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Chapter 12
Kinetics of Particles:
Newton’s Second Law
1
Kinetics (จลนศาสตร์)
• เป็ นการศึกษาการเคลื่อนที่ของวัตถุ โดยพิจารณาถึงแรงที่กระทาต่อ
วัตถุน้ นั
• เราใช้ Kinetics ในการทานายการเคลื่อนที่ของวัตถุเมื่อมีแรงกระทา
หรื อใช้ในการคานวณแรงเพื่อให้วตั ถุน้ นั เคลื่อนที่ตามที่ตอ้ งการ
2
Newton’s Second Law of Motion
• ถ้าแรงลัพธ์ที่กระทากับอนุภาคมีค่าไม่เท่ากับศูนย์ อนุภาคจะเคลื่อนที่ดว้ ย
ความเร่ งที่เป็ นสัดส่ วนโดยตรงกับแรงลัพธ์ และมีทิศทางเดียวกับทิศทาง
ของแรงลัพธ์ โดยอัตราส่ วนของแรงกับขนาดของความเร่ งจะมีค่าคงที่
• Consider a particle subjected to constant forces,
F1 F2 F3


   constant  mass, m
a1 a2 a3

• When a particle of mass m is acted upon by a force F ,
the acceleration of the particle must satisfy


F  ma
3
Linear Momentum of a Particle
หากพิจารณา กฎข้อที่สองของนิวตัน เมื่อมวลของวัตถุคงที่


F

m
a



d
v
F = m
dt

d
d
L
= ( m v ) =
dt
dt

L = linear momentum of the particle
• Linear Momentum Conservation Principle:
If the resultant force on a particle is zero, the linear
momentum of the particle remains constant in both
magnitude and direction.
4
Systems of Units
• Of the units for the four primary dimensions (force,
mass, length, and time), three may be chosen arbitrarily.
The fourth must be compatible with Newton’s 2nd Law.
• International System of Units (SI Units): base units are
the units of length (m), mass (kg), and time (second).
The unit of force is derived,
kg  m
 m
1 N  1 kg 1 2   1 2
 s 
s
5
Eqs of Motion in Rectangular Components
• Newton’s second law provides


F

m
a

• Solution for particle motion is facilitated by resolving
vector equation into scalar component equations, e.g.,
for rectangular components,






 Fx i  Fy j  Fz k   ma x i  a y j  a z k 
 Fx  max  Fy  ma y  Fz  maz
 Fx  mx  Fy  my  Fz  mz
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Eqs of Motion in Tangential and
Normal Components
• For tangential and normal components,
 F t  mat
dv
F

m
 t
dt
 F n  man
Fn  m
v2

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Eqs of Motion in Radial & Transverse
Components
• Consider particle at r and , in polar coordinates,
 Fr  mar  mr  r 2 
 F  ma  mr  2r 
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Dynamic Equilibrium
• Alternate expression of Newton’s second law,


F

m
a
0


 ma  inertial vector
• With the inclusion of the inertial vector, the system
of forces acting on the particle is equivalent to
zero. The particle is in dynamic equilibrium.
• Methods developed for particles in static
equilibrium may be applied, e.g., coplanar forces
may be represented with a closed vector polygon.
• Inertia vectors are often called inertial forces as
they measure the resistance that particles offer to
changes in motion, i.e., changes in speed or
direction.
• Inertial forces may be conceptually useful but are
not like the contact and gravitational forces found
in statics.
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Sample Problem 12.1
SOLUTION:
• Resolve the equation of motion for the
block into two rectangular component
equations.
A 200-N block rests on a horizontal
plane. Find the magnitude of the force
P required to give the block an acceleration or 10 m/s2 to the right. The coefficient of kinetic friction between the
block and plane is mk  0.25.
• Unknowns consist of the applied force
P and the normal reaction N from the
plane. The two equations may be
solved for these unknowns.
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Sample Problem 12.1
SOLUTION:
• Resolve the equation of motion for the block
into two rectangular component equations.
+
 Fx  ma :
P cos 30  0.25N  20.3910  203.9
y
O
x
W
200 N

g 9.81 m s 2
 20.39 kg
F  mk N
m
 0.25 N
+
 Fy  0 :
N  P sin 30  200 N  0
• Unknowns consist of the applied force P and
the normal reaction N from the plane. The two
equations may be solved for these unknowns.
N  P sin 30  200 N
P cos 30  0.25P sin 30  200 N   203.9
P  343.11 N
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Sample Problem 12.3
SOLUTION:
• Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
• Write the equations of motion for the
blocks and pulley.
• Combine the kinematic relationships
with the equations of motion to solve for
the accelerations and cord tension.
The two blocks shown start from rest.
The horizontal plane and the pulley are
frictionless, and the pulley is assumed
to be of negligible mass. Determine
the acceleration of each block and the
tension in the cord.
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Sample Problem 12.3
O
x
y
SOLUTION:
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
y B  12 x A
a B  12 a A
• Write equations of motion for blocks and pulley.
+ F m a :

x
A A
T1  100 kg a A
+
 Fy  mB aB :
mB g  T2  mB a B
300 kg 9.81m s 2  T2  300 kg a B
T2  2940N - 300 kg a B
+
 Fy  mC aC  0 :
T2  2T1  0
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Sample Problem 12.3
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
O
x
y
y B  12 x A
a B  12 a A
T1  100 kg a A
T2  2940N - 300 kg a B

 2940N - 300 kg  12 a A

T2  2T1  0
2940 N  150 kg a A  2100 kg a A  0
a A  8.40 m s 2
a B  12 a A  4.20 m s 2
T1  100 kg a A  840 N
T2  2T1  1680 N
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Sample Problem 12.4
SOLUTION:
• The block is constrained to slide down
the wedge. Therefore, their motions are
dependent. Express the acceleration of
block as the acceleration of wedge plus
the acceleration of the block relative to
the wedge.
• Write the equations of motion for the
wedge and block.
The 5.4 kg block B starts from rest and
slides on the 13.6 kg wedge A, which is • Solve for the accelerations.
supported by a horizontal surface.
Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
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Sample Problem 12.4
SOLUTION:
• The block is constrained to slide down the
wedge. Therefore, their motions are dependent.



aB  a A  aB A
• Write equations of motion for wedge and block.
+ F m a :

x
A A
N1 sin 30  mA a A
y
0.5 N1  mA a A
+
x
 Fx  mB a x  mB a A cos 30  aB A :
 mB g sin 30  mB a A cos 30  aB A 
aB A  a A cos 30  g sin 30
+
 Fy  mB a y  mB  a A sin 30 :
N1  mB g cos 30  mB aA sin 30
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Sample Problem 12.4
• Solve for the accelerations.
0.5N1  mAaA , N1  2mAaA
N1  mB g cos 30  mB a A sin 30
2m A a A  mB g cos 30  mB a A sin 30
aA 
aA 
mB g cos 30
2m A  mB sin 30
5.49.81 cos 30
213.6   5.4 sin 30
a A  1.54 m s 2
aB A  a A cos 30  g sin 30
aB A  1.54 cos 30  9.81 sin 30
aB A  6.24 m s 2
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Sample Problem 12.5
SOLUTION:
• Resolve the equation of motion for the
bob into tangential and normal
components.
• Solve the component equations for the
normal and tangential accelerations.
The bob of a 2-m pendulum describes
an arc of a circle in a vertical plane. If
the tension in the cord is 2.5 times the
weight of the bob for the position
shown, find the velocity and acceleration of the bob in that position.
• Solve for the velocity in terms of the
normal acceleration.
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Sample Problem 12.5
SOLUTION:
• Resolve the equation of motion for the bob into
tangential and normal components.
• Solve the component equations for the normal and
tangential accelerations.
+
mg sin 30  mat
 Ft  mat :
at  g sin 30
+
 Fn  man :
at  4.9 m s 2
2.5mg  mg cos 30  man
an  g 2.5  cos 30
an  16.03 m s 2
• Solve for velocity in terms of normal acceleration.
an 
v2

v  an 
2 m 16.03 m s 2 
v  5.66 m s
19
Sample Problem 12.6
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the car
are its weight and a normal reaction
from the road surface.
Determine the rated speed of a
highway curve of radius  = 122 m
banked through an angle  = 18o. The
rated speed of a banked highway curve
is the speed at which a car should
travel if no lateral friction force is to
be exerted at its wheels.
• Resolve the equation of motion for
the car into vertical and normal
components.
• Solve for the vehicle speed.
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Sample Problem 12.6
• Resolve the equation of motion for
the car into vertical and normal
components.
R cos  W  0
 Fy  0 :
W
R
cos
 Fn  man : R sin  
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the
car are its weight and a normal
reaction from the road surface.
W
an
g
W
W v2
sin  
cos
g 
• Solve for the vehicle speed.
v 2  g tan
 9.81122tan18
v  19.72 m / s
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Angular Momentum of a Particle



• H O  r  mV  moment of momentum or the angular
momentum of the particle about O.



• H O is perpendicular to plane containing r and mV



H O  rmV sin 
i
j
k

 rm v
H  x
y
z
O
 mr 2
mv x
mv y
mvz
• Derivative of angular momentum with respect to time,

 
 
 


H O  r  mV  r  mV  V  mV  r  ma

 rF

  MO
• It follows from Newton’s second law that the sum of
the moments about O of the forces acting on the
particle is equal to the rate of change of the angular
momentum of the particle about O.
22
Conservation of Angular Momentum
• When only force acting on particle is directed
toward or away from a fixed point O, the particle
is said to be moving under a central force.
• Since the line of action of the central force passes
through O,  M O  H O  0 and
 

r  mV  H O  constant
• Position vector and motion
 of particle are in a
plane perpendicular to H O .
• Magnitude of angular momentum,
H O  rmV sin   constant
 r0 mV0 sin 0
or
H O  mr 2  constant
HO
angular momentum
 r 2  h 
m
unit mass
23
Newton’s Law of Gravitation
• Gravitational force exerted by the sun on a planet or by
the earth on a satellite is an important example of
gravitational force.
• Newton’s law of universal gravitation - two particles of
mass M and m attract each other with equal and opposite
force directed along the line connecting the particles,
Mm
F G 2
r
G  constant of gravitatio n
 66.73  10
12
m3
kg  s
2
 34.4  10
9
ft 4
lb  s 4
• For particle of mass m on the earth’s surface,
MG
m
ft
W  m 2  mg g  9.81 2  32.2 2
R
s
s
24
Sample Problem 12.8
SOLUTION:
• Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular momentum
at A and B and solve for the velocity at B.
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 18820 km/h from
an altitude of 240 km. Determine the
velocity of the satellite as it reaches it
maximum altitude of 2340 km. The
radius of the earth is 6345 km.
25
Sample Problem 12.8
SOLUTION:
• Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular momentum
at A and B and solve for the velocity at B.
rm v sin   H O  constant
rA m v A  rB m v B
vB  v A
rA
rB

6345  240 km
 18820 km/h 
6345  2340 km
v B  14269 .4 km/h
26