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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 30 "Classical Mechanics with \+
Maple" }}{PARA 256 "" 0 "" {TEXT -1 36 "Section 3.1: Newton's Laws of \+
Motion" }}{PARA 19 "" 0 "" {TEXT -1 41 "Dr. Harald Kammerer\nmaple@jad
emountain.de" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Initialisation" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 57 "libname:=\"C:/mylib/m6dynlib\",\"C:/mylib
/m6dynfig\",libname:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "wit
h(linalg):with(plots):with(plottools):with(dynamics);with(figures_chap
ter_3);" }}}}{SECT 0 {PARA 3 "" 0 "3.1" {TEXT -1 17 "3.1 Newton's Laws
" }}{PARA 0 "" 0 "" {TEXT 265 18 "Newton's first law" }{TEXT -1 312 " \+
of motion states that a particle's velocity will not change unless a f
orce is applied to the particle. This means that a body that is at a r
est doesn't move until a force is applied. A moving body continues its
 motion with constant velocity unless a force is applied. This first l
aw is a qualitative statement. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "newton 2" {TEXT 266 19 "Newton's second law" }{TEXT -1 
113 " gets into a quantitative relationship between force and motion. \+
In the form of an equation, the second law reads" }}{PARA 257 "" 0 "" 
{TEXT -1 0 "" }{TEXT 260 1 "F" }{TEXT -1 3 " = " }{TEXT 262 1 "m" }
{TEXT -1 1 " " }{TEXT 261 1 "a" }}{PARA 0 "" 0 "" {TEXT -1 29 "with th
e acceleration vector " }{TEXT 256 1 "a" }{TEXT -1 11 ", the mass " }
{TEXT 257 1 "m" }{TEXT -1 56 " and the resultant of all acting forces \+
on the particle " }{TEXT 258 1 "F" }{TEXT 263 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 13 "According to " }{TEXT 270 10 "D'Alembert" }{TEXT -1 21 " \+
we can write this as" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 271 1 "
F" }{TEXT -1 3 " - " }{TEXT 273 1 "m" }{TEXT -1 1 " " }{TEXT 272 5 "a \+
= 0" }}{PARA 0 "" 0 "" {TEXT -1 10 "The term -" }{TEXT 274 1 "m" }
{TEXT -1 0 "" }{TEXT 275 1 "a" }{TEXT -1 41 " can be regarded as a fic
tive force, the " }{TEXT 276 14 "inertial force" }{TEXT -1 181 ". This
 inertial force acts in the opposite direction as the acceleration. Th
en this second law takes the same shape as the equilibrium conditions \+
in the solution of static problems." }}{PARA 0 "" 0 "" {TEXT -1 104 "T
he kilogram [kg] is used as the unit of mass. The unit of length and s
o the displacement is the meter [" }{XPPEDIT 18 0 "m" "6#%\"mG" }
{TEXT -1 248 "]. Velocity is calculated as displacement divided by tim
e and so its units are meters per second [m/s]. At last acceleration i
s calculated as velocity divided by time and so its units are meters p
er second per second, or meters per second squared [" }{XPPEDIT 18 0 "
m/s**2" "6#*&%\"mG\"\"\"*$%\"sG\"\"#!\"\"" }{TEXT -1 105 "]. The force
 sufficient to accelerate one kilogram by one meter per second squared
 is called one Newton [" }{XPPEDIT 18 0 "N" "6#%\"NG" }{TEXT -1 2 "].
" }}{PARA 0 "" 0 "" {TEXT -1 43 "Verbally spoken, Newton's second law \+
says: " }{TEXT 259 86 "\"the resultant of all forces on a particle is \+
in equilibrium with the inertial force\"." }}{PARA 0 "" 0 "" {TEXT -1 
161 "To solve practical exercises it is usual to write three coordinat
e equations instead of the one vector equation. So we write for exampl
e in cartesian coordinates" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }
{XPPEDIT 18 0 "Fx=m*xdd:" "6#/%#FxG*&%\"mG\"\"\"%$xddGF'" }}{PARA 257 
"" 0 "" {XPPEDIT 18 0 "Fy=m*ydd:" "6#/%#FyG*&%\"mG\"\"\"%$yddGF'" }}
{PARA 257 "" 0 "" {XPPEDIT 18 0 "Fz=m*zdd:" "6#/%#FzG*&%\"mG\"\"\"%$zd
dGF'" }}{PARA 0 "" 0 "" {TEXT -1 8 "(As in  " }{HYPERLNK 17 "section 2
," 1 "sec-02.mws" "" }{TEXT -1 210 " the derivative of a variable with
 respect to time is ususly denoted by a dot above the variable. Here i
t is denoted by an appended letter d. So we write for the coordinates \+
of the acceleration in x-direction: " }{TEXT 264 3 "xdd" }{TEXT -1 58 
". Same is valid for the components in y- and z-direction.)" }}{PARA 
0 "" 0 "" {TEXT -1 64 "Desscriptions in polar or natural coordinates a
re also possible." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 267 18 "Newton's third law" }{TEXT -1 249 " addresses the nature
 of forces.  This third law states that the force resulting from the i
nteraction of two bodies acts with equal magnitude on both of them and
 in the opposite directions. For every action, there is an equal and o
pposite reaction. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "pa
ssive forces" {TEXT -1 109 "Forces that act on the particle due to geo
metrical boundaries that restrict the particle's motion are called " }
{TEXT 268 14 "passive forces" }{TEXT -1 1 "." }}{PARA 0 "" 0 "active f
orces" {TEXT -1 64 "Forces  that are completely described by a force l
aw are called " }{TEXT 269 13 "active forces" }{TEXT -1 98 ". Such for
ces are for example the weight force, the spring force and the sliding
 (friction) force." }}{SECT 0 {PARA 0 "" 0 "Example 1" {TEXT 329 34 "E
xample: The Mathematical Pendulum" }}{PARA 0 "" 0 "" {TEXT -1 76 "In t
his example we will analyse the motion of a mass particle with the mas
s " }{TEXT 278 1 "m" }{TEXT -1 52 " that hangs by a non-stretchable th
read with length " }{TEXT 277 1 "L" }{TEXT -1 199 ". The motion of the
 mass particle is forced on a circlular trajectory around the suspensi
on point. Here it makes sense to describe the motion with polar coordi
nates. In Fig. 1 the situation is shown." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 66 "display(Fig_3_1(),scaling=constrained,axes=none,title
=\"Figure 1\");" }}}{PARA 0 "" 0 "" {TEXT -1 102 "Friction and the mas
s of the thread are neglected. The motion is compledly describable wit
h the angle " }{TEXT 279 1 "f" }{TEXT -1 111 " (in the figures: phi). \+
So the system has one degree of freedom (dof). It is  also usual to ca
ll such systems: " }{TEXT 280 32 "single degree of freedom systems" }
{TEXT -1 8 " (sdof)." }}{PARA 0 "" 0 "" {TEXT -1 131 "At first we cut \+
the mass particle free and and consider all (active and passive) force
s. The only active force is the weight force " }{TEXT 282 1 "G" }
{TEXT -1 13 " of the mass " }{TEXT 283 1 "m" }{TEXT -1 46 ".  The only
 passive force is the thread force " }{TEXT 281 1 "S" }{TEXT -1 27 ". \+
We know the direction of " }{TEXT 284 1 "S" }{TEXT -1 67 ", because th
e thread can only transmit forces in its own direction." }}{PARA 0 "" 
0 "" {TEXT -1 58 "Now we consider the mass particle in a displaced pos
ition " }{TEXT 285 1 "f" }{TEXT -1 64 ". The trajectory of the motion \+
is known as a circle with radius " }{TEXT 286 1 "L" }{TEXT -1 25 ". Fr
om the kinematics in " }{HYPERLNK 17 "section 2" 1 "sec-02.mws" "" }
{TEXT -1 50 " we know for the acceleration in polar coordinates" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "a:=polar_acc(L,phi(t),t);" }
}}{PARA 0 "" 0 "" {TEXT -1 115 "The radial component of the accelerati
on points from the mass particle to the suspension point. The inertial
 force " }{TEXT 287 1 "Z" }{TEXT -1 89 " belonging to it acts in the o
pposite direction, away from the suspension point. We write" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Z:=-m*a[1];" }}}{PARA 0 "" 0 "" 
{TEXT -1 69 "The tangential component of the acceleration points in th
e direction " }{TEXT 288 1 "e" }{TEXT 289 1 "f" }{TEXT -1 104 " (ephi \+
in the figures). Thus the inertial force belonging to it acts in the o
pposite direction. We write" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "T:=-m*a[2];" }}}{PARA 0 "" 0 "" {TEXT -1 57 "In Fig. 2 we see all \+
acting forces on the mass particle. " }}{PARA 0 "" 0 "" {TEXT 290 7 "N
otice:" }{TEXT 291 0 "" }{TEXT 292 0 "" }{TEXT -1 26 " In the figure t
he forces " }{TEXT 321 1 "Z" }{TEXT -1 5 " and " }{TEXT 322 1 "T" }
{TEXT -1 71 "  are drawn in the positive coordinate direction. In real
ity the force " }{TEXT 293 1 "T" }{TEXT -1 121 " shows in the opposite
 direction as drawn in the figure, but this is respected by the negati
ve sign in the definition of " }{TEXT 323 1 "T" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "display(Fig_3_2(),scaling=co
nstrained,axes=none,title=\"Figure 2\");" }}}{PARA 0 "" 0 "" {TEXT -1 
57 "The equilibrium in tangential and radial direction yields" }}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQt:=-G*sin(phi(t))+T=0;" "6#>
%$EQtG/,&*&%\"GG\"\"\"-%$sinG6#-%$phiG6#%\"tGF)!\"\"%\"TGF)\"\"!" }}}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQr:=G*cos(phi(t))-S+Z=0;" "6#
>%$EQrG/,(*&%\"GG\"\"\"-%$cosG6#-%$phiG6#%\"tGF)F)%\"SG!\"\"%\"ZGF)\"
\"!" }}}{PARA 0 "" 0 "" {TEXT -1 37 "The force law for the weight forc
e is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "FLG:=G=m*g:" "6#>%$FLG
G/%\"GG*&%\"mG\"\"\"%\"gGF)" }}}{PARA 0 "" 0 "" {TEXT 294 1 "G" }
{TEXT -1 109 " always points downwards. This is taken into account in \+
Fig 2, so there is no negative sign in the force law." }}{PARA 0 "" 0 
"" {TEXT -1 66 "Substituting the force law into both equilibrium condi
tions yields" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "EQt:=subs(FL
G,EQt);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "EQr:=subs(FLG,EQ
r);" }}}{PARA 0 "" 0 "" {TEXT -1 4 "For " }{XPPEDIT 18 0 "m <> 0" "6#0
%\"mG\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "L <> 0;" "6#0%\"LG\"\"
!" }{TEXT -1 21 " we get from EQt the " }{TEXT 328 18 "equation of mot
ion" }{TEXT -1 15 " for the system" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "EOM:=expand(-EQt/(m*L));" }}}{PARA 0 "" 0 "" {TEXT 
-1 70 "This is an ordinary differential equation of degree 2 with resp
ect to " }{TEXT 295 1 "f" }{TEXT -1 1 "(" }{TEXT 296 1 "t" }{TEXT -1 
2 ")." }}{PARA 0 "" 0 "" {TEXT -1 26 "From EQr we get the force " }
{TEXT 297 1 "S" }{TEXT -1 17 " in the thread as" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 16 "S:=solve(EQr,S);" }}}{PARA 0 "" 0 "" {TEXT -1 
64 "After solving the equation of motion we can calculate the force " 
}{TEXT 299 1 "S" }{TEXT -1 27 " in dependence of the time " }{TEXT 
298 1 "t" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 95 "The derived e
quation of motion  is nonlinear and cannot be easily solved. But for s
mall angles " }{TEXT 300 1 "f" }{TEXT -1 31 " we can linarize the equa
tion. " }}{PARA 0 "" 0 "" {TEXT -1 4 "For " }{TEXT 301 1 "f" }{TEXT 
-1 31 " << 1 we make the substitutions" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "subst1:=sin(phi(t))=phi(t):" "6#>%'subst1G/-%$sinG6#-%$
phiG6#%\"tG-F*6#F," }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "subst2:
=cos(phi(t))=1:" "6#>%'subst2G/-%$cosG6#-%$phiG6#%\"tG\"\"\"" }}}
{PARA 0 "" 0 "" {TEXT -1 41 "One more substitution is usual. We write \+
" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "g/L=omega**2:" "6#
/*&%\"gG\"\"\"%\"LG!\"\"*$%&omegaG\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 
53 "or in equivalent but more practicable manner in MAPLE" }}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "subst3:=g=omega**2*L:" "6#>%'subst3G/
%\"gG*&%&omegaG\"\"#%\"LG\"\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 43 "So we
 get the linearized equation of motion" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "LEOM:=subs(\{subst1,subst2,subst3\},EOM);" }}}{PARA 
0 "" 0 "" {TEXT -1 35 "Now we get for the solution of LEOM" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sol:=dsolve(LEOM,phi(t));" }}}
{PARA 0 "" 0 "" {TEXT -1 26 "The integration constants " }{TEXT 303 3 
"_C1" }{TEXT -1 5 " and " }{TEXT 302 3 "_C2" }{TEXT -1 34 " depend on \+
the initial conditions." }}{PARA 0 "" 0 "" {TEXT -1 8 "At time " }
{TEXT 304 3 "t=0" }{TEXT -1 11 " the angle " }{TEXT 305 1 "f" }{TEXT 
-1 14 "(t) should be " }{TEXT 306 2 "f0" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "IC1:=phi(0)=phi0;" }}}{PARA 0 "" 0 "" {TEXT -1 30 "an
d the angular velocity diff(" }{TEXT 307 1 "f" }{TEXT -1 17 "(t),t) sh
ould be " }{TEXT 308 2 "f_" }{TEXT -1 1 "d" }{TEXT 309 2 "_0" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "IC2:=D(phi)(0)=phi_d_0;" }}}
{PARA 0 "" 0 "" {TEXT -1 35 "So we get for the complete solution" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Csol:=dsolve(\{LEOM,IC1,IC2
\},phi(t));" }}}{PARA 0 "" 0 "" {TEXT -1 15 "The expression " }{TEXT 
310 1 "w" }{TEXT -1 11 " is called " }{TEXT 311 23 "natural angle freq
uency" }{TEXT -1 71 ". It is an important measure to describe periodic
 motions. The unit of " }{TEXT 316 1 "w" }{TEXT -1 4 " is " }{TEXT 
317 3 "1/s" }{TEXT -1 16 " (1 per second)." }}{PARA 0 "" 0 "" {TEXT 
-1 23 "Sometimes one uses the " }{TEXT 312 17 "natural frequency" }}
{PARA 257 "" 0 "" {XPPEDIT 18 0 "f=omega/2/Pi" "6#/%\"fG*(%&omegaG\"\"
\"\"\"#!\"\"%#PiGF)" }}{PARA 0 "" 0 "" {TEXT -1 24 "to describe the mo
tion. " }{TEXT 313 1 "f" }{TEXT -1 68 " is the number of cycles of a m
otion within one second. The unit of " }{TEXT 314 1 "f" }{TEXT -1 4 " \+
is " }{TEXT 315 4 "1 Hz" }{TEXT -1 9 " (Hertz)." }}{PARA 0 "" 0 "" 
{TEXT -1 24 "The reciprocal value of " }{TEXT 318 1 "f" }{TEXT -1 15 "
 is called the " }{TEXT 324 7 "period." }}{PARA 257 "" 0 "" {XPPEDIT 
18 0 "T=1/f:" "6#/%\"TG*&\"\"\"F&%\"fG!\"\"" }}{PARA 0 "" 0 "" {TEXT 
-1 2 "or" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "T=2*Pi/omega" "6#/%\"TG*(
\"\"#\"\"\"%#PiGF'%&omegaG!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 28 "In our
 example the period is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "T:
=expand(subs(1/omega=sqrt(L/g),2*Pi/omega));" }}}{PARA 0 "" 0 "" 
{TEXT -1 36 "Here we make the third substitution " }{TEXT 319 6 "subst
3" }{TEXT -1 6 " back." }}{PARA 0 "" 0 "" {TEXT -1 99 "We see that in \+
this system the periodic time is independent of the mass and the initi
al conditions." }}{PARA 0 "" 0 "" {TEXT -1 108 "Finally we consider th
e pendulum's motion with some concrete values. We know for the acceler
ation of gravity" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "g:=9.81:" 
"6#>%\"gG-%&FloatG6$\"$\")*!\"#" }}}{PARA 0 "" 0 "" {TEXT -1 139 "The \+
following values for the mass and the length of the thread can be chan
ged. But don't forget that the solution is only valid for angles " }
{TEXT 320 1 "f" }{TEXT -1 6 " << 1." }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "L:=10:" "6#>%\"LG\"#5" }}}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "m:=10:" "6#>%\"mG\"#5" }}}{PARA 0 "" 0 "" {TEXT -1 24 "
So we get for the motion" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "
assign(subs(1/omega=sqrt(L/g),Csol)):" }}}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "phi(t);" "6#-%$phiG6#%\"tG" }}}{PARA 0 "" 0 "" {TEXT 
-1 44 "The natural angle frequency of the motion is" }}{EXCHG {PARA 0 
"> " 0 "" {XPPEDIT 19 1 "omega:=sqrt(g/L);" "6#>%&omegaG-%%sqrtG6#*&%
\"gG\"\"\"%\"LG!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 28 "and the natural \+
frequency is" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "f:=evalf(omega
/2/Pi);" "6#>%\"fG-%&evalfG6#*(%&omegaG\"\"\"\"\"#!\"\"%#PiGF," }}}
{PARA 0 "" 0 "" {TEXT -1 27 "The period of the motion is" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(T);" }}}{PARA 0 "" 0 "" {TEXT 
-1 100 "In Fig. 3 we see the time history of the motion over 10 period
s for the following initial conditions" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "phi0 := .1:" "6#>%%phi0G-%&FloatG6$\"\"\"!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "phi_d_0:=0:" "6#>%(phi_d_0G\"
\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "P1:=plot(phi(t),t=0.
.10*T, color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "disp
lay(P1,title=\"Figure 3\");" }}}{PARA 0 "" 0 "" {TEXT -1 43 "Fig. 4 sh
ows the time history of the force " }{TEXT 325 1 "S" }{TEXT -1 1 "." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "P2:=plot(S,t=0..10*T, 0..m*
g*1.2, color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "dis
play(P2,title=\"Figure 4\");" }}}{PARA 0 "" 0 "" {TEXT -1 111 "We see \+
that the force in the thread is always positive.  Further we see that \+
the frequency of the thread force " }{TEXT 326 1 "S" }{TEXT -1 41 " is
 two times the frequency of the angle " }{TEXT 327 1 "f" }{TEXT -1 96 
". We don't discuss this phenomen further at this point because it is \+
not subject of this course." }}{PARA 0 "" 0 "" {TEXT -1 245 "Fig. 5 sh
ows a little animation of the motion of the pendulum. You can try what
 happens when you change the initial conditions or the mass or the len
gth of the pendulum. To start the animation click on the picture and t
he \"play\" in the menu bar." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
105 "display(Fig_3_3(L,m,eval(phi(t)),t,10*T),insequence=true,scaling=
constrained,axes=none,title=\"Figure 5\");" }}}}{SECT 0 {PARA 5 "" 0 "
" {TEXT -1 46 "General Strategy to Get The Equation of Motion" }}
{PARA 0 "" 0 "" {TEXT -1 122 "The above example shows the strategy to \+
get the equation of motion for a dynamic system. Here we itemize the s
ingle steps:" }}{PARA 0 "" 0 "" {TEXT -1 86 "1. Define an appropriate \+
coordinate system and choose the relevant degrees of freedom." }}
{PARA 0 "" 0 "" {TEXT -1 51 "2. Cut the mass particle free in a genera
l position" }}{PARA 0 "" 0 "" {TEXT -1 68 "3. Fix all passive forces w
ith their known geometrical appointments." }}{PARA 0 "" 0 "" {TEXT -1 
49 "4. Formulate the force law for all active forces." }}{PARA 0 "" 0 
"" {TEXT -1 89 "5. Add the inertial forces against the direction of th
e positive acceleration components." }}{PARA 0 "" 0 "" {TEXT -1 72 "6.
 Formulate the equilibrium conditions in form of coordinate equations.
" }}{PARA 0 "" 0 "" {TEXT -1 103 "7. Mathematical analyses of the equa
tion of motion by integration and suited to the initial conditions." }
}{PARA 0 "" 0 "" {TEXT -1 28 "8. Discussion of the result." }}}}}
{MARK "0 0 0" 16 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 
2 33 1 1 }