Download Kinetics of a Particle

The design of conveyors for a bottling plant requires knowledge of the forces that act
on them
the ability
predict
the motion
of the Education,
bottles they
transport.
Unpublished Work © 2007 by R. C. Hibbeler.
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Pearson
Prentice
Hall, Pearson
Inc.,
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Kinetics of a Particle:
Force and Acceleration
CHAPTER OBJECTIVES
• To state Newton’s Laws of Motion and Gravitational Attraction and to
define mass and weight.
• To analyze the accelerated motion of a particle using the equation of
motion with different coordinate systems.
• To investigate central-force motion and apply it to problems in space
mechanics.
13.1 Newton’s Laws of Motion
Many of the earlier notions about dynamics were dispelled after 1590
when Galileo performed experiments to study the motions of pendulums
and falling bodies. The conclusions drawn from these experiments gave
some insight as to the effects of forces acting on bodies in motion. The
general laws of motion of a body subjected to forces were not known,
however, until 1687, when Isaac Newton first presented three basic laws
governing the motion of a particle. In a slightly reworded form, Newton’s
three laws of motion are
First Law: A particle originally at rest, or moving in a straight line with
a constant velocity, will remain in this state provided the particle is not
subjected to an unbalanced force.
Second Law: A particle acted upon by an unbalanced force F
experiences an acceleration a that has the same direction as the force
and a magnitude that is directly proportional to the force.*
Third Law: The mutual forces of action and reaction between two
particles are equal, opposite, and collinear.
*Stated another way, the unbalanced force acting on the particle is proportional to the
time rate of change
the particle’s
linear momentum.
See footnote
† on next
page. Hall, Pearson Education, Inc., Upper Saddle River,101
Unpublished
Work ©of2007
by R. C. Hibbeler.
To be published
by Pearson
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102
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
The first and third laws were used extensively in developing the
concepts of statics. Although these laws are also considered in dynamics,
it is Newton’s second law of motion that forms the basis for most of this
study, since this law relates the accelerated motion of a particle to the
forces that act on it.
Measurements of force and acceleration can be recorded in a
laboratory so that in accordance with the second law, if a known
unbalanced force F is applied to a particle, the acceleration a of the
particle may be measured. Since the force and acceleration are directly
proportional, the constant of proportionality, m, may be determined
from the ratio m = F>a.* The positive scalar m is called the mass of the
particle. Being constant during any acceleration, m provides a
quantitative measure of the resistance of the particle to a change in
its velocity.
If the mass of the particle is m, Newton’s second law of motion may be
written in mathematical form as
F = ma
This equation, which is referred to as the equation of motion, is one of
the most important formulations in mechanics.* As previously stated, its
validity is based solely on experimental evidence. In 1905, however,
Albert Einstein developed the theory of relativity and placed limitations
on the use of Newton’s second law for describing general particle
motion. Through experiments it was proven that time is not an absolute
quantity as assumed by Newton; as a result, the equation of motion fails
to predict the exact behavior of a particle, especially when the particle’s
speed approaches the speed of light 10.3 Gm>s2. Developments of the
theory of quantum mechanics by Erwin Schrödinger and others indicate
further that conclusions drawn from using this equation are also invalid
when particles are the size of an atom and move close to one another.
For the most part, however, these requirements regarding particle speed
and size are not encountered in engineering problems, so their effects
will not be considered in this book.
*Recall that the units of force in the SI system and mass in the FPS system are derived
from this equation, where N = kg # m>s2 and slug = lb # s2>ft (see Sec. 1.3 of Statics). If,
however, the units of force, mass, length, and time were all selected arbitrarily, then it is
necessary to write F = kma, where k (a dimensionless constant) would have to be
determined experimentally in order to preserve the equality.
*Since m is constant, we can also write F = d1mv2>dt, where mv is the particle’s linear
momentum.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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13.1 NEWTON’S LAWS OF MOTION
103
Newton’s Law of Gravitational Attraction. Shortly after
formulating his three laws of motion, Newton postulated a law governing
the mutual attraction between any two particles. In mathematical form
this law can be expressed as
F = G
m1m2
r2
(13–1)
where
F = force of attraction between the two particles
G = universal constant of gravitation; according to
experimental evidence G = 66.73110 -122 m3>1kg # s22
m1 , m2 = mass of each of the two particles
r = distance between the centers of the two particles
Any two particles or bodies have a mutually attractive gravitational
force acting between them. In the case of a particle located at or near the
surface of the earth, however, the only gravitational force having any
sizable magnitude is that between the earth and the particle. This force is
termed the “weight” and, for our purpose, it will be the only gravitational
force considered.
Mass and Weight. Mass is a property of matter by which we can
compare the response of one body with that of another. As indicated
above, this property manifests itself as a gravitational attraction between
two bodies and provides a quantitative measure of the resistance of matter
to a change in velocity. It is an absolute quantity since the measurement
of mass can be made at any location. The weight of a body, however, is
not absolute since it is measured in a gravitational field, and hence its
magnitude depends on where the measurement is made. From Eq. 13–1,
we can develop a general expression for finding the weight W of a particle
having a mass m1 = m. Let m2 be the mass of the earth and r the distance
between the earth’s center and the particle. Then, if g = Gm2>r2, we have
W = mg
By comparison with F = ma, we term g the acceleration due to gravity.
For most engineering calculations g is measured at a point on the surface
of the earth at sea level, and at a latitude of 45°, which is considered the
“standard location.”
The mass and weight of a body are measured differently in the SI and
FPS systems of units, and the method of defining these units should be
thoroughly understood.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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104
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
m$ W
g (slug)
m (kg)
a $ g (ft/s2)
a $ g (m/s2)
W (lb)
W $ mg (N)
SI system
FPS system
(a)
(b)
Fig. 13–1
SI System of Units. In the SI system the mass of the body is
specified in kilograms, and the weight must be calculated using the
equation of motion, F = ma. Hence, if a body has a mass m (kg) and is
located at a point where the acceleration due to gravity is g 1m>s22, then
the weight is expressed in newtons as W = mg 1N2, Fig. 13–1a. In
particular, if the body is located at the “standard location,” the acceleration
due to gravity is g = 9.80665 m>s2. For calculations, the value
g = 9.81 m>s2 will be used, so that
W = mg 1N2
1g = 9.81 m>s22
(13–2)
Therefore, a body of mass 1 kg has a weight of 9.81 N; a 2-kg body weighs
19.62 N; and so on.
FPS System of Units. In the FPS system the weight of the body is
specified in pounds, and the mass must be calculated from F = ma. Hence,
if a body has a weight W (lb) and is located at a point where the
acceleration due to gravity is g 1ft>s22 then the mass is expressed in slugs
as m = W>g (slug), Fig. 13–1b. Since the acceleration of gravity at the
standard location is approximately 32.2 ft>s2 1= 9.81 m>s22, the mass of
the body measured in slugs is
m =
W
1slug2 1g = 32.2 ft>s22
g
(13–3)
Therefore, a body weighing 32.2 lb has a mass of 1 slug; a 64.4-lb body
has a mass of 2 slugs; and so on.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13.2 THE EQUATION OF MOTION
105
13.2 The Equation of Motion
When more than one force acts on a particle, the resultant force is
determined by a vector summation of all the forces; i.e., FR = ©F. For
this more general case, the equation of motion may be written as
©F = ma
(13–4)
To illustrate application of this equation, consider the particle P shown
in Fig. 13–2a, which has a mass m and is subjected to the action of two
forces, F1 and F2 . We can graphically account for the magnitude and
direction of each force acting on the particle by drawing the particle’s
free-body diagram, Fig. 13–2b. Since the resultant of these forces
produces the vector ma, its magnitude and direction can be represented
graphically on the kinetic diagram, shown in Fig. 13–2c.* The equal sign
written between the diagrams symbolizes the graphical equivalency
between the free-body diagram and the kinetic diagram; i.e., ©F = ma.†*
In particular, note that if FR = ©F = 0, then the acceleration is also
zero, so that the particle will either remain at rest or move along a
straight-line path with constant velocity. Such are the conditions of static
equilibrium, Newton’s first law of motion.
F2
a
P
F1
(a)
F2
FR $ (F
P
$
ma
F1
Free-body
diagram
Kinetic
diagram
(b)
(c)
Fig. 13–2
*Recall the free-body diagram considers the particle to be free of its surroundings and
shows all the forces acting on the particle. The kinetic diagram pertains to the particle’s
motion as caused by the forces.
†The equation of motion can also be rewritten in the form ©F - ma = 0. The vector
-ma is referred to as the inertia force vector. If it is treated in the same way as a “force
vector,” then the state of “equilibrium” created is referred to as dynamic equilibrium. This
method for application is often referred to as the D’Alembert principle, named after the
French mathematician Jean le Rond d’Alembert.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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106
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
Inertial Frame of Reference. Whenever the equation of motion
is applied, it is required that measurements of the acceleration be made
from a Newtonian or inertial frame of reference. Such a coordinate system
does not rotate and is either fixed or translates in a given direction with a
constant velocity (zero acceleration). This definition ensures that the
particle’s acceleration measured by observers in two different inertial
frames of reference will always be the same. For example, consider the
particle P moving with an absolute acceleration a P along a straight path
as shown in Fig. 13–3. If the observer is fixed in the inertial x, y frame of
reference, this acceleration, a P , will be measured by the observer
regardless of the direction and magnitude of the velocity vO of the frame
of reference. On the other hand, if the observer is fixed in the noninertial
x¿, y¿ frame of reference, Fig. 13–3, the observer will not measure the
particle’s acceleration as a P . Instead, if the frame is accelerating at a O¿ the
particle will appear to have an acceleration of a P>O¿ = a P - a O¿ . Also, if
the frame is rotating, as indicated by the curl, then the particle will appear
to move along a curved path, in which case it will appear to have other
components of acceleration (see Sec. 16.8). In any case, the measured
acceleration from this observer cannot be used in Newton’s law of motion
to determine the forces acting on the particle.
When studying the motions of rockets and satellites, it is justifiable to
consider the inertial reference frame as fixed to the stars, whereas
dynamics problems concerned with motions on or near the surface of the
earth may be solved by using an inertial frame which is assumed fixed to
the earth. Even though the earth both rotates about its own axis and
revolves about the sun, the accelerations created by these rotations are
relatively small and can be neglected in most computations.
y¿
y
aO¿
x¿
Noninertial frame
of reference
O¿
aP
P
Path of particle
vO
O
Inertial frame of reference
x
Fig. 13–3
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13.2 THE EQUATION OF MOTION
107
We are all familiar with the sensation one feels when sitting in a car that is subjected to a forward acceleration. Often people
think this is caused by a “force” which acts on them and tends to push them back in their seats; however, this is not the case.
Instead, this sensation occurs due to their inertia or the resistance of their mass to a change in velocity.
Consider the passenger who is strapped to the seat of a rocket sled. Provided the sled is at rest or is moving with constant
velocity, then no force is exerted on his back as shown on his free-body diagram.
W
N2
N1
At rest or constant velocity
When the thrust of the rocket engine causes the sled to accelerate, then the seat upon which he is sitting exerts a force F on him
which pushes him forward with the sled. In the photo, notice that the inertia of his head resists this change in motion
(acceleration), and so his head moves back against the seat and his face, which is nonrigid, tends to distort.
F
W
N2
N1
Acceleration
Upon deceleration the force of the seatbelt F¿ tends to pull his body to a stop, but his head leaves contact with the back of the
seat and his face distorts forward, again due to his inertia or tendency to continue to move forward. No force is pulling him
forward, although this is the sensation he receives.
F¿
W
N2
N1
Deceleration
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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108
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
13.3 Equation of Motion for a System
of Particles
The equation of motion will now be extended to include a system of n
particles isolated within an enclosed region in space, as shown in Fig. 13–4a.
In particular, there is no restriction in the way the particles are connected,
so the following analysis applies equally well to the motion of a solid,
liquid, or gas system. At the instant considered, the arbitrary ith particle,
having a mass mi , is subjected to a system of internal forces and a resultant
external force. The resultant internal force, represented symbolically as fi ,
is determined from the forces which the other particles exert on the ith
particle. Usually these forces are developed by direct contact, although
the summation extends over all n particles within the dashed boundary.
The resultant external force Fi represents, for example, the effect of
gravitational, electrical, magnetic, or contact forces between the ith particle
and adjacent bodies or particles not included within the system.
The free-body and kinetic diagrams for the ith particle are shown in
Fig. 13–4b. Applying the equation of motion yields
©F = ma;
Fi + fi = mia i
When the equation of motion is applied to each of the other particles of
the system, similar equations will result. If all these equations are added
together vectorially, we obtain
©Fi + ©fi = ©mia i
z
Fi
fi
i
ri
G
rG
Fi
fi
y
x
$
Free-body
diagram
Inertial coordinate
system
mi ai
Kinetic
diagram
(b)
(a)
Fig. 13–4
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13.3 EQUATION OF MOTION FOR A SYSTEM OF PARTICLES
109
The summation of the internal forces, if carried out, will equal zero, since
internal forces between particles all occur in equal but opposite
collinear pairs. Consequently, only the sum of the external forces will
remain, and therefore the equation of motion, written for the system of
particles, becomes
©Fi = ©mia i
(13–5)
If rG is a position vector which locates the center of mass G of the
particles, Fig. 13–4a, then by definition of the center of mass,
mrG = ©miri , where m = ©mi is the total mass of all the particles.
Differentiating this equation twice with respect to time, assuming that no
mass is entering or leaving the system, yields
ma G = ©mia i
Substituting this result into Eq. 13–5, we obtain
©F = ma G
(13–6)
Hence, the sum of the external forces acting on the system of particles is
equal to the total mass of the particles times the acceleration of its center
of mass G. Since in reality all particles must have a finite size to possess
mass, Eq. 13–6 justifies application of the equation of motion to a body
that is represented as a single particle.
Important Points
• The equation of motion is based on experimental evidence and
•
•
•
•
is valid only when applied from an inertial frame of reference.
The equation of motion states that the unbalanced force on a
particle causes it to accelerate.
An inertial frame of reference does not rotate, rather it has
axes that either translate with constant velocity or are at rest.
Mass is a property of matter that provides a quantitative
measure of its resistance to a change in velocity. It is an
absolute quantity.
Weight is a force that is caused by the earth’s gravitation. It is
not absolute; rather it depends on the altitude of the mass from
the earth’s surface.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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110
C H A P T E R 13
KINETICS
PA R T I C L E : F O R C E
OF A
AND
A C C E L E R AT I O N
13.4 Equations of Motion: Rectangular
Coordinates
When a particle is moving relative to an inertial x, y, z frame of reference,
the forces acting on the particle, as well as its acceleration, may be
expressed in terms of their i, j, k components, Fig. 13–5. Applying the
equation of motion, we have
z
©F = ma
Fz
P
Fx
Fy
z
x
y
Fig. 13–5
For this equation to be satisfied, the respective i, j, k components on the
left side must equal the corresponding components on the right side.
Consequently, we may write the following three scalar equations:
y
x
©Fx i + ©Fy j + ©Fz k = m1ax i + ay j + az k2
©Fx = max
©Fy = may
©Fz = maz
(13–7)
In particular, if the particle is constrained to move only in the x–y plane,
then the first two of these equations are used to specify the motion.
Procedure for Analysis
The equations of motion are used to solve problems which require a
relationship between the forces acting on a particle and the accelerated
motion they cause.
Free-Body Diagram.
• Select the inertial coordinate system. Most often, rectangular or x, y, z
coordinates are chosen to analyze problems for which the particle has
rectilinear motion.
• Once the coordinates are established, draw the particle’s free-body
diagram. Drawing this diagram is very important since it provides a
graphical representation that accounts for all the forces 1©F2 which act
on the particle, and thereby makes it possible to resolve these forces
into their x, y, z components.
• The direction and sense of the particle’s acceleration a should also be
established. If the senses of its components are unknown, for
mathematical convenience assume that they are in the same direction
as the positive inertial coordinate axes.
• The acceleration may be represented as the ma vector on the kinetic
diagram.*
• Identify the unknowns in the problem.
*It is a convention in this text always to use the kinetic diagram as a graphical aid when
developing the proofs and theory. The particle’s acceleration or its components will be
shown as blue colored vectors near the free-body diagram in the examples.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
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13.4 EQUATIONS OF MOTION: RECTANGULAR COORDINATES
111
Equations of Motion.
• If the forces can be resolved directly from the free-body diagram, apply
•
•
•
the equations of motion in their scalar component form.
If the geometry of the problem appears complicated, which often occurs
in three dimensions, Cartesian vector analysis can be used for the
solution.
Friction. If a moving particle contacts a rough surface, it may be
necessary to use the frictional equation, which relates the coefficient of
kinetic friction mk to the magnitudes of the frictional and normal forces
Ff and N acting at the surfaces of contact, i.e., Ff = mkN. Remember
that Ff always acts on the free-body diagram such that it opposes the
motion of the particle relative to the surface it contacts. If the particle
is on the verge of relative motion then the coefficient of static friction
should be used.
Spring. If the particle is connected to an elastic spring having negligible
mass, the spring force Fs can be related to the deformation of the spring
by the equation Fs = ks. Here k is the spring’s stiffness measured as a
force per unit length, and s is the stretch or compression defined as the
difference between the deformed length l and the undeformed length
l0 , i.e., s = l - l0 .
Kinematics.
• If the velocity or position of the particle is to be found, it will be
•
•
•
•
•
•
necessary to apply the proper kinematic equations once the particle’s
acceleration is determined from ©F = ma.
If acceleration is a function of time, use a = dv>dt and v = ds>dt which,
when integrated, yield the particle’s velocity and position.
If acceleration is a function of displacement, integrate a ds = v dv to
obtain the velocity as a function of position.
If acceleration is constant, use v = v0 + ac t, s = s0 + v0t + 21 ac t2,
v2 = v20 + 2ac1s - s02 to determine the velocity or position of the
particle.
If the problem involves the dependent motion of several particles, use
the method outlined in Sec. 12.9 to relate their accelerations.
In all cases, make sure the positive inertial coordinate directions used
for writing the kinematic equations are the same as those used for
writing the equations of motion; otherwise, simultaneous solution of
the equations will result in errors.
If the solution for an unknown vector component yields a negative
scalar, it indicates that the component acts in the direction opposite to
that which was assumed.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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112
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
EXAMPLE 13.1
The 50-kg crate shown in Fig. 13–6a rests on a horizontal plane for
which the coefficient of kinetic friction is mk = 0.3. If the crate is
subjected to a 400-N towing force as shown, determine the velocity of
the crate in 3 s starting from rest.
P $ 400 N
30'
SOLUTION
Using the equations of motion, we can relate the crate’s acceleration
to the force causing the motion. The crate’s velocity can then be
determined using kinematics.
(a)
Free-Body Diagram. The weight of the crate is W = mg =
50 kg 19.81 m>s22 = 490.5 N. As shown in Fig. 13–6b, the frictional
force has a magnitude F = mkNC and acts to the left, since it opposes
the motion of the crate. The acceleration a is assumed to act
horizontally, in the positive x direction. There are two unknowns,
namely NC and a.
y
a
x
490.5 N
400 N
30'
F $ 0.3 NC
NC
(b)
Fig. 13–6
Equations of Motion. Using the data shown on the free-body
diagram, we have
+ ©F = ma ;
:
400 cos 30° - 0.3NC = 50a
(1)
x
x
+ c ©Fy = may ;
NC - 490.5 + 400 sin 30° = 0
(2)
Solving Eq. 2 for NC , substituting the result into Eq. 1, and solving
for a yields
NC = 290.5 N
a = 5.19 m>s2
Kinematics. Note that the acceleration is constant, since the applied
force P is constant. Since the initial velocity is zero, the velocity of the
crate in 3 s is
+ 2
1:
v = v0 + act = 0 + 5.19132
Ans.
= 15.6 m>s :
490.5 N
400 N
30'
$
50a
F $ 0.3 NC
NC
(c)
NOTE: We can also use the alternative procedure of drawing the
crate’s free-body and kinetic diagrams, Fig. 13–6c, prior to applying
the equations of motion.
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113
13.4 EQUATIONS OF MOTION: RECTANGULAR COORDINATES
EXAMPLE 13.2
A 10-kg projectile is fired vertically upward from the ground, with an
initial velocity of 50 m>s, Fig. 13–7a. Determine the maximum height
to which it will travel if (a) atmospheric resistance is neglected; and
(b) atmospheric resistance is measured as FD = 10.01v22 N, where v is
the speed at any instant, measured in m>s.
z
SOLUTION
In both cases the known force on the projectile can be related to its
acceleration using the equation of motion. Kinematics can then be
used to relate the projectile’s acceleration to its position.
Part (a) Free-Body Diagram. As shown in Fig. 13–7b, the projectile’s
weight is W = mg = 1019.812 = 98.1 N. We will assume the unknown
acceleration a acts upward in the positive z direction.
(a)
Equation of Motion.
+ c ©Fz = maz ;
z
2
-98.1 = 10a,
a = -9.81 m>s
The result indicates that the projectile, like every object having
free-flight motion near the earth’s surface, is subjected to a constant
downward acceleration of 9.81 m>s2.
a
98.1 N
Kinematics. Initially, z0 = 0 and v0 = 50 m>s, and at the maximum
height z = h, v = 0. Since the acceleration is constant, then
(b)
v2 = v20 + 2ac1z - z02
1+ c 2
0 = 15022 + 21-9.8121h - 02
h = 127 m
Ans.
2
Part (b) Free-Body Diagram. Since the force FD = 10.01v 2 N tends
to retard the upward motion of the projectile, it acts downward as
shown on the free-body diagram, Fig. 13–7c.
Equation of Motion.
+ c ©Fz = maz ;
a
98.1 N
-0.01v2 - 98.1 = 10a,
a = -0.001v2 - 9.81
Kinematics. Here the acceleration is not constant since FD depends
on the velocity. Since a = f1v2, we can relate a to position using
1+ c 2 a dz = v dv;
z
FD
1-0.001v2 - 9.812 dz = v dv
(c)
Fig. 13–7
Separating the variables and integrating, realizing that initially z0 = 0,
v0 = 50 m>s (positive upward), and at z = h, v = 0, we have
h
0
0
v dv
= -500 ln1v2 + 98102 `
2
L0
L50 0.001v + 9.81
50
h = 114 m
dz = -
Ans.
NOTE: The answer indicates a lower elevation than that obtained in
part (a) due to atmospheric resistance.
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114
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
EXAMPLE 13.3
A
B
The baggage truck A shown in the photo has a weight of 900 lb and
tows a 550-lb cart B and a 325-lb cart C. For a short time the driving
frictional force developed at the wheels of the truck is FA = 140t2 lb,
where t is in seconds. If the truck starts from rest, determine its speed
in 2 seconds. Also, what is the horizontal force acting on the coupling
between the truck and cart B at this instant? Neglect the size of the
truck and carts.
C
900 lb
550 lb
325 lb
FA
NA
NC
NB
(a)
SOLUTION
Free-Body Diagram. As shown in Fig. 13–8a, it is the frictional
driving force that gives both the truck and carts an acceleration. Here
we have considered all three vehicles.
Equation of Motion. Only motion in the horizontal direction has to
be considered.
900 + 550 + 325
+ ©F = ma ;
;
40t = a
ba
x
x
32.2
a = 0.7256t
Kinematics. Since the acceleration is a function of time, the velocity
of the truck is obtained using a = dv>dt with the initial condition that
v0 = 0 at t = 0. We have
v
L0
900 lb
T
FA
NA
(b)
Fig. 13–8
2
dv =
L0
0.7256t dt;
2
v = 0.3628t2 ` = 1.45 ft>s
0
Ans.
Free-Body Diagram. In order to determine the force between the
truck and cart B, we can consider a free-body diagram of the truck so
that we can “expose” the coupling force T as external to the free-body
diagram, Fig. 13–8b.
Equation of Motion. When t = 2 s, then
900
+ ©F = ma ;
;
40122 - T = a
b[0.7256122]
x
x
32.2
T = 39.4 lb
Ans.
NOTE: Try and obtain this same result by considering a free-body
diagram of carts B and C.
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115
13.4 EQUATIONS OF MOTION: RECTANGULAR COORDINATES
EXAMPLE 13.4
A smooth 2-kg collar C, shown in Fig. 13–9a, is attached to a spring
having a stiffness k = 3 N>m and an unstretched length of 0.75 m. If
the collar is released from rest at A, determine its acceleration and the
normal force of the rod on the collar at the instant y = 1 m.
SOLUTION
Free-Body Diagram. The free-body diagram of the collar when it is
located at the arbitrary position y is shown in Fig. 13–9b. Note that the
weight is W = 219.812 = 19.62 N. Furthermore, the collar is assumed
to be accelerating so that “a” acts downward in the positive y direction.
There are four unknowns, namely, NC , Fs , a, and u.
-NC + Fs cos u = 0
(1)
+ T ©Fy = may ;
19.62 - Fs sin u = 2a
(2)
From Eq. 2 it is seen that the acceleration depends on the magnitude
and direction of the spring force. Solution for NC and a is possible
once Fs and u are known.
The magnitude of the spring force is a function of the stretch s of the
spring; i.e., Fs = ks. Here the unstretched length is AB = 0.75 m,
Fig. 13–9a; therefore,
Since k = 3 N>m, then
s = CB - AB = 4y2 + 10.7522 - 0.75.
Fs = ks = 3 A 4y2 + 10.7522 - 0.75 B
u
y
B
k $ 3 N/m
C
Equations of Motion.
+ ©F = ma ;
:
x
x
0.75 m
A
(a)
a
x
19.62 N
y
(b)
Fs
u
NC
Fig. 13–9
(3)
From Fig. 13–9a, the angle u is related to y by trigonometry.
tan u =
y
0.75
(4)
Substituting y = 1 m into Eqs. 3 and 4 yields Fs = 1.50 N and
u = 53.1°. Substituting these results into Eqs. 1 and 2, we obtain
NC = 0.900 N
Ans.
2
a = 9.21 m>s T
Ans.
NOTE: This is not a case of constant acceleration, since the spring force
changes both its magnitude and direction as the collar moves downward.
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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116
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
EXAMPLE 13.5
Datum
sA
C
sB
A
B
(a)
The 100-kg block A shown in Fig. 13–10a is released from rest. If the
masses of the pulleys and the cord are neglected, determine the speed
of the 20-kg block B in 2 s.
SOLUTION
Free-Body Diagrams. Since the mass of the pulleys is neglected, then
for pulley C, ma = 0 and we can apply ©Fy = 0 as shown in
Fig. 13–10b. The free-body diagrams for blocks A and B are shown
in Fig. 13–10c and d, respectively. One can see that for A to remain
static requires T = 490.5 N, whereas for B to remain static requires
T = 196.2 N. Hence A will move down while B moves up. Here we will
assume both blocks accelerate downward, in the direction of +sA and
+sB . The three unknowns are T, aA , and aB .
Equations of Motion. Block A (Fig. 13–10c):
T T
+ T ©Fy = may ;
981 - 2T = 100aA
(1)
196.2 - T = 20aB
(2)
Block B (Fig. 13–10d):
+ T ©Fy = may ;
2T
(b)
2T
aA
sA 981 N
(c)
T
aB
sB
196.2 N
(d)
Fig. 13–10
Kinematics. The necessary third equation is obtained by relating aA
to aB using a dependent motion analysis. Using the technique developed
in Sec. 12.9, the coordinates sA and sB measure the positions of A
and B from the fixed datum, Fig. 13–10a. It is seen that
2sA + sB = l
where l is constant and represents the total vertical length of cord.
Differentiating this expression twice with respect to time yields
(3)
2aA = -aB
Notice that in writing Eqs. 1 to 3, the positive direction was always
assumed downward. It is very important to be consistent in this
assumption since we are seeking a simultaneous solution of equations.
The solution yields
T = 327.0 N
aA = 3.27 m>s2
aB = -6.54 m>s2
Hence when block A accelerates downward, block B accelerates
upward. Since aB is constant, the velocity of block B in 2 s is thus
1+ T2
v = v0 + aBt
= 0 + 1-6.542122
= -13.1 m>s
The negative sign indicates that block B is moving upward.
Ans.
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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PROBLEMS
117
PROBLEMS
13–1. Determine the gravitational attraction between two
spheres which are just touching each other. Each sphere has
a mass of 10 kg and a radius of 200 mm.
13–2. By using an inclined plane to retard the motion of a
falling object, and thus make the observations more
accurate, Galileo was able to determine experimentally
that the distance through which an object moves in free
fall is proportional to the square of the time for travel.
Show that this is the case, i.e., s r t2, by determining the
time tB , tC , and tD needed for a block of mass m to slide
from rest at A to points B, C, and D, respectively. Neglect
the effects of friction.
sD $
9m
sC $
80 km/h
4m
2m
sB $ A
B
Prob. 13–4
C
D
*13–4. A crate having a mass of 60 kg falls horizontally off
the back of a truck which is traveling at 80 km>h.
Determine the coefficient of kinetic friction between the
road and the crate if the crate slides 45 m on the ground
with no tumbling along the road before coming to rest.
Assume that the initial speed of the crate along the road is
80 km>h.
20'
Probs. 13–1/2
13–3. The 300-kg bar B, originally at rest, is being towed
over a series of small rollers. Determine the force in the
cable when t = 5 s, if the motor M is drawing in the cable
for a short time at a rate of v = 10.4t22 m>s, where t is in
seconds 10 … t … 6 s2. How far does the bar move in 5 s?
Neglect the mass of the cable, pulley, and the rollers.
13–5. The crane lifts the 700-kg bin with an initial
acceleration of 3 m>s2. Determine the force in each of the
supporting cables due to this motion.
M
B
5
A
v
5
4 4
3
3
C
B
Prob. 13–3
Prob. 13–5
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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118
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–6. The baggage truck A has a mass of 800 kg and is
used to pull the two cars, each with mass 300 kg. If the
tractive force F on the truck is F = 480 N, determine the
initial acceleration of the truck. What is the acceleration of
the truck if the coupling at C suddenly fails? The car wheels
are free to roll. Neglect the mass of the wheels.
AND
A C C E L E R AT I O N
*13–8. The 200-kg crate is suspended from the cable of a
crane. Determine the force in the cable when t = 2 s if the
crate is moving upward with (a) a constant velocity of 2 m>s,
and (b) a speed of v = 10.2t2 + 22 m>s, where t is in seconds.
A
C
B
F
v
Prob. 13–6
Prob. 13–8
13–7. The 500-kg fuel assembly for a nuclear reactor is
being lifted out from the core of the nuclear reactor using
the pulley system shown. It is hoisted upward with a
constant acceleration such that s = 0 and v = 0 when
t = 0, and s = 2.5 m when t = 1.5 s. Determine the tension
in the cable at A during the motion.
13–9. The elevator E has a mass of 500 kg, and the
counterweight at A has a mass of 150 kg. If the motor
supplies a constant force of 5 kN on the cable at B,
determine the speed of the elevator when t = 3 s, starting
from rest. Neglect the mass of the pulleys and cable.
13–10. The elevator E has a mass of 500 kg and the
counterweight at A has a mass of 150 kg. If the elevator
attains a speed of 10 m>s after it rises 40 m, determine the
constant force developed in the cable at B. Neglect the mass
of the pulleys and cable.
A
s
E
B
A
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Prob.
13–7 This publication is protected by Copyright and written permission
Probs. should
13–9/10
New Jersey. All rights
reserved.
be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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PROBLEMS
13–11. The water-park ride consists of an 800-lb sled
which slides from rest down the incline and then into the
pool. If the frictional resistance on the incline is Fr = 30 lb,
and in the pool for a short distance Fr = 80 lb, determine
how fast the sled is traveling when s = 5 ft.
119
13–14. Each of the two blocks has a mass m. The
coefficient of kinetic friction at all surfaces of contact is m. If
a horizontal force P moves the bottom block, determine the
acceleration of the bottom block in each case.
B
P
A
(a)
B
100 ft
P
A
(b)
Prob. 13–14
Prob. 13–11
100 ft
s
*13–12. A car of mass m is traveling at a slow velocity
v0 . If it is subjected to the drag resistance of the wind,
which is proportional to its velocity, i.e., FD = kv,
determine the distance and the time the car will travel
before its velocity becomes 0.5v0 . Assume no other
frictional forces act on the car.
13–15. The driver attempts to tow the crate using a rope
that has a tensile strength of 200 lb. If the crate is originally
at rest and has a weight of 500 lb, determine the greatest
acceleration it can have if the coefficient of static friction
between the crate and the road is ms = 0.4, and the
coefficient of kinetic friction is mk = 0.3.
30'
v
FD
Prob. 13–12
13–13. Determine the normal force the 10-kg crate A
exerts on the smooth cart if the cart is given an acceleration
of a = 2 m>s2 down the plane. Also, what is the acceleration
of the crate? Set u = 30°.
Prob. 13–15
*13–16. The 3.5-Mg engine is suspended from a 500-kg
spreader beam and hoisted by a crane which gives it an
acceleration of 4 m>s2 when it has a velocity of 2 m>s.
Determine the force in chains AC and AD during the lift.
C
60'
A
60'
A
B
D
E
a
u
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Prob. 13–13
Prob.
13–16
New Jersey. All rights reserved.
This publication is protected by Copyright and written permission
should
be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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120
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–17. The bullet of mass m is given a velocity due to gas
pressure caused by the burning of powder within the
chamber of the gun. Assuming this pressure creates a force
of F = F0 sin1pt>t02 on the bullet, determine the velocity of
the bullet at any instant it is in the barrel. What is the
bullet’s maximum velocity? Also, determine the position of
the bullet in the barrel as a function of time.
AND
A C C E L E R AT I O N
13–19. A 40-lb suitcase slides from rest 20 ft down the
smooth ramp. Determine the point where it strikes the
ground at C. How long does it take to go from A to C?
*13–20. Solve Prob. 13–19 if the suitcase has an initial
velocity down the ramp of vA = 10 ft>s and the coefficient
of kinetic friction along AB is mk = 0.2.
A
20 ft
F
30'
F0
4 ft
t0
t
C
R
Prob. 13–17
13–18. The 400-lb cylinder at A is hoisted using the motor
and the pulley system shown. If the speed of point B on the
cable is increased at a constant rate from zero to
vB = 10 ft>s in t = 5 s, determine the tension in the cable
at B to cause the motion.
B
Probs. 13–19/20
13–21. The winding drum D is drawing in the cable at an
accelerated rate of 5 m>s2. Determine the cable tension if
the suspended crate has a mass of 800 kg.
B
D
vA
A
Prob. 13–18
Prob. 13–21
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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PROBLEMS
13–22. At a given instant the 5-lb weight A is moving
downward with a speed of 4 ft>s. Determine its speed 2 s
later. Block B has a weight of 6 lb, and the coefficient of
kinetic friction between it and the horizontal plane is
mk = 0.3. Neglect the mass of the pulleys and cord.
B
121
*13–24. At a given instant the 10-lb block A is moving
downward with a speed of 6 ft>s. Determine its speed 2 s
later. Block B has a weight of 4 lb, and the coefficient of
kinetic friction between it and the horizontal plane is
mk = 0.2. Neglect the mass of the pulleys and cord.
B
A
A
Prob. 13–22
Prob. 13–24
13–23. A force F = 15 lb is applied to the cord.
Determine how high the 30-lb block A rises in 2 s starting
from rest. Neglect the weight of the pulleys and cord.
13–25. A freight elevator, including its load, has a mass of
500 kg. It is prevented from rotating due to the track and
wheels mounted along its sides. If the motor M develops a
constant tension T = 1.50 kN in its attached cable,
determine the velocity of the elevator when it has moved
upward 3 m starting from rest. Neglect the mass of the
pulleys and cables.
B
C
M
F
A
Prob. 13–23
Prob. 13–25
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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122
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–26. At the instant shown the 100-lb block A is moving
down the plane at 5 ft>s while being attached to the 50-lb
block B. If the coefficient of kinetic friction is mk = 0.2,
determine the acceleration of A and the distance A slides
before it stops. Neglect the mass of the pulleys and cables.
A
5
4
AND
A C C E L E R AT I O N
*13–28. The 400-kg mine car is hoisted up the incline
using the cable and motor M. For a short time, the force in
the cable is F = 13200t22 N, where t is in seconds. If the car
has an initial velocity v1 = 2 m>s when t = 0, determine its
velocity when t = 2 s.
13–29. The 400-kg mine car is hoisted up the incline using
the cable and motor M. For a short time, the force in the
cable is F = 13200t22 N, where t is in seconds. If the car has
an initial velocity v1 = 2 m>s at s = 0 and t = 0, determine
the distance it moves up the plane when t = 2 s.
3
M
C
v1 $ 2 m/s
D
B
17
8
15
Prob. 13–26
Probs. 13–28/29
13–27. The safe S has a weight of 200 lb and is supported
by the rope and pulley arrangement shown. If the end of the
rope is given to a boy B of weight 90 lb, determine his
acceleration if in the confusion he doesn’t let go of the rope.
Neglect the mass of the pulleys and rope.
"13–30. The tanker has a weight of 80011062 lb and is
traveling forward at v0 = 3 ft>s in still water when the
engines are shut off. If the drag resistance of the water is
proportional to the speed of the tanker at any instant and
can be approximated by FD = 140011032v2 lb, where v is in
ft>s, determine the time needed for the tanker’s speed to
become 1.5 ft>s. Given the initial velocity of v0 = 3 ft>s
through what distance must the tanker travel before it stops?
S
v
B
Prob. 13–27
FD
Prob. 13–30
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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PROBLEMS
13–31. The spring mechanism is used as a shock absorber
for railroad cars. Determine the maximum compression of
spring HI if the fixed bumper R of a 5-Mg railroad car,
rolling freely at 2 m>s, strikes the plate P. Bar AB
slides along the guide paths CE and DF. The ends of all
springs are attached to their respective members and are
originally unstretched.
123
13–34. The boy having a weight of 80 lb hangs uniformly
from the bar. Determine the force in each of his arms in
t = 2 s if the bar is moving upward with (a) a constant
velocity of 3 ft>s, and (b) a speed of v = 14t22 ft>s, where t
is in seconds.
k $ 80 kN/m
B
F
D
I
H
P
R
C
E
A
k¿ $ 160 kN/m
Prob. 13–34
2 m/s
k $ 80 kN/m
Prob. 13–31
*13–32. The 2-kg collar C is free to slide along the
smooth shaft AB. Determine the acceleration of collar C if
(a) the shaft is fixed from moving, (b) collar A, which is
fixed to shaft AB, moves downward at constant velocity
along the vertical rod, and (c) collar A is subjected to a
downward acceleration of 2 m>s2. In all cases, the collar
moves in the plane.
13–35. The 10-kg block A rests on the 50-kg plate B in the
position shown. Neglecting the mass of the rope and pulley,
and using the coefficients of kinetic friction indicated,
determine the time needed for block A to slide 0.5 m on the
plate when the system is released from rest.
13–33. The 2-kg collar C is free to slide along the smooth
shaft AB. Determine the acceleration of collar C if collar A
is subjected to an upward acceleration of 4 m>s2.
B
45'
0.5 m
C
mAB $ 0.2
A
C
A
B
mBC $ 0.1
30'
Probs. 13–32/33
Prob. 13–35
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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124
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
*13–36. Determine the acceleration of block A when the
system is released. The coefficient of kinetic friction and
the weight of each block are indicated. Neglect the mass
of the pulleys and cord.
AND
A C C E L E R AT I O N
13–38. An electron of mass m is discharged with an initial
horizontal velocity of v0 . If it is subjected to two fields of
force for which Fx = F0 and Fy = 0.3F0 , where F0 is
constant, determine the equation of the path, and the speed
of the electron at any time t.
y
++++++++++++++
++++++++++++++
A
v0
20 lb B
80 lb
mk $ 0.2
60'
Prob. 13–36
13–37. The conveyor belt is moving at 4 m>s. If the
coefficient of static friction between the conveyor and
the 10-kg package B is ms = 0.2, determine the shortest
time the belt can stop so that the package does not slide on
the belt.
x
Prob. 13–38
13–39. The conveyor belt delivers each 12-kg crate to the
ramp at A such that the crate’s speed is vA = 2.5 m>s,
directed down along the ramp. If the coefficient of kinetic
friction between each crate and the ramp is mk = 0.3,
determine the speed at which each crate slides off the ramp
at B. Assume that no tipping occurs. Take u = 30°.
vA $ 2.5 m/s
B
A
3m
u
Prob. 13–37
B
Prob. 13–39
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125
PROBLEMS
*13–40. A parachutist having a mass m opens his
parachute from an at-rest position at a very high altitude. If
the atmospheric drag resistance is FD = kv2, where k is a
constant, determine his velocity when he has fallen for a
time t. What is his velocity when he lands on the ground?
This velocity is referred to as the terminal velocity, which is
found by letting the time of fall t : q .
13–43. Blocks A and B each have a mass m. Determine the
largest horizontal force P which can be applied to B so that
A will not slip up B.The coefficient of static friction between
A and B is ms . Neglect any friction between B and C.
A
P
u B
C
FD
Probs. 13–42/43
v
*13–44. Each of the three plates has a mass of 10 kg. If the
coefficients of static and kinetic friction at each surface of
contact are ms = 0.3 and mk = 0.2, respectively, determine
the acceleration of each plate when the three horizontal
forces are applied.
Prob. 13–40
18 N
13–41. Block B rests on a smooth surface. If the
coefficients of static and kinetic friction between A and B
are ms = 0.4 and mk = 0.3, respectively, determine the
acceleration of each block if someone pushes horizontally
on block A with a force of (a) F = 6 lb, and (b) F = 50 lb.
D
100 N
C
15 N
B
A
Prob. 13–44
F
A
20 lb
B
30 lb
13–45. Crate B has a mass m and is released from rest
when it is on top of cart A, which has a mass 3m. Determine
the tension in cord CD needed to hold the cart from moving
while B is sliding down A. Neglect friction.
B
Prob. 13–41
13–42. Blocks A and B each have a mass m. Determine
the largest horizontal force P which can be applied to B so
that A will not move relative to B. All surfaces are smooth.
D
C
u
A
Prob. 13–45
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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126
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
"13–46. The tractor is used to lift the 150-kg load B with the
24-m-long rope, boom, and pulley system. If the tractor is
traveling to the right at a constant speed of 4 m>s, determine
the tension in the rope when sA = 5 m. When sA = 0, sB = 0.
13–47. The tractor is used to lift the 150-kg load B with the
24-m-long rope, boom, and pulley system. If the tractor is
traveling to the right with an acceleration of 3 m>s2 and has
a velocity of 4 m>s at the instant sA = 5 m, determine the
tension in the rope at this instant. When sA = 0, sB = 0.
AND
A C C E L E R AT I O N
13–49. Block A has a mass mA and is attached to a spring
having a stiffness k and unstretched length l0 . If another
block B, having a mass mB , is pressed against A so that the
spring deforms a distance d, determine the distance both
blocks slide on the smooth surface before they begin to
separate. What is their velocity at this instant?
13–50. Block A has a mass mA and is attached to a spring
having a stiffness k and unstretched length l0 . If another
block B, having a mass mB , is pressed against A so that the
spring deforms a distance d, show that for separation to
occur it is necessary that d 7 2mkg1mA + mB2>k, where mk
is the coefficient of kinetic friction between the blocks and
the ground. Also, what is the distance the blocks slide on the
surface before they separate?
12 m
sB
k
B
A
B
A
sA
Probs. 13–49/50
Probs. 13–46/47
*13–48. Block B has a mass m and is hoisted using the
cord and pulley system shown. Determine the magnitude of
force F as a function of the block’s vertical position y so that
when F is applied the block rises with a constant
acceleration aB . Neglect the mass of the cord and pulleys.
13–51. The block A has a mass mA and rests on the pan B,
which has a mass mB . Both are supported by a spring
having a stiffness k that is attached to the bottom of the pan
and to the ground. Determine the distance d the pan should
be pushed down from the equilibrium position and then
released from rest so that separation of the block will take
place from the surface of the pan at the instant the spring
becomes unstretched.
2m
1m
A
B
B
y d
k
A
Prob. 13–48
Prob. 13–51
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13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
127
13.5 Equations of Motion:
Normal and Tangential Coordinates
When a particle moves over a curved path which is known, the equation
of motion for the particle may be written in the tangential, normal, and
binormal directions. We have
©F = ma
©Ftut + ©Fnun + ©Fbub = ma t + ma n
Here ©Ft , ©Fn , ©Fb represent the sums of all the force components
acting on the particle in the tangential, normal, and binormal directions,
respectively, Fig. 13–11. Note that there is no motion of the particle in the
binormal direction, since the particle is constrained to move along the
path. The above equation is satisfied provided
©Ft = mat
©Fn = man
©Fb = 0
(13–8)
Recall that at 1= dv>dt2 represents the time rate of change in the
magnitude of velocity. Consequently, if ©Ft acts in the direction of
motion, the particle’s speed will increase, whereas if it acts in the
opposite direction, the particle will slow down. Likewise, an 1= v2>r2
represents the time rate of change in the velocity’s direction. Since this
vector always acts in the positive n direction, i.e., toward the path’s
center of curvature, then ©Fn , which causes a n , also acts in this direction.
For example, when the particle is constrained to travel in a circular path
with a constant speed, there is a normal force exerted on the particle by
the constraint in order to change the direction of the particle’s velocity
1a n2. Since this force is always directed toward the center of the path, it is
often referred to as the centripetal force.
The centrifuge is used to subject a
passenger to very large normal
accelerations caused by high rotations.
Realize that these accelerations are
caused by the unbalanced normal force
exerted on the passenger by the seat of the
centrifuge.
b
(Fbub
O
t
n
(Fnun
(Ftut
P
Inertial coordinate
system
Fig. 13–11
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128
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
Procedure for Analysis
When a problem involves the motion of a particle along a known
curved path, normal and tangential coordinates should be
considered for the analysis since the acceleration components can
be readily formulated. The method for applying the equations of
motion, which relate the forces to the acceleration, has been
outlined in the procedure given in Sec. 13.4. Specifically, for t, n, b
coordinates it may be stated as follows:
Free-Body Diagram.
• Establish the inertial t, n, b coordinate system at the particle and
draw the particle’s free-body diagram.
• The particle’s normal acceleration an always acts in the positive
•
•
n direction.
If the tangential acceleration a t is unknown, assume it acts in
the positive t direction.
Identify the unknowns in the problem.
Equations of Motion.
• Apply the equations of motion, Eqs. 13–8.
Kinematics.
• Formulate the tangential and normal components of
•
acceleration; i.e., at = dv>dt or at = v dv>ds and an = v2>r.
If the path is defined as y = f1x2, the radius of curvature at the
point where the particle is located can be obtained from
r = [1 + 1dy>dx22]3>2> ƒ d2y>dx2 ƒ .
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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129
13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EXAMPLE 13.6
Determine the banking angle u for the race track so that the wheels of
the racing cars shown in Fig. 13–12a will not have to depend upon
friction to prevent any car from sliding up or down the track. Assume
the cars have negligible size, a mass m, and travel around the curve of
radius r with a speed v.
u
(a)
b
SOLUTION
Before looking at the following solution, give some thought as to why
it should be solved using t, n, b coordinates.
Free-Body Diagram. As shown in Fig. 13–12b, and as stated in the
problem, no frictional force acts on the car. Here NC represents the
resultant of the ground on all four wheels. Since an can be calculated,
the unknowns are NC and u.
Equations of Motion. Using the n, b axes shown,
+ ©F = ma ;
:
n
n
+ c ©Fb = 0;
an
NC
n
u
W $ mg
v2
NC sin u = m
r
(1)
NC cos u - mg = 0
(2)
(b)
Fig. 13–12
Eliminating NC and m from these equations by dividing Eq. 1 by
Eq. 2, we obtain
v2
tan u =
gr
v2
u = tan-1 a b
Ans.
gr
NOTE: The result is independent of the mass of the car. Also, a force
summation in the tangential direction is of no consequence to the
solution. If it were considered, then at = dv>dt = 0, since the car
moves with constant speed. A further analysis of this problem is
discussed in Prob. 21–48.
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130
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
EXAMPLE 13.7
The 3-kg disk D is attached to the end of a cord as shown in
Fig. 13–13a. The other end of the cord is attached to a ball-and-socket
joint located at the center of a platform. If the platform is rotating
rapidly, and the disk is placed on it and released from rest as shown,
determine the time it takes for the disk to reach a speed great enough
to break the cord. The maximum tension the cord can sustain is 100 N,
and the coefficient of kinetic friction between the disk and the
platform is mk = 0.1.
Motion of
platform
D
1m
(a)
SOLUTION
Free-Body Diagram. The frictional force has a magnitude
F = mkND = 0.1ND and a sense of direction that opposes the relative
motion of the disk with respect to the platform. It is this force that gives
the disk a tangential component of acceleration causing v to increase,
thereby causing T to increase until it reaches 100 N. The weight of the
disk is W = 319.812 = 29.43 N. Since an can be related to v, the
unknowns are ND , at , and v.
b
29.43 N
F $ 0.1 ND
t
at
Equations of Motion.
T
ND
(b)
Fig. 13–13
©Fn = man ;
an
n
©Ft = mat ;
©Fb = 0;
v2
b
1
0.1ND = 3at
T = 3a
ND - 29.43 = 0
(1)
(2)
(3)
Setting T = 100 N, Eq. 1 can be solved for the critical speed vcr of the
disk needed to break the cord. Solving all the equations, we obtain
ND = 29.43 N
at = 0.981 m>s2
vcr = 5.77 m>s
Kinematics. Since at is constant, the time needed to break the cord is
vcr = v0 + att
5.77 = 0 + 10.9812t
t = 5.89 s
Ans.
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131
13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EXAMPLE 13.8
Design of the ski jump shown in the photo requires knowing the type of
forces that will be exerted on the skier and his approximate trajectory. If
in this case the jump can be approximated by the parabola shown in
Fig. 13–14a, determine the normal force on the 150-lb skier the instant he
arrives at the end of the jump, point A, where his velocity is 65 ft>s. Also,
what is his acceleration at this point?
SOLUTION
Why consider using n, t coordinates to solve this problem?
y
y$
Free-Body Diagram. The free-body diagram for the skier when he
is at A is shown in Fig. 13–14b. Since the path is curved, there are two
components of acceleration, a n and a t . Since an can be calculated, the
unknowns are at and NA .
x
200 ft
A
Equations of Motion.
+ c ©Fn = man ;
(a)
2
NA - 150 =
+ ©F = ma ;
;
t
t
0 =
150 1652
a
b
r
32.2
(1)
150
a
32.2 t
(2)
[1 + 1dy>dx22]3>2
ƒ d2y>dx2 ƒ
`
x=0
=
n
an
The radius of curvature r for the path must be determined at point
1
1
1
A(0, -200 ft). Here y = 200
x2 - 200, dy>dx = 100
x, d2y>dx2 = 100
,
so that at x = 0,
r =
1 x2 # 200
200
[1 + 1022]3>2
1
ƒ 100
ƒ
t
at
= 100 ft
Substituting into Eq. 1 and solving for NA , we have
NA = 347 lb
150 lb
(b)
NA
Ans.
Fig. 13–14
Kinematics. From Eq. 2,
at = 0
Thus,
an =
16522
v2
=
= 42.2 ft>s2
r
100
aA = an = 42.2 ft>s2 c
Ans.
NOTE: Apply the equation of motion in the y direction and show that
when the skier is in mid air the acceleration is 32.2 ft>s2.
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132
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
A C C E L E R AT I O N
AND
EXAMPLE 13.9
v0 $ 1 m/s
ds $ 0.5 du
SOLUTION
Free-Body Diagram. The free-body diagram for a package, when it
is located at the general position u, is shown in Fig. 13–15b.The package
must have a tangential acceleration a t , since its speed is always
increasing as it slides downward.The weight is W = 219.812 = 19.62 N.
Specify the three unknowns.
Equations of Motion.
v2
+b©Fn = man;
-NB + 19.62 cos u = 2
(1)
0.5
+R©Ft = mat;
19.62 sin u = 2at
(2)
u
du r $ 0.5 m
(a)
NB
an
at
u
19.62 N
t
n
(b)
Fig. 13–15
Packages, each having a mass of 2 kg, are delivered from a conveyor to
a smooth circular ramp with a velocity of v0 = 1 m>s as shown in
Fig. 13–15a. If the effective radius of the ramp is 0.5 m, determine the
angle u = umax at which each package begins to leave the surface.
At the instant u = umax , the package leaves the surface of the ramp so
that NB = 0. Therefore, there are three unknowns, v, at , and u.
Kinematics. The third equation for the solution is obtained by noting
that the magnitude of tangential acceleration at may be related to the
speed of the package v and the angle u. Since at ds = v dv and
ds = r du = 0.5 du, Fig. 13–15a, we have
v dv
at =
(3)
0.5 du
To solve, substitute Eq. 3 into Eq. 2 and separate the variables. This gives
v dv = 4.905 sin u du
Integrate both sides, realizing that when u = 0°, v0 = 1 m>s.
v
L1
v dv = 4.905
u
L0°
sin u du
u
v2 v
` = -4.905 cos u ` ; v2 = 9.8111 - cos u2 + 1
2 1
0°
Substituting into Eq. 1 with NB = 0 and solving for cos umax yields
2
19.62 cos umax =
[9.8111 - cos umax2 + 1]
0.5
43.24
cos umax =
58.86
umax = 42.7°
Ans.
NOTE: The speed of the package is increasing because its tangential
acceleration is increasing with u, Eq. 2.
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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133
PROBLEMS
PROBLEMS
*13–52. Determine the mass of the sun, knowing that the
distance from the earth to the sun is 149.611062 km. Hint: Use
Eq. 13–1 to represent the force of gravity acting on the earth.
13–53. The 1.40-Mg helicopter is traveling at a constant speed
of 40 m>s along the horizontal curved path while banking
at u = 30°. Determine the force acting normal to the blade,
i.e., in the y¿ direction, and the radius of curvature of the path.
13–57. The 600-kg wrecking ball is suspended from the
crane by a cable having a negligible mass. If the ball has a
speed v = 8 m>s at the instant it is at its lowest point, u = 0°,
determine the tension in the cable at this instant. Also,
determine the angle u to which the ball swings before it stops.
13–54. The 1.40-Mg helicopter is traveling at a constant
speed of 33 m>s along the horizontal curved path having a
radius of curvature of r = 300 m. Determine the force the
blade exerts on the frame and the bank angle u.
y
y¿
u
u
12 m
r
Probs. 13–53/54
13–55. The plane is traveling at a constant speed of 800
ft/s along the curve y = 20110-62x2 + 5000, where x and y
are in feet. If the pilot has a weight of 180 lb, determine the
normal and tangential components of the force the seat
exerts on the pilot when the plane is at its lowest point.
*13–56. The jet plane is traveling at a constant speed of
1000 ft>s along the curve y = 20110-62x2 + 5000, where x
and y are in feet. If the pilot has a weight of 180 lb,
determine the normal and tangential components of the
force the seat exerts on the pilot when y = 10 000 ft.
Prob. 13–57
13–58. Prove that if the block is released from rest at point B
of a smooth path of arbitrary shape, the speed it attains when it
reaches point A is equal to the speed it attains when it falls
freely through a distance h; i.e., v = 22gh.
y
B
y $ 20(10#6)x2 " 5000
h
A
x
Probs. 13–55/56
Prob. 13–58
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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134
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–59. The sled and rider have a total mass of 80 kg and
start from rest at A(10 m, 0). If the sled descends the smooth
slope, which may be approximated by a parabola, determine
the normal force that the ground exerts on the sled at the
instant it arrives at point B. Neglect the size of the sled and
rider. Hint: Use the result of Prob. 13–58.
*13–60. The sled and rider have a total mass of 80 kg and
start from rest at A(10 m, 0). If the sled descends the smooth
slope, which may be approximated by a parabola, determine
the normal force that the ground exerts on the sled at the
instant it arrives at point C. Neglect the size of the sled and
rider. Hint: Use the result of Prob. 13–58.
AND
A C C E L E R AT I O N
13–63. If the crest of the hill has a radius of curvature
r = 200 ft, determine the maximum constant speed at
which the car can travel over it without leaving the surface
of the road. Neglect the size of the car in the calculation.
The car has a weight of 3500 lb.
v
r $ 200 ft
y
y$
1
20
x2 # 5
A
5m
C
2m
x
Prob. 13–63
B
10 m
Probs. 13–59/60
13–61. At the instant u = 60°, the boy’s center of mass G
has a downward speed vG = 15 ft>s. Determine the rate of
increase in his speed and the tension in each of the two
supporting cords of the swing at this instant. The boy has a
weight of 60 lb. Neglect his size and the mass of the seat
and cords.
*13–64. The airplane, traveling at a constant speed of
50 m>s, is executing a horizontal turn. If the plane is banked
at u = 15°, when the pilot experiences only a normal force
on the seat of the plane, determine the radius of curvature r
of the turn. Also, what is the normal force of the seat on the
pilot if he has a mass of 70 kg.
13–62. At the instant u = 60°, the boy’s center of mass G
is momentarily at rest. Determine his speed and the tension
in each of the two supporting cords of the swing when
u = 90°. The boy has a weight of 60 lb. Neglect his size and
the mass of the seat and cords.
u
10 ft
u
r
G
Unpublished Work © 2007 by R. C. Hibbeler.
To be13–61/62
published by Pearson Prentice Hall, PearsonProb.
Education,
Probs.
13–64Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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135
PROBLEMS
13–65. The 150-lb man lies against the cushion for which
the coefficient of static friction is ms = 0.5. Determine the
resultant normal and frictional forces the cushion exerts on
him if, due to rotation about the z axis, he has a constant
speed v = 20 ft>s. Neglect the size of the man.Take u = 60°.
*13–68. The 200-kg snowmobile with passenger is traveling
down the hill at a constant speed of 6 m>s. Determine the
resultant normal force and the resultant frictional force
exerted on the tracks at the instant it reaches point A. Neglect
the size of the snowmobile.
13–66. The 150-lb man lies against the cushion for which
the coefficient of static friction is ms = 0.5. If he rotates
about the z axis with a constant speed v = 30 ft>s,
determine the smallest angle u of the cushion at which he
will begin to slip off.
13–69. The 200-kg snowmobile with passenger is traveling
down the hill such that when it is at point A, it is traveling at
4 m>s and increasing its speed at 2 m>s2. Determine the
resultant normal force and the resultant frictional force
exerted on the tracks at this instant. Neglect the size of the
snowmobile.
z
y
x
8 ft
G
5m
A
u
10 m
y $ #5(10#3)x3
Probs. 13–68/69
Probs. 13–65/66
13–67. Determine the constant speed of the passengers on
the amusement-park ride if it is observed that the
supporting cables are directed at u = 30° from the vertical.
Each chair including its passenger has a mass of 80 kg. Also,
what are the components of force in the n, t, and b
directions which the chair exerts on a 50-kg passenger
during the motion?
13–70. A collar having a mass of 0.75 kg and negligible size
slides over the surface of a horizontal circular rod for which
the coefficient of kinetic friction is mk = 0.3. If the collar is
given a speed of 4 m>s and then released at u = 0°, determine
how far, s, it slides on the rod before coming to rest.
z
4m
b
u
6m
100 mm
t
y
u
n
s
x
v
Unpublished Work © 2007 by
R. C.13–67
Hibbeler. To be published by Pearson Prentice Hall, Pearson Education,
Inc., Upper Saddle River,
Prob.
Prob. 13–70
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136
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–71. The roller coaster car and passenger have a total
weight of 600 lb and starting from rest at A travel down the
track that has the shape shown. Determine the normal force
of the tracks on the car when the car is at point B, where it
has a velocity of 15 ft>s. Neglect friction and the size of the
car and passenger.
AND
A C C E L E R AT I O N
13–73. The 5-kg pendulum bob B is released from rest
when u = 0°. Determine the initial tension in the cord and
also at the instant the bob reaches point D, u = 45°. Neglect
the size of the bob.
y
A
B
A
u $ 45'
p
y $ 40 cos ( ––
40 x)
40 ft
B
2m
x
D
40 ft
Prob. 13–73
20 ft
Prob. 13–71
*13–72. The smooth block B, having a mass of 0.2 kg, is
attached to the vertex A of the right circular cone using a
light cord. The cone is rotating at a constant angular rate
about the z axis such that the block attains a speed of
0.5 m>s. At this speed, determine the tension in the cord
and the reaction which the cone exerts on the block.
Neglect the size of the block.
13–74. A ball having a mass of 2 kg and negligible size
moves within a smooth vertical circular slot. If it is released
from rest when u = 10°, determine the force of the slot on
the ball when the ball arrives at points A and B.
10'
z
u
A
200 mm
r $ 0.8 m
A
B
400 mm
300 mm
10'
B
Prob. 13–72
Prob. 13–74
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137
PROBLEMS
13–75. The rotational speed of the disk is controlled by a
30-g smooth contact arm AB which is spring-mounted on
the disk. When the disk is at rest, the center of mass G of the
arm is located 150 mm from the center O, and the preset
compression in the spring is 20 mm. If the initial gap
between B and the contact at C is 10 mm, determine the
(controlling) speed vG of the arm’s mass center, G, which
will close the gap. The disk rotates in the horizontal plane.
The spring has a stiffness of k = 50 N>m, and its ends are
attached to the contact arm at D and to the disk at E.
13–77. The 35-kg box has a speed of 2 m>s when it is at A
on the smooth ramp. If the surface is in the shape of a
parabola, determine the normal force on the box at the
instant x = 3 m. Also, what is the rate of increase in its
speed at this instant?
y
2 m/s
O
k $ 50 N/m
DG
1
y $ 4 # –– x2
9
BC
A
150 mm
A
x
E
10 mm
Prob. 13–77
Prob. 13–75
*13–76. The 2-kg spool S fits loosely on the inclined rod
for which the coefficient of static friction is ms = 0.2. If the
spool is located 0.25 m from A, determine the minimum
constant speed the spool can have so that it does not slip
down the rod.
z
5
13–78. The man has a mass of 80 kg and sits 3 m from the
center of the rotating platform. Due to the rotation his
#
speed is increased from rest by v = 0.4 m>s2. If the
coefficient of static friction between his clothes and the
platform is ms = 0.3, determine the time required to cause
him to slip.
3
4
S
0.25 m
3m
10 m
A
Prob. 13–76
Prob. 13–78
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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138
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–79. The collar A, having a mass of 0.75 kg, is attached
to a spring having a stiffness of k = 200 N>m. When rod BC
rotates about the vertical axis, the collar slides outward
along the smooth rod DE. If the spring is unstretched when
s = 0, determine the constant speed of the collar in order
that s = 100 mm. Also, what is the normal force of the rod
on the collar? Neglect the size of the collar.
AND
A C C E L E R AT I O N
13–81. If the bicycle and rider have a total weight of 180 lb,
determine the resultant normal force acting on the bicycle
when it is at point A while it is freely coasting at vA = 6 ft>s.
Also, compute the increase in the bicyclist’s speed at this
point. Neglect the resistance due to the wind and the size of
the bicycle and rider.
y
C
k $ 200 N/m
D
A
vA $ 6 ft/s
5 ft
20 ft
p
y $ 20 cos —
x
20
A
E
s
x
B
Prob. 13–79
Prob. 13–81
*13–80. The block has a weight of 2 lb and it is free to
move along the smooth slot in the rotating disk. The spring
has a stiffness of 2.5 lb>ft and an unstretched length of 1.25 ft.
Determine the force of the spring on the block and the
tangential component of force which the slot exerts on the
side of the block, when the block is at rest with respect to
the disk and is traveling with a constant speed of 12 ft>s.
13–82. The 5-lb packages ride on the surface of the
conveyor belt. If the belt starts from rest and increases to a
constant speed of 12 ft>s in 2 s, determine the maximum
angle u so that none of the packages slip on the inclined
surface AB of the belt. The coefficient of static friction
between the belt and a package is ms = 0.3. At what angle f
do the packages first begin to slip off the surface of the belt
after the belt is moving at its constant speed of 12 ft>s?
Neglect the size of the packages.
B
6 in.
f
k $ 2.5 lb/ft
A
Prob. 13–80
u
Prob. 13–82
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13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
13.6 Equations of Motion:
139
(Fzuz
Cylindrical Coordinates
(Fuuu
When all the forces acting on a particle are resolved into cylindrical
components, i.e., along the unit-vector directions ur , uu , uz , Fig. 13–16,
the equation of motion may be expressed as
©F = ma
©Frur + ©Fuuu + ©Fzuz = marur + mauuu + mazuz
To satisfy this equation, the respective ur , uu , uz components on the left
side must equal the corresponding components on the right side.
Consequently, we may write the following three scalar equations of motion:
©Fr = mar
P
(Frur
z
O
u
r
Inertial coordinate system
©Fu = mau
©Fz = maz
(13–9)
Fig. 13–16
If the particle is constrained to move only in the r–u plane, then only the
first two of Eqs. 13–9 are used to specify the motion.
Tangential and Normal Forces. The most straightforward type
of problem involving cylindrical coordinates requires the determination
of the resultant force components ©Fr , ©Fu , ©Fz which cause a particle
to move with a known acceleration. If, however, the particle’s accelerated
motion is not completely specified at the given instant, then some
information regarding the directions or magnitudes of the forces acting on
the particle must be known or computed in order to solve Eqs. 13–9. For
example, the force P causes the particle in Fig. 13–17a to move along a
path r = f1u2. The normal force N which the path exerts on the particle
is always perpendicular to the tangent of the path, whereas the frictional
force F always acts along the tangent in the opposite direction of motion.
The directions of N and F can be specified relative to the radial coordinate
by using the angle c (psi), Fig. 13–17b, which is defined between the
extended radial line and the tangent to the curve.
r $ f (u)
r $ f (u)
Tangent
Tangent
c
N
r
O
r
F
u P
O
(a)
Fig. 13–17
u
(b)
As the car of weight W descends the
spiral track, the resultant normal
force which the track exerts on the
car can be represented by its three
cylindrical components. -Nr creates
a radial acceleration, -ar , Nu creates
a transverse acceleration au , and the
difference, W - Nz creates an
azimuthal acceleration -az .
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140
C H A P T E R 13
KINETICS
Tangent
dr
du
c
rc
c $ 75'
2a
(13–10)
Procedure for Analysis
Tangent
u $ 30'
r
dr>du
If c is calculated as a positive quantity, it is measured from the extended
radial line to the tangent in a counterclockwise sense or in the positive
direction of u. If it is negative, it is measured in the opposite direction to
positive u. For example, consider the cardioid r = a11 + cos u2, shown
in Fig. 13–18. Because dr>du = -a sin u, then when u = 30°,
tan c = a11 + cos 30°2>1-a sin 30°2 = -3.732, or c = -75°, measured
clockwise, as shown in the figure.
Fig. 13–17 (cont.)
r
A C C E L E R AT I O N
tan c =
(c)
O
AND
ds
u
O
PA R T I C L E : F O R C E
This angle can be obtained by noting that when the particle is displaced
a distance ds along the path, Fig. 13–17c, the component of displacement
in the radial direction is dr and the component of displacement in the
transverse direction is r du. Since these two components are mutually
perpendicular, the angle c can be determined from tan c = r du>dr, or
r $ f (u)
r du
OF A
Cylindrical or polar coordinates are a suitable choice for the analysis of
a problem for which data regarding the angular motion of the radial
line r are given, or in cases where the path can be conveniently
expressed in terms of these coordinates. Once these coordinates have
been established, the equations of motion can be applied in order to
relate the forces acting on the particle to its acceleration components.
The method for doing this has been outlined in the procedure for
analysis given in Sec. 13.4.The following is a summary of this procedure.
Free-Body Diagram.
• Establish the r, u, z inertial coordinate system and draw the
Fig. 13–18
•
•
particle’s free-body diagram.
Assume that a r , a u , a z act in the positive directions of r, u, z if they
are unknown.
Identify all the unknowns in the problem.
Equations of Motion.
• Apply the equations of motion, Eqs. 13–9.
Kinematics.
• Use the methods# of$ Sec. 12.8 to determine r and the time
# $
$
derivatives r, r, u, u, #z, and then
evaluate
the acceleration
$
$
##
$
components ar = r - ru2, au = ru + 2ru, az = z.
• If any of the acceleration components is computed as a negative
quantity, it indicates that it acts in its negative coordinate direction.
• When taking the time derivatives of r = f1u2, it is very important
to use the chain rule of calculus.
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141
13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
EXAMPLE 13.10
The 2-lb block in Fig. 13–19a moves on a smooth horizontal track,
such that its path is specified in polar coordinates by the parametric
equations r = 110t22 ft and u = 10.5t2 rad, where t is in seconds.
Determine the magnitude of the tangential force F causing the
motion at the instant t = 1 s.
SOLUTION
F
Free-Body Diagram. As shown on the block’s free-body diagram,
Fig. 13–19b,the normal force of the track on the block,N,and the tangential
force F are located at an angle c from the r and u axes. This angle can be
obtained from Eq. 13–10. To do so, we must first express the path as
r = f1u2 by eliminating the parameter t between r and u. This yields
r = 40u2. Also, when t = 1 s, u = 0.511 s2 = 0.5 rad. Thus,
r
40u2
=
= 0.25
`
dr>du
4012u2 u = 0.5 rad
c = 14.04°
r
u
(a)
tan c =
Because c is a positive quantity, it is measured counterclockwise from
the r axis to the tangent (the same direction as u) as shown in
Fig. 13–19b. There are presently four unknowns: F, N, ar and au .
Equations of Motion.
2
+ T ©Fr = mar ;
F cos 14.04° - N sin 14.04° =
a
(1)
32.2 r
2
c + ©Fu = mau ;
F sin 14.04° + N cos 14.04° =
a
(2)
32.2 u
Kinematics. Since the motion is specified, the coordinates and the
required time derivatives can be calculated and evaluated at t = 1 s.
r = 10t2 `
#
r = 20t `
t=1 s
t=1 s
= 10 ft
u = 0.5t `
t=1 s
u
14.04'
N
Tangent
au
F
r
c = 14.04'
ar
u
(b)
Fig. 13–19
= 0.5 rad
#
= 20 ft>s u = 0.5 rad>s
$
$
r = 20 ft>s2
u = 0
#
$
ar = r - ru2 = 20 - 1010.522 = 17.5 ft>s2
$
##
au = ru + 2ru = 10102 + 2120210.52 = 20 ft>s2
Substituting into Eqs. 1 and 2 and solving, we get
F = 1.36 lb
Ans.
N = 0.942 lb
NOTE: The tangential axis is in the direction of F, and the normal axis
is in the direction of N.
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142
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
EXAMPLE 13.11
The smooth 2-kg cylinder C in Fig. 13–20a has a peg P through its
center which passes through the slot# in arm OA. If the arm rotates in
the vertical plane at a constant rate u = 0.5 rad>s, determine the force
that the arm exerts on the peg at the instant u = 60°.
u
0.4 m
O
·
u $ 0.5 rad/s
r
C
SOLUTION
Why is it a good idea to use polar coordinates to solve this problem?
Free-Body Diagram. The free-body diagram for the cylinder is
shown in Fig. 13–20b. The force on the peg, FP , acts perpendicular to
the slot in the arm. As usual, a r and a u are assumed to act in the
directions of positive r and u, respectively. Identify the four unknowns.
Equations of Motion. Using the data in Fig. 13–20b, we have
+R©Fr = mar ;
+b©Fu = mau ;
P
(1)
(2)
Kinematics. From Fig. 13–20a, r can be related to u by the equation
A
r =
(a)
FP
u
u
NC
r
(b)
Fig. 13–20
0.4
= 0.4 csc u
sin u
Since d1csc u2 = -1csc u cot u2 du and d1cot u2 = -1csc2 u2 du, then
r and the necessary time derivatives become
#
u = 0.5
r = 0.4 csc u
$
#
#
u = 0
r = -0.41csc u cot u2u
19.62 N
ar
19.62 sin u - NC sin u = 2ar
19.62 cos u + FP - NC cos u = 2au
= -0.2 csc u cot u
#
#
$
r = -0.21-csc u cot u21u2 cot u - 0.2 csc u1-csc2 u2u
au
= 0.1 csc u1cot2 u + csc2 u2
u
Evaluating these formulas at u = 60°, we get
#
u = 0.5
r = 0.462
$
#
u = 0
r = -0.133
$
r = 0.192
#
$
ar = r - ru2 = 0.192 - 0.46210.522 = 0.0770
$
##
au = ru + 2ru = 0 + 21-0.133210.52 = -0.133
Substituting these results into Eqs. 1 and 2 with u = 60° and
solving yields
NC = 19.4 N
FP = -0.356 N
Ans.
The negative sign indicates that FP acts opposite to the direction
shown in Fig. 13–20b.
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143
13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
EXAMPLE 13.12
A can C, having a mass of 0.5 kg, moves along a grooved horizontal slot
shown in Fig. 13–21a.The slot is in the form of a spiral, which is defined by
the equation r = 10.1u2 m, where u is in radians. If the arm OA is
#
rotating at a constant rate u = 4 rad>s in the horizontal plane, determine
the force it exerts on the can at the instant u = p rad. Neglect friction
and the size of the can.
SOLUTION
Free-Body Diagram. The driving force FC acts perpendicular to the
arm OA, whereas the normal force of the wall of the slot on the can, NC ,
acts perpendicular to the tangent to the curve at u = p rad, Fig. 13–21b.
As usual, ar and a u are assumed to act in the positive directions of r and
u, respectively. Since the path is specified, the angle c which the
extended radial line r makes with the tangent, Fig. 13–21c, can be
determined from Eq. 13–10. We have r = 0.1u, so that dr>du = 0.1,
and therefore
r
0.1u
tan c =
=
= u
dr>du
0.1
u
r $ 0.1 u
r
A
O
·
u $ 4 rad/s
C
(a)
FC
ar
r
f
NC
f
au
Tangent
-1
When u = p, c = tan p = 72.3°, so that f = 90° - c = 17.7°, as
shown in Fig. 13–21c. Identify the four unknowns in Fig. 13–21b.
Equations of Motion.
Fig. 13–21b, we have
+ ©F = ma ;
;
r
r
+ T ©Fu = mau ;
u
Using f = 17.7° and the data shown in
NC cos 17.7° = 0.5ar
FC - NC sin 17.7° = 0.5au
Kinematics. The time derivatives of r and u are
#
u = 4 rad>s
r = 0.1u
$
#
#
u = 0
r = 0.1u = 0.1142 = 0.4 m>s
$
$
r = 0.1u = 0
At the instant u = p rad,
#
$
ar = r - ru2 = 0 - 0.11p21422 = -5.03 m>s2
$
##
au = ru + 2ru = 0 + 210.42142 = 3.20 m>s2
(b)
(1)
(2)
r $ 0.1 u
u$p
r
c
f
Tangent
u
(c)
Fig. 13–21
Substituting these results into Eqs. 1 and 2 and solving yields
NC = -2.64 N
FC = 0.800 N
Ans.
What does the negative sign for NC indicate?
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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144
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
PROBLEMS
13–83. A particle, having a mass of 1.5 kg, moves along a
path defined by the equations r = 14 + 3t2 m,
u = 1t2 + 22 rad, and z = 16 - t32 m, where t is in seconds.
Determine the r, u, and z components of force which the
path exerts on the particle when t = 2 s.
*13–84. The path of motion of a 5-lb particle in the
horizontal plane is described in terms of polar coordinates
as r = 12t + 12 ft and u = 10.5t2 - t2 rad, where t is in
seconds. Determine the magnitude of the unbalanced force
acting on the particle when t = 2 s.
13–87. The 4-kg spool slides along the rotating rod. At the
instant
shown, the angular rate of rotation
of the rod is
#
$
u = 6 rad>s, which is increasing at u = 2 rad>s2. At this
same instant, the spool is moving outward along the rod at
3 m>s, which is increasing at 1 m>s2 when r = 0.5 m.
Determine the radial frictional force and the normal force
of the rod on the spool at this instant.
r $ 0.5 m
••
u $ 2 rad/s2
O
13–85. The spring-held follower AB has a weight of 0.75 lb
and moves back and forth as its end rolls on the contoured
surface of the cam, where r = 0.2 ft and z = 10.1 sin u2 ft. If
the cam is rotating at a constant rate of 6 rad>s, determine
the force at the end A of the follower when u = 90°. In this
position the spring is compressed 0.4 ft. Neglect friction at
the bearing C.
•
u $ 6 rad/s
vs $ 3 m/s
as $ 1 m/s2
Prob. 13–87
13–86. The spring-held follower AB has a weight of 0.75 lb
and moves back and forth as its end rolls on the contoured
surface of the cam, where r = 0.2 ft and z = 10.1 sin 2u2 ft. If
the cam is rotating at a constant rate of 6 rad>s, determine the
maximum and minimum force the follower exerts on the cam
if the spring is compressed 0.2 ft when u = 45°.
*13–88. The boy of mass 40 kg is sliding down the spiral
slide at a constant speed such that his position, measured
from the top of the chute, has components r = 1.5 m,
u = 10.7t2 rad, and z = 1-0.5t2 m, where t is in seconds.
Determine the components of force Fr , Fu , and Fz which
the slide exerts on him at the instant t = 2 s. Neglect the
size of the boy.
z
u
z $ 0.1 sin 2u
z
·
u $ 6 rad/s
0.2 ft
r $ 1.5 m
C
B
A
k $ 12 lb/ft
z
Probs. 13–85/86
Prob. 13–88
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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145
PROBLEMS
13–89. The girl has a mass of 50 kg. She is seated on the
horse of the merry-go-round
which undergoes constant
#
rotational motion u = 1.5 rad>s. If the path of the horse is
defined by r = 4 m, z = 10.5 sin u2 m, determine the
maximum and minimum force Fz the horse exerts on her
during the motion.
13–91. The particle has a mass of 0.5 kg and is confined to
move along the smooth horizontal slot due to the rotation
of the arm OA. Determine the force of the rod on the
particle and the normal force of the slot on the particle
when u =# 30°. The rod is rotating with a constant angular
velocity u = 2 rad>s. Assume the particle contacts only one
side of the slot at any instant.
*13–92. Solve $Problem 13–91 if# the arm has an angular
acceleration of u = 3 rad>s2 and u = 2 rad>s at this instant.
Assume the particle contacts only one side of the slot at
any instant.
z
A
u r
·
u $ 2 rad/s
r
0.5 m
u
O
Prob. 13–89
13–90. The 0.5-lb particle is guided along the circular path
using
the slotted arm guide. If the arm has $an angular velocity
#
u = 4 rad>s and an angular acceleration u = 8 rad>s2 at the
instant u = 30°, determine the force of the guide on the
particle. Motion occurs in the horizontal plane.
Probs. 13–91/92
13–93. A smooth can C, having a mass of 3 kg, is lifted
from a feed at A to a ramp at B by a rotating rod.
If the rod
#
maintains a constant angular velocity of u = 0.5 rad>s,
determine the force which the rod exerts on the can at the
instant u = 30°. Neglect the effects of friction in the
calculation and the size of the can so that r = 11.2 cos u2 m.
The ramp from A to B is circular, having a radius of 600 mm.
B
r
u
0.5 ft
·
u $ 0.5 rad/s
u
0.5 ft
Prob. 13–90
C
r
600 mm
A
600 mm
Prob. 13–93
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146
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–94. The 2-lb collar slides along the smooth horizontal
spiral rod, r = 12u2 ft, where u is in radians. If its angular rate
#
of rotation is constant and equals u = 4 rad>s, determine the
tangential force P needed to cause the motion and the normal
force that the rod exerts on the spool at the instant u = 90°.
13–95.
Solve Prob. 13–94 if the spiral rod is vertical.
AND
A C C E L E R AT I O N
13–97. The smooth particle has a mass of 80 g. It is
attached to an elastic cord extending from O to P and due
to the slotted arm guide moves along the horizontal circular
path r = 10.8 sin u2 m. If the cord has a stiffness
k = 30 N>m and an unstretched length of 0.25 m, determine
the force of the guide on the particle
# when u = 60°. The
guide has a constant angular velocity u = 5 rad>s.
$
#
13–98. Solve Prob. 13–97 if u = 2 rad>s2 when u = 5 rad>s
and u = 60°.
P
P
r $ 2u
r
r
·
u $ 4 rad/s
·
u $ 5 rad/s
0.4 m
u
O
u
Probs. 13–94/95
O
Probs. 13–97/98
*13–96. The forked rod is used to move the smooth 2-lb
particle around the horizontal path in the shape of a
limaçon, r = 12 + cos u2 ft. If u = 10.5t22 rad, where t is in
seconds, determine the force which the rod exerts on the
particle at the instant t = 1 s. The fork and path contact the
particle on only one side.
13–99. Determine the normal and frictional driving forces
that the partial spiral track# exerts on the 200-kg
$ motorcycle
at the instant u = 53 p rad, u = 0.4 rad>s, and u = 0.8 rad>s2.
Neglect the size of the motorcycle.
r $ (5u) m
2 ft
r
u
r
u
·
u
3 ft
Prob. 13–96
Prob. 13–99
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PROBLEMS
*13–100. Using a forked rod, a smooth cylinder C having a
mass of 0.5 kg is forced to move along the vertical slotted
path r = 10.5u2 m, where u is in radians. If the angular
position of the arm is u = 10.5t22 rad, where t is in seconds,
determine the force of the rod on the cylinder and the
normal force of the slot on the cylinder at the instant
t = 2 s. The cylinder is in contact with only one edge of the
rod and slot at any instant.
147
13–102. The smooth surface of the vertical cam is defined in
part by the curve r = 10.2 cos u + 0.32 m. If the forked rod is
#
rotating with a constant angular velocity of u = 4 rad>s,
determine the force the cam and the rod exert on the 2-kg
roller when u = 30°. The attached spring has a stiffness
k = 30 N>m and an unstretched length of 0.1 m.
r
C
u
u
r $ 0.5 u
Prob. 13–100
Prob. 13–102
13–101. The ball has a mass of 2 kg and a negligible size. It is
originally traveling around the horizontal circular path of
radius
r0 = 0.5 m such that the angular rate of rotation is
#
u0 = 1 rad>s. If the attached cord ABC is drawn down
through the hole at a constant speed of 0.2 m>s, determine the
tension the cord exerts on the ball at the instant r = 0.25 m.
Also, compute the angular velocity of the ball at this instant.
Neglect the effects of friction between the ball and horizontal
plane. Hint: First show that$ the equation
of motion
# in the u
##
direction yields au = ru + 2ru = 11>r21d1r2u2>dt2 = 0.
#
When integrated, r2u = c, where the constant c is determined
from the problem data.
13–103. The collar has a mass of 2 kg and travels along
the smooth horizontal rod defined by the equiangular
spiral r = 1eu2 m, where u is in radians. Determine the
tangential force F and the normal force N acting on the
collar when u = 90°, if the force F maintains a constant
#
angular motion u = 2 rad>s.
A
B
u
F
r
r0
·
u0
r
0.2 m/s
r $ eu
u
C
F
Prob. 13–101
Prob. 13–103
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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148
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
*13–104. The smooth surface of the vertical cam is defined in
part by the curve r = 10.2 cos u + 0.32 m. The forked rod is
$
rotating with an angular acceleration of u = 2 rad>s2, and
#
when u = 45° the angular velocity is u = 6 rad>s. Determine
the force the cam and the rod exert on the 2-kg roller at this
instant. The attached spring has a stiffness k = 100 N>m and
an unstretched length of 0.1 m.
AND
A C C E L E R AT I O N
13–106. Using air pressure, the 0.5-kg ball is forced to
move through the tube lying in the horizontal plane and
having the shape of a logarithmic spiral. If the tangential
force exerted on the ball due to the air is 6 N, determine the
rate of increase in the ball’s speed at the instant u = p>2.
What direction does it act in?
r
r $ 0.2e0.1u
u
r
F$6N
u
Prob. 13–106
Prob. 13–104
13–105. The pilot of an airplane executes a vertical loop
which in part follows the path of a “four-leaved rose,”
r = 1-600 cos 2u2 ft, where u is in radians. If his speed at A
is a constant vP = 80 ft>s, determine the vertical reaction
the seat of the plane exerts on the pilot when the plane is
at A. He weighs 130 lb. Hint: To determine the time derivatives
necessary to compute the acceleration components ar
and au , take the first and second time derivatives of
r = 40011 + cos u2. Then, for further information, use
#
Eq. 12–26 to determine u. Also, take the time
derivative of
$
#
Eq. 12–26, noting that vC = 0, to determine u.
80 ft/s
13–107. Solve Prob. 13–106 if the tube lies in a vertical plane.
r $ 0.2e0.1u
A
r
r $#600 cos 2u
r
u
F$6N
u
Prob. 13–105
Prob. 13–107
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149
PROBLEMS
#
*13–108.
The arm is rotating at a rate of u = 5 rad>s when
$
u = 2 rad>s2 and u = 90°. Determine the normal force it
must exert on the 0.5-kg particle if the particle is confined
to move along the slotted path defined by the horizontal
hyperbolic spiral ru = 0.2 m.
13–110. The tube
# rotates in the horizontal plane at a
constant rate of u = 4 rad>s. If a 0.2-kg ball B starts at the
#
origin O with an initial radial velocity of r = 1.5 m>s and
moves outward through the tube, determine the radial
and transverse components of the ball’s velocity at the
instant it leaves the outer end at C, r = 0.5 m. Hint: Show
that the equation of motion in the r direction is
$
r - 16r = 0. The solution is of the form r = Ae -4t + Be4t.
Evaluate the integration constants A and B, and determine
the time t when r = 0.5 m. Proceed to obtain vr and vu .
z
0.2
r $—
u
·
u $ 4 rad/s
O
r u· $ 5 rad/s, u·· $ 2 rad/s2
u
r
0.5 m
u $ 90'
y
B
x
C
Prob. 13–108
Prob. 13–110
13–109. The collar, which has a weight of 3 lb, slides along
the smooth rod lying in the horizontal plane and having the
shape of a parabola r = 4>11 - cos u2, where u is in radians
and r is #in feet. If the collar’s angular rate is constant and
equals u = 4 rad>s, determine the tangential retarding
force P needed to cause the motion and the normal force
that the collar exerts on the rod at the instant u = 90°.
13–111. A 0.2-kg spool slides down along a smooth rod.
If
# the rod has a constant angular rate of rotation
u = 2 rad>s in the vertical plane, show that the equations of
$
motion for the spool are r - 4r - 9.81 sin u = 0 and
#
0.8r + Ns - 1.962 cos u = 0, where Ns is the magnitude of
the normal force of the rod on the spool. Using the
methods of differential equations, it can be shown that
the solution of the first of these equations is
#
r = C1e -2t + C2e2t - 19.81>82 sin 2t. If r, r, and u are
zero when t = 0, evaluate the constants C1 and C2 to
determine r at the instant u = p>4 rad.
P
r
u
Prob. 13–109
u
u $ 2 rad/s
r
Prob. 13–111
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150
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
A C C E L E R AT I O N
AND
*13.7 Central-Force Motion
and Space Mechanics
P
F
du
r
u
1
dA $ — r 2du
2
O
(a)
If a particle is moving only under the influence of a force having a line of
action which is always directed toward a fixed point, the motion is called
central-force motion. This type of motion is commonly caused by
electrostatic and gravitational forces.
In order to determine the motion, we will consider the particle P
shown in Fig. 13–22a, which has a mass m and is acted upon only by the
central force F. The free-body diagram for the particle is shown in
Fig. 13–22b. Using polar coordinates (r, u), the equations of motion,
Eqs. 13–9, become
-F = mc
d2r
du 2
ra
b d
dt
dt2
0 = mar
d2u
dr du
+ 2
b
dt dt
dt2
(13–11)
The second of these equations may be written in the form
1 d 2 du
c ar
bd = 0
r dt
dt
so that integrating yields
u
r
u
F
(b)
Fig. 13–22
r2
du
= h
dt
(13–12)
Here h is a constant of integration. From Fig. 13–22a notice that the
shaded area described by the radius r, as r moves through an angle du, is
dA = 12 r2 du. If the areal velocity is defined as
dA
1 du
h
= r2
=
dt
2 dt
2
(13–13)
then, it is seen that the areal velocity for a particle subjected to centralforce motion is constant. In other words, the particle will sweep out equal
segments of area per unit of time as it travels along the path. To obtain
the path of motion, r = f1u2, the independent variable t must be
eliminated from Eqs. 13–11. Using the chain rule of calculus and
Eq. 13–12, the time derivatives of Eqs. 13–11 may be replaced by
dr
dr du
h dr
=
= 2
dt
du dt
r du
2
d h dr
d h dr du
d h dr
h
dr
=
a
b =
a
b
= c a 2
bd
dt r2 du
du r2 du dt
du r du r2
dt2
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151
13.7 CENTRAL-FORCE MOTION AND SPACE MECHANICS
Substituting a new dependent variable (xi) j = 1>r into the second
equation, we have
d2j
d2r
= -h2j2 2
2
dt
du
Also, the square of Eq. 13–12 becomes
a
du 2
b = h2j4
dt
Substituting these last two equations into the first of Eqs. 13–11 yields
-h2j2
or
d2j
du2
d2j
du2
- h2j3 = -
+ j =
F
mh2j2
This satellite is subjected to a central force
and its orbital motion can be closely
predicted using the equations developed
in this section.
F
m
(13–14)
This differential equation defines the path over which the particle travels
when it is subjected to the central force F. *
For application, the force of gravitational attraction will be considered.
Some common examples of central-force systems which depend on
gravitation include the motion of the moon and artificial satellites about
the earth, and the motion of the planets about the sun. As a typical
problem in space mechanics, consider the trajectory of a space satellite
or space vehicle launched into free-flight orbit with an initial velocity v0 ,
Fig. 13–23. It will be assumed that this velocity is initially parallel to the
tangent at the surface of the earth, as shown in the figure.†*Just after the
satellite is released into free flight, the only force acting on it is the
gravitational force of the earth. (Gravitational attractions involving
other bodies such as the moon or sun will be neglected, since for orbits
close to the earth their effect is small in comparison with the earth’s
gravitation.) According to Newton’s law of gravitation, force F will
always act between the mass centers of the earth and the satellite,
Fig. 13–23. From Eq. 13–1, this force of attraction has a magnitude of
F = G
Free-flight
trajectory
v0
Satellite
F
F
r $ r0
Power-flight
trajectory
Launching
Fig. 13–23
Mem
r2
where Me and m represent the mass of the earth and the satellite,
respectively, G is the gravitational constant, and r is the distance between
*In the derivation, F is considered positive when it is directed toward point O. If F is
oppositely directed, the right side of Eq. 13–14 should be negative.
†The case where v0 acts at some initial angle u to the tangent is best described using the
conservation of angular momentum (see Prob. 15–103).
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152
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
the mass centers. To obtain the orbital path, we set j = 1>r in the
foregoing equation and substitute the result into Eq. 13–14. We obtain
d2j
2
du
+ j =
GMe
h2
(13–15)
This second-order ordinary differential equation has constant coefficients
and is nonhomogeneous. The solution is represented as the sum of the
complementary and particular solutions. The complementary solution is
obtained when the term on the right is equal to zero. It is
jc = C cos1u - f2
where C and f are constants of integration. The particular solution is
jp =
GMe
h2
Thus, the complete solution to Eq. 13–15 is
j = jc + jp
GMe
1
= = C cos1u - f2 +
r
h2
D
directrix
P
A
u#f
r
x
F
f
u
x¿
p
D
Fig. 13–24
focus
(13–16)
The validity of this result may be checked by substitution into Eq. 13–15.
Equation 13–16 represents the free-flight trajectory of the satellite. It is
the equation of a conic section expressed in terms of polar coordinates.
As shown in Fig. 13–24, a conic section is defined as the locus of point P,
which moves in a plane in such a way that the ratio of its distance from a
fixed point F to its distance from a fixed line is constant. The fixed point
is called the focus, and the fixed line DD is called the directrix. The
constant ratio is called the eccentricity of the conic section and is denoted
by e. Thus,
FP
e =
PA
which may be written in the form
FP = r = e1PA2 = e[p - r cos1u - f2]
or
1
1
1
= cos1u - f2 +
r
p
ep
Comparing this equation with Eq. 13–16, it is seen that the eccentricity of
the conic section for the trajectory is
e =
Ch2
GMe
(13–17)
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13.7 CENTRAL-FORCE MOTION AND SPACE MECHANICS
153
and the fixed distance from the focus to the directrix is
p =
1
C
(13–18)
Provided the polar angle u is measured from the x axis (an axis of
symmetry since it is perpendicular to the directrix), the angle f is zero,
Fig. 13–24, and therefore Eq. 13–16 reduces to
GMe
1
= C cos u +
r
h2
(13–19)
The constants h and C are determined from the data obtained for the
position and velocity of the satellite at the end of the power-flight
trajectory. For example, if the initial height or distance to the space
vehicle is r0 (measured from the center of the earth) and its initial speed
is v0 at the beginning of its free flight, Fig. 13–25, then the constant h may
be obtained from Eq. 13–12. When u = f = 0°, the velocity v0 has no
radial component; therefore, from Eq. 12–25, v0 = r01du>dt2, so that
h = r20
or
du
dt
h = r0v0
(13–20)
To determine C, use Eq. 13–19 with u = 0°, r = r0 , and substitute
Eq. 13–20 for h:
C =
GMe
1
a1 b
r0
r0v20
(13–21)
The equation for the free-flight trajectory therefore becomes
GMe
GMe
1
1
b cos u + 2 2
= a1 2
r
r0
r0v0
r0v0
(13–22)
The type of path taken by the satellite is determined from the value of
the eccentricity of the conic section as given by Eq. 13–17. If
e = 0
free-flight trajectory is a circle
e = 1
free-flight trajectory is a parabola
e 6 1
e 7 1
free-flight trajectory is an ellipse
free-flight trajectory is a hyperbola
(13–23)
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154
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
Each of these trajectories is shown in Fig. 13–25. From the curves it is
seen that when the satellite follows a parabolic path, it is “on the
border” of never returning to its initial starting point. The initial launch
velocity, v0 , required for the satellite to follow a parabolic path is called
the escape velocity. The speed, ve , can be determined by using the
second of Eqs. 13–23 with Eqs. 13–17, 13–20, and 13–21. It is left as an
exercise to show that
ve =
2GMe
A r0
(13–24)
The speed vc required to launch a satellite into a circular orbit can be
found using the first of Eqs. 13–23. Since e is related to h and C,
Eq. 13–17, C must be zero to satisfy this equation (from Eq. 13–20, h
cannot be zero); and therefore, using Eq. 13–21, we have
vc =
GMe
A r0
(13–25)
Provided r0 represents a minimum height for launching, in which
frictional resistance from the atmosphere is neglected, speeds at launch
which are less than vc will cause the satellite to reenter the earth’s
atmosphere and either burn up or crash, Fig. 13–25.
Hyperbolic trajectory
e&1
Parabolic trajectory
e$1
Elliptical trajectory
e%1
v0
Crash
trajectory
Circular
trajectory
e$0
v 0 % vc
r0
Fig. 13–25
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13.7 CENTRAL-FORCE MOTION AND SPACE MECHANICS
155
b
rp O
ra
b
a
a
Fig. 13–26
All the trajectories attained by planets and most satellites are
elliptical, Fig. 13–26. For a satellite’s orbit about the earth, the minimum
distance from the orbit to the center of the earth O (which is located at
one of the foci of the ellipse) is rp and can be found using Eq. 13–22 with
u = 0°. Therefore;
rp = r0
(13–26)
This minimum distance is called the perigee of the orbit. The apogee or
maximum distance ra can be found using Eq. 13–22 with u = 180°.* Thus,
ra =
r0
12GMe>r0v202 - 1
(13–27)
With reference to Fig. 13–26, the semimajor axis a of the ellipse is
a =
rp + ra
(13–28)
2
Using analytical geometry, it can be shown that the minor axis b is
determined from the equation
b = 2rpra
(13–29)
*Actually, the terminology perigee and apogee pertains only to orbits about the earth. If
any other heavenly body is located at the focus of an elliptical orbit, the minimum and
maximum distances are referred to respectively as the periapsis and apoapsis of the orbit.
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156
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
Furthermore, by direct integration, the area of an ellipse is
A = pab =
p
1r + ra22rpra
2 p
(13–30)
The areal velocity has been defined by Eq. 13–13, dA>dt = h>2.
Integrating yields A = hT>2, where T is the period of time required to
make one orbital revolution. From Eq. 13–30, the period is
T =
p
1r + ra22rpra
h p
(13–31)
In addition to predicting the orbital trajectory of earth satellites, the
theory developed in this section is valid, to a surprisingly close
approximation, at predicting the actual motion of the planets traveling
around the sun. In this case the mass of the sun, Ms , should be
substituted for Me when the appropriate formulas are used.
The fact that the planets do indeed follow elliptic orbits about the sun
was discovered by the German astronomer Johannes Kepler in the early
seventeenth century. His discovery was made before Newton had
developed the laws of motion and the law of gravitation, and so at the
time it provided important proof as to the validity of these laws. Kepler’s
laws, developed after 20 years of planetary observation, are summarized
as follows:
1. Every planet moves in its orbit such that the line joining it to the
sun sweeps over equal areas in equal intervals of time, whatever the
line’s length.
2. The orbit of every planet is an ellipse with the sun placed at one
of its foci.
3. The square of the period of any planet is directly proportional to the
cube of the minor axis of its orbit.
A mathematical statement of the first and second laws is given by
Eqs. 13–13 and 13–22, respectively. The third law can be shown from
Eq. 13–31 using Eqs. 13–19, 13–28, and 13–29.
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13.7 CENTRAL-FORCE MOTION AND SPACE MECHANICS
157
EXAMPLE 13.13
A satellite is launched 600 km from the surface of the earth, with an initial
velocity of 30 Mm>h acting parallel to the tangent at the surface of the
earth, Fig. 13–27. Assuming that the radius of the earth is 6378 km and
that its mass is 5.976110242 kg, determine (a) the eccentricity of the
orbital path, and (b) the velocity of the satellite at apogee.
SOLUTION
Part (a). The eccentricity of the orbit is obtained using Eq. 13–17.The
constants h and C are first determined from Eqs. 13–20 and 13–21. Since
rp = r0 = 6378 km + 600 km = 6.97811062 m
v0 = 30 Mm>h = 8333.3 m>s
v0 $ 30 Mm/h
ra
rp
then
h = rpv0 = 6.9781106218333.32 = 58.1511092 m2>s
O
A
vA
GMe
1
C =
a1 b
rp
rpv20
600 km
66.73110 -122[5.976110242]
1
f = 25.4110 -92 m-1
=
e
1
6.97811062
6.9781106218333.322
Hence,
2.54110 -82[58.1511092]2
Ch2
e =
=
= 0.215 6 1
GMe
66.73110 -122[5.976110242]
Fig. 13–27
Ans.
From Eq. 13–23, observe that the orbit is an ellipse.
Part (b). If the satellite were launched at the apogee A shown in
Fig. 13–27, with a velocity vA , the same orbit would be maintained
provided that
h = rpv0 = ravA = 58.1511092 m2>s
Using Eq. 13–27, we have
rp
6.97811062
ra =
=
= 10.80411062
2GMe
2[66.73110 -122][5.976110242]
-1
-1
rpv20
6.9781106218333.322
Thus,
vA =
58.1511092
10.80411062
= 5382.2 m>s = 19.4 Mm>h
Ans.
NOTE: The farther the satellite is from the earth, the slower it moves,
which is to be expected since h is constant.
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158
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
PROBLEMS
In the following problems, except where otherwise indicated,
assume that the radius of the earth is 6378 km, the earth’s mass
is 5.976110242 kg, the mass of the sun is 1.99110302 kg, and the
gravitational constant is G = 66.73110-122 m3>1kg # s22.
*13–112. The rocket is in circular orbit about the earth at
an altitude of h = 4 Mm. Determine the minimum
increment in speed it must have in order to escape the
earth’s gravitational field.
13–115. The rocket is traveling in free flight along an
elliptical trajectory A¿A. The planet has a mass 0.60 times
that of the earth’s. If the rocket has an apoapsis and
periapsis as shown in the figure, determine the speed of the
rocket when it is at point A.
r $ 3.20 Mm
A
B
O
6.40 Mm
A¿
16 Mm
Prob. 13–115
h $ 4 Mm
*13–116. An elliptical path of a satellite has an
eccentricity e = 0.130. If it has a speed of 15 Mm>h when it
is at perigee, P, determine its speed when it arrives at
apogee, A. Also, how far is it from the earth’s surface when
it is at A?
Prob. 13–112
13–113. Prove Kepler’s third law of motion. Hint: Use
Eqs. 13–19, 13–28, 13–29, and 13–31.
13–114. A satellite is to be placed into an elliptical orbit
about the earth such that at the perigee of its orbit it has an
altitude of 800 km, and at apogee its altitude is 2400 km.
Determine its required launch velocity tangent to the
earth’s surface at perigee and the period of its orbit.
A
P
Prob. 13–116
2400 km
Prob. 13–114
800 km
13–117. A communications satellite is to be placed into an
equatorial circular orbit around the earth so that it always
remains directly over a point on the earth’s surface. If this
requires the period to be 24 hours (approximately),
determine the radius of the orbit and the satellite’s velocity.
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PROBLEMS
13–118. The rocket is traveling in free flight along an
elliptical trajectory A¿A. The planet has no atmosphere, and
its mass is 0.60 times that of the earth’s. If the rocket has the
apogee and perigee shown, determine the rocket’s velocity
when it is at point A. Take G = 34.4110 -921lb # ft22>slug 2,
Me = 409110212 slug, 1 mi = 5280 ft.
A
B
O
r $ 2000 mi
13–121. The rocket is traveling in free flight along an
elliptical trajectory A¿A. The planet has no atmosphere, and
its mass is 0.70 times that of the earth’s. If the rocket has an
apoapsis and periapsis as shown in the figure, determine the
speed of the rocket when it is at point A.
A¿
r $ 3 Mm
A
4000 mi
O
A¿
6 Mm
13–119. If the rocket is to land on the surface of the planet,
determine the required free-flight speed it must have at A¿
so that the landing occurs at B. How long does it take for
the rocket to land, in going from A¿ to B? The planet has
no atmosphere, and its mass is 0.6 times that of the earth’s.
Take G = 34.4110 -921lb # ft22>slug 2, Me = 409110212 slug,
1 mi = 5280 ft.
B
B
10 000 mi
Prob. 13–118
A
159
O
r $ 2000 mi
9 Mm
Prob. 13–121
13–122. If the rocket in Prob. 13–121 is to land on the
surface of the planet, determine the required free-flight
speed it must have at A¿ so that it strikes the planet at B.
How long does it take for the rocket to land, going from A¿
to B along an elliptical path?
A¿
r $ 3 Mm
A
4000 mi
B
O
A¿
10 000 mi
Prob. 13–119
*13–120. The speed of a satellite launched into a circular
orbit about the earth is given by Eq. 13–25. Determine the
speed of a satellite launched parallel to the surface of the
earth so that it travels in a circular orbit 800 km from the
earth’s surface.
6 Mm
9 Mm
Prob. 13–122
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160
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
13–123. A satellite S travels in a circular orbit around the
earth. A rocket is located at the apogee of its elliptical orbit
for which e = 0.58. Determine the sudden change in speed
that must occur at A so that the rocket can enter the
satellite’s orbit while in free flight along the blue elliptical
trajectory. When it arrives at B, determine the sudden
adjustment in speed that must be given to the rocket in
order to maintain the circular orbit.
AND
A C C E L E R AT I O N
13–126. The rocket is traveling in a free-flight elliptical
orbit about the earth such that e = 0.76 and its perigee is
9 Mm as shown. Determine its speed when it is at point B.
Also determine the sudden decrease in speed the rocket
must experience at A in order to travel in a circular orbit
about the earth.
A
120 Mm
B
B
A
S
10 Mm
9 Mm
Prob. 13–126
Prob. 13–123
*13–124. An asteroid is in an elliptical orbit about the
sun such that its perihelion distance is 9.3011092 km.
If the eccentricity of the orbit is e = 0.073, determine
the aphelion distance of the orbit.
13–125. A satellite is in an elliptical orbit around the earth
such that e = 0.156. If its perigee is 5 Mm, determine its
velocity at this point and also the distance OB when it is at
point B, located 135° away as shown.
13–127. A rocket is in free-flight elliptical orbit around
the planet Venus. Knowing that the periapsis and apoapsis
of the orbit are 8 Mm and 26 Mm, respectively, determine
(a) the speed of the rocket at point A¿, (b) the required
speed it must attain at A just after braking so that it
undergoes an 8-Mm free-flight circular orbit around Venus,
and (c) the periods of both the circular and elliptical orbits.
The mass of Venus is 0.816 times the mass of the earth.
B
135'
P
5 Mm
O
O
A¿
A
8 Mm
18 Mm
Prob. 13–125
Prob. 13–127
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13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
161
Design Project
13–1D. Design of a Ramp Catapult
The block B has a mass of 20 kg and is to be catapulted from the table.
Design a catapulting mechanism that can be attached to the table and to
the container of the block, using cables and pulleys. Neglect the mass of
the container, assume the operator can exert a constant tension of 120 N
on a single cable during operation, and that the maximum movement of
his arm is 0.5 m. The coefficient of kinetic friction between the table and
container is mk = 0.2. Submit a drawing of your design, and calculate the
maximum range R to where the block will strike the ground. Compare
your value with that of others in the class.
B
0.75 m
R
Fig. 13–1D
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New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
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162
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
Chapter Review
Kinetics
Kinetics is the study of the relationship
between forces and the acceleration
they cause. This relationship is based
on Newton’s second law of motion,
expressed mathematically as ©F = ma.
(See page 102.)
The mass m is the proportionality constant
between the resultant force ©F acting on
the particle and the acceleration a caused
by this resultant. The mass represents the
quantity of matter contained within the
particle. It measures the resistance to a
change in its motion.
Before applying the equation of motion, it
is important to first draw the particle’s
free-body diagram in order to account for
all of the forces that act on the particle.
Graphically, this diagram is equal to
the kinetic diagram, which shows the result
of the forces, that is, the ma vector.
(See page 105.)
F2
FR = (F
P
$
ma
F1
Free-body
diagram
Kinetic
diagram
Inertial Coordinate Systems
When applying the equation of motion, it
is important to make measurements of the
acceleration from an inertial coordinate
system. This system has axes that do not
rotate but are either fixed or translate with
a constant velocity. Various types of
inertial coordinate systems can be used to
apply ©F = ma in component form.
(See page 106.)
Rectangular x, y, z axes are used to
describe rectilinear motion along each of
the axes.
Normal and tangential n, t axes are often
used when the path is known. Recall that
a n is always directed in the +n direction. It
indicates the change in the velocity
direction. Also recall that a t is tangent to
the path. It indicates the change in the
velocity magnitude.
a
P
Path of particle
vO
O
Inertial frame of reference
©Fx = max , ©Fy = may , ©Fz = maz
©Ft = mat , ©Fn = man , ©Fb = 0
at = dv>dt or a t = v dv>ds, and a n = v2>r
e =
[1 + 1dy>dx22]3>2
ƒ d2y>dx2 ƒ
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CHAPTER REVIEW
Cylindrical coordinates are useful when
angular motion of the radial coordinate r
is specified or when the path can
conveniently be described with these
coordinates. For some problems, the
direction of the forces on the free-body
diagram will require finding the angle c
between the extended radial coordinate
and the tangent to the curve.
(See pages 106, 123, and 135.)
#
$
$
# #
$
©Fr = m1r - ru 22, ©Fu = m1ru + 2r u2, ©Fz = mz
tan c =
Central-Force Motion
When a single force acts upon a particle,
such as during the free-flight trajectory of a
satellite in a gravitational field, then the
motion is referred to as central-force
motion. The orbit depends upon the
eccentricity e; and as a result, the trajectory
can either be circular, parabolic, elliptical,
or hyperbolic.
(See page 147.)
163
r
dr>du
Free flight
trajectory
v0
Satellite
F
F
r $ r0
Power flight
trajectory
Launching
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