phys+110-chapter+5

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Physics 110
1435-1436 H
Instructor: Dr. Alaa Imam
E-mail: [email protected]
Chapter 5
FORCE AND MOTION -I
Sections 5-2, 5-3, 5-4, 5-5, 5-6
Newtonian Mechanics
Newton’s First Law
Force
Mass
Newton’s Second Law
— Important skills from this lecture:
1.
2.
3.
4.
5.
6.
Explain Newton’s first law
Define the force and its unit
Resolve forces and find the resultant along x & y axes
Define the mass and its relation to force
Explain newton second law
Relate the force component along an axis to its
acceleration
7. Define Newton unit
8. Draw free-body diagram
9. Apply Newton 2nd law in one & two dimensions
Newtonian Mechanics
— The study of the relation between force & acceleration is
called Newtonian mechanics
— Newtonian mechanics does not apply to all situations;
— If the speeds of the interacting bodies are very large (near
the speed of light) à Einstein’s special theory of relativity
applied
— If the interacting bodies are on the scale of atomic structure,
e.g. electrons à quantum mechanics is applied
— Newtonian mechanics is applied to the motion of objects
ranging in size from the very small to astronomical
objects
Send a puck sliding over the ice of a skating rink, however, and it goes a
lot farther. You can imagine longer and more slippery surfaces, over which the
puck would slide farther and farther. In the limit you can think of a long, extremely slippery surface (said to be a frictionless surface), over which the
puck would hardly slow. (We can in fact come close to this situation by sending a puck sliding over a horizontal air table, across which it moves on a film
— When there is no force acting on a body:
of air.)
From— these
we can
conclude
that a body will keep moving with
If theobservations,
body is at rest,
it stays
at rest
constant—velocity
no force
acts on itit.continues
That leads to
us move
to the first
Newton’s three
If the if
body
is moving,
withofthe
laws of motion:
same velocity (same magnitude & direction)
Newton’s First Law
Newton’s First Law: If no force acts on a body, the body’s velocity cannot change;
that is, the body cannot accelerate.
In other words, if the body is at rest, it stays at rest. If it is moving, it continues to
move with the same velocity (same magnitude and same direction).
5-4 Force
We now wish to define the unit of force. We know that a force can cause the
the vector rules
of Chapter 3.
d so on.
Thus
in magnitude
general, if our
antity,
with
both
and standard
direction.body
Is force also a vector
This
means
thatdirection
when two
magnitude
know that
force
F must
be the
easily assigna,awe
direction
to aaforce
(just
assign
of or more forces act on a body, w
force,
or
resultant
force, bythat
adding the individual forces vector
of
force
(in sufficient.
newtons) is
equal
the
magbutthe
that
is not
We
musttoprove
by experiment
that
has forces
the magnitude
direction of the net force has the
s per second
per second).
quantities.
Actually,
that has been
done:
are indeedand
vector
body
asacceleraallcombine
the individual
forces
e acceleration
produces.
However,
ave
magnitudesitand
directions,
and they
according
to together. This fact is called the
position
forces. The world would be quite strange if, for e
magnitude
also afor
vector
Chapter 3.and direction. Is force
rection
a force
(just
assign
the
friend
werewe
toofcan
pullfind
on their
the standard
body in the same direction
hat
whento
two
or more
forces
act
on direction
a body,
net
— the
A force
isbymeasured
by
the
acceleration
produces
ufficient.
must
prove
experiment
that
of
1 N, and
yet
somehow
netitpull
was 14 N.
force, by We
adding
individual
forces
vectorially.
A singlethe
force
y, thatand
has direction
been done:
forces
indeed
vector
this
book,
forces
most
— of
Acceleration
isInahas
vector
quantity
àare
force
is aoften
vectorrepresented
quantity with a vec
nitude
the netare
force
the
same
effect
on
the
:
:
nd
directions,
and they
combine
according
to is represented
( ) This
F, is
and
a net
ividual
forces together.
fact
called
theforce
principle
of super- with the vector symbol Fnet. As
aisforce
orfor
a (N)
net
force can
s. The world would
quite
strange
if,
example,
you have
and acomponents along coordinate ax
— be
Force
unit
Newton
more
act on body
a body,
find
theiranet
ll on forces
the standard
in we
thecan
same
direction,
each with
forceare single-component forces. The
only
along
single
axis, athey
the individual
forces
vectorially.
Aprinciple
single force
Superposition
for forces:
mehow
the net —
pull
was
14 N.
on
of the
force
thetwo
same
the act
forces
are net
most
often
represented
with aon
vector
symbol
astheir net force or
— has
When
or effect
more
forces
on asuch
body,
:
ether.
This factwith
is called
the principle
superresultant
force ( Fof
are
theother
vector
addition of the
is
represented
the
vector
symbol
with
vectors,
net.)As
ld be
if, foralong
example,
a When forces act
individual
forcesyou and
rce
canquite
have strange
components
coordinate
axes.
theare
same
eachforces.
with
force
ebody
axis, in
they
single-component
Then
can drop the
— direction,
A single force
that ahas
thewe
magnitude
& direction of the net
ll was 14 N.
force has the same effect on the body as all the individual
ften represented with
a vector
symbol such as
forces
together
:
h the vector symbol
other
vectors,
— Fnet. As
canwith
have
many
components along coordinate axes
ponents along coordinate
When
act a single axis, they are single— Whenaxes.
forces
act forces
only along
ngle-component forces.
Then weforces
can drop
thearrows could be replaced by signs
component
à the
to indicate the forces directions
Force
of the forces along that axis.
Instead of the wording used in Section 5-3, the more proper statement of
5 FORCENewton’s
AND MOTION—I
Law
is in terms
of a net on
force:
— If First
many
forces
are acting
a body, and their net
force is zeroà the body cannot accelerate
are made in, say, an elevator that is accelerating relative to the ground, then
: and the results can be
the measurements are being made in a noninertial frame
Newton’s
First
Law:
If
no
net
force
acts
on
a
body
(F
net ! 0), the body’s velocity
surprising.
cannot change; that is, the body cannot accelerate.
CHECKPOINT 1
:
Which of the figure’s six arrangements correctly show the vector addition of forces F1
:
:
yield the thirdforces
vector, which
is meant
Fnet?net force
and F
may
be
acting
ontoarepresent
body,their
butnetifforce
their
2 tomultiple
There
body cannot accelerate.
F1
Inertial Reference Frames
F
F1
is zero, the
F1
ü
F
F
2
2
2
(b)
(c)
Newton’s(a) first law
is not true
in all reference
frames,
but
we can always find
reference framesF 2 in which it (as well as the rest of Newtonian mechanics) is
true. Such special frames are referred to as inertial reference frames, or simply
F1
F1
F1
inertial frames.
ü
(d)
ü
(e)
F2
(f )
F2
An inertial reference frame is one in which Newton’s laws hold.
Mass
— Force produces different magnitudes of acceleration for different
bodies
— e.g., if a baseball
& a bowling ball
are given the
same kick
à baseball acceleration > bowling ball?
— Because the mass of the baseball differs from the mass of the bowling
— A less massive baseball receives a greater acceleration than a more
massive bowling ball when the same force is applied to both
— The mass of a body is the characteristic that relates a force on the
body to the resulting acceleration
— Mass is:
— an intrinsic characteristic of a body
— a scalar quantity
E
All the definitions, experiments, and observations we have discussed so far can be
This equation
is statement:
simple, but we must use it cautiously. First, we must b
summarized
in one neat
:
5-6
Newton’s
Second
Law
certain about which body we are applying it to. Then Fnet must be the vector sum
of all
the
act
body.
Only
forces
that
act on
that
All
theforces
definitions,
we
discussed
sobody’s
farbody
can beare to b
Newton’s
Secondthat
Law:experiments,
Theon
net that
forceand
on aobservations
body
is equal
to have
the
product
of the
summarized
one neatsum,
statement:
included
inacceleration.
thein vector
not forces acting on other bodies that might b
mass
and its
involved in the given situation. For example, if you are in a rugby scrum, the ne
force on
you isSecond
the vector
of all
and
pulls on
your
body. It doe
Newton’s
Law: Thesum
net force
on athe
bodypushes
is equal to
the product
of the
body’s
In equation
form,
mass andany
its acceleration.
not include
push or pull on another player from you or from anyone els
:
Every time you work F:
a force
problem,
your
firstlaw).
step is to clearly state
the body t
(Newton’s
second
(5-1)
!
ma
net
which
you
are
applying
Newton’s law.
In equation form,
Likeequation
equations,
is component
equivalent
toFirst,
three
This
is simple,
but weEq.
must
use
it cautiously.
wecomponent
must be equa
—other
This vector
equation
is equivalent
to5-1
three
equations,
one
:
:
:coordinate
for
eachbody
axis of
an
xyz
coordinate
system
tions,
for
each
axis
an
xyz
system:
certainone
about
which
we
are
applying
it
to.
Then
F
must be the vector(5-1)
sum
(Newton’s second net
law).
Fnet ! ma
Newton’s Second Law
of all the forces that act on that body. Only forces that act on that body are to be
Fnet,isxsum,
! ma
Fnet,y
!
mause
and
Fnet,z
!
ma
. might
includedThis
in the
vector
not
forces
acting
other
bodies
thatzwe
equation
simple,
we must
cautiously.
First,
must be
be (5-2)
x,but
y,onit
:
involved
in about
the given
situation.
if to.
you
are Fin
rugby
the sum
net
certain
which
body weFor
areexample,
applying it
Then
bescrum,
the vector
netamust
forceofon
the
vector
of all
theOnly
pushes
andthat
pulls
body.are
Itan
does
allyou
the is
forces
that actsum
on
that
body.
acton
onyour
that along
body
toaxis
be to th
Each
of
these
equations
relates
the
net forces
force
component
AND
MOTION—I
not include
push
or pull
on not
another
player
from
you
or
from
anyone
else.
included
inalong
the
vector
sum,
forces
acting
on
other
bodies
that
might
be us tha
— any
Each
of these
equations
relates
the
net
force
component
along
acceleration
that
same
axis.
For
example,
the
first
equation
tells
an
to
acceleration
along
that
same
axis
Every
time
work
a the
force
problem,
first
step
isxare
to
clearly
state
thexbody
to
involved
in axis
the
situation.
For your
example,
if
you
in acauses
rugby
scrum,
the
net
the
sum
ofyou
all
thegiven
force
components
along
the
axis
the
component
a
force
on
you
is the component
vector
sum
of
all
the no
pushes
and pulls
body.
It zdoes
which
you
areacceleration
applying
Newton’s
law.
The
along
a given
axisacceleration
is caused
only byon
the
sum
force
of
the
body’s
acceleration,
but
causes
inyour
theof
y the
and
direction
not include
any
push
pullaxis,
on
another
player
from
orcomponent
from
anyone
else.
Like
other vector
equations,
Eq.
5-1
to you
three
equaalong
thatorsame
and
notisbyequivalent
force
components
along
any
other
Turnedcomponents
around, the
acceleration
component
ax is caused
only
byaxis.
the sum of th
time
youaxis
work
force
your
first step is to clearly state the body to
tions,Every
one for
each
ofaan
xyzproblem,
coordinate
system:
forcewhich
components
alongNewton’s
the x axis.
you are applying
law.In general,
Fnet,
ma
Fnet,yifEq.
!the
ma
ma
. component
(5-2)
x, that
y,is and
net,za !
zis
Equation
5-1x !
tells
us
force Fon
body
zero, the body’s
Like
other
vector
equations,
5-1net
equivalent
to
three
equa-
OTION—I


— From Newton’s
law, if Fnet = 0 ⇒ a = 0
The acceleration
a given
by theat
sum
of the force
— Ifcomponent
the body along
is at rest
(v axis
= 0 is&caused
a = 0),only
it stays
rest
components along that same axis, and not by force components along any other axis.
— If the body is moving (v ≠ 0 & a = 0), it continues to
move at constant velocity
on such
balance
another,
andthe body’s
Equation —
5-1Any
tellsforces
us that
if thebody
net force
on one
a body
is zero,
thebody
forces
and
the
body at
are
said
in
acceleration :
a ! both
is at
rest,
it stays
rest;
if ittoisbe
moving,
it continues
0. If the
equilibrium
to move at constant
velocity. In such cases, any forces on the body balance one
Thethe
forces
could
one another,
another, and —
both
forces
andcancel
the body
are said to be in equilibrium.
but still
the
Commonly, the forces
are act
alsoon
said
tobody
cancel one another, but the term “cancel” is
2nd
tricky. It does not mean that the forces cease to exist (canceling forces is not like
canceling dinner reservations). The forces still act on the body.
For SI units,
tells us that
— InEq.
SI 5-1
units
1 N ! (1 kg)(1 m/s2) ! 1 kg " m/s2.
(5-3)
Some force units in other systems of units are given in Table 5-1 and Appendix D.
Table 5-1
For SI units, Eq. 5-1 tells us that
1 N ! (1 kg)(1 m/s2) ! 1 kg " m/s2.
(5-3)
Some force—units
in other
units are
given of
in Table
Appendix D.
Some
forcesystems
units inofother
systems
units 5-1
areand
given
in Table 5-1
Table 5-1
Units in Newton’s Second Law (Eqs. 5-1 and 5-2)
System
Force
Mass
Acceleration
SI
CGSa
Britishb
newton (N)
dyne
pound (lb)
kilogram (kg)
gram (g)
slug
m/s2
cm/s2
ft/s2
1 dyne ! 1 g " cm/s2.
a
1 lb ! 1 slug " ft/s2.
b
To solve problems with Newton’s second law, we often draw a free-body
diagram in which the only body shown is the one for which we are summing
forces. A sketch of the body itself is preferred by some teachers but, to save space
accelerationEveryday
for different
bodies. Put a baseball and a bowling ball on the floor
experience tells us that a given force produces different magnitudes of
and give both
the same
kick.
EvenPut
if you
don’tand
actually
do this,
youtheknow
acceleration
forsharp
different
bodies.
a baseball
a bowling
ball on
floor
the result: The
noticeably
larger
acceleration
the you
bowling
and baseball
give both receives
the same a
sharp
kick. Even
if you
don’t actuallythan
do this,
know
ball. The two
differ
because
the mass
of acceleration
the baseball
differs
from
the accelerations
result: The baseball
receives
a noticeably
larger
than
the bowling
The
two accelerations
differ exactly,
because is
the
mass of the
the mass ofball.
the
bowling
ball —
butunknown
what,
mass?
Calculate
an
mass
bybaseball differs from
mass ofhow
the bowling
ball —mass
but what,
exactly, is mass?
We canthe
explain
to measure
by imagining
a series of experiments in
knowing
another
mass
and
their
We can
to measure mass
by imagining
series
of experiments
in
an inertial frame.
In explain
the firsthow
experiment
we exert
a forcea on
a standard
body,
an inertial frame. In the first experiment we exert a force on a standard body,
whose masswhose
m0 ismass
defined
to
be 1.0
kg.
Suppose
that that
the the
standard
body
acceleraccelerations
m
is
defined
to
be
1.0
kg.
Suppose
standard
body
acceler0
2
ates at 1.0 m/s
.
We
can
then
say
the
force
on
that
body
is
1.0
N.
ates at 1.0 m/s2. We can then say the force on that body is 1.0 N.
We next apply
thatapply
same
force
would
needneed
some
wayway
of of
being
We next
that
same(we
force
(we would
some
beingcertain
certain itit
is the sameisforce)
to aforce)
second
whose
massmass
is not
known.
the same
to a body,
secondbody
body,X,
body
X, whose
is not
known.Suppose
Suppose
If afind
= 1body
N isXapplied
on at
2 0.25
bodies,
standard
body,
2 m/s2.aWe
know
that
a2 lessmassive
massive
we
that Fthis
accelerates
. We know
that
a less
we find that
thisforce
body
X
accelerates
at 0.25
m/s
whose
mass
m
=
1.0
kg,
and
acceleration
a
=
1.0
m/s
,
and
ogreater acceleration than a more massive
o
baseballareceives
aacceleration
bowlingball
ballwhen
when
baseball receives
greater
than
a
more
massive
bowling
nd
thesame
2 force
body
X whose
mass
(mX)Let
is not
known,
and
its
the
(kick)
is
applied
to
both.
us
then
make
the
following
conjecthe same force
(kick)
is
applied
to
both.
Let
us
then
make
the
following
conjec2
acceleration
= masses
0.25 of
m/s
To findis m
X : to the inverse of the ratio of
ture:
The ratio ofaX
the
two. bodies
equal
ture: The ratio of the masses of two bodies is equal to the inverse of the ratio of
their accelerations when the same force is applied to both. For body X and the
their accelerations
whenthis
thetells
same
force is applied to both. For body X and the
standard body,
us that
Fo = Fx
standard body, this tells us that
mX
a0
mX ma0 0 ! aX .
!
.
Solving for mX yields
m0
aX
Solving for mX yields
a0
1.0 m/s2
mX ! m0
aX
! (1.0 kg)
2 m/s2
0.25
1.0 m/s
! 4.0 kg.
a0
! (1.0 kg)
! 4.0 kg.
mX ! m0
2
Our conjecture
be
useful,
of
course,
only
if it continues to hold when
awill
0.25
m/s
X
we change the applied force to other values. For example, if we apply an 8.0 N force
Free-body Diagram
— To solve problems with Newton’s second law, a freebody diagram is drawn
— A free-body diagram is a diagram that contains the
only body under the forces
— In the free-body diagram, the body is represented
with a dot, and each force on the body is drawn as a
vector arrow with its tail on the body
tion is if
a vector
quantity,
with both magnitude and direction. Is force also a ve
, and so on. Thus in general,
our standard
body
quantity?
canFeasily
a direction to a force (just assign the directio
n of magnitude a, we know
that aWe
force
must assign
be
the acceleration),
but magthat is not sufficient. We must prove by experiment
ude of the force (in newtons)
is equal to the
forces are vector quantities. Actually, that has been done: forces are indeed v
eters per second per second).
quantities;
they have
magnitudes
directions, and they combine accordin
— Ait system
consists
ofacceleraone
or moreand
bodies
y the acceleration
produces.
However,
the vector
rulesalso
of Chapter
th magnitude and direction.
Is force
a vector3.
means
that
whenof
two or
more forces act on a body, we can find thei
Types
ofThis
forces
that
affect
any
system:
a direction to a—force
(just
assign
the
direction
force,
resultant
force,that
by adding the individual forces vectorially. A single
ot sufficient. We must
proveorby
experiment
— External
force:
any force
on the bodies that comes from
that
has
the
magnitude
and
ually, that has been done:
forces
indeed vector direction of the net force has the same effect on
outside
theare
system
as all the according
individual to
forces together. This fact is called the principle of su
es and directions, and body
they combine
— Internal
forces
two bodies
inside
the if, for example, you a
positionforces:
for forces.
Thebetween
world would
be quite
strange
system were to pull on the standard body in the same direction, each with a f
or more forces act on afriend
body, we can find their net
1 N,
and yet
net pull
14 N.connected to
If of
the
bodies
making
athe
system
arewas
rigidly
ng the individual —
forces
vectorially.
Asomehow
singleup
force
In
book,
forces
are most often represented with a vector symbol su
one
ection of the net force
hasanother,
thethis
same
effect
on the
:
:
F
,
and
a
net
force
is
represented
with
the
vector
symbol
F
à
the
system
is
treated
as
one
composite
body,
and
net. As with other vec
together. This fact is called the principle of supera if,
force
or
a netisforce
can
components
alongforces
coordinate axes. When force
on
the
the
of all external
would be quite strange
forsystem
example,
youvector
andhave
a sum
onlya along
a single
axis,
they are
single-component
ard body in the same
direction,
each
a force
— e.g.,
system
ofwith
a connected
railroad
engine & car forces. Then we can drop
pull was 14 N.
If a tow line pulls on the front of the engine
t often represented with
a vector
symbol
as
à tow
force acts
on such
the whole
engine – car system
:
with the vector symbol
other vectors,
à Fnet. As
on with
the system
can be related to its
omponents along coordinate
axes. When
act 2nd law
acceleration
usingforces
Newton’s
e single-component forces.
Then
drop
theof the system
(m) will
be we
thecan
total
mass
law, Fnet ! ma , where m is the total mass of the system.
CHECKPOINT 2
The figure here shows two horizontal forces acting on a block on a frictionless floor. If a
:
third horizontal force F3 also acts on the block,
:
3N
5N
what are the magnitude and direction of F3 when
the block is (a) stationary and (b) moving to the
left with a constant speed of 5 m/s?
F1 = 5N, F2 = −3N
From Newton’s 2nd law, Fnet = ma
(a) The block is stationary à v = 0, a = 0, à Fnet = 0
F3 + F1 + F2 = 0
à F3 + 5 −3 =0
à F3 = −2N
(b) The body is moving with a constant velocity à a = 0
From Newton’s 2nd law, Fnet = 0
à F3 = −2N
Sample Problem
acceleration.
Situation A: For Fig. 5-3b, where only on
93
5-6 KNEWTON’S
SECOND
LAW
(a) acts, Eq. 5-4
Sample
Problem
gives us
EY IDEA
One- and two-dimensional forces, puck
This is a free-body
F1 ! max,
In each situation we can relate the acceleration :
a to the net Puck F1 x diagram.
:
axis,
we with
can simplify
situation
writing the second
Parts A, B, and C of Fig. 5-3 show three force
situations
which
acting
on the
puck
Newton’seach
second
law, by which,
Fnet in
with given data, yields
:
:
law forthe
x components
only:only the x
one or two forces
act on a puckProblem
that moves
frictionless
Sample
Fnetover
. However, because
motion is along
! ma
(b)
F1
4.0 N
ice along an x axis, in one-dimensional motion. The puck’s
:
:
Fnet, x ! main
a(5-4)
! can simpli
! 20 m
x. which
x ! we
axis,
Parts
A,
B,
and
C
of
Fig.
5-3
show
three
situations
mass is m ! 0.20 kg. Forces F1 and F2 are directed alongAthe
m
0.20 kg
B
The free-body
diagrams
for the
three situations are
alsofor x componen
axis and have magnitudes
N and
F2 ! 2.0
N. Force
! 4.0
law
oneF1 or
two
forces
act
on a puck
that
moves
over
frictionless
:
The
positive
answer
indicates
The
horizontal
force
These
forces
compete.that the ac
by
F3 is directed at angle u ! 30° and has magnitude F3 ! 1.0 F1 given in Fig. 5-3, with the puck represented
F2
F1 a dot.
positive
direction
thecauses
x axis.
causes a horizontal
iceis the
along
an x ofaxis,
in one-dimensional
motion. The puck’s
Their net of
force
x
In each situation,
acceleration
the
puck?
axis, we
can
simplifyxSituation
situation
by
writing
the
second
hreeN.situations
in what
which
:each
:
acceleration.
a horizontal
5-3b, wherealong
only onethe
horizontal
force acceleration.
F
mass is m ! 0.20 kg. Forces
and FA:2 For
areFig.directed
1
(a)
Situation
B:
In
Fig. 5-3d, two horizonta
(c)
law for x componentsacts,
only:
moves over frictionless
Eq. 5-4 gives us
:
KEY IDEA
puck, F1 in the
positive
direction of xdia
and
The
free-body
axis and have magnitudes
F ! 4.0 N and F2 ! 2.0 N. Force
Puck F1 1 This is a free-body
onal Inmotion.
The
puck’s
F
F
:
:
F
!
ma
,
2x
1
direction.
NowisEq.
5-4 gives us
This
a free-body
1
each situation we can
relate
the acceleration
a to the net
x
diagram.
x (5-4)
:
Fand
! ma
given
in Fig. 5-3, wi
is
directed
at
angle
u
!
30°
magnitude
F
!
1.0
F
:
net, x has
x.
3
3
diagram.
directed
along
F2 are
force
on thethe
puck with Newton’s second law,
Fnet acting
which,
with
given
data,
yields
F1 " F2 ! max,
:
:
N.
In
each
situation,
what
Fnet ! ma
. However, because
the
motion
is along only
the
x is the acceleration of the puck?
(b)
(d)
The free-body diagrams for the three
situations
are
N and F2 ! 2.0 N. Force
F
4.0
N
which,
withalso
given data, yields
ax ! 1 !
! 20 m/s2.
(Answer)
Situation A: For F
m
0.20 kg
given in Fig. 5-3,B with the puck represented
by a Cdot.
has magnitudeAF3 ! 1.0
F1 " F2
4.0 N " 2.0 N
Eq.
a x ! acts,
! 5-4 gives u
positive
indicates that the acceleration
is in
The horizontal force
K EFYTheIThese
D
E forces
A answer
compete.
mhorizontal 0.20 kg
eration of the puck?
Only
thethe
F1
F2
F2
1
positive
direction
of
the
x
axis.
causes a horizontal
Their net force causes
component of F3
x
x
x force
Situation
A:
For
Fig.
5-3b,
where
only
one
horizontal
:
θ
acceleration.
horizontal
acceleration.
F3 Thus,
with F2.
net force accelerates
the puck in
In each
situation we can relatea the
acceleration
netthe competes
a to the
: acts, Eq. 5-4 gives
(a)
forces
act
on
the
(c) us Situation B: In Fig. 5-3d, two horizontal
(e) tion of the x axis.
:
:
force
acting
on
the
puck
with
Newton’s
second
law,
F
which, with given d
F
F
puck,
in
the
positive
direction
of
x
and
net
1
2 in the negative
:
Puck F1 : This is a free-body
:
F2
F1 direction.
F
ThisFismotion
aNow
free-body
This
free-body
Eq. 5-4
gives
us 2only the
Situation
C: is
InaFig.
5-3f, force F3 is not
:
x netdiagram.
!
ma
,
F
.
However,
because
is
along
x
!
ma
x the
x
1
x
θ
cceleration a to the net
diagram.
diagram.
the puck’s acceleration; only
F1 " F2 ! maxF,3 direction of
:
a the!m
Newton’s second
law,
(b)
(d) data, yields
( f ) is. (Force F3 is two-dimensional but x
which, with
given
which, with given data, yields
A
dimensional.) Thus, we write Eq. 5-4 as
One- and two-dimensional forces, puc
One- and two-dimensional forces, puck
otion is along only
the x
B
ontal force
horizontal
on.
F2
F1
x
(c)
F2
F1
(d)
x
Fig. 5-3
In three situations, forces act on a puck that moves
F1 " F2
4.0 N " 2.0 N
F3,shown.
F1 horizontal
x " F2 ! max.
along
are also
a x4.0
! N
! an x axis.
! 10diagrams
m/s2. The
2 Free-body
positive
answe
The
force
.
!
(Answer)
These forces compete. F F1a x !
Only
the horizontal
m ! 20 m/s
0.20 kg
2
From the figure, we see that F3,x ! F3 cos
m
0.20
Their net force causes
component
ofkg
F3
positive direction o
(Answer)
a horizontal
x
x θ causes
acceleration and substituting for F3,x yield
F3
a horizontal acceleration.
competes
with Faccelerates
Thus,
the net force
the puck in the positiveAdditional
direc2.
examples, vid
acceleration.
(e)
The positive answer
indicates
that the accelerationa is! in
F3,xthe
F3 cos $ " F2
" F2
tion
of the x axis.
!
x
(a)
Situation
m
mB: In Fi
of the
x
axis.
:
F2
:
This is apositive
free-body direction
This
is
a
free-body
along
the 30#) " 2.0 N
xSituation C: In Fig. 5-3f, force F3 is not directed
(1.0
N)(cos
F1 in the!pos
puck,
θ
diagram.
diagram.
"5.7
!
This
is aof
free-body
direction
the puck’s acceleration; only x component
F3, x
Puck F1F3
:
0.20 kg
direction.
Now
Eq.
F3 is two-dimensional
is.5-3d,
(Force two
but the motion
is only
(f x
) Fig.
Situation B: In
horizontal forces
act on
theonediagram.
C
:
dimensional.) Thus, we write Eq.
:5-4 as
Kthe
E Yacceleration
IDEA
N. In each situation, what is
of the puck?
:
Situation
A: For Fig
eachsituation
situationwe
wecan
canrelate
relatethe
theacceleration
acceleration:
thenet
net
InIneach
aatotothe
:
:
acts, Eq. 5-4
giveswith
us
forceFFnetacting
actingon
onK
the
puck
withNewton’s
Newton’ssecond
secondlaw,
law,
E Ypuck
IDEA
which,
force
the
with
which,
with
net :
::
: . However, because the motion is
Fnet!!ma
alongonly
only the x
ma
FIn
. However,
the
motion is:
neteach situation
we canbecause
relate the
acceleration
to the net the x
aalong
o
force F F
acting
on
the puck
with
Newton’s
second law,
m = 0.20kg, F1 = 4 N, F2 = 2N,
=
1N,
θ
=
30
3 AA
:
Fnet
:
net
:
! ma . However, because the motion is along only the x
Fnet ,x = max
1
(a)
(a)
F1 = max
F
4N
ax = 1 =
= 20m / s 2
m 0.20Kg
F1
Puck
Puck
x
causes a horizontal
acceleration.
This is a free-body
(a)Puck
F1
F1F1 x This is a free-body
diagram.
x This
diagram.
is a free-body
diagram.
x
(b)
(b)
Theseforces
forcescompete.
compete.
F1 These
These
forces
compete.
F
F1 1 x
Their net force causes
Their
causes
net net
forceforce
causes
x x Their
F2
FF22
a horizontal acceleration.
a horizontal
acceleration.
a horizontal
acceleration.
(c)
(c)
(c)
F2 F2
F1 F1 This This
is a free-body
is a free-body
xF1 x This is a free-body
F2
x diagram.
diagram.
diagram.
(d)(d)
(d)
F2
F3,x − F2
F cosθ − F2
= 3
=
m
m
(1)cos(−30 o ) − 2
ax =
= −5.7m / s 2
0.20kg
F
F2 2
x
θ
F3
(e)F3 θ
ax =
F3
(e)
(e)
F2
F2
F2
F3
θ
(f ) θ
F3
ax !
F1 a"ax xF
!
m
Thus, the n
Thus,
the ne
Thus, the
net force
ac
tion
of
the
tion of the
x
axis.
tion of the x
Situation
Situation
C: In Fig.C
Situation
directiondirection
of the puck
o
:
direction
of:
F3(Force
is. (Forceis.
is two-d
:F
is. (Force
3
dimensional.)
Thus,Fw
dimensiona
F
C
C
F3,x − F2 = max
direction. Now Eq. 5-
dimensiona
C
c) Fnet ,x = max
Situation
: B
:F in
puck,
1in t
F1 Fig.
puck,
Situation
B: In
: direction. N
N
puck, F1direction.
in the positi
which, with given dat
B
B
B
The positive answer
positive direction
of t
Situation
which,with
with
which,
(b)
b) Fnet ,x = max
F1 − F2 = max
F − F2
4 N − 2N
ax = 1
=
= 10m / s 2
m
0.20Kg
F
ax !
Thepositive
positiv
The
m
positivedire
dir
positive
F F1
A
a)
Thehorizontal
horizontalforce
force
The
causes a horizontal
x x causes a horizontal
The acceleration.
horizontal force
acceleration.
which, with given dat
θ
θ
x
Only the horizontal
Only the Fhorizontal
component
Only theofhorizontal
3
component
of F
xcompetes
with
F
2. F 3
component of
competes with3F .
competes with F2.2
This is a free-body
diagram.
This is a free-body
x
x
x
This is a free-body
diagram.
diagram.
From the figure, we s
From the
fi
acceleration
Fromand
thesubs
fig
acceleratio
acceleration
F3,x " F2
ax !
F!3
m
a x ! F3,x
!
a x N)(cos
(1.0
3
!
0.20
(1
! (1.0
!
F3
Thus, the net force a
f)
In three (situations,
forces act on a puck that moves
(f )
rection of the x axis.
along an x axis. Free-body diagrams are also shown.
Thus, the n
Fig. 5-3 In three situations, forces act on a puck that moves
Thus, the ne
Fig. 5-3 In three situations, forces act on a puck that moves
rection of t
Fig. 5-3
along an x axis. Free-body diagrams are also shown.
along an x axis. Free-body diagrams are also shown.
rection of th
:
F1 #
:
F2
#
:
F3
:
! ma ,
Sample
93 Problem
5-6 (5-6)
NEWTON’S" SECOND
LAW
(20 N) sin 90°
! (2.0 kg)(3.0 m/s ) sin 50° " (10 N) sin("150°)
which gives us
! "10.4 N.
:
F3
:
! ma "
:
F1
"
:
F 2.
Sample Problem
(5-7)
Two-dimensional forces, cook
Vector: In unit-vector notation, we can write
:
F3 ! F3, x î # F3, y ĵ ! (12.5 N)î " (10.4 N) ĵ
! (13 N)î "
(Answer)
:(10 N) ĵ.
In the overhead view of Fig. 5-4a, a 2.0 kg cookie tin is accelerated at 3.0 m/s in the direction shown by a , over a frictionless horizontal surface. The acceleration is caused by
three horizontal forces, only two of which are shown: F of
magnitude 10 N and F of magnitude 20 N. What is the third
force F in unit-vector notation and in magnitude-angle
notation?
x compone
Calculations: Because this is a two-dimensional problem,
:
One-weand
two-dimensional
forces,
cannot
find F 3 merely by substituting
magnitudes
2 the puck
for the vector quantities on the right side of Eq. 5-7. Instead,
We can now use a vector-capable calculator to get the mag-3,x
:
:
:
:
we must vectorially add ma , "F 1 (the reverse of F 1), and
nitude and the angle of F 3 . We can also use Eq. 3-6 to obtain
:
situation
by writing the second
ree situations
in which: axis, we can simplify eachthe
"F 2 (the reverse of F 2), as shown in Fig. 5-4b. This addition
magnitude and the angle (from the:
positive direction of
law for x components
only:the x axis) as
moves over
can befrictionless
done directly on a vector-capable
calculator because
1
:
2
2
we
know
both
magnitude
and
angle
for
all
three
vectors.
F3 ! 2F 3,x # F 3,y ! 16 N
nal motion. The puck’s
2 Eq. 5-7 inF
However, here we shall
evaluate
the
right
side
of
(5-4)
net, x ! max.
F3,y
"1
are directed
the :
terms ofalong
components,
first along the x axis and then along
$ ! tan
! "40%.
and
(Answer)
3
F3, x
3,x
the
y
axis.
The free-body diagrams for the three situations are also
N and F ! 2.0 N. Force
2
as magnitude F3 ! 1.0
eration of the puck?
given in Fig. 5-3, with the puck represented by a dot.
y
We draw the product
F2
y
This
is
the
resulting
TheseSituation
are two
A: ForKFig.
oneand
horizontal
of mass
acceleration
E Y5-3b,
I Dwhere
E A only
horizontal acceleration
of the three
as a vector.
acts,
Eq.
vector. us
horizontal
force
: 5-4 gives
– F1
vectors.
net a
F1 !
: max,
– F2
net
force
The net force F on the tin is the sum of the three forces
celeration :
a to the
50°
a via Newton’s
and is related to the acceleration
second law
ma
:
x
:
Newton’s second
law,! ma
which,
with givenx data, yields
(Fnet
).30°Thus,
tion is along only the x
ntal force
orizontal
on.
F
F1
4.0 N
F3
: a ! : 1 ! : (b)
:
! 20 m/s2.
x
F 1 # Fm
ma ,Then we can add the(Answer)
three(5-6)
2 # F0.20
3 !kg
(a)
which gives
Theuspositive answer indicates that
vectors to find the missing
third
force vector.
the
acceleration
is in
positive direction
of the
: x axis.
:
Fig. 5-4 (a) An overhead
forces
:view of two of three horizontal
: that act on a cookie tin,
:
the
F
! ma
! m(
Then, subst
F
! (2.
"
! 12
y compone
F3, y ! may
! m(a
! (2.0
"(
! "10
(5-7)
Vector: In
Situation B: In Fig. 5-3d, two horizontal forces act on the
:
F ! ma " F " F .
:
:
resulting in acceleration :
, "F1, and
a . F3 is not shown. (b) An arrangement of vectors ma
3
2
1
:
:
"F2 to find force F3.
:
:
F
m = 2kg, a
F = ma
F1 + F2 + F3 = ma
for the vector quantities on the right side of Eq. 5-7. Instead,
:
:
:
we must vectorially add ma
, "F 1 (the reverse of F 1), and
:
:
"F 2 (the reverse of F 2), as shown in Fig. 5-4b. This addition
can be done directly on a vector-capable calculator because
know
magnitude
all three vectors.
=we3m
/ s 2both
, F
= 10Nand
, angle
F2 =for20N
1
However, here we shall evaluate the right side of Eq. 5-7 in
terms of components, first along the x axis and then along
the y axis.
F3 = ma − F1 − F2
For x component
F3,x = max − F1,x − F2,x
F3,x = m(a cos50) − (F1 cos(30 −180)) − (F2 cos90)
and
y
These are two
of the three
horizontal force
vectors.
F2
y
This is the resulting
horizontal acceleration
vector.
a
50°
F3,x = 2(3cos50) −10cos(−150) − 20cos90
30°
F3,x = 12.5N
For y component
We can now u
nitude and th
the magnitud
the x axis) as
x
F1
(a)
(b)
F3,y = may − F1,y − F2,y
F3,y = m(a sin 50) − (F1 sin(−150)) − (F2 sin 90) Fig. 5-4
F3,y = 2(3sin 50) − 10 sin(−150) − 20 sin 90
F3,y = −10.4 N
(a) An overhead view of two of three horizontal force
:
resulting in acceleration :
a . F3 is not shown. (b) An arrangement
:
:
"F2 to find force F3.
Additional examples, video, and practice
F3 in unit vector

F3 = F3,x iˆ + F3,y ĵ

F3 = 12.5iˆ − 10.4 ĵ
The magnitude
F3 = (12.502 + (−10.4) 2
F3 = 16N
The angle
θ = tan −1
θ = −40o
F3. y
F3,x
= tan −1
−10.4
12.5
Examples:
Q.1: One Newton equals:
(a) Kg.m
(b) kg.m/s2
(c) kg/s2
(d) m/s2
(c) kg/s2
(d) m/s2
F = ma à N = kg.m/s2
Q.2: The basic SI unit of the force is:
(a) Kg.m
(b) kg.m/s2
Q.3: A box is moving with a constant speed of 24.7 m/s. The net
force on the box is:
(a) Zero
(b) 4N
If ΣF = 0 à a = 0
(c) 5N
(d) 45N
Q.4: Three forces act on a particle of mass m, F1 = 80i + 60j,
F2 = 40i + 100j. If the particle moves with a constant speed of
4m/s, then F3 is:
(a) 80i + 60j (b) 80i − 60j (c) −80i + 60j (d) −120i−160j
v constant à a = 0 àΣF = 0
F1 + F2 + F3 = 0 à F3 = − (F1 + F2 ) = −120i − 160j
Q.5: Two forces act on a particle that moves with constant
velocity. If F1 = 6i − 2j, then F2 is:
(a) 6i − 2k
(b) − 2i + 6k (c) − 6i + 2j
(d) − 2i + 6j
v constant à a = 0 à ΣF = 0
F1 + F2 = 0 à F2 = − F1 = − 6i + 2j
Chapter 5
FORCE AND MOTION -I
Sections 5-7
Some Particular Forces
— Important skills from this lecture:
1. Define the gravitational force & write it in unit vector
notation and magnitude and angle notation
2. Define the weight and differentiate between mass and
weight
3. Define the normal force and calculate it
4. Define the frictional force
5. Calculate the tension force
these
early
chapters,
we
do
not
nature
of this
force
a second
body.
In the
these
early
chapters,
we
do not discuss
the nature
th
und.
We
shall
assume
that
the discuss
ground
is an
inertial
frame.
SOM
y axis
along
the
body’s
path,
with
the
positive
direction
upward.
Forof
th
and
usually
consider
situations
in which
the
second
body
is Earth. Thus,
when5-7
we
and
usuallybody
consider
situations
in which
thewe
second body is Earth. Thus, w
: is Earth.
der
situations
inofwhich
the
second
Thus,
when
: usually
Suppose
a body
of
mass
m
is
in
free
fall
with
the
free-fall
acceleration
o
! ma
which,
Newton’s
second
law
can
be
written
in
the
form
Fanet,y
F
speak
the
gravitational
force
on
a
body,
we
mean
force
thaty ,pulls
:
g
F
speak
of
the
gravitational
force
on
a
body,
we
usually
mean
a
force
th
g
Fdirectly
tational g.
force
onwe
a body,
we
usually
mean
a the
force
that
pulls
g if
on
it5-7
toward
the
center
of
Earth
—
that
is,
directly
down
toward
the
situation,
becomes
nitude
Then,
neglect
the
effects
of
air,
the
only
force
acting
ontowa
th
on
it
directly
toward
the
center
of
Earth
— that
is, directly
down
Some
Particular
Forces
:
ward the center
of Earth
— thatthat
is,shall
directly
down
toward
the
ground. We
shall assume
the
ground
is
anthe
inertial
frame.
ground.
We
assume
that
ground
is
andownward
inertial frame. force and
F
y is the gravitational
force
.
We
can
relate
this
g
"F
!mm("g)
Suppose
a is
body
of
mass frame.
m
is in of
free
with
acceleration of
gfall
assume that the
ground
an inertial
The
Gravitational
Force
Suppose
a body
mass
is :
inthe
freefree-fall
fall
: with the free-fall accelera
(F
!
).force
nward
acceleration
with
second
law
We
place
a the
vertica
magnitude
g.free
Then,
if Newton’s
we
neglect
the
effects
of
the
air,
thema
only
acting
on
: g. Then,
magnitude
if
we
neglect
the
effects
of
the
air,
the
only
force
ody
of mass
m
is
in
fall
with
the
free-fall
acceleration
of
:
A
gravitational force
force Fgon
on aa body:
body isaaforce
certain
type
of pull
that
is
directed towardacting
: pulls
—body
A
gravitational
that
on
the
body
or body’s
mg.
Fpositive
is the gravitational
force
can
relate
thiscan
downward
force
and axis
g!
Fg.discuss
body
is the
gravitational
force
We
relate
this
downward
for
g. We F
the
path,
with
the
direction
upward.
For
this
a
second
body.
In
these
early
chapters,
we
do
not
the
nature
of
this
force
:
n,isifalong
we neglect
the
effects
of
the
air,
the
only
force
acting
on
the
:
directly toward the center of Earth (downward ê)
:
:
(F
!
ma
)
downward
with
Newton’s
second
law
.
We
place
a
vertical
: acceleration
(F
!
ma
). We we
downward
acceleration
with
Newton’s
second
law
place a
and
usually
consider
situations
in
which
the
second
body
is
Earth.
Thus,
when
!
ma
,
which,
inForou
wton’s
second
law
can
be
written
in
the
form
F
Fg. the
itational yIn
force
We
can
relate
this
downward
force
and
net,y
yto
:
words,
magnitude
of
the
gravitational
force
is
equal
the
product
y
axis
along
the
body’s
path,
with
the
positive
direction
upward.
th
axis
along
the
body’s
path,
with
the
positive
direction
upward.
For
this
axis,
F:g onof
of the
gravitational
force
a body,
usually mean
a force
that pulls m
— Inspeak
free fall
motion,
when :
the
effects
air is we
neglected,
the only
force
ation,
(F
!written
ma
).gravitational
ration becomes
withNewton’s
Newton’s
second
law
We
place
a force
vertical
! in
ma
which,
Newton’s
second
law
can
bethe
indirectly
the
form
Fnet,y
, which,
our
second
law
bem
in
the
form
Fnet,y
y ,the
ydown
This
same
gravitational
force,
with
same
magnitude,
still
acts
on th
on
it on
directly
toward
the
center
of
Earth
—written
that
is,
toward
acting
a body
of can
mass
is
the
F! ma
The Gravitational Force
g
situation,
becomes
situation,
becomes
body’s path,
with
the
positive
direction
upward.
For
this
axis,
ground.
We
shall
assume
thatin
the
ground
isbut
an
inertial
frame.
even
when
the
body
is
not
free
fall
is,
say,
at
rest on a pool table or m
"F
!
m("g)
g m is
nd form
Suppose
body
of à
mass
in
free
fall
with
the
free-fall acceleration of
ma
,
which,
in
our
law can across
beFrom
written
in athe
Fgravitational
Newton’s
2(For
law
,
or
"F
!
m("g)
net,y
y
g
"F!
!
m("g)
the
table.
the
force
to
disappear,
Earth would h
g
magnitude g. Then, if we neglect the
effects of the air, the only force acting on the
s
:
disappear.)
or
Fg ! this
mg. downward force
F
body
is
the
gravitational
force
and
Fg !Fmg.
(5-8)
g. We can relate
or
(5-8)
:
g ! mg.
:
We
write
Newton’s
second law
forlaw
the(Fgravitational
force
in these
! ma ). We place
downward
acceleration
with Newton’s
second
a vertical
"Fcan
g ! m("g)
In words,
the
magnitude
the
gravitational
force
is
equal
the product
m
y axisthe
along
the
body’s
path,
with theofpositive
direction
upward.
Fortomg.
this
axis,
forms:
In
words,
magnitude
of
the
gravitational
force
is
equal
to
the
product
ords, the magnitude
of theThis
gravitational
force
isform
equal
to!the
product
mg.
same
gravitational
force,
with
the
same
magnitude,
still
acts
ma
,
which,
in
our on th
Newton’s
second
law
can
be
written
in
the
F
net,y
y
This same
gravitational
force,
with
the
same
magnitude,
still
acts
on
the
body
:
Fg !
mg.
(5-8)
: at rest on a pool table or
even
whenwith
the body
is"F
not in
free
fall but
is,mg
say,
—even
Using
unit
vector
notation:
situation,
becomes
This same
gravitational
force,
the
same
magnitude,
still
actsoron
the bod
F
, table
!
ĵ
!
"mg
ĵ
!
g say, at rest on a pool
when the body is not in free gfall but is,
moving
across the table. (For the gravitational force to disappear, Earth would h
across
the
table.
(For
theisgravitational
to disappear,
would
have
to
!product
m("g)
nnitude
whenofthe
body
is
not
in
free
fall
but
is,
say,
at
rest
ontheaaEarth
pool
or away
movin
gforce
gravitational
force
equal
to"F
the
mg.
—the
The
gravitational
force
acts
on
the
body
even
when
body
not in
disappear.)
where
ĵ is the unit
vector
that
points
upward
along
y axis,istable
directly
fro
disappear.)
:
theor
free
fall
situation
We
can write
Newton’s
second
law
for
thea
gravitational
force
in dow
these
ss the table.
(For
the
gravitational
force
to
disappear,
Earth
would
have
to
vitational
force,
with
the
same
magnitude,
still
acts
on
the
body
F
!
mg.
(5-8)
g
ground,
and
is
the
free-fall
acceleration
(written
as
vector),
directed
g
We can write forms:
Newton’s second law for the gravitational force in these vector
ppear.)
dy is not in
free
fallgravitational
but is, say, at
resttoon
a pool table
or
moving
—forms:
For
the
force
disappear,
Earth
has
to disappear
:
In words, the magnitude of the gravitational force is equal
to the product
mg.
:
F
mg
,
!
"F
ĵ
!
"mgĵ
!
g
g
For
the
gravitational
force
to
disappear,
Earth
would
have
to
We can write Newton’s
secondF:law
for
the
force
vecto
:
This same gravitational
force,
with
thegravitational
same
magnitude,
still actsinonthese
the(5-9)
body
mg
,
!
"F
ĵ
!
"mg
ĵ
!
g
g
where
ĵ
is
the
unit
vector
thatis,points
a ytable
axis,or
directly
away fr
even
when
the
body
is
not
in
free
fall but
say, atupward
rest onalong
a pool
moving
ms:
Weight
where ĵ is the unit vector that points upward along a y axis, directly away from the
:
The weight W of a body is the magnitude of the net force required to preven
The weight W of a body
is the
magnitude
of the
net forcebyrequired
theFor examp
body
from
falling freely,
as measured
someoneto
onprevent
the ground.
body from falling freely,
as measured
theyou
ground.
example,
keep
a ball at restbyinsomeone
your handon
while
stand For
on the
ground, to
you must pro
an upward
forceyou
to stand
balanceonthe
the ball from E
keep a ball at rest in your
hand while
thegravitational
ground, youforce
muston
provide
Suppose the
of theforce
gravitational
2.0 N. Then
the magnitu
an upward force to balance
the magnitude
gravitational
on theforce
ballisfrom
Earth.
your upward
force mustforce
be 2.0isN,
and
thus
thethe
weight
W of theofball is 2.0 N
Suppose—the
magnitude
gravitational
2.0
N.net
Then
magnitude
The
weight (W)ofofthe
a body:
is the magnitude
of
the
force
required
also say that the ball weighs 2.0 N and speak about the ball weighing 2.0 N.
your upward
force must
be 2.0
N, falling
and thus
theasweight
W of
ball ison
2.0 N. We
to prevent
the body
from
freely,
measured
bythe
someone
A
ball
with
a
weight
of
3.0
N
would
require
a
greater
force from y
TERalso
5 FORCE
AND
MOTION—I
the
ground
say that
the
ball
weighs
2.0
N
and
speak
about
the
ball
weighing
2.0
N.
namely, a 3.0 N force — to keep it at rest. The reason is that the gravitational
A ball with a weight
of 3.0
N would
a greater—force
from you —say that this
you must
has a require
greater
magnitude
— e.g., if the magnitude
ofbalance
the gravitational
force
on a ball is namely,
2.0 N. 3.0 N. We
W
namely, a 3.0
N
force
—
to
keep
it
at
rest.
The
reason
is
that
the
gravitational
force
ond
ball
is
heavier
than
the
first
ball.
To keep a ball
magnitude
of 2.0
N
or at rest, an upward
W !force
Fg with
(weight,
with ground
as inertial
frame).
5-7
you must balance
has
a greater
magnitude
— namely,
3.0 N.
We sayathat
Now
let
us generalize
the situation.
Consider
bodythis
thatsechas an acceler
has to be
applied
to balance
the gravitational
one
: equation tells us (assuming the ground is an inertial frame) that
This
athe
relative
to N
the ground, which we again assume to be an inertial fr
ond ball is heavier
than
first
à the weight
Wofofzero
the ball.
ball is 2.0
:
F
Two
forces
act
on
the
body:
a
downward
gravitational
force
Now let us generalize the situation. Consider a body that has an acceleration
g and a balan
:
— relative
If another ball
requires
a greater
force toW.
keep
at write
rest Newton’s secondFglaw for a verti
upward
force
of magnitude
We itcan
a of zero
to
the
ground,
which
we
again
assume
to
be an inertial frame.
nd ball is heavier than the 1st one
:
à
the
2
axis,
theW
positive
direction
as F
Thewith
weight
of a body
is equalupward,
to theforce
magnitude
Fg of the gravitational force
mR
Force
E AND
MOTION—I
Two
forces
act onThe
theGravitational
body:
a downward
gravitational
g and a balancing
body.
:
! type
malaw
Fbalancing
upward —force
of forces
magnitude
We
can
write
Newton’s
second
vertical
y
If two
act onW.
a body;
net,y
y.upward
Fg(downward)
A gravitational
force
on a body
is&a acertain
offor
pulla that
is directed
towa
force
ofa magnitude
W upward,
axis, with the
positive
direction
asbecomes
second
body.
In these
early
chapters, we do not discuss the nature of this for
In
our
situation,
this
or
W
!
F
(
weight,
with
ground as inertial frame).
(5-11)
nd
g
APTER 5 FORCE
AND
MOTION—I
à fromand
Newton’s
2
law:
usually consider
situations
inEq.
which
the
second
body is Earth. Thus, when w
Substituting
mg
for
F
from
5-8,
we
find
:
g
!
ma
.
F
à net,y
àan
W
Fgframe)
!we
m(0)
y is
tellsof
usthe
(assuming
the ground
inertial
that mean a force that pu(
F
speak
gravitational
force
a"
body,
usually
g on
FgR =This
mRg equation
on it
directly toward the center ofWEarth
— that
is, directly down toward th
In our situation, this
becomes
mgground
(weight),
or
W ! Fg à
(weight,!with
as inertial frame).
(5
à
l-arm balance. When
ground. We shall assume that the ground is an inertial frame.
Watells
"
Fus
!
m(0)
equation
(assuming
ground
is an inertial
frame)
that
nce, the gravitational
body
m is to
in
free
with
theforce
free-fall
The weight W Suppose
ofThis
a body
isaequal
toof
the
magnitude
Fthe
of
thefall
gravitational
on(5-10)
theacceleration
gmass
gits
which
relates
body’s
weight
mass.
y being weighed
(on
body.
magnitude
Then,aifbody
we neglect
effects of the air, the only force acting on th
Tog.weigh
meansthe
to
: measure its weight. One way to do this is
e total gravitational
body is the gravitational force Fg. We can relate this downward force an
:
the body on one of the pans of an equal-arm balance
(Fig. 5-5) and then p
:
rence bodies m
(on
the
(F
!
ma
). We placeforce
downward
acceleration
with
Newton’s
second
law
a vertic
The
weight
W
of
a
body
is
equal
to
the
magnitude
F
of
the
gravitational
on t
g
R
Weight
5-7 Some Particular Forces
:
force FgL on the body being weighed (on
the left pan) and the total gravitational
:
force FgR on the reference bodies (on the
right pan) are equal. Thus, the mass mL of
the body being weighed is equal to the total
mass mR of the reference bodies.
How to weight a body
CHAPTER 5 FORCE AND
96
Two ways for weighting a body:
1. Equal-arm balance: the body is placed on one of
the pans, a reference bodies (with known masses)
is placed on the other pan
when a balance is obtained:
à the gravitational forces on the two sides match
à the masses on the pans match
à we know the mass of the body
mL
mR
FgL = mLg
FgR = mRg
An equal-arm balance. When
the device is in balance, the gravitational
:
force FgL on the body being weighed (on
the left pan) and the total gravitational
:
force FgR on the reference
Scalebodies
marked(on the
in
either
right pan) are equal. Thus, the mass mL of
or to the total
the body being weighedweight
is equal
mass units
mass mR of the reference bodies.
Fig. 5-5
1. Spring scale: the body stretches a spring, which
causes a movement of a pointer along a scale. The
scale has been calibrated & marked in either mass
or weight units
It is accurate only when the value of g is the same
as when the scale was calibrated
Fg = mg
Fig. 5-6 A spring scale. The reading is
proportional to the weight of the object on
which r
To
the bod
erence
ance (s
pans th
the loca
We
a sprin
MOTION
either m
way an
or un
mass
was
cal
This
Th
vertica
scale Th
in
mentbod
w
weight
weight.
Subs
Ca
and is r
of g is
weight
whic
on EarT
butthe
theb
eren
ance
The
No
pans
the lo
If you
s
The reaW
spry
upaon
eithe
buckled
wayd
floor
mass
floor
co
was
Thc
comes T
— The weight of a body must be measured when the body is not
accelerating vertically relative to the ground
— A body’s weight is not its mass
— Weight is the magnitude of a force
— Weight is related to mass by Newton’s 2nd law
— If a body is moved to a point where the value of g is
different:
— The body’s mass will not differ
— The body’s weight will differ
e.g., for a bowling ball of mass 0.3 kg
On Earth
On Moon
M = 0.3 Kg
M = 0.3 Kg
g = 9.8 m/s2
g = 1.6 m/s2
W= (0.3) (9.8) = 2.9 N
W= (0.3) (1.6) = 0.49 N
is due
to reason
Earth'sis that the mattress, because
The
it deforms
downward due to you, pushes
he gravitational
The
forces
x
Fg
x a floor,
on you.
it deforms (it is compressed, bent, or
downward
pull. Similarly,Fgif you
orce
on
theupblock
balance.
(a) stand on
(b)
e gravitational
The
forces
Fg up on you.
buckled
ever
so
slightly)
and
pushes
The
gravitational
TheEven
forcesa seemingly rigid concrete
F
due
Earth's
Fgbalance. :
g
ce
ontothe
block
ravitational
The
forces
FFg gsitting directly onbalance.
floor
does
this
(iftable
it isexperiences
not
theFground,
enoughto
people on the
force
on
the
block
7
(a)
A
block
resting
on
a
a
normal
force
perpendicular
F
ownward
pull.
(a)
(b)
g
N
on
the
block
balance.
due to Earth's
floor
could break it).
is
due
to
Earth's
etop.
(b)
The
free-body
for the block.
:
to Earth's
wnward
pull. The pushdiagram
(a)
(b) :or floor is a normal force FN . The name
on
you
from
the
mattress
is block
downward
A
one.g.,
a table
experiences
a normal
FN(b)
perpendicular to
ward
pull.restingpull.
(a)
(b) orforce
if you
stand(a)
on a
floor
a mattress
The Normal Force
comes from the mathematical term normal, meaning perpendicular: The force on
t on
: buckled slightly)
b)block
The free-body
diagram
for
the
block.
à
they
deform
(compressed,
bent,
or
you
from,
say,
the
floor
is
perpendicular
to
the
floor. downtoon a table,
:
:
resting
on
a
table
experiences
a
normal
force
F
perpendicular
ture
if resting
5-7a
shows
an
example.
A
block
of
mass
m
presses
N
ock
on
a
table
experiences
a
normal
force
F
perpendicular
(a) A block resting
on pulls
a table
experiences
normal
force
FNto
perpendicular to
Earth
you
down, andathey
pushes
you
up
N
:
The
free-body
diagram
for
the
block.
e free-body
diagram
theremain
block.
ing
it somewhat
because
ofstationary
the gravitational force Fg on the block. The
àfor
you
op.
e (b) The free-body diagram for the block. :
7a
shows anthe
example.
A
block
of mass
m Fpresses
down
on a diagram
table, for the
ushes
block
with
normal
force
.
The
free-body
N
the up onWhen
:
a body presses
: against:a surface, the surface (even a seemingly rigid
: The
somewhat
because
of
the
gravitational
force
on
block.
Fonly
s
gaon
FN body
sahows
given
Fig.
5-7b.
Forces
are
the
two
forces
on the
block
Fg and
aninexample.
A block
ofpushes
mass
m:
presses
down
athe
table,
one)
deforms
and
on
the
with
normal
force
F
that
is perpendicular
to
shows
an
example.
A
block
of
mass
m
presses
down
on
a
Ntable,
:
re
5-7a
shows
an
example.
A
block
of
mass
m
presses
down
on
a
table,
:
on
up
on the
block
normal
force
free-body
diagram
Fblock
mewhat
because
ofwith
the Thus,
gravitational
on can
the write
block.
The for the
F we
the
surface.
N . The
: Newton’s
y are
both
vertical.
for
theforce
second law
g
omewhat
because
of:theofgravitational
force
block.
The The
Fg onFthe
:
:
g
it
somewhat
because
the
gravitational
force
on
the
block.
:
g
theFig.
block
with
diagram
foron
the
nnin
5-7b.
Forces
and FF
areyfree-body
only
two
forces
the block
F(F
NN. The
gforce
: free-body
ositive-upward
ynormal
!
ma
)the
as
:axis
:net, y force
pFig.
on
the
block
with
normal
.
The
diagram
for the
F
N two
hes
up
onForces
theThus,
block
with
normal
force
. The
free-body
diagram
Fwrite
are
theonly
only
forces
on Newton’s
the
block
the
on
the
block
Fg and
Nforces
:F
:
N are
both5-7b.
vertical.
for
the
block
we
can
second
lawfor the
:
:
Fand
Fig.in5-7b.
Forces
are
the
only
two second
forces
on the
block
Fblock
they
are
both
vertical
g and
N"
hin
vertical.
Thus,
for
the
we
can
write
Newton’s
law
F
given
Fig.
5-7b.
Forces
are
the
only
two
forces
on
the block
F
F
F
!
ma
.
g
N
N
g
y
-upward y axis (F
!
ma
)
as
net,
y
y
nd
à Newton’s
2thelaw
in ycan
axis:
The
force
oth
Thus,
fory) the
we
Newton’s
second
law
ward
y axis vertical.
(Fnet,
asforblock
Normal
forceFFlaw
Thenormal
normal
force
are vertical.
both
Thus,
block
wewrite
can
write
Newton’s
second
y ! ma
N
Normal
force
(from
the
table)N
q.
5-8,
we
substitute
mg
for
F
,
finding
is
the
force
on
is the force on
F!
"
F!
.
upward
y axisy(F
ma
asgma
Nnet,
yas
net,
y(F
yg)!
tive-upward
ma
)
y
y
Faxis
"
F
!
ma
.
N
g
y
the
theblock
blockfrom
fromthe
the
mg. ! masupporting
.
N"
,substitute
we substitute
mg
finding
table.
FNF
!
ma
g, F
mg for
Fgfor
,F
finding
F
NF"
g"
y may. ysupportingtable.
Block
Block
g!
he
magnitude
of
normal
mg
may.is
Fmg
"
ma
FNàthe
we
substitute
mg
for
, finding
NF"
. 5-8,
we substitute
mg
for
Fgy!
,.force
finding
g!
ude
of the
force
is !mg
mg
ma
#gravitational
ay)
à
The
gnitude
ofnormal
the normal
force
is!#
N
y ! m(g
ma
FF
mg
!
N "F
y. may. The gravitational
N"
y
y
FNFN
Block
Block
x x
F
(5-13)Fg
forceononthe
theblock
block
Fgg
force
F
!
mg
#
ma
!
m(g
#
a
)
(5-13)
(duebe
to Earth's
pull) accelN
y#
y # and
offorce
table
(they
in an
vertical
acceleration
ayma
duetotoEarth's
Earth's might
FNthe
! normal
mgforce
! m(g
ay)is isblock
(5-13)
nitude
of the
isythe
due
magnitude
ofnormal
is
downward
pull.accel- relative
(a)
elevator).ayIfof the
tableand
and
block
notbeaccelerating
to the
the table
block
(theyare
might
inpull.
an
cceleration
downward
(a)
cal acceleration a of the table and block (they might be in an accel-
The
forces
The
forces
balance.
balance.
Fg
(b)
(b)
Fg
g and F are the
downward
on the
block
balance.
(a)
(b)only
:
ockforce
is given
inpull.
Fig.
5-7b.
Forces
two
forces
on the block
F
F
g
N
g
force onbecause
the block of the gravitational
balance.
deforming
it
somewhat
force
on
the block. The
F
xg
is
due
to
Earth's
: write
nd they are both vertical.
for the block we can
Newton’s second law
:
is dueThus,
to Earth's
table
pushes
up
on
the
block
with
normal
force
.
The
free-body
diagram
forLAR
theFORCE S
F
A block resting
a table experiences
a normal force
to
N FN perpendicular
5-7forces
SOM
E PARTICU
downward
pull. on
(a)
(b)
pull.
:
:
The
gravitational
The
r a(a)
positive-upward
ydownward
axis
(Fnet,
!
ma
)
as
(a)
(b)
9
y
y
Fg
5-7forces
SOME PARTICULAR
FORCES
top.
(b) The
free-body
diagram
for
the
block.
F
block
is given
in Fig.
5-7b.
Forces
and
are
the
only
two
on
the
block
F
FNg
g
force on the block
balance.
:
:
y
F
"
F
!
ma
.
Fig.
5-7
A
block
resting
on
a
table
experiences
a
normal
force
F
to
and
are
both(a)
vertical.
Thus,
for
the
block
we
can
write
Newton’s
law
y
(a)
Athey
block
resting
on
a
table
experiences
a
normal
force
F
perpendicular
to second
is due to Earth'sN
g
y
N perpendicular
N
The normal
force
the tabletop.
(b)The
The
free-body
diagram
the block.
Normal
force
FNfor
normal
force
downward
pull.
(a)
(b)
for
a
positive-upward
y
axis
(F
!
ma
top.
(b)
The
free-body
diagram
for
the
block.
Normal
force
F) as
net,
y
y
ure
5-7a
shows
an
example.
A
block
of
mass
m
presses
down
on a table,
is
the
force
on
om Eq. 5-8, we substitute
mg onfor Fg, finding
is the force
:
the
blockthe
from
thefrom
block
ng it somewhat
because
of
thethegravitational
force
theF: block.
The to
Fg on
F
"
F
!
ma
.
Fig. 5-7
(a)
A
block
resting
on
a
table
experiences
a
normal
force
perpendicular
N
g
y
N
:
supporting
table.
supporting
table.
Figure 5-7a shows an
A ma
block
Fof mass m presses down on a table,
"
mg F
!
.Nfree-body
FNexample.
Block yF
Block
:
Block
ushes
up
on
the
block
with
normal
force
.
The
diagram
for
the
the
tabletop.
(b)
The
free-body
diagram
for
the
block.
Block
N
ure 5-7a shows
an example.
m presses
down
on
a
table,
deforming
it somewhat
because
the gravitational
force
on
the
block.
The
F
:A block
: ofofmass
g
:
x
From
Eq.
5-8,
we
substitute
mg
for
F
,
finding
:
x forces on the block
g the only
Fwith
given
inmagnitude
Fig.
5-7b.
Forces
and
gthe
N are
ng
itthe
somewhat
because
ofF
gravitational
forcetwo
the block.
Thefor the
table
pushes
up on
the
block
force
. The
free-body
diagram
FNF
hen
of
the
normal
force
isnormal
g on
:
:
:
The
gravitational
The
forces
yshes
areup
both
vertical.
Thus,
for
the
block
we
can
write
Newton’s
second
law
F mass
5-7a
an example.
A
block
of
m
presses
down
on
a table,
F
is
given
inshows
Fig.
5-7b.
Forces
are
the
only
two
forces
on
the
block
F
The
gravitational
The
forces
on block
theFigure
block
with
normal
.
The
free-body
diagram
for
the
F
FNand
g
N
"
mg
!
ma
.
F
force on
the block force
balance.
F
NF
y
g
:
:
:
gforthe
Fblock
mg
#
!
#two
abalance.
) forces
(5-13)
force
the
sitive-upward
ythey
axis
(F
!!
ma
asma
deforming
iton
somewhat
because
of
gravitational
force
theblock
block.
The
and
are
both
vertical.
Thus,
them(g
block
we
write F
Newton’s
second
law
isnet,
due
to
Earth's
y the
ycan
yN
y)
g on
F
given in Fig.
5-7b.
Forces
and
are
only
on
the
F
:
g
N
is due toup
Earth's
downward
(a)
table
onthe
thepull.
block
with
FN . The free-body diagram for the
for
a pushes
positive-upward
ynormal
axis (F
! ma
as (b)
Then
the
magnitude
of
force
is
net,
ynormal
y)force
:. can
:write
yr are
vertical.
Thus,pull.
for
the
block
we
second
downward
(a)
(b) Newton’s
F
"
F
!
ma
of
the
table
and
block
might
beon
inlaw
anblock
accelanyboth
vertical
acceleration
a
gForces
y
block Fig.
is given
Fig.Nresting
5-7b.
theF(they
only
two
forces
the
Fyg anda F
5-7 (a)in
A block
on
a table experiences
normal
force
perpendicular
to
N are
Fthe
"
Fg!
!not
may.accelerating
Nma
:
sitive-upward
y(a)
axis
(F
!
)Thus,
asblock
the
tabletop.
(b) The
free-body
for
block.
Fma
!
mg
#
m(g
#
awrite
(5-13)
ating
elevator).
If
the
table
and
are
relative
tolaw
the
±
net,
y vertical.
ydiagram
Ntable
y
y)
are
both
for
the
block
we
can
Newton’s
second
Fig.and
5-7 they
A
block
resting
on
a
experiences
a
normal
force
F
−
N perpendicular to
q. 5-8, we substitute mg for Fg, finding
the tabletop.
(b)
The
diagram
for
the
block.
From
Eq.
5-8,free-body
we substitute
mgnet,
for
Fgma
, finding
for
positive-upward
y axis
(F
ound, then
aya!
0 Figure
and
Eq.
5-13
yields
y!
y) as
F
"
F
!
ma
.
5-7a shows
an a
example.
A
block
of
mass
m presses
down
on a table,
of
the
table
and
block
(they
might
be in an accelfor any 1.
vertical
acceleration
N
g
y
y !not
If the
table
and
block
are
accelerating
relative
to
the
ground,
"
mg
ma
.
F
deforming it somewhat
because of the
gravitational
force
the block. The
N
y" F
mg
!
ma
. F onaccelerating
FNblock
ma
g!
y.ynot
àtable
ay =pushes
0Ifà
erating
elevator).
the
table
and
are
relative
to the
F
!
mg.
(5-14)
Figure
5-7a
shows
an
example.
A
block
of
mass
m
presses
down
on afortable,
up
on F
the,block
withN
normal force F . The free-body diagram
the
q. 5-8, we substitute
mg
for
finding
:
g
isnormal
given in because
Fig.force
5-7b.
Forces
andforce
are theforce
only two
the block
Fgravitational
deforming
it somewhat
ofis
the
on theonblock.
The
Fg forces
e magnitude
ofblock
the
Then
the
magnitude
of the
normal
From
Eq.
5-8,
we
substitute
mg
for
FgF, finding
ground,
then
a
!
0
and
Eq.
5-13
yields
: is
y
andup
they
both
vertical.
Thus,
for the
block
can
write Newton’s
second
table
onare
the
block
withare
normal
force
The
free-body
diagram
forlaw
the ground,
FN .we
2. pushes
If the
table
and
block
accelerating
up-ward
relative
to
the
:
:
"
mg
!
ma
.
F
for
a
positive-upward
y
axis
(F
!
ma
)
as
N
y
FN
block is à
given
theay!
only
two
FFgyNand
!F
mg
#!
ma
m(g
# ay) on the block (5-13)(5-13)
"
mg
FFig.
!5-7b.
mg
# ma
!
m(g
#
)ma
ay in
is
+ve
à Forces
N are
y. forces
N
y!
F
mg.
(5-14)
N
F block
" F !we
macan
. write Newton’s second law
and they are both
vertical. Thus, for the
CHECKPOINT
3
evertical
magnitude
ofany
thevertical
normal
force
isnormal
table
and block
(they
in an accelfor
acceleration
a y)ofasthe
the
magnitude
of
the
force
is: (they
for
aThen
positive-upward
axis
(Fnet,
the
table
block
might
bemight
in anbeaccelacceleration
a yof
y ! mayand
The Normal Force
N
N
g
g
:
N
:
g
net, y
:
N
:
N
:
g
y
N
g
y
3. the
If the table and
are accelerating
relative to the ground,
y
g
FNdown-ward
n Fig. 5-7, is
magnitude
ofblock
the
normal
force
greater
than,
less than,
or equal
to
erating
elevator).
If
the
table
and
block
are
not
accelerating
relative
to the
ay F
is
−ve
à
elevator). If à
the
table
and
block
are
not
accelerating
relative
to
the
F
"
F
!
ma
.
"
mg
!
ma
.
F
−mayayy!
F
!Ngmg
#y#
m(g # ay)
(5-13)
Ny
mg
#an
ma
m(g
) upward
(5-13)
N!
N!
mg if the
block
and
table
are
in
elevator
moving
(a)
at
constant
speed
and
ground,
then
a
!
0
and
Eq.
5-13
yields
y
3
Then
the
magnitude
of
the
normal
force
is
thenCHECKPOINT
ay From
! for
0 Eq.
and
Eq.
5-13
yields
5-8, we substitute mg for Fg, finding
of the
table(they
and
(they
might
beaccelin an accelany
vertical
acceleration
ayand
b)
at
increasing
speed?
: block
of
the
table
block
might
be
in
an
vertical
acceleration
a
F
!
mg.
(5-14)
y
F
!
mg
#
ma
!
m(g
#
a
)
(5-13)
In Fig. 5-7,
is the
magnitude
ofFthe
normal
forceare
N " mg
y N
yFN greater than, less than, or equal to
! ma
erating
elevator).
If the
table
and
block
not accelerating relative to the
N
y.
From Eq. 5-8, we substitute mg for F , finding
FN ! mg.
(5-14)
(5-14)
FN ! mg.
CHECKPOINT 3
:
In Fig. 5-7, is the magnitude of the normal force FN greater than, less than, or equal to
5-7 SOM E PARTICU LAR FORCE S
mg if the block and table are in an elevator moving upward (a) at constant speed and
5-7 SOME PARTICULAR FORCES
(b) at increasing speed?
P
9
97
y
y
normal force
Normal force FN
The
normal
force
(from theNormal
table)force F
e force on
N
FN = mg + ma y
a)
is
the
force
on
block from the
the block from the
to slide a body over a surface, the motion is resisted
porting table.If we either slide or attempt
F
a =0
Friction
N
Block
supportingby
table.
y
F Block
a bonding between
the body
and the surface. (We discuss
this bonding more in
Block
N
Block
:
the next chapter.) The resistance isx considered to be a F
single
force f , called either
= mg
x
N
the frictional force or simply friction. This force is directed along the surface, opgravitational
TheThe
forces
The gravitational
forces (Fig. 5-8). Sometimes, to simplify a sitposite the direction of the Fintended
motion
g Fg
F
F = ma y + mg
g
e onforce
the on
block
Fg
the
block(due
balance. (the surface Nis frictionless).
to Earth's pull)
uation,
friction
is assumed
to be balance.
negligible
b)
ue to
is Earth's
due to Earth's
nward
pull. pull.
downward
Tension
(a)
(a)
(b) (b)
: :
a y = +ve
F > mg
N
(a) Aresting
block resting
on a experiences
table experiences
a normal
force
FN perpendicular
block
on a table
a normal
force
FN perpendicular
toto
op. (b)
The free-body
diagram
forblock.
the block.
The
free-body
diagram
the
When
a for
cord
(or
a rope, cable, or other such object) is attached to a body and
:
pulled taut, the cord pulls on the body with a force T directed away from the
ure
5-7a shows
an example.
A block
of mass
m presses
down on a table,
shows
an example.
A block
of mass
m presses
:down on a table,
: F on the block. The
ng
it somewhat
because
the gravitational
force
g
mewhat
because
of theofgravitational
force
F
:
g on the block. The
: FN . The free-body diagram for the
shes up on the block with normal force
: force:FN . The free-body diagram for the
on the block with normal
:
:
FN are the only two forces on the block
given in Fig. 5-7b. Forces
Fg and
Fig. 5
attem
upward (a) at constant speed and
CHECKPOINT 3
:
In Fig. 5-7, is the magnitude of the normal force FN greater than, less than, or equal to
mg if the block and table are in an elevator moving upward (a) at constant speed and
(b) at increasing speed?
Friction
rriction
a surface, the motion is resisted
— If
a body is slid over a surface, the motion is resisted by a
(We discuss this
bonding
more in
: to slide
we either slide or attempt
a body over
a surface,
motion
is resisted
bonding
between
the
body the
and
the surface
be
a
single
force
called
either
f
,
y a bonding between the body and the surface. (We discuss this bonding more in
:
isnext
directed
the resistance
surface,
ophe
chapter.)along
The
is considered
a single
force f, ,called
called either
— resistance
This
is to
a be
single
force
frictional force or
he
frictional
force or simply
friction.
This
force is directed along the surface, op5-8).
Sometimes,
to simplify
a sitsimply
friction
osite the direction of the intended motion (Fig. 5-8). Sometimes, to simplify a siturface
is frictionless).
Direction of
ation, friction is assumed to be negligible (the surface is frictionless).
attempted
slide
ension
Direction of
attempted
slide
f
f
:
When a cord (or a —rope,
cable,
or
other
such
object)
is
attached
to
a
body
and
Fig.
5-8
A
frictional
force
f
opposes the
:
Frictional force is directed
along the surface, opposite
the
:
bject)
is
attached
to
a
body
and
Fig.
5-8 A
frictional
f opposes
the
slide of
a body over a surface.
ulled taut, the cord pulls on the body with a force
T directed
away
from theforceattempted
:
direction of the motion
attempted slide of a body over a surface.
force T directed away from the
— To simplify a situation, sometimes a friction is assumed to be
negligible (the surface is frictionless)
friction.
This
force
is directed
along the surface, op:
dmply
to
be
negligible
(the
surface
is
frictionless).
the next chapter.) The resistance is considered to be a single force f , called eitherDirection of
he intended motion (Fig. 5-8). Sometimes, to simplify a sitthe frictional force or simply friction. This98force CHAPTE
is directed
along
surface, op-attempted
R 5 FORCE
AN Dthe
MOTION—I
slide
ed
beanegligible
(the surface
pt toto
slide
body over a surface,
the motionisisfrictionless).
resisted
Direction of
posite
direction
of the
motion
(Fig. 5-8). Sometimes, to simplify a sitbody
and the
the surface.
(We discuss
thisintended
bonding more
in
attempted
f
:
stance
is considered
a single force
either
, called
uation,
frictiontoisbeassumed
tof be
negligible
(the surface is frictionless).
slide
98
:
CHAPTER
5 FORCE
ply friction. This force is directed along
the surface,
op- AND MOTION—I
T
T
cable,
or
other
such
object)
is
attached
to
a
body
and
Fig.
5-8
A
frictional
force
f
opposes the
f
intended motion (Fig. 5-8). Sometimes, to
:simplify a sit—(theTension
If a cordaway
is attached
to a body
and
attempted
slide of a body over a surface.
ls
onnegligible
the body
with ais force
force( T):directed
from the
to be
surface
frictionless).
Direction of
:
pulled,
theobject)
cord pulls
on the
body
tension
force
attempted
e,Tension
cable, or other
such
is98attached
to a5with
body
and
Fig. 5-8 A frictional force f opposes
CHAPTER
FORCE
AND MOTION—I
:
slide
forces at the two ends of
5-9 away
(a)
The
cord,
pulled
taut,
is
— The
along theFig.
cord,
the
body
attempted
slide TofThe
a
body
over a surface
Tfrom
T
ulls on the body
withdirection
a forceof T isdirected
away
from
the
the cord are equal in magnitude
under
tension.
If
its
mass
is
negligible,
f
When a cord (or a rope, cable, or other such
object) is and
attached
to a body and
Fig. 5-8 A
the cord pulls on the body
the hand
:
:
—such
Tension
is on
the
magnitude
of
force
with
force
, even
if the
runs f opposes
attempted sl
cable,
or other
object)
attached
to athe
body
and with
pulled
taut,
the
cord:is (T):
pulls
body
a Tforce
Tcorddirected
away
Fig.the
5-8
A frictional
force
the from the
Tension
:
a massless, frictionless pulley as T
attempted
slide of a body over a surface.
if T has
magnitude
= 50around
N
on the body with e.g.,
a force
directed
away fromTthe
in (b) and (c).
T
T
T
(a)
The forces at the two ends of
T
Fig. 5-9 (a) The cord, pulled taut, is
à the
tension in the cord is 50 N
the cord are equal in magnitude.
under tension. If its mass is negligible,
the cord pulls on the body and the hand
body and along the cord (Fig. 5-9a). The
:
T
with
force
,
even
if
the
cord
runs
because the cord is said to be in a state of
— The cord properties:
The forces at the two ends of
T
Fig. 5-9pulley
(a) The
around a massless, frictionless
as cord, pulled taut, is
means that it is being pulled taut. The tensi
the cord are equal
(b ) in magnitude.
(c )
under tension. If its mass is negligible,(a)
in (b) and (c).
— Unstretchable
force on the
body. For example, if the force
the cord pulls on the body and the hand
tude T ! 50 N, the tension in the cord is 50
:
T , even if the cord
force
runs
98
—CHAPTE
Massless
(itsAN
mass
is negligible
compared
to the body’s mass)
R 5 FORCE
D with
MOTION—I
cordforce
is often
said tocalled
be massless
(m
body
andasalong the cord (Fig. 5-9a).AThe
is often
a tension
around a massless, frictionless
pulley
the of
body’s
mass)
unstretchable.
The
because
cord is said to be(a)
in a to
state
tension
(or and
to be
under
(b ) tension),
— Exists only as a inconnection
between
2 the
bodies
(b) and (c).
between
two
on both bod
means that it is being pulled taut. The
tension
in bodies.
the cordItispulls
the magnitude
T
even
if
the
bodies
and
the
cord
are
acceler
— Pulls on both bodies with the same
force
magnitude
T
force on the body. For example, if the force on the body from the cord has m
body and along
the cordfrictionless
(Fig. 5-9a).
The (Figs.
force 5-9b
is ofte
a massless,
pulley
an
tude T ! 50 N, the tension
in the cord is 50 N.
T
T
T
because the cord
is said totobe
a stateand
of tension
(orfric
to
compared
theinbodies
negligible
A cord
is often
said
to be
massless (meaning its mass is negligible com
— If the cord wraps halfway around a pulley,
the
netmeans
force
on
cord
wrapstaut.
halfway
aroundin
a pulley,
as
that it isthe
being
pulled
The tension
the cord
to the body’s mass) and unstretchable. The cord then exists only as a conn
from
theexample,
cord hasifthe
magnitude
2T.body f
the pulley from the cord has the magnitude
2T force on the body.
For
force on the
Tthe
between two bodies. It pulls on both bodies with the
same force magnit
tude T ! 50 N, the tension
T in the cord is 50 N.
The forces
at the
two bodies
ends of and the cordT are accelerating
even
if the
and even if the cord runs a
Fig. 5-9 (a) The cord, pulled taut, is
A
cord
is
often
said
to
be
massless (meaning its mas
the cord aare
equal in magnitude.
CHECKPOINT
under tension. If its mass is negligible,
massless,
frictionless pulley (Figs. 5-9b
and c). Such a4pulley has negligibl
to the body’s mass) and unstretchable. The cord then ex
the cord pulls on the body and the hand
N.
The friction
suspended
in Fig.
5-9c weighs
compared to the bodies and negligible
onbody
its axle
opposing
its 75
rota
:
between two bodies.
It
pulls
on
both
bodies
with
the
with force T , even if the cord runs
N when
body5-9c,
is moving
upward
cos
the cord wraps halfway around a pulley,
as the
in Fig.
the net
force(a)
onatthe
around a massless, frictionless pulley as
even if the bodies
the cord speed?
are accelerating and eve
(c)and
at decreasing
from
the
cord
has
the
magnitude
2T.
(a)
(b
)
(c
)
in (b) and (c).
a massless, frictionless pulley (Figs. 5-9b and c). Such a p
the cord wraps halfway around a pulley, as in Fig. 5-9c, the net force on the pulley
from the cord has the magnitude 2T.
CHECKPOINT 4
The suspended body in Fig. 5-9c weighs 75 N. Is T equal to, greater than, or less than 75
N when the body is moving upward (a) at constant speed, (b) at increasing speed, and
(c) at decreasing speed?
T
5-8 Newton’s Third Law
T
Two bodies are said to interact when they
T push or pull on each other — that is,
when a force acts on each
body due
to the other body. For example, suppose you
T
The forces at the two ends of
T
position
a book B so it leans against a crate C (Fig. 5-10a). Then the book and
the cord are equal
in magnitude.
:
crate interact: There is a horizontal force FBC on the book from the crate (or due
:
to the crate) and a horizontal force FCB on the crate from the book (or due to the
book). This pair
5-10b. Newton’s third law states that
(a)
(b ) of forces is shown in(cFig.
)
F
=
ma
a) Fy = ma y
c) Fy = may
b)
y
y
T − W = ma
T − W = m(−a)
− W =force
ma
along the cord (Fig. 5-9a). The force is often called a Ttension
a a=state
0 ofThird
he cord is said to be in
tensionLaw:
(or to
be under
which
Newton’s
When
twotension),
bodies
from
each
T =W
− ma
T = W interact,
+
ma the forces on the bodies
at it is being pulled taut.
tension
cordinismagnitude
the magnitude
of the
T The
−W
=always
0m in the
other
are
equal
andTopposite
in direction.
T = 75 − ma
nheBbody. For example,
T
=
75
+
ma
if the force on the body from the cord has magniT =W
s50the
same
T <W
N, the
tension in the cord is 50 N.
T
>
W
T = 75 N
is often said to be massless
(meaning its mass is negligible compared
sd the
For
the
book
and crate, we can write this law as the scalar relation
dy’s to
mass)
due
B. and unstretchable. The cord then exists only as a connection
two bodies. It pulls on both bodies with the same
force
T, magnitudes)
FBC
! Fmagnitude
(equal
CB
e bodies
and the cord are accelerating and even if the cord runs around
ainst
crate
or(Figs.
as the
vector
s,efrictionless
pulley
5-9b
and c).relation
Such a pulley has negligible mass
book
Examples:
Q.1: In which figure of the following the y-component of the net force is zero?
(a)
(b)
6N
3N
2N
(c)
6N
3N
2N
3N
2N
3N
2N
7N
3N
2N
2N
2N
4N
4N
(d)
5N
4N
4N
Q.2: In which figure of the following the particle moves with a constant velocity?
6N
6N
(a)
(b)
3N
2N
(c)
3N
2N
3N
2N
4N
ΣFx = 0, ΣFy = 0
7N
5N
(d)
3N
3N
2N
1N
4N
3N
2N
4N
4N
Q.3: A particle of mass 2kg at a point where g = 9.8m/s2, the weight of this
particle at point where g = 0 is:
(a) 49N
(b) 98N
(c) zero
(d) 9.8N
W = mg = 2 X 0 = 0
Q.4: The direction of the acceleration of a body is:
(a) Opposite to the net force
(b) The same direction of the net force
(c) Perpendicular to the direction of the net force
(d) The same of the initial velocity
Q.5: In which figure of the following the particle moves up if it starts from rest?
(a)
(b)
6N
3N
6N
5N
5N
3N
2N
4N
ΣFx = 0, ΣFy=+ve
(c)
8N
3N
3N
7N
3N
2N
2N
1N
4N
(d)
5N
4N
4N
Q.6: In which figure of the following the acceleration of the particle
moves to right?
6N
(a)
(b)
3N
2N
6N
(c)
2N
2N
3N
2N
4N
5N
(d)
3N
3N
3N
5N
2N
1N
4N
7N
4N
4N
(a) ΣFx = 1, ΣFy = 0
(b) ΣFx = 0, ΣFy = −1
(c) ΣFx = 0, ΣFy = 0
(d) ΣFx = −2, ΣFy = 1
the correct answer is (a) because the direction of a has to be as the
same direction as Fnet
Q.7: In the figure, the net force on the block is:
(a) 1N right
(b) 6N up
(c) 2N left
(d) 4N down
6N
ΣFx = 3 − 2 = 1N, ΣFy = 6 − 2 − 4 = 0N
3N
2N
2N
4N
Q.8: When a force of 10N is applied to a body, the body accelerates
with 2m/s2. The mass of the body is:
(a) 20kg
(b) 10kg
(c) 0.5kg
(d) 5kg
m = F/a = 10/2 = 5kg
Q.9: From the figure, the acceleration of the block of mass = 0.5kg
moving along the x-axis on a horizontal frictionless table is:
(a) 10m/s2
(b) − 10m/s2
(c) − 6.3m/s2
(d) −
8.3m/s2
Motion direction
a = F/m =− 5/0.5 = −10m/s2
F=5N
Q.10: A force of 7N is applied to a mass of 7kg, the resulting
acceleration is:
(a) 3m/s2
(b) 1m/s2
(c) 2m/s2
(d) 4m/s2
a = F/m =7/7 = 1m/s2
Q.11: A force accelerate a 5kg particle from rest to a speed of
12m/s in 4s. The magnitude of this force is:
(a) 10N
(b) zero
(c) 20N
(d) 15N
M = 5kg, vo = 0, v = 12m / s, t = 4s
v = vo + at ⇒ a = 3m / s 2
F = Ma = 3(5) = 15N
Q.12: A body of mass 1kg is accelerating by a = 3i + 4jm/s2, the
magnitude of the acting force F on the body is:
(a) 2.5N
(b) 7.5N
(c) 12N
(d) 5N


F = ma = (1)(3i + 4 j) = 3i + 4 j

F = 9 + 16 = 5N
Q.13: A net force of 15N acts on a body of weight 29.4N. The
acceleration of the body is:
(a) 9.8m/s2
(b) 5m/s2
(c) 6.5m/s2
(d) 2.4m/s2
W = mg ⇒ m =
a=
W 29.4
=
= 3kg
g
9.8
F 15
=
= 5m / s 2
m 3
Q.14: Only two forces are acing on a particle of mass 2kg that moves
with an acceleration of 3m/s2 in the positive direction of y axis. If F1
= 8i N, the magnitude of F2 is:
(a) 12N
(b) 10N
(c) 17N
(d) 15N
 
F1 + F2 = ma


8i + F2 = 2(3 j) ⇒ F2 = −8i + 6 j

F2 = 64 + 36 = 10N
Chapter 5
FORCE AND MOTION -I
Sections 5-8, 5-9
Newton’s Third Law
Applying Newton’s Laws
— Important skills from this lecture:
1. Explain Newton’s third law and apply it to different
cases
2. Apply Newton’s laws to solve problems for one and
two body system
when a force
on has
each
due2T.
to the other body. For example, suppose you
from acts
the cord
thebody
magnitude
:
ts on
each
body
due
toLaw
the
other
body.
Forthe
example,
suppose
you
from the
position
a
book
B
so
it
leans
against
a
crate
C
(Fig.
5-10a).
Then
the
book
and
nteract:
There
is
a
horizontal
force
on
book
from
the
crate
(or
due
F
5-8
Newton’s
Third
BC
ewton’s
Third
Law
CHECKPOINT
4a crate:C (Fig.
:
B
so
it
leans
against
5-10a).
Then
the
book
anddue
crate
interact:
There
is
a
horizontal
force
on
the
book
from
thetocrate
F
rate)
and
a
horizontal
force
F
on
the
crate
from
the
book
(or
the (or dueCHEC
BC
:
CHECKPOINT
4
CB
N. IsorT:
equalon
to,each
greater
than,—orthat
less is,
than 75
The suspended
body
in Fig.when
5-9c weighs
Two bodies
are said to
interact
they 75
push
pull
other
here
aforce
horizontal
force
on
the
book
from
the
crate
(or
due
Fupward
dies
areissaid
to interact
when
they
push
orFig.
pull
on
each
other
—
that
is,to,from
the
crate)
and
a horizontal
force
FNewton’s
on
the
the
BC
when
theon
body
is
moving
at
constant
speed,
increasing
speed,
andbook
N.(b)
Is at
T crate
equal
greater
than,
orthat
less(or
thandue
75 to the
The
suspended
body
in(a)
5-9c
weighs
75
CB
This
pair
of
forces
is
shown
in
5-10b.
third
law
states
when
atoN
acts
each
body
due
toFig.
the
other
body.
For
example,
suppose
you
:
The susp
acts book).
on
each
body
due
toof
the
other
body.
For
example,
suppose
you due
N
when
the
body
is
moving
upward
(a)
at
constant
speed,
(b)
at
increasing
speed,
and
(c)
at
decreasing
speed?
dorce
aposition
horizontal
force
F
on
the
crate
from
the
book
(or
to
the
This
pair
forces
is
shown
in
Fig.
5-10b.
Newton’s
third
law
states
that
a book B so it leans
N when
CB against a crate C (Fig. 5-10a). Then the book and
Newton’s Third Law
: 5-10a). Then the book and
(c) at decreasing
a book
B
so it leans
against
a crate speed?
C (Fig.
crate
interact:
There
is
a
horizontal
force
from
thestates
crate (orthat
due
F
:
BC on the book
of forces
isashown
in force
Fig. 5-10b.
Newton’s
third
law
: the
eract:
There
is
horizontal
on
book
from
the
crate
(or
due
F
BC F
to the crate) and a horizontal
force
:
CB on the crate from the book (or due to the
ate)
and
aThis
horizontal
force Fistwo
onbodies
the
fromNewton’s
thethe
book
(or due
to
thebodies
CB
book).
pair ofWhen
forces
shown
in crate
Fig.interact,
5-10b.
third
law
states
that
ton’s
Third
Law:
forces
on
the
from
eachfrom each
Newton’s
Third
Law:
When
two
bodies
interact,
the
forces
on the
bodies
his pair of forces is shown in Fig. 5-10b. Newton’s third law states that
are alwaysother
equal
magnitude
and
oppositeand
in direction.
areinalways
equal in
magnitude
opposite in direction.
Two
bodies
are
said
to
interact
when
they
push
or pull
on each
other
— that is,
d Law: When two Two
bodies
interact,
the
forces on
the
bodies
from
each
bodies
are
said
to
interact
when
they
push
or
pull
on
each other — that is,
Newton’s
Third
Law:
When
two bodies
interact,
the other
forces body.
on theFor
bodies
from each
ame
when
a
force
acts
on
each
body
due
to
the
example,
suppose
youB suppose you
Book
Crate C
sCrate
equal
in
magnitude
and
opposite
in
direction.
when
a force
acts
on
each
body
due
to the
other
body.
For example,
other
are
always
equal
in
magnitude
and
opposite
in direction.
on’s
Third
Law:
When
two
bodies
interact,
the
forces
on
the
bodies
from
each
C
—
Both
book
&
crate
in
the
figure
interact:
Book B
Crate C
position a book
B soaitbook
leansBagainst
a crate
C (Fig.
5-10a).
Then
the
book
and the book and
position
so
it law
leans
against
alaw
crate
Crelation
(Fig.
5-10a).
Then
:as
For
the
book
and
crate,
we
can
write
this
as
the
scalar
relation
ebook
always and
equal
in
magnitude
and
opposite
in
direction.
crate,
we
can
write
this
the
scalar
:
There
is a
horizontal
thebook
book
from
crate interact:
There
is a horizontal
force
on force
the
from
the crate (or due
FBC on
crate
interact:
There isforce
a
horizontal
F
B.
BC on the book from the crate
(a) (or due
:
:
the crate)
and
horizontal
force
FCB!
the
crate
from
the
book
(or due
the (or due to the
(a) to
For thetobook
and crate,
can write
law
ason
the
scalar
relation
to
theawe
crate)
athis
horizontal
force
FCB
on the
crate
from
the
book
the
crate
and
aand
horizontal
force
on
the
crate
F
F
(equal
magnitudes)
BC
CB
F
!pair
F
(equal
magnitudes)
d crate,
we can
write
this
law
as
the
scalar
relation
BC
CB
book).This
pair
of
forces
is
shown
in
Fig.
5-10b.
Newton’s
third
law states
thatlaw states that
book).
This
of
forces
is
shown
in
Fig. 5-10b.
Newton’s
third
ate
book
and crate,
we
can
write
this
law
as
the
scalar
relation
from the book
FCB
FBC
FBC ! FCB
(equal magnitudes)
5-8 Newton’s
Third LawThird Law
5-8 Newton’s
or as the vector relation
eFCBvector
or as the relation
vector
relation
!
F F!
FF
(c) at dec
5-8 Ne
Two bodi
when a fo
position a
crate inte
to the cra
book). Th
FCB
FBC
magnitudes)
(equal
magnitudes)
B
C
:
CB : (equal
BC BC CB
B
C
e
"F
(equal
magnitudes
and
opposite
directions),
(5-15) Newto
:
:FBC !
:Newton’s
:Newton’s
(b) each
CB When
Third
Law:
two bodies
interact,
the
forces
on the
bodies from
Third
Law:
When
two
bodies
interact,
the
forces
on the
bodies
from
each
vector
relation
The
force
on B other are
!
"F
F
(equal
magnitudes
and
opposite
directions),
(5-15)
(b)
BC
CB
!
"F
F
(equal
magnitudes
and
opposite
directions),
(5-15)
gnirelation
BC
CB
The
force
on
B
other
are
always
equal
in
magnitude
and
opposite
in
direction.
force on: B other:are always equal in magnitude and opposite in direction.
due to C has the same
where
the
minus
sign
means
that
these
two
forces
are
in
opposite
directions.
due
to
C
has
the same We
where
the
minus
sign
means
that
these
two
forces
are
in
opposite
directions.
We
!
"F
F
(equal
magnitudes
and
opposite
directions),
(5-15)
—
These
two
forces
are
called
action
&
reaction
forces
:to
C hasBC
the:sameCB
magnitude as the
same
magnitude
asWe
the
the
minus
sign
means
that
these
forces
are in
opposite
directions.
!
"F
(equal
magnitudes
andtwo
opposite
(5-15)
can
call
the
forces
between
two
interacting
bodies
a When
third-law
force
pair.
When
the
forces
between
two
interacting
bodies
a directions),
third-law
force
pair.
nitude
ascall
the
BC can
CB
the bo
force
on C
dueFor
to B.
For
the
book
and
crate,
we
can
write
this
law
as
the
scalar
relation
e
he minus sign
means
that
these
two
forces
are
in
opposite
directions.
We
force
on
C
due
to
B.
For
the
book
and
crate,
we
can
write
this
law
as
the
scalar
relation
el on
C due
to—B. A
the
forces
between
two
interacting
bodies
a pair.
third-law
pair.
When
third-law
force
pair: athe
forces
between
two force
othe
B.
forces
between
two
interacting
bodies
third-law
force
When
Fig.
5-10
(a)
Book
B leans against crate
F
!
F
(equal
magnitudes)
s sign means that these two
forces
are
in
opposite
directions.
We
BC
CB
: against crate
Fig.
5-10
(a)
Book
B
leans
FBC ! FCB
(equal magnitudes)
C.
leans against crateinteracting bodies
: (b) Forces FBC (the force on the book
or as the v
C. (b) Forces F (the force on the:book
or as the vector relation
crate
ces
bodies a third-law force
pair.from
When
the crate) and F (the force on the
rce
onbetween
the book
or as thetwo
vectorinteracting
relation
from the crate) and F (the force on the
BC
ok force on the —
(the
thethe same magnive
magnidirection.
:
CB
CB
When any two bodies
interact(equal
in any
situationopposite
(stationary,
crate from the book) (5-15)
have the same magni!
magnitudes and
directions),
:
:
crate from the
book) have the same magnitude(5-15)
and are opposite in direction.
"FCB
FBC !accelerating),
(equal a
magnitudes
and force
oppositepair
directions),
moving,
third-law
is present
where the
tude
and are opposite
in direction.
:
FBC
:
"FCB
where the minus sign means that these two forces are in opposite directions. We
where the minus
signthe
means
thatbetween
these two
forces
are in opposite
can call
forces
two
interacting
bodies adirections.
third-law We
force pair. When
can call the forces between two interacting bodies a third-law force pair. When
can call t
eand
interaction.
focus
on the
forces acting on the cantaloupeCantaloupe
(Fig.
Force THIRD LAW
5-8 NEWTON’S
:
:
( b ) 5-11b).
forceLet’s
is the
F first
:the cantaloupe
:
interaction.
The third-law
force pair for the
cantaloupe
– table
interaction.
CE
( afor
) the cantaloupe
are (d)
– Earth
FCT and
CE. (c) The third-law force pair
is the normal
force Fon
the cantaloupe from the
table, and force
FCT
FCE is the
interaction.force
(d) The5-11
third-law
pair for thelies
cantaloupe
– table
hey a third-law
(a)force
A cantaloupe
onEarth.
a table
thatinteraction.
stands
on
Earth. (b)force
The forces on
Cantaloupe
gravitationalFig.
force
on
the
cantaloupe
due
to
Are
they
a
Fthird-law
:
:
CE
( cCantaloupe
)
C
the
cantaloupe
are FCT and FCE. (c) The third-law force pair for the cantaloupe
– Earth
FFCT
from tab
ntaloupe,
and
not
on
EC (normal force
These
forces
pair?
No,
because
they
are
forces
on
a
single
body,
the
cantaloupe,
and
not
on
F CT
rce pair isany
present.
The interact
two
bodies
in any
situation,
third-law
force
is present. The
interaction.
(d) The third-law
force
pair forathe
cantaloupe
– tablepair
interaction.
F
just
happen
CE
These
are
two
interacting
bodies.
FECpresent. The
Cantaloupe
C
any
two
bodies
interact
in
any
situation,
a
third-law
force
pair
is
force
table)
F
(gravitational
force)
lawfrom
would
still
hold
if
Table
T
CE
book
and
crate in Fig. 5-10a are stationary,
but tothe
third law would
still
So on
are the
these.
be balanced.
third-law
force hold if
ntaloupe
but
on
the
To
find
a
third-law
pair,
we
must
focus
not
cantaloupe
but
on
the
book and crate in Fig. 5-10a are stationary,
but the third law would
still holdpairs.
if
F TC
(b)
Earth
they
were
moving
and
even
if
they
were
accelerating.
These
are
F
interaction
between
the
onesituation,
other body.
InEC the cantaloupe
–is present. T
ional
In force)
the
cantaloupe
– two
they
were moving
and
even
if they
wereand
accelerating.
Table
T cantaloupe
any
bodies
interact
in
any
a
third-law
force
pair
third-law
force
Fig.
5-11
A cantaloupe
lies onpairs
a table that
stands on the
Earth.can(b) T
Asthe
another
example,
let(the
us
find
the(a)
third-law
force
involving
pairs
involving
can: pairs :
Earth
interaction
(Fig.
5-11c),
Earth
pulls
on
the
cantaloupe
with
a
gravitational
)
d
)
Earth
E
As
another
example,
let
us
find
third-law
force
involving
the
canEarth
e with a gravitational
pairs.
book and crate in Fig. 5-10a
are stationary,
third
law force
would
hold
the cantaloupe
are FCT andbut
FCE.the
(c) The
third-law
pairstill
for the
can
:
:
taloupe
in
Fig.
5-11a,
which
lies
on
a
table
that
stands
on
Earth.
The
cantaloupe
:
Earth.
The
cantaloupe
force
the
cantaloupe
pulls
Earth
with
aThe
gravitational
Are
FCE
taloupe
inand
Fig.
5-11a,
which
lies
on aon
table
that
stands
on
Earth.force
Theforce
cantaloupe
interaction.
(d)
third-law
pair
forFthe
– table inte
EC.cantaloupe
n Earth.
(b) The
forces
on
they
were
moving
and
even
if
they
were
accelerating.
tional
force
.
Are
F
EC
) the
( con
)
Earth
E ( atable
interacts
with
table
and
with
Earth
(this
time,
there
are whose
three bodies whose
these
forces
a third-law
force
pair?
Yes,
because
they
are
two
interactinteracts
with
the
and
with
Earth
(this
time,
there
areforces
three
bodies
ir for the
cantaloupe
– Earth
are
three
bodies
whose
As
another
example,
let
us
find
the
third-law
force
pairs
involving
F CT the ca
rces
on
two
interacte – table
interaction.
interactions
we
must
sortto
out).
ing
bodies,
the
force
on
each
due
the
other.
Thus,
by
Newton’s
third law,
interactions
we
must
sort
out).
taloupe
in
Fig.
5-11a,
which
lies
on
a
table
that
stands
on Earth.
The cantalou
F
(normal
force
from
table)
any
two
bodies
interact
in
any
situation,
a third-law
force pS
(
a
)
(
c
)
CT
on’s third law,
These
forces
Let’s
first
focus
on
the
forces
acting
on
the
cantaloupe
(Fig.
5-11b).
Force
Let’s
first
focus
on
the
forces
acting
on
the
cantaloupe
(Fig.
5-11b).
Force
:
:
— Theinteracts
cantaloupe
interacts
with
the
table
&
with
Earth
Fthird
with
the
table
and
with
Earth
(this
time,
there
are
three
bodies
who
F
book
and
crate
in
Fig.
5-10a
are
stationary,
but
the
law
TC
:
:
CT
upe (Fig.
5-11b).
Force
:
:
(cantaloupe ! Earth interaction).
F
"
!F
just
happen
CE
EC
isCT
the
force on
the
cantaloupe
from
the
table,
and
isaccelerating.
the
FCT
Fwere
F CE
(gravitational
force)
is:normal
the normal
force
onsort
the
cantaloupe
from
theforce
table,
force FCE is the
F
CEand
they
were
moving
and
even
if
they
F
(normal
force
from
table)
interactions
we
must
out).
aw
force
pair
is
present.
The
to
be
balanced.
CT
n).
le, andgravitational
force
is
the
F
So
These
forces
(Earth.
b)
(d)
CEcantaloupe–table
force
on
the
cantaloupe
due
to
Are
they
a
third-law
force
—
The
interaction:
As
another
example,
let
us
find
the
third-law
force
pairs
gravitational
force
on
the
cantaloupe
due
to
Earth.
Are
they
a
third-law
force
F
Next,
in
the
cantaloupe
–
table
interaction,
the
force
on
the
cantaloupe
from
e third law would still hold
ifjustfirst
TC
Let’s
focus on the forces acting on
cantaloupe (Fig.
5-11b).
For
happen
:
:
:
:
they
a
third-law
force
pair?
No,
because
they
are
forces
on
single
body,
the
cantaloupe,
onthat
F CEa(gravitational
force)
Fig.
5-11
(a) A
cantaloupe
lies on
a table
that
stands
on
Earth.
(b)
The
forces
on
taloupe
in
Fig.
5-11a,
which
lies and
on
a not
table
stands
on Ear
g.
the
table
is
and,
conversely,
the
force
on
the
from
thetable)
cantaloupe
is
F
FTCforce
pair?
No,
because
they
are
forces
on
atable
single
body,
the
cantaloupe,
and Fnot
ont
to
balanced.
—
(normal
force
on
the
cantaloupe
from
the
: be
:
is
the
normal
force
on
the
cantaloupe
from
the
table,
and
is
F
CT
the cantaloupe
from
CT
CE
the
cantaloupe
are
and
F
.
(c)
The
third-law
force
pair
for
the
cantaloupe
–
Earth
F
(b)
( d ) there are t
CT
CE
interacts
withpair,
the and
tableso
and with Earth (this time,
two5-11d).
interacting
bodies.
:the
force (Fig.
pairs
involving
cancantaloupe,
and
not
on
These
forces
are
also
a
third-law
force
two
interacting
bodies.
gravitational
force
on
the
cantaloupe
due
to
Earth.
Are
they
a
third-law
for
interaction.
(d)
The
third-law
force
pair
for
the
cantaloupe
–
table
interaction.
—
(force on
the
table
from
the
cantaloupe)
the
cantaloupe
is
F
interactions
we
must
sort
out).
TC
To
find
a
third-law
pair,
we
must
focus
not
on
the
cantaloupe
but
on
the
Fig.
5-11
(a)
A
cantaloupe
lies
on
a
table
that
stands
on
Earth.
(b)
The
forces
on
nds on Earth. The cantaloupe
: No,
:
: third-law
:
To
find
a
pair,
we pair
must
focus
not
on
thethe
cantaloupe
onnot
the
pair?
because
they are
forces
on
a single
body,
cantaloupe,
and
Let’s
first
focus
on
the
forces
acting
on
thebut
cantaloupe
the
cantaloupe
are
and
F
The third-law
force
for
the
cantaloupe
–In
Earth
F
CT
CE
so
interaction
between
cantaloupe
and
one
other
body.
the
cantaloupe
–
there are
three bodies
whose
(cantaloupe
–
table
interaction).
FCTthe
". (c)
!F
:
TC
interaction.
(d)
The
third-law
forceinpair
forsituation,
the cantaloupe
–the
table
interaction.
two
interacting
bodies.
cantaloupe
but
on
the
interaction
between
the
cantaloupe
and
one
other
cantaloupe
–
is the
normal
force
on the
cantaloupe
from
the table, an
F
any
two
bodies
interact
any
aon
third-law
force
pair
iswith
present.
The In the
Earth interaction (Fig. 5-11c), Earth pullsCT
cantaloupe
abody.
gravitational
book
and
crate
in
Fig.(Fig.
areinteraction:
stationary,
but
the
third
law
would
still
holdthe
if: cantaloupe
gravitational
force
on
the
cantaloupe
due to
Areon
they
Tocantaloupe
find
a5-10a
third-law
pair,
we
must
focus
not
on
but
t
Earth
interaction
5-11c),
Earth
pulls
on
the
cantaloupe
with
a Earth.
gravitational
—F:
The
cantaloupe–Earth
dy.
the
cantaloupe
–
antaloupe
(Fig.
5-11b).
Force
). In force
and
the
pulls
on
Earth
with
a
gravitational
force
.
Are
F
CE
EC
:
:the canta
:
they were
moving and between
even if they were
accelerating.
pair?
No,force
because
they
are
forces
on a single
body,
interaction
the
cantaloupe
and
one
other
body.
In
the
cantaloup
he
table,
and
force
is
the
F
any
two
bodies
interact
in
any
situation,
a
third-law
pair
is
present.
The
force
and
the
cantaloupe
pulls
on
Earth
with
a
gravitational
force
F
F
CE
these
third-law
force
pair?
Yes,
because
they
are
forces
on
upe with
a forces
gravitational
EC. Are
— AsaCE
(Earth
pulls
the
cantaloupe
with
a involving
gravitational
force)
another
example,
leton
us find
the
third-law
force
pairs
thetwo
can-interactCHECKPOINT
5
two
interacting
bodies.
:
and crate
in Fig.
5-10a
are lies
stationary,
butthat
theEarth
third law
would
still
hold
if
h. Are they
a third-law
force
Earth
interaction
(Fig.
pulls
onThe
the
cantaloupe
with
a gravitation
taloupe
inforce
Fig.
5-11a,
which
onthe
a 5-11c),
table
stands
on
Earth.
cantaloupe
these
forces
a
third-law
force
pair?
Yes,
because
they
are
forces
on
two
ingbook
bodies,
the
on
each
due
to
other.
Thus,
by
Newton’s
third
law,
:
: cant
itational
force
.
Are
F
To
find
a
third-law
pair,
we
must
focus
not
oninteractthe
—
(cantaloupe
pulls
on
Earth
with
a
gravitational
force)
they were
moving
and
even
if
they
were
accelerating.
EC
Suppose
that
the
cantaloupe
and
table
of
Fig.
5-11
are
in
an
elevator
cab
that
begins
to
y, the cantaloupe,
and
not
on
force
and
the
cantaloupe
pulls
on
Earth
with
a
gravitational
force
F
interacts
with F
the
table
and
with
Earth
(this
time,
there
are
three
bodies
whose
EC. A
: the other.
:
ing
bodies,
the
force
on
each
due
to
Thus,
by
Newton’s
third
law,
:CE
:find
interaction
between
the
cantaloupe
and
one
other
body.
In
As
another
example,
let
us
the
third-law
force
pairs
involving
the
canforcesaccelerate
on two
interact(a)
Do
the
magnitudes
of FTCpair?
increase,
decrease,
or stay
the on two intera
upward.
FYes,
interactions
we
must
sort
out).
CT
(cantaloupe
!and
Earth
interaction).
F
"
!F
these
forces
a
third-law
force
because
they
are
forces
CE
EC
Earth
interaction
5-11c),
Earth
pulls on
the cantaloupe w
taloupe inLet’s
Fig. 5-11a,
which
lies
a table
stands
on Earth. (Fig.
The
cantaloupe
:
:that
firstthe
focus
on
theon
forces
acting
on
the
cantaloupe
(Fig.
5-11b).in
Force
same?
(b)
Are
those
two
forces
still
equal
in
magnitude
and
opposite
direction?
(c)
on
the cantaloupe
but
on
:
wton’s
third
law,
: with
: by Newton’s third law,
bodies,
the
force
on(this
each
due
toand
the
other.
Thus,
: and
: Earth
interacts
the
table
with
time,
there
are
three
bodies
whose
(cantaloupe
!
Earth
interaction).
F
"increase,
!F
the
cantaloupe
pulls
Earth with a gravitatio
FCE
CE
ECforce
is ing
the
normal
force
on
the
cantaloupe
from
the
table,
and
force
the on
FCTcantaloupe
F(d)
ator
cabDo
that
begins
to
CE isAre
Next,
in
the
cantaloupe
–
table
interaction,
the
force
on
the
cantaloupe
from
the
magnitudes
of
and
F
decrease,
or
stay
the
same?
those
two
F
er body.
In
the
–
CE
EC
interactions we must sort out).
:
:
these
forcesAre
a third-law
force pair?
they are forc
gravitational
force on the cantaloupe
due
to
they a third-law
force Yes, because
:
: Earth.
Example:
Q.1: A book rests on a table, exerting a downward force on it. The
reaction to this force is:
(a) Force from the Earth on the table
(b) Force from the book on Earth
(c) Force from the Earth on the book
(d) Force from the the table on the book
Applying Newton’s Laws
tal of five forces act on the blocks, as shown in Fig. 5-13:
the cord does not stretch
certainblock
time, block
S mov
table,
hanging
time. This means that th
accelerations have the sam
1. The cord pulls to the right on sliding block S with
a forceon
Block
of magnitude
T.
Sample
Problem
2. The
pulls upward
on S
hanging
block H with
a force
Figure
5-12cord
shows
a block
(the sliding
block)
with mass
of the
same
This to
upward
force
keepsablock
Q How do I classify this
Block on
block
hanging
M table,
! 3.3
kg.
Themagnitude
block isT.free
move
along
horizontal
ticular law of physics
H from falling freely.
frictionless surface and connected, by a cord that wraps over
Yes. Forces, masses, a
3. Earth pulls down on block S with the gravitational force
g block) witha mass
:
frictionless
pulley,
to
a
second
block
H
(the
hanging
they should suggest Newt
FgS, which has a magnitude equal to Mg. FN
Block S
along a horizontal
:
block),
with
mass
m
!
2.1
kg.
The
cord
and
pulley
have
neg. That is our starting Ke
ma
4. Earth pulls down on block H with the gravitational force
T
ord that wraps
overmasses
:
ligible
compared
to
the
blocks
(they
are
“massless”).
FgH, which has a magnitude equal to mg. M
Q If I apply Newton’s se
k H (the hanging
:
body should I apply it
The hanging
block
as the
sliding
block
5. The table
pushesHupfalls
on block
S with
a normal
force S
FNaccelerates
.
nd pulley have
to negthe right. Find (a) the acceleration of block S, (b) the ac- We focus on two bodie
block. Although they ar
ey are “massless”).
FgScord.
celeration of block H, and (c) the tension in the
points), we
T can still treat
block S accelerates
Sliding
every part
of it5-13
moves Th
in e
Fig.
block S
block S, (b) the
acQ What is this problem all about?
macting
Idea is to
apply
Newton’s
Block
onHthe tws
in the cord.
M
You are given two
bodies — sliding block and hanging
Q Whatblocks
about the
pulley
of Fig.
5block — butFig.
must5-13
also consider
Earth, which pulls on both We Fcannot
represent
The forces
gH
different parts of it move
bodies. (Without
nothing would happen here.) A toactingEarth,
on Frictionless
the two
There is a
rotation,
we
shall
deal wit
surface
block and hanging
tal of five forces
actofon
the
blocks, as shown in Fig. 5-13:
blocks
Fig.
5-12.
does
eliminatethe
thecord
pulley
fro
Hanging
m
hich pulls on both
certain
time,
b
H S with a force
mass to be
negligible
com
1. The cord pulls to the right on sliding block
block
appen here.) A toIts
only This
function
There is another thing you should note.blocks.
We assume
thatmei
time.
of magnitude T.
wn in Fig. 5-13:
the cord does not stretch, so that if block H falls accelerations
1 mm in a
2. The cord pulls upward on hanging block H with a forceQ OK. Now how do I ap
certain
block
S moves
1 mmofto
the rightRepresent
in that block
sameS as
block time,
S of mass
is connected
to a block
mass
ock S with a force
ofFig.
the5-12
sameAmagnitude
T.MThis
upward
forceHkeeps
block
Q How do I
Block on table, block hanging
s a block S (the sliding block) with mass
block is free to move along a horizontal
— by
Wea cord
needthat
to find
Fnetover
in x & y directions for both
e and connected,
wraps
blocks of masses M & m
lley, to a second block H (the hanging
m ! 2.1 kg. The cord and pulley have neg— For block S of mass M:
mpared to the blocks (they are “massless”).
Fnet ,x S= accelerates
Max ⇒ T = Max = Ma (1)
k H falls as the sliding block
(a) the acceleration of block
S, Ma
(b) the
acFnet ,y =
y ⇒ FN − FgS = M (0) = 0
k H, and (c) the tension in the cord.
⇒ FN = FgS = Mg
(2)
Fig. 5-13 The forces
FN
Block S
T
M
FgS
T
m
Block H
F
gH
— For block H of mass m:
roblem all about?
acting on the two
n two bodies — sliding block
and
hanging
Fnet ,x = 0
blocks of Fig. 5-12.
also consider Earth, which pulls on both
Fnet ,y here.)
= mayA⇒to-T − FgS = may
y
Earth, nothing would happen
There is another thing you should note. We assume that
ct on the blocks, as shown in Fig. 5-13: ⇒ T − mg
y 1 mm in a
the=cord
−madoes
(3)not stretch, so that if block H falls
certainintime,
S moves 1 mm to the right in that same
−a block
because
block
H accelerates
the block
negative
to the right on The
sliding
S with
a force
a
direction of the y axis
FN
time. This means that the blocks
move together and their
T.
T
accelerations have the same
a.
T
M magnitude
m
upward on hanging
block
H
with
a
force
x
x
Both blocks M & m accelerate with the same
gnitude T. Thismagnitude
upward force
Q How do I classify this problem?
Should it suggestHanging
a para keeps block
Sliding
FgH block H
a
ticular law of physics to Fme?
freely.
gS
block S
By substituting
1 into
3:
Yes. Forces, masses, and accelerations are involved, and
wn on block S with
the gravitational
force
:
m
they
should
suggest
Newton’s
second
law
of
motion,
F
a magnitude equal to Mg.Ma − mg = −ma ⇒ a =
net !
g
(4)
:
(a)
(b)
M .+That
m is our starting Key Idea.
ma
wn on block H with the gravitational force
a magnitude equal to mg.
:
es up on block S with a normal force FN.
5-14 (a) A free-body diagram for block S of Fig. 5-12.
Q If I applyFig.
Newton’s
second law to this problem, to which
(b) A free-body diagram for block H of Fig. 5-12.
body should I apply it?
Now note that Eqs. 5-18 and 5-20 are simultaneous equations with the same two unknowns, T and a. Subtracting
these equations eliminates T. Then solving for a yields
By substituting 4 into 1:
m
(5-21)
a"
g.
M$m
Mm
T =5-18 yieldsg
Substituting this result into Eq.
M +m
12.
Mm
T"
g.
(5-22)
M
$
m
Putting in the values of M, m, and g from the problem, we obtain:
Putting in the numbers gives, for these two quantities,
m
2.1 kg
g"
(9.8 m/s2)
M$m
3.3 kg $ 2.1 kg
" 3.8 m/s2
(Answer)
a"
es sense
ction in
and T "
o apply
s explain
ea: The
an write
(5-16)
f the net
mponent
block S
Mm
(3.3 kg)(2.1 kg)
g"
(9.8 m/s2)
M$m
3.3 kg $ 2.1 kg
" 13 N.
(Answer)
Q The problem is now solved, right?
That’s a fair question, but the problem is not really finished until we have examined the results to see whether they
make sense. (If you made these calculations on the job,
wouldn’t you want to see whether they made sense before
you turned them in?)
Look first at Eq. 5-21. Note that it is dimensionally
correct and that the acceleration a will always be less than g.
This is as it must be, because the hanging block is not in free
tem and a free-body diagra
not determine acceleration along the plane.
The acceleration along the plane is set by the
force
compo:
Sample
Problem
102
Cord accelerates blockFrom
up aFig.
ramp
positive
direction
CH
A
P of
Tthe
ExR
F
5-15h,
we
write
Newton’s
second
law
(
net "
nents along the plane (not by force components perpendicular
:
from the cord is up the plan
a ) for motion along the x axis as
m
:
to
expressed
by aNewton’s
second
lawup
(Eq.
5-1).a
In the
Fig.plane),
5-15a, as
a cord
pulls on
box of sea
biscuits
along
component
along
the
plane
is
down
the
plane
and
has mag- yforce F
N. The
gravitational
g
Cord accelerates block up aFig.
ramp
5-15
(a)
A
box
is
pull
2
Sample
Problem
Tindicated
! mg sin in
u "Fig.
ma.5-15g.tude
(5-24)
(To
see
why
that
nitude
mg
sin
u
as
frictionless
plane
inclined
at
u
"
30°.
The
box
has
mass
m
"
mg
" (5.00
Cord
Calculation: For convenience, we draw a coordinate sysFig. 5-15 (a)(b)
A box
isthree
pulled
upkg)(9.8
a acting
planem/s
by
The
forces
on
trig
function
is
involved,
we
go
through
the
steps
of
Figs.
5.00
kg,
and
the
force
from
the
cord
has
magnitude
T
"
25.0
Norm
: the
In
Fig.
a
cord
p
Substituting
andasolving
for three
a,5-15a,
we find
tem and a free-body
shown
in Fig.
In Fig. diagram
5-15a, aascord
pulls
on a5-15b.
box The
of:sea biscuits
updata
along
along
plane
is
down
(b)component
The
forces
acting
on
the
box:
the
cor
force
forc
T
, the gravitational
Fig.
5-15
(a)
A
box
is
pulled
up
a
plane
by
a
cord.
5-15c
to
h
to
relate
the
given
the
force
compo:
: angle to
frictionless
plane
inc
N.
What
is
the
box’s
acceleration
component
a
along
the
inpositive direction
of the x plane
axis isinclined
up the plane.
Force
force
gravitational
force
andin
the
nof
T
, the
FFinding
2 mg
nitude
sinAcan
uF:as
indicated
Fig.
frictionless
at u "
30°.TThe box
hasTo
mass
m"
5.00
kg,
and
the
The
box
accelerates.
g,force
FNthe
,
(Answer)
a
"
0.100
m/s
force
nents.)
indicate
the
direction,
we
write
the
clined
plane?
:
N. (c)–(i)
(b)
three
forces
acting
box: the
N. What
is
the
box’s
fromThe
the cord
is up
the plane
and on
has the
magnitude
T "cord’s
25.0
force
(c)–(i)
Finding
the
component
.plane?
F!mg
Nfunction
Cordforce
:the force from
trig
is
involved,
we go th
:
:
5.00
kg,
and
the
cord
has
magnitude
T
"
25.0
the
plane
and
perpendicular
down-the-plane
component
as
sin
u.
The
normal
force
clined
where
the
positive
result
indicates
that
the
box
accelerates
N. The T
gravitational
force Fg is downward
and
has
magni:
force
force
and
the
normal
, the gravitational
,
F
the plane
plane (Fig.
and perpendicular
to does
it.
g
KisEthe
Y I D E A acceleration
T
FN the
isaperpendicular
and thus
5-15c
to h5-15i)
to relate
the
given ang
N. kg)(9.8
What
component
along the in-to the
tude mg:
(5.00
m/s2) "box’s
" with
49.0 N. More important,
its
up
plane.
block)
mass
force
Finding
the force components along
FN. (c)–(i)
5-15 (a) A boxacceleration
is pulledhe
upBlock
a plane bynents.)
a cord.
notFig.
determine
along
the
acceleration
alon
θ boxplane.
To
indicate
the
directio
clined
The acceleration
alongplane?
the plane is set by the force compo:
FT
S
Fig.
5-15
(a)
A
box
N
The
accel
e
rates.
nents
along
the
plane
along
a horizontal
From
Fig.
5-15h,
we
write
Newton’s second law (Fnet " is pulle
the
and
perpendicular
to it. perpendicular
nentsplane
along
the
plane
(not by force components
(b)
The
three
forces
acting
on
the
box:the
cord’
s
down-the-plane
component
as
!mg
to
the
plane),
as
expr
(b) The
three
forces
acting
onC
:
:
a
m
)
for
motion
along
the
x
axis
as
:
:
:
y
Athe
to
plane),
as expressed
lawI(Eq.
gravitational
force
, the
TF
Thisforce
is aT:right
Calculation:
co
rd
that
wraps
overby Newton’s second
KEY
D E 5-1).
A
is
perpendicular
toFor
the plane
(F
force T, the gravitational
force
and
the
normal
,
F
N
This
is
a
right
g
tem
and
a
free-body
Normal
force
force
(c)–(i)
Finding
the
fo
.
F
M
T
!
mg
sin
u
"
ma.
(5-24)
N
:
(a) acceleration
90°along
−θ
triangle.
positive
direction
o
Calculation:
For
convenience, we draw a coordinate sysnot
determine
th
H (the The
hanging
F
the
perpendicular
− θplane and
This
is galsto
triangle.
forceofforce
Finding
theand
force components
along
FN.x(c)–(i)axes
x is
acceleration
along the
plane
set bySubstituting
the
compoThe
box is moving
in the
+veisdirection
from
theFN 90°
cord
up
θ
θ
data
solving
for
a,
we
find
tem and a free-body diagram as shown in Fig. 5-15b. The
Cord
N. The
gravitational
From
Fig. 5-15h, we write NewG
:
dpositive
pulley
have
nentsofnegalong
by force
components
perpendicular
tude
mg
"T (5.00 kg)
the
plane
and
perpendicular
to
it.
direction
the xthe
axisplane
is up (not
the plane.
Force
T
:
2
Perpendicular
Fig. 5-15 (a) A box is pulled up a plane by a cord.
m am/s
) for
along
the x axis as
, motion
(Answer)
a " 0.100
This
is
a
right
The
acceleration
component
along
the
inclined
plain
The
box
accelerates.
to
the
plane),
as
expressed
by
Newton’s
second
law
(Eq.
5-1).
from
the
cord
is
up
the
plane
and
has
magnitude
T
25.0
"
fo
θ
ey are
Cord's pull
(b) The“massless”).
three forces acting on:the box: the cord’s
F
A
θ
gS
:
:
component
of
where
boxisaccelerates
N.force
The Tgravitational
force
and
hasismagniright
is required
− θthe positive result indicates that the This
90°force
−Fθg Fisg, downward
This
also. 90°
Ta !
mg sin u " m
triangle.
and the normal
, the gravitational
:
2
(a)
T
Calculation:
For
convenience,
we
draw
a
coordinate
syslock
S
accelerates
tude
mg
(5.00
kg)(9.8
m/s
"
)
"
49.0
N.
More
important,
its
90° − θ (b)
F
up
the
plane.
triangle.
force FN. (c)–(i) Finding
theneed
force components
Fgg
à We
to find Falong
θ
net in the x directionθ
Gravitational
A
This is a right
triangle. 90° −θ This
Block on table, block hanging
θ
Substituting
data and(a)
solving
for a, w
θ
(c) A bo
5-15
force
m
(c)
(b) The three
forces
Block
H Hy
2
force yT
the
gravitati
,
,
a
"
0.100
m/s
(b)
Perpendi
c
ul
a
r
θ
F
force F
. (c)–(i) Find
and
a free-body
diagram as shown in Fig. 5-15b.
The
the plane
andtem
perpendicular
to it.
θ
lock
S, (b)
the
ac:
positive direction
of
the
x
axis
is
up
the
plane.
Force
T
Fnet ,x = max
nA
the cord.
This is a right(a)T " 25.0
θ from the cord is up the plane and has magnitude
Fig.
:
g
:
N
(c)
⇒ T − mgsin
component
of indicatesA
where
result
N. The gravitational
forceθF=g ma
is downward and
has
Perpendicular
90° − θ theFNpositive
90°magni−θ
Thi
s
i
s
al
s
o.
tri
a
ngl
e
.
x
This is a right
F
Fig.
5-13
The
forces
gH
T
2
Cord
Adjacent
leg(c)
component
T
tude
kg)(9.8
" 49.0 N. More important,
itsof up the plane.Parallel
90°)− θ
90° mg
− θ " (5.00This
is also.m/s
triangle.
(u
F
T
:
a = plane
− gsin
θ the two
Fg
(a) A box is pulled up aacting
by on
a cord.
m
(c)
(d)
The
box accelerates.
θ
θ
(b) The three forces acting on the box:
the cord’s
: blocks of Fig. 5-12.
:
25the normal
force T, the
gravitational force Fg, and
F
A
: θ
=
(9.8)sin
30 = 0.1m / s 2 g θ
force FN. (c)–(i) Finding the force components
along
θ
the plane and perpendicular to it. 5
Fig. 5-15
block and hanging
hich pulls on both
appen here.) A to(c)
n in Fig. 5-13:
Normal force
the plane and perpen
This is a
triangle.
(e)
θ
Hypotenuse
θ
g
right
(use cos θ )
mg
Gravitational
(c )
Opposite leg
force
y
θ mg cos θ
(f )
(b)
mg(use
sin θsin θ )mg mg sin θ
Cord
mg
fo
−
th
θ Cord's pull of
θ
component
θ mg cos θ
Hypotenuse
θFgmg cos θ
θ
y
FF
θ
90°
g
g
Parallel
There(d)is another
you should note. We assume that
The net thing
of(e)these
component
of
(a)
the cord does not
stretch,
so
1 lemm
in a O
l
Fg that if block H xfalls Paral
forces
determines
(a) certain
A box is pulled
up ablock
plane bySa cord.
Perpendicular
(c) 1 mm
(d) right (e)in component
(f)θ (u
mgFsin
time,
moves
to
the
thatof same
y accelerates.
N
The
box
T
The net of these the acceleration.
θ
These forces
( g)
Adjacent leg
component of
5-15
a righta Fig.
ockThis
S iswith
force
(b) The three forces acting on the box: the cord’s
us, F2 ! 2.00 N and
e x axis.
Eq. 5-25 leads to
ax ! #2.00 m/s2.
Sample Problem
(Answer)
Forces within an elevator c
Sample Problem
In Fig. 5-17a, a passenger of mass m ! 72.2 kg stands on
Block on
table, block
hanging
a
platform
scale
in an elevator cab. We are concerned with
Sample Problem
the scale readings when the cab is stationary and when it is
g block) with mass
movingan
upelevator
or down.cab
Forces
within
FN
Block S
along
a horizontal
y
T whatever
ord
that
wraps
over
(a)
Find
a
general
solution
for
the
scale
reading,
! 72.2 kg stands on
M
k
H
(the
hanging
vertical motion of the cab.
are concernedthe
with
FN
nd
pulley
have it
negnary
and when
is
ey are “massless”).
FgS
KEY IDEAS
T
block S accelerates
eblock
reading,
whatever
(1) The
S, (b)
the
ac- reading is equal to the magnitude of the normal force m
:
Block H
other force actin the cord. FN on the passenger from the scale. The only
assenger
Passenger
:
ing on the passenger is the gravitational force Fg, as shown in
F
Fig. 5-13 The forces
These
the free-body
diagram of Fig. 5-17b. (2) We
can forces
relate the gH
acting on the two
:
compete.
forces
on
the
passenger
to
his
acceleration
a by using
of
the
normal
force
block and hanging
Fig. 5-17
blocks of Fig.:
5-12.
F
g
:
Their
net force
Newton’s
second law (F net ! ma ). However,
recall
that we
only
force
actdicates eit
hichother
pulls
on
both
:
causes
a vertical
can
use this lawThere
only in
an
inertial
frame.
If
the
cab
accelerorce
as shown
Fg, here.)
body
diag
appen
A in
tois another
thing
you
should
note.
We
assume
that
acceleration.
( a )frame. So( bwe
) choose
We
can
relate
the
on him fro
ates,
then
it
is
not
an
inertial
the ground
wn in Fig.
5-13:
the cord does not stretch, so that if block
H falls 1 mm
in a
:
ration a by using
Fig. 5-17
(a)block
A passenger
stands
a platform
scale that
in- same
certain
time,
S moves
1 on
mm
to the right
in that
the
readings (F
when the
cab is stationary and when it is
for yscale
components
net, y " may) to get
moving up or down.
FN ! Fg " ma
(a) Find a general solution for the scale reading, whatever
or
F " F # ma.
(5-27)
the vertical motion of theNcab. g
This tells us that the scale reading, which is equal to FN,
depends
on the
vertical
acceleration.
Substituting
mg for Fg
Fnet ,y = ma
⇒
F
−
F
=
ma
⇒
F
=
mg
+
ma
K
E
Y
I
D
E
A
S
y
N
g
N
gives us
⇒
FN = m(a
g)normal force
(1) The reading is equal
to m(g
the magnitude
of +
the
# a)
(Answer)
(5-28)
FN "
:
FN on the passenger from the scale. The only other force act:
for on
anythe
choice
of acceleration
a.
ing
passenger
is the gravitational
force Fg, as shown in
the
diagram
of Fig.
5-17b.
(2)cab
We is
canstationary
relate theor
(b) free-body
What does
the scale
read
if the
:
forces
the passenger
to 0.50
his acceleration
a by using
movingon
upward
at a constant
m/s?
:
:
Newton’s second law (F net ! ma
). However, recall that we
When the cab is stationary or moving with a constant speed, a = 0
can use this law only in an
inertial frame. If the cab accelerKEY IDEA
ates, then it is not an inertial frame. So we choose the ground
For any constant velocity (zero or otherwise), the accelera= m(a
+ g) → isFNzero.
= mg = (72.2)(9.8) = 708N
tion a F
ofN the
passenger
Calculation: Substituting this and other known values into
Eq. 5-28, we find
FN " (72.2 kg)(9.8 m/s2 # 0) " 708 N.
(Answer)
This is the weight of the passenger and is equal to the mag-
FN " (7
" 47
For an upward acce
is increasing or its d
reading is greater th
is a measurement o
in a noninertial fram
decreasing upward
the scale reading is l
(a)
(d) During the upw
Fig. 5-17 (a)
pass
th
magnitude
FnetAof
dicates
either his weigh
the
magnitude
ap,cab
body diagram
theDp
frame
of the for
cab?
on him from the scale a
Calculation: The m
on the passenger do
senger or the cab; so
the magnitude FN of
the upward accelerat
the net force on the p
Fnet " FN ! Fg "
during the upward
passenger
ong the y
w written
FN " (72.2 kg)(9.8 m/s2 # 3.20 m/s2)
This is the weight of the passenger and is equal to the mag" 939 N,
(Answer)
nitude Fg of the gravitational force on him.
and for a " !3.20 m/s2, it gives
(c) What does the scale read if
the cab accelerates
upward
2
2
" (72.2
kg)(9.8 at
m/s3.20
! 3.20
at 3.20 m/sF2 Nand
downward
m/s2m/s
? )
" 477 N.
(5-27)
ual to FN,
mg for Fg
(5-28)
ionary or
(Answer)
alues into
(Answer)
during the upwar
ap,cab relative to th
inertial frame of
map,cab, and Newto
For an upward acceleration (either the cab’s upward speed
isAccelerates
increasing upward
or its downward
the scale
FN =speed
m(a +isg)decreasing),
→ FN = 72.2(3.2
+ 9.8) = 939N
reading is greater than the passenger’s weight. That reading
is a measurement of an apparent weight, because it is made
Sample Problem
in
a noninertial
frame. For
(either + 9.8) = 476.52N
Accelerates
downward
FNa =downward
m(−a + g)acceleration
→ FN = 72.2(−3.2
decreasing upward speed or increasing downward
speed), of block pushing on bloc
Acceleration
the scale reading is less than the passenger’s weight.
:
In Fig.
5-18a,
constant
horizontal
force
(d)
During
the aupward
acceleration
in part
(c),Fwhat
the
app ofismagnitude
" 4.0
20 N is applied
A the
of passenger,
mass mAand
force on
whatkg,
is which
magnitude
Fnet of thetonetblock
the
magnitude
ap,cab
of hisBacceleration
asB measured
the blocks
pushes
against
block
of :mass m
" 6.0 kg.inThe
:
" ma p,cab
frame
of theacab?
Does Fnetsurface,
?
slide over
frictionless
along
an x axis.
accelera-
Fnet " FN ! F
(a) What is the
Calculation:
Theacceleration
magnitude Fgofofthe
theblocks?
gravitational force
F
=
F
−
F
net doesNnot depend
g
: motion of the pason the passenger
on the
Serious Error: Because force Fapp is applied directly
senger or theFcab;
so, from part (b), Fg .is8708
) N. From part (c),
net = 939 − (72 .2)(9
to
block
A,
we
use
Newton’s
second
law toduring
relate that
the magnitude FN of the normal
force
on
the
passenger
Facceleration
= 231N :
net
a ofNblock
force
to
the
the motion
the upward acceleration is the 939
readingA.
onBecause
the scale.Thus,
is along
x axis,
we isuse that law for x components
the
net forcethe
on the
passenger
(Fnet, x " max), writing it as
Fnet " FN ! Fg " 939 N ! 708 N " 231 N,
(Answer)
only horizontal fo
:
force FAB from blo
Dead-End Solut
ing, again for the x
(We use the min
Because FAB is a
equation for a.
Successful Solu
Sample Problem
Acceleration of block pushing o
Sample Problem
:
only horiz
In Fig. 5-18a, a constant horizontal force Fapp of magnitude
:
Block on20table,
force FAB fr
N is block
appliedhanging
to block A of mass mA " 4.0 kg, which
pushes against block B of mass mB " 6.0 kg. The blocks
g block) with mass
Dead-End
slide over a frictionless surface, along anFNx axis.
Block
S
along a horizontal
ing,
again
f
"
total
mass
m
A
B
What is the acceleration ofThis
theforce
blocks?
ord that wraps(a)
over
causes the T
values, we find
M
Fapp
A
:
acceleration of the full
k H (the hanging
Fapp system.
Serious Error: Becausex force
is applied directly
Fa
two-block
nd pulley havetonega !use t
(We
block A, we use
(a) Newton’s second law to relate that
mA "
:
ey are “massless”).
F
Because F
gS
force to the acceleration a of block A. Because
the motion
T equation fo
block S accelerates
two x
forces
is along the x axis, we use These
that are
lawthefor
components Thus,
the acceler
Fapp
FAB
block S, (b) the
acacting
on
just
block
A.
A
m
(Fnet, x " max), writing it asx
in the
positive
Block
H dir
Their net force causes
in the cord.
Successfu
2.0 m/s2. :
(b)
Fapp " its
macceleration.
Aa.
force Fapp i
F
:
(b)gH What is the
Fig. 5-13 The forces
system.
We
However, this is seriously wrong because Fapp is not the block
A (Fig. 5-18
acting on the two
B
block and hanging
This is the only force
blocks of Fig. 5-12.
FBA
causing the acceleration
hich pulls on both
x
of block B.
appen here.) A toThere
that the
(c) is another thing you should note. We assume
We can relate
wn in Fig. 5-13:
not
stretch,
soF:that
if block H fallseration
1 mmwith
in New
a
(a) Adoes
constant
horizontal
force
Fig. the
5-18cord
app is applied to block
A, which
pushes
againstblock
block B.
Two horizontal
act on
certain
time,
S(b)
moves
1 mm forces
to the
right in that same
to block
s act on
re, once
pushes against block B of mass mB " 6.0 kg.
The blocks
PA R T 1
slide over a frictionless surface, along an x axis.
105
REVIEW & SUMMARY
B
(a) What is the acceleration of the blocks?
F
A
:
app
F
=
20N,
m
=
4Kg,
m
=
6Kg
app
A
B
Serious
Error:
Because
force Fapp is
applied directly
total mass m " m . Solving for a and substituting
known
A
Dead-End
ing, again
x
B
(a) that
to
block
A, we use Newton’s second law to relate
values,
we find
:
Fnetto
(mAacceleration
+ mB )a
,x =the
force
the motion
Fapp
20 a
N of block A. Because
!(m we
! 2.0law
m/s2. for x components
is along
the
x
axis,
use
that
Fappa +!Fm
−
F
=
+
m
)a
4.0
BAA " m
ABB
A kg "B 6.0 kg
Fapp
FAB
A
),6)a
writing it as
(Fnet,
(Answer)
20x+"
0 =ma
(4x+
Thus, the acceleration
of the system
and
ofa.each block is
F
"
m
2
app
A
a =positive
2m / sdirection of the x axis and has the magnitude:
in the
2.0 m/s2.
However,
this is seriously wrong because Fapp
(b) What is the (horizontal) force
block A (Fig. 5-18c)?
KEY IDEA
:
FBA
FBA = mB a = 6 × 2 = 12N
on block B from
(b)
is not the
B
FBA
x
(c)
This force ca
acceleration
two-block sys
(We use t
Because F
These are the
equation
f
x
acting on jus
Their net forc
Successf
:
its acceleratio
force Fapp
system. We
This is the on
causing the a
of block B.
:
Fig. 5-18 (a) A constant horizontal force F
We can relate the net force on block B to the block’s accelA, which pushes against block B. (b) Two hori
eration with Newton’s second law.
block A. (c) Only one horizontal force acts on
Calculation: Here we can write that law, still for components along the x axis, as
eration of the system with Newton’s seco
FBA ! mBa,
again for the x axis, we can write that law a
which, with known values, gives
Examples:
Q.1: A constant force of 46N is applied at an angle
of 60o to a block A of a mass 10kg as shown in the F
figure. Block A pushes another block B of mass
36kg. Assuming a frictionless surface, the total
acceleration of the blocks along the x-axes is:
(a) 1.5m/s2
(b) 0.25m/s2 (c) 0.5m/s2
(d) 2m/s2
A
B
FNA
Fnet ,x = (mA + mB )a
a=
θ
F cosθ
46 cos 60
=
= 0.5m / s 2
mA + mB
10 + 36
FNB
o
F
Fcosθ
Fsinθ
A
B
mAg mBg
Motion
direction
Q.2: A 3kg box is placed in the top of a 10kg box.
The bottom box is pushed with a force F. The two
boxes moves together with an acceleration of
2m/s2. The horizontal force F is:
(a) 3N
(b) 26N
(c) 1N
(d) 5N
A
F
Motion
direction
B
Fnet ,x = (mA + mB )a = (10 + 3)2 = 26N
Q.3 In the figure, two blocks are connected and pulled
on a horizontal table by a force with a magnitude of
20N. If the m1 =3kg, m2 = 2kg, then T and a are:
(a) 5N, 4m/s2
(b) 8N, 4m/s2 (c) 5N, 4m/s2
(d) 10N, 3m/s2
Fnet ,x = (m1 + m2 )a
F
20
⇒a=
=
= 4m / s 2
m1 + m2
5
To find T we apply Newton's 2nd law on one of the mass:
F − T = m1a ⇒ T = F − m1a = 20 − 12 = 8N
or T = m2 a = 2(4) = 8N
Motion direction
m2
T
T
m1
F
Q.4: An elevator of total mass 2000kg moves upward. The tension
in the cable pulling it is 24000N. The acceleration of the elevator
is:
T
(a) 2.2m/s2
(b) 9.8m/s2
(c) 12m/s2
(d) 3.6m/s2
Fnet ,y = ma
T − mg = ma
a
⇒a=
T − mg 24000 − 2000(9.8)
=
= 2.2m / s 2
m
2000
mg
Q.5: A 70kg man stands on a spring scale in an elevator that has a
downward acceleration of 2.8m/s2. The scale will read:
(a) 980N
(b) 680N
(c) 490N
(d) 343N
T
a
Fnet ,y = ma
T − mg = −ma
⇒ T = m(g − a) = 70(9.8 − 2.8) = 490N
mg
Q.6: An elevator has a body of 10kg. The tension in the cable when
the elevator is moving upward at a constant speed of 10m/s is:
(a) zero
(b) 98N
(c) 1.5N
(d) 7.3N
T
Fnet ,y = ma
T − mg = 0
v
⇒ T = ma = 10(9.8) = 98N
mg
Q.7: Two masses m1 = 4kg and m2 = 6kg are connected to a rope of a
negligible mass. An upward force of 198N is applied as shown. The
magnitude of the acceleration of the system is:
(a) 10m/s2
(b) 40.2m/s2
(c) 50.2m/s2
(d) 70.2/s2
T
Fnet,y = Ma
T − (m1 + m2 )g = (m1 + m2 )a
=
198 − (10)9.8
= 10m / s 2
10
T − (m1 + m2 )g
⇒a=
m1 + m2
a
Mg
Q.8: A block slides down a frictionless inclined plane with an
acceleration of 4.9m/s2. The angle between the plane and the
horizontal is:
(a) 30o (b) 26o (c) 21.55o
(d)14.32o
FN
Fnet ,x = max
⇒ −mgsin θ = −ma ⇒ a = gsin θ
a
⇒ θ = sin −1 = 30 o
g
mg sinθ
θ
mg cosθ
Q.9: A 40kg crate is held at rest on a frictionless incline by a force
parallel to the incline. If the incline is 30o above the horizontal, the
magnitude of the applied force is:
FN
F
(a) 20N
(b) 40N
(c) 23.5N
(d) 10N
Fnet,x = max
mg sinθ
at rest à a = 0
30o
⇒ F − mgsin θ = 0 ⇒ F = mgsin θ = 40(9.8)sin 30 = 20N
o
mg cosθ
Q.10: A block of mass 4kg is pushed up a smooth 30o inclined
plane by a constant force of magnitude 40N and parallel to the
incline. The magnitude of the acceleration of the block is:
(a) zero
(b) 9.8m/s2
(c) 1.2m/s2
(d) 5.1m/s2
Fnet ,x = max ⇒ F − mgsin θ = ma
F − mgsin θ 40 − 4(9.8)sin 30 o
⇒a=
=
= 5.1m / s 2
m
4
FN
mg sinθ
30o
Q.11: If the mass of the block is 5kg. Find T if
the block moves with a constant velocity upward
the smooth inclined plane (or at rest).
(a) 45N
(b) 24.5N
(c) 42N
(d) 25N
Fnet ,x = max ⇒ T − mgsin θ = 0
⇒ T = mgsin θ = 5(9.8)sin 30 = 24.5N
o
F
FN
mg cosθ
F
mg sinθ
30o
mg cosθ
-15)
conyou
able
Eqs.
x!
ese
Acceleration
Equation
Missing
Number
Equation
Quantity
Q.12: A 5kg block
is pushed upward
30o inclined
plane with initial velocity of 14m/s. The distance
2-11
! v0 is:
" at
x # x0
that the blockvgoes
1 2
!
v
t
"
x
#
x
0
0
(a)
20m
(b)
10m
(c) 18m(d)
24m
2 at
2-15
v
2-16
t
v 2 ! v 20 " 2a(x # x0)
1
2-17
x
#
x
!
v)t θ a
0
0 "
2 (v
Fnet ,x = max ⇒
ma
= −mgsin
1 2
x
#
x
!
vt
#
2-18
v0
0
2 at / s 2
⇒ a = −gsin θ = −4.4m
FN
mg sinθ
30o
mg cosθ
vo =the
14macceleration
/ s, v = 0 is indeed
Make sure that
2
constant before using
in this table.
vo2 the14equations
Δx = −
=
= 20m
a 4.4
a
Q.13: From the figure, the normal force FN on a
block of weight 60N sliding down a frictionless
plane is:
(a) 50N
(b) 30N (c) 25N (d) 40N
Fnet ,y = may ⇒ FN − mg cosθ = 0
⇒ FN = mg cosθ = 60 cos 60 = 30N
o
FN
mg sinθ
60o
mg cosθ
Q.14: Show the correct direction of the tension force T:
B
X
B
A
(a)
B
X
A
X
(c)
Q.15: A block of mass m is connected to a block of
mass M as shown. The normal force on the block m is:
(a) mg − T
(b) mg
(c) Mg − T
(d) Mg
A
(d)
m
X
M
FN
T
X
T
mg
X
A
(b)
m
B
M
Mg
Q.16: Referring to the last example, if the block M is moving
downward, the net force acting on it is:
(a) Ma − T = Mg
(b) T = Ma (c)T = Mg (d) T − Mg = − Ma
Fnet ,y = Ma
⇒ T − Mg = −Ma
FN
m
T
X
T
mg
M
Mg
Motion
direction
!!"#$%&"'((")*(+,$*-."/%0%"!!
!!"#$%&"'((")*(+,$*-."/%0%"!!
•1 Only
two horizontal forces act on a 3.0 kg body that can move
sec. 5-6 Newton’s Second Law
over a!!"#$%&"'((")*(+,$*-."/%0%"!!
frictionless floor. One force is 9.0 N, acting due east, and the
•1
Only
two
forces
act
on a 3.0
kg body
can move
sec. 5-6
Newton’s
Second
other
is 8.0
N,horizontal
acting 62°
northLaw
of west.
What
is thethat
magnitude
of
SSM Worked-out solution available in Student Solutions Manual
WWW Worked-out solution is at
http://w
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
–
Number
of
dots
indicates
level
of
problem
difficulty
ILW
Interactive
solution
is
at
•
•••
sec. 5-6 Newton’s Second Law
••8 A 2.00 kg object is subjec
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
a!
eration :
over
a frictionless
floor. One force is 9.0 N, acting due east, and the
Problems:
the
acceleration?
•1 body’s
Only two
horizontal forces act on a 3.0 kg body that can move
other is 8.0 N, acting 62° north of west. What is the magnitude of
overTwo
a frictionless
floor.
One
acting due
east,that
andcan
the
•2
horizontal
forces
actforce
on a is
2.09.0kgN,chopping
block
the
body’s
acceleration?
otherover
is 8.0
N, acting 62°
north counter,
of west. What
theinmagnitude
of
slide
a frictionless
kitchen
which islies
an xy plane.
:
•2
forces
on aN)ĵ.
2.0 Find
kg chopping
block that
the Two
body’s
One
forcehorizontal
isacceleration?
the acceleration
of can
the
F1 ! (3.0
N)î act
# (4.0
slide
over
a
frictionless
kitchen
counter,
which
lies
in
an
xy
plane.
chopping
block
the other
force
is can
(a)
•2 Two horizontal
forces actnotation
on a 2.0 when
kg chopping
block
that
: in unit-vector
:
:
One
force
is
Find
the
acceleration
of
the
F
!
(3.0
N)î
#
(4.0
N)ĵ.
1 # ("4.0 kitchen
and
Fslide
N)ĵ, (b)counter,
F2 ! ("3.0
over aN)î
frictionless
whichN)î
lies#in(4.0
an N)ĵ,
xy plane.
2 ! ("3.0
:
: in unit-vector notation when the other force is (a)
chopping
block
(c)
! (3.0
F2 force
("4.0
One
is N)î
F1 !#(3.0
N)îN)ĵ
# .(4.0 :
N)ĵ. Find the acceleration of the
:
Fchopping
N)î #in("4.0
N)ĵ, (b)
F2 ! ("3.0
# (4.0force
N)ĵ, isand
2 ! ("3.0block
unit-vector
notation
when N)î
the other
:
2 (a)
•3
If
the
1
kg
standard
body
has
an
acceleration
of
2.00
m/s
at
:
:
(c)
(3.0N)î
N)î #
# ("4.0
("4.0N)ĵ,
N)ĵ . (b) F ! ("3.0 N)î # (4.0 N)ĵ, and
F2 F!2 !
("3.0
2
20.0°:to the positive direction of an x axis, what are (a) the x com(c) F
! (3.0
N)î # ("4.0body
N)ĵ . has an acceleration of 2.00 m/s2 at
•3
If2 the
ponent
and1 kg
(b)standard
the y component
of the net force acting on the
20.0°
to
the
positive
direction
of
an
axis,
what notation?
are (a)
the xm/s
com2
•3 Ifand
the(c)
1 kg
standard
body
has
acceleration
of 2.00
at
body,
what
is the net
force
in xan
unit-vector
ponent
and
(b)
the
y
component
of
the
net
force
acting
on
the
20.0° to the positive direction of an x axis, what are (a) the x com••4
While
forces
actnet
on force
it, a parbody,
andand
(c)two
what
is the
in of
unit-vector
notation?
y acting on the
ponent
(b) the
y component
the net force
ticle is to move at the constant velocbody,While
and (c)
is the on
net force
in unit-vector notation?
••4
twowhat
forces
a par" (4 act
ity :
One
of
m/s)ĵ. it,
v ! (3 m/s)î
y
: at the constant velocticle
is
to
move
••4:forces
While
forces
act#on("6
it, aN)ĵ.
parthe
is two
F1 !
(2 N)î
y
"
ity
One
of
!
(3
m/s)î
(4
m/s)ĵ.
v
ticle isisto
at
the
constant
velocWhat
themove
other
force?
:
F1
:
the
forces
is
F
!
(2
N)î
#
("6
N)ĵ.
1
ity v ! (3 m/s)î " (4 m/s)ĵ. One of
θ1
: astronauts,
••5
Three
propelled
x
What
is
the
other
force?
F1
the forces is F1 ! (2 N)î # ("6 N)ĵ.
θ3
by jet backpacks, push and guide a
θ1 F2
What
is
the
other
force?
••5
Three
astronauts,
propelled
x
F1
120 kg asteroid toward a processing
θ3 θ1 F2
by
backpacks,
push
andshown
guide in
a
••5jet exerting
Threethe
astronauts,
propelled
F3
dock,
forces
x
120
kg
asteroid
toward
a
processing
θ3
by jet
backpacks,
guide
F2
Fig.
5-29,
with F1 !push
32 N,and
F2 !
55 N,a
F
dock,
exerting
the
forces
shown
in
3
asteroid
toward
a processing
F120
41 N,
u1 ! 30°,
and
u3 ! 60°.
3 !kg
Fig. 5-29 Problem
5.
Fig.
5-29,
with
F
!
32
N,
F
!
55
N,
1
2
F
dock,isexerting
the forces
shown(a)in
What
the asteroid’s
acceleration
"(8.00 m/s2)î # (
: 2.00 kg object is subjec
••8
are FA
N
1 !
: (30.0 N)î # (16.0
2
a ! "(8.00
eration
m/s )î # (
find
force.
••8 the
2.00 kg
object is subjec
:A third
are F1 !: (30.0 N)î # (16.0
N
a0.340
eration
! "(8.00
m/s2)î #m
••9
A
kg
particle
: third force.
find the
are
F
(30.0 N)î
# (16.0
N
to x(t)1!!"15.00
# 2.00t
" 4.0
••9
0.340
kg
particle
m
find the
third
force.
with
xAand
y in
meters
and t in
to
x(t) ! "15.00
# 2.00t
" an
4.0
the
(b) the
••9 magnitude
A 0.340 and
kg particle
m
with
x
and
y
in
meters
and
t
in
of
axis)
of the
force
tothe
x(t)x!
"15.00
# net
2.00t
" 4.o
the
magnitude
and direction
(b) the ano
gle
of
particle’s
with
xthe
and
y in meters
and t i
of
the
x
axis)
of
the
net
the magnitude
and
(b) force
the ano
••10
A
0.150
kg
particle
m
gle
of
the
particle’s
direction
o
of x(t)
the x!axis)
of the
net force
to
"13.00
# 2.00t
# 4.0
••10
kg direction
particle
m
gle of A
theIn0.150
particle’s
seconds.
unit-vector
notatio
to
x(t) !
# s?
2.00t
# 4.0
particle
t ! 3.40
••10
Aat"13.00
0.150
kg
particle
seconds.
unit-vector
notatio
to x(t)A!In
"13.00
# 2.00t
# 4.
••11
2.0
kg
particle
moves
particle
3.40 s? notati
seconds.atforce
Int !
unit-vector
variable
directed along
particle
at
t
!
3.40
s?ct 2moves
••11
A (4.0
2.0 kg
particle
3.0 m #
m/s)t
#
" (2.
variable
force
••11 AThe
2.0 factor
kgdirected
particle
move
seconds.
c is 2aalong
const
3.0
mhas
# (4.0
m/s)t
# ct
" (2.
variable
directed
ticle
aforce
magnitude
of along
36
N
seconds.
The
factor
c
is
a
2 const
3.0 m
# (4.0
m/s)t # ct " (2
axis.
What
is c?
ticle
has
a
magnitude
ofa36
N
seconds.
The
factor
c
is
cons
•••12What Two
forc
axis.
is
c? horizontal
ticle
has
a
magnitude
of
36
slides over frictionless ice,No
axis. What
is c?
•••12
Two
horizontalsyste
forc
which
an xy
coordinate
:
slides
frictionless
horizontal
is•••12
laid over
out.Two
Force
inforto
F1 is ice,
which
xy frictionless
coordinate
syste
slides an
over
positive
direction
of
x ax
: the ice,
is
laid
out.
Force
in
F1 is
which
coordinate
and
hasan
a xy
magnitude
of syste
7.0t
:
: direction of the x ax
positive
is laidF2out.
F1 is in
Force
has Force
a magnitude
of 9t
e cab and its
en that occuis 8.00 m/s2
from a height of 10.0 m by holding onto a rope that
runs over a frictionless pulley to a 65 kg sandbag.
With what speed does the man hit the ground if he
started from rest?
heir humanom a cannon
ersion of the
els to land in
non and at a
or 5.2 m and
e underwent
magnitude of
ough it were
••53 In Fig. 5-48, three connected blocks are pulled to the right on
a horizontal frictionless table by a force of magnitude T3 ! 65.0 N.
If m1 ! 12.0 kg, m2 ! 24.0 kg, and m3 ! 31.0 kg, calculate (a) the
magnitude of the system’s acceleration, (b) the tension T1, and (c)
the tension T2.
onnected by
by the cable
m1
T1
m2
Fig. 5-48
'((")*(+,$*-."/%0%"!!
T2
m3
Problem 53.
T3
and 65.
are F1 ! (30.0 N)î # (16.0 N
over a frictionless floor. One force is 9.0 N, acting due east, and the
find the third force.
other is 8.0 N, acting 62° north of west. What is the magnitude of
1. We are only concerned with horizontal forces in this problem (gravity plays no direct
the body’s
acceleration?
••9 A 0.340 kg particle
role).
We take East as the +x direction and North as +y. This calculation is efficiently
on forces
a vector-capable
using magnitude-angle
notation
•2 Twoimplemented
horizontal
act on acalculator,
2.0 kg chopping
block that
can
units understood).
slide over a frictionless kitchen counter, which lies in an xy plane.
:
One force is F1 ! (3.0 N)î F# (4.0
0 g Find
118 acceleration
b9.0 N)ĵ.
b8.0 the
g b2.9 53 g of the
a
m notation3.0when the other force is (a)
chopping block in unit-vector
:
:
F2 ! ("3.0 N)î # ("4.0 N)ĵ, (b) F2 ! ("3.0 N)î
# (4.0 N)ĵ, and
Therefore, the acceleration has a magnitude of 2.9 m/s2.
:
(c) F2 ! (3.0 N)î # ("4.0 N)ĵ .
(withtoSIx(t)
! "15.00 # 2.00t " 4
with x and y in meters and t i
the magnitude and (b) the an
of the x axis) of the net force
gle of the particle’s direction
••10 A 0.150 kg particle
2. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2).2The net force
to x(t) ! "13.00 # 2.00t # 4.
•3 If the
1 kg
has
2.00addition
m/s isatdone using
applied
on standard
the choppingbody
block is
FnetanF1acceleration
F2 , where theof
vector
seconds. In unit-vector notat
20.0° to unit-vector
the positive
direction
of an of
x axis,
what
are by
(a)a the
x comnotation.
The acceleration
the block
is given
F
F
/
m
.
d 1 2i
particle at t ! 3.40 s?
ponent and (b) the y component of the net force acting on the
body, and
what
is the net force in unit-vector notation?
(a) (c)
In the
first case
••4 While two forces
on
it, aˆi parF1 Fact
3.0N
4.0N
2
ticle is to move at the constant velocity :
(3am/s)î
v ! so
0 . " (4 m/s)ĵ. One of
:
the forces is F1 ! (2 N)î # ("6 N)ĵ.
(b) In the second case, the acceleration a
What is the
other force?
ˆj
equals
3.0N ˆi y
4.0N ˆj
F1
0
ˆi
3.0Npropelled
4.0N ˆj
3.0N ˆi
4.0N ˆj θ1
••5
ThreeFastronauts,
x
F
1
2
(4.0m/s 2 )ˆj.
θ3
by jet backpacks,m push and guide a 2.0kg
F2
120 kg asteroid toward a processing
(c) In this final situation, a is
F3
dock, exerting
the forces shown in
Fig. 5-29, with F1 ! 32 N,
F ! 55
N,
3.0N 2 ˆi
4.0N ˆj
3.0N ˆi
4.0N ˆj
F1 F2
ˆ
(3.05.
m/s 2 )i.
F3 ! 41 N, u1 ! 30°, and u3 ! 60°.
Fig.
5-29
Problem
2.0 kg
m
What is the asteroid’s acceleration (a)
in unit-vector
notation
and
as law
(b)(specifically,
a
3. We apply
Newton’s
second
Eq. 5-2).
Alex
magnitude and (c) a direction relative
(a) We find the x component of the force is
Charles
••11 A 2.0 kg particle move
variable force directed along
3.0 m # (4.0 m/s)t # ct 2 " (2
seconds. The factor c is a cons
ticle has a magnitude of 36 N
axis. What is c?
•••12
Two horizontal for
slides over frictionless ice,
which an xy coordinate syst
:
is laid out. Force F1 is in
positive direction of the x a
and has a magnitude of 7.0
:
Force F2 has a magnitude of
N. Figure 5-32 gives the x co
ponent vx of the velocity of
disk as a function of time t d
One force is F1 ! (3.0 N)i # (4.0 N)j. Find the acceleration of the
ˆi
ˆj
3.0N ˆi notation
4.0N ˆj when3.0N
4.0N
of2 the
F1 F2 in unit-vector
chopping block
the other
force is
(a)
ˆj. x axis) of the net force o
(4.0m/s
)
:
:
gle of the particle’s direction o
F2 ! ("3.0 N)î
("3.0 N)î # (4.0 N)ĵ, and
m # ("4.0 N)ĵ, (b) F2 !
2.0kg
:
(c) F2 ! (3.0 N)î # ("4.0 N)ĵ .
••10 A 0.150 kg particle m
to x(t) ! "13.00 # 2.00t # 4.0
is has an acceleration of 2.00 m/s2 at
In the
this final
•3(c) If
1 kg situation,
standard abody
seconds. In unit-vector notatio
20.0° to the positive direction of an x axis, what are (a) the x comˆi
ˆi
particle at t ! 3.40 s?
4.0Nofˆj the net
3.0N
4.0Nonˆj the
ponent and
y component
force
acting
F1 (b)
F2 the 3.0N
2 ˆ
(3.0 m/s
)i. A 2.0 kg particle moves
••11
body, and (c)mwhat is the net force in unit-vector
notation?
2.0 kg
variable force directed along
••4 While two forces act on it, a pary
3.0 m # (4.0 m/s)t # ct 2 " (2.
3. We
apply
Newton’s
second law
(specifically, Eq. 5-2).
ticle
is to
move
at the constant
velocseconds. The factor c is a const
ity :
v ! (3 m/s)î " (4 m/s)ĵ. One of
: x component of the force is
ticle has a magnitude of 36 N
(a)forces
We find
the
is the
F1 !
(2 N)î # ("6 N)ĵ.
axis. What is c?
What is the other force?
F2 1
F
ma
ma
cos
20.0
1.00kg
2.00m/s
cos 20.0 1.88N.
•••12
x
x
172
CHAPTER
5 Two horizontal forc
θ1
••5
Three astronauts, propelled
x
slides over frictionless ice, o
θ3
by jet backpacks, push and guide a
F2
which an xy coordinate syste
(b)kg
Theasteroid
y component
of the
force is
:
120
toward
a processing
is laid out. Force F1 is in t
F3
dock, exerting the forces shown in
2
171
Fy F ma
ma
sin
20.0
1.0kg 2.00m/s sin 20.0 0.684N.positive direction of the x ax
y
Fig. 5-29, with
!
32
N,
F
!
55
N,
1
2
and has a magnitude of 7.0
:
F3 ! 41 N, u1 ! 30°, and u3 ! 60°.
Fig. 5-29 Problem 5.
Force F2 has a magnitude of 9
(c) Inisunit-vector
notation,
the force(a)
vector is
What
the asteroid’s
acceleration
N. Figure 5-32 gives the x com
in unit-vector notation and as (b) a
ponent vx of the velocity of t
ˆ F ˆj (1.88 N)i
ˆ
ˆ
F Frelative
Alex (0.684 N) j .
magnitude and (c) a direction
xi
y
disk as a function of time t du
Charles
to the positive direction of the x axis?
ing the sliding. What is the a
4. Since v = constant, we have a = 0, which implies
gle between the constant dire
••6 In a two-dimensional tug-of:
:
tions of forces F1 and F2?
war, Alex, Betty, and Charles Fpull F F ma 0 .
net
1
2
horizontally on an automobile tire
sec. 5-7 Some Particular F
atThus,
the angles
shown
in
the
overhead
the other force must be
•13 Figure 5-33 shows an arr
view of Fig. 5-30. The tire remains
20.0° to the positive direction of an x axis, what are (a) the x com172 (b) the y component of the net force acting on the
ponent and
body, and (c) what is the net force in unit-vector notation?
(b) The
component
ofit,
thea force
••4 While
twoyforces
act on
par- is
y
ticle is to move at the constant velocFy ma y ma sin 20.0
1.0kg 2.00m/s 2 sin 20.0
:
ity v ! (3 m/s)î " (4 m/s)ĵ. One of
:
the forces is F1 ! (2 N)î # ("6 N)ĵ.
(c) In unit-vector notation, the force vector is
What is the other force?
F
1
particle
at t5! 3.4
CHAPTER
••11 A 2.0 kg p
variable force di
3.0 m # (4.0 m/s
seconds. The fact
0.684N.
ticle has a magni
axis. What is c?
•••12
Two h
slides over frict
which an xy coo
is laid out. Forc
positive directio
and has a magn
:
Force F2 has a m
N. Figure 5-32 g
ponent vx of the
disk as a functio
the vector
5. The net force applied on the chopping block is Fnet F1 F2 F3 , where
ing the
sliding. W
is done using unit-vector
given bythe
gleisbetween
••6 In addition
a two-dimensional
tug-of- notation. The acceleration of the block
:
a Betty,
F1 F2andF3Charles
/ m.
tions of forces F1
war, Alex,
pull
horizontally on an automobile tire
sec. 5-7 Some
at the angles
shown
the overhead
(a) The
forcesin
exerted
by the three astronauts can be expressed in unit-vector notation as
•13 Figure 5-33
view of follows:
Fig. 5-30. The tire remains
Betty
137°
pended by cords
stationary in spite of the three pulls.
θ1 N)ˆj .
F Fx ˆi Fy ˆj (1.88 N)iˆ (0.684
••5
Three astronauts, propelled
x
θ3
by jet backpacks, push and guide a
F2
= constant,
we have a = 0, which implies
4. Since vtoward
120 kg asteroid
a processing
F3
dock, exerting the forces shown in
Fig. 5-29, with F1 ! 32 N, F2 ! 55 N,Fnet F1 F2 ma 0 .
F3 ! 41 N, u1 ! 30°, and u3 ! 60°.
Fig. 5-29 Problem 5.
Thus, the other force must be
What is the asteroid’s acceleration (a)
in unit-vector notation and as (b) a
F2
F1 ( 2 N) ˆi ( 6 N) ˆj .
Alex
magnitude and (c) a direction relative
Charles
to the positive direction of the x axis?
d
i
en that occuis 8.00 m/s2
heir humanom a cannon
ersion of the
els to land in
non and at a
or 5.2 m and
e underwent
magnitude of
ough it were
onnected by
by the cable
With what speed does the man hit the ground if he
started from rest?
••53 In Fig. 5-48, three connected blocks are pulled to the right on
a horizontal frictionless table by a force of magnitude T3 ! 65.0 N.
If m1 ! 12.0 kg, m2 ! 24.0 kg, and m3 ! 31.0 kg, calculate (a) the
magnitude of the system’s acceleration, (b) the tension T1, and (c)
the tension T2.
m1
T1
m2
Fig. 5-48
'((")*(+,$*-."/%0%"!!
T2
m3
Problem 53.
T3
v
2(1.3 m/s 2 )(10 m)
2a y
5.1 m/s.
53. We apply Newton’s second law first to the three blocks as a single system and then to
the individual blocks. The +x direction is to the right in Fig. 5-48.
(a) With msys = m1 + m2 + m3 = 67.0 kg, we apply Eq. 5-2 to the x motion of the system,
in which case, there is only one force T3
T3 i . Therefore,
T3
msys a
65.0 N
(67.0 kg)a
which yields a = 0.970 m/s2 for the system (and for each of the blocks individually).
(b) Applying Eq. 5-2 to block 1, we find
T1
m1a
12.0kg 0.970m/s 2
11.6N.
(c) In order to find T2, we can either analyze the forces on block 3 or we can treat blocks
1 and 2 as a system and examine its forces. We choose the latter.
T2
m1 m2 a
12.0 kg 24.0 kg 0.970 m/s 2
34.9 N .
54. First, we consider all the penguins (1 through 4, counting left to right) as one system,
to which we apply Newton’s second law:
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