Chapter 5 - Chapter
5
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Unformatted text preview: Chapter
5
 Addi-onal
Applica-ons
of
Newton’s
 Laws
 October
1
and
October
6,
2009
 Fric-on
   Fric-on
   Opposes
mo-on
between
systems
in
contact
   Parallel
to
the
contact
surface
   Depends
on
the
force
holding
the
surfaces
 together
(the
normal
force
N)
   Sta-c
fric-on
   Force
required
to
move
a
sta-onary
object
  fs
is
less
than
or
equal
to
µs
N
   Kine-c
fric-on
   Fric-onal
force
on
an
object
in
mo-on
  Can
be
less
than
sta-c
fric-on
 f 
=
µk
N
   k Proper-es
of
Surface
Fric-on 
 These relations are all useful APPROXIMATIONS to messy reality. Newton’s
laws:
example 
 T=50 N µk = 0.2 M = 5 kg Find acceleration of block Newton’s
laws:
example 
 T=50 N µk = 0.2 M = 5 kg Find acceleration of block Frictional force: f = µkMg opposes the tension T Net force: Fnet = T - f = T - µkMg Acceleration: a = Fnet / M = ((50 - 0.2 x 5 x 9.8) / 5) m/s2 Answer: 8.04 m/s2 Newton’s
laws:
another
example 
 T=50 N θ=500 µk = 0.2 M = 5 kg Find acceleration of block Newton’s
laws:
another
example 
 T=50 N θ=500 µk = 0.2 M = 5 kg Find acceleration of block Hints: Resolve T in x and y components: Tx=T cosθ, Ty=T sinθ Draw free body diagram Solve for y-component of force, and note that y-acceleration is zero (Obtain relationship between T and N) Solve for x-component of force, then use ax=Fx/m Fnet Tx − µk N Tx − µk ( Mg − Ty ) T cos θ − µk ( Mg − T sin θ) a= = = = M M M M Answer: acceleration of block is 6.0 m/s2 in +x direction Rolling
Fric-on:
Car
Tires
   Fric-on
keeps
the
car
wheels
from
spinning
in
place
   You
want
the
-res
to
roll
   You
want
the
fric-on
to
be
high
   The
contact
point
is
at
rest
‐
although
the
car
is
in
mo-on
 »  What
maXers
is
the
coefficient
of
sta-c
fric-on!
 Consider Newton’s 3rd law: Froad on car Fcar on road Froad on car is the actual force ON the car. Static Friction µsN is its maximum value weight Maximum Static friction ( > Fcar on road for car not to spin in place!) Surface
Fric-on… 
   Fric-on
is
caused
by
the
“microscopic”
interac-ons
between
the
two
surfaces:
 Ques-on
1
 You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about a) four times as great b) twice as great c) equally as great e) one-fourth as great as the force required before you changed the crate orientation. Ques-on
1
 You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about a) four times as great b) twice as great c) equally as great e) one-fourth as great as the force required before you changed the crate orientation. f s ≤ µs N f k = µk N Frictional force does not depend on the area of contact. It depends only on the normal force and the coefficient of friction for the contact. Surface
fric-on… 
 friction force Inclined
Plane
with
Fric-on
(1) 
 What is the acceleration of the block? θ Inclined
Plane
with
Fric-on
(2) 
 Draw free-body diagram for block: ma µKN N θ mg θ Inclined
Plane
with
Fric-on
(3) 
 Consider i and j components of FNET = ma i component : mg sinθ - µKN = ma j component : N = mg cosθ µKN ma N θ € mg sin θ − µKmg cos θ = ma θ mg mg sin θ mg cos θ € a = sin θ − µK cos θ g Sta-c
fric-on 
 •  Previous
problems
considered
fric-on
ac-ng
when
 the
two
surfaces
move
rela-ve
to
each
other
–
ie.,
 when
the
slide.
 –  We
also
know
that
fric-on
acts
when
the
surfaces
are
 stuck
together:
the
“sta-c
case.”
 •  When
surfaces
do
not
move
rela-ve
to
each
other,
 the
fric-on
force
depends
on
the
OTHER
forces
on
 the
parts
of
the
system.
 fF Sta-c
Fric-on… 
 (with
one
surface
fixed) 
   The
maximum
possible
force
that
the
fric-on
between
two
 objects
can
provide
is
fMAX
=
µSN,
where
µs
is
the
“coefficient
of
 sta-c
fric-on.”
   So
fF
≤

µS
N.
   As
one
increases
F,
fF
gets
bigger
un-l
fF
=
µSN
and
the
object
 starts
to
move.
 fF Sta-c
Fric-on… 
   Just
like
in
the
sliding
case
except
a
=
0.
 i
:
 
 
F
- fF
=
0
 j
:
 
 
N
=
mg
   While the block is static: fF = F fF Sta-c
Fric-on… 
     The
“coefficient
of
sta8c
fric8on,”
µS,
determines
maximum
sta-c
fric-onal
 force,
µSN,
that
the
contact
between
the
objects
can
provide.

 µS
is
discovered
by
increasing
F
un-l
the
object
starts
to
slide: 
 
FMAX
‐
µSN
=
0
 
 
N
=
mg
 

 
 

 

 
FMAX
=
µS
mg

























µS
= FMAX
/
mg
 µS N Sta-c
Fric-on… 
   We
can
also
consider
µS
on
an
inclined
plane.
 θ   In
this
case,
the
force
provided
by
fric-on
will
depend
on
the
angle
θ
of
the
 plane.
 Sta-c
Fric-on… 
   The
force
provided
by
fric-on,
fF
,
depends
on
θ.
 fF ma = 0 (block is not moving) mg sin θ - ff = 0 N θ (Newton’s 2nd Law along x-axis) mg θ Sta-c
Fric-on… 
   We
can
find
µs
by
increasing
the
ramp
angle
un-l
the
block
slides:
 mg sin θ - ff = 0 In this case: ff = µSN = µSmg cos θM µSN mg sin θM - µSmg cos θM = 0 N θM mg θ µS = tan θM Problem:

Sta-c
Fric-on 
 A
box
of
mass
m
=10.21
kg
is
at
rest
on
a
floor.

The
coefficient
of
 sta-c
fric-on
between
the
floor
and
the
box
is
µs
=
0.4.
 A
rope
is
aXached
to
the
box
and
pulled
at
an
angle
of
θ
=
30o
above
 horizontal
with
tension
T
=
40
N.
   Does
the
box
move?
     
 
(A)

yes




(B)
no


(C)
too
close
to
call
 T static friction (µs = 0.4 ) m θ Problem:

Sta-c
Fric-on
‐‐
Solu-on 
 y m =10.21 kg, µs = 0.4, θ = 30o, T = 40 N     x Pick
axes
&
draw

FBD
of
box:
 Apply FNET = ma y: N + T sin θ - mg = maY = 0 N = mg - T sin θ N = 80 N T x: T cos θ - fFR = maX The box will move if T cos θ - fFR > 0 fFR θ m mg Problem:

Sta-c
Fric-on
‐‐
Solu-on 
 y m =10.21 kg, µs = 0.4, θ = 30o, T = 40 N x y: N = 80 N x: T cos θ - fFR = maX The box will move if T cos θ - fFR > 0 N T cos θ = 34.6 N fMAX = µsN = (.4)(80N) = 32 N T fMAX = µsN So T cos θ > fMAX and the box does move θ m mg An-lock
Brakes 
   An-‐lock
brakes
work
by
making
sure
the
wheels
roll
without
 slipping.

This
maximizes
the
fric-onal
force
slowing
the
car
since
 µS

>

µK
.
   The
driver
of
a
car
moving
with
speed
vo
slams
on
the
brakes.

 The
coefficient
of
sta-c
fric-on
between
the
wheels
and
the
road
 is
µS=0.5
and
the
coefficient
of
kine-c
fric-on
between
the
car
 and
the
road
is
µK=0.4.

What
is
the
stopping
distance
D
for
(a)
 an-lock
brakes
and
(b)
locked
wheels?
 ab vo v=0 D An-lock
Brakes 
 If wheels are turning, the tire surface is stationary with respect to the road, and the friction is static friction. So magnitude of maximum friction force here is µSN=0.5mg. If wheels are sliding, the friction is kinetic friction, with magnitude µKN=0.4mg. Find stopping distance from v2 = v2 x + 2a x D x 0 v2 x When vx = 0, D = - 0 2a x vo ab v=0 € D Since |ax| = µSg, the ratio of the stopping distances is µS/µK Drag
Forces 
 •  Objects
moving
through
a
fluid
such
as
air
or
 water
experience
a
drag
force
that
opposes
 the
mo-on
of
the
object.

The
magnitude
of
 the
drag
force
increases
as
the
speed
 increases
(unlike
the
kine-c
fric-on
force!).

 Empirically
it
is
typically
found
that
 where b is a constant and n is typically 1 or 2. Terminal
speed
of
falling
object 
 •  The
terminal
speed
vT
is
the
 speed
at
which
the
drag
force
 bvn
exactly
balances
the
force
of
 gravity
mg.
 € € bv n = mg T mg 1/ n vT = b Ques-on
–
sky
diver 
 A
sky
diver
jumps
out
of
an
airplane
at
5000m
al-tude.
She
reaches
terminal
 velocity
(due
to
the
drag
of
the
air)
aner
about
six
seconds.
 If
a
box
of
steel
parts
that
has
the
same
weight
as
the
diver
is
dropped
 simultaneously,
the
box
will
fall:
 






faster
than
the
diver
 






slower
than
the
diver
 






the
same
as
the
diver
 Ques-on
–
sky
diver 
 A
sky
diver
jumps
out
of
an
airplane
at
5000m
al-tude.
She
reaches
terminal
 velocity
(due
to
the
drag
of
the
air)
aner
about
six
seconds.
 If
a
box
of
steel
parts
that
has
the
same
weight
as
the
diver
is
dropped
 simultaneously,
the
box
will
fall:
 






faster
than
the
diver
 correct 






slower
than
the
diver
 






the
same
as
the
diver
 The force of gravity is proportional to the mass of an object, and the drag coefficient is proportional to the cross-section of the object. The box of steel is much more dense than a human body (whose density is about that of water), so the gravitational force on it is much larger compared to the drag force than for the person. (In other words, the terminal velocity vT=(mg/b)1/n, which is larger for the box of steel than it is for the person.) Mo-on
along
a
curved
path 
 •  Centripetal
accelera-on
is
the
accelera-on
 perpendicular
to
the
velocity
that
occurs
when
 a
par-cle
is
moving
on
a
curved
path.
 •  The
force
associated
with
this
centripetal
 accelera-on
is
directed
towards
the
center
of
 the
circle
defined
by
the
radius
of
curvature,
 and
has
magnitude
v2/R.
 Centripetal
Accelera-on
 
 Derivation of a=v2/R: Centripetal
Accelera-on
 
 Derivation of a=v2/R: Centripetal
Force 
       The
force
causing
the
centripetal
accelera-on
is
 some-mes
called
the
centripetal
force
 This
is
not
a
new
force,
it
is
a
new
role
for
a
force
 It
is
a
force
ac8ng
in
the
role
of
a
force
that
causes
 a
circular
mo8on
 Uniform
Circular
Mo-on 
 •  If
an
object
is
moving
along
a
curved
path,
it
is
 accelera-ng.
 •  Newton’s
second
law
⇒
a
force
must
be
ac-ng 
 on
the
object.
 •  If
object
is
moving
with
constant
speed
on
the
 circle,
 Uniform
Circular
Mo-on 
       A
force
causing
a
 centripetal
 accelera-on
acts
 toward
the
center
of
 the
circle
 It
causes
a
change
in
 the
direc-on
of
the
 velocity
vector
 If
the
force
vanishes,
 the
object
would
 move
in
a
straight‐line
 path
tangent
to
the
 circle
 Mo-on
in
a
Horizontal
Circle 
     The
speed
at
which
the
object
moves
depends
on
 the
mass
of
the
object
and
the
tension
in
the
cord
 The
centripetal
force
is
supplied
by
the
tension
 2 mv T= ⇒ r Ques-on 
 An
object
of
mass
m
is
suspended
from
a
point
in
the
 ceiling
on
a
string
of
length
L.

The
object
revolves
 with
constant
speed
v
in
a
horizontal
circle
of
radius
r.
 
 (The
string
makes
an
angle
θ
with
the
ver-cal).
 The
speed
v
is
given
by
the
expression:
 θ L Ques-on 
 An
object
of
mass
m
is
suspended
from
a
point
in
the
 ceiling
on
a
string
of
length
L.

The
object
revolves
 with
constant
speed
v
in
a
horizontal
circle
of
radius
r.
 
 (The
string
makes
an
angle
θ
with
the
ver-cal).
 θ L The
speed
v
is
given
by
the
expression:
 y correct Fnet = ma v2 x : T sin θ = m r y : T cos θ − mg = 0 v2 ⇒ tan θ = ⇒ v = rg tan θ rg Horizontal
(Flat)
Curve 
     The
force
of
sta-c
fric-on
 supplies
the
centripetal
 force
 The
maximum
speed
at
 which
the
car
can
 nego-ate
the
curve
is
 v2 fs = m r f max s 2 vmax =m = µ s mg r Note, this does not depend on the mass of the car Banked
Curve 
     These
are
designed
to
be
 navigable
when
there
is
 no
fric-on
 There
is
a
component
of
 the
normal
force
that
 supplies
the
centripetal
 force
 n cosθ = mg v2 n sin θ = m r Non‐Uniform
Circular
Mo-on 
         The
accelera-on
and
 force
have
tangen-al
 components
 Fr
produces
the
 centripetal
accelera-on
 Ft
produces
the
tangen-al
 accelera-on
 ΣF
=
ΣFr
+
ΣFt Ver-cal
Circle
With
Non‐Uniform
Speed 
   The
gravita-onal
force
 exerts
a
tangen-al
force
 on
the
object
     Look
at
the
components
of
 Fg
 The
tension
at
any
point
 can
be
found
 Top
and
BoXom
of
Circle 
       The
tension
at
the
boXom
 is
a
maximum
 The
tension
at
the
top
is
a
 minimum
 If
Ttop
=
0,
then
 The
Center
of
Mass
 •  Defini-on
of
center
of
mass:
 Where For a continuous object (e.g., a solid sphere) Center
of
Mass
(2) 
 •  Newton’s
Laws
for
a
collec-on
of
objects:
 The acceleration of the center of mass is determined entirely by the net force on the object. A
horizontal
force
is
used
to
push
an
object
of
mass
m
 up
an
inclined
plane.

The
angle
between
the
plane
and
 the
horizontal
is
θ.

The
normal
reac-on
force
of
the
 plane
ac-ng
on
the
mass
m
is

 A. 
mg
cos
θ
+
F
cos
θ.
 B. 
mg
cos
θ.

 C. 
mg
cos
θ
+
F
sin
θ.

 E. impossible
to
determine
because
the
 coefficient
of
fric-on
is
not
given.

 A
horizontal
force

is
used
to
push
an
object
of
mass
m
 up
an
inclined
plane.

The
angle
between
the
plane
and
 the
horizontal
is
θ.

The
normal
reac-on
force
of
the
 plane
ac-ng
on
the
mass
m
is

 A
car
going
around
a
curve
of
radius
R
at
a
speed
V
 experiences
a
centripetal
accelera-on
ac.

What
is
its accelera-on
if
it
goes
around
a
curve
of
radius
3R
at
a
 speed
of
2V?

 A. (2/3)ac





 B. (4/3)ac





 C. (2/9)ac





 





 E. (3/2)ac

 A
car
going
around
a
curve
of
radius
R
at
a
speed
V
 experiences
a
centripetal
accelera-on
ac.

What
is
its
 accelera-on
if
it
goes
around
a
curve
of
radius
3R
at
a
 speed
of
2V?

 The
figure
shows
a
top
view
of
a
ball
on
the
end
of
a
 string
traveling
counterclockwise
in
a
circular
path.
The
 speed
of
the
ball
is
constant.
If
the
string
should
break
 at
the
instant
shown,
the
path
that
the
ball
would
 follow
is 

 A. 1





 B. 2





 C. 3





 





 E. impossible
to
tell
from
the
given
informa-on.

 The
figure
shows
a
top
view
of
a
ball
on
the
end
of
a
 string
traveling
counterclockwise
in
a
circular
path.
The
 speed
of
the
ball
is
constant.
If
the
string
should
break
 at
the
instant
shown,
the
path
that
the
ball
would
 follow
is 

 ...
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This note was uploaded on 09/01/2011 for the course PHY 303 taught by Professor Erskine/tsoi during the Spring '08 term at University of Texas at Austin.

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