Transcript F - Erwin Sitompul

Lecture 8
Ch6. Friction
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Solution of Homework 6: The Traffic Light
A traffic light weighing 122 N hangs from a cable tied to two
other cables fastened to a support, as in the figure below. The
upper cables make angles of 37° and 53° with the horizontal.
These upper cables are not as strong as the vertical cable and
will break if the tension in them exceeds 100N.
Will the traffic light remain hanging in this situation, or will one
of the cables break?
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Solution of Homework 6: The Traffic Light
Free-Body Diagram
for the Traffic Light
Free-Body Diagram
for the Knot
T3  mg
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Solution of Homework 6: The Traffic Light
Forces along the x axis:
Fnet  ma  0 • Why zero?
Fnet, x  T1 cos 37  T2 cos 53  0
T1 cos37  T2 cos53
cos 53
T1 
T2  0.7536T2
cos 37
Forces along the y axis:
Fnet, y  T1 sin 37  T2 sin 53  T3  0 • Why zero?
(0.7536T2 )sin 37  T2 sin 53  122 N
0.4535T2  0.7986T2  122 N
1.2521T2  122 N
T2  97.4363 N
Both values of T1 and T2 are less
than 100 N. So, the cables will
not break.
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T1  0.7536  97.4363 N
T1  73.4280 N
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Friction
 Frictional forces are unavoidable in our daily lives.
 If we were not able to counteract them, they would stop every
moving object and bring every rotating shaft to a halt. About
20% of the gasoline used in a car is needed to counteract
friction.
 On the other hand, if friction were totally absent, we could not
get an automobile to go anywhere, and we could not walk or
ride a bicycle. Nobody can hold a pencil and it would not
write.
 Here we deal with the frictional forces that exist between dry
solid surfaces, either stationary relative to each other (static)
or moving across each other at slow speeds (kinetic).
 The friction is a passive force. When a force is applied to an
object, the friction force arises against it.
→
fs
fk
→
: static frictional force
: kinetic frictional force
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Friction
(a)
• A block rests on a tabletop with the
gravitational force balanced by a normal force
• There is no friction force
FN   Fg
No
motion
(b) (c) (d)
• An increasing force is applied to the block,
attempting to pull it to the left
• The block does not move. An equal and
opposite static frictional force arises between
the object and the surface, exactly balancing
the applied force
F   fs
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Friction
Acceleration
(e)
• Eventually the object will be accelerated
when the applied force reaches a
certain magnitude
• The frictional force that oppose the
motion is now the kinetic frictional force
F  f k  ma
Constant
velocity
(f)
• The magnitude of the kinetic frictional
force, which acts when there is motion,
is less than the maximum magnitude of
the static frictional force, which acts
when there is no motion
• To move the block across the surface
with a constant speed, the magnitude
of the applied force must be decreased
once the block begins to move
F  fk
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Friction
(a) (b) (c) (d)
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(e)
(f)
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Friction
 When two ordinary surfaces are placed
together, only the high points touch each
other.
 Some contact points do cold-welt together.
These welds produce static friction when an
applied force attempts to slide the surface
relative to each other.
 If the applied force is great enough to pull one
surface across the other, there is first a tearing
of welds (at breakaway) and then a
continuous re-forming and tearing of welds as
movement occurs.
 If the two surfaces are pressed together
harder, many more points cold-weld. Sliding
the surfaces relative to each other requires a
greater applied force.
Two surfaces
In Contact
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Properties of Friction
 Experiment shows that
→ when a body presses against a
surface and a force F attempts to slide the body along the
surface, the resulting frictional force has three properties.
 Property 1
→
If the body does not move,
then the static frictional force fs
→
and the component of F that is parallel to the surface balance
each other. They are equal in magnitude, and oppose in
direction.
 Property 2
→
The magnitude of fs has a maximum value fs,max that is given
by
fs,max  s FN
μs : coefficient of static friction
 Property 3
If he body begins to slide along the surface, the magnitude of
the frictional force rapidly decreases to a value fk given by
f k  k FN
μk : coefficient of kinetic friction
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Properties of Friction
 Friction depends on how strongly the surfaces are pressed
together.
→ This pressing strength is represented by the normal
force FN.
 It is easier to keep an object sliding than to get it starts sliding.
 The coefficients μs and μk are dimensionless and must be
determined experimentally.
 The value of μs and μk depends
on certain properties of both the
body and the surface. Therefore,
they are usually referred to with
the preposition “between.”
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Properties of Friction
A block lies on a floor.
(a) What is the magnitude of the frictional force on it from the
floor? Zero
(b) If a horizontal force of 5 N is now applied to the block, but
the block does not move, what is the magnitude of the
frictional force on it? 5 N
(c) If the maximum value of fs,max of the static fictional force on
the block is 10 N, will the block move if the magnitude of the
horizontally applied force is 8 N? No
(d) What about if it is 12 N? Yes
(e) What is the magnitude of the frictional force in part (c)? 8 N
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Example: Jaguar
If a car’s wheels are “locked” (kept
from rolling) during emergency
braking, the car slides along the
road. Ripped-off bits of tire and
small melted sections of road form
the “skid marks.”
The record for the longest skid
marks on a public road was
reportedly set in 1960 by a Jaguar
on the M1 highway in England –
the marks were 290 m long.
Assuming that μk = 0.6 and the
car’s acceleration was constant
during the braking, how fast was
the car going when the wheels
became locked?
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Example: Jaguar
Fnet,x  m ax
 fk  m a
 k mg  m a
a   k g
v 2  v02  2a ( x  x0 )
(0)2  (v0 ) 2  2a( x  x0 )
v0  2a( x  x0 )
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v0  2k g ( x  x0 )
 2(0.6)(9.8)(290)
 58.40 m s
 210.24 km h
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Checkpoint
→
F1=
10 N is applied
to a box on a floor, but the box does not
→
slide. Then, F2 is increased from zero. Before the box begins to
slide, do the following quantities increase, decrease, or stay the
same?
The same, 10 N
(a) The magnitude of the frictional force on the box.
(b) The magnitude of the normal force on the box Decrease
from the floor.
(c) The maximum value fs,max of the static frictional
force on the box.Decrease
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Example: Amber Block
A 2.5 kg block is→initially at rest on a horizontal surface. A→
horizontal force F of magnitude 6 N and a vertical force P are
then applied to the block (see below). The coefficients of friction
for the block and surface are μs = 0.4 and μk = 0.25.
Determine the magnitude→of the frictional force acting on the
block if the magnitude of P is
(a) 8 N
(b) 10 N
(c) 12 N
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Example: Amber Block
s  0.4
→
FN
k  0.25
F
P 8 N
fs,max  0.4  (2.5)(9.8)  8 
 6.6  F
The box does not move
f  fs  F  6 N
→
f
→
Fg
Forces along the y axis:
FN  P  Fg  0
FN  Fg  P
 mg  P
On the verge of sliding, the
maximum static frictional force is:
fs,max  s FN
 s (mg  P)
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P  10 N 
fs,max  0.4  (2.5)(9.8)  10 
 5.8  F
The box moves
f  f k  k FN
 0.25  (2.5)(9.8)  10 
 3.625 N
P  12 N 
The box moves
f  f k  k FN
 0.25  (2.5)(9.8)  12 
 3.125 N
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Final Homework 7: 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.
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