Transcript document

Lecture 3
Chapter 6
Force and Motion
II
Friction
More Circular motion
Apparent Weight
Drag Forces
Numerical integration
Derivatives/Integration
(Lecture 1)
Tension between Blocks
(Lecture 2)
Misconceptions
Friction
You are standing still, then begin to walk.
What was the external forced that caused
you to accelerate?
Hint: It is very hard to start walking if you
are standing on ice.
What force causes a car to accelerate when
a traffic light turns green?
Friction
Fixed block
N
mg
F
fs
Static Friction: Push with a force F and block does not move because
fs = F. The force of friction varies from 0 up to some maximum. The
maximum value equals fs = msN, where N is the normal force. Above we
would have fs = msmg. The coefficient of static friction ranges from 0 to 1.2
Kinetic Friction: If we increase F until the block starts to move, the friction
force decreases to fk= mkN and remains constant throughout the motion.
Frictional Forces
Friction is an attractive force between two surfaces that is a
result of the vector sum of many electrical forces between
the surface atoms of the two different bodies.
Only about 10-4 of the surface atoms actually contribute.
Model of dry friction 3 or 4 asperites support
top block. Temporarily weld together
The Friction and Lubrication of Solids
F. P. Bowden and D. Tabor, Oxford University Press
1964
Models of friction
See Chabay and Sherwood
Matter and Ineractions
Volume 1
ISBN 0-471-35491-0
Problem Solving with Newton’s 2nd Law
involving friction
• Vector sum of external forces in x direction
= max
• Vector sum of external forces in y direction
= may
• If no acceleration, then set sum equal to 0
Mass on Inclined plane
Question: How high can we tilt
the book before the coin slides
down the block?
Apply Newton's 2nd law in the x and
y directions and using fs = ms N
x direction
fs  mgsin  0
ms N  mgsin  0
y direction
N  mg cos  0
N  mg cos
ms 
mg sin 
h
 tan  
mg cos
d
Mass on Inclined plane
Question: How high can we tilt
the book before the coin slides
down the block?
Apply Newton's 2nd law in the x
and y directions and using
fs = ms N
Take x direction along the plane
fs  mgsin   0
ms N  mgsin  0
F
Mass On Inclined Plane
x direction
fk  mgsin  F  0
mk N  mgsin  F  0
y direction
F
N  mg cos  0
N  mg cos 
F  mg sin   m k mg cos 
Free Body Diagram of an accelerating
system: Atwood’s machine with friction.
Find a and T.
N
f
T
Frictionless pulley
Mg
T
fk
T
-----
T
T
mg
N  Mg
fk  m k N
T  f  Ma
T  Ma  m k Mg
T  mg  ma
Ma  m k Mg  mg  ma
mg mk Mg
a
Mm
T  (1 mk )
-y

M
mg
Mm
Problem 13 Chapter 6 edition 6 and 7
Question: What is the minimum magnitude force required to start the crate moving?
+y
m  68kg
f  15 deg rees
m s  0.5
N
f
Tcos f
fs
mg
x components
T cos f  fs  0
T
+x
y components
T sin f  N  mg  0
N  mg  T sin f
T cos f  m s N  m s (mg  T sin f ), Solve for T
m s mg
(0.5)(68)(9.8)
T

 304N
(cos f  m s sin f ) cos(15)  0.5 sin(15)
Problem 13 continued
Question: What is the initial acceleration if mk0.35?
m  68kg
f  15 deg rees
m s  0.35
+y
N
f
Tcos f
fs
T
+x
mg
Newton's 2nd law
x components
T cos f  fk  ma
y components
T sin f  N  mg  0
fk  m k N  m k (mg  T sin f )
ma  T cos f  m k (mg  T sin f )
304 cos(15)  0.35sin(15)
T (cos f  m k sin f )
a
 mk g 
 (0.35)(9.8)
m
68
a  1.3m / s 2
Inertial Drag Force and Terminal Velocity
Drag force: Whenever you have a body like a ball
moving through a medium that behaves like a fluid,
there will be a drag force opposing the motion.
D
1
C  Av 2
2
Imagine a falling ball slowed down due to elastic collisions
with air molecules. Simply pushing the air out of the way.
Hand waving argument
dp d(mv)
dv
dm

m v
dt
dt
dt
dt
dm   Ady
dm
dy
  A   Av
dt
dt
F  v Av   Av 2
F
ball
v
.
A
dy
air molecules
Inertial drag
Terminal speeds in air
Using Newton’s 2nd law,
1
ma  mg  C  Av 2
2
where m is the mass of the falling ball
1
mg  C  Av0 2  0
2
Solve for v0
2mg
v0 
CA
Stokes-Napier Law
TERMINAL SPEEDS IN AIR
Object `
Feather
Snowflake
BB
Mouse
Tennis ball
Baseball
Sky diver
Cannonball
Speed (m/s) Speed (mph)
0.4
0.9
1
9
13
31
42
60 -120
250
Show demo of falling feather in vacuum
2.2
20
29
66
86
134 -268
560
How to solve this equation?
Two ways
1
F  Mg  C  Av 2
2
F  Mg  bv 2
One way is to use a spread sheet in Exel.
Use Newtons 2nd Law
Initial component of momentum:
Initial force on ball:
Using 2nd Law
Find new p
p1  Mv1
F  Mg  bv12
dp p p2  p1
F


dt t
t
p2  p1  Ft
p2  p1  Ft
Substitute in for p1 and F
Newtons 2nd Law
p2  p1  Ft
p2  p1  (Mg  bv12 )t
p2
b 2
v2 
 v1  [g  ( )v1 ]t
M
M
v1  v2
x2  x1  v2 t  x1  (
)t
2
Go to Excel Spread Sheet
delta_t=
g=
m_1=
b_1=
v_init=
0.05
10
0.75
0.25
0
time
gravity
mass
drag coefficient
initial downward velocity
velocity
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0
0.5
0.995833
1.479305
1.942833
2.379923
2.785522
3.156203
3.490176
3.787154
4.048112
distance
0
0.0125
0.049896
0.111774
0.197328
0.305397
0.434533
0.583076
0.749235
0.931169
1.12705
b _1


vi  vi 1   g  (
)vi 12  t
m _1


vi 1  vi
xi  xi 1  (
)t
2
631 Website: Lecture 3 Materials
=C16+(g-(b_1/m_1)*C16*C16)*delta_t
=D16+1/2*(C16+C17)*delta_t
Velocity vs. Time
10
9
Velocity (m/s)
8
7
6
Drag
5
No Drag
4
reduce mass by 4
3
2
1
0
0
0.5
1
Time (s)
1.5
2
Displacement vrs Time
Displacement (m)
30
Displacement w/ drag
20
Displacement w/out
drag
mass reduced by 4
10
0
0
0.5
1
Time (s)
1.5
2
We can also solve the equation to get the velocity as a function of
time before it reaches terminal velocity. Let b = 1/2CA
dv
m
 mg  bv 2
dt
dv
m /b
 mg/b  v 2
dt
let v 0 =
mg
b
dv
2
m /b
 v0  v 2
dt
1
dv  b /mdt
2
2
v0  v
Solving equation continued
1
dv  b /mdt
2
2
v0  v
1
1
1
1
( 0
 0
)
2
2
v  v v  v 2v 0
v0  v
dv
dv
2v 0b


dt
v0  v v0  v
m
v

0
dv

v0  v
v

0
dv
2v 0b

v0  v
m
t
 dt
0

v

0
dv

v0  v
v

dv
2v 0b

v0  v
m
t
 dt
0
This can be integrated, and fixing the constant of
integration by the requirement that the velocity be zero at t
= 0 , which is the case for free fall we find:
2v b

 0 t 
1 e m 
v  v 0 
2v 0 b 

t 

m
1 e

Now show comparison of this solution with numerical integration
with Excel.
Comparison
True Terminal Velocity Function
60
Velocity (m/s)
50
40
TRUE
Veloc ity Squared
30
20
10
0
0
5
10
15
20
Time (s)
The curve modeled by velocity squared for terminal velocity
Differs from the true equation due to a large delta t.
When delta t becomes small enough the two curves are
Indistinguishable.
Water Resistance and Drag Forces
Whenever you have a body moving through a liquid
there will be a drag force opposing the motion. Here
the drag force is proportional to - kv. Viscous drag.
Stokes Law: terminal velocity is proportional to mass
A 1000 km boat in the water shuts off its engine
at 90 km/hr. Find the time required to
slow down to 45 km/hr due to a water
drag force equal to -70v, where v is
the speed of the boat. Let k = 70.
dv
  kv
dt
mdv   kvdt
m
dv
k
  dt
v
m
v
dv
k t
 v   m  dt
v0
0
ln v v  ln v  ln v0
v
ma = - kv
v/v0 = 45/90 =1/2
t = m/k ln 2=1000/70 ln 2 = 9.9 s
0
v
k
ln   t
v0
m
v
 mk t
e
v0
Uniform circular motion
T  Ma
2
v
TM
R
VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
ac
ac
mg
N
N
At the top:
mv 2
N  mg  m(ac )  
r
mv 2
N
 mg
r
Minimum v for N = 0:
(apparent weightlessness)
2
mg
mv

 mg
r
v  rg
VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
ac
ac
mg
N
At the bottom:
N
mv
N  Mg  mac 
r
2
2
Apparent weight = N = + mv /r + mg
mg
What do we mean by Fictitious Forces
Ff = - ma
(the fictitious force always acts in the opposite
direction of acceleration)
T
ma
T
mg
T

mg
ma
ma
tan  
mg
a  g tan 
Example of fictitious force (Ff = - ma)
In a vertically accelerated reference frame, eg. an elevator,
what is your apparent weight? Apparent weight = N
Upwards is positive
Downwards is negative
Upward acceleration
N  mg  ma
N  mg  ma
Downward acceleration
N  mg  ma
N  mg  ma
Free fall a = g
N = 0 Weightless condition
Tug-of-war demo illustrates how a small sideways
force can produce a large horizontal force
Problem 13-10
Suppose two guys in the tug of war are at 4.5 meters
apart and I pull the rope out 0.15 meters. Then f4degrees
T
0.15
tan f 
2.25
f  4o
T
2.25 m
T
f
F
2
F
sin f 
2T
F
T
2 sin f
If F  100lbs thenT  700lbs
Therefore,
F
F
T

 7F
o
2sin(4 ) (2)(0.07)
The smaller the angle the larger
the magnification
ConcepTest 4.6 Force and Two Masses
A force F acts on mass m1 giving acceleration a1. The same
force acts on a different mass m2 giving acceleration a2 =
2a1. If m1 and m2 are glued together and the same force F
acts on this combination, what is the resulting acceleration?
1) 3/4 a1
2) 3/2 a1
3) 1/2 a1
4) 4/3 a1
5) 2/3 a1
F
F
F
m1
a1
m2
m2 m1
a2 = 2a1
a
3
ConcepTest 4.6 Force and Two Masses
A force F acts on mass m1 giving acceleration a1. The same
force acts on a different mass m2 giving acceleration a2 =
2a1. If m1 and m2 are glued together and the same force F
acts on this combination, what is the resulting acceleration?
1) 3/4 a1
2) 3/2 a1
3) 1/2 a1
4) 4/3 a1
5) 2/3 a1
F
m1
a1
F = m1 a1
a2 = 2a1
F
m2
F = m2 a2 = (1/2 m1 )(2a1 )
Mass m2 must be (1/2)m1 because its
acceleration was 2a1 with the same
force. Adding the two masses
together gives (3/2)m1, leading to an
F
m2 m1
acceleration of (2/3)a1 for the same
a
3
F = (3/2)m1 a3 => a3 = (2/3) a1
applied force.
ConcepTest 4.7 Climbing the Rope
When you climb up a rope, the first
1) this slows your initial velocity which
is already upward
thing you do is pull down on the
2) you don’t go up, you’re too heavy
rope. How do you manage to go
3) you’re not really pulling down – it
just seems that way
up the rope by doing that??
4) the rope actually pulls you up
5) you are pulling the ceiling down
ConcepTest 4.7 Climbing the Rope
When you climb up a rope, the first
1) this slows your initial velocity which
is already upward
thing you do is pull down on the
2) you don’t go up, you’re too heavy
rope. How do you manage to go
3) you’re not really pulling down – it
just seems that way
up the rope by doing that??
4) the rope actually pulls you up
5) you are pulling the ceiling down
When you pull down on the rope, the rope pulls up on
you!! It is actually this upward force by the rope that
makes you move up! This is the “reaction” force (by the
rope on you) to the force that you exerted on the rope.
And voilá, this is Newton’s 3rd Law.
ConcepTest 4.8a Bowling vs. Ping-Pong I
In outer space, a bowling ball
and a ping-pong ball attract
each other due to gravitational
forces. How do the
magnitudes of these attractive
forces compare?
1) The bowling ball exerts a greater
force on the ping-pong ball
2) The ping-pong ball exerts a greater
force on the bowling ball
3) The forces are equal
4) The forces are zero because they
cancel out
5) There are actually no forces at all
F12
F21
ConcepTest 4.8a Bowling vs. Ping-Pong I
In outer space, a bowling ball and
a ping-pong ball attract each other
due to gravitational forces. How
do the magnitudes of these
1) The bowling ball exerts a greater
force on the ping-pong ball
2) The ping-pong ball exerts a greater
force on the bowling ball
3) The forces are equal
attractive forces compare?
4) The forces are zero because they
cancel out
5) There are actually no forces at all
The forces are equal and
opposite by Newton’s 3rd
Law!
F12
F21
ConcepTest 4.10a Contact Force I
If you push with force F on either the
heavy box (m1) or the light box (m2),
in which of the two cases below is
the contact force between the two
boxes larger?
1) case A
2) case B
3) same in both cases
A
m2
F
m1
B
m2
m1
F
ConcepTest 4.10a Contact Force I
If you push with force F on either the
heavy box (m1) or the light box (m2),
in which of the two cases below is
the contact force between the two
boxes larger?
1) case A
2) case B
3) same in both cases
The acceleration of both masses together
A
is the same in either case. But the contact
force is the only force that accelerates m1
m2
F
m1
in case A (or m2 in case B). Since m1 is the
B
larger mass, it requires the larger contact
force to achieve the same acceleration.
Follow-up: What is the accel. of each mass?
m2
m1
F
ConcepTest 5.3b Tension II
Two tug-of-war opponents each pull
with a force of 100 N on opposite
ends of a rope. What is the tension
in the rope?
1) 0 N
2) 50 N
3) 100 N
4) 150 N
5) 200 N
ConcepTest 5.3b Tension II
Two tug-of-war opponents each pull
1) 0 N
with a force of 100 N on opposite
2) 50 N
ends of a rope. What is the tension
3) 100 N
in the rope?
4) 150 N
5) 200 N
This is literally the identical situation to the
previous question. The tension is not 200 N !!
Whether the other end of the rope is pulled by a
person, or pulled by a tree, the tension in the rope
is still 100 N !!
ConcepTest 5.4 Three Blocks
Three blocks of mass 3m, 2m, and m
1) T1 > T2 > T3
are connected by strings and pulled
2) T1 < T2 < T3
with constant acceleration a. What is
3) T1 = T2 = T3
the relationship between the tension in
4) all tensions are zero
each of the strings?
5) tensions are random
a
3m
T3
2m
T2
m
T1
ConcepTest 5.4 Three Blocks
Three blocks of mass 3m, 2m, and m
1) T1 > T2 > T3
are connected by strings and pulled
2) T1 < T2 < T3
with constant acceleration a. What is
3) T1 = T2 = T3
the relationship between the tension in
4) all tensions are zero
each of the strings?
5) tensions are random
T1 pulls the whole set
of blocks along, so it
a
must be the largest.
T2 pulls the last two
3m
T3
2m
T2
m
T1
masses, but T3 only
pulls the last mass.
Follow-up: What is T1 in terms of m and a?
ConcepTest 5.5 Over the Edge
In which case does block m experience a
1) case 1
larger acceleration? In (1) there is a 10 kg
2) acceleration is zero
mass hanging from a rope and falling. In
3) both cases are the same
(2) a hand is providing a constant
downward force of 98 N. Assume massless 4) depends on value of m
ropes.
5) case 2
m
m
10kg
a
a
F = 98 N
Case (1)
Case (2)
ConcepTest 5.5 Over the Edge
In which case does block m experience a
1) case 1
larger acceleration? In (1) there is a 10 kg
2) acceleration is zero
mass hanging from a rope and falling. In
3) both cases are the same
(2) a hand is providing a constant
downward force of 98 N. Assume massless 4) depends on value of m
ropes.
5) case 2
In (2) the tension is 98 N
due to the hand. In (1)
the tension is less than
m
m
10kg
a
98 N because the block
a
F = 98 N
is accelerating down.
Only if the block were at
rest would the tension
be equal to 98 N.
Case (1)
Case (2)
ConcepTest 5.12 Will it Budge?
A box of weight 100 N is at rest on
a floor where ms = 0.5. A rope is
attached to the box and pulled
horizontally with tension T = 30 N.
Which way does the box move?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
Static friction
(ms = 0.4 )
m
T
ConcepTest 5.12 Will it Budge?
A box of weight 100 N is at rest on
a floor where ms = 0.4. A rope is
attached to the box and pulled
horizontally with tension T = 30 N.
Which way does the box move?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
The static friction force has a
maximum of msN = 40 N. The
tension in the rope is only 30 N.
Static friction
(ms = 0.4 )
m
T
So the pulling force is not big
enough to overcome friction.
Follow-up: What happens if the tension is 35 N? What about 45 N?
ConcepTest 5.19c Going in Circles III
You swing a ball at the end of string in a
1) Fc = T – mg
vertical circle. Since the ball is in
2) Fc = T + N – mg
circular motion there has to be a
3) Fc = T + mg
centripetal force. At the top of the ball’s
path, what is Fc equal to?
4) Fc = T
5) Fc = mg
top
v
R
ConcepTest 5.19c Going in Circles III
You swing a ball at the end of string in a
1) Fc = T – mg
vertical circle. Since the ball is in
2) Fc = T + N – mg
circular motion there has to be a
centripetal force. At the top of the ball’s
3) Fc = T + mg
path, what is Fc equal to?
4) Fc = T
5) Fc = mg
Fc points toward the center of the
circle, i.e. downward in this case. The
v
mg
T
weight vector points down and the
tension (exerted by the string) also
points down. The magnitude of the
net force, therefore, is:
Fc = T + mg
Follow-up: What is Fc at the bottom of the ball’s path?
R