Download 1. Newton`s first law of motion Any system in mechanical equilibrium

§5.1 Newton’s first law of motion and the
concept of force
1. Newton’s first law of motion
Any system in mechanical equilibrium
remains in mechanical equilibrium unless
compelled to change that state by a nonzero
total force acting on the system.
1system—a particle or particles whose motion
is to be studied.
2 Mechanical equilibrium—a system with a
constant velocity is in equilibrium.
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§5.1 Newton’s first law of motion and the
concept of force
3 The causes of acceleration of a system--The
first law of motion states that zero total force
exists when the system has a constant velocity;
a nonzero total force on the system causes its
acceleration.
r
a≠0 ⇔
r
Ftotal ≠ 0
4Inertial reference frames—the reference
frames in which Newton’s first law is valid
§5.1 Newton’s first law of motion and the
concept of force
2. The concept of force
1force is a vector quantity—the vector
r sum of
all the forces on a system is the force Ftotal on
the system.
2 The measurement of force and the unit of
force-- N (Newton) (§5.4)
3. The fundamental forces of nature
Gravitational force
Electromagnetic force
Strong force
Weak force
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§5.1 Newton’s first law of motion and the
concept of force
4. Some common forces
1 Normal force of a surface. (§5.10)
2 Tensions in ropes, strings, and cables;(§5.11)
3 Static friction and kinetic friction; (§5.12~5.13)
§5.2 Newton’s second law and third law
of motion
§5.2 Newton’s second law of motion and
Newton’s third law of motion
1. Newton’s second law of motion and its aspects
⎧ Fx ,total = ma x
r
r ⎧F
= ma n
r ⎪
Ftotal = m a ⇒ ⎨ F y ,total = ma y or F = ⎨ n ,total
⎩ Ft ,total = ma t
⎪F
ma
=
z
P176-177 ⎩ z ,total
A force of magnitude 1 N on the standard
kilogram produces an acceleration of
magnitude 1m/s2.
1 N=1 kg·m/s2
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§5.2 Newton’s second law and third law
of motion
2. Newton’s third law of motion
If a system A exerts a force on another system
B, then B exerts a force of the same magnitude
on A but in the opposite direction.
r
r
FA on B = − FB on A
A third law force pair are equal magnitude
and opposite direction , but act on different
systems.
§5.3 the limitation to applying Newton’s
law of motion
§5.3 The limitation to applying Newton’s
law of motion
1. Reference frames
Newton’s law can only be used in inertial
reference frames (that are not being
accelerated).
2. Speed limits
Newtonian mechanics is a good approximation
as long as the speed of the system is much less
than the speed of light.
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§5.3 the limitation to applying Newton’s
law of motion
3. Quantum mechanics
Newtonian mechanics can not describe or
account for many phenomena on the atomic
and nuclear scale.
4. Force propagation
Force somehow take time to propagate from
one place to another. Newton’s third law
does not account for such propagation
delays.
§5.3 the limitation to applying Newton’s
law of motion
5. Chaotic—nonlinear system
One of the hallmarks of Newton’s laws is their
ability to predict the future behavior of a
system,if we know the forces that act and the
initial motion.
One of the particular theme of chaotic
dynamics is that tiny changes in the initial
conditions of a problem can be greatly
amplified and can cause substantial differences
in the predicted outcomes.
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§5.3 the limitation to applying Newton’s
law of motion
The trajectory of the Cassini mission to Saturn
§5.4 Some topics of discussion
1. Do inertial frames really exist?
Inertial reference frame is an ideal model.
The earth spins on its axis. The centripetal
acceleration is less than 3.4×10-2m/s2.
The surface of the earth is a approximate
inertial reference frame.
The earth moves around the sun. The
centripetal acceleration is 6×10-3m/s2.
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§5.4 Some topics of discussion
2. Noninertial reference frames
N
Alice
r
a0
Bob
?
A
mg
Newton’s law is not valid in the car.
§5.4 Some topics of discussion
A
B
N
m
mg
B
r
a0 N
r
a0
m
mg
A
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§5.4 Some topics of discussion
r
τr
F
B
A
r
s m
n
ω
ω
Noninertial reference
frames are accelerated
reference frames.
r
r
O
F
r
F0
r
G
§5.4 Some topics of discussion
3. Applying Newton’s second law of motion in
noninertial reference frames
r r
r = R + r′
r
r′
r
r
r
dr dR dr ′
=
+
dt dt dt
O’
r
R
r
r
r
vPO = vPO′ + vO′O
r
r
r
aPO = aPO′ + aO′O
r r r
a = a ′ + a0
P
r
r
O
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§5.4 Some topics of discussion
r
r
r
r
Ftotal = m a = m a ′ + m a0
r
r
r
Ftotal + ( − m a0 ) = m a ′
r
r
dictate Fpseudo = − m a0
called pseudo force.
The pseudo force arise only because of the
acceleration of the noninertial reference frame.
Without the pseudo force term, the accelerated
frame cannot properly describe the motion of
the particle using Newton’s law.
§5.5 applications of Newton’s law
How to apply Newton’s laws of motion
1identifies the system;
2 draw the free-body diagrams and illustrates
all forces acting on the system with their
direction indicated explicitly in the diagram;
3choose a coordinate system;
4make appropriate vector sum of all the
forces and describe the motion by using
Newton’s second law of motion.
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§5.5 applications of Newton’s law
Example 1: A submarine is
sinking in ther sea. If the
buoyancy is
, the kinetic
r F
r
friction is f = − kAv , where A
is the area of the cross
section of the submarine.
Find the function of the
speed of the submarine with
respect to the time.
r
F
c
r
W
r
f
o
y
§5.5 applications of Newton’s law
Solution: F y , total = − F − kAv + mg = ma y
mg − F − kAv = m
dv
dt
v
t
∫
m dv
=
mg − F − kAv
v
−
0
∫
0
−
∫ dt
0
m
d( mg-F-kAv
kA
mg − F − kAv
m
ln( mg-F-kAv
kA
)
t
=
∫ dt
0
)
v
0
= t
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§5.5 applications of Newton’s law
−
v
vm
t
mg − F − kAv
mg − F
= e
= t
−
kA
t
m
kA
−
mg − F ⎛
⎜1 − e m
v =
⎜
kA
⎝
o
t = 0 v = 0,
m
mg-F-kAv
ln
kA
mg − F
t↑ v↑
t → ∞ v = vmax =
t
⎞
⎟
⎟
⎠
dv
↓,
dt
mg − F
= constant --terminal speed
kA
§5.5 applications of Newton’s law
Example 2: Block m1 in the Figure has a mass of
4.20kg and block m2 has a mass of 2.30kg. The
coefficient of kinetic friction between m2 and the
horizontal plane is 0.47. The inclined plane is
frictionless. Find (a) the acceleration of the
blocks and (b) the tension in the string.
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§5.5 applications of Newton’s law
y
x
r
N1
r
T1
y
x
r
T2
θ
θr
r
W1 = m1 g
⎧m1 g sin θ − T = m1a
⎨
⎩ N 1 − m1 g cosθ = 0
r
N2
r
fk
r
r
W2 = m 2 g
r
r
T1 = T2 = T
fk = µk N 2
a1 x = a 2 x = a
θ = 27 o
⎧T − µ k N 2 = m 2 a
⎨
⎩ N 2 − m2 g = 0
§5.5 applications of Newton’s law
Exercise 1 : As shown in figure,
the block B weighs 94kg and
block A weighs 29 kg. Between
block B and the plane the
coefficient of static friction is
0.56 and the coefficient of the
kinetic friction is 0.25. (a) find
the acceleration if B is moving
up the plane. (b) what is the
acceleration if B is moving
down the plane?
B
42o
A
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§5.5 applications of Newton’s law
Exercise 2: in the figure, A is a 4.4 kg block and
B is a 2.6 kg block. The coefficients of static and
kinetic friction between A and the table are 0.18
and 0.15. (a) Determine the minimum mass of
the block C that must be placed on A to keep it
from sliding. (b) Block C is suddenly lifted off A.
What is the acceleration of block A?
C
A
B
§5.5 applications of Newton’s law
Exerscise 3: an ideal pendulum with a
pear of mass m and an inflexible,
massless thread of length l. Keep the
pendulum at horizontal initially, and
then drop it from rest. When the angle
between the thread and the horizontal
is θ , find the speed of the pear and the
tension in the thread.
O
θ
l
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§5.5 applications of Newton’s law
Solution:
mg cos α = maτ = m
θ
O
dυ
dt
α
dα
r
T
l
ds
dυ
ds = m dυ
mg cos αds = m
dt
dt
d s = ld α ,
∫
θ
0
ds
=υ
dt
r
v
r
mg
g cos αdα = υdυ
υθ
g cosαdα = ∫ υdυ
0
T − mg sinθ = man = m
ds
υθ2
l
υθ = 2gl sin θ
Tθ = 3mg sin θ
§5.5 applications of Newton’s law
Example 3: A block of mass m is placed on a
frictionless inclined plane of a triangle block of
mass M which is on the frictionless horizontal
plane , the inclined plane is at the angle θ to the
horizontal. Find (a)the magnitude of the normal
force of the inclined plane on the block;(b)the
acceleration of m with respect to M.
m
θ
M
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§5.5 applications of Newton’s law
Solution:
Draw the forces diagram
Choose triangle block as the
reference frame
y
r
a′
x
r
N
θ
Fx = mgsinθ = ma′
m
r
r
W = mg
Fy = N − mgcosθ = 0
The results:
N = mg cosθ
a ′ = gsinθ
Are these results right?
§5.5 applications of Newton’s law
Choose the ground as the reference frame:
For the triangle block
r
Q
y
x
θ
θ
M
r
W
r
aM
r
r
N′ = −N
F x = N sin θ = Ma M
(1 )
F y = Q − Mg − N cos θ = 0
(2)
a M ≠ 0 , M is not inertial reference frame.
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§5.5 applications of Newton’s law
Choose the ground as the reference frame:
For block of mass m
y
x
r
N
r
a′
m
r
am r
r
aM
W
r
r
r
a mG = a mM + a MG
r
r r
am = a ′ + a M
amx = a′ − a M cosθ
amy = − a M sin θ
Fx = mgsinθ = mamx = m(a′ − aM cosθ )
(3)
Fy = N − mgcosθ = mamy = −maM sinθ
(4)
§5.5 applications of Newton’s law
M:
m:
Fx = N sinθ = MaM
(1)
Fy = Q − Mg − N cosθ = 0 (2)
Fx = mgsinθ = mamx = m(a′ − aM cosθ )
(3)
Fy = N − mgcosθ = mamy = −maM sinθ
(4)
F x = N sin θ = Ma M
Fx = mg sinθ = mamx = m(a′ − aM cosθ )
Fy = N − mg cosθ = mamy = −maM sinθ
(1 )
(3)
(4)
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§5.5 applications of Newton’s law
The results are
Mmg cosθ
M + m sin 2 θ
mg cosθ sinθ
aM =
M + m sin 2 θ
(M + m )g sinθ
a′ =
M + m sin 2 θ
r
N
y
N=
x
r
a′
m r
aM
r
am r
W
§5.5 applications of Newton’s law
N=
(M + m )g sinθ
Mmg cosθ
mg cosθ sinθ
; aM =
; a′ =
2
2
M + m sin θ
M + m sin θ
M + m sin 2 θ
θ = 0
aM
m
θ = π
0
m
M → ∞
2
M 0
0
m
M
a′
N
0
g
g sin θ
mg
0
mg cos θ
M
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§5.5 applications of Newton’s law
Choose horizontal plane as reference frame:
r
Q
y
x
r
aM
M
θ
v
r r
′
N
=
−
N
W
For triangle block of mass M
Fx = N sin θ = Ma M
(1)
F y = Q − Mg − N cos θ = 0
( 2)
§5.5 applications of Newton’s law
Choose M as the reference frame:
y
r x
r
F0 = − m a M
r
a′
r
N
m
r
W
For block of mass m
Fx = mg sin θ + ma M cosθ = m a ′
( 3)
F y = N − mg cos θ + ma M sin θ = 0
( 4)
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§5.5 applications of Newton’s law
A
B
m2 m3
m1
Exercise 4: There are three
blocks of mass m1, m2, m3
respectively, and m1>m2+m3 .
Pulleys and strings are
massless, the kinetic friction
forces between the pulley and
the strings are ignored. (a)
find the accelerations of m1,
m2 and m3. (b) what are the
tensions of the two strings.
§5.5 applications of Newton’s law
r
T1
Solution:
(a) The
free-body
diagrams
r
T1
r
T2
r
a1
r
a1
B
r
T3
r
T3
r
T2
m1
r
m1 g
m2
r
a 2′
r
m 2 a1
r
m2 g
m3
r
a 3′
r
m 3 a1
r
m3 g
+
m1 g − T1 = m1a1
m 2 g + m 2 a1 − T2 = m 2 a 2′
m 3 g + m 3 a1 − T3 = − m 3 a ′3
T1 = T2 + T3
T2 = T3
a ′2 = a′3
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§5.5 applications of Newton’s law
(b)
r
T1
r
T2
r
a1
B
r
T3
r
r
r
r
r r
or
a 2 = a 2′ + a1
r
a
If the direction of 2 is downward
then a 2 = a ′2 − a1
(1)
r
r r
because a 3 = a 3′ + a1
r
Assume a 3 is upward
then − a3 = −a′3 − a1 (2)
because a m G = a m
′ 2p + apG
2
From equations (1) and (2) ,we can get a2 an a3.
§5.5 applications of Newton’s law
Exercise 5:A small metal tube is
sliding down along the string
with the acceleration a with
respect to the string as shown in
Figure. The mass of the block is
m1, and the mass of the tube is
m2, the string is massless, the
kinetic friction force between the
pulley and the string is ignored.
Find the accelerations of the
block and the tube with respect
to the ground, the tension of the
string and the friction force
between the string and the tube.
A
o
m2
m1
The string is
not a inertial
reference
frame!
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