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» Instructor's Manual to Accompany
; Vector Mechanics
“for Engineers ii ao oa
* Ferdinand P, Beer DIES
E. Russell Johnston, gr.
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Instructor's Manual to Accompany
Vector Mechanics
for Engineers | pynamics
Fifth Edition
FERDINAND P. BEER
Lenigh University
E. RUSSELL JOHNSTON, gr.
University of Connecticut
McGraw-Hill Book Company
NewYork St Louis San Francisco Auckiand Bogotá Caracas
Colorado Springs Hamburg Lisbon London Madrid Mexico
Milan Montreal NewDeihi Oklahoma City Panama Paris
San Juan São Paulo Singapore Sydney Tokyo Toronto
ET ção x = sr»
DESCRIPTION OF THE MATERIAL CONTAINED IN
'VECTOR MECHANICS FOR ENGINEERS: DYNAMICS", Fifth Edition
11. Kinematics of Particles
In this chapter, the motion of bodies is studied without regard to their
size; all bodies are assumed to reduce to single particles. The analysis of
the effect of the size of a body and the study of the relative motion of the
various particles forming a given body are postponed until Chapter 15. In order
to present the simpler topics first, Chapter 11 has been divided into two parts:
rectilinear motion of particles, and curvilinear motion of particles.
In Section 11.2, position, velocity anã acceleration are defined for a
particle in rectilinear motion. They are defined as quantities which may be
either positive or negative and students should be wvarned not to confuse position
coordinate and distance traveled, or velocity and speed. The significance of
positive and negative acceleration should be stressed. Negative ácceleration may
indicate a loss in speed in the positive direction or a gain in speed in the
negative direction.
As they begin the study of dynamics, many students are under the belief
that the motion of a particle must be either uniform or uniformly accelerated.
To destroy this misconception, the motion of a particle is first described under
very general conditions, assuming a variable acceleration which may depend upon
the time, the position, or the velocity of the particle (Sec. 11.3). To facili-
tate the handling of the initial conditions, definite integrals, rather than in-
definite integrals, are used in the integration of the equations of motion.
The special equations releting to uniform anã uniformly accelerated motion
are derived in Sections 11.4 anã 11.5. The students should be warned to check
carefully, before using these equations, that the motion under consideration is
actually a uniform or a uniformly accelerated motion.
Two important concepts are introduced in Section 11.6: (1) the concept of.
relative motion, which will be developed further in Sections 11.12 and 15.5, (2)
the concept of dependent motions anã of degrees of freedom.
The first part of Chapter 11 ends with the presentation of several graph-
ical methoês of solution of rectilinear-motion problems (Secs. 12.7 and 11.8).
This material is optional and may be omitted. Several problems in which the data
are given in graphical form have been included (e.g., Probs. 21.72 - 11.75).
The second part of the chapter begins with the introduction of the vectors
defining the position, velocity and aeceleration of a particle in curvilinear
motion. The derivative of a vector function is defined anã introduced st this
point (Sec. 11.10). The motion of a particle is first studied in terms of rec-
tangular components (Sec. 11.11); it is shown that in many cases (for example,
projectiles) the study of curvilinear motion may be reduced to that of two in-
dependent rectilinear motions. The concept of fixed anã moving frames of ref-
erence is introduceã in Section 11.12 and immediately used to trest the relative
motion of perticles.
Sections 11.13 anã 11.1) introduce respectively the use of tangential and
normal components, and of radial anã transverse components. Both systems of
coordinates are first introduced in two dimensions and then extended to include
three-dimensional space.
iii
12. Kinetics of Particles: Newton's Second Law
As indicateã earlier, this chapter and the following two are concerned only
with the kinetics of particles and systems of particles. They neglect the effect
of the size of the bodies considered and ignore the rotation of the bodies about
their mass center. The effect of size will be taken into account in Chapters 16
through 18, which deal with the kinetics of rigid bodies.
Section 12.2 presents Newton's second law of motion and introduces the con-
cept of a newtonian frame of reference. In Section 12.3 the concept of linear
momentum of e particle is introduced, anã Newton's second law is expressed in its
alternate form, which states that the resultant of the forces acting on & perticle
is equal to the rate of change of the linear momentum of the particle. Section
12.4 reviews the two systems of units used in this text and previously discussed
in Sec. 1.3, namely the SI metric units and the U.S. customary units. This
section also emphasizes the difference between an absolute and a gravitational
system of units.
A number of problems with two degrees of freedom have been included
(Problems 12.26 through 12.33), some of which require a careful analysis of the
accelerations involved (see Sample Problem 12.1).
Section 12.5 applies Newton's second law to the study of the motion of a
particle in terms of rectangular components end tangential and normal components.
In Section 12.6, dynamic equilibrium is presented as an alternate way of express-
ing Newton's second law of motion, although it will not be used in any of the
Sample Problems in this text. Note that the term inertia vector is used in pref-
erence to inertia force or reversed effective force to avoid any possible con-
fusion with actual forces.
In Section 12.7 the concept of angular momentum of a particle is introduced,
and Newton's second lew is used to show that the sum of the moments about a point
O of the forces acting on a particle is equal to the rate of change of the an-
gular momentum of the particle about O. Section 12.8 analyzes the motion of a
particle in terms of radial anã transverse components, and Section 12.9 considers
the particular case of the motion of a particle under a central force. The early
introduction of the concept of angular momentum greatly facilitates the discussion
of this motion. Section 12.10 presents Newton's law of gravitation aná its appli-
cation to the study of the motion of earth satellites.
Sections 12.11 through 12.13 are optional. Section 12.11 derives the differ-
ential equation of the trajectory of a particle under a central force, while
Section 12.12 discusses the trajectories of satellites and other space vehicles
under the gravitational attraction of the earth. While the general equation of
orbital motion is derived (Eg. 12.39), its application is restricted to launchings
in which the velocity at burnout is parallel to the surface of the earth.
(Oblique launchings are considered in Section 13.9.) The periodic time is founã
directly from the fundamental definition of areal velocity rather than by formulas
requiring a previous knowledge of the properties of conic sections. The instruct-
or will note that he may omit Sections 12.11 through 12.13 and yet assign a number
of interesting space mechanics problems to his students after they have reached
Section 13.9.
iv
13. Kinetics of Particles: Energy aná Momentum Methods
After a brief introduction designeã to give to the students some motivation
for the study of this chapter, the concept of work of a force is introduced in
Section 13.2. Attention is called to the fact that the term work is always used
in connection with a well-defined force. Three examples considered are the work
of a weight (i.e., the work of the force exerted by the earth on a given body),
the work of the force exerteã by a spring on a given body, anã the work of a
gravitational force. Confusing statements such as the work done on a body, or
the work done on a spring are avoided.
The concept of kinetic energy is introduced in Section 13.3 aná the prin-
ciple of work and energy is derived by integration of Newton's equation of
motion. In aspplying the principle of work and energy, the students should te
encoursgeã to âraw separate sketches representing the initial and final positions
of the body (Sec. 13.4). Section 13.5 introduces the concepts of power and
efficiency.
Sections 13.6 through 13.8 are Gevoted to the concepts of conservative
forces and potential energy, and to the principle of conservation of energy.
Potential energy shoulã always be associated with a given conservative force act-
ing on a body. By avoiding statements such as "the energy contained in a spring"
a clearer presentation of the subject is obtained, which will not confiiet with
the more advanced concepts that the students may encounter in later courses, In
applying the principle of conservation of energy, the students should again be en-
couraged to draw separate sketches representing the initial and final positions
of the body considered.
In Section 13.9, the principles of conservation of energy and of conser-
vation of angular momentum are applied jointly to the solution of problems in-
volving conservative central forces. A large number of problems of this type,
dealing with the motion of satellites and other space vehicles, are available
for nomevork assignment. As noted earlier, these problems (except the last two,
indicated by asterisks) may be solved even if Sections 12.11 through 12.13 have
teen omitted,
The second part of Chapter 13 is devoted to the principle of impulse anã
momentum and to its application to the study of the motion of a particle.
Section 13.10 introduces the concept of linear impulse and derives the principle
of impulse and momentum from Newton's second law. The instructor should emphe-
size the fect that impulses and momenta are vector quantities. He should en-
courage his students to draw three separate sketches when applying the principle
of impulse and momentum and show clearly the vectors representing respectively
the initial momentum, the impulses, and the final momentum. It is only after
the concept of impulsive force hes been presented that the students will begin
to appreciate the effectiveness of the method of impulse and momentum
(Section 13.11).
Direct central impact and oblique central impact are studied in Sections
13.12 through 13.14. Note that the coefficient of restitution is defined as the
ratio of the impulses during the period of restitution and the period of defor-
mation. This more basic approach will make it possible in Section 17.12 to ex-
tend to the case of eccentric impact the results obtained here for central im-
pact. Emphasis should be placeã on the fact that, except for perfectly elastic
impact, energy is not conserved. Note that the discussion of oblique central
impact in Section 13.1) has been expanded to cover the case when one or both of
the colliding bodies is constrained in its motion.
x6. Plane Motion of Rigid Bodies: Forces and Accelerations
This chapter is devoted to the plane motion of rigid bodies which consist of
Plane slabs or which are symmetrical with respect to the reference plane. Cases
involving the plane motion of nonsymmetrical bodies and, more generally, the
motion of rigid bodies in three dimensions are considered in Chapter 18. If the
determination of mass moments of inertia has not been covered in the previous
stetics course, the instructor should include material from Sections 9.11 through
9.15 of Appendix B (or from the second part of Chap. 9) at this point.
In Section 16.2 the fundsmental relations derived in Chapter 1h for a system
of perticles are used to show that the external forces acting on a rigid tody are
equipollent to the vector ma attached at the mass center G of the body and the
couple of moment Ha. This result, which is illustrateã in Fig. 16.3, is valiã in
the most general case of motion of a rigid body (three-dimensional as well as
plane motion).
It is shown in Section 16.3 that, in the case of the plane motion of a slab
or symmetrical body, the angular momentum E, reduces to Tu and its rate of change
to Ta. Section 16.) is devoted to D'Alembest's principle. It is shown that the
external forces acting on a rigid body are aetualiy equivalent to the effective
forces representeã by the vector mã and the couple ia. As noted in Section 16.5,
this result is obtained independently of the principle of transmissibility (Sec.
3,18) anê may be used to derive this principle from the other axioms of mechanics.
At this point the students will have reacheà the climax of the study of
rigid-body motion in two dimensions. Indeed they mey solve any problem by draw-
ing two sketches - one showing the external forces, the other the vector mã and
the couple la - and then expressing that the two systems of vectors shown are
equivalent. To avoid drawing two separate sketches, the method of dynamic equi-
librium may be useã, with a single sketch showing the external forces, the in-
ertia vector - mã and the inertisa couple - Ty (Sec. 16.6). However, to facilitate
the transition to the study of three-dimensionel motion (Chap. 18), the two-sketch
method showing separately the external forces and the effective forces will de
used in all sample problems.
The various types of plane-motion problems have been grouped according to
their kinematic characteristics. The translation, the centroidal rotation, and
the plane motion consisting of a translation and of an unrelateã centroidal ro-
tation are considered first, since they are the simplest ones to analyze, They
are followed by plane motions with various kinematic constraints: non-centroidal
rotation, rolling motion, aná other types of plane motion. Problems involving
systems of rigid bodies have been included at the enã of this chapter, with
either one degree of freedom (Problems 16.14! through 16.117) or two degrees
of freedom (Problems 16.148 through 16.155). The instructor should stress the
fact that, in spite of the different kinematic characteristics of these
various motions, the approach to the Kinetics of the motion is consistently
the same: all problems are solved by drawing two sketches - one showing the ex-
ternal forces, the other the vector mã and the couple io - anã then expressing
that the two systems of vectors shown are equivalent.
viii
Since the approach used in this text differs from others by the emphasis
placeã on the direct application of D'Alembert's principle, rather than on
specialized formulas, it might be appropriate at this point to sumarize the ad-
vantages derived from this approach.
(1) A single methoã is useã, which applies to all cases of plane motion,
regardless of their kinematic characteristics, and may be used safely under any
conditions. This is in contrast with the equation EM = Ia which is limited in
its applications, as pointeá out in Problem 16.105.
(2) By stressing the use of the free-body diagram, this method provides a
better understanding of the kinetics of the motion. There will be little danger,
for example, in the solution of a problem of non-centroidal rotation, that the
students forget the effect of the acceleration of the mass center on the reaction
at the fixed point, a mistake which oceurs frequently when the specialized formula
EM, = Toa is used.
(3) The method used divides the solution of a problem into two main parts,
one in which the Kinematic and kinetic characteristics of the problem are con-
sidered (separately if necessary), anã the other in which the methods of statics
are used. In this way the techniques of each separate field may be used most
efficiently. For example, moment equations may pe written to eliminate unvanted
reactions, just as it was done in statics; this may pe done independently of the
kKinetice characteristics of the problem.
(bh) By resolving every plane motion (even a non-centroidal rotation) into
a translstion and a centroidsl rotation, a unified approach is obtained, which
will aiso be used in Chapter 17 with the method of work end energy and with the
method of impulse and momentum, and which will be extended in Chapter 18 to the
study of the three-dimensional motion of a rigia body. This approach is a pasic
one, which may be applied effectively throughout the study of mechanics, in ad-
vanced courses as well as in elementary ones.
17. Plane Motion of Rigid Bodies: Energy and Momentum Methods
The first portion of the chapter extends the method of work and energy,
already used in Chapter 13, to the study of the plane motion of rigid bodies.
The expressions for the work of a couple and for the ginetic energy of a rigid
body are derived in Secs. 17.3 and 17.4. Using the results obtained in Sec. 14,7,
the kinetic energy of a rigiã body is separated into a translational part and a
rotational part (about the mass center). The authors believe that, while it may
lead to slightly longer solutions, this method is more fundamental and should be
used in preference to special formulas. Indeed, it follows the basic idea of re-
solving every plane motion into a translation and a centroidal rotation.
It is shown in Section 17.5 that the method of work and energy is especially
effective in the case of systems of rigid bodies connected by pins, inextensible
corãs, and meshed gears. In Section 17.6 the principle of conservation of energy
is used to analyze the plane motion of rigid bodies.
ix
In the second part of Chapter 17, the method of impulse and momentum is ex-
tended to the study of the motion of rigid bodies. The approach used is different
from that of most elementary textbooks. Ready-to-use formulas are evoided; in-
stead, the students are taught to express the general principle of impulse and
momentum by means of free-body diagrams and to write the equations most appro-
priate to the solution of the problem considered.
The results obtaineá in Section 1h.9 for a system Of particles are directly
applicable to the system of particles forming a rigiã body anã may be used to
analyze the plane motion of rigiã bodies. It is shom in Section 17.8 that the
momenta of the various particles forming a rigid body reduce to a vector nv and a
couple To in the most general case of plane motion.
While the principle of conservation of angular momentum is discussed in
Section 27,10 because of its physical and historical significance, it is not
actually used in the solution of problems, To solve any problem, regardless of
the type of motion, and whether it involves constant forces oí finite magnitude
applied for & finite time, or impulsive forces applied for a very short time in-
terval, the students are told to draw three separate sketches showing respectively
the initial momenta, the impulses of the external forces, and the final momenta.
The momenta of a rigid body are represented in the most general case by a momentum
vector my attached at the mass center and a mgmentum couple Iw. If the students
consider then the components of the vectors involved, they obtain relations be-
tween linear impulses and linear momenta. If they consider the moments of the
seme vectors, they obtain angular impulses and angular momenta. If, by equating
moments about a point such as a pivot, they obtain an equation which does not in-
volve any of the external forces, they will have automatically established conser-
vation of angular momentum about that point.
The advantages derived from this approach may be summarized as follows:
(1) The students leern only one method of solution, a method based directly
on a fundamental principle, a methoá which may be used safely under any conditions.
This is in contrast with the equation EMt = Twp-u,), which is limited in its
applications (see Problem 17.66).
(2) The methoa stresses the use of free-body disgrams and thus provides a
better understanding of the kinetics of the motion. It is unlikely, for example,
that the students will forget an impulsive reaction at a fixeà support.
(3) The students use the basic tools they learned in Statics: reduction of
a system of vectors to a vector and a couple, equations relating the components or
the moments of these vectors.
(4) Again, the same unified approach is used: every plane motion is re-
solveê into a translation and a centroidal rotation. And in Chapter 18 this
approach will be extended to the solution of problems involving the three-dimen-
sional motion of rigid bodies.
Some teachers may fear that the inclusion of momentum vectors and momentum
couples in the same diagrams may lead to confusion. This will not be the case,
however, if the students are instructed to write separate equstions involviag
either components or moments, as they did in Statics. The first equations will
contain linear impulses and linear momenta expressed in Nes or lbºs, anã the
latter angular impulses and. angular momenta expressed in Nemes or lbeftes.
Section 17.12 is devoted to the eccentric impact of tvo rigid bodies, a
topic seldom included in an elementary text, No special difficulty will be
encountered, however, if separate sketches are useê as indicated above,
18. Kinetics of Rigid Bodies in Three Dimensions
In this chapter the restrictions imposed in preceding chapters (e-g., plane
motion, syrmmetrical todies) are lifted and the student proceeãs to the analysis of
more general (and more difficult) problems, such as the rotation of nonsymmetrical
bodies about fixed axes and the motion of gyroscopes.
In Section 18,1, the general result obtaineã in Section 16.2 is recalled,
namely, that the external forces acting on a rigid body are equipollent to the
vector mã attached at the mass center G of the body and the couple of moment Kg.
It is also pointed out that the main feature of the impulse-momentum method,
namely, the reduction of the momenta of the particles of a rigid body to a linear
momentum vector my attached-at G and an angular momentum E, remains valiã, and
that the vork-energy principle and the principle of conserYation of energy still
apply in the case of the motion of a rigid body in three dimensions. The dif-
ficulties encountereã in the study of the three-dimensional motion of a rigiã body
are relateã to the determination of the angular momentum Hg, of its rate of change
Hg» and of the kinetic energy of the body.
The determination of the angular momentum H,. of a rigid body from its an-
gular velocity w is discussed in Section 18.2. Since this requires the use of
mass products of inertia, as well as of mass- moments of inertia, the instructor
should cover Sections 9.16 and 9.17 from Appendix B (or from the seconã part of
Chapter 9) if this material has not been included in the previous statics course.
Section 18.3 is devoted to the application of the impulse-momentum principle
to the three-dimensional motion of a rigid body, and Section 18.4 to the deter-
mination of its Kinetic energy.
In Sections 18.5 and 18.6, the rate of change of the angular momentum Hg is
computed and the equations of motion for a rigid body in three dimensions are de-
rived. D'Alembert's principle is extended to the case of three-dimensional motion
ty showing that the external forces are actually equivalent to the effective forces
represented by the vector má and the couple Hç. Sections 18.7 and 18.8 are de-
voted, respectively, to the particular cases of the motion of a rigid body about
a fixed point and of the rotation of a rigid body about a fixed axis, with appli-
cations to the balancing of rotating shafts.
While Euler's equations of motion have been derived on page 907, it should
te noted that the more fundamental vector relations represented by Equetions
(18.22), (18.23), and (18.28) are used in the actual solution of problems.
xi
The remaining portion of this Chepter (Secs. 18,9 through 18.11) is de-
signed for advanced students and, in general, should be omitted for ordinary
classes. In Sections 18.9 and 18.10 the motion of a gyroscope is considered. At
this point Eulerian angles are introduced. It should be carefully noted that the
rotating system of axes Oxyz is attacheã to the inner gimbal; these axes are prin-
cipal axes of inertia and they follow the precession and nutation of the gyro=
scope; they do not however spin with the gyroscope. The special case of steady
precession is considered in Section 18,10. Several problems dealing with the
steady precession of a top are included and, in one of them (Prob. 18.89, it is
shown that the formula usually given in introduetory texts is only approximate.
Two problems dealing with the general motion of a top have also been included
(Probs. 18,108 end 18.109).
The motion cf an axisymmetrical body under no force (Sec. 18.11) introduces
the student to one of the most interesting aspects of classical dynamics - an
aspect which nes gained widespread attention in recent years due to the interest
in space vehicles and artificial satellites, In this connection, it shoulãá be
pointed out that the instructor may cover the Poinsot theory of the motion of a
nonsymmetrical body under no force by assigning three problems (Probs. 18.11h
through 18,116), An adãitionsl problem (Prob. 18.117) relates to the stability
of the rotation of a nonsymuetrical body about a príncipal axis.
19. Mechanical Vibrations
Tnis chapter provides an introduction to the study of mechanical vibrations.
While only one-degree-of-freedom systems are included, all the basic principles
are presented. The various topics covered are as follows:
(a) Free, undamped vibrations of a particle (Sec. 19.2). The differential
equation characterizing simple harmonic motion is derived and 822 basic terms,
such as period, frequency, anê amplitude, are defineã, Both the analytical and
the geometrical methods of solution are described. It is shown in Section 19.3
that the motion of a simple pendulum may be approximated by a simple harmonic
motion. Section 19.4, which is optional, shows how an exact solution may be ob-
tained for the period of oscillations of a simple pendulum.
(b) Free, undamped vibrations of a rigid body. The principle of equiv-
alence of the systems of applied and effective forces is first used to determine
the frequency and the period of oscillations of a rigid tody (Sec. 19.5). Note
that the same positive sense is assumed for the angular acceleration and dis-
placement; this results in an apparently unrealistic assumption for the sense of
the vector mã and the couple Ia. The principle of conservation of energy is then
used to solve the same type of problems (Sec. 19.6).
(c) Forced, undamped vibrations of a particle (Sec. 19.7). This section
introduces the students to the concepts of natural and forced frequencies, tran-
sient and steady-state vibrations, and resonance. While all students will be
able to understand this section, those with a knowledge of elementary differential
equations will derive a greater benefit from it since it provides a direct
application of the solution of linear nonhomogeneous equations with constant co-
efficients.
xii
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Problem Number?
SI Unite U.S. Units Problem Description
CHAPTER 11: KINHMATICS OF PARTICLES
RECTILINEAR MOTION OR PARTICLES
Analyze motion of a particle, given
11.3,4 position = f(t)
also find total distance traveled
11.11,8 acceleration = f(t)
11.15,1h acceleration = f(x)
11.19,18 seceleration = flv)
11.20
net 11.25 velocity = f(x)
11.26 1.27 accelerstion v L/r2
21.29 1.28 Simple hermonic motion
Uniformly accelerated motion
11.31,30 2.33,32 motion of one particle
11.35,34
11.39,36 12.37,38 motion of two particles
Relative motion of particles
systems with one degree qr. freedom and
1.k5 ko 11.h2 constant velocities
11.b1,bh ai.h3 constant uccelerations
1h h6 21.h9,18 systems with tun degrees of Ireedon
11.51,*50
Graphical methods
11.55,54 12.53.52 analyze motion using a given motion curve and inílial condilions
11.57,56 11.592,58 motion of one particle
11.61,60
aL.6p 11.6h,63 motion of two partícies
11,67,66 11.65
11.68 211.69 problems involving average velocity
nn 12.70 and rate of change of Acceleration
11.73,74 n.7 problems solveã by numerical integration
11.75 problem solved by use of v-x curve
1.77,78 11.79,76 nroblems solveã by moment-area method
” CIMVILTIHEAR MOTION OF PARTICLES
11.83,82 11.81,80 Anstyze motion of a particle given
4 1.85 x and y as functions of t
11.86 position vecter in two dimensions
position vector in three dimensions
Projectiles
1.91,88 11.89.90 with horizontal Initiul velocity
11.95,9h 11.93 with initial velocity of given mugnitude and direction
12.97,96 11.99,92 riven direction, find magnitude 07 initial velocity
11.98
11.103,*102 *11.101,100 | given magnitude, find direction of initial velocity (solution of
11.105,104 quadratir equation required)
Relative motion of partácles with
11.107,108 11.109,106 constant velovities
11.111,110 veloecities and accelerations
11.133,112 given relative ncccleration
12.115,114 constent accelerations
2.N7,116 q1.119,218 two given observatious of relative motion
*Problems which do not involve any specific system cf units rave been indicated ty underlining their number.
TABLE TI: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Numbert
SI Units U.S. Units Problem Description
Wethod of impulse and momentum applied to problems involving
13.113,1H 13.115,112 single body
13.116
13.127,120 13.119,28 two or more connected bodies
13.121,122
13,123,124 13.125,126 force defined by a graph
13.127,130 13.329,128 impulsive motion of A single body
13.131,132 13.135,136 impulsive motion of two »edies
13.133,134 .
13.137 13.138 impulsive motion vith different supports
13.1h1,3h2 13.139,10 Direct central impact,
13.143 2h
47,148 33.145,146 Oblique central impact
51,152 13.149,150 Impact against Fixed surfaçes
13.155,156 13.153,15
13.159,160 13.157,158 Oblique central impact with one body constrained to move in a
13.161 given direction
13.165,16: 13.163,162 Problems involvíng conservation of energy, conservation of
13.167,166 momentum and impact
13.171,170 13.169,168
13.172 13.173 and oue body constraineã to move in s given direction
“13.175 13.174 — Epecial problems
IHITLITO 13.179,180 Review problems
13.181,178
13.183,18) 13.187,182
13.185,186
13.01,03 13,ce,eh Computer problems
CHAPTER 1h: SYSTEMS OF PARTICLES
14.3,4 1h.1,2 Conservation of linear momentim (one dimension)
1h.5,6
14.9,10 14.7,8 Computation of angular momentum
1h.12,14 14,13,12 Motion of mass center
1h,15
1:.19,18 1h,17,16 Conservation c? linear momentum (2 or 3 dinensions!
14.23.,20
14,25 24 1h.23,22 — Nerivations and proots
11,27 11.26 Conservatiou vT linear momentum and computations of cnange
14.29,28 24.31,30 in kinetic energy
a 1h.37,32 Conservation of linear momentum und conservation of energy
135,3 1i.39,38
TEhl ho aWh3,h0 Conservation of linesr and anguler momentum
15.h5,h 1h.h9,h8 Conservation of linesr momentun, angular momentum, and energy
21.47,16
+Problems which do not involve any specific system of units
xviii
have been Indicated by underlining their number.
TABLE IT: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Hunber?
SI Units U.S. Units Problem Description
1h.51,50 1Y.52 Thrust Causcd by diverted flou
2h,53,5h 1b.55,56
14,59,56 ab.sT,6o | Reactions at supports of vane or conveyor
14.61,62 -
14.632,66 1h.65,6h Thrust developed by a fan, propeller, or jet engine.
1).69,68 1h.67,70
Wo Ui 73,7) Power and efficiency
*4.75,M76; TT Theery problems from fluid mechanica
1.78 — Motion of sprinkler
14.79,80 14.81,82 Motion of chains or of cars of variable mass
11,85,86 10.03,8h
15-89,90 1h.87,88 Thrust and acceleration of rocketa
1h.91,92 14.93,94 Yelacity of rocket for given fuel usage
14.97.98 14.95.96 Distance traveled by rocket
14.99 14.100 Derivetion of efficiency of jet engine or rocket
14.107,106 24,2101,102 Review problems
14. 109,108 1.103,10)
30.711,110 15.105,112
15.ca cl 1h,C2,03 Computer problems
CHAPTER 15: KINEMATICS OF RIGID BODIES
Rotation about a Tixed axis
15.3, uniformly accelerated rotation
15.11,8 rotation about akew axis
15.13,12 motion of the earth
15.19,18 motion ín two dimensions
15.021,20 rolling contact vith no slipping
linear and angular motion
15.29,28 disks brought into contact, slípping occurs
*25.30 special problem
General planc motion - velacíties
15.31,3h 15.33,32 motion of single rigid body
15.35.38 Da application of vector algebra to motion of a plate.
15.39,h0 15.42 planetary gear systems
15.43,
motion of 8 rod connected to:
15.145,18 15.47 ,h6 a crank and a sliding block
15.149,50 15.51,52 two cranks
15.53,5h & rolling vheel anã a horizontal surface
“15.55 special problem
Instantaneous center Gf rotation
15.59,58 15.57,56 problems involving parallel velocities
biido 15.63,62 single rod: angle between controlling velocities = 90º
15.65,
15.67,66 single rod: angle between controlling velocities É 90º
15.69,68 single plate guidcá by pins fitted into slots
15.70 15.71 problems involving tuo instantaneous centers
15.173,72
15.75 15.7h space and body centrodes
15.79,76 15.717,78 previous problems solved using instantaneous center
*Problems which do not
involve any specific system of units have beer indicuted by underlining their number.
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number
ST Units U.8. Units Problem Description
General plane motion - accelerations
15.821,80 find reletions among aecelerations (w = 0)
15.863,82
15.85,8h 15.89,88 rolling motion
15.87,86
motion of a body controlled by
15,93,90 15.91,92 one crank vith constent engular velocity
15.95,9h 15.97,96 two eranks
15.99,98 15.101,100 a crank and & sliding block
15.102 special problem
*15.103,%104 *15.105,*106 Analysis of plane motion in terms of a parameter
»15.107,*108 *15.109,*110
+15 131,9112 415.113,%11h
Plane motion of a particle relstive to a rotating frame
25.117,16 15.115,118 velocity of a partícle with respect to a rotatirg frame
15.119,120
Coriolis ecceleration
15.121,122 15.123,124 accelcration of particle moving ut a constant speed with
15.125 ,126 respect to a frame rotating with a constant anguler velocily
15.127,128 15,129,130
15.131,132 15.133,134 accelerstion of « particle moving with respert to a frame
15.137.138 15.135,136 rotating with an angular acceleratioa
*15.139,41h0 415.1h1,*1L2 special problems
Motion «bout a fixed point. General motion
15.1h45,246 15.153 ,14k relation betveen linear aná angular velocity for a rigiã body
with a fixed point
15.119,15€. angular scceleration of body rotating sirmltaneously about
two axes
vetocity and/or acceleration of a point on a buly rotating
about a fixed point
15.153,15h 15.155,156 angular velocities ang rate of cuasge o? angular velccitics given
15.159 157,158
15.161,160 163,162 velocities of points on tody specified
15,164 65 cone rolling on fixcd surface
15.169,165 15.167,166 rod attacheã to moving collers: find velccities of coliara
15.171,170 (angular velocity is indeterminate)
15.173,172 universal Jointo
15.175 15.174 rod with clevis attached at one end
*15,179,ML78 R1S.17T,*ITO rod attached to moving collars: “ind acceieretions o” cotLaro
(angular velocity anã acecieration are indeterminato)
ivo to E rotaling ramo.
Thres dimensional! motion ol a particle re
: forjolis accelerati
15.183,18 Q=0andã=0
15.185 Bfosna ufa
15.187,186 15.189,188 previous problems to be solveá usink rotating frames
15.193,192 25.191,190 axis of º in plane of mechanism
15.195,19
15.197,198 15.199,196 frame of reterence in gencral motion
15,203,20P 15.202,200
«208 15,205,204 Review problems
5.213,212 15.209,206
.215,21b 11,210
«C1,Ch 3,02 Computer problems
*Problems which do not involve any specific system of units nave D ed ly underlini:
xx
g their nutber.
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number* .
SI Units U.S. Units Problem Description
17.131,130 17.133,132 Review problems
17.137,136 17.135,13
17.139,138 AT.JuL,iho
17.C1,03,Ch 1T.c2 Computer problems
CHAPTER 18: KINETICS OF RIGID BODIES
IN THREE DIMENSTONS
18.1,2 18,3,4 Computation of angular momentum (no product of inertia)
18.7,6 18.5 ..
18.9,8 18.211,10 tise of equation HE = PXTV+H
18.12
18.13,14 Computation of angular momentum (using products of inertia)
18.15
18,19,18 Impulse and momentur in three dimensions
19.25 ,24
18.727,26.
15.29,30 Impact o? meteorite on saLelliLe
18.35,38 Computation of kinetic energy
18.37,36,
18.h3,h2, 18.h1,44 Change in kinetic energy due Lo impact
18.15 ,46 1á.L7,L8 Computaticn of derivative of angular momentum
18.51,50
Dynamic reactions on roteting shaft for
18.55,56. 18.53,5h constant w
18.59,58 18.57,60 m=Oandazfo
18.61 18,62 wand az o
18,63,6h 18.65,66 Simple gyroscopic cffect (axes of rotation perpendicular)
12.67,68 28.69
Relative equilibrium under rotation about fixed axis
Iynamic reactions
18.73,72 28.712,74 for combincã rotations (constant rete)
28.75.76 +18 for combined rotations with angular acceleration
*18.79 *18,90 Advanced problems
18.851,82 18.853,04 Steady precession. Simple problems
18.85,86 18,87,88 Steady precession. More advanced problems
18,90 18.89
18.91 Spevial problem (ayrocompass)
18.95,94 18.093,92 Frecession under no force (theory)
18.599,98 18.971,96)
96 Precession under no force (applications)
28.103,102 78.101,100 .
18.10h
Conservation of energy and angular momentum for bodics
*18.105,*106 *18,107 witr precession sná nutation
*20.115,4110 *18.109, with precession, nutation, and apin
+18.112 418.113
*18.115,*1]h “18.117,16 Poinsot analysis
18.121,120 18.119,118 Review prótlems
18.125,122 2B.123,124 .
26.127,126 18.129,128
18.C1,Ch 18.03,c2 Computer proble-s
“Problems which do not invulve any specific
em of units have been indicated by underlicing their number.
xxiii
TABLE IT: CLASSIFICATION AND DESCRIPTION OF PROBLEMS ( CONTINURD)
Problem Humber*
81 Units U.S. Units ' Problem Description
CHAPTER 29: MECHANICAL VIBRATIONS
VIBRATIONS WITHOUT DAMPING
Free víbrations of a particle
19.1,2 19.544 problems involving amplitude, frequency, maximum velocity
19.3,6 x anã aeceleration
19,9,8 39.11,10 problems invclving velocity anã acceleration at q given tnstant
19.215,14 19.13,12 aprings in parallel or series
19.19,18 19.17,16 change mass or stiffness of system
19.21,22 19.23,20 special problems
*19.27,426 *19,25,%ah simple pendulum with large amplitude
Free vibrations of rigid bodies
systems with elastic restoring forces
19.31,28 19.29,30 rolling or noncentroidsl rotation, also find vg
19.33,32
19.35,36. 19.37,3h special problems
systems vith restoring forces due to gravity
19.38 19.39,40 derivetions for motion of compounã perdulua
19. hz 19.43,h4 find period or frequeney of compound pendulum
19.h5,h6 19.47
19.49.18 19.51,50
19.53,52 19.57,56 torsional vibrations
19.55,5h
Conservation of energy used Lo analyze the notion of
19,61,60 particles
19.65,6h single rigid body: pendulum
19.67.58
19.69,70 single rigid body; elastic restorins forces
19.73,76 comected rígid bodies
15.78
19.83,80 special problems
19, BL
Forccd vibrations
. 19.89,88 problems involving periodic applied furve cr periadic
19.90 support motion
19.914,92 19.93,94 problems involving rotating machinery
19.95,96 .
19.97,98 19.99,100 special problems
19.103,10 19.101,102
19.105 10.197,106
*Problemns vhich do not involve ary
axiv
specific system of units have been indicated by underlining their number.
TABLE IT: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number *
51 Units U.S. Unite Problem Description
DAMPED VIBRATIONS
Damped fres vibrations
19.109,108 19.111,110 proofs concerning damped free vibrations
19.112
19.113,11 problems involving logarithmic decrement
19.115,116 eritically damped system
Demped forced vibrationa
19.117 19.118 proofs concerning damped forced vibration
19-.121,122 19.119,120 problems involving rotating machinery
19.123,12h 19.125,126 problems involving periodic applied force
“19.127 *19.128 special problems
19.131,132 slectrical analogues
19.134
19.135,138 19.137,136 Review problems
29.2h1,1h4 29.139,2h0
19.143,146 19.145,1h2
19.C1,C2 19€3,Ch Computer problems
“Problems which do not involve any specific system of? units have been indicated by underlining their number.
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e (ata) dpi co rasateatoae (2)
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dep lno bn te dação DS rs ae glad (4)
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la)me none Posirwe EnnEcTIOM Deconmrro
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c “a
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to tmrt a
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(elê indo) + 0o Ala rindo
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aja SUR RT ih IO dolo = EP ddr
E) petociry or E 15 SAME AS tetoaITy or Es
Ur tstide = Sé od 4
P ul 8
LEU A8 din do
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INE CHOOSE THE POSITIVE DIR TMN TO THE EdFT
0
bt sum
E
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;
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ap t+ 2Zagros
REATINE MO TOM: For VA o THE LEFT, Age delas EU
«aguas
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e e AR Êo a
WE MTE THAT gpa MUST BE To ANT, CM, Ugo — 950 meosls
mos unem É=ÉS: -R5O mmlso-ZUa
Mnsã00 conto
AEE UpE AD medo
(o) SprcE RECEEMRADIA ARE CONSTA,
Mp = Qt! 4300 mujrza,(és,
2 z
SO mente Lg Smnde "a «q
450 m/s = Ag (és)
Ag =-RSamfs”
A Emo! mA
28 an mf 8 )=-2000mmlr
: ag=800 pls +
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MOTES POSITIVE DIRETO 4$ Ta LET.
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agr Bufo
LEBRE US DF CONSTANT LEA
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apS AS info! = Sto fe A
(DD) ame A sr8ers From RE, (ttal=2, aee fra] zo
AT É=-S8i Ag binho E Oesp ts)
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(6) sEmeTA OF Perry GE CpgeE CRE,
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eim
dora Vg= LO ommfr
ipr2 20 muli)-o
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emsTasT
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be) Ritative vetar op O Wir CESPET To E
We Ugo Ly
Wa emos Ugo +00 twin af = 400 omoaf
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8
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BIÉ Cove E haLUE OF.O = Stopa OF Ul coma
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56
=
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51 6)
MÉ CURVE: CHANGE IJ AREA UNA É CURVE, Es dé mm
Erro
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Kay ge Elm k 65) = 18 Etta
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d=2os To Erbos! Kyo ans CEmÁsNIO 5) = "BOM Kg 2%
SINCE AN aura Pagverina 12S ms UItL ARENE AZ.
Siena. 8 JUST AS 47 TORNE EEE, THE TIME
Cen Sétiil E JUANES BREEM 45
É = DistâNtE dg 25
TO IRS ek 12.5 mfs
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uma RÉ LS E
- 263
rênt)
EÉxI2bs
Ay ta 325 m
HOUSE E HSM ME) = 324
-D,75 4) +37E —zas do
(E =-Ws8s Joe telotes
MEX SPEED: = HSÉ, EhSÍIO4Z) = 1:63 20;
GSE hm A
MOTDEIST EBOULO MCCELICATE FOR MO GES My
THEM Co TIME AT SRED OF SEI fomh
PISTANCE TREE OA
DERAM BATOM OE DG) FROM Srrrun TRANS
Eita)
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avo fes
» dé taeto) WO aoas
/ ,
Loss “2 “3500 injs
tome, “ee ij
A(Rjo +
(a) wrem É =245, MAX VALUE OF FOSITIOM COML =45Pm 270 += pm == a trromb ace)
(b) Posso Coon Sem IpHEN É=I8S quo b=308 4 ps! Dna 26) Estroin, É
ta) , &
MES Pesre Pias MSF For az et quad cons, Vutã) TO mn açtas
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f=0 Tu É=%os TUR LISTÍME TEGVELEO =88m » Fe Lise)
th) Dezreminma o é: Eno suma rramereS Deca nove Pins q -270 = 978 20007]
Dis Or IO (amo Nim emos) Bs A “e 02% -Ó0R” 0-0
Rj ELLBIM
HS?
WE CAGE POBITIVE TO THE IEtaMT, THU rare
e tO-ês do pés
a o For x=cyt=Soró = 365 4
16 | (SOUTO aprs=(06 mb)E=iZnho 5 6=185
õ8
ze)
mw
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voto) ue Ag SHam)i sm
:
E aa —
am
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300 media mitos ) ria nisfE,)+42om Los /4s PLAMER AMoves = 30 in, fo RAT: FIO FE H6A,AS bes
Tera TIME = E ebgrlço Seitas fo = TES | Ponce muesosoin orem! caos Dto E ROSS
ia ETIML TIME = ISPESHAS E LS FOSSA ASEIOES Em E ES
COMECE OF PeRMSTÇLE UBACITAS (5 ARtM)s E We Era
Simet pecstemamom 15 - BinÃO, or 260. [5%
THE SLOHE THE É COEVE MUST ALSO RE Eri ls) DALbin/S
afiado)
&
e) mês
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j AT
a som. e
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H
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ps tag
ÁREA UNDER WE curve = Z80m. For sie
ACCELERATION AN DECELSRATIOM CHOOSE CONSTANT
ACCLLEIATIA AMZ CONSTANT DECELERATIAM HE CURE AMENITURE,
Almjs?) sa
o Zu
artomte)
nr A
5 eye fe)
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Br É[1 mt + (45) = 4840 om
Comin TRA nin 18 Ani
mia
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as Sms os
sas
a-tousm 4
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AREA VDER VIÉ CURVE = J80m. WE CHOOSE
COMSIAMT DECELERATIOS AMO CONSTANT RECELER A TIDAS,
Nozê: PéceemaTos + RitLttAIaM OE O 6 mist
palmos)
+a= 0,6 msn
ô z
“o UR
Zês
vii
ms, + Aga io
DANI AM / Aa “49
É 1 E :
k
RBS
Aa beso m: - US MI Tay VZ8 5) = 280 m
oi Eae BF 18 toh A
For fereunanos Fans : aro np A
AS + OE mito sinfe pr PE comete
4 e mÊs
96 po Fatos
e
É ME.667S
É HE 86Is = 284
z =// 3385
É 4É,228s 5
for Decertepren Base:
Lo= SOME CE V-E CORVE
as Pam] h&mis Some — pSrmde
o & ZA 2325
ap - 0 880m/e* 4
MEZ] ren. Da= Et dmho 4Em]s , aço
Avzo: Cas Pi Bmlh = 25 of
3 tanto) auto
Rs
1
») “o
so Ee
For Hevck Yg=!5 E
£or Avto: (t>i0s) ugralzs kar ZE (é -t0)
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tirasse 4
Xe “8AS Á
tomem Pct, Tg %g?
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(&) AuiO STBRTS 35 AFTER PeVCk PMESES AUTO
atom)
AUTO
AS|--— —
48 ç
1
. ls)
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Es Zos
%="300 m
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AA
Vr= LE Us mbtos)
1.63 | PRAZIVE 70 TRUE, COR MusT move &
DISTANCE! furl rHOrSa Ino [PE EE
ALLOIARIE INCIEATE 1H SEED: AN, 250 “BC ads mido= 2RTES
cae
(ré pe
fd zo fe ise
Cm [ci
mM
5 rpuere 46)
t— E —+& +44
Aecerenanos Punstr Er He = as As
Dececegamem funse : Ego" Bhonh!S A
(22N4%) = 88.946
fakas) = 424 fé
BET: AVEM AA LHE TEMP HIDE SIL) EP 387S
Compal UPE HE AME PETAAS IPI tras 4
LES | RELATIVE TO TRUCK, CAR MUST move a
Distance: ax JátyosSatro =/ vá PE
vfiets) car
ANE O
a de [DO (rata
Try T a
“o Preto ets)
ps —>—+ 4 A
4 SE = 20h53 = FE,
Are Ada Srt ft = lim tsto)
sete Gisa NE Ade)
Ba HR Zy= é. 8885
Esorad Etta 6 RICAS
Abr SE ES(ESM)= SHIB PET RES mito
Somee Das BSmilo, EI IZE SM; HIS ao att
12 =
[1765
ml Rat A |
Memes Rena er.
CoRLISIOM 15 AUST...
AVMDER. THU WevEM
PISTANCE BE TwEEM.
ma ANO TRUCK IS ZERO TUE
, o) TRIM AMD AUTO MUST
HAVE SAME VELGUTL Ve
| Spade MEGA = 4x = 0%
É lua tefsK os) + É (18 fio) £ = 30 FE
AY Creu Net EJA USAS HA s22n)= SO Ftjs
Bog asas Ava (E)
-30ft- Agyml1 3335) Agyro
(2) ab=somi/h = Be/t%
Genasss
=-27,5 ft
=SE ls
we Dra vt curve (assine Bog is ese) LET commnuro
“vo - Vímio cena sm
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OVER TAES FRENHT
REUITOR nem fracas
UNDER THE É Curi
RE ESURL, Du7 6,
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PEPRECENTS OIPPRENCE
/ JO MESAS) 5 Bo,
mos Ab> 88, Axo
| TIME REG UBEO IS
| E-sratsees
W 1a
Feranr;
gr as at
Disrames ranvemo s
a=Bet(Sakkiso)
d' = 705
eae FAN = TEEM - 30 Aka
E) = HE anil =é6 th
66
Venal Cotar dE Vtls Both = 26 Fis
for Bora) pres) TRE Eme peleciry 45 PassTrê
AS WE MINE ASSUME, THES For gera te) ame
DECEERA TOM = LES FU?
Amit)
dio
FP REEMT LULEVATOR MEVES UPWARO AT
CONSTANT VEL CITY
Y =E mf Gp=o
Poscemem EsivATOR STPEA Fem Ee ano
SINO vue rarzt
= 42%?
THREE SECAS MITER IT AS FISSO EF
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Def Tue ai aro gt colour
(correm)
mee | aos agro “|
ob fronea) 5 sm) | de
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jnrfm/s) eres 4 PELATIV Marim “so
br) 0244 5 irei Guto 40 cómçtm Lo 4 OL jo
md caes EMITE =, O
—— 1 E “ “
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Hm om; 6» doxue=s
tre2s
pa É apro rOME.. His pts
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A 1 o < +
a id E) 2 (O [ | bsmls
Gs Ap 10%, dn [026 E NR
TUE MARA BOL smnores Mpweep y M ARE of
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dl rectas DE NÉ Dinsmaa IS: Judas - =dxa ga
FúGee LA Zen
Cross) tons dfo NS adsipao) = +0XCanje
e(s)
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b é,
39)
tos) os
0)
25 sã
Tess) [
too) |
(a) ugeu = ne: SD «|
Ka (60mte K 200) (suor Area)
a 200% 8%» Cpo 8
(b) tram É
AX= (60 muy s— Es)” (smapão 6208)
= fevendehas)- mz P8Om
Áx caapã
Cavernas” de” Ss ema E PAS ms
13
UPR | ue aroma THE or eve or rue
SERIES OF HORIBONTAL EOSMEU LINES Srta
- PY THE FIBORE, FOR ENCH INTERVAL THE
ACC ELERATIOS à PSSUMED Comsrin
Ava a af or abr Ya
axe varrza(né)?
= 90 Bm/h= 25 mes
Reta 1= PS quis TO S= 201,
ASA ms AVE ZO-M ="5 mft
= AY
Aé=
axa mstatalas) (ES 285) HEINZ zo)
ESSRSPm SP m Áx= SES A,
FiZoMA S= 20 m/s TO = domêbs:
A PE mf AT=/O-20 = — dom
st= SE. das AS
Ax= sé +dalat)= Comtio) Hasta)"
dx=
= 80%» 70m dom
FR = Sms Tp EzO
A Sms
Axe bate do dafatj= mto Dá (e Fm fik 2.828 x
= DEI3m gm AT AEE m
For ENTIRE MT» VeRE m/s 79 vo
= 1
HE. TE [o Re mtnitA(é- 2)
al y . = Bt Ut dat NÉ)
3 e
EA £ Vnnitepa st? «
amis 3) -
A 1
so E E 6733
h OT Z/4..0.8)
0) ga)
10
E
ses
RE) since Ar=0 WeEm É=0 Aro lira É=d, THE
CHANGE JM AT RETUEEM É=O AMO É=-A 45 ER,
TÉLS NEEM Umgan a-f cveve ds ZERO
Aria tão
Leste dlritolos)+ E (-rofa-os)=0
9-2-Sg p40 g 285 4
(8) Postzom unem tet= 248
EK, A,/E- 02) P Ap(E,-0:733) + Aa (EXE -o)
Oro dakas- aaja dns genas as] Ztay-8)
= Bm LEMA ps 766 m
HIS] sec soro or Peg (184 poe at core
Etmey tnios , ds
(a)é=F0t= 2885 +4s+2235 A=Qys afmis?) tes dos RE,
(8) x CEM= SELIM táom tibs xt CA EIITA tre 2 ato)
of » B
UTE | ta) AT E Por Compésporame ro fa So,
X=3ip, we DES TRMSENT DO mx cueve (A) Fosrmem Wuin Esdos -Hels
“otindo) a WE UEM mens: X= Kore + TREE)
= AEB mis =-tm + femtotzos so eso fr litasUs)
zo dr. 3 ade =emasom 424% = 2m RE Hom
e A (6) Minumoura Posrmis Cmmemmare occups Wam ds26
Ev: arde SEE É Corvê AMO NOTE THT Vga, OéCUAI We COPAS
=lisens), me) sos =-14M
am GA ij" | tm
io)
lar A bin,
(8) Br que Bermr Corr estorno To V=ó infos
INE Dfw THE TANSENZ TO THE VÊ cuers,
» Vinho) WE THEN MEnSORE: mma YEm gm -/a%m Ena SPO
' .
das Sh
mA 1.78 | E as
”/ Bur &s E 8! E do O
2 7 -48 4%
“pokes) 2 ah Ne
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o a=- SE tus a 2] K
i VA
o.
AO. IM ORQER To SE TRE LET
Soovem tm Fars (ÉS TRE SAME
STALE MUST BE MED ON TÉ A pp x
PIES, BY LISA THE BOSE RRLATRrIS AA
ATL NÉ MAS MEASURE 41 Ar Ala
CUANDO OF TRE SEALES USED
Bo
Ema Yoga E Ko! mt TA(Eç E)
Roi 16 Entendemos) male) foe)
Lypes duiys:
= Rr FA)
= cc tisrenyinaa to restos) ze) + le tutekas)os)
+fsamfag ze)
EA LB m + IE m LED LO
Xu, = 1/04 40
16
MOO | nu 283 g4* =22s24
Br So Et oe la = se 12
A Su rIZt Je Ape Ha = é
(0) pmem E=is:
[184] qo fee po
: = gem Ap rip é
to % no “Ape pe
vsma & me mar toras
é <E xi oá
= ly Epesçé
o
as = Npc pls ção
Bon. tysé Weldtih zo 6 4 yo o pEfA cosa )o pise gs Le de
+ agaré As TU mA ES - cf P(A css pe) = as" y
(6) mdep tos? “8 Jr nrês Ê
CB apa O 18 int « Eae Dis atous rum de
A 18 A a, E IES «4 “Ss OrgrersD Alone OP. Homer,
fe) me “= 3 9 a , > AsxPRESSIoAS FOR Gy GO AG Shots
em Lx 25% >
= Há vga 44 Testes Per a “al qse a o
aero agrté esmo” AB) afcaiçaz = eras calor Eostab) O
7487 é nero 67 bo Pag a(A Cntot + EE sris pt) o
h mer Foriro que Enocuer Ke Orverne E er (D, mengo gr memger
=97 (e) Lite) Te 4
4
«a
GEMA MIS A Hrtrega « rar emplr us amr
2 1,85 =s6lm- e TE) g= Esse
g=(é+) g- HtaerE nda = ture Uj = 1257 cas STÉ
Be 24) WE-B(E mi 4 ae = pre agr OW = -samsnvré
mA Ag: até” (2) Wpem Eixos
Epa fo: º = +47 =+07º U=82lntediao A
Wet ty Ve sab quo | az agro Er VS dn do ca -«
e+2 au rap A CR CEA 882” E] (2) mem toorers= ps:
(bi pm Eros, Dz tre MO = 939% mr -
ts d,=-2.37 82 fa TIE “ E mam eastema)= O Vo ssdindo «
It? AGU Bam A EU? “| a cute stages ir +, as ssintirms A
= norteia) == 03 JR O
ML82 | ge sigam grosttaze-s És
va glretra =2+2 11 86 = [E sis péli retj HReospo)
ae Sigde = =Cya-1 W=/Epcopll + cj -(Rponpi)B
van = (ze Rena Be (Rent) — (potes pa)
a MARGIITINE CP VEl oerry
deem A1 do drnerno, 115 mes peniana; Trus, mê aprno | AS tg nO Lespeesgi) 4 + Re pesin poe) O
dE) afae- 23) nMtra=2(88-4 442) =O te aaa Piceeç a
=2(tt-mj-o E- 2.805 « dudértdE 00 Gees st ted Om fecrc evito 2
tetos = 28 drarrajra;- Cp ls) ol pasa) = Re
x=(2ataea)+? Z--28% «a
amostras se Sm 4/1487 87] r= até AGE res
Ga atas) - -8 WE Age) cr deasº A URL, quase) peee?
ESP ess o 2
“yr Z8+2 Micrnao S Veserad « Ve corram Taro qr (agoe)- -agg?
183 z For B=| 82: ati ratgr ze
pe, XE ve ci fre, =45é 7 sas oe b
= ce = amar cas dt roda » soé a sej + &
=. =“ =,
Car ciaõom de Mu vão at ege Dada? w=f/a sed Paises
im tete AR alyne eis
ve som! Rs) pum 4 velmto = 4ms* «4
ae veem o AQ (6) item EA ri ER eo] 3 feresado
o axPrsuia)Tem = 7a 3 onde
(8) sue tio 25 + + Fer se 1 SDEUTIM OF Pesa MRE, WE compre
453 44 cos = 247 Te Muy hs] Sra => A vga Ep 22
dy” dofa)= do - aê r” é tee Vis aa EEE
a rosa 2= so mafl | me Ez! “ro «1
ay= de bre b=2s*
de routers) 1 tt opopmpt d
"2 santo ?
IZ
1188 |
(2) veericas mega
(um, Accer. Morrer)
joe, Caro
ge fat
20 = flop *
das
Hotrzantae porto (ur Fetm
x (dt
HOm [NAZI E)
tc dert mts
(6) For he Zêm amo fbs bl Pomnis, me mave no
geést? monHONa)E trrors
Cefult= (1) ImisN2OR 5) x"527m
EP
7 / Ass
Lg 2m E la
A: -57
tm Ea
o
amomípe 1
mui o y
GAME HITS CS LRRIEST VALUE SF,
Ka ESOM de= 1om
Dom Elos me?
ES Da = VU HEIS 6)
Exp usas s
ys 294”
G= 587 uts
Ke WE
SEND HITS 83 SMALLEST vaLVE OE Vo
Rar Am Gg E 4350
tu
Je 7a” AI SLB) Eco sam 5
me mê Lmeblesms) age 28tmts <d
FOR SAND To ENTER CHUTES 281 m/s SUL EBIm/; A
(a) Bau grs Cr: Ee/S E, ge=3ft
Bair durs i
qrógtl art=fiszame)t
RE Wt ME tt=W(om8r7s)
0) Bmemes 8: Ya SME, Gac 28, (romena)
Get apeilzrane! Pro Bsoss
ug bÊ NS fé = 05 (0.3826) = Hab Ft);
BAIL HITS DE KpT US /= Le gp2 EEE from)
Gorági? art=Álmarrei E=04217 5
Yo GÊ mer = fonte) apr 32% HE]
FoR BRL FO STOME COBUPR RCD3 BLA CUL
fr, Ret =o.78901,
(1, Ja = Yo smzo o, 220 dy
1492 x sk -— a
3&
Eros
1sR
5 xa 66
Enzezs fis ATE: AnsO VaLvs or Us jade 3 FE
st PE DES E=ESS/ a
= Ly z
stt = 0 ao é +Elsrade (asas
SENTE For És Bafo 300) SESI À
FELH3+ ES) menti A
vas RPE ATE: sra est VALUE Or Dente,
CRE=OSLES E rERES, dg ts
astro ar mtv (sec? et
“asse fo gem pras) 16 E)" Ls essttts
FOR SOMEeE TO EnTER PIRES JERTEIA SIG E MS A
93
HI0
| SEE swerem or soturmon or Prog 4) 89
WE Noto Have dz RE
(a) eat miTs E: Hs Pitt, ge SR
gesto afacb(gazp)o Ecos
Tb 28ft-Wlomers) esto
6) BALL ITS ES Rg RPE “as 28%
Yo Ego? ate blaza po d=oatães
%g" Ut as tem ilassass 70.9 AAA]
Ve) BR MITS DE MASI up st
getos am-dlempe” Leonsos
Rel Hft=Blov3s) Ur ssErt) <T
DRA BUU Ta STR Cone BED! SEL USUI
set rt]
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Cosgtnaos tele
WATER ETIOXES RB: SMMEEST HETOBT b
Kg PO Fé gb
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o=h ilinzsricosme o)-d(22:2 210 NO PE <)
0=h+/40025-/5.2553 h=4359 +E
Érova6és
WATER STekES CS toReET HEHE by
Ve RPE Goch
RU RIfE-Boratak)t é=2/095
Yes Goat Sae 2
D= halimesreapne)-2(222 Jin s)
o» venia) * We gal de
For WATER To EnTER BET
A3SIt cheese
18
5 E
102] rstomj oiêess (ajetbaina [MOS PR
3 HoRiItoNTaL Motrem
Eidedad (Exportar Méier 4G=/6 E
a B - EU = (8 cosa Je Soe/6 nbs 2
ATE: Yp=/8m = (18 case t E
Sgt « Qrt233 1) A E
Í Cosa
heltça sem To
VER TRA. Mo Troy =,
gude dg é? pn =>
9-8 soa) - -A(asa * Far partimos Gg, BALL RENCHES CELINE WITY
ATE: dg =IRm=(18 sir DAI sigo (OE ops A Honsromzal vecocery.
; VERTICAL LAO Tras Ré UNE Omi AEERATED)
SUBSTITUTE cos E secim = /4 bon? (uy +2ay
12= 15 Cano — Py0 (Je bata ) ATE: vo, q am? 02 fu) -2(cbrmf ia
Fara = phor Land + 4$73= O é (UT 989 cais
Lana = 23 tanus/ 434 Flom A oC Es Cia EE - comes
a = Taz PARE 9
Sretma MTS RB FOR a=dtPor apr MÁ] pra sy =) elo
Vito] vom) Celossra tg C)aemss mto
HORIBONTA MO TIass
hotsm =, = (1235 mo, 90985) = 12,
A BS mohSm a 2om KoF (Ed, Exa é mio. 95005) em
E gm Z= 49% E =4BM-S240m = S66 m
= From E TOR? t
os /6 mf; (UE tb cosa Cad 6 sina £ e X - Som Ss pas
sm = 43 ce fl VAI mk
Lemon torne m=(ULE=(H cost) xja
AT RI YaF18m= (focos) £= LES (1) Ven qreõe taozom -
Zago ,
MEZTGA AvOT: g-lnt-tgê g=29 tie alga minho sau a) > = 0:36 m
ALR: Ugo Pm=(/e sinat- ELA! FRoM SETE WE MTE Zur Garbo
: p Als - -
use Eol): 2= (e sino fito) acl E h= Gai = ESPPm asa
! EA
suasmrom Expj SECAS Iara (h106 | Mata = 4% 4 Uatulideas 130º
2 dB rana 62079 (rttana) p= RE Pais
Enio PAPO ond + Lageo= DR Sin tão!
am O ABS Land + hIZar= o Ses = Sosa
and = 7,3320 ÉanA TONE a
= ég80" e =224+º B=19.7 ,
LHES enETuer Bate Mo7S Cpremkg = Puro tis IRD
oe onde” fm) = Mg sin dio? = ep 204 fel 4,
vg (Ur st=/406 PRE = de Wi= ss 2(sks) cos v208
Ve gt emoe cps posts dE SE calota) cos
FOR ha* MEET Va O = 18,20 =p.até E=hrses dos 777, = 2 339 bs
2 ” “Aa Ta
t-2gé ah= =
t-ég À Va é fede E HEIs PAES éys
= mm jé = Strst spo
duto Etr bases 9a E UR A ass)” PZzo0 yr ty Rr 270º AM
Nisa E MO27%m 6) í . x
2. sm À
Seneé Crreneê E 47 rum AtovEd, ig SE rtn0) 2(chsoro
Fel A=UAN Brel HITS Ceni. E VC = Mto
For à=298%º (ae VE sin 2USH = 2E8ES Fe A ao E? Soy sinto!
“ado gt=2888-p7/é 8=189 “BIO ftys SLB fts
Fem MAY item g TESES RUE E O FOU)S a =s2s"
g- Cut -Lget= tee (e). loane , à | Dtd=do!; LESTdedo fue! Mato Bd <«“
mer ESTE ABES: Jomeç? 2MEI(O BOM) alrmfo NA) | (e) pm dos pnrervars Lee Lap 000 é em yes
Goma 3/72 om so ADD E SUA Tio
SME csprenmê v5 AP = hm AEOME A, ms Teo) maia EA eira
Bate wtee giro Porn B For AESA] TREE Desa E PO
e Vo tda Dee ema em! Va
£!
NORTE
LAW OF cospar
1pã e (520) (são) 2lsakczojeess"
Vain = E HO era dh
LAW DE sines
E
sda SK
persas
Ve” Sofa, 82" A
“VE 00 Rao/h,
x
Ê +
o p= Sl
.— - Le
227 ta
df = 904% cratera"
1.108
Uestobn/h
Lpw GE cos puts
nas? = aoot+700"- 2 [imo 700) cos B
cosa = asas
p= 382
vaso = A 48.7º
Va
Dhaceror or Amrre Bis GE we or scwm M
UE More pHAT
Ye
14,107 dos dé A Va Wmiln= é FG
vd ad 7 30 meia = “pttls
tu 0 cosnEs
E te A Up do! 9" a(s Naglcos do*
MPis & Vaga PO. 99 Life
Lita DF simes
BU, "Loja Sep o EM gua
ss 20.39
rm Pop ndo! gas “202?
(2) RELATIVA VELoeiTr: Ea o figa 8
(6) cnguee Ip Posirom cor At=as,
AY ajp” Veg dE = [ORAGO Cap Sss e PO
(c) DosmaMcE GENTEIN AUTOS 2 Secad AFTER
Auto À NOS MESED INTERSEE Troty
BuTo A Temes For As.
As 66 ij 4! 20º,
For UE =/6 Erk)(os)= 3218
SBR FE do DO
Lg te -
“" 7 Auzo 8 =p deh
ter / To GE s(otilolss)= 200 dé À
Tm di CL Costner .
Cao” (132220) = afpsziesalcos tro
Ca? 292280
Je lat aa Liomecg provem Guros= 293 EAR
uito LEMETH QE CABLE = CONSTANE
L= RH consraer
BAU to
Ag tZcap -0 (a
Dt
Para: = AS indo? me Pk)
cast)mata) 272% gj e ap
45 =-Zga 7 =-2 Aga
Cear IE ineo agr PE dado
Lj VE dnfo? Ca GOO E 4 vin fg Pes 0º
fa)lggocr E 8 Ag Vem
LB CF CosmES
2 “
toa tas 2000) cos 50
= 7013 m/s
Ler OF Sins
E se
Ea “203 f
à = 90 pasa qa! = dorsét
= 0) info 406
(b) peceteRAmoo GER: Ag MAL IB couro er
Usinê FIMÂLYSIS SIMINEAR Ta
THAT USED GROVE pros Lo
AN MTERNRTE MAGIA (7
Rg Eu * Lema
= dE int) 4 DELE! Dom dr
=-15 9 «(25005 sei +(1s mta
= g = EME É MEUS
a TE E “0.19j
a
É
AL
a
Iê
Y
1
1 40.027
|
= ME bj pés A
ZA
Hm
CC) For E=205 gr Pis qe
ME
Às So mis ree decr:
&
o Mat Vea
tau Of cCosmes,
“ A, a(s)" UXAS)cosdo!
po Não Gs ="gporass
. Sto LA or Eures
É SE EIS qu 7?
25 go fé
Vos HOY infs bag a
CB) Fon = 120, Cor VS tdo 730”
Law SE Cossnes
PADUA) cos te
uz 10.892 tm
LOW SE Eiras
Sbg sitios aroma!
25 70857
sas
Acdop-sar as =
= 00h Vós A
- 2
La = Emenda .
Emar 420 on mprwl do
TED EI7 77
Le“ Ga "Qu
CE) = Cimo) comeram o é t8fs =
Sec smo enris EFE, Core pec, Mazur)
Cu Cs mo Fes» =Fataaz fps)
= 1294 Etts À
CrLoTvE VegoesTy Do Como EF Tm 15º
Ve = “o Usgo
Vo 8
LE
E Tape CEPE
E / g/8m*
= =p -2"/848
“ to), Mori qejo-g A
Y= US
! =P)
my om = 18.926
Sp
£LAv OF costs
z Es
“small mmXB)eos (15º ms)
Vo= 70,08 Pg
Zi SE Sus Sa
WHEN be ZE: = Agtas)= (826% mmleNas)
Vo E ASAS conhe
ADE ASA mf; Cap
Lap como At LAw of comes sol) sntent) lara! io [IE UAI
Zoe ot 2 a 2 É Coca + Cm “OAB.
ue ai-(se) tiro) eloelicerasão! | SRA 20.08 File Er 20.08 Ri
aço SELL mn fe . = 243º
Low DE Comes $ Us fetiã 2?
se 120" . Vo pasame cm ADA
= = [ZE - .
20 me EREE rente (E8.6 . Vsk = 20186 N 88) «q
A = 180 -v286 = Sp Poa rAsoçã a
(0) Pecmemamers E! ESPE mf A SAS A
(3) peecesry of & KE Ag= comzar? RO Nh=6 WMEnÊcO NS
men d= 85 K= toa) ALeSMÃa5)=182. 6 aonko DO | rum Poor HS pro Herr mus cstrm
luz 1826 nho Et! A - 15 Perito 2748
te (ed Per Soy Te 86 veria
MS “ag LE Var
Mus] x = Bom” ro! E es, nda
« Zoe Lap” !Romnlos —m /s a pá Dad
Ag” 24 tLgya g + 15 ta
ps Som. Lt 05 Costner Ha Se am PET unteço
Es age (20) ao ala eçzos see pre Ma =a2r2tm
É .
Anjos nET Aga 52.82 mens * raso Gnt6s9 .
ZA OF sims . Med A
CSBEO sirms = 2448 6) For Mirra Vajg, WE mate Go Ng,
Dom” SPLL NÃ g .
o pe cinza"
(0) heceutprav ar E Cp LES men? GEE” aj % : E mera em 3.484
(6) tecoerry 0E E; me Ageconsaaro tro NG=O tones É =, = LIS Ff
amo 1$º= 249º
K persa”
e dose tie mis!
E
23
PHLIEG CONTINDED
W Acer manos
WE MA
ver” Ba sbS vols? —s
Ag = 03232 00/68 Tem PULO
2 * Rafa
Su — 2%
agr An E 4
am = =
ar Sms “a peafç 4
2 4 au (Bond; =230m
ow ce cosmes apyg(o3r22) + (he) doresalisjeasfas"-12) anos E vainto «(lord fo 8
Aitg7 338 eomts +
Hh 132
due ee a
aros Une Prodbernd is Fraso )
“e ams 9 Bltos 308) = rio an/e
x Usontscação”
Eu ASS Om
AT: (rmangarorm desta) apa 90 RB
Law OF Sines Sink Sins us a Vossas mp)"
di MST” 2338 v= es a-8º + 1 RUmpi= Pa =>2480m 4
2ajge IES À fe
1,130 E e bo) festa jose cones | 1) /33 | a) ser some peos dA
ATA S Fela
A
A= 372 ride
mes a-g= 82%
de ans (SAR sin 88" e 26.829
«58 N n uÊ
E Pa
(6) Lotizonça sro Tom 15 umiform: constam?
Wocrry A a a e
= Vence
E o
RU = oe
: assa
z eva ms 2
2.39 eds (ari ane E ; e
R a
um qe 227 8 LC istecsa)E 2
AT A= 3 so cs = 3 A
5
+ (Pose - BTAIA=O, cosd=1 E
A usa a tar 4 Pon” 3 p= foi Ga) «
=
TE HS] cana: acer é fm
Ri .
tr
Dib — lE)= 08%; eme ele, ate (E) “E m
Ba ES From Plog ne: a-Rp?, U=[eisRão piso Geo
eo, som (RpPco (Case
fa) 47 ds 3» =2%» y
armado eee en É «
aço A Ê 7 Rp?
a Ef a uÊ 3 z
Ed 4 HP) = geo Kb /25) 20.139 asas Fm
em ts 2
N RV = pre smcg €, Len olof$z PE) )
No p= 16573 fe Na ES mf AM
Gram - See soco 0E Fur 487 Rot uatues OF à, 5 fo Pilde
era sl ) (02) mute tros am ponl S=tm)s, ouldéso
: co Bros. = — a?
TB: UC testo A3RSP anfe Eot): famoso o + [gar]
ag tar) a = sed prot
é trmfeio BZ)” favas mag) (lihen deas: a se nfs VT nd, Pta
s
sato: (sm)* (encsjº | gal É
SU «(STEPS VERAS 3 peletIm af
Ré
= EMO mê — ZO3P x10t
4: 322 eds
* ooo rugas a sãos rd
abre rot
VE 2$ 501 life =/ 739706
Ra
Fe IT To milho,
4
= [BI6O mé 4 MO mi )= Hbowê = AddBNIO ft
136 | ar= 16000 mi tiigo r= 6ot% qoel
= milh=23.%4,
ilh=? or . EA É estos - god?
o fasso x 5280) Peso -sgot
am E ara ao
Bu tudu É=ds: p= dotortoim.
P=sta-tos Bomje
: 252 Feszo-moro
É) vecowrr org: ”
r= 322 tensas = AS sua not fe E Verê
(23,46 )* q
r=gerz me mos,
«=. $
Here agove Susrace = VEW.3 330 = PBL3 mí depris dom “tt?
hs tm 4 . Prapeo ne nos id
M.137 | Na = (tombo vlitomiles MM
Ds
(4) Recermeprom or &:
2 gr ape Pré O (toi Kt radfe em do dn)
eo Ag 2 TB raTE = (von ferrado) e 2bomiforade) = +45 1/8)
«
SO fa) me mou:
a, = PA creio
(e) Aececmeamon E E RELATIVE “PO g0b
a=éraVE = POSA wide *
Y= sa eas = vaso
a=-leroinde +oinide, A
algo 1) soe E In" O «
SETIME Ae? =. er Aq 5 -
508) 3 (9 Hs r= nasttoge
Duos vs PROBOLTANA Po . SED, =2.5p-274"
. , Fozs-cré
lb) temem 68 cre omez: = 27 same Wuem P=/67 q =)28-09=0,38m
Premo (26 Tirati For eme org, ÉS (2) f 7=-09mjf
fot: ves E v=R/E Fones -arms
% (2) uesociry qr 8: .
fofa) dez AEr am LA “Ozmls Aga Veras dead
Rap R ga hs =A10ô Í
MEGLECIME MIMOSINERE, WE cimose Va = Tp E=/85+ta; ie =2529º
. 27 4 A
mma gu La ra E = LB m/s
- - é o
R= 4370-Rm = suo - p Vas LS mfo, pros? «
Emnrenã ente, exi agy = Suas
Emir Etr minides | (W)pccmensro» or &:
[1.139 “A Phi) (2800 irradh) = ceu8stemf6E
Ago TB +2Ê = (002 Cm rindo ati 2 ms redhs)
x =-RL9P+ALTEO = 2047 ampe *
Ser Sotoram or Frog 4/3) Fe OR past 20
Petro, o Est) , Y=(80Wtas! Eat = sora
(a S7x16m) ——————
=(RBh, a 4
ss 2 a ção am Celtas leo c sta
ae posmério? aT= 4008 m/s Es cá
AT = SUS Rmfh nr= SEO Lnjo Ago ET mk, ums, A
— NOTE SincH DiAMOpIER OF MDOM *Pr$Am, TH (O) neceeratios 0E E REBTNE TP TREROD
MOD MOVES DELATIVE TO THE EltT4 A Drag sm
PEbo7 ema movr Tejo RIO nfs Y- 80; E
RT
asrôsró o sabotar) = 407
bo; Velimmider
(0) men tro: ds 0a ke,
ADO, Aga HOTELES De (J2S7 im fo des AA
Per Ramo E see Pres igidom,
(6) mms bos
Vos 4D di is e ró=beibJar= sm 48.7 nb
puro ras EM = 46 5 VE = JB, dg
Va jo mise, +USVindolãa,
a ="tor (a)a or E - anjo!
= 207 tar SÃO = 218º
as (287 mile +28 md eo
Aga 0OT= 1857 44,
Anadia = UE ndo?
Hrmê Po 26 cout Saut Hg r=2&li+cos jrê)
rtmebat ra FonTê ade
(2) tetoerry P= fm etre
Vo Pe cAbw sinal Gar6s Ibicaswê geoerr Um Pao-Db safra q
atado (280) Fcis usPe (cos co) = foéu) marés Mó (r+ecs £re) «a
dcesuraros ae téc geerecamy a =Y-r6*
= F- rd casacos ut (bes at Xi)? Lengua - do Blrrcos fre)
Ce rates ata) Cro cg TI + 2 cas fr) a
sebitsia ut : ae rôsará
eatrabo (pa?) cadi + sinui)= (44) = 0 +2( 28 sintré£S)
ar sbu* «q as -mEsintré <]
(U) snee Ve fbw- constmir, ageo (a) waiy d=o: r=28(n)=H8 0
Fes: Ag A= vam sado vezury Wo Gs bl) fia
az Ês (5a, disse er bo p=b uezrber 4
P Foe TUE FRIA p= comeram | Rerum as FZ
o e THUS, PRIR IS A CIRCLE “q
b
43 fmcmpazocam SPA CElmien este: vozêlmos Ejaz4
A jo e=ané setoery = -DbsinTo-p&
ie É ram = Dera) DE
lemocrry rcÊ=to mis ape rô= (ot)n)= 2ont V=- DBer + TÉ Es «
Aeegreamom astro o esta) =-s07% fectresôpor q Lnglages if)
agr e PL AT -28
«
a
-L9BeVEo
MIBE | see soro or Pane lJAS Poe 50, O,10s
fodpdem t=dss p=2birasE)samib, aus
nbs EDITA Vas meureos 2)= s702 0h
«a a -omme e +10M8£, A
he zaã) a 5
Selic szonrt as Pibeint= 030074
+
2
aA=-1207 ze EDIT ÊL
-«
(E)wmem é=2s vao e=7 =u8*
. P=rvass s-Mbsero =nblitesn)co — V=O «
= Begins J fecfs ram e-77 as fiBtiszen= Los agr -PAsnT=o
o ' AÍ a-inbeo M
Srisga' —— Z
= SRF mms” ?
Sigo
| Aos
RE dj geo!
D
fé=e, ro, or0p a Boto ve; (820
R8
HIS8 =8 S=2nt E smABnê
ê zo &=27 Ê 20» 8 cos 2rnt
Ê-o 6-0 E 97W'6 ain amnt
Vercerry (em dhr)
Vc Ret ROC, + ER
V= HALO Es tam nE cos 2né R
v=grfa*+ ntelcosdemnt E)
Peteenmoy (EG $0)
a=(Ê-RSTE pres abb est Eb
2- sine MW din crnt A
-«
az sra q nº et sintemme
1159 contivuro
qe zrfnirs cosa
dE VAIIRTaRO
é n2s nã cado
since AT, 9=0
Zuus
smce nz (BD, Ga fy
1.159 | From Frog hIS3 Fóre n=! aro Es É
Ve ADA E tim Bes6A= a0[ A E, + Bens E]
Eine £]
A urAto- 7 Esme bi = 47 [her
Diptera or ElhormaL 45 MERLLAL TO EYQ
From Fr MR, WE OBSWEE THAT
Core Cr B=€p Brte= Ea
"
traz em[iforêcs e 6] sato Eme £|
= emfai “AB cosO E, AR Sine el (1)
WE SMAIL EXPRESS PrT vsime Tue M9,8 Commona ra
Ep" CosE E + Ina G
-smer +cosod
SUBSATLTE InTe gel):
ara=str [AR -ABf-cos o sino i + cos 3)
dr 97 Esin6 cego
visar) acara o gé E
A
a= o Prim 5 fu as
SINCE JHt=0, WE MAY Were
assa ante
Aa xira PrgrUS DE Cpavarpes cecues arD
dar
P(A ve) fel
SMA [Mem À
MiMIMMvM SEADIDS DE CLRVATURE Oceves n7 BD
; E
tesao a=trVateg? do.
Tavsr
st e ut smaE
asas 5 E”
mm latçer (ui vãs e
«
1.160 | Pohtto gre ache
= bh temp 6-47 é=h
fo é-o Z-o
Ve Re rRÓR HER = h dans Corhifimplere, rh
ve hy tas € +47%44 daria +
ve ham AÍ I77 7) =
as-(Ê-csjar(rã ab ER
=(-ht Emekem de, + (2h bana)ae,
= Com" bt tenJta (uh tema E
«
-AB(sinGcaçã E + sinta á)]
= sr [66 ma (todo voimto);]
va a amS [RR - Ag 5]
(a) We mozE THAT SINCE A amo E Me CONSTANT
THE WICECTOM DE THE EMORMpE( PRRACLEL
7º TRA) 15 ConsSTAM7 AND THUT THE
Ap MEMO IS ContameD tm ha franeo A
SIMEE THE INTER SECOM
OF A RÁNE Ano A
CretnpeR ts Are EULIPsE,
Farm 15 au Ersrose
(Db) Baon sr Convarono mes Pesg dhiog For ne)
fececenozas (EG ps)
a n(Ê RE ep t(RE RE) ER
=[ne tag «eu leç+l
os20 etjane, - 06%
“ER
O = urh tema Vrtore* «
HLlé! | Prab-et)
=heE
Werocrry (Ea 95)
v= Rep + Rós, à
U=A0 "er ame De, +actk
lejymen so: Rest; Rea 88 velhos 4
Umert-cistero qzmhe, e LA «4
ne q0f0%r 6 cod aca A2 TEA smianE add azfitrendavar 4
ge =ant3) Zigfcos ame (nsinapaJam 47 Ra inicos int gumes É =c0; e te
. Varia Tcoçtage Et Eca art &= -s9"4€, assra o!
ContinveD) o ”
s/
HLI6Z
ULI6M | mem sorira or Pers 1.86 E
RA O=zmt east
Rizo á=27 DA sin vê cost
. . DA sinznt
R=0 80 =R2.3A cas art
vELocTy (EG 4.49)
Ve Rep trocar IR = LmAegrirA ch Amt
ve rA/Ey sbiz7e «
AcemetoTeN Legitso) '
| astE-Ra Dep +(RôseRó)e +84
=-sraLe 2a cosare R
aspmalo + costeze «
REZAR cost e
11.163
É
Re-Zgamt é
FP =-gRcost &
ta) fara cr PáezeeE offer
o 9
Cxmstose PRoJELTIAM 06 PH
9 Plane,
K=Pos8 =2R code
= 2A(iVircos 28)
(u-R)=A cos ze
g= FEMO = 2R205€ sin)
g=2R(4) sino
g=A snze
WE MoTe quer (x Aegto Alca? tar sitrejo At
THIS (8 EGUATIOS DE À CIRELE OF RAEUS À yisza
Center 47 RR, jo sinee te vetoarr
Cortona =p “5 COMSTINT, THE ERTH OS RB?
ELI, VERTICAL ÁS 7 670 RÃ «
(é) vescerry ano aecestrnzrer
La map! U= REptRÕE 1a
VE( RA cintleg ticos rpR
SImcE Ep Ea, R nam fr oRTHOSONAL TRAD
reger = Ver sine) PER cost) p?
«Vet tE + cost) rp? veto!
WE NOTE THAT NT = CONSTANT, TEUS dufft =o
EQuso: a-(E-réDE, «RÉ cera, + ER
=(akcost het) ep + (0-4 Rsint)es + O
= -(shcostle, - (eh sinto,
as Wairar rap = 4Bflade + shPé aê 4
(c) apos ot corvarurE we tetra Tur ISO,
LO has
Te (Pecas pl ref -CRp sin pa)
m=fpspoo fas pe)
SMEE BOT VAMO O dE ly THE OZMATIM PLANE,
dum NOZE THAT que furcerios 15 BRALLE! TO JAR,
£ á
ares lRprospt) epginçe)
Apsiapt) o ARfcsspt)
=-etptcospê é rep" singt det RS Hsipéreos pt)
Aua= Rp-ecospti +Rpj re siopê 4]
dez : = Que BETUEEY Cy ANO TRE 6 ASS
cr, frxaeg rp lre]
heat) Pp ie cos parcele sim pa)?
cos = Fr
Vigia?
Ever Lincama! L OscuLATINE PANE, THE ANSLE
BETWEEN OSculaTINS PANE Art) THE q As
Ss E= 90%
Lan =dan Go) cota fÊ qi dE q
*
q R
aê bar ps É ? Basa
+1).165
From soLorns or Prom UV:
Er tTrstR
REj Ae R
SincE BOA SO AMO BLUE db LHE OSCULATINE PANE,
WE MOTE THAT SINORMAL 15 BREBLLEA TE VAR,
td 8
quad qe geo RE E 4 1208
a ur 4h
Cade TST MEC” = 9208
EIMOEMAC IS unit vEtTOR Epi
tra
-B6c-8j +29.8]
= i
cos É » ops Se asa -«
cos Te = -ooras ar 223º «ad
a
= sã = aanar S H.0 4
Mibá
PANEL pro TIO (erica Bececem ATA)
AT 70, WE SLhm/h= AS mts
AT S=iom Ve Ando = PE nfs
E abieza (5,5)
(25)*=(15)"+ zas(iso ve)
AT E=/oom: = Ut 2a (ss)
ur = ces)" e 2(us28)(100 0) =4H67
ve APLD cale
= 2047 mh*
ap=hasz et *
Ui zon?
PO 2
TOPA! ACELERAR
a =f0.223) Eizo)? ar
a «fara
acdvini: 4
169
CARLIA der be =+00005
vem ax (ga vt)
” e =9+R%)
õ
! ; 1 Au A
bold dlordiio 4 apt ade)
DATAS iz cmol, US ds fe, Some ad?
| fogoosk1ço)*
PA E = 4448
he a dale EE =). E er mis (it trm68)
b = (sos0Xo640) b=HBtm A
LHS
(E sa BEE -2cos ave) d
(87 cos DEE H(47 sm ari)
r
Y
a=Emeinnado «(Bfitos ané)5
E we Érrgs
eriti 4
a =-femsignj +87 Eos em)4
e: o tem ar (EM) E)
(4) farm or perco
% ; s
| de
ns
LeRriene, PIO TI Leds EEE ALELIRA DES
ME MOTO qua MAO Merss = ÉE runas
AT y= Yi, Vys so:
Va “ta +2alg-u)
os tuá 2227.2662) 404)
2 ;
(orzsz ho (urigos td
Since = 4OfE)s, We mera
ia tiirto;
a ,
(so Pu) = DZ + zera tdo
Cedz seem Rg
LET Eq TIME For WATER TO RENA For? A
ATAS ao Cy ta)-3 Ém
= 1605-322 é,
Za O. s98p 5
For wazem vo Eepem E: Ez 2 099 S
err Or MobDO mo fUNIF orar aa 700)
DEL) ba (26.444 Novas) = 36,634
= 36.4 FE
«
SE NOVELGS ; xr deent ur-zesanê
Been THAT cos Ze= sm Amp wire
ya Ausarts (1 2swtnt) ty
Bu, Simee, “= sm 123
OR smirt = %s EQ) Peer
3]
I— = Ho
sir ss
Praga GEP, -«
33
vPWaRO memo |O
(Posre Sense 1)
Dobun vago Aga Tra
Possrive Simoes +)
emp rsss - Pb
Peoerem HC!
a= vas = -& + A a?)
lg bela) = h n=2gd (2:22)
A
as Va =
Carchr
GR)
[LO PRRRFRANAREEANIIEREOACLRIEALS ASAS RLENEARUAA PESAREANEIREAEERRARRAS: | CRIMAISREEARRIAS CREAM ARRAES SIR a
Ea PROBLEM 12.C1 ajoe PROBLEM 11.01 .
aa vs BASTE VERSION «jr FORTRAM VERSION -
pree ce ermuteçe :
e
Definition of Symbols c Definition of Synbols
c
VO = initial velucity of rocket (m/s) c vo = initial yetac. -» of rocket (o/a!
H = narinia height reached by rocket (mr c H = maximus height reached by rocket (n)
À = Geceleration of rocket tn/8) e A = aeceleration ef rocket (nei)
e
Data REAL 4 40
Ka 0808 e Data
6 = e.6100m . En p.s005
6=9.8
PRINT vo H wo e
PRINT" tuts) tm em cpRINT MRNE Co ttamado > vo H vei
MRITE (eo COBGBNIS + (m/s) cm ca
FORVO = TO 400 STEF ES WRITE des! C/s BRANDO
e
Hacioum height reached by rocket DO 19 1 = 6,400,25
H= CSPROLDEC Lea t90-2)/0) w=1
e
velocsty of rocket as t hits ground c Haximia height reached by rockek
VE = SQRIB/GÕA — EXPI=BEHMO 1) Ha 49.SAOSLOBC eita (UDEAE) 6)
e
PRINT USING" a savO E Weiocity of rocket as it hite around
PRINT USING 0 Mad miau UE = SORTCA/KOLLO — ENPI-B MME)
NEST e
WRITE o28) vOsHyUF
va w 28 FORMAT BKPS-l IBPT ERA?
18 CONTINUE
qm/s3 em E
9 a.eo
es as as
so nsi9
a 252.86 o É Ba
128 ULI.96 eis o
125 ses:
1 tam anão asi
175 quan aé 199.3 :
119158 vaias
eo nica 11.78 Ligas au
2es 178.46 19.0 acesa ds. 12
eso tása és s82/28 : e
s85.79 gaia
es 1878.96 12e.gi
7-$3.97 102.38
ae regia? 126.92
Bia 199.36
E 1858.78 1êg.e8
= asie sia mus.as t1u.7a
1275.44 nela
as ergo.1s 131.22 ai ss ja 29
soe sara.as ts2.28 1579.98 124.81
ias Lu? 18s.se
1e53.78 189.49
980.12 raglga
2
132.20
36
00C0000000000000000000000000000000000000000000006
, LP eoerem
Fhrw or Szpepro or Wehrer
5 mt
É=o o | T
1 afio
42 do E
enem E * TT
gre
gta sinto é - fgu*
ReMCRANHEs AÍ Rs do!) fifa)
» Z (EB cia dot +5 (a 12)-o
cr
deerrenta Mezom (DiiLorm)
Us U casdo*
VERTICRE POTRO (otros AECEERATES)
7 = UG sr bo! —
9 9
O
z=B =À) coso)é = e. sf
g
+0:0? g =. SE nindo'» AE Ja 52
3
cá (9:42) £=05(-8 EEE ) S
N, fr to, Pheqreee or hunzer Lemes A
Wire velcery Vs AND Srs
N Tht Eleoueas or Eur Ding
N 87 TumE É,
Wy
ÀS MEME Of Vi INCRERSES WWE BAVES
Esp JO Que cu CH, Ware
EPuIMRES Glrua 67 4; dice -
ATT
X= Sm Qro GALÉ, mAze
DTRMES tobil RT; Las
E
= é duo U<SIZM Wuarer
ETwmis Poor Br; Foste
2—
Fort GTO, WATER LTAKES Llovmi ATA,
Fo
FósÊ
E=--VE os Cala: 2%
use E) 70 cmo É mem use (PD) 30 Ermo Ne,
Fropo o GETOSGÊ, Emo É= IA
ve (2) ve cer gy,
E=T mb) Crolêrata = 26/8
se (E) ve Eme E, rmem use Do FIND Xe; THEM BD 214%
Be VT los Co luafa = Lp
vsE (E) Te come É, Titio ese (D Demo
10 eresaas essi emmmaners | CS e er
Bo PROBLEM 11.02 + | ce PROBLEM 11.02 x
a a BASIC VERSION + Lo FORTRAM VERSIDA +
“a va eres Re sRMASM Eat se seem | E: REMO NIE NI ASIA CTA NA LERALA MSM SAMA Latinas REM
sa. e
o XwY = coordinates: origin at the fest pf the firefignter | C Definition of Symbols
7” VD = initial velocity of the watar c À
o: T > tive for mater to gove from the noztle to c XY = coorginates: origin at the feet of the fireman
sa. where 1t strikes the ground or the building c VO e initial velocity of water
106 * c 7» time for water to aovo from the nozzla to
c where 1t strikes the geaunt or the building
126 * y foras angle of 48 degress vit c B+C = coefficient in quadratic formula
136 + horizentatl therefore
1a + Vo Babe = Vileos 48 = vDxa,5 sem vo
1 VD sub VOssin de o
1=6
es —
1-2
176 rs
168 *
198 PRINT “Noter Strikes ground at à distance 4 from firefighter” MRITE (e ML/SBRBANICA “Water stei ,
ess * BRITE teccranro àrefigirter
e Note = 1725 4,5 metera, mater mill ntrike buildirg.
c
COM TINurD Com tra
S7
Potes tema LLC ConNTiseer
Frosram Base Vizesso Continigts
Fecigterra FORTRAM VERSO CONTINUE
zes FOR VO = 4 TD dé DO IG M = 4,16
esa Bm SOR3I0/D: Ca -2.4/6 vD=N
Bos T = &SWC-BASORID'O MUCH): Sm TAVD/E B = -SORTIa.Jav0/6
ese WHILE X <4.5 Er tis
eo PRINT USINE "Vo = 88 0/5 “IVO Tm S.58-B+ERRTIDEE — 4.265)
E PRINT USING "x = 88.888 ASIA * = TWaia,
coa sora sig LE CI.LT.6.5) THEN
Pa a VRITE Cu,86) VD,
288 fa let IFI>1 GOD 5 ea FORMAT E9ks Vo = PaF3.D afataOho! x = MoFbD 87)
E PRINT nero eo
ze PRINT "Witar atrikes tno wall at à distance y above ground"!
E Note: x = 4.50 and y< é metara º BOTO MG
e Ta ev ENDIF
= Y = 1.2 + BAEEDAVDAT -.sage(TIE L
E WILEY Cd ao 12141
E PRINT USING" Vo ké m/s 405 JE E.s7.1) BOTO Soo
sa ORINTUGIHS = de déa WRITE to" (//4BKBANH? "Mater strikes the mail at”,
3a BOTO s18 + * distance y abeve ground”
sea END WRITE tes t(/,AAM)
“o ses. c Note nn 4Sa am yr da
o DES MES cora asa as T=9.80
asa PRINT "Mater strikes the roof at a distance BD from B” e V= LB + O B6AOSVONT — O.5HGRITAER)
Er Note: Height of roof ía ya & a; x < iB meters.
se B =BOR(BHVO/6: E» 9,609081/5 e TE tao vos
cao = 58 ox. x -
fil QuE UPE IPDE ctmenha o nbavoa E) FORMAT CORO PO a ARS, mM SAPO! PD
«so BD=x-4.5 BOTO ste
“98 PRINT USING * Vo = Mk m/0 *9V9I e
309 PRINT USINE "BD = AMAS aº;8D Boto aa
sa SOTO 18 ENDIF
E e Co gn3+s
“Me -3+
Ei Ena O Me TEMA ora ado TF (3.GT.L1 GUTO 450
ss PRINT MRITE (as! (Z7/BXSBAV) O Uator strikes tne roof at”,
“Water strikes grouna at à aLetanca » from firafighter! . a distance BD from Rº
ses + date: Water ques completely over Ouildino MRITE Gesrirrano
578 BD = SORIDAVD/O: Cm -2.4/6 c Note : Height of roof is ye bm. 1x (I2a
ses T= SM-DASORIBPE ARCH): = TMMO/D 50 Dm cSURTIS.MAVOIG
E PRINT USING" Vo = Mt m/s 140) S= abro
28 PRINT USING «yu 00.808 arsã Tm 0 5M-BASORTADeE — 4,40)
s19 NET X= 6.5 + Der
SF EXLT.12.> THEN
RM E 24-45
Mater atrikes ground at a distance x from firerighter WRITE (+48) VD,BD
Vo = 4m/s *= 1$8a “a FORMAT (Me Vo = sF3.6 m/s Mo BD = PoFhss!
voa Sais xe 276 a GOTO &15
Voz 4 m/s += 27404 ELSE
Goto 53
Slater strikes the mall at a distance y above ground ENDIF
Vo = 7a/s y= 6.886 e c
v- Ga/s v* 276 a se F=k+t
vo e y= 689 a 1F (E.61.1) GOTO 578
vo = y= Dela WRITE (4,7 (//,BKBAU") "Water strikes ground at”,
Vo ve Sta + * a distance x from firefighter”
MRITE tea teZrANdoO
Water atrikes the roof at a distance BD from E c Noto 1 Mater goes conplatsly over building
vo = 12 8/6 BD= 4.138 a mo & = -EORI(INWOIG
Vo = 19 0/5 Bb = 6.761 a € = BA
7» 0. 5eI-B+BRTIBRE — 4,60)
Mater strikes ground at a distance x froa firefighter * = Tavares
Vo = 16 m/s = 179% à MRITE (9,56) VOA
Vo = 13 m/s y = 28.593 m SE FORMAT CS, Vo é EBD, M/S! = TaPR Dm)
Vo = 16 m/s y- B3.07 e SB CONTINUE
END
Water strikes ground at a distante x fron firarignter
vo o 4. ms +» toma
ve = 5 mis “= Eb a
vo = ma “= 37
Mater strikes the mall at distance y above ground
vo + 7 ais * = Bs
vo = 8. m/s “= Ema
vo = 9. a/s * = 4a
Vo = 18. m/s * = Sela
Wo = 1 m/s + = Sim
Mater strikes the roof at a distance BD fros E
vo = 12, ais BD = 4.138 m
vo = 13. m/s BN = 67%
Water strikes ground at a Gistance x from firetigntar
vo = 16. 0/5 * =17.96m
vo = 15. m/s x = 26.88 m
Vo = ió. m/s * = B32m
ss
Flrogenra Fi: Ensu Vesostar
Projectile mith drag
Constant ralates to gragr K = 2
7 v vr
: as mts
a "mes 771.35
5 mos st.as
19 91725 unas
15 9.85 2aB.35
28 19.5 127.35
es qi9.es 28.68
a vi9.05 196.48
35º q19.85 -S8s.65
“a ques 56.45
es o19.25 877.68
T= 08.00» NED= -s9.met
Range = GE neters
Tine at impact = 87.9) »
Projectite with drag
Constant related to drag: E = 7
T va vv
s mis ts
2º mos ms
5 08% dee.34
18 eov.ê7 v35.81
15 esg.a 271.30
Bo sseso 128.49
es BB -Bp.8p
zo gana -2t2,72
E BIS
sm esta 587.86
os Buaze ABR és
T=47.50s YE = -Seg.5s2
Range « 41328 sotera
Tine at impact = 47.08 5
Projeckile nitn drag
Constant related to drag: k = 7
7 va w
s as ms
2 gas Ti
5 seg.m s7t.9%4
10 67.16 386.98
15 Tot e12.s
2» at? a
E) 13.58
a Jo6.ze -BéB.1p
as aaa 6,15
“8 ess 557.88
Tr a4S6s YNEG e 179.808
Range = 33466 meters
Time at impact = 44,84
Projectite with dra
Conatant related to rag: k = ?
r we w
s mis na
D ge mas
5 Eeom sazize
10 251.56 235.45
15º 698.49 135.92
eo ag “11.46
as ago.46 17.2
3» 571.66 -B19,29
a Sauas aus. 05
“o 46.95 573.33
T=oLSOs YNEB= 13.669
Rargo a 27615 meters
Tine at lepact = 61.48 4
8
as96
193
13789
15908
est
EA
34770
1346
eos
2E-&
x
a
usp
E
12564
1796
eos
oa
Bose
E
E
Pos
10E-6
x
2
usa
E
12u8
16439
epaas
Bessa
e7se7
so7se
vPOs =
BOE-4,
eua
esses
veos =
Y
a438
sgas
77nS
sa
asse
seis
E
“sao
mes
seems
asa,393
Fhossera MC 3 commp
Prosena Rom: ferra Vezes
Projectile uith drag
Constant related to drag 1 K » &
T vx w x Y
- ms ais m a
2 9.25 8. 5.
5. ago asno. Basa,
1. mi9,as Pa, aos,
15 99.25 asg2s 19709. 798.
2. 9.25 eras 1aBAS. g987,
Es. 9,05 965 apo. qem.
8 GI9.B5 9445 BOSTA. gêso,
3. MI9.25 asmos aos. 7e75-
“5. 939.85 516,65 Sb77D. Saça.
450 OI9.25 MT7OS Giz6s. BI9B,
T 00.995. NMES » 49,682 vPDS » 21,994
Range = 44BE. a
Tine at impact = 47,9] 5,
Projectile with drag
Conatant related to drag : K =» EE-&
T vx uy x Y
5 ais ns ” a
e moas ras a 9.
5 S08.74 abas esta. quam,
1. So) assai 9009. assa.
15 OMbL BYLLDO 13664 7795.
Bm GEBS4 IRIS 17h 83
es. 7.50 seg qegog, Bog,
3. G7.iS ceiaite Poma pela,
35 E59.Bb sra a1O4a. 4757.
dB: GSI) -Sar.Bb anaas. asgo.
45. BUBLE3 ÓBD.As DOSTO. tune.
T =47.50a. VNES = -spg.529 vos = 42,097
Range = 41329, a.
Tine at impact = 47.98 6.
erojactila with drag
Constant related to day : K = 108-6
T va w x Y
s ara aís n a
e migas ras a. ..
5. eme sm auto. qast.
19. 87.16 386.99 esa. sms,
mo 99% Bus ion. 228.
Ba. MI? 95.08 L6b3g. 7082.
Bs. 788.06 3,98 PASS. 270.
Je. Josze aéBiP BaMhI. s75D.
55. 67H43 culhlS p74S7, 890.
a8. 4B99 557.35 9754 cume.
Tess, VHES = et9.897 vos = 163.289
Range = 32468. a.
Tine at impact = 46.84 8.
Projectile with drag
Constant related to drag : K = Z0E-6
T ur w x r
s aís as a a
9 om rms o 5.
S. cam Sage ugo%, 3055,
19. mt.50 335.65 ses. só,
18. a9sat ISSO! 11H86. bhúbr
BB. 497 ciiio6 IUBU6. TR69.
ES. éMM4s 167.83 18903. asup.
J S/l.ié -Bi325 BIB. 534.
35 BMth -MQ.05 eaIto. sa.
vB. a96.95 333 eMBo, Ba
Temos. vNES = 19.471 veos = ee3.a01
Range = E74]5, e.
Tine at pat = 41,48 5.
41
Eecguem Wc
aq
td ô 2 alo? mx? q
As ají+a, A E —
2 é? > p soco FE
fes)t= afras
” A =Açtth a Vas) - a; <q
Ave a, dé <
For comam d Dorme (My TemvaL dt:
- 4 2.
Ax= VAL +7adÉ =
= VAL + Elavjat
ax = (rrgav)at <]
-u +
apa pt A <a
10» errenas Ereem
8 PROBLEM 11.04 [e PRIBLEM 11.C4 .
ag e BASIC VERSION +| 6 FORTRON VERSION +
“o cassa sentantass | O asscassrentao esses ese
see e Definition of Syabols
“o Definition of Symbols e
% e ve velocity tfe/s)
o» wa velocity trtzas c X = distance travaled by train (ft)
> X + distanca travoles by train tft) e AA] = components of acceleration (ft/n452)
106 ANsAT = componente of accaleration (FtfarE) e Ta time since brakes var first applied (seconds)
no X = tine since braes mero first applied (amconds) c DELT = tine interval aê uhich brakes are raset [1 second)
126 * DELT = time interval at which brakes are reset (4 second) e DT - final time interval until V becomes zero (seconds)
ap + DT = final 2106 Anterval until V becones zero (seconds) | C
ag * E initial velocity: 4 mi/h = de ste.
186 V = 685 PInitiul velocity: 46 aih» 8 Fte” v= 68.
149 DELT = 1: “Brake setting interval in seconds! c Brake sekting interval
170 » DELT = 1.
185 *Caleulate U and X at ore-gecond intervais until c
sa - ng veineity becomes negativr 87.6.) THEN
Boo WHILE V > & = VerE/ugaa.
21 A = Verao > -SORTCR. Sine - Axo)
ses ar = -SoR(Z 502 = Amro) = ATSDELT
2a De = tATMDELT a tv + DUB. MADELT
E DX É (44 DVJBIADELT =veD
esa Va v+Dy K= 34 DX
asa Xe xe Ta T+ DEI
ao T=7+ DEI ELSE
28 vem soro 24
ema» Eni
E “Asmune uniforaly aceelerated notion during ons-secord esta 18
310 "intarval uhen velocity b9coons zero. e
2 TaT- DEDO X=X-0R Va pr c brsune uniforaiy secelarated notion during the one-second
388 DT» V/MRSLA £ Anfervai Man VHOCIty Dacoda dera.
do Ox = (rar 20 Te T- BAT
3a Tetebl Ka xD xa x>0x
E v=v- De
378 PRINT "Total tame required to stop trains” DT = VIAS!)
Das PRINT UBING * T = 84.44 seconds; TiPRINT DX = (vz. tupT
É 4 PRINT “Total distance traveled by the train: TeT+BT
noo PRINT USING * 1 OMG.MO Feet1z x= 45X
e
WRITE (6,38) T
am à FORMAT (/sSks "Total tino raquiros ta StOp Erain 1ºs/420X6!T = MF
Total time required to stop train: 45.85 seconda?)
Tn 38.39 amcorea SRITE te,48) X
“8 FORNAT (7,8X, "Total distance traveled by the train 25/520,X =
Total distance traveled by the tram +P7.ês! fonte)
X = [780.55 fast em
Total tine required to stop train
T = 29.39 seconds
Total castanco traveled by the train
= 788.55 feet
+42
DO0000000000000900000000000000000000000090000000%
[2.1 =32,09(1+ 0,0053 sin' de : [2.4] Mass of satellite 15 independent ot quavby:
ê ( soa) Mb me S00 Rg
= 32.09 fys* vy=26. X 102 Rem dh 5
g = Is sete” = (26.1 x10%m/hX is )= Z2Sx rim
fo 22.26 Rfs L= mar= (500 RX 225 x 1025)
Maçã: At al latifude m= od b L=3,63x0 “ego aé
A 00 db cy ibe
Cds Mm = CSSH bit ATE STE E
Weight: Wo mg
PO W=(bIssy bs/eifaz00m6h) = H9R/L A
Pedi tl (050% JAP EA) = 500016 q] O Ny É
ba: Wefoleserjersrampi= son al
2.2) 6) Hego wWomg = CANTE me) |
W= 824 N e
“o Mac * Sose os en modo: mo = 2h Dar?
TZ
frmg A=(sta
"Wjel Drive sisp=o:
E pl) = pd
+ 2homa H+k ma
MW = Ma
LR = pa EA = Pg =080(88))
[12.3 ja
do 10 seole irdoeçier És
| fre foue Fexertod | (b) Front -Whvel Drive
or bhe sprink nc prq fo te w Fi ma
cr to 4 0- bg frase: E a — LN DE z
S : = So w) =
F F =(IDRAnBImf) = 1962 N Fr ? ora = cuz
Free Bed N-04W N=06W - as dé
Fa g624N Tere Eedy: Boy Tom Lo
Est . ma - 06 Rge oslosoyaB)
= ma = AN mg a. poa
ima (isto =1mAN-(58 Kim) . Ê
os a = 308m- Bm! vie zen (Km con)= 565) mis?
a =82mjs* + 4 W=l3IIMh =8Sbmlh
E Lero Lente É) Reoc Wheel Drive
Assu. unegue ) fever arms w F = ma
ob op cos DEM=o: — me MD4W)= ma
| =] bBh-ch=o0 o F a OEbW ora
E MN -
É É Epoca =06M > =orig- autos)
Elevatr af rest = 3.139 m/s*
ma NÉ, - W=0
Frhe AMA MM
a= 1848 m/s”
(a8t8 miss om )= 76 EE
V=3069 mk =H05 Am/h «ad
Substituto into CI):
algra)= E (Saga) mt) O)
Tomparme Edna (3), we havem =. Thus fhe Same
[pass is oued fo balame bar. Mass indicated is still 15 figa
am oh Substitae indo (1) | TS POR ( 16 be Om) = 171 ih
mg = e (rtgs = m/s = 69.3 tm 6
amgt teia 5 2.6 |60) Umtemdy aereteroted motion
to arte : derated m-€(i5tg) (2) “p= 198 fenloz 3 ms; V=0
va lar ULelera
mia E Cio tigla v'ewtsrax: “5. o)! raa(rs)
=Ha o a -- 600 mist
a z G =6.00.D/8* e
mt (tg)g - V=wtrat: 0=% -600t E=500s «df
* Ficuf= ma Elisio - (sta lá, “ Eçre
E = milgra) “E clstalgra) & = avo
7 a
2
FouN fu=W ma . bOQNh
t M Gg O RBImp
M= 0,612 -«
H3