Lecture 10
Ch6. Friction, Drag, and Centripetal Force
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2013
Homework 8: Coin On A Book
The figure below shows a coin of mass m at rest on a book that
has been tilted at an angle θ with the horizontal. By
experimenting, you find that when θ is increased to 13°, the coin
is on the verge of sliding down the book, which means that even
a slight increase beyond 13° produces sliding.
What is the coefficient of static friction μs between the coin and
the book?
Hint: Draw the free-body diagram of the coin first.
Erwin Sitompul
University Physics: Mechanics
10/2
Solution of Homework 8: Coin On A Book
Forces along the y axis:
Fnet,y  m a y
FN  Fg cos   0 • Why zero?
FN  Fg cos 
FN  mg cos 
Forces along the x axis:
Fnet,x  m ax
Fg sin   fs  0 • Why zero?
Fg sin   s FN  0
mg sin   s mg cos   0
sin 
h
s 
 tan  
cos 
d
So, the coefficient of static friction is:
s  tan13  0.231
Erwin Sitompul
University Physics: Mechanics
10/3
Virtual Experiment: Determining μk
θ
An object is kept in rest on an inclined surface. The angle θ is
26°, which is greater than the critical angle θc (μs = tanθc).
Upon release, the object directly move and slide down to the
bottom. It requires 4.29 s to reach the bottom, which is 18 m
away from the initial point.
Determine the coefficient of kinetic friction μk between the object
and the surface.
Erwin Sitompul
University Physics: Mechanics
10/4
Example: Blue Block
→
A block of mass m = 3 kg slides along a floor while a force F of
magnitude 12 N is applied to it at an upward angle θ. The
coefficient of kinetic friction between the block and the floor is
μk = 0.4. We can vary θ from 0 to 90° (with the block remains on
the floor.
What θ gives the maximum value of the block’s acceleration
magnitude a?
Erwin Sitompul
University Physics: Mechanics
10/5
Example: Blue Block
Forces along the y axis:
Fnet,y  m a y
FN  Fy  Fg  0
FN  mg  F sin 
Forces along the x axis:
Fnet,x  m ax
Fx  f k  m a
F cos   k FN  m a
F
F

 • What θ gives the
a  cos   k  g  sin   maximum value of a?
m
m


• da/dθ = 0
Erwin Sitompul
University Physics: Mechanics
10/6
Example: Blue Block
If a is given by
F
F


a  cos   k  g  sin  
m
m


then, the derivative of a with respect to θ is
da
F
F
  sin   k cos   0
d
m
m
tan   k
  tan 1  k
 tan 1 (0.4)
 21.80
Erwin Sitompul
University Physics: Mechanics
10/7
Example: Two Blocks
Block B in the figure below weighs 711 N. The coefficient of
static friction between block and table is 0.25; angle θ is 30°.
Assume that the cord between B and the knot is horizontal.
Find the maximum weight of block A for which the system will
be stationary.
Erwin Sitompul
University Physics: Mechanics
10/8
Example: Two Blocks
→
TW
→
TB
Block B
→
TB
→
FgB
Knot
→
TW
→
TA
→
TA
→
FNB
→
fs,max
Wall
Block A
→
FgA
→
TW
→
fs,max
Knot
→
FgA
Erwin Sitompul
University Physics: Mechanics
10/9
Example: Two Blocks
Forces along the y axis:
Fnet,y  0
TWy  FgA  0
TW sin   mA g
→
TW
TWy
θ
→
fs,max
Knot
TWx
Forces along the x axis:
Fnet,x  0
TWx  fs,max  0
TW cos   s FNB
TW cos   s mB g
mA g s mB g

sin 
cos 
sWB
WA

sin  cos 
→
FgA
WA  sWB tan 
 (0.25)(711) tan 30
 102.624 N
Erwin Sitompul
University Physics: Mechanics 10/10
Example: Multiple Objects
A block of mass m1 on a rough, horizontal surface is connected
to a ball of mass m2 by a lightweight cord over a lightweight,
frictionless pulley as shown in the figure below.
A force of magnitude F at an angle θ with the horizontal is
applied to the block as shown and the block slides to the right.
The coefficient of kinetic friction between the block and surface
is μk.
Find the magnitude of acceleration of the two objects.
Erwin Sitompul
University Physics: Mechanics 10/11
Example: Multiple Objects
→
FN
Fy
T
Forces in m2
m2
→
Fg2
→
fk
m1
→
Fg1
→
Fnet,y  m2 a2 y
T  Fg2  m2 a
T  m2 ( g  a)
F
θ
→
T
→
Fx
Forces in m1
Fnet,x  m1a1x
Fx  T  f k  m1a
F cos   T  k FN  m1a
T  F cos   m1a  k FN
Fnet,y  0
Fy  FN  Fg1  0
FN  Fg1  Fy
FN  m1 g  F sin 
Erwin Sitompul
University Physics: Mechanics 10/12
Example: Multiple Objects
T  m2 ( g  a)
T  F cos   m1a  k FN
FN  m1 g  F sin 
m2 ( g  a)  F cos   m1a  k (m1 g  F sin  )
m1a  m2 a  F cos   k F sin   k m1 g  m2 g
(m1  m2 )a  F (cos   k sin  )  (k m1  m2 ) g
a
Erwin Sitompul
F (cos   k sin  )  ( k m1  m2 ) g
m1  m2
University Physics: Mechanics 10/13
Example: Trio Blocks
When the three blocks in the figure below are released from rest,
they accelerate with a magnitude of 0.5 m/s2. Block 1 has mass
M, block 2 has 2M, and block 3 has 2M.
What is the coefficient of kinetic friction between block 2 and the
table?
Erwin Sitompul
University Physics: Mechanics 10/14
Example: Trio Blocks
a
Forces in m1
Fnet,y  m1a1 y
a T1  Fg1  Ma
T1  M ( g  a)
a
→
FN
→
T1
→
T1
m1
→
fk
→
Fg1
Erwin Sitompul
→
T2
→
T2
m2
→
Fg2
m3
→
Fg3
Forces in m2
Fnet,x  m2 a2 x
T2  T1  f k  2Ma
T2  T1  k FN  2Ma
Fnet,y  m2 a2 y
FN  Fg2  0
FN  2Mg
Forces in m3
Fnet,y  m3a3 y
T2  Fg3  2M (a)
T2  2M ( g  a)
University Physics: Mechanics 10/15
Example: Trio Blocks
T1  M ( g  a)
T2  T1  k FN  2Ma
FN  2Mg
T2  2M ( g  a)
 2M ( g  a)    M ( g  a)   k  2Mg   2Ma
k  2Mg    2M ( g  a)    M ( g  a)   2Ma
2M ( g  a)    M ( g  a)   2Ma

k 
Mg  5Ma
k 
2Mg
g  5a

2g
(9.8)  5(0.5)

2(9.8)
 0.372 m s2
Erwin Sitompul
2Mg
University Physics: Mechanics 10/16
The Drag Force and Terminal Speed
 A fluid is anything that can flow – generally a gas or a liquid.
 When there is a relative velocity between a fluid and a body
(either because the body moves through the fluid or because
the fluid
→ moves past the body), the body experiences a drag
force D that opposes the relative motion.
 Here we examine only cases in which air is the fluid, the body
is blunt rather than slender, and the relative motion is fast
enough so that the air becomes turbulent (breaks up into
swirls) behind the body.
 In such cases, the magnitude of the drag force is related to
the relative speed by an experimentally determined drag
coefficient C according to
D  12 C  Av 2
Erwin Sitompul
ρ : air specific density
A : effective cross-sectional area of the body
C : drag coefficient
University Physics: Mechanics 10/17
The Drag Force and Terminal Speed
What is the meaning of the asterixs???
Erwin Sitompul
University Physics: Mechanics 10/18
The Drag Force and Terminal Speed
 When a blunt body falls from rest through
air, the drag force D
→
is directed upward. This upward
→ force D opposes the
downward gravitational force Fg on the body.
D  Fg  ma
 If the body falls long enough, D eventually equals Fg. This
means that a = 0, and so the body’s speed no longer
increases. The body then falls at a constant speed, called the
terminal speed vt.
1
2
C  Avt2  Fg  0
vt 
Erwin Sitompul
2 Fg
C A
University Physics: Mechanics 10/19
The Drag Force and Terminal Speed
• Cyclists and downhill
speed skiers try to
maximize terminal
speeds by reducing
effective crosssectional area
Erwin Sitompul
University Physics: Mechanics 10/20
Example: Falling Cat
If a falling cat reaches a first terminal speed of 97 km/h while it
is wrapped up and then stretches out, doubling A, how fast is it
falling when it reaches a new terminal speed
vt ,1 
2 Fg
C A
 95km h
2 Fg
95
vt ,2 

 67.18km h
C 2A
2
Erwin Sitompul
University Physics: Mechanics 10/21
Example: Raindrop
A raindrop with radius R=1.5 mm falls from a cloud that is at
height h=1200 m above the ground. The drag coefficient C for
the drop is 0.6. Assume that the drop is spherical throughout its
fall. The density of water ρw is 1000 kg/m3, and the density of air
ρa is 1.2 kg/m3.
(a) What is the terminal speed of the drop?
(b) What would be the drop’s speed just before impact if there
were no drag force?
2(mw g )
2(  w 43  r 3 g )
8  w rg
2(  wvw g )



(a) vt 

2
C  a ( r )
C a Aw
3 C a
C a Aw
C A
2 Fg
8 (1000)(1.5 103 )(9.8)
 7.38 m s  26.56 km h

3
(0.6)(1.2)
(b) vy2  v0,2 y  2 g ( y  y0 )  vy  2(9.8)(0  1200)
 153.36 m s  552.10 km h
Erwin Sitompul
University Physics: Mechanics 10/22
Homework 9A
In the next figure, blocks A and B have weights of 44 N and
22 N, respectively.
(a) Determine the minimum weight of block C to keep A from
sliding if μs, between A and the table is 0.20.
(b) Block C suddenly is lifted off A. What is the acceleration of
block A if μk between A and the table is 0.15?
Erwin Sitompul
University Physics: Mechanics 10/23
Homework 9B
1. The figure shows a 1.0-kg University Physics book
connected to a 500-g tea mug. The book is pushed up
the slope and reach a speed of 3.0 m/s before being
released. The coefficients of friction are μs = 0.50 and
μk = 0.20.
(a) How far will the book slide upwards?
(b) After the book reaches the highest point, will the
book stick to the surface, or will it slide back down?
2. In downhill speed skiing, a skier is retarded by both
the air drag force on the body and the kinetic
frictional force on the skis. Suppose the slope angle
is θ = 40.0°, the snow is dry with a coefficient of
kinetic friction μk = 0.04, the mass of the skier and
equipment is m = 85.0 kg, the cross-sectional area
of the (tucked) skier is A = 1.30 m2, the drag
D
coefficient is C = 0.150, and the air density is 1.20
kg/m3.
fk
(a) What is the terminal speed? (b) If a skier can
vary C by a slight amount dC by adjusting, say, the
hand positions, what is the corresponding variation
in the terminal speed?
Erwin Sitompul
University Physics: Mechanics 10/24