun to follow a curved path. To
make you go in the curved path, the seat (friction) or the door of the car (direct
contact) exerts a force on you (Fig. 5—16). The car itself must have an inward forcc
exerted on it if it is to move in a curve. On a flat road, this force is supplied by fric
tion between the tires and the pavement.
If the wheels and tires of the car are rolling normally without slipping or slid
ing. the bottom of te tire is at rest against the road at each instant: so the friction
force the road exerts on the tires is static friction. But if the static friction force is
not great enough, as under icy conditions, sufficient friction force cannot be applied
and the car will skid out of a circular path into a more nearly straight path. Sec
Fig. 5—17. Once a car skids or slides, the friction force becomes kinetic friction
which is less than static friction.
= man
—
In
—
(1000 kg)
2
(14m/s)
(50m)
=
3900 N.
rally we hope the maximum total friction force (the sum of the friction
s acting on each of the four tires) will be at least this large. For (a),
0.60, and the maximum friction force attainable (recall from Section 5—1
I)is
(r)ma\
(0.60)(9800N)
=
=
FIGURE 5—17 Race car heading
down into a curve. From the tire
5900 N.
a force of only 3900 N is needed, and that is, in fact, how much will he
d by the road as a static friction force, the car can make the turn fine. But
the maximum friction force possible is
(Ftr)rni
,LLSFN
2500N.
(0.25)(9800N)
marks we can see that most cars
experienced a sufficient friction
force to give them the needed
centripetal acceleration for rounding
the curve safely. But, we can also see
a few tire tracks of cars on which
ar will skid because the ground cannot exert sufficient force (390() N is
d) to keep it moving in a curve of radius 50 m.
there was not sufficient force—and
which followed more nearly straight
line paths.
F
FIGURE 5—16 The road exerts an inward force (friction against the
tires) on a car to make it moe in a circle; and the car exerts an inward
force on the passenger.
F
Ffr
0 = mg
F
(b)
(a)
Force on car
(sum of friction forces
acting on each tire)
Tendency for
passenger to
go straight
Forceon
passenger
118
CHAPTER 5
Further Applications of Newton’s Laws
FIGURE 5—18 Forces on a car
rounding a curve on a flat road.
Example 5 12. (a) Front view,
(b) top view.
ituation is worse if the wheels lock (stop rotating) when the brakes are
d too hard. When the tires are rolling, static friction exists. But if the wheels
the tires slide and the friction force, which is now kinetic friction, is less.
)ver, when the road is wet or icy, locking of the wheels occurs with less force
brake pedal since there is less road friction to keep the wheels turning
than sliding. Antilock brakes (ABS) are designed to limit brake pressure
fore the point where sliding would occur, by means of delicate sensors and a
mputer.
SECTION 5—3
Highway Curves, Banked and Unbanked
119
The banking of curves can reduce the chance of skidding because the normc
force of the road, acting perpendicular to the road, will have a component towal
the center of the circle (Fig. 5—19). thus reducing the reliance on friction. For
given banking angle. 0. there will be one speed for which no friction at all
required. This will he the case when the horizontal component of the normal fore
toward the center of the curve. F\ sin 0 (see Fig. 5—19). is just equal to the fore
required to give a vehicle its centripetal acceleration—that is. when
V
I
Nonuniform Circular Motion
peed of a particle revolving in a circle is changing. there will be a tangential
ation, ataii. as well as the radial (centripetal) acceleration. tifl. The tangential
F ation arises from the change in the magnitude of the ve1ocitv
(Ic
atan
EN
F sin 0
FN
i
R
The banking angle of a road. 0, is chosen so that this condition holds for a particu
lar speed, called the “design speed.”
the radial acceleration arises from the change in direction of the velocity.
we have already seen, has magnitude
aR
FIGURE 5—19 Normal force on a
car rounding a banked curve.
resolved into its horizontal and
ertical components. \ote that the
ceniripetal acceleration is horizontal
(and not parallel to the sloping
road). The friction force on the tires
is not shown. It could point up or
down the slope, depending on the
car’s speed. The friction force will he
zero for one particular speed.
‘() for a particle moving counterclockwise. If the speed is decreasing.
a
ing
ataii
atati
0.
+
+(i
11
(ij
mstant when the speed is a
[Note in this case that F\
mg since cos 1)
F\ into the equation for the horizontal motion.
1.1
We substitute this relation fo
UTION
(35 rn/s
j’
in
—
cos 0
-
aR
in
=
15 rn/s. and again when
i’
30 rn/s.
—
Orn/s)
(15 rn/s)
2
=
(500m)
0.45
rn/s2
second instant,
rngtanO=rn
aR
=
so
—
2
(30 rn/s)
(500rn)
1.8rn/s
idial acceleration continually increases, although the tangential acceleration
a
2
rg
constant.
This is the formula for the banking angle 0.
(b) Forr = 50rn and a
50 km/h (or 14m/s),
hese concepts can be used for an object moving along any curved path. such
shown in Fig. 5—2 1 We can treat any portion of the curve as an arc of a cir
F Ii a radius of curvature a. The velocity at any point is always tangent to the
5°
0
=
(14 rn/s)
2
=
-
—
(0 m)(9.S rn/s-)
22
0.40.
FIGURE 5—21 Particle following a
curved path. At point P the path has
a radius of curvature r: the particle
has elocity v, tangential
acceleration a (it is increasing in
speed), and radial (centripetal)
/r pointing
2
v
acceleration tiR
toward the center of curvature C.
P
F
.
tan0
FIGURE 5—20 For nonunilorm
circular motion, the acceleration has
a tangential (a) as ell as a radial
(tiR) component.
us
—
or
tan0
7
the same for both instants.
the earlier instant.
and obtain
sin0
/
/
7
(a) anin is constant, of magnitude
(It
mg
\
a circular track of radius 500 m. Assuming constant tangential
ation, find (a) the tangential acceleration, and (b) the radial acceleration,
rn
0
aR
(5—3)
a.
is
\
/
i )ving on
cos 0
5fl
,‘,,
Two components of acceleration. A racing car starts
st in the pit area and accelerates at a uniform rate to a speed of 35 rn/s in
Thus,
F\
/
1
a
are always perpendicular to each other, the magnitude of a at an’
a
In the vertical direction, the forces are FN cosO upward (Fig. 5—19) and the weigi
of the car (rng) downward. Since there is no vertical motion, the y component t
the acceleration is zero, so
ma gives us
F=
and
.i,.
t
J/cosO
tic
intiparaflel to v. In either case. a
111 and a are always perpendicular to each
nd their directions change continually as the particle moves along its circu
h. The total vector acceleration, a. is the sum of these two:
We choose our x and v axes as horizontal and vertical so that ar
which is horizontal, is along the x axis. The components of F\ are as shown Vi
Fig. 5—19. (a) For the horizontal direction.
gives
2
,11r
‘_;“%
/
gential acceleration always points in a direction tangent to the circle, and is
direction of motion (parallel to v) if the speed is increasing, as shown in
SOLUTION
fNsinO
.——,
r
/
Banking angle. (a) For a car traveling with speed c aroun
a curve of radius r, determine a formula for the angle at which a road should b
banked SO that no friction is required. (b) What is this angle for an expresswa
off-ramp curve of radius 50 m at a design speed of 50 km/h?
mg
(5—2)
(It
—
m
—
lie acceleration can be written, in general. as a sum of two components. the
component
and the radial (centripetal) component
= di’/dt.
ataiF
/
C
itial
1’
120
CHAPTER 5
Further Applications of Newton’s Laws
*SECTIQN 5_4
Nonuniform Circular Motion
121
Ftan
,,a
-%
—
/
‘F
/
L___
\
FR
\
/
/
FIGURE 5—22 The speed of a
particle moving in a circle changes if
the net force on it has a tangential
component.
Circular motion at constant speed occurs when the net force on an object
exerted toward the center of the circle, If the net force is not directed toward tF
center, but acts at an angle as shown in Fig. 5—22, then the force has two comp
nents. The component directed toward the center of the circle, FR, gives rise to ft
centripetal acceleration, an, and keeps the object moving in a circle. The comp
nent Ftan, tangential to the circle, acts to increase (or decrease) the speed and thu
gives rise to the tangential acceleration (Fig. 5—20).
When you first start revolving a ball on the end of a string around your hea
you must give it tangential acceleration. You do this by pulling on the string wi
your hand displaced from the center of the circle. In athletics, a hammer throwc
accelerates the hammer tangentially in a similar way so that it reaches a hig
speed before release.
=
0
F
—
-
dv/dt
g. but as time goes on dv dt
(= slope of curve) decreases because
of FJ). Eventually. v approaches a
maximum value, v
1 the terminal
0 has
velocity, which occurs hen F
magnitude equal to ing.
mg
The minus sign is necessary since the drag force opposes the motion. Here b is
constant (approximately) that depends on the viscosity of the fluid and on the siz
and shape of the object. This works well for small objects moving at low speed in
viscous liquid. It also works for very small objects moving in air at very low speed
such as dust particles. For objects moving at high speeds, such as an airplane, a sk
diver, a baseball, or an automobile, the force of air resistance can be better appro
imated as being proportional to v
:
2
by
= 0.
(5—6)
(5—7)
—
b
resistive force is assumed proportional to v
, or an even higher power of v,
2
quence of events is similar and a terminal velocity reached, although it will
given by Eq. 5—7.
Force proportional to velocity. Determine the velocity as
tion of time for a body falling vertically from rest when there is a resistive
linearly proportional to v.
‘
UTION
We start with Eq. 5—5 and
are two variables, v and t. We collect variables of the same type on one or
er side of the equation:
dv
—dt
h
—v
g
m
dv
mg
h
or
.
)
1
F
= —hV
rng
(a)
From Newton’s second law F
—
—
ilf
mg
bv=rn
(
lnv
b
}
/
—ln—
ln
have
dv
dt
(5—5
Jo
mg
b
m
b
v—mg/h
—mg/b
=
h
m
t.
each side to the exponential [note that the natural log and the exponen
x, or In (e’)
x] and obtain
inverse operations of each other:
—
h
we
dt
In
se
—
ma,
Pt
b
b
fflg\\
—
—
h
di.
in
= 0 at t = 0:
e can integrate, remembering v
0 x v
F
.
2
0
mg + F
mg
by.
=—
—
—
For accurate calculations, however, more complicated forms and numerical intc
gration generally need to be used, For objects moving through liquids. Eq. S
works well for everyday objects at normal speeds (e.g.. a boat in water).
Let us consider an object that falls from rest, through air or other fluid, undc
the action of gravity and a resistive force proportional to v. The forces acting o
the object are the force of gravity, mg, acting downward, and the drag force, —b
acting upward (Fig. 5—23a). Since the velocity v points downward, let us take th
positive direction as downward. Then the net force on the object can be written
write
h
dv
v.
=g—
m
dt
(5—4
by.
—
Oand the object no longer increases in speed. It has reached
s point dv/dt
minal velocity and continues to fall at this constant velocity until it hits the
1. This sequence of events is shown in the graph of Fig. 5—23b. The value of
minal velocity VT can be obtained from Eq. 5—6:
Velocity-Dependent Forces; Terminal Velocity
When an object slides along a surface, the force of friction acting on the object
nearly independent of how fast the object is moving. But other types of resisti
forces do depend on the object’s velocity. The most important example is for a
object moving through a liquid or gas, such as air. The fluid offers resistance to ti
motion of the object, and this resistive force, or drag force, depends on the veloci
of the object.
The way the drag force varies with velocity is complicated in general. But &
small objects at very low speeds, a good approximation can often be made f
assuming that the drag force, FJ), is directly proportional to the velocity, v:
FIGURE 5—23 (a) Forces acting on
an object faIling downward.
(b) Graph of the velocity of a body
falling in air when air resistance drag
force is FD
by. Initially, v = 0 and
itude of the resistive force, —by, approaches that of the gravitational force, mg;
the two are equal, we have
mg
or finally,
= h
0
(1
e
ml)
elation gives the velocity v as a function of time and corresponds to the
of Fig. 5—23b. As a check, note that at t = 0, v = 0 and
m(b\\
h
dv
ml)
=
1
e
a(t = 0) =
b
dt
b dt
a
t
approaches zero, so v approaches
pected (see also Eq. 5—5). At large t, e m
which is the terminal velocity, v, as we saw earlier. If we set r = in/b,
e 7-), So r = ni/b is the time required for the velocity to reach
VT(1
cent of the terminal velocity (since e = 0.37).
—
—
—
(b)
where we have written the acceleration according to its definition as rate o
change of velocity, a = dv/dt. At t = 0. v = 0 and the acceleration dv/dt = g. Bu
as the object falls and increases in speed, the resistive force increases, and thi
reduces the acceleration, dv/dt (see Fig. 5—23h). The velocity continues t
increase, but at a slower rate. Eventually, the velocity becomes so large that th
—
Any buoyant force (Chapter 13) is ignored in this Section.
122
CHAPTER 5
Further Applications of Newton’s Laws
*SECTION 5—5
Velocity-Dependent Forces; Terminal Velocity
123
I
t
Summary
When two bodies slide over one another, the force of fric
tion that each exerts on the other can he written approxi
mately as Er =
E, where E is the normal force (the
force each body exerts on the other perpendicular to their
contact surface), and ,u.k is the coefficient of kinetic fric
tion, if the bodies are at rest relative to each other, then
Fir is just large enough to hold them at rest and satisfies
the inequality Fir
where
is the coefficient o
static friction,
A particle revolving in a circle of radius r with constan
speed v must have a net force acting on it that is directe
toward the center of the circle at every moment. Th
magnitude of this net force must equal the product of th
particle’s mass in and its centripetal acceleration i’
/r.
2
ii
cribe all the forces acting on a child riding a horse on a
iy-go-round. Which of these forces provides the cen
cta1 acceleration of the child?
‘ do bicycle riders lean in when rounding a curve at high
ucket of water can be whirled in a vertical circle without
sater spilling out een at the top of the circle when the
ket is upside down. Explain.
onauts who spend long periods in outer space could he
rsely affected by weightlessness. One way to simulate
I
gravity is to shape the spaceship like a bicycle wheel that
rotates about an axis just like a wheel, with the astronauts
walking on the inside of the “tire.” Explain how this simu
lates gravity. Consider (a) how objects fall. (b) the force e
feel on our feet. and (c) any other aspects of gravity you can
think of.
17. Why do airplanes bank when they turn? how would you
compute the banking angle gisen the airspeed and radius of
the turn?
Oitional material is generally not included in the Summary
Problems
Questions
1. Cross-country skiers prefer their skis to have a large coeffi
cient of static friction but a small coefficient of kinetic fric
tion. Explain why. hunt: Think of uphill and downhill.]
2, When you must brake your car very quickly. why is it safer
if the wheels don’t lock? When driving on slick roads. why is
it advisable to apply the brakes slowly?
3. Why is the stopping distance of a truck much shorter than
for a train going the same speed?
4. Can a coefficient of friction exceed 1.0?
5. A block is given a push so that it slides up a ramp. When the
block reaches its highest point, it slides hack down, Why is
its acceleration less on the descent than on the ascent?
6. A heavy crate rests on the bed of a flatbed truck. When
the truck accelerates, the crate remains where it is on the
truck, so it, too, accelerates. What force causes the crate to
accelerate?
7. The game of tetherhall is plaed with a ball tied to a pole
with a string. When the ball is struck, it hirls around the
pole as shown in Fig. 5—24. In what direction is the accelera
Hon of the ball, and what causes the acceleration?
8. \Vhen attempting to stop a car quickly on dry pas emern
which of the following methods will stop the car in the leas
time? (a) Slam on the brakes as hard as possible, lockin
the wheels and skidding to a stop. (b) Press the brakes a
hard as possible without locking the wheels and rolling t
a stop. Explain.
9. Sometimes it is said that water is removed from clothes in
spin dryer by centrifugal force throwing the water outward
Is this correct? Discuss.
10. lèehnical reports often specify only the rpm for centrifug
experiments. Why is this inadequate?
11. Suppose a car moves at constant speed along a mountair
road. At which of the following places does it exert th
greatest and the least forces on the road: (a) At the top of
hill. (b) at a dip between two hills. (c) on a level stretch nea
the bottom of a hill?
12. A rider on a Ferris wheel moses in a vertical circle of radiu
r at constant speed v (Fig. 5 25). Is the normal force tha
the seat exerts on the rider at the top of the circle (a) les
than. (b) more than. or (c) the same as. the force the sea
exerts at the bottom of the circle? Explain.
.
If the coefficient of kinetic friction between a 12.0-kg
and the floor is 0.30, what horizontal force is required
ove the crate at a steady speed across the floor? What
iontal force is required if p is 7ero?
\ force of 25.() N is required to start a 6.0-kg box mos
wross a horizontal concrete floor. ( a ) V’hat is the coeffi—
of static friction between the box and the floor? (b ) If
50-N force continues, the box accelerates at 0.50 m/s.
is the coefficient of kinetic friction?
A box sits at rest on a rough 37 inclined plane.
the free-bods diagram. showing all the forces acting
he box. ( b ) Flow would the diagram change if the box
sliding do’ n the plane. ( ) How would it change if the
sere sliding up the plane after an initial shove?
i)
he coefficient of friction between hard rubber and nortreet pavement is about 0.8. On boss steep a hill (max
a angle) can VC)u leave a car parked?
tippose that ou are standing on a train accelerating at
What minimum coefficient of static friction must
between sour feet and the floor it you are not to slide?
hat is the maximum acceleration a car can undergo if
oefficient of static friction between the tires and the
ad is 0.80?
\ 15.0-kg box is released on a 30 incline and accelerates
a the incline at 0.30 m/s. Find the friction force imped—
l’s motion. What is the coefficient of kinetic friction?
in g
\ car can decelerate at 4.80 m/s’ without skidding
coming to rest on a level road. What would its dcccl
in be if the road were inclined at 13 uphill? Assume
me static friction force.
.%‘
FN
For the system shown in Fig. 5—7 (Example 5—5). how
mass must box I have to prevent an motion from
= 0.25.
rring’? Assume
mg
\ wet bar of soap slides frcel down a ramp 9.0 m long
ied at 8.0 How long does it take to reach the bottom?
tme ,LL = 0.060.
.
FIGURE 5—24
124
CHAPTER 5
Question 7.
Further Applications of Newton’s Laws
FIGURE 5—25
Question 12.
(11) A box is given a push so that it slides across the floor.
How far will it go. given that the coefficient of kinetic fric
tion is 0.25 and the push imparts an initial speed of 2.5 m/s?
(II) Determine a formula for the acceleration of the sstem
shown in Fig. 5 7 in terms of ni ni,, and the mass of the
cord. i11( Define an other variables needed.
.
.
.
( II ) j\ 1 00(1—kg car Pulls a 350—kg trailer. The car exerts a
horizontal force of 3.5 X 10’ N against the ground in order
to accelerate. What force does the car exert on the trailer?
The trailer wheels are not friction-free. so estimate the net
friction force on the trailer. using an effective friction coeffi
cient of 0.15.
( I I ) ( a ) Show that the niiniinuin stopping distance for an
automobile traveling at speed v is equal to i’/2jrg. where
/L iS the coefficient of static friction between the tires and
the road. and g is the acceleration of grasity. (h) What is this
0.75?
distance for a 1200-kg car traveling 95 km/h if
(c ) \Vhat would it be if the car were on the Moon ( the
acceleration of grasits on the Nloon is about g 6) but all
else stayed the same?
•
(II) Piles of snow on slippery roofs can become dangerous
projectiles as they melt. Consider a chunk of snow at the
ridge of a roof with a pitch of 3() (a) What is the minimum
value of the coefficient of static friction needed to keep the
sf055 from sliding rlo ii? ( b ) As the snow begins to melt the
coefficient of static friction decreases and the snow finally
slips. Assuming that the distance from the chunk to the edge
of the roof is 5.0 ni and the coefficient of kinetic friction is
0.20, calculate the speed of the snow chunk svhen it slides off
the roof. (c) If the edge of the roof is 10.0 m above ground,
\s hat is the speed of the snow when it hits the ground?
.
(II) Police lieutenants, examining the scene of an accident
invols ing two cars, measure the skid marks of one of the
cars. which nearly came to a stop before colliding, to he
80 m long. The coefficient of kinetic friction between rubber
and the pavement is about 0.8. Estimate the initial speed of
that car.
Problems
125
(II) Two crates, of mass 80 kg and 210 kg, are in contact and
at rest on a horizontal surface (Fig. 5—26). A 750-N force is
exerted on the 80-kg crate. If the coefficient of kinetic fric
tion is 0,12, calculate (a) the acceleration of the system, and
(b) the force that each crate exerts on the other. (c) Repeat
with the crates reversed.
What is the acceleration of the system shown in
5—30 if the coefficient of kinetic friction is 0.10? Assume
1 = 5.0 kg and
t the blocks start from rest and that (a) in
1 = 2.0kg.
in
2
in
in
(III) A small block of mass in rests on the rough. sloping
side of a triangular block of mass M which itself rests on a
horizontal frictionless table as shown in Fig. 5—32. If the
coefficient of static friction is p, determine the minimum
horizontal force F applied to Al that will cause the small
block ni to start moving up the incline.
in
750N
FIGURE 5—28
80kg
Problems 21. 22. and 23.
210kg
F
5.0kg
FIGURE 5—26
Problem 17.
(11) The block shown in Fig. 5—27 lies on a plane tilted at
0.17.
22.0 to the horizontal, with k
an angle 0
(a) Determine the acceleration of the block as it slides
down the plane. (b) If the block starts from rest 9.3 m up
the plane from its base, what will be the block’s speed when
it reaches the bottom of the incline?
=
‘2
0!
—
=-
(II) For two blocks, connected by a cord and sliding dow
the incline shown in Fig. 5—28 (see Problem 21 ), descrif
> j. (c) Dete
the motion (a) if Pi < i. and (b) if
mine a formula for the acceleration of each block and ty
, m
1
. and 0: iriterpr
2
tension F[ in the cord in terms of in
‘our results in light of your answers to (a) and (h).
(11) Do Problem 21, hut assume the two blocks sliding (
the l)laIe (Fig. 5—28) are connected by a rigid rod instead
a cord.
(II) In Fig. 5—29 the coefficient of static friction hetwee
mass ln and the table is (1.40. whereas the coefficient i’
1
kinetic friction is 0.30 (a) What minimum value of in
keep the system from starting to move’? (b) What value(’
of ui
1 will keep the system moving at constant speed’?
FIGURE 5—32
30°
FIGURE 5—30
Problems 27 and 28.
What are the minimum and maximum values of ni
1 in
5—30 to keep the system from accelerating’? Take
0.50.
I’k
H.A. child slides down a slide with a 2$ incline. and at the
tom her speed is precisely half what it would have been
IC slide had been frictionless. Calculate the coefficient of
tic friction between the slide and the child.
1
Boxes are moved on a conveyor belt from where they
tilled to the packing station 10 m away. The belt is mistationary and must finish with zero speed. The most
I
mt
84
FIGURE 5—27 Block on inclined plane.
Problems 18 and 19.
ifl’)
2.0 kg
(II) Radio engineers are erecting a communications tower
that is 18 m high. During the installation they stabilize the
tower with 30-rn-long cables running from the top of the
tower to the ground. The anchors consist of concrete blocks
to which the cables can be secured. Each block weighs
1600 N. If the coefficient of static friction between a block
and the ground is 0.80. what is the maximum tension that
can exist in a cable before that cable’s anchor is in danger of
moving?
(II) Two blocks made of different materials connected
together by a thin cord, slide down a plane ramp inclined at
an angle 0 to the horizontal as shown in Fig. 5—28 (block 2
is above block I). The masses of the blocks are ,n and
1 and i-. if
in, and the coefficients of friction are i
126
CHAPTER 5
Further Applications of Newton’s Laws
I transit is accomplished if the belt accelerates for half
distance, then decelerates for the final half of the trip. If
oefficient of static friction between a box and the belt
0, what is the minimum transit time for each box?
A bicyclist can coast down a 7.0 hill at a steady
rn/h. If the drag force is proportional to the square of
speed v, so that f = cc
, calculate (a) the value of the
2
stant c and (b) the average force that must he applied in
hr to descend the hill at 25 km/h. The mass of the cyclist
bicycle is $0.0 kg. Ignore other types of friction.
A 4.0-kg block is stacked on top of a 12.0-kg
L. which is accelerating along a horizontal table at
2 (Fig. 5—3 1). Let I’k
.2 rn/s
Is = r. (a) What minirn coefficient of friction i between the two blocks ill
cnt the 3.0-kg block from sliding off’? (b) if ,u is only
1 this minimum value, what is the acceleration of the
Kg block with respect to the table. and (c) with respect
ic 12.0-kg block’? (d) What is the force that must be
ed to the 12.0-kg block in (a) and in (h). assuming that
ible is frictionless’?
-
FIGURE 5—29
Problem 24.
0u
(11) A small block of mass in is given an initial speed v
a ramp inclined at angle 0 to the horizontal. It travels a di
tance d up the ramp and comes to rest. (a) Determine a ft
mula for the coefficient of kinetic friction between bloc
and ramp. (b) What can you say about the value of the coc
ficient of static friction?
(Ii) On an icy day. you worry about parking your car
your drieway, which has an incline of 12. Your neighb
Ralph’s driveway has an incline of 9.0 and Bonnie’s drivL
way across the street has one of 6.0 The coefficient of st,
tic friction between tire rubber and ice is 0.15. Whic
driveway(s) will be safe to park in’?
.
Problem 33.
V
‘
(II) A block is given an initial speed of 3.0 rn/s up the 22.0
plane shown in Fig. 5—27. (a) How far up the plane will it
go’? (b) How much time elapses before it returns to its start
ing point’? Assume I’k = 0.17.
M
-
-
1
0.20 and p
0.30. determin
5.0 kg. and [L
(a) the acceleration of the blocks and (b) the tension in tI
3(1
cord, for an angle 0
Jfl
(I) Calculate the centripetal acceleration of the Earth in its
orbit around the Sun and the net force exerted on the
Earth. What exerts this force on the Earth’? Assume the
Earth’s orbit is a circle of radius I .50 >< 1011
(I) What is the maximum speed ith which a 1200-kg car
can round a turn of radius 80.0 m on a flat road if the coef
ficient of friction between tires and road is 0.55’? Is this
result independent of the mass of the car’?
(1) A horizontal force of 60(1 N is exerted on a 2.00-kg dis
cus as it is rotated uniformly in a horizontal circle (at arm’s
length) of radius 1.00 in. Calculate the speed of the discus.
(1) A child moves with a speed of 1.50rn/s when 9.Om
from the center of a merry-go-around. Calculate (a) the
centripetal acceleration of the child and (b) the net horizon
tal force exerted on the child (mass
25 kg).
(II) How fast (in rpm) must a centrifuge rotate if a particle
9.0 cm from the axis of rotation is to experience an acceler
ation of 100,000 g’s?
(II) Is it possible to whirl a bucket of water fast enough in a
ertical circle so that the water won’t fall out’? If so. what is
the minimum speed’? Define all quantities needed.
(II) At what minimum speed must a roller coaster he tray
cling when upside down at the top of a circle (Fig. 5—33) if
the passengers are not to fall out’? Assume a radius of cur
ature of 8.0 m.
4.0 kg
12.0kg
a
=
5.2 m/s
2
.
FIGURE 5—31
Problem 32.
FIGURE 5—33
Problem 40.
Problems
127
(II) A coin is placed 12.0 cm from the axis of a rotating
turntable of variable speed. When the speed of the turntable
is slowl increased, the coin remains fixed Ofl the turntable
until a rate of 50 rpm is reached, at which point the coin
slides oft. What is the coefficient of static friction between
the coin and the turntable?
I In Problem 52 assume the tangential acceleration is
nstant and determine the components of the instanta“us acceleration at (a) t = 0.0, (b) r = 1.0) s, and
‘ I = 2.0 s.
1 and in-,, connected to each other and t(
(II) Two masses, in
a central post by cords as shown in Fig. 5—35, rotate abon
the Post at frequency f (revolutions per second) on a fric
1 and r from th
tionless horizontal surface at distances r
t
post. l)erive an algebraic expression for the tension in eac
segment of the cord.
1
-
(II) The design of a new road includes a straight stretch
that is horizontal and flat hut that suddenly dips down a
steep hill at 22 The transition should he rounded with what
minimum radius so that cars traveling 90 km/h will not
leave the road (Fig. 5—34)’?
‘II,
.
———___:‘-
A body moves in a circle of radius 20 m with its speed
by t’ = 3.6 ± I .5t
. with i’ in meters per second and t
2
seconds. At t — 3.0 s. find (a) the tangential acceleration
d (b) the radial acceleration.
1
‘
/
‘
FIGURE 5—35
,
1 ) A particle rotates in a circle of radius 3.60 in. At a
ticular instant its acceleration is 0.210 g in a direction
makes an angle of 28.0) to its direction of motion.
, tcrmine its speed (a) at this moment and (h) 2.OOs later.
Llming constant tangential acceleration.
-
\
(II) The drag force on large objects such as cars. planes, and
sky divers moving through air is more nearly FD = —be
.
2
(a) For this quadratic dependence on v, determine a formula
for the terminal velocity t’ of a vertically falling object.
(b) A 75-kg sky diver has a terminal velocity of about
60 rn/s: determine the value of the constant h. (c) Draw a
i.2. For the
curve like that of Fig. 5—23 for this case of FD
same terminal velocity. would this curve lie above or below
that in Fig. 5—23? Explain why.
(Ill) Two drag forces act on a bicycle and rider: F due to
rolling resistance. which is essentially velocity independent:
and F)2 due to air resistance. which is proportional to i,2. For
a specific bike plus rider of total mass 8(1 kg. FDI
4.0 N:
and for a speed of 2.2 rn/s. “1)2
1.0 N. (a) Show that the
total drag force is
Problem 48.
“O
II) An object of mass m is constrained to move in a circle
adius r. Its tangential acceleration as a function of time
riven by atan
b + ct
, shere b and c are constants. If
2
Co at t — 0, determine the tangential and radial compots of the force.
and F acting on the object at any
JIC t > 0.
)
1
F
,
(Ill) A thin circular horizontal hoop of mass rn and radiu
I? rotates at frequency f about a vertical axis through it
center (Fig. 5 36). Determine the tension within the boo1
[Iliiit: Consider a tiny section of the hoop.j
‘
‘1’
-
FT
I
,
In Example 5—15. use dimensional analysis to determine
‘ie time constant r is T
ui/h or r
b/ui.
The terminal velocity of a 3 X 1(1 -kg raindrop is
9 m. s. Assuming a drag force ED
determine
the value of the constant 1’ and ( b ) the time required for
h a drop. starting from rest. to reach 63 percent of termi
I ‘e1ocity.
)
‘Ut
FIGURE 5—34
Problem -12.
An object moving vertically has v = v
11 at t = 0. Detera formula for its velocity as a function of time assum
a resistive force F — —be as well as gravity for two
11 is downward and (b) v is upward.
5: (a) v
(II) On an ice rink two skaters of equal mass grab hands
and spin in a mutual circle once every three seconds. If we
assume their arms are each 0.80 m long, how hard are they
pulling on one another, assuming their individual masses are
bOo kg?
(II) Firzan plans to cross a gorge by swinging in an arc
from a hanging vine. If his arms are capable of exerting a
force of 1300 N on the rope. what is the maximum speed he
can tolerate at the lowest point of his swing? His mass is
80kg and the sine is 4.8 m long.
(II) Redo Example 5—9. precisely this time. liv not ignoring
the weight of the hall. In particular. find the magnitude of
1 and the angle it makes with ihe horizontal.
F
(II) A 0.35-kg ball, attached to the end of a horizontal cord,
is rotated in a circle of radius 1.0 m on a frictionless hori
zontal surface. If the cord will break when the tension in it
exceeds 80 N, what is the maximum speed the ball can has e?
How would your answer he affected if there were friction’?
(II) A 1000-kg sports car moving at 20 m/s crosses the
lOOm). Determine (a) the
rounded top of a hill (radius
normal force on the car. (b) the normal force on the 70-kg
driver. and (c) the car speed at which the normal force
is zero.
‘
where v is in m/s, and Ff) is in N and opposes the motion,
(h) Determine at what slope angle 0 the bike and rider can
coast downhill at a constant speed of 10 m/s.
(III) Determine a formula for the position and acceleration
of a falling object as a function of time if the object starts
from rest at I = 0) and undergoes a resistive force F = be.
as in Example 5—15.
WI ) A motorboat traveling at a speed of 2.4 rn/s shuts off
its eneines at I = (I. How far does it travel before coming to
rest ilit is noted that after 3.0 s its speed has dropped to half
its original value’? Assume that the drag force of the water is
proportional to i’.
([II) A block slides along a horizontal surface lubricated
with a thick oil which provides a drag force proportional to
the square root of velocity:
2C
FIGURE 5—36
Problem 49.
(Ill) If a curve with a radius of 65 m properly banked fo
a car tra cling 70 km/h. what must he the coefficient of sta
tic friction for a car not to skid when traveling at 10(1 km/h
(III) A curse of radius 6Dm is banked for a design speed o
90 km/h. If the coefficient of static friction is 0.30 (we’
paveineit ). at hat range of speeds can a car salelv make
the curve?
is
(II) A particle starting from rest revolves with uniforml
increasing speed in a clockwise circle in the xy plane. Th
center of the circle is at the origin of an xy coordinate sys
0.0. v - 2.0 m. A
0. the particle is at x
tern. At t
2.0 s. it has made one-quarter of a revolution and is a
x = 2.0 m, v = 0.0. Determine (a its speed at r = 2.0 s
(h) the aerage velocity vector. and (c) the average acceler
ation vector during this interval.
4.0) + 0.21 C
,
2
2
j;j
If
i’
=
e at t
—
=
0. determine
i’
and x as functions of time,
General Problems
/0-kg silverware drawer does not slide readily. The owner
-dually pulls svith more and more force. and when the
-plied force reaches 8.0 N. the drawer suddenly opens.
I owing all the utensils to the floor. What is the coefficient
.tiitic friction between the drawer and the cabinet’?
“ ise a method using
an inclined plane to measure the
(ficient of static friction.
hetsseen two surfaces.
ag race tires in contact with an asphalt surface probably
‘,e one of the highest coefficients of static friction in the
ryday world. Assuming a constant acceleration and no
ping of tires, estimate the coefficient of static friction for
rag racer that covers the quarter mile in 6.0 5.
‘.
offee cup on the dashboard of a car slides forward on the
Ish when the driver decelerates from 45 km/h to rest in 3.5 s
less, but not if he decelerates in a longer time. What is the
‘cfficient of static friction between the cup and the dash?
‘
-
‘.
,
-
An 18.0-kg box is released on a 37.0 incline and accelerates
down the incline at 0.270 rn/s
. Find the friction force
2
impeding its motion. How large is the coefficient of friction’?
A tiathed truck is carrying a heavy- crate. The coefficient of
static friction between the crate and the bed of the truck is
01.75. What is the maximum rate at which the driver can dccclcrate and still avoid having the crate slide against the cab?
‘
-
In an earthquake, the ground is found to accelerate with a
maximum value of amax . (a) If an object is going to “hold its
place” on the ground, show that it must have a coefficient of
static friction with respect to the ground of at least
= a\/g.
(h) Numerically, the famous Loma Prieta
Earthquake that stopped the 1989 World Series produced
maximum ground accelerations of up to 4.0 m/s
2 in the San
Francisco
A 1000-kg car rounds a curve of radius 80 m banked at al
angle of 14. If the car is traveling at 80 km/h. will a frictior
force be required? If so, how much and in what direction?
A roller coaster reaches the top of the steepest hill with a
speed of 6.1) km/h. It then descends the hill, which is at an
average angle of 45 and is 45.0 in long. What will its speed
= 0.12.
be when it reaches the bottom? Assume
A motorcyclist is coasting with the engine off at a steady
speed of 20.0 m/s but enters a sandy stretch where the coef
ficient of kinetic friction is 0.80. Will the cyclist emerge from
the sandy stretch without having to start the engine if the
sand lasts for 15 m? If so, what will be the speed upon
emerging?
A flat puck (mass Al) is revolved in a circle on an air
hockey (frictionless) table top. and is held in this orbit by a
light cord which is connected to a dangling mass (mass in)
through the central hole as shown in Fig. 5—37. Show that
the speed of the puck is given by
i’
small mass in is set on the surface of a sphere. Fig. 5—41.
ihe coefficient of static friction is P’s = 0.60. at what angle
sould the mass start sliding?
A car maintains a constant speed r’ as it traverses the hil
and valley shown in Fig. 5—39. Both the hill and valley hat
a radius of curvature R. (a) How do the normal forces act
ing on the car at A, B, and C compare? (Which is largest
Smallest?) Explain. (b) Where would the driver feel hea
est? Lightest? Explain. (c) How fast can the car go withou
losing contact with the road at A?
flu
F
F
/
/
0
FIGURE 5—43
ImgR
i —k-—.
FIGURE 5—39
Problem 74.
A device for training astronauts and jet fighter pilots is
designed to rotate the trainee in a horizontal circle of radius
10.1) m. If the force felt by the trainee is 7.75 times her own
weight. how fast is she rotating? Express your answer in
both rn/s and re /s.
In a Rotor-ride” at a carnival, people pay money to be
rotated in a vertical cylindrically walled “room. (See
Fig. 5—38). If the room radius is 5.0 rn. and the rotation fre
quencv is 0.50 revolutions per second when the floor drops
out. what is the minimum coefficient of static friction so that
the people will not slip down? People describe this ride by
saying they were being “pressed against the wall.” Is this
true? Is there really an outward force pressing them against
the wall? If so. what is its source? If not, what is the proper
description of their situation (besides nausea)?
Problem 79.
A car at the Indianapolis 500 accelerates uniformly from th.
pit area. going from rest to 320 km/h in a semicircular ar
with a radius of 200 m. Determine the tangential and radin
acceleration of the car when it is halfway through the turn
assuming constant acceleration. If the curve were flat. whn
would the coefficient of static friction have to he between
the tires and the roadbed to provide this acceleration with
no slipping or skidding?
iii
*
130
CHAPTER 5
Further Applications of Newton’s Laws
55.5s5L****55,
FIGURE 5—40
Problem 82.
the mass of sand added to the bucket. (h) Calculate the
acceleration of the system.
An airplane traveling at 520km/h needs to reserse its
course. ‘I’ he pilot decides to accomplish this by banking the
wings at an angle of 38 (a) Find the time needed to reverse
course. (h) What additional force will the passengers
experience during the turn? [H/ut: Assume an aerodynamic
“lift” force that acts perpendicularly to the flat ‘ings: see
Fig. —44.1
.
Lift
force
FIGURE 5—44
Figaro the cat (5.0 kg) is hanging on the tablecloth. pullin
Cleo’s fishhowl (11 kg) toward the edge of the tabi
(Fig. 5—40). The coefficient of kinetic friction beteen tin
tablecloth (ignore its mass) under the fishhowl and the tahi
is 0.44. (a) What is the acceleration of Figaro and the fisl
bowl? (b) If the fishbowl is 0.90 m from the edge of tI
table, how much time does it take for Figaro to pull Cleo o
the table?
.snF*W*s
Problem 86.
38°
A particle revolves in a horizontal circle of radius 2.70 m.!
a particular instant, its acceleration is 1.05 m/s?, in a dire
don that makes an angle of 32.0 to its direction of motion
Determine its speed (a) at this moment, and (b) 2.00s later
assuming constant tangential acceleration.
Problem 76.
Determine the tangential and centripetal components of the
net force exerted on a car (by the ground) when its speed is
30 m/s. and it has accelerated to this speed from rest in 9.0 s
on a curve of radius 450 m. The car’s mass is 11)01) kg.
Problem 83.
70.0-kg climber in Fig. 5—42 is supported in the “chim
by the friction forces exerted on his shoes and back.
static coefficients of friction between his shoes and the
and between his back and the wall. are 0.80 and 0.60.
cctively. What is the minimum normal force he must
t? Assume the walls are vertical and that the static frie
lorces are both at their maximum. Fir = P’s F.
FIGURE 5—42
*
/
/
FIGURE 5—41
FIGURE 5—38
4)
I
A
(tv)
FIGURE 5—37
28.0 kg
Problem 84.
und throws ou a baseball with an initial horizontal
lv of 30 rn/s. Its path is parabolic. hut at any point we
letermine a radius of curvature. (a) Define a method
Ltermining the radius of curvature. (b) What is this
is of curvature for the baseball immediately after it left
friend’s hand?
1) kg block is connected to an empty 1.00-kg bucket by
id running over a frictionless pulley (Fig. 5—43). The
n.ient of static friction between the table and the block
{() and the coefficient of kinetic friction between the
nd the block is 0.320. Sand is gradually added to the
st until the system just begins to move. (a) Calculate
Problem 87.
A circular curve of radius R in a new highway is designed so
that a car traveling at speed v can negotiate the turn safely
on glare ice (zero friction). If a ear travels too slowly, then it
will slip toward the center of the circle. If it travels too fast.
then it will slip away from the center of the circle. As the
coefficient of static friction increases, it becomes possible
for a car to stay on the road while traveling at any speed
ssithin a range from vmjn to 1
max Dense formulas for vmjn
and 1
functions
as
and R.
of
v.
’max
P’s’
A train traveling at a constant speed rounds a curse of
radius 275 m. A pendulum suspended from the ceiling
swings out to an angle of 17.5 throughout the turn. What is
the speed of the train?
The force of air resistance on a rapidly falling body has the
form F = —kv. so that Newton’s second law applied to
such an ob)ect is
in
dv
dt
ung
2
ku’
sshen the downward direction is considered positive. Deter
mine the speed and position at 2.0-s intervals, up to 20.0 s, of
a 75-kg person (sky diver) who starts from rest, assuming
k
0.22 kg/in. Also, show that the body eventual1 reaches
a steady speed. the terminal speed. and explain why this
happens.
—
General Problems
131
2 acts during the upward
kr
Assume a net force F = —nig
vertical motion of a 250-kg rocket, starting at the moment
(t = 0) when the fuel has burned out and the rocket has an
upward speed of 120 rn/s. Let k = 0.65 kg/rn. Calculate the
maximum height reached by the rocket. Compare to freeflight conditions sithout air resistance (k = 0).
A small bead of mass m is constrained to slide without fric
tion inside a circular vertical hoop of radius r which rotates
about a vertical axis (Fig. 5-45) at a frequency f. (a) Deter
mine the angle 0 where the bead will he in equilibrium—
that is. where it will have no tendency to move up or down
along the hoop. (h) If f = 4.0 rev/s and r = 20 cm, what is
0? (c) Can the bead ride as high as the center of the circle
(0 = 90 )? Explain.
—
$
with the vertical. \
The sides of a cone make an angle
small mass in is placed on the inside of the cone and
cone, with its point down, is revolved at a frequency f (r
olutions per second) about its symmetry axis. If the coei
cient of static friction is p. at what positions on the co
can the mass be placed without sliding on the cone? (G
the maximum and minimum distance.r, from the axis).
A ball of mass rn = 1.0 kg at the end of a thin cord of lent
r = 0.80 m revolves in a vertical circle about point 0,
shown in Fig. 5—46. During the time we observe it. the ol
forces acting on the ball are gravity and the tension in t
cord. The motion is circular hut not uniform because of
force of gravity. The ball increases in speed as it descer
and decelerates as it rises on the other side of the circle. \
the moment the cord makes an angle 0 = 30 below
horizontal, the ball’s speed is 6.0 m/s. At this point. deL
mine the tangential acceleration, the radial acceleration, a
the tension in the cord, F. Take 0 increasing downward
shown.
The astronauts in the upper left of
this photo are working on the space
shuttle. As they orbit the Earth—at a
pretty high speed—they experience
apparent weightlessness. The Moon.
in the background. also is orbiting
the Earth at high speed. What keeps
the Moon and the space shuttle (and
its astronauts) from moving off in a
straight line away from Earth? It is
the force of gravity. According to
Newton’s law of universal
gravitation, all objects attract all
other objects with a force
proportional to their masses and
inversely proportional to the sqtLare
of the distance between them.
I
-
\
4
/
70
°*c
/
‘I
I
\
/
in
__c
izg
1
i
Sin
0
//
lflgCOsO
/
/
/
ing
FIGURE 5—45
Problem 92.
FIGURE 5—46
Problem 94.
Gravitation and Newton’s Synthesis
t‘
3
,
saac Newton was not only the inventor of the three great laws of motion
I serve as the foundation for the study of dynamics. He also conceived of
ther great law to describe one of the basic forces in nature. gravitation.
Ihiulied it to understand the motion of the planets. This new law. published
ii his great hook Philosophiac Naturalis Principia Matheniatica (the Pun
short), is called Newton’s law of universal gravitation. It was the capstone
n’s analysis of the physical world. Indeed. Newtonian mechanics, with its
s of motion and the law of universal gravitation, was accepted for cen
mechanical basis for the way the universe works.
Newton’s Law of Universal Gravitation
uS many great accomplishments. Sir Isaac Newton examined the motion
uvenly bodies—the planets and the Moon. In particular, he wondered
nature of the force that must act to keep the Moon in its nearly circular
umd the Earth.
132
CHAPTER 5
Further Applications of Newton’s Laws
133