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```PHY 113 A General Physics I
11 AM-12:15 PM MWF Olin 101
Plan for Lecture 6:
Chapters 5 & 6 – More applications of
Newton’s laws
1. Friction and other dissipative forces
2. Forces in circular motion
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PHY 113 C Fall 2013 -- Lecture 6
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PHY 113 C Fall 2013 -- Lecture 6
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Isaac Newton, English physicist and mathematician
(1642—1727)
1. In the absence of a
net force, an object
remains at constant
velocity or at rest.
2. In the presence of a
net force F, the
motion of an object of
mass m is described
by the form F=ma.
3. F12 =– F21.
http://www.newton.ac.uk/newton.html
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PHY 113 C Fall 2013 -- Lecture 6
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Webassign question from assignment #5
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PHY 113 C Fall 2013 -- Lecture 6
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Webassign question from assignment #5 – continued
Two forces, F1 = (2.80ˆi + 4.45ˆj ) N and F2 = (2.70ˆi + 6.30ˆj ) N,
act on a particle of m  1.90 kg that is initially at rest at
F1  F2  Fnet  m a
ri  (1.60ˆi + 4.40ˆj )m.
F1
F2
m
Fnet
a
m
vt   v i  at
r t   ri  v i t  12 at 2
F1  F2 = (2.80ˆi + 4.45ˆj )  (2.70ˆi + 6.30ˆj ) N,
 2.80  2.70 ˆi  4.45  6.30 ˆj N
 5.50ˆi  10.75ˆj N

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
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PHY 113 C Fall 2013 -- Lecture 6
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Webassign question from assignment #5 – continued
F1  F2  Fnet  m a
F1
Fnet
a
m
vt   v i  at
F2
m

r t   ri  v i t  12 at 2

 
F1  F2 = 5.50ˆi  10.75ˆj N
5.50ˆi  10.75ˆj N
a
 2.895ˆi  5.658ˆj m / s 2
1.90kg
vt   0  2.895  t ˆi  5.658  t ˆj m / s


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PHY 113 C Fall 2013 -- Lecture 6
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Webassign question from Assignment #4
a
a
T
mg
T
mg sin q
T  m1 g  m1a
T  m2 g sin q  m2 a
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PHY 113 C Fall 2013 -- Lecture 6
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Fnet  m a
Newton’s second law
Forces we have known :
Gravity near Earth' s surface : F   gˆj
Force due to tension in massless rope : F  T
Support force normal to a surface : F  n  Fapplied
n
If surface is frictionless :
T
Fnet  T  m a
mg
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Another example of motion along a frictionless surface
Fnet  T  n  mgˆj
T  T cos q ˆi  T sin q ˆj
n  nˆj
F  T cos q  ˆi  T sin q  n  mg ˆj
net
n
If T sin q  n  mg  0
F  T cos q  ˆi
T
net
mg
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PHY 113 C Fall 2013 -- Lecture 6
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iclicker exercise:
n
T
mg
If T sin q  n  mg  0
F  T cos q  ˆi
Assuming T>0, how does
the suitcase move along
surface?
A. It moves at constant
velocity.
B. It accelerates.
C. Whether it moves or
not depends on
magnitude of T.
net
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In presence of friction force:
Fnet  T  n  mgˆj  fˆi
F  T cos q  f  ˆi 
net
T sin q  n  mg ˆj
n
T
f
mg
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Friction forces
The term “friction” is used to describe the category of forces
that oppose motion. One example is surface friction which
acts on two touching solid objects. Another example is air
friction. There are several reasonable models to quantify
these phenomena.
 Fapplied
Surface friction: f  
Normal force between
 N
surfaces
Material-dependent
coefficient
at low speed
 Kv
D
Air friction:
2
at high speed
 K v
K and K’ are materials and
shape dependent constants
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PHY 113 C Fall 2013 -- Lecture 6
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surface friction force
Models of surface friction forces
fs,max=sn
(applied force)
Coefficients s , k depend on the surfaces; usually, s > k
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PHY 113 C Fall 2013 -- Lecture 6
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Surface friction models
Static friction case:
F-fs = 0 if F < sn=smg
Kinetic friction case:
if F>sn=smg , then F-fk=ma (fk= kmg)
F   k mg
a
m
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Consider a stationary block on an incline:
Forces normal to surface :
n  mg cos q  0  n  mg cos q
Forces along surface :
f  mg sin q  0  f  mg sin q
n
mg cos q
mg
mg sin q
q
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Consider a stationary block on an incline:
What happens when
f>fs,max?
n
mg cos q
mg
mg sin q
q
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Consider a stationary block on an incline:
n  mg cos q  0  n  mg cos q
What happens when
f=fs,max?
n
f  mg sin q  0  f  mg sin q
If f  f S ,max   S n   S mg cos q
Then  S mg cos q  mg sin q
 S  tan q
mg cos q
mg
mg sin q
q
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iclicker exercise:
Suppose you place a box on an inclined surface as
shown in the figure and you notice that the box slides
down the incline at constant velocity V. Which of the
following best explains the phenomenon:
A. There is no net force acting on the box.
B. There is a net force acting on the box.
q
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Consider a block sliding down an inclined surface;
constant velocity case
n  mg cos q  0  n  mg cos q
n
f  mg sin q  0  f  mg sin q
If f   K n   K mg cos q
Then  K mg cos q  mg sin q
f=kn
 K  tan q
mg
q
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Summary
 K  tan q when block moves at constant velocity
 S  tan q when block is just about to slip
n
mg cos q
mg
mg sin q
q
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A block of mass 3 kg is pushed
up against a wall by a force P
that makes an angle of q=50o
with the horizontal. s=0.25.
Determine the possible values
for the magnitude of P that allow
the block to remain stationary.
f
N
Vertical forces :  s N  P sin q  mg  0
Horizontal forces : P cos q  N  0
f
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mg
PHY 113 C Fall 2013 -- Lecture 6
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f
N
Vertical forces :  s N  P sin q  mg  0
Horizontal forces : P cos q  N  0
f
mg
N  P cos q
 s P cos q  P sin q  mg  0
mg
Solving for P : P 
sin q  s cos q
3  9.8 N
P
sin 50o  0.25 cos 50o
31.72 N

48.57 N
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Models of air friction forces
bv
For small velocities : Fair  bv
For larger velocities : Fair   Dv
2
mg
Denoting up direction as  and assuming v  0 :
 bv  mg  ma
Solution to differential equation :
dv
 bv  mg  m
dt
mg
v(t )  
1  e bt / m
b

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
PHY 113 C Fall 2013 -- Lecture 6
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Recall:
Uniform circular motion:
animation from
http://mathworld.wolfram.com/UniformCircularMotion.html
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Uniform circular motion – continued
If vi  v f  v, then the acceleration
in the radial direction and the
centripetal acceleration is :
v2
a c   rˆ
r
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Uniform circular motion – continued
r
v2
a c   rˆ
r
 2  ˆ
a c  
 rr
 T 
2
a c  2f  rrˆ
2
In terms of time period T for one cycle:
2r
v
T
In terms of the frequency f of complete cycles:
1
f  ;
T
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v  2πfr
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Uniform circular motion and Newton’s second law
r
F  ma
v2
a c   rˆ
r
iclicker exercise:
For uniform circular motion
A. Newton’s laws are repealed
B. There is a force pointing radially outward from
the circle
C. There is a force pointing radially inward to the
circle
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Example of uniform circular motion:
http://earthobservatory.nasa.gov/Features/OrbitsCatalog/page1.php
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Example of uniform circular motion:
Consider the moon in orbit about the Earth
M M  7.35 10 22 kg
Mass of Moon :
Distance from center of Earth : RM  3.84 108 m
Rotational period of Moon :
T  27.3 days  2.36 106 s
 2 
3
2
ac  
 RM  2.72 10 m / s
T 
Newton' s Second Law  F  M M ac  2.0 10 20 N
2
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Example of uniform circular motion:
Vertical component of
Newton' s second law :
T cos q  mg  0
of Newton' s second law :
rˆ
mv 2
mv 2
T sin q  mac 

r
L sin q
v2
a c   rˆ
r
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Example of uniform circular motion:
of Newton' s second law :
Vertical component of
Newton' s second law :
n  mg  0
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n  mg
mv 2
f  mac 
r
Maximum condition :
mv 2
μs n  μs mg 
 vmax  μs gr
r
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vmax  μs gr
Example : μs  0.5
r  35m
vmax  μs gr  0.5  9.8  35m / s
 13.1m/s  29mi / hr
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Mass on a swing:
L
q T
mg
iclicker exercise:
Which of these statements
about the tension T in the rope
is true?
A. T is the same for all q.
B. T is smallest for q0.
C. T is largest for q0.
Newton' s laws in the direction along the rope :
v2
T  mg cos q  m
L
For q  0 :
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v2
T  mg  m
L
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Newton’s law in accelerating train car
Vertical direction :
T cos q  mg  0
Horizontal direction : T sin θ  ma
a
 tan q 
g
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Notion of fictitious forces
iclicker question
A. Fiction is for English class and not for
physics class.
B. The concept of fictitious forces is a bad
idea.
C. The concept of fictitious forces is
necessary for analyzing certain types of
problems.
In analyzing motion in an inertial frame of reference,
the notion of “fictitious force” does not appear.
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Example of analysis of forces
Ferris wheel moving at constant speed v
ntop  mg  mac
ntop  m g  ac 
ac
nbottom  mg  mac
ac
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nbottom  m g  ac 
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Example of analysis of forces
Mass moving in vertical circle
Analysis similar to Ferris
wheel except that the
speed v varies and the
magnitude of the
centripetal acceleration
varies accordingly.
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