13.1 Newton’s law of motion1.Newton’s 2nd law of motion

(1) A particle subjected to an unbalanced force F

experiences an acceleration a

having the same

direction as F

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’slinear momentum.

F

am

vdt

dm

dt

vdm

vmdt

d

)(

(if m=constant)

2. Newton’s Law of Gravitational Attraction

r

m1 m2221

r

mmGF

G = universal constant of gravitation = 2312 skg/m10x73.66

r = distance between centers of two particles

Weight of a particle with mass m1 = m

22

r

mmGF

)r

Gm(m

22

=mg

m2 : mass of the earth

r = distance between the earth center and the particle

22

r

Gmg=

= acceleration due to gravity

=9.81 2sm measured at a point on the surface of the

earth at sea level and at a latitude of 045

1F

2F

a

p

13-2 The equations of motion

1.Equations of motion of a particle subjected to more than one force.

amFFR

2F

1F

FFR p

amKinetic diagram of particle p.

p

amFFR

………...equation of motion

Free body diagram of particle p.

D’A lembert principle

:

0

am

amFR

inertia force vector

Dynamic equilibrium diagram

RF

amp

( 慣性力 )

0am若 則此狀態為靜平衡

amF

0amF

0F

R

R

+

)0(

/

dt

vdaa

aaa

oop

opop

(1) Inertial frame

pathp

ap

x

y

o

0v

2. Inertial frame of reference (newtonian)A coordinate system is either fixed or translates in a given direction with a constant velocity.

(2) Noninertial frame

a

pathp

p

o

y

x

0a

0/ aaa pop

irif

iFz

y

x

i

Equation of motion of particle i.Dynamic equilibrium diagram of particle i.

iii fF

iiam

13-3 Equation of motion for a system of particle

xyz: Inertial Coordinate System

iiii amfF

i

i

f

F

resultant external force

resultant internal force

n

ijj

ijf1

Equation of motion of a system of particles.

............

......)(......)()(

2211

2211

ii

ii

amamam

fFfFfF

iii

i

iiii

amF

f

amfF

0

By definition of the center of mass for a systemof particles.

iiG rmrm

i

G

mm

r

Position vector of the center of mass G.

Total mass of all particles.

)()(2

2

2

2

iiG rmdt

drm

dt

d

Assume that no mass is entering or leavingthe system.

iiG

ii

G

amam

dt

rdm

dt

rdm

2

2

2

2

Hence：Gi amF

This equation justifies the application of the equation of motion to a body that is representedas a single particle.

amF

RectangularCoordinatesystem.

z

y

x

zF

path

xF yF

Equation of motion of particle P.

In rectangular components

kmajmaimakFjFiF zyxzyx

13-4 Equations of motion ： Rectangular

Coordinate

zz

yy

xx

maF

maF

maF

scalar eqns.

Analysis procedure1. 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.

amF

or

zz

yy

xx

maF

maF

maF

(2) Friction force

NF kf (3) Spring force

ksFs

3. Equations of kinematics Apply vdvads

dt

dsv

dt

dva 、、 for the solutions

Curve path of motion of a particle is known.

tu

nu

bu

=Tangential unit vector

=Normal unit vector

=Binormal unit vector

= tn uu

13.5 Equation of Motion:Normal and Tangential

Coordinates

P

nb

t

Curve

tunu

bu

F

path

amF

Or scalar form

tF

uF

bF

tam

nam

=

=

= 0

ntbbnntt amamuFuFuF

Equation of motion

dtdvat /

2van

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.

Analysis procedure

ds/vdvdt/dva t

2van

3/2

22

2

/

/1

dxyd

dxdy

r

z

zzuF uF

rruF

amF

13.6 Equation of Motion :Cylindrical coordinate

Equation of motion in cylindrical coordinates

zrzzrramamamuFuFuF

zz

rr

maF

maF

maF

za

rra

rra

z

r

2and

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 forceIf 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

dr :radial componentrd :transverse componentds:distance

The directions of F and N can be specified relative to the radial coordinate r by computing the angle . Angle is defined between the extended radial line and the tangent to the path.

dr

rd

ddr

r

dr

rdtan

+ positive direction of

- negative direction of