Download 6 Force and Motion II

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„
Force and Motion II
In this chapter we focus on the physics of three
common forces: frictional (摩擦) force, drag (拖曳)
force, and centripetal (向心) force.
When will they occur?
Frictional force: relative motion with a rough
and rigid surface.
Drag force: relative motion with fluid (air,
water..)
Centripetal force: turning
* Each type of force obeys a special force formula.
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(定性分析)
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Disadvantages:
Frictional
forces
are
unavoidable in our daily lives. If we were not
able to counteract (對抗) them, they would stop
every moving object. About 20% of the
gasoline used in an automobile is needed to
counteract friction in the engine.
Advantages: If friction were totally absent, we
could not get an automobile to go anywhere,
and we could not walk or ride a bicycle. We
could not hold a pencil easily, and, if we could,
it would not write. Nails and screws would be
useless.
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[Fig. 6-1a] A block rests on a
tabletop, gravitational force Fg
balance normal force FN.
„
[Fig. 6-1b] You exert a force F
on the block, attempting to pull
it to the left but not really move.
In response, a force directed to
the right must exist, called the
static frictional force fs, to
exactly balance your force F.
The block does not move.
F = fs
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[Figures 6-1c & 6-1d]
As F ↑, fs ↑ and the block remains at rest (still
∑F=0). When F reaches a certain magnitude
(fs,max, 最大靜摩擦力),
最大靜摩擦力 however, the block “breaks
away” from ( 從 ... 掙 脫 ) its intimate ( 緊 密 の )
contact with the tabletop and accelerates leftward
[Fig. 6-1d→e].
The frictional force that opposes the motion
becomes so-called the kinetic frictional force fk
[Fig. 6-1e].
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„
Usually, fk<fs,max. If you wish
the block to move across the
surface with a constant speed,
you must usually decrease the
magnitude of F once the block
begins to move [Fig. 6-1f].
fs,max
fs
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A frictional force is the vector sum
of many forces acting between the
surface atoms & molecules of one
body and those of another body.
* When two ordinary surfaces are placed together,
only the high points touch each other. The actual
microscopic area of contact is much less than the
apparent macroscopic contact area, perhaps by a
factor of 104.
* Nonetheless (但是), many contact points do coldweld (冷焊接) together (微觀上，表面分子與分子
間交互作用力引起). These welds produce static
friction when an applied force attempts to slide
the surfaces.
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* If the applied force is great enough to pull one
surface across the other, there is first a tearing (撕
開) of welds [at breakaway (掙脫點)] and then a
continuous re-forming ( 再 形 成 ) and tearing of
welds as movement occurs.
* The kinetic frictional force fk
that opposes the motion is
the vector sum of the forces
at those many chance
contacts.
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* Sliding motion of one surface over another is “jerky”
(顛簸) because the two surfaces alternately (交替
地 ) stick ( 黏 住 ) together and then slip. Such
repetitive stick-and-slip can produce squeaking(吱
吱聲) or squealing(長而尖的聲音), as when tires
skid (打滑) on dry pavement, fingernails scratch
along a chalkboard.
* If the two surfaces are pressed together harder (FN
↑), many more points cold-weld. Now getting the
surfaces to slide requires a greater applied force:
(1) fs has a greater maximum value.
(2) Once the surfaces are sliding, there are many
more points of momentary (短暫) cold-welding, so
the kinetic frictional force fk also has a greater
magnitude.
¨ fs,max & fk both depend on FN.
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(定量分析)
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Experiment shows that when a dry body presses
against a surface and a force F attempts to slide
the body along the surface, the resulting frictional
force has 3 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 fs is directed opposite
that component of F.
F
f
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Property 2. The magnitude of fs has a maximum value
fs,max that is given by
where μs is the coefficient of static friction and FN is
the normal force on the body from the surface.
Property 3. If the body begins to slide along the
surface, the magnitude of the frictional force
rapidly decreases to a value fk given by
where μk is the coefficient of kinetic friction.
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The direction of fs or fk is always parallel to the
surface and opposed to the attempted sliding, and
the normal force FN is perpendicular to the surface.
The coefficients μs and μk are dimensionless and
must be determined experimentally. Their values
depend on certain properties (ex: roughness,
material) of both the body and the surface.
F
f
摩擦力~兩表面分子之間吸引力~彈簧模型: f與F反向
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A car sliding to the right and finally stopping after
a displacement of 290m. Assuming that μk=0.60
and the car’s acceleration was constant (a=const.)
during the braking, how fast was the car going
when the wheels became locked?
開始剎車
a=constant
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Free-body diagram
∑Fx=max ¨
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雪橇
v=const. a=0
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∑F=ma=0 ¨
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Free-body diagram
∑Fx=max=0 ¨
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∑Fy=may=0 ¨
Combining Eqs. (6-6) with (6-7), we can solve
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By Eq. (6-7), we know
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瀕於
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Why?
Because the coin
“attempts” to slide
down the book.
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∑F=ma=0 ¨
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Free-body diagram
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∑Fx=max=0 ¨
∑Fy=may=0 ¨
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A fluid is anything that can flow — generally
either a gas or a liquid. When there is a relative
velocity between a fluid and a body, the body
experiences a drag force D that opposes the
relative motion and points in the direction in which
the fluid flows relative to the body.
Here we examine only cases in which air is the
fluid. In such cases, the magnitude of the drag
force is related to the relative speed v by an
experimentally determined drag coefficient C
according to
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where ρ is the air density (mass per volume) and A
is the effective cross-sectional area of the body
(the area of a cross section taken perpendicular to
v ). The typical values range of the drag coefficient
C is from 0.4 to 1.0.
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Skiers know well that drag
depends on A and v2. To reach
high speeds a skier must reduce
D as much as possible by, for
example, riding the skis in the
“egg position” to minimize A.
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When a body falls from rest through
air, the drag force D is directed
upward; its magnitude gradually
increases as the speed of the body
increases.
We can relate these forces to the
body’s
acceleration
by
writing
Newton’s 2nd law for a vertical y axis
(Fnet,y=may) as,
D ∝ v2
a
Fg
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D=½CρAv2
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„
(6-14)
As suggested in Fig. 6-7, if the body falls long
enough, D=Fg eventually. From Eq. 6-15, 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 (終端速率).
To find vt, we set a=0 in Eq. 6-15 and substitute
for D from Eq. 6-14, obtaining
which gives
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v=vt
time
Fig. 6-7
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鉛球
高空跳傘者
m越小，vt越小
A越大，D越大，vt越小 (Fig. 6-8)
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raindrop
ground
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raindrop
y=h
ground y=0
(y-y0)
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„
Recall that when a body moves in a circle at
constant speed v, it is said to be in uniform
circular motion. Also recall that the body has a
centripetal acceleration of constant magnitude
given by
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Ex
1. Rounding a curve (car)
The centripetal force is a frictional force on the
tires from the road. People inside the car…
2. Orbiting Earth (moon)
The centripetal forces are gravitational pulls
exerted by Earth and directed radially inward,
toward the center of Earth.
3. A hockey puck moving around in a circle at
constant speed v while tied to a string looped
around a central peg. The centripetal force is the
radially inward pull on the puck from the string.
Without that force, the puck would slide off in a
straight line instead of moving in a circle.
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Ex. 1
Ex. 2
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Note again that a centripetal force
kind of force. The name merely
direction of the force. It can, in fact,
force, a gravitational force, the force
or any other force.
Ex. 3
is not a new
indicates the
be a frictional
from a string,
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For any situation:
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From Newton’s second law and Eq. 6-17 (a=v2/R),
we can write the magnitude F of a centripetal force
as
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驚人表演
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∑Fy=may ¨
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帆布
y
r
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∑Fy=may ¨
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∑Fr=mar
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FL
k
c
a
r
t
flat
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(注意: 在徑向無滑動)
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From Newton’s second law
∑Fr=mar
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∑Fy=may=0
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FL,90 = FL,28.6(90/28.6)2 = 663.7(90/28.6)2 = 6572 N
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FN
R
θ
Fg
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∑Fr=mar ¨
∑Fy=may ¨
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Homeworks
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Ans: (a) 11 N (b) 0.14 m/s2
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Ans: 0.53
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(a) 6 N (b) 3.6 N (c) 3.1 N
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Homeworks
(a) 12.1 m/s (b) 19.4 m/s
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Ans: (a)
(b)
(c)
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Ans: (a) 8.6 N (b) 46 N (c) 39 N
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Ans: 1.0×102 N
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Ans: 8.5 N
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Ans: 0.37
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Ans: (a) 147 N
(b) As remarked at the end of our solution to part (a), the result does
not depend on the frictional parameters. The answer here is the same
as in part (a).
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Ans: mB = 3.3 kg
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Ans:
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Ans: (a) 66 N (b) 2.3 m/s2
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Ans: 488 N
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Ans: (a) -6.1 m/s2 i (b) -0.98 m/s2 i
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Ans:
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Ans:
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Ans:
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Ans:
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Ans: (a) 0.4 N (b) 1.9 s
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Ans: (a) 7.5 m/s2
(b) down the slop
(c) 9.5 m/s2
(d) down the slop
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Ans:
(a)
(b) 55o (d) 59o
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Ans:
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Ans: (a) 27 N (b) 3.0 m/s2
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Ans: 9.4 N
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Ans:
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Ans: (a) 1.05 N
(b) 3.62 m/s2
(c) Reversing the blocks is equivalent to switching the labels. We see from our
algebraic result in part (a) that this gives a negative value for T. Thus, the
situation is as it was before except that the rod is now in a state of compression.
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Ans: 9.4 N
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Ans: 118 N
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Ans: (a)
(b)
(c)
(d)
(e)
(f)
(g) The block moves
up the wall in case
(d) where a > 0
(h) The block moves
down the wall in
case (f) where a < 0
(i) The frictional force
is directed down in
cases (a), (c) and (d).
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Ans: 20￮
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Ans: (a) 2.2 m/s2
(b) 53 N
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Ans: (a) 0.34 (b) 0.24
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Ans: 178 km/h
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Ans: 3.4 m/s2
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Ans: (a) 2.1m/s2
(b) down
(c) 3.9m
(d) stays at that location
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Ans: (b) 240 N (c) 60%
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Ans: (a) 74 N
(b)
(c) 23°
(d) 70 N
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