AP Physics Chapter 6
Force and Motion – II
1
AP Physics




2
Turn in Chapter 5 Homework, Worksheet &
Lab Report
Take Quiz 6
Lecture
Q&A
Review on Chapter 5



Newton’s Second Law: F = ma
Newton’s Third Law: Action and Reaction
Forces
Forces




3
Weight
Normal Force
Tension
Apparent Weight
Four Fundamental (Basic) Forces in
Nature




4
Gravitational Force
Electromagnetic Force
Strong (Nuclear) Force
Weak (Nuclear) Force
Gravitational Force

Gravity or Weight (Gravitational force Earth
pulling on objects around it): W
W  mg

Gravitational force always exist, but value of g
can be different at different location.
–
–
–
5
On surface of earth, g = 9.81 m/s2.
Higher elevation, smaller g.
Higher latitude, larger g.
Electromagnetic Force

Almost all other forces we encounter in daily
life are electromagnetic force in nature.






6
Electric force: F
Magnetic force: F
Tension (pull, string): T
Push: F
Normal (support): N (not to be confused with
Newton, the unit of force)
Friction: ƒ (ƒs or ƒk)
Strong (Nuclear) Force



7
Holds protons and neutrons within the nucleus
Exist in very short distance only ( 10-15 m)
Stable nucleus
Weak (Nuclear) Force




8
Also holds protons and neutrons within the
nucleus
Also very short range ( 10-15 m)
Not strong enough. Protons and neutrons can
escape  nuclear reaction
Unstable nucleus
Weakest
9
Strong
Weak
Electromagnetic
Gravitational
Relative Strength of Forces
Strongest
Force at Distance

Contact force:
 Force giver and receiver must be in contact
 Tension, push, normal force, friction, …

Field force / Distant force:
 Force giver and receiver do not have to be in contact
 Gravity, electric force, magnetic force, …
10
Unification of Forces


Electromagnetic and weak forces have been
unified into (understood as) one force:
Electroweak force
It is believed that all four forces are different
aspects of a single force.
–
–
11
Grand Unification Theories (GUTs) and
Supersymmetric theories
Not yet successful
Friction

12
Friction: force opposing relative motion or
tendency of relative motion between two rough
surfaces in contact
 Smooth surface  No friction
 No normal force  No friction
v
Two kinds of frictions

Static friction:




k: coefficient of kinetic friction
Kinetic friction force is constant.
 is a constant depending on the properties of the two
surfaces


13
s: coefficient of static friction
Static friction force is not constant, it has a maximum.
Kinetic (or sliding) friction: f k  k N


f s ,max  s N  f s  s N

 = 0 when one of the surfaces is smooth (frictionless.)
s > k for same surfaces.
 has no unit.
Example:
N
What happens to the frictional
force as you increase the force
pushing on a table on the floor?
Fapp
f
W
But what if the applied force increases
just slightly?
f
 sN
•
• Once table is moving, a smaller
constant kinetic friction. fk < fs, max
•
•
•
0
14
 sN
Fapp
Is friction always opposite to
motion?
When you step on the pedal
4
v
v
v
1
7
v
15
2
3
f1
6
f2
5
v
Example:
A trunk with a weight of 220N rests on the floor. The
coefficient of static friction between the trunk and the
floor is 0.41, while the coefficient of kinetic friction is
0.32.
a). What is the minimum magnitude for a horizontal force
with which a person must push on the trunk to start in
moving?
b) Once the trunk is moving, what magnitude of
horizontal force must the person apply to keep it
moving with constant velocity?
c) If the person continued to push with the force used to
start the motion, what would be the acceleration of the
trunk?
16
Solution
s  0.41, k  0.32,W  220 N
a)a  0, Fmin  ?
 F  Fmin  f s ,max  m a  0
y
 Fmin  f s ,max  s N  sW
N =W
 0.41  220 N  90 N
b)a  0, Fmin  ?
f
F
Similarly,
Fmin  f k  k N  kW  0.32  220 N  70 N
17
W
x
s  0.41, k  0.32,W  220 N
Solution
c)a  ?
 F  F  f k  ma
F  fk
F f
a

m
W/g
90 N  70 N

220 N /  9.8m / s 2 
y
N =W
f
F
 0.89m / s 2
W
18
x
Example: Pg135-57
The coefficient of kinetic friction in
between the incline and the
block is 0.20, and angle  is
60o. What is the acceleration
of the block if
a) It is sliding down the slope and
b) It has been given an upward
shove and is still sliding up the
slope?
19

Wx  W sin 

Wy  W cos
Solution
  0.20,   60.0o a ) a  ?
 Fx  Wx  f  ma
y
f
N
 mg sin    N  ma
 Fy  N  Wy  0
Why?
 N  Wy  mg cos 

Wx
x
 mg sin    mg cos   ma
 a  g sin    g cos   g  sin    cos  
m
m
o
o
 9.8 2  sin 60  0.20 cos 60   7.5 2
s
s
20

W
Wy
Wx  W sin 

Wy  W cos 
Solution (2)
  0.20,   60o b)a  ?
 Fx  Wx  f  ma
y
N
 mg sin    N  ma
 Fy  N  Wy  0
 N  Wy  mg cos 
f
Wx
x
 mg sin    mg cos   ma
 a  g sin    g cos   g  sin    cos 
m
m
o
o
 9.8 2  sin 60  0.20 cos 60   9.5 2
s
s
21


W
Wy
Practice:
An 11 kg block of steel is at rest on a horizontal table.
The coefficient of static friction between block and
table is 0.52.
a) What is the magnitude of the horizontal force that will
just start the block moving?
b) What is the magnitude of a force acting upward 60o
from the horizontal that will just start the block
moving?
c)
If the force acts down at 60o from the horizontal, how
large can its magnitude be without causing the block
to move?
22
m  11kg ,   0.52
Solution
 Fx  F cos 
F 
 y F sin 
a)
N
Fmin  f s ,ma x   N   mg  0.52  11kg  9.8m / s  56 N
2
f
b)
 Fx  Fx  f  F cos   f  0  F cos    N





 Fy  Fy  N  W  F sin   N  W  0  N  mg  F sin 
 F cos    F sin    mg
 mg
0.52  11kg  9.8m / s 2
F

 59 N
o
o
 cos    sin   cos 60  0.52 sin 60
23
W
y
 F cos    mg  F sin     mg   F sin 
 F  cos    sin     mg
F
N
f
Fy F

Fx
W
x
Solution (2)
c)
Similarly to Part b), we have
 mg
0.52  11kg  9.8m / s 2
3


1.1

10
N
F
o
o
 cos    sin   cos  60   0.52sin  60 
Another approach is to use the same equation as in b)
but change the angel to –60o.
24
 mg
0.52  11kg  9.8m / s 2
3
F


1.1

10
N
o
o
 cos    sin   cos  60   0.52sin  60 
Practice: Pg133-21
Block B in the diagram
weighs 711 N. The
coefficient of static friction
between block and
horizontal surface is 0.25.
Find the maximum weight
of block A for which the
system will be stationary?
25
30o
B
A
Solution: Pg133-21
P:
y
T
 Fpx  TA cos   T  0  TA 
cos 
 Fpy  TA sin   WA  0
 WA  TA sin 
 T

 cos 

  sin   T tan 

y
N
f
T
B
T


P
WB
mB :
 FBx  T  f  0  T  f   N 
  T  WB
 FBy  N  WB  0  N  WB

26
x
o
 WA   WB  tan   0.25  711N  tan 30  103 N
WA
TA
x
Drag Force
Drag Force:
1
D  C  Av 2
2

C: drag constant (not a true constant)

: density of fluid (air or liquid)

A: cross-sectional area of object moving in fluid

v: speed of object relative to fluid
Source: fluid in relative motion pushing on object
Direction: opposing relative motion of object relative
to fluid, same direction as relative motion of fluid
27
D
1
C  Av 2
2
Terminal velocity
Initially, v: small  D: small
W _____
> D
D
Fnet: downward  a: downward
+
 v: Increases  D: increases
Eventually, D ___
= W:  Fnet = 0
W
 a = 0 :  v: constant
1
2
F  0  W  D  0  D  W  C  Av  mg  v 
2
28
2mg
C A
Practice: Pg133-33
Calculate the ratio of the drag force on a jet flying with a speed of 1000
km/h at an altitude of 10 km to the drag force on a prop-driven transport
flying at half the speed and half the altitude. The density of air is 0.38
kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes
have the same effective cross-sectional area and the same drag
coefficient C.
10 km  0.38
D
kg
kg
1
,


0.67
,
v

v10 km
5 km
5 km
3
3
m
m
2
1
C  Av 2
2
D10 km

D5 km
1
C 10 Av10 2
2
1
C  5 Av5 2
2
10 v10 2

5v52
kg
 v10 2
3
4  0.38
0.38  v10 2
m
 2.3



2
2
0.67
v
kg  v 
0.67  10
0.67 3   10 
4
m  2 
0.38
29
Review: Uniform Circular Motion
and Centripetal Acceleration
v2
a
r


30
, where

v: speed of particle
r: radius of circle or circular arc
v
Direction of acceleration is
always toward the center of
a
a
circle (or circular arc)
a
 Centripetal
v
av
for uniform circular
motion at any time.
v
Centripetal Force
v2
Fc  mac  m
r


31
Centripetal force is in general not a single physical
force; rather, it is in general the net force.
Do not draw centripetal force on force diagram (Free
Body Diagram)
Rear View
N
N
f

W
W
Examples of centripetal forces

Rounding a curve in a car
–
Flat curve: Static friction provides the centripetal force (when
no skidding)
vmax  s Rg
–
Max velocity w/o skidding
Banked curve: a component of Normal force (net force)
v2
v  Rg tan     tan
Rg
1

Orbiting the Earth (Sun or other object)
–
32
No friction
Gravity
Example: Pg138-87
A car weighing 10.7 kN and traveling at 13.4 m/s attempts to round an
unbanked curve with a radius of 61.0 m.
a) What force of friction is required to keep the car on its circular path?
b) If the coefficient of static friction between the tires and road is 0.35,
is the attempt at taking the curve successful?
y
W 10.7  103 N

 1.09  103 kg
W  mg  m 
m
g
9.8 2
s
N
a) f  ?
x
2
 F  f  Fc  m
v
R
 1.09  10 kg 
3
13.4m / s 
61.0m
2
 3.21  10 N
3
f
W
b) f s ,max  ?
f s ,max  s  N   s  W  0.35  10.7  103 N   3.75  103 N  3.21  103 N
Yes, successful.
33
Practice: Pg135-50
A banked circular highway curve is designed for traffic moving at 60
km/h. The radius of the curve is 200 m. Traffic is moving along the
highway at 40 km/h on a rainy day. What is the minimum
coefficient of friction between tires and road that will allow cars to
negotiate the turn without sliding off the road?
vo  60
km 1000m
h
m


 16.7 , v  40 km  11.1 m , r  200m
h
km 3600s
s
h
s
vo 2
16.7m / s   8.1o
  tan
 tan 1
Rg
200m  9.8m / s 2
y
N
2
1
Fx  N sin   f cos   maC  N sin    N cos   m
 N  sin    cos    m
x
2
v
R
Ny
fy f
fx
Nx

v2
mv 2
 N
R
R  sin    cos 
W
 Fy  N cos   f sin   W  0  N cos   N sin   W
 N  cos   sin   W  N 
mv 2
mg


R  sin    cos    cos    sin  
 N x  N sin

 N y  N cos
mg
 cos    sin  
 v 2  cos   sin   gR sin   cos   v 2 cos   v 2  sin   gR sin   gR cos 
2
2
 v 2  sin  gR cos  gR sin  v 2 cos    v sin   gR cos   gR sin   v cos
9.8m / s 2   200m  sin8.1o  11.1m / s  cos8.1o
gR sin   v 2 cos
 2

 0.079
 v sin  gR cos  11.1m / s 2 sin8.1o  9.8m / s2   200m  cos8.1o
2
34
 f x  f cos

 f y  f sin 
Practice: Pg134-45 (Modified wording)
A student of weight 667 N rides a steadily rotating Ferris wheel (the student
sits upright). At the highest point, the apparent weight of the student is 556
N. (a) Does the student feel “light’ or “heavy” there? (b) What is the
apparent weight at the lowest point? If the wheel’s speed is doubled, what is
the apparent weight at the (c) highest point and (d) lowest point?
W  667 N , Ntop  556 N
Ntop
N
a) Wapp < W
W
 Feel “light”
b) Nbottom  ?
Top:
35
 F  W  Ntop
Nbottom
v2
 m
R
v2
 m  W  Ntop  667 N  556 N  111N
R
2
v
Bottom:  F  Nbottom  W  m
R
2
v
 Nbottom  W  m
 667 N  111N  778 N
r
W
W
Practice: Pg134-45 (Modified wording)
A student of weight 667 N rides a steadily rotating Ferris wheel (the student
sits upright). At the highest point, the apparent weight of the student is 556
N. (a) Does the student feel “light’ or “heavy” there? (b) What is the
apparent weight at the lowest point? If the wheel’s speed is doubled, what is
the apparent weight at the (c) highest point and (d) lowest point?
Ntop’
W  667 N , Ntop  556 N , v '  2v
c) N top '  ?
2v 

v2
v'
 Ftop  W  N top '  m
m
 4m
R
R
R
2
v
 N top '  W  4m
R
 667 N  4  111N   223N
2
2
d ) Nbottom '  ?
2
 Fbottom
36
2
v
v'
4
m

 Nbottom ' W  m
R
R
W
Nbottom’
W
v2
 Nbottom '  W  4m  667 N  4  111N   1111N
R
Practice: Pg139-107
A certain string can withstand a maximum tension of 40 N without
breaking. A child ties a 0.37 kg stone to one end and, holding the
other end, whirls the stone in a vertical circle of radius 0.91 m, slowly
increasing the speed until the string breaks.
a) Where is the stone on its path when the string break?
b) What is the speed of the stone as the string breaks?
a) In order to keep the same speed,
and therefore the same
centripetal force, the tension
required is largest when the rock
is at the bottom.
•T
a
T
aT
•
a
b)T  40 N , m  0.37kg , r  0.91m, v  ?
W
2
v
v2
v2
 F  T  W  ma  m  T  W  m  mg  m
r
r
r
2
T
v
T
 g
 v 2  r   g 
m
r
m

37
W
 40 N
m
m
T

0.91
m


9.8

9.5
r

g

v



2 
0.37
kg
s
s
m




•
W
Practice: Pg133-29
Two blocks (m = 16 kg and M = 88 kg) are not attached to each other.
The coefficient of static friction between the blocks is s = 0.38, but
the surface beneath the larger block is frictionless. What is the
minimum magnitude of the horizontal force F required to keep the
smaller block from slipping down the larger block?
m:
 Fx  F  N  ma  F  N  ma
  F  f W   f 
0
W  mg
 y
mg
  N  mg  N 

F
M:
38
mg

F
m
M
y
x
f
 ma
mg
mg
 Fx  N  Ma 
 Ma  a 

M
m
16kg  9.8 2
mg 
m
mg
mg
s

1

F
m

0.38
  M 

M
 488N
F
N
N2
W
 16kg 
 1 

88
kg


N
f
W2
Practice:
A little girl of mass m1 = 10 kg sits on a slab of mass m2 = 20 kg on a
frozen lake. Between girl and slab, the coefficient of static friction is
0.60, and the coefficient of static friction between the sled and lake is
010. The sled is pulled forward by a horizontal force F. What is the
maximum magnitude of F such that the girl stays on the sled?
m1  10kg , m2  20kg , 1  0.60, 2  0.10, Fmax  ?
m1:
 Fx  f1  m1a  1 N1  m1a
F
 Fy  N1  W1  0  N1  W1  m1 g
N1
f1
 1 m1 g  m1 a  a  1 g
m2:
W1
 Fx  F  f1  f 2  m2 a
N2
 F  f1  f 2  m2 a  1 N1  2 N 2  m2 a
 Fy  N 2  W2  N1  0
 N 2  W2  N1  m2 g  m1 g   m1  m2  g
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f1
F
f2
N1
W2
Practice (Continued)
F  1 N1  2 N 2  m2 a
F
 1m1 g  2  m1  m2  g  m2 1 g
N1
 1m1 g  m2 1 g  2  m1  m2  g
f1
 1 g  m1  m2   2  m1  m2  g
W1
  1  2  m2  m1  g
m
  0.10  0.60 10kg  20kg   9.8 2
s
 210 N
N2
f1
F
f2
N1
W2
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