Lecture 5
Chapter 5
Using Newton’s Laws
Friction
Circular motion
Drag Forces
Numerical integration
Misconceptions
Assignment 2
• Chp 4 4-52, 4-67, 4-75
• Chp 5 5-30, 5-44
Due Friday 12:00 Mar 11
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?
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
No motion and no horizontal forces
What is the free body diagram without friction?
No Motion
What is the free body diagram
with friction but no motion?
Still No Motion with larger Force F
Increase F more and now you get
acceleration
Static friction force decreases to kinetic friction
and its value remans constant and now you get
acceleration
Constant velocity
How is the frictional force related to the
normal?
Fixed block
N
mg
F
fs
fs = ms N
The maximum value equals is
where N is the normal force. Above we would have
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 = mk N
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 incline plane at rest with impending
motion. Find the coefficient of static friction
What is the free body diagram?
+y
N

fs

3/24/2016
d
mg

Physics 631
h
+x
.
Free Body Diagram
+y
Apply Newton's 2nd law in
the x directions and using fs =
s N
N

mgsinq
+x
fs
x direction
fs - mgsinq = 0
ms N - mgsinq = 0
 mgcosq
3/24/2016
mg
Physics 631
.
+y
Free Body Diagram
Apply Newton's 2nd law in
the y direction
N

+x
fs
 mgcosq
N - mgcosq = 0
N = mgcosq
ms N - mgsinq = 0
Apply Newtons 2nd Law
F = ms mgcosq - mgsinq = 0
3/24/2016
mg
Physics 631
.
+y
F = ms mgcosq - mgsinq = 0
sin q
ms = mgsinq / mgcosq =
cosq
N

mgsinq
+x
fs
 mgcosq
3/24/2016
mg
Physics 631
ms = tanq
Find a and T for the Atwood’s machine
with friction between M and surface.
T
fk
T
T
Find a and T for the Atwood’s machine with
friction between M and surface.
+y
N
f
+x
T
T - fk = Ma ---------x direction
T = Ma + mk Mg
T
Body 2
T - mg = -ma ------y direction
Ma + mk Mg - mg = -ma
T
T
mg
-y
N = Mg ------------y direction
fk = m k N
Mg
T
fk
Body 1
Solve for a and then T
Find a and T for the Atwood’s machine
with friction between M and surface.
mg - mk Mg
a=
M+m
T
fk
T = (1+ mk )
T
T
M
mg
M+m
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

Tcosφ
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 - ms N = ms (mg - T sin f ), Solve for T
T=
ms mg
(0.5)(68)(9.8)
=
= 304 N
(cos f + ms sin f ) cos(15) + 0.5 sin(15)
Question: What is the initial acceleration if
 k=0.35?
m = 68kg
+y
f = 15deg rees
mk = 0.35
N

fs
Tcos 
T
+x
mg
Newton's 2nd law
x components
T cos f - fk = ma
y components
T sin f + N - mg = 0
fk = mk N = mk (mg - T sin f )
ma = T cos f - m k (mg - T sin f )
304 [ cos(15) + 0.35 sin(15)]
T (cos f + m k sin f )
- mk g =
- (0.35)(9.8)
m
68
a = 1.3m / s 2
a=
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.
1
D = C r Av 2
2
Imagine a falling ball slowed down due to elastic collisions
with air molecules. Simply pushing the air out of the way.
dv
F = ma = m when mass is constant
dt
when mass is chnanging
d(mv)
dv
dm
=
=m +v
dt
dt
dt
ball
v
.
A
dy
air molecules
Inertial drag
Inertial Drag Force and Terminal Velocity
dm
F=v
dt
dm = r Ady
dm
dy
= r A = r Av
dt
dt
F = vr Av = r Av 2
ball
v
1
2
D = C r Av
2
.
A
dy
air molecules
Inertial drag
Terminal speeds in air
Using Newton’s 2nd law,
1
ma = mg - C r Av2
2
where m is the mass of the falling ball
1
mg - C r Av0 2 = 0
2
Solve for v0
2mg
v0 =
CrA
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
1
9
13
31
42
60 -120
250
Show demo of falling feather in vacuum
0.9
2.2
20
29
66
86
134 -268
560
How to solve this equation?
Two ways
Ball
1
F = Mg - C r Av 2
2
F = Mg - bv 2
One way is to use a spread sheet in Exel.
Air
Gravity
Use Newtons 2nd Law
Initial component of momentum:
Initial force on ball:
Using 2nd Law
Find new p
p1 = Mv1
F = Mg - bv12
dp Dp p2 - p1
F=
=
=
dt Dt
Dt
p2 - p1 = FDt
p2 = p1 + FDt
Substitute in for p1 and F
Newtons 2nd Law
p2 = p1 + FDt
p2 = p1 + (Mg - bv12 )Dt
p2
b 2
v2 =
= v1 + [g - ( )v1 ]Dt
M
M
v1 + v2
x2 = x1 + v2 Dt = x1 + (
)Dt
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
é
ù
vi = vi -1 + ê g - (
)vi -12 ú Dt
m _1
ë
û
v +v
xi = xi -1 + ( i -1 i )Dt
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 = 1/2CrA
dv
m
= mg - bv 2
dt
dv
2
m /b = mg /b - v
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
v
t
dv'
dv'
2v0 b
+
=
dt
'
ò0 v0 - v' ò0 v0 + v' m ò0
v
v
t
dv'
dv'
2v0 b
ò0 v0 - v' + ò v0 + v' = m ò0 dt '
2v0 b
[- ln(v0 - v') + ln(v0 + v')] =
t
m
v0 + v
2v0 b
ln(
)=
t
v0 - v
m
v
0
v0 + v
=e
v0 - v
2v0 b
t
m
v0 + v = (v0 - v)e
2v0 b
t
m
v0 + v = (v0 - v)e
v(1+ e
2v0 b
t
m
2v0 b
t
m
= v0 e
) = v0 (-1+ e
2v0 b
t
m
2v0 b
t
m
-
2v0 b
t
m
- ve
)
2v0 b
t
m
v -1+ e
1- e
=
=
2v0 b
2v0 b
t
t
v0
1+ e m
1+ e m
2v b
æ
- 0 t ö
ç 1- e m ÷
v = v 0ç
2v 0 b ÷
t÷
ç
m
1+
e
è
ø
Now show comparison of this solution with
numerical integration with Excel.
2v0 b
t
m
Comparison
True Terminal Velocity Function
60
Velocity (m/s)
50
40
TRUE
Velocity 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.
ma = - kv
A 1000 kg 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 Newtons,
where v is the speed of the boat. Let k = 70 N/s/m.
dv
m = - kv
dt
mdv = - kvdt
dv
k
= - dt
v
m
v
t
dv
k
ò v = - m ò dt
v0
0
v
ln v v = ln v - ln v0
0
v
k
ln = - t
v0
m
v
=e
v0
-k
m
t
v/v0 = 45/90 =1/2
t = m/k ln 2=1000/70 ln 2 = 9.9 s
UNIFORM CIRCULAR MOTION
• Centripetal Acceleration: accelerates a body by
changing the direction of the body’s velocity without
changing the speed. There must be a force also
pointing radially inward to make this motion.
• Examples:
– Ball on a string : show demo: Force is produced by the
weight of the mass and transmitted by the tension in the
string.
– Moon in orbit around the earth: gravitational force
– A car making a sharp turn: friction
– A carousel; friction and contact forces
• Demo: pushing bowling ball with broom in a circle
39
Uniform circular motion
T = Mac
v2
ac =
R
Centripetal force is really not a new force like gravity, tension, friction.
Motion of earth around sun – centripetal force is a result of gravity
Rock whirled around on a string – centripetal force is a result of tension
Sometimes it is a result of friction or the normal force
CENTRIPETAL ACCELERATION:
ac = Dv Dt
q
r0
r
.
v points radially inward
Find Dv
v0
Dv = v - v0
v
Dv
v0
q
v
41
CENTRIPETAL ACCELERATION
q
r0
r
.
Dr
r
r0
q
Dr = r - r0
v0
Dv Dv = v - v0
v
v0
q
v
Triangles are similar
42
Centripetal Acceleration
Magnitudes are related by Dv
And, Dr = vDt
Dv (vDt)
=
v
r
Dv v 2
=
Dt
r
v =
Dr
r
due to similar triangles
ac
so
r
.
v
v2
Magnitude of ac =
r
Period of the motion
2p r
T=
v
43
What is the magnitude of ac and its
direction for a radius of r = 0.5 m and a
period of T= 2 s,
v
• ac =
r
2
Need to find v
2p r 2p * 0.5
v=
=
= 1.57m / s
T
2
v 2 1.57 2
ac = =
= 4.92m / s 2
r
0.5
• What is the direction of ac ?
INWARD
44
QUALITATIVE QUIZ
A ball is being whirled around on a string.
The string breaks. Which path does the
ball take?
c
d
b
e
a
v
45
VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
mv 2
- mg
At the top: Apparent weight= N =
r
N
N
r
Minimum v for N = 0:
(apparent weightlessness)
mv 2
= mg
r
v = rg
At the bottom:
Apparent weight = N = mv2/r + mg
Weigh more
What do we mean by Fictitious Forces
Ff = - ma
(the fictitious force always acts in the opposite
direction of acceleration)
T

ma
T
mg
Ff
mg

T
ma
tan q =
mg
a = g tan q
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
Figure 5.34
Find the T in the cord between
A and B and find the
acceleration a
mA a = mA gsin q - T - m A mA cosq
mB a = mB gsin q + T - mB mB cosq
m A < mB
m A = mB
Wednesday Agenda
• Any questions about homework Assignment 2
• See Section 2-8 page 38 and problem 5-102 on
numerical integration
• Also see 5-87, 5-92, 5-100
• More about centripetal acceleration
• Conceptual questions from Chapter 5
• Go on to Chapter 7- Skip Chapter 6 for now
Figure 5.18
Let T =FT1
mg + T = mac
T = mac - mg
T - mg = mac
T = mac + mg
What happens in between?
Figure 5.56
What is the tension T in the cord and the tangential at and
centripetal ac acceleration as a function of m,g, r, v and theta?
T - mgsin q = mac
T = mac + mgsin q
ac = mv 2 / r
T = mv / r + mgsin q
2
at = gcosq
Figure 5.20
Find v velocity and T period
in terms of g, l, and
angle theta.
lgsin 2 q
v=
cosq
l cosq
T = 2p
g
FT cosq = mg
Newtons 2nd law in y direction
FT sin q = mv 2 / r
Newtons 2nd Law in radial direction
Figure 5.24
y direction
FN cosq - mg - f sin q = 0
x direction
mv 2
FN sin q + f cosq =
r
f = m FN
f
Conceptual Questions
Chapter 5
100. The radial acceleration is a R =
v2
, and so aR =
v2
=
( 6.0 m s )
2
= 45 m s 2 .
r
0.80 m
r
The tension force has no tangential component, and so the tangential force is seen from the
diagram to be Ftang = mg cos q .
(
)
Ftang = mg cos q = matang ® atang = g cos q = 9.80 m s 2 cos 30° = 8.5 m s 2
The tension force can be found from the net radial force.
v2
FR = FT - mg sin q = m
®
r
v2 ö
æ
FT = m ç g sin q + ÷ = (1.0 kg ) ( ( 9.80 m s 2 ) sin 30° + 45 m s2 ) = 50 N
r ø
è
Note that the answer has 2 significant figures.