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13.1 Newton’s law of motion
1.Newton’s 2nd law of motion
(1) A particle subjected to an unbalanced force
experiences an acceleration
direction as

F

a

F
having the same
and a magnitude that is directly
proportional to the force.

F

=m a
m = mass of a particle
=a quantitative measure of the resistance of the
particle to a change in its velocity.
(2) The unbalanced force

F
acting on the particle is
proportional to the time rate of change of the particle’s
linear momentum.
 d 
F  ( mv )
dt

dv dm 
m 
v
dt dt

 ma (if m=constant)
2. Newton’s Law of Gravitational Attraction
m1 m 2
FG 2
r
r
m1
m2
G = universal constant of gravitation
= 66.73x10 12 m3 / kg  s 2
r = distance between centers of two particles
Weight of a particle with mass m1 = m
mm 2
FG 2
r
Gm 2
 m( 2 )
r
=mg
m2 : mass of the earth
r = distance between the earth center and the particle
g=
Gm 2
2
r
= acceleration due to gravity
m
=9.81 s 2 measured at a point on the surface of the
earth at sea level and at a latitude of 450
13-2 The equations of motion
1. Equations of motion of a particle subjected
to more than one force.


FR   F  ma

a

F1

F2
p
Free body diagram of particle p.

F1
 p

FR   F

F2
Kinetic diagram of particle p.

ma
p


 FR   F  ma ………...equation of motion
D’A lembert principle


FR  ma  0

 ma : inertia force vector
Dynamic equilibrium diagram

FR
p

ma
(慣性力)

+
 F  0 

 FR  ma  0


FR  ma

若 ma  0則此狀態為靜平衡
2. Inertial frame of reference (newtonian)
A coordinate system is either fixed or translates in a
given direction with a constant velocity.
(1) Inertial frame

ap
y

p v
0
path
x
o



a p / o  a p  ao

dvo
 
 a p ( ao 
 0)
dt
(2) Noninertial frame

ap
y
o
p

a0
x
path

 
a p / o  a p  a0
13-3 Equation of motion for a system of particle

Fi
z

fi

ri
i
y
xyz: Inertial Coordinate System
x
Equation of motion of particle i.
Dynamic equilibrium diagram of particle i.
 
Fi  f i
i

mi ai
 

 Fi  f i  mi ai

Fi  resultant external force
n 

f i  resultant internal force   f ij
j 1
j i
Equation of motion of a system of particles.
 
 
 
( F1  f1 )  ( F2  f 2 )  ......  ( Fi  f i )  ......



 m1a1  m2 a2  ......  mi ai  ......


 Fi   fi   mi ai
 fi  0
  Fi   mi ai
By definition of the center of mass for a system
of particles.


mrG   mi ri

rG  Position vector of the center of mass G.
m   mi  Total mass of all particles.
d2
d2


 2 (mrG )  2 ( mi ri )
dt
dt
Assume that no mass is entering or leaving
the system.

2
d rG
d ri
 m 2   mi 2
dt
dt


maG   mi ai
2
Hence：


F

m
a
 i
G
This equation justifies the application of the
equation of motion to a body that is represented
as a single particle.
13-4 Equations of motion：Rectangular
Coordinate

Fz
z

Fx
x
path

Fy
y
Rectangular
Coordinate
system.
Equation of motion of particle P.


 F  ma
In rectangular components






 Fxi   Fy j   Fz k  maxi  may j  maz k
F
F
F
x
 max
y
 may
z
 maz
scalar eqns.
Analysis procedure
1. Free Body Diagram.
(1) Select the proper inertial coordinate system.
(2) Draw the particle’s F.B.D.
2. Equation of motion
(1) Apply the equations of motion in scalar form
or vector form.
 Fx  max


 F  ma
or
F
F
y
 may
z
 maz
(2) Friction force
Ff   k N
(3) Spring force
Fs  ks
3. Equations of kinematics
dv
ds
Apply a  、v  、ads  vdv for the solutions
dt
dt
13.5 Equation of Motion:Normal and Tangential
Coordinates
Curve path of motion of a particle is known.
Curve path
b
n
un

t
ub
ut
P
F
ut =Tangential unit vector
u n =Normal unit vector
u b =Binormal unit vector
= un  ut
 F  ma
Equation of motion





 Ftut   Fnun   Fbub  mat  man
Or scalar form
 Ft = mat
F
 Fb
u
= m an
=
0
at  dv / dt
an 
v2

Analysis procedure
1. Free body diagram
Identify the unknowns in the problem.
2. Equation of motion
Apply the equations of motion using
normal and tangential coordinates.
3. Kinematics
Formulate the tangential and normal
components of acceleration.
v2
an 
a t  dv / dt  vdv / ds

2 2/3

1  dy / dx  
 
d 2 y / dx 2
13.6 Equation of Motion :Cylindrical coordinate
F u
z
z

z

F u
F u
r

r
r
Equation of motion in cylindrical coordinates
 F  ma






 Fr u r   Fu    Fz u z  ma r  ma   ma z
 F  ma
 F   ma
 F  ma
r
z

ar  r  r 
r
z


and
 
a  r   2 r 

az  z
Cylindrical or polar coordinates are suitable for a problem for
which Data regarding the angular motion of the radial line r are
given, or in Cases where the path can be conveniently expressed
in terms of these coordinates.
Normal and Tangential force
If the particle’s accelerated motion is not completely specified,
then information regarding the directions or magnitudes of the
forces acting on the particle must be known or computed.
Now, consider the case in which the force P causes the particle
to move along the path r=f() as shown in the following figure.
r=f() :path of motion of particle
P:External force on the particle
F:Friction force along the tangent
N:Normal force perpendicular to
tangent of path
Direction of F & N
The directions of F and N can be specified relative to the
radial coordinate r by computing the angle y.
Angle y is defined between the extended radial line and
the tangent to the path.
dr :radial component
rd :transverse component
ds:distance
rd

dr
rd
r
tan y 

dr dr d
y
+ positive direction of 
- negative direction of 