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PROPRIETARY A N I) ( ONM DKM'IA
I
,
Th is Manual is the proprietary property of The McGraw - 1 1 i 1 1 Companies, Inc. ( "Mc ( J ravv - 1 1 i 11"
)
and protected by copyright and other state and federal laws. By open in li and using this Manual (he
user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the
Manual should be promptly returned unopened lo McGraw-Hill: This Manual is being provided
only to authorized professors and instructors for use in preparing for the classes using the
affiliated textbook. No other use or distribution of this Manual is permitted. This Manual
may not he sold and may not be distributed to or used by any student or other third party.
No part of this Manual may be reproduced, displayed or distributed in any form or by any
means, electronic or otherwise, without the prior written permission of McGraw-Hill.
Instructor's and Solutions ManualVolume 1 , Chapters 2-5
to accompany
VECTOR MECHANICSFOR ENGINEERS
Instructor's and Solutions Manualto accompany
Vector Mechanics for
Engineers, Statics
Ninth Edition
Volume 1, Chapters 2-5
Ferdinand P. BeerLate ofLehigh University
E. Russell Johnston, Jr.University ofConnecticut
David F. MazurekUnited States Coast Guard Academy
Elliot EisenbergThe Pennsylvania State University
Prepared by
Amy MazurekWilliams Memorial Institute
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property ofThe McGraw-Hill Companies, Inc. ("McGraw-Hill") and protected by copyright and
other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and ifthe recipient
does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being
provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook.
No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or
used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any formor by any means, electronic or otherwise, without the prior written permission of the McGraw-Hill.
Higher Education
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Instructor's and Solutions Manual, Volume 1 to accompany
VECTOR MECHANICS FOR ENGINEERS, STATICS, NINTH EDITIONFerdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek, and Elliot Eisenberg
Published by McGraw-Hill Higher Education, an imprint ofThe McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,
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ISBN: 978-0-07-724918-2
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TABLE OF CONTENTS
TO THE INSTRUCTOR v
DESCRIPTION OF THE MATERIAL CONTAINED IN
VECTOR MECHANICSFOR ENGINEERS: STATICS, NINTH EDITION vii
TABLE I: LIST OF THE TOPICS COVERED IN
VECTOR MECHANICSFOR ENGINEERS: STATICS xiv
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS xv
TABLE III: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN STATICS(50% of Problems in SI Units and 50% of Problems in U.S. Customary Units) xxviii
TABLE IV: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN STATICS(75% ofProblems in SI Units and 25% ofProblems in U.S. Customary Units) xxxix
TABLE V: SAMPLE ASSIGNMENT SCHEDULE FOR A COMBINED COURSEIN STATICS AND DYNAMICS (50% of Problems in SI Units and 50%of Problems in U.S. Customary Units) xxx
PROBLEM SOLUTIONS I
TO THE INSTRUCTOR
As indicated in its preface, Vector Mechanics
for Engineers: Statics is designed for the first
course in statics offered in the sophomore
year of college. New concepts have, therefore,
been presented in simple terms and every step
has been explained in detail. However, because
of the large number of optional sections
which have been included and the maturity
of approach which has been achieved, this
text can also be used to teach a course
which will challenge the more advanced
student.
The text has been divided into units, each
corresponding to a well-defined topic and
consisting of one or several theory sections,
one or several Sample Problems, a section
entitled Solving Problems on Your Own, and a
large number of problems to be assigned. Toassist instructors in making up a schedule of
assignments that will best fit their classes, the
various topics covered in the text have been
listed in Table I and a suggested number of
periods to be spent on each topic has been
indicated. Both a minimum and a maximumnumber of periods have been suggested, and
the topics which form the standard basic
course in statics have been separated from
those which are optional. The total number of
periods required to teach the basic material
varies from 26 to 39, while covering the entire
text would require from 41 to 65 periods. If
allowance is made for the time spent for
review and exams, it is seen that this text is
equally suitable for teaching a basic statics
course to students with limited preparation
(since this can be done in 39 periods or less)
and for teaching a more complete statics
course to advanced students (since 41 periods
or more are necessary to cover the entire text).
In most instances, of course, the instructor
will want to include some, but not all, of the
additional material presented in the text. In
addition, it is noted that the text is suitable for
teaching an abridged course in statics which
can be used as an introduction to the study of
dynamics (see Table J).
The problems have been grouped according to
the portions of material they illustrate and have
been arranged in order of increasing difficulty,
with problems requiring special attention
indicated by asterisks. We note that, in most
cases, problems have been arranged in groups
of six or more, all problems of the same group
being closely related. This means that instructors
will easily find additional problems to amplify a
particular point which they may have brought
up in discussing a problem assigned for
homework. A group of problems designed to
be solved with computational software can be
found at the end of each chapter. Solutions for
these problems, including analyses of the
problems and problem solutions and output for
the most widely used computational programs,
are provided at the instructor s edition of the
text's website:
http://www .mhhe.com/beerjohnston
.
To assist in the preparation of homework
assignments, Table II provides a brief
description of all groups of problems and a
classification of the problems in each group
according to the units used. It should also be
noted that the answers to all problems are
given at the end of the text, except for those
with a number in italic. Because of the large
number of problems available in both systems
of units, the instructor has the choice of
assigning problems using SI units and problems
using U.S. customary units in whatever
proportion is found to be most desirable for a
given class. To illustrate this point, sample
lesson schedules are shown in Tables III, IV,
and V, together with various alternative lists
of assigned homework problems. Half of the
problems in each of the six lists suggested in
Table 311 and Table V are stated in SI. units
and half in U.S. customary units. On the other
hand, 75% of the problems in the four lists
suggested in Table TV are stated in SI units
and 25% in U.S. customary units.
Since the approach used in this text differs in
a number of respects from the approach used
in other books, instructors will be well advi-
sed to read the preface to Vector Mechanics
for Engineers, in which the authors have
outlined their general philosophy. In addition,
instructors will find in the following pages a
description, chapter by chapter, of the more
significant features of this text. It is hoped
that this material will help instructors in
organizing their courses to best fit the needs
of their students. The authors wish to
acknowledge and thank Amy Mazurek of
Williams Memorial Institute for her careful
preparation of the solutions contained in this
manual.
E. Russell Johnston, Jr.
David Mazurek
Elliot R Eisenberg
VI
DESCRIPTION OF THE MATERIAL CONTAINED INVECTOR MECHANICS FOR ENGINEERS: STATICS, Ninth Edition
Chapter 1
Introduction
The material in this chapter can be used as a
first assignment or for later reference. The six
fundamental principles listed in Sec. 1.2 are
introduced separately and are discussed at
greater length in the following chapters.
Section 1.3 deals with the two systems of
units used in the text. The SI metric units are
discussed first. The base units are defined and
the use of multiples and submultiples is
explained. The various SI prefixes are
presented in Table 1.1, while the principal SI
units used in statics and dynamics are listed in
Table 1.2. In the second part of Sec. 1.3, the
base U.S. customary units used in mechanics
are defined, and in Sec. 1.4, it is shown hownumerical data stated in U.S. customary units
can be converted into SI units, and vice versa.
The SI equivalents of the principal U.S.
customary units used in statics and dynamics
are listed in Table 1.3.
The instructor's attention is called to the fact
that the various rules relating to the use of SI
units have been observed throughout the text.
For instance, multiples and submultiples
(such as kN and mm) are used whenever
possible to avoid writing more than four digits
to the left of the decimal point or zeros to the
right of the decimal point. When 5-digit or
larger numbers involving SI units are used,
spaces rather than commas are utilized to
separate digits into groups of three (for
example, 20 000 km). Also, prefixes are never
used in the denominator of derived units; for
example, the constant of a spring which
stretches 20 mm under a load of 100 N is
expressed as 5 kN/m, not as 5 N/mm.
in order to achieve as much uniformity as
possible between results expressed respectively
in SI and U.S. customary units, a center point,
rather than a hyphen, has been used to combine
the symbols representing U.S. customary
units (for example, 10 lb • ft); furthermore, the
unit of time has been represented by the
symbol s, rather than sec, whether SI or U.S.
customary units are involved (for example, 5 s,
50 ft/s, 15 m/s). However, the traditional use
of commas to separate digits into groups of
three has been maintained for 5-digit and
larger numbers involving U.S. customary
units.
Chapter 2
Statics of Particles
This is the first of two chapters dealing with
the fundamental properties of force systems.
A simple, intuitive classification of forces has
been used: forces acting on a particle (Chap. 2)
and forces acting on a rigid body (Chap. 3).
Chapter 2 begins with the parallelogram law
of addition of forces and with the introduction
of the fundamental properties of vectors. In
the text, forces and other vector quantities are
always shown in bold-face type. Thus, a force
F (boldface), which is a vector quantity, is
clearly distinguished from the magnitude F(italic) of the force, which is a scalar quantity.
On the blackboard and in handwritten work,
where bold-face lettering is not practical, vector
quantities can be indicated by underlining.
Both the magnitude and the direction of a
vector quantity must be given to completely
define that quantity. Thus, a force F of
magnitude F = 280 lb, directed upward to the
right at an angle of 25° with the horizontal, is
indicated as F - 280 lb ^fL 25° when printed
or as F - 280 lb ^L 25° when handwritten.
Unit vectors i and j are introduced in Sec. 2.7,
where the rectangular components of forces
are considered.
VI!
In the early sections of Chap. 2 the following
basic topics are presented: the equilibrium
of a particle, Newton's first law, and the
concept of the free-body diagram. These first
sections provide a review of the methods
of plane trigonometry and familiarize the
students with the proper use of a calculator.
A general procedure for the solution of
problems involving concurrent forces is
given: when a problem involves only three
forces, the use of a force triangle and a
trigonometric solution is preferred; when a
problem involves more than three forces, the
forces should be resolved into rectangular
components and the equations £F.v = 0, T,Fy =
should be used.
The second part of Chap. 2 deals with forces
in space and with the equilibrium of particles
in space. Unit vectors are used and forces are
expressed in the form F = FA + Fyj + Fzk =
FA, where i, j, and k are the unit vectors
directed respectively along the x, y, and z
axes, and X is the unit vector directed along
the line of action of F.
Note that since this chapter deals only with
particles or bodies which can be considered as
particles, problems involving compression
members have been postponed with only a
few exceptions until Chap. 4, where students
will learn to handle rigid-body problems in a
uniform fashion and will not be tempted to
erroneously assume that forces are concurrent
or that reactions are directed along members.
It should be observed that when SI units are
used a body is generally specified by its mass
expressed in kilograms. The weight of the
body, however, should be expressed in newtons.
Therefore, in many equilibrium problems
involving SI units, an additional calculation is
required before a free-body diagram can be
drawn (compare the example in Sec. 2.11 and
Sample Probs. 2.5 and 2.9). This apparent
disadvantage of the SI system of units, when
compared to the U.S. customary units, will be
offset in dynamics, where the mass of a body
expressed in kilograms can be entered directly
into the equation F = ma, whereas with U.S.
customary units the mass of the body must
first be determined in lb • s7ft (or slugs) from
its weight in pounds.
Chapter 3
Rigid Bodies:
Equivalent Systems of Forces
The principle of transmissibility is presented
as the basic assumption of the statics of rigid
bodies. However, it is pointed out that
this principle can be derived from Newton s
three laws of motion, (see Sec. 16.5 of
Dynamics). The vector product is then
introduced and used to define the moment
of a force about a point. The convenience
of using the determinant form (Eqs. 3.19
and 3.21) to express the moment of a force
about a point should be noted. The scalar
product and the mixed triple product are
introduced and used to define the moment of
a force about an axis. Again, the convenience
of using the determinant form (Eqs. 3.43
and 3.46) should be noted. The amount of
time which should be assigned to this part
of the chapter will depend on the extent to
which vector algebra has been considered
and used in prerequisite mathematics and
physics courses. It is felt that, even with no
previous knowledge of vector algebra, a
maximum of four periods is adequate (see
Table 1).
In Sees. 3.12 through 3.15 couples are
introduced, and it is proved that couples are
equivalent if they have the same moment.
While this fundamental property of couples is
often taken for granted, the authors believe
that its rigorous and logical proof is necessary
if rigor and logic are to be demanded of the
students in the solution of their mechanics
problems.
Iii Sections 3.16 through 3.20, the concept of
equivalent systems of forces is carefully
presented. This concept is made more intuitive
through the extensive use offree-body-diagram
equations (see Figs. 3.39 through 3.46). Note
that the moment of a force is either not shown
or is represented by a green vector (Figs. 3.12
and 3.27). A red vector with the symbol ) is
used only to represent a couple, that is, an
actual system consisting of two forces (Figs.
3.38 through 3.46). Section 3.21 is optional; it
introduces the concept of a wrench and shows
how the most general system of forces in space
can be reduced to this combination of a force
and a couple with the same line of action.
Since one of the purposes of Chap. 3 is to
familiarize students with the fundamental
operations of vector algebra, students should
be encouraged to solve all problems in this
chapter (two-dimensional as well as three-
dimensional) using the methods of vector
algebra. However, many students may be
expected to develop solutions of their own,
particularly in the case of two-dimensional
problems, based on the direct computation of
the moment of a force about a given point as
the product of the magnitude of the force and
the perpendicular distance to the point-
considered. Such alternative solutions mayoccasionally be indicated by the instructor (as
in Sample Prob. 3.9), who may then wish to
compare the solutions of the sample problems
of this chapter with the solutions of the same
sample problems given in Chaps. 3 and 4 of
the parallel text Mechanics for Engineers. It
should be pointed out that in later chapters the
use of vector products will generally be
reserved for the solution of three-dimensional
problems.
Chapter 4
Equilibrium of Rigid Bodies
In the first part of this chapter, problems
involving the equilibrium of rigid bodies in
two dimensions are considered and solved
using ordinary algebra, while problems
involving three dimensions and requiring the
full use of vector algebra are discussed in
the second part of the chapter. Particular
emphasis is placed on the correct drawing and
use of free-body diagrams and on the types of
reactions produced by various supports and
connections (see Figs. 4.1 and 4.10). Note that
a distinction is made between hinges used
in pairs and hinges used alone; in the first
case the reactions consist only of force
components, while in the second case the
reactions may, if necessary, include couples.
For a rigid body in two dimensions, it is
shown (Sec. 4.4) that no more than three
independent equations can be written for a
given free body, so that a problem involving
the equilibrium of a single rigid body can be
solved for no more than three unknowns. It is
also shown that it is possible to choose
equilibrium equations containing only one
unknown to avoid the necessity of solving
simultaneous equations. Section 4.5 introduces
the concepts of statical indeterminacy and
partial constraints. Sections 4.6 and 4.7 are
devoted to the equilibrium of two- and three-
force bodies; it is shown how these concepts
can be used to simplify the solution of certain
problems. This topic is presented only after
the general case of equilibrium of a rigid body
to lessen the possibility of students misusing
this particular method of solution.
The equilibrium of a rigid, body in three
dimensions is considered with full emphasis
placed on the free-body diagram. While the
tool of vector algebra is freely used to
simplify the computations involved, vector
algebra does not, and indeed cannot, replace
the free-body diagram as the focal point of an
equilibrium problem. Therefore, the solution
of every sample problem in this section
begins with a reference to the drawing
of a free-body diagram. Emphasis is also
placed on. the fact that the number of
unknowns and the number of equations must
be equal if a structure is to be statically
determinate and completely constrained.
Chapter 5
Distributed Forces:
Centroids and Centers of Gravity
Chapter 5 starts by defining the center of
gravity of a body as the point of application of
the resultant of the weights of the various
particles forming the body. This definition is
then used to establish the concept of the
centroid of an area or line. Section 5.4
introduces the concept of the first moment of
an area or line, a concept fundamental to the
analysis of shearing stresses in beams in a
later study of mechanics of materials. All
problems assigned for the first period involve
only areas and lines made of simple geometric
shapes; thus, they can be solved without using
calculus.
Section 5.6 explains the use of differential
elements in the determination of centroids by
integration. The theorems of Pappus-Guldinus
are given in Sec. 5.7. Sections 5.8 and 5.9 are
optional; they show how the resultant of
a distributed load can be determined by
evaluating an area and by locating its
centroid. Sections 5.10 through 5.12 deal with
centers of gravity and centroids of volumes.
Here again the determination of the centroids
of composite shapes precedes the calculation
of centroids by integration.
It should be noted that when SI units are used,
a given material is generally characterized by
its density (mass per unit volume, expressed
in kg/m ), rather than by its specific weight
(weight per unit volume, expressed in N/nv).
The specific weight of the material can
then be obtained by multiplying its density
by g = 9.81 m/s (see footnote, page 222 of
the text).
Chapter 6
Analysis of Structures
In this chapter students learn to determine the
interna! forces exerted on the members of pin-
connected structures. The chapter starts
with the statement of Newton s third law
(action and reaction) and is divided into two
parts: (a) trusses, that is, structures consisting
of two-force members only, (b) frames
and machines, that is, structures involving
multiforce members.
After trusses and simple trusses have been
defined in Sees. 6.2 and 6.3, the method of
joints and the method of sections are
explained in detail in Sec. 6.4 and Sec. 6.7,
respectively. Since a discussion of Maxwell's
diagram is not included in this text, the use of
Bow's notation has been avoided, and a
uniform notation has been used in presenting
the method of joints and the method of
sections.
In the method of joints, a free-body diagram
should be drawn for each pin. Since all forces
are of known direction, their magnitudes,
rather than their components, should be used
as unknowns. Following the general
procedure outlined in Chap. 2, joints
involving only three forces are solved using a
force triangle, while joints involving more
than three forces are solved by summing x
and y components. Sections 6.5 and 6.6 are
optional. It is shown in Sec. 6.5 how the
analysis of certain trusses can be expedited by
recognizing joints under special loading
conditions, while in Sec. 6.6 the method of
joints is applied to the solution of three-
dimensional trusses.
It is pointed out in Sec. 6.4 that forces in a
simple truss can be determined by analyzing
the truss joint by joint and that joints can
always be found that involve only two
unknown forces. The method of sections
(Sec. 6.7) should be used (a) if only the forces
in a few members are desired, or (b) if the
truss is not a simple truss and if the solution
of simultaneous equations is to be avoided
(for example, Fink truss). Students should be
urged to draw a separate free-body diagram
for each section used. The free body obtained
should be emphasized by shading and. the
intersected members should be removed and
replaced by the forces they exerted on the free
body. .It is shown that, through a judicious
choice of equilibrium equations, the force in
any given member can be obtained in most
cases by solving a single equation. Section 6.8
is optional; it deals with the trusses obtained
by combining several, simple trusses and
discusses the statical determinacy of such
structures as well as the completeness of their
constraints.
Structures involving multiforce members are
separated into frames and machines. Frames
are designed to support loads, while machines
are designed to transmit and modify forces. It
is shown that while some frames remain rigid
after they have boon detached from their
supports, others will collapse (Sec. 6.11). In
the latter case, the equations obtained by
considering the entire frame as a free body
provide necessary but not sufficient
conditions for the equilibrium of the frame. It
is then necessary to dismember the frame and
to consider the equilibrium of its component
parts in order to determine the reactions at the
external supports. The same procedure is
necessary with most machines in order to
determine the output force Q from the input
force P or inversely (Sec. 6.12).
Students should be urged to resolve a force of
unknown magnitude and direction into two
components but to represent a force of knowndirection by a single unknown, namely its
magnitude. While this rule may sometimes
result in slightly more complicated arithmetic,
it has the advantage of matching the numbers
of equations and unknowns and thus makes it
possible for students to know at any time
during the computations what is known and
what is yet to be determined.
Chapter 7
Forces in Beams and Cables
This chapter consists of five groups of
sections, all of which are optional. The first
three groups deal with forces in beams and
the last two groups with forces in cables.
Most likely the instructor will not have time
to cover the entire chapter and will have to
choose between beams and cables.
Section 7.2 defines the internal forces in a
member. While these forces are limited to
tension or compression in a straight two-force
member, they include a shearing force and a
bending couple in the case of multiforce
members or curved two-force members.
Problems in this section do not make use of
sign conventions for shear and bending
moment and answers should specify which
part of the member is used as the free body.
In Sees. 7.3 through 7.5 the usual sign
conventions are introduced and shear and
bending-moment diagrams are drawn. All
problems in these sections should, be solved
by drawing the free-body diagrams of the
various portions of the beams.
The relations among load, shear, and bending
moment are introduced in Sec. 7.6. Problems
in this section should be solved by evaluating
areas under load and shear curves or by
formal integration (as in Probs. 7.87 and
7.88). Some instructors may feel that the
special methods used in this section detract
from the unity achieved in the rest of the text
through the use of the free-body diagram, and
they may wish to omit Sec. 7.6. Others will
feel that the study of shear and bending-
moment diagrams is incomplete without this
XI
section, and they will want to include it. The
latter view is particularly justified when the
course in statics is immediately followed by a
course in mechanics of materials.
Sections 7.7 through 7.9 are devoted to
cables, first with concentrated loads and then
with distributed loads. In both cases, the
analysis is based on free-body diagrams. The
differential-equation approach, is considered in
the last problems of this group (Probs. 7.124
through 7.126). Section 7.10 is devoted to
catenaries and requires the use of hyperbolic
functions.
Chapter 8
Friction
This chapter not only introduces the general
topic of friction but also provides an
opportunity for students to consolidate their
knowledge of the methods of analysis presented
in Chaps. 2, 3, 4, and 6. It is recommended
that each course in statics include at least a
portion of this chapter.
The first group of sections (Sees. 8. 1 through
8.4) is devoted to the presentation of the laws
of dry friction and to their application to
various problems. The different cases which
can be encountered are illustrated by
diagrams in Figs. 8.2, 8.3, and 8.4. Particular
emphasis is placed on the fact that no relation
exists between the friction force and the normal
force except when motion is impending or
when motion is actually taking place.
Following the general procedure outlined in
Chap. 2, problems involving only three forces
are solved by a force triangle, while problems
involving more than three forces are solved
by summing x and y components. In the first
case the reaction of the surface of contact
should be represented by the resultant R of
the friction force and normal force, while in
the second case it should be resolved into its
components F and N.
Special applications of friction are considered
in Sees. 8.5 through 8.10. They are divided
into the following groups: wedges and screws
(Sees. 8.5 and 8.6); axle and disk friction,
rolling resistance (Sees. 8.7 through 8.9); belt
friction (Sec. 8.10). The sections on axle and
disk friction and on rolling resistance are not
essential to the understanding of the rest of
the text and thus may be omitted.
Chapter 9
Distributed Forces Moments of Inertia
The purpose of Sec. 9.2 is to give motivation
to the study of moments of inertia of areas.
Two examples are considered: one deals with
the pure bending of a beam and the other with
the hydrostatic forces exerted on a submerged
circular gate. It is shown in each case that
the solution of the problem reduces to the
computation of the moment of inertia of
an area. The other sections in the first
assignment are devoted to the definition and
the computation of rectangular moments of
inertia, polar moments of inertia, and the
corresponding radii of gyration. It is shown
how the same differential element can be used
to determine the moment of inertia of an area
about each of the two coordinate axes.
Sections 9.6 and 9.7 introduce the parallel-
axis theorem and its application to the
determination of moments of inertia of
composite areas. Particular emphasis is placed
on the proper use of the parallel-axis theorem
(see Sample Prob. 9.5). Sections 9.8 through
9.10 are optional; they are devoted to
products of inertia and to the determination of
principal axes of inertia.
Sections 9.11 through 9.18 deal with the
moments of inertia of masses. Particular
emphasis is placed on the moments of inertia
of thin plates (Sec. 9.13) and on the use
of these plates as differential elements in
the computation of moments of inertia
XI
1
of symmetrical three-dimensional bodies
(Sec. 9.14). Sections 9.16 through 9.18
are optional but should be used whenever the
following dynamics course includes the
motion of rigid bodies in three dimensions.
Sections 9.16 and 9.17 introduce the moment
of inertia of a body with respect to an
arbitrary axis as well as the concepts of mass
products of inertia and principal axes of
inertia. Section 9.18 discusses the determination
of the principal axes and principal moments
of inertia of a body of arbitrary shape.
When solving many of the problems of Chap. 5,
information on the specific weight of a mate-
rial was generally required. This information
was readily available in problems stated in
U.S. customary units, while it had to be
obtained from the density of the material in
problems stated in SI units (see the last
paragraph of our discussion of Chap. 5). In
Chap. 9, when SI units are used, the mass and
mass moment of inertia of a given body are
respectively obtained in kg and kg mdirectly from the dimensions of the body in
meters and from its density in kg/m . However,
if U.S. customary units are used, the density
of the body must first be calculated from its
specific weight or, alternatively, the weight of
the body can be obtained from its dimensions
and specific weight and then converted into
the corresponding mass expressed in lb s /ft
(or slugs). The mass moment of inertia of the
body is then obtained in lb • ft • s (or slug ft ).
Sample Problem 9.1.2 provides an example of
such a computation. Attention is also called to
the footnote on page of the 513 regarding the
conversion of mass moments of inertia from
U.S. customary units to SI units.
Chapter 10
Method of Virtual Work
While this chapter is optional, the instructor
should give serious consideration to its inclu-
sion in the basic statics course. Indeed, students
who learn the method of virtual work in their
first course in mechanics will remember it as
a fundamental and natural principle. They
may, on the other hand, consider it as an
artificial device if its presentation is postponed
to a more advanced course.
The first group of sections (Sees. 10.2 through
10.5) is devoted to the derivation of the prin-
ciple of virtual work and to its direct application
to the solution of equilibrium problems. The
second group of sections (Sees. 10.6 through
10.9) introduces the concept of potential energy
and shows that equilibrium requires that the
derivative of the potential energy be zero.
Section 10.5 defines the mechanical efficiency
of a machine and Sec. 10.9 discusses the
stability of equilibrium.
The first groups of problems in each assign-
ment utilize the principle of virtual work as an
alternative method for the computation of
unknown forces. Subsequent problems call for
the determination of positions of equilibrium,
while other problems combine the conventional
methods of statics with the method of virtual
work to determine displacements (Probs. 10.55
through 10.58).
XI 11
TABLE I: LIST OF THE TOPICS COVERED IN VECTOR MECHANICS FOR ENGINEERS: STATICS
Sections Topics Basic Course
1. INTRODUCTION1 .16 This material may be used for the first assignment
or for later reference
2. STATICS OF PARTICLES2.1—6 Addition and Resolution of Forces
2.7-8 Rectangular Components2.9-11 Equ i Iibrium of a Particle
2.12-14 Forces in Space
2.15 Equilibrium in Space
3. RIGID BODIES: EQUIVALENT SYSTEMS OF FORCES3 . 1 -8 Vector Product ; Moment of a Force about a Point
3.9-1 1 Scalar Product; Moment of a Force about an Axis
3.12-16 Couples
3 . 1 7-20 Equ ivalen I Systems of Forces* 3
.
2 1 Reduction ofa Wrench
4. EQUILIBRIUM OF RIGID BODIES4 . 1 4 Equi Iibrium in Two Dimensions
4.5 Indeterminate Reactions; Partial Constraints
4.6-7 Two- and Three-Force Bodies
4.8 9 Equi librium in Three Dimensions
5. CENTROIDS AND CENTERS OF GRAVITY5. 1 5 Centroids and First Moments ofAreas and Lines
5.6—7 Centroids by Integration
* 5.8 9 Beams and Submerged Surfaces
5 . 10- 1
2
Centroids ofVolumes
6. ANALYSIS OF STRUCTURES6.1-4
'6.5
= 6.6
6.7
6.9-1
6.12
Trusses by Method of Joints
Joints under Special Loading Conditions
Space Trusses
Trusses by Method of Sections
Combined Trusses
Frames
Machines
7. FORCES IN BEAMS AND CABLES* 7.12 Internal Forces in Members* 7.3 5 Shear and Moment Diagrams by FB Diagram* 7.6 Shear and Moment Diagrams by Integration
* 7.7 9 Cables with Concentrated Loads; Parabolic Cable* 7. 10 Catenary
8. FRICTION8 . 1-4 Laws of Friction and Applications
8.5-6 Wedges and Screws* 8.7—9 Axle and Disk Friction; Rolling Resistance
8.10 Beit Friction
9. MOMENTS OF INERTIA9 . 1
--5 Moments of Inertia 1 ofAreas
9.6-7 Composite Areas* 9 .
8-9 Products of Inertia ; Principal Axes*9.10 Mohr's Circle
9.11 --1 5 Moments of Inert ia ofMasses^
* 9. 1 6- 1 8 Mass Products of Inertia; Principal Axes and Principal
Moments of Inertia
1 0. METHOD OF VIRTUAL WORK10.1-4 Princ ipie of Virtua 1 Work
Mechanical Efficiency
Potential Energy; Stability
10.5
10.6-9
0.5 1
0.5-1
1-2
1-2
1
1-1.5
1.5-2
0.5-1
1-2
1-2
1-2
1-1.5
1-2
2-3
1-2
1-2
I
3
1-2
Suggested Number of Periods
Additional Abridged Course to be
Topics used as an introduction
to dynamics*
0.5-1
Total Number of Periods 26-39
1-1.5
0.25-0.5
0.5-1
0.25-0.5
1
1-2
1-2
1-2
i
1-2
1-2
1
1-2
1-2
1-2
0.5-1
1-1.5
15-26
0.5-1
0.5-1
1
1
!
1-2
1-2
1
1-1.5
1 .5-2
1-2
0.5-1 .5
1-2
14-21
+ A sample assignment schedule for a course in dynamics including this minimum amount of introductory material in statics is given Table V. ll is
recommended that a more complete statics course, such as the one outlined in Tables III and IV of this manual, be used in curricula which include
the study ofmechanics of materials.
# Mass moments of inertia have not been included in the basic statics course since this material is often taught in dynamics.
xiv
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Problem Number*SI Units U.S. Units Problem Description
CHAPTER 2: STATICS OF PARTICLES
FORCES IN A PLANE
Resultant ofconcurrent forces
2.1,4 2.2,3 graphical method
2.7,8 2.5,6 law of sines
2.9, 10 2.11,72
2.13 2.14 special problems
2.17,/* 2.15, 16 laws ofcosines and sines
2.19,20
2.21, 24 2.22, 23
2.26, 27 2.25,30
2.2*, 29
2.32, 34 2.31,33
2.35, 36 2.37, 38
2.39, 40 2.4L 42
Rectangular components of force
simple problems
more advanced problems
Resultant by ZF. = 0, XF:,=
Select force so that resultant has a given direction
2.43, 44 2.47, 48
2.45, 46
2.51,52 2.49, 50
2.55, 56 2.53, 54
2.57, 60 2.58, 59
2,61,62 2.63, 64
2.65, 66 2.67, 68
2.69, 70
Equilibrium. Free-Body Diagram
equilibrium of 3 forces
equilibrium of4 forces
find parameter to satisfy specified conditions
special problems
2.71,72 2.75, 76
2.73, 74 2.77, 78
2.79, 80 2.81,82
2.83, 84
2.87, 88 2.85, 86
2.89, 90
2.91,92 2.93, 94
2.95, 96 2.97, 98
FORCES IN SPACE
Rectangular components of a force in space
given F, 0, and 0, find components and direction angles
relations between components and direction angles
direction of force defined by two points on its line of action
resultant of two or three forces
2.99, 100
2.101,102
2.107, 108
2.111, 112
2.115, 116
2.117, 118
2.103,104
2.105, 106
2.109,7/0
2.113,114
2.119, 120
Equilibrium of a particle in space
load applied to three cables, introductory problems
intermediate problems
advanced problems
* Problems which do not involve any specific system ofunits have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xv
TABLE U: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
2.121, 122
2.123,724
2.125, 126
problems involving cable through ring
special problems
2.131,732
2.135, 136
2.137, 138
2.127, 128
2.129, 130
2.133, 134
Review problems
2.CLC32.C4
2.C2
2.C5
Computer problems
CHAPTER 3; RIGID BODIES; EQUIVALENT SYSTEMS OF FORCES
3.1,2
3.3,4
3.9, .10
3.11,14
3.15
3.16, j7
3.19
3.21,22
3.23, 25
3.27, 28
3.3 L 32
3.5, 6
3.7, 8
3.12, 13
3.18
3.20
3.24, 26
3.29, 30
3.33, 34
Moment of a force about a point: Two dimensions
introductory problems
direction of a force defined by two points on its line of action
derivation ofa formula
applications of the vector product
Moment ofa force about a point: Three dimensions
computingM = r x F, introductory problems
computing M = r x F, more involved problems
usingM to find the perpendicular distance from a point to a line
3.35
3.37, 38
3.4), 42
3.45
3.47, 48
3.49, 50
3.55, 56
3.57, 58
"3.64, *65
*3.66, *67
3.36
3.39, 40
3.43, 44
3.46
3.51,52
3.53, 54
3.59,60
3.6], 62
3.63
3.68, *69
Scalar Product
Finding the angle between two lines
Mixed triple product
Moment of a force about the coordinate axes
Moment ofa force about an oblique axis
Finding the perpendicular distance between two lines
3.70, 72
3.76
3,79, 80
3.83, 84
3.85, 86
3.87, 88
3.91, 92
3.93, 94
3.97, 98
3.71, 73
3.74
3.75, 77
3.78
3.81,82
3.89, 90
3.95, 96
3.99, 100
Couples in two dimensions
Couples in three dimensions
Replacing a force by an equivalent force-couple system: two dimensions
Replacing a force-couple system by an equivalent force or forces
Replacing a force by an equivalent force-couple system: three dimensions
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xvi
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
3.101,102
3.103
3.107
3.111, 112
3.115,116
3.118
3.119, 120
3.124, 125
3.127, 128
"3.131, *132
3. 133. *13S
*3.137
3.739. 140
3.141
*3.J43,*144
3.104
3.105, 106
3.108,109
3.110, 113
3.114,117
3.1.21, 122
3.123, 126
3.129, 130
3.134. *136
*3.13S
"3.142
3.145. * 146
Equivalent force-couple systems
Finding the resultant of parallel forces: two dimensions
Finding the resultant and its line of action: two dimensions
Reducing a three-dimensional system offorces to a single force-couple system
Finding the resultant of parallel forces: three dimensions
Reducing three-dimensional systems offerees or forces and couples to a wrench
axis ofwrench is parallel to a coordinate axis or passes through Oforce-couple system parallel to the coordinate axes
general, three-dimensional case
special cases where the wrench reduces to a single force
special, more advanced problems
3.147, 148
3.150, 154
3.155,157
3.149, 151
3.152, 153
3.156, 158
Review problems
3 .CI, C43.C5
3.C2,C3
3.C6
Computer problems
CHAPTER 4 : EQUILIBRIUM OF RIGID BODIES
EQUILIBRIUM IN TWO DIMENSIONS
Parallel forces
Parallel forces, find range ofvalues of loads to satisfy multiple criteria
Rigid bodies with one reaction of unknown direction and. one ofknown direction
4.2,3 4.1,4
4.5,
6
4.7,8
4.9, 10 4.11, 14
4.12, 13
4.15,76 4.17, 18
4.19,20
4.21,26 4.22, 23
4.27, 28 4.24, 25
4.30, 31 4.29, 33
4.^2, 34
4.37, 38 4.35, 36
4.41,42 4.39, 40
4.43, 46 4.44, 45
4.49, 50 4.47, 48
4.5]., 53 4-52, 54
4.55, 56 4.57, 58
4.59 4.60
Rigid bodies with three reactions of known direction
Rigid bodies with a couple included in the reactions
Find position of rigid body in equilibrium
Partial constraints, statical indeterminacy
Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xvn
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
4.61,62 4.63, 64
4.66, 67 4.65, 68
4.69, 70 4.71,74
4.72, 73
4.76, 77 4.75,81
4.75,79 4.82
4.80
4.83,84 4.86, 87
4.85, 88 4.89, 90
Three-force bodies
simple geometry, solution ofa right triangle required
simple geometry, frame includes a two-force membermore involved geometry
find position ofequilibrium
EQUILIBRIUM IN THREE DIMENSIONS
4.92, 93 4.9 1 , 94 Rigid bodies with two hinges along a coordinate axis and
4.95, 96 an additional reaction parallel to another coordinate axis
4.97, 98 4.99, 1 00 Rigid bodies supported by three vertical wires or by vertical reactions
4.101,102 4.103,104
4.106, 107 4.105, 108 Derrick and boom, problems involving unknown tension in two cables
4.109,110 4.111,112
4.1 13, 1 14 4. 117, 1 1
8
Rigid bodies with two hinges along a coordinate axis and an additional
4.1 15, 116 reaction not parallel to a coordinate axis
4. 1 1 9, 1 22 4. 1 20, 121 Problems involving couples as part of the reaction at a hinge
4.125,126 4.123, 124 Advanced problems
4.129,130 4.127,128
4.131, 132
4.135, 1.36 4. 133, 334 Problems involving taking moments about an oblique line passing
4.140, 141 4. 1 37, 1 38 through two supports
4. J 39
4.142,143 4.144,146 Review problems
4.145, 149 4.747,148
4.150, 151 4.152. 153
4,C2 , C5 4 .C 1 , C3 Computer problems
4.C6 4.C4
CHAPTER 5: DISTRIBUTED FORCES: CENTROIDS AND CENTERS OF GRAVITY
Centroid ofan area formed by combining
rectangles and triangles
rectangles, triangles, and portions of circular areas
triangles, portions of circular or elliptical areas, and areas of analytical
functions
Derive expression for location ofcentroid
Find ratio of dimensions so that centroid is at a given point
First moment of an area
Center ofgravity ofa wire figure
Equilibrium ofwire figures
Find dimension to maximize distance to centroid
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xviii
5.1,2 5.3,
4
5.6,9 5.5, 7
5.8
5.10, 12 5.11, 14
5.13, 15
5.16 5.17
5.19 5.18
5.20, 23 5.21,22
5.24, 25 5.26, 27
5.29, 30 5.28,U5.32,33
5.35,M 5.34
5-i&39 5.37
5.40 5.41
5.42
5.43, 44
5.45, 46 *5.47
*5.48, *49
5-50, 51.
5.52, 53 5.54, 55
5.56, 57
5.59, 6
J
5.58, 60
5.62, *65 5.63, 64
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
Use integration to find centroid of
simple areas
areas obtained by combining shapes of Fig. 5.8A
parabolic area
areas defined by different functions over the interval of interest
homogeneous wires
areas defined by exponential or cosine functions
areas defined by a hyperbola
Find areas or volumes by Pappus - Guldinus
rotate simple geometric figures
practical applications
Distributed load on beams
resultant of loading
reactions at supports
special problems
Forces on submerged surfaces
reactions on dams or vertical gates
reactions on non-vertical gates
special applications
Centroids and centers ofgravity ofthree-dimensional bodies
composite bodies formed from two common shapes
composite bodies formed from three or more elements
composite bodies formed from a material ofuniform thickness
composite bodies formed from a wire or structural shape of uniform cross
section
composite bodies made oftwo different materials
use integration to locate the centroid of
standard shapes: single integration
bodies of revolution: single integration
special applications: single integration
special applications: double integration
bodies formed by cutting a standard shape with an oblique
plane: single integration
5.67 5.66
5.68,71 5.69, 70
5.73 5.72
5.74, 75 5.76, 77
5.78, 79
5.80, 82 5.81,84
5.83, 86 5.85
5.87
5.88, 89 5.90,9)
5.93, 94 5.92
5.95
5.96, 99 5.97,98
5.100, 101 5.103, 104
5.102, 105
5.106, 107 5.108,709
5.110, 112 5.111, 113
5.114, 115 5.116, 117
5.118, 121 5.119, 1 20
5.123, 124 5.122
5.126. 127 5.125
5.128. *129
5.132 *5.130./3i
5.133,134
5.135. »136
Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xix
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
5.137,739 5.138,140 Review problems
5.143, /4* 5.141. 142
5.147,148 5.145, 146
5.C2, C3 5.C1, C4 Computer problems
5.C5, C6 *5.C7
CHAPTER 6: ANALYSIS OF STRUCTURES
TRUSSES
Method of joints
6.1,3 6.2, 4 simple problems
6.6,7 6.5,8
6.9, 10 6.11,12 problems ofaverage diffl
6.13,14 6.15, 16
6.19,20 6.17,18 more advanced problem?
6.23, 24 6.21,22
6.25, 26 6.27
6.28
6.29 6.30 designate simple trusses
6.31, 32 find zero-force members
6.33, 34
6.36, *37 *6.35, *38 space trusses
*6.39, *40 *6.41, *42
6.45, 46 6.43, 44
6.47, 48
6.49, 50 6.51,52
6.53, 54 6.57, 58
6.55, 56 6.59, 60
6.61,62 6.63, 64
6.65, 66 6.67, 68
6.70, 7J. 6-69, 72
(>.73,U
Method ofsections
two ofthe members cut are parallel
none of the members cut are parallel
K-type trusses
trusses with counters
Classify trusses according to constraints
FRAMES AND MACHINES
Analysis of Frames
6.75, 76 6.77, 78 easy problems
6.81,82 6.79,80
6.83, 84 6.87, 88 problems where internal forces are changed by repositioning a couple or by
6.85, 86 6.89 moving a force along its line of action
6.90 replacement of pulleys by equivalent loadings
6.91, 92 6.93, 94 analysis offrames supporting pulleys or pipes
6.95, 96 analysis ofhighway vehicles
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type,
xx
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
6.97, 98 6.99, 100
6.101, .102 6.103, 104
6.105, 106
6.107, 108 6.109. 110
6.111, 112 6.115. 116
6.113,7/4
6.119,720 6.117,118
6.121
analysis offrames consisting of miiltiforce members
problems involving the solution ofsimultaneous equations
unusual floor systems
6.124, 125
6.126, 127
6.128
6.129,130
6.133,134
6.137, 138
6.139, 140
6.143, 144
6.146, 148
6.151, 154
6.152, 153
6.159, 160
6.163
Analysis ofMachines
6.122, 123 toggle-type mach ines
6.131, 132 machines involving cranks
6.135, 136 machines involving a crank with a collar
robotic machines
6.141, 142 tongs
6. 145, 147 pliers, boltcutters, pruning shears
6. 149, 1 50 find force to maintain position oftoggle
garden shears, force in hydraulic cylinder
6.155, 156 large mechanical equipment
6.157, 158
*6. 161,*162 gears and universal joints
special longs
6.165, 166
6.167, 170
6.172,774
6.164, 168
6.169, 171
6.173, 175
Review problems
6.C2, C46.C6
6.C1.C3
6.C5
Computer problems
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
CHAPTER 7: FORCES IN BEAMS AND CABLES
Internal forces in members
7.3,4 7.1,2 simple frames
7.5, 6
7.7,8 7.9,10 curved members7.13,14 7,11,12
7.15, 1
6
7.19, 20 frames with pulleys or pipes
7.17, 18
7.27,22 effect of supports
7.23, 25 7.24, 28 bending moment in circular rods due to their own weight
7.26,27
BEAMS
Shear and bending-moment diagrams using portions of beam as free bodies
problems involving no numerical values
beams with concentrated loads
beams with mixed loads
beams resting on the ground
beams subjected to forces and couples
find value ofparameter to minimize absolute value of bending moment
Shear and bending-moment diagrams using relations among co, V, andMproblems involving no numerical values
problems involving numerical values
find magnitude and location ofmaximum bending moment
Determine Fand Mby integrating ft)twice
Find values ofloads for which Ml is as small as possible
CABLES
Cables with concentrated loads
7.93, 94 7.95, 96 vertical loads
7.97, 98 1.99, 100
7. 1 03, 1 04 7. 1 1 , 1 02 horizontal and vertical loads
7.105, 106
Parabolic cables
7. 1 07, 108 7.1 09, 1 1 supports at the same elevation
7.111,112 7.113,114
7. 1.1 7, 1 1
8
7.115,116 supports at different elevations
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xxii
7.29,30 7.3L32
7.31 34
7.35, 36 7.37, 38
7.39, 40 7.41,42
7.43, 44 7.45, 46
7.4148
7.49, 50 7.52, 53
7.51, 54
7.55, 56 7.58, 59
7.57, *62 7.60, 61
7.63,64 7.65,66
7.6Z687.69, 70 7.73, 74
7.71,72 7.75, 76
7.77,78 7.79, 80
7.83, 84 7.81. 82
1-87,88 7.85,86
7.89, 90 *7.91,*92
TABLE ii: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
*7.U9
7.120, 121
*7.124
7.127, 129
1.131, 132
7.133, 135
7.139, 140
7.143, 144
7.145, 146
*7.151.*152
*7.153
Derive analogy between a beam and a cable
1.122, 123 use analogy to solve previous problems*7. 125,
*126 Derive or use d'yldx - w(.v) T()
Catenary
7.128, 130 given length ofcable and sag or Tm, find span of cable
7. 1 34, 3 36 given span and length of cable, find sag and/or weight
1.137, 138 given span, T„„ and w, find sag
7.141, 142 given T , w, and sag or slope, find span or sag
7.147,* 148
1. 149, *150 special problems
7.154, 155
7.159,161
7A 64, 165
7.C2, C47.C5, C6
7.156, 157
7./5A', 160
7.162,163
7.C1.C3
Review problems
Computer problems
8.3,4 8.1.2
8.5,
6
8.*, 10
8.7, 9
8.11. 12
8. 13, 14
8.15, 16
8.19,20 8-17,18
8.21,22 8.23, 24
8.26, 28 8.25, 27
8.29, 32 8.30,31
$.33
8.36, 37 8.34, 35
8.38 8.39, 40
8.42, 43 8.41,45
8.44
8.48, 49 8.46, 47
8.50, 51 8.52
8.53,54
8.55
8.57, 60 8.56, 58
8.61 8.59, 62
*8.64, *65 8.63
8.66 8.(57, 68
8.69, 70 8.72, 73
8.7/, 74
CHAPTER 8: FRICTION
For given loading, determine whether block is in equilibrium and find friction force
Find minimum force required to start, maintain, or prevent motion.
Analyze motion of several blocks
Sliding and/or tipping ofa rigid body
Problems involving wheels and cylinders
Problems involving rods
Analysis ofmechanisms with friction
Analysis of more advanced rod and beam problems
Analysis of systems with possibility of slippage for various loadings
Wedges, introductory problems
Wedges, more advanced problems
Square-threaded screws
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xxm
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
8.76, 77 8.75, 81 Axle friction
9.78, 79 8.82, 84
8.80, 83 8.85, 86
8.88, 89 8.87
8.91, *92 8.90, 95 Disk friction
*8.93, *94
8.96, 99 8.97. 98 Rolling resistance
8.100
Belt friction
8. 1 02, 1 03 8. 1 1,1 04 belt passing over fixed drum
8.105, 106
8 . 1 07, 1 08 8. 1 1 0, 1 1 1. transmission belts and band brakes
8.109,112 8.113,114
8.115
8.1 16, 117 8.775, 119 advanced problems
8.122,123 8.120,121
8.126, 727 8.124,125
8. 130, 1 3
1
8.L28, 129 derivations, V belts
8. 1.32, 134 8. 1 33, 1 36 Review problems
8.135. 139 8. 13 7, 138
8.142,74? 8.140,747
8 .C 1 , C3 8 .C2 , C5 Computer problems
8.C4, C6 8.C7
8.C8
9-i, 3 9.2,
4
9.6, 7 9.5,8
9.9, 10 9.11, 129.13, 14
9-15,17 9-16, 18
9.19, 9.20
9-21,23 9.22, 25
9.24, 28 9.26, 21*9.29 *9.30
9.31,33 9.32, 34
935, 36
9.37, 38 9.39, 40
9.41,42 9.43, 44
CHAPTER 9: DISTRIBUTED FORCES: MOMENTS OF INERTIA
MOMENTS OF INERTIA OF AREAS
Find by direct integration
moments of inertia ofan area
moments of inertia and radii ofgyration of an area
polar moments of inertia and polar radii ofgyration ofan area
Special problems
Parallel-axis theorem applied to composite areas to find
moment of inertia and radius ofgyration
centroidal moment of inertia, given / or Jcentro ida I moments of inerl ia
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xxiv
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
9-45, 46 9.47, 48
9.49, 5
1
9.50, 52
9.53, 55 9.54, 56
9.57, 58 9.59, *9.60
9.67, 9.62
*9.64 *9.63
*9.65 *9.66
cenlroidal polar moment of inertia
centroidal moments of inertia ofcomposite areas consisting of rolled-steel
shapes:
symmetrical composite areas
singly-symmetrical composite areas (first locale centroid of area)Center ofpressure
for end panel ofa trough
for a submerged verlieai gate or cover
used to locate centroid ofa volume
special problems
9.67, 68 9.69, 70
9.71,72 9.73, 74
9.75, 78 9.76, 77
9.79, 80 9.81,83
9.82, 84
9.85, 86 9.87, 89
9.88, 90
Products of inertia ofareas found by
direct integration
parallel-axis theorem
Using the equations defining the moments and products of inertia with respect to
rotated axes to find
'*> A-1' (vy f°r a giyen angle of rotation
principal axes and principal moments of inertia
9.91, 92 9.93, 95
9.94, 96
9.97, 98 9.99, 100
9.101, 102
9.104, 105 9.103,* 106
9.107, 108 9.109, 110
Using Mohr's circle to find
ly,, 1 ,, I., for a given angle of rotation
principal axes and principal moments of inertia
Special problems
9.1 1 L 1.14 9.112, 113
9.115, 116 9.1 17. 1.18
9.119. 122 9.120, 12.1
9.123. 126 9.124.* 1259.127 9.128
9.129. 130 9.752, 133
9.131. 134
9.135. 136 9.137, 1.38
9.139, "140
9.141, 144 9.142, 143
9.145, 148 9.146, 147
9.149, 150 9.151, 152
9.153, 154
9.155, 156
9.157, 159 9.158. 160
9.162 9.161
MOMENTS OF INERTIA OF MASSES
Mass moment of inertia
of thin plates: two-dimensions
ofsimple geometric shapes by direct single integration
and radius ofgyration ofcomposite bodies
special problems using the parallel-axis theorem
of bodies formed ofsheet metal or of thin plates: three-dimensions
ofmachine elements and of bodies formed ofhomogeneous wire
Mass products of inertia
ofmachine elements
ofbodies formed ofsheet metal or of thin plates
of bodies formed ofhomogeneous wire
Derivation and special problem
* Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
XXV
TABLE II; CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description
9. 3 65, 1 66 9.163 , 164 Mass moment of inertia ofa body with respect to a skew axis
9. 169J 70 9.167, 168
9.171, 172
9,173. 174 9.175. 176 Ellipsoid of inertia and special problems
9.777. J 78
*9.I79, *380 *9. 1 82 .* S83 Principal mass moments of inertia and principal axes of inertia
"9.181, *184
9.185. 186 9.187. 188 Review problems
9.189, 190 9.191, .192
9.WSJ96 9.193. 194
9.C3, C5 9.C 1 , C2 Computer problems
*9.C7, *C8 9.C4, C6
CHAPTER 10: METHOD OF VIRTUAL WORK
For linkages and simple machines, find
force or couple required for equilibrium (linear relations among displacements)
force required for equilibrium (trigonometric relations among displacements)
couple required for equilibrium (trigonometric relations among displacements)
force or couple required for equilibrium
Find position of equilibrium
for numerical values ofloads
linear springs included in mechanism
torsional spring included in mechanism
Problems requiring the use of the law ofcosines
Problems involving the effect of friction
Use method of virtual work to find:
reactions ofabeaminternal forces in a mechanism
movement ofa truss joint
Potential energy method used to
1 0.6 1, 63 1 0.59, 60 solve problems from Sec. 1 0.4
10.65,66 10.62,64
10.67 10.68 establish that equilibrium is neutral
find position ofequilibrium and determine its stability for a system involving:
1 0.71, 72 1 0.69, 70 gears and drums
10.73, 74 torsional springs
10.77,79 10.75,7(5 linear springs
Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xxvi
10.1,3 10.2,4
10.7,8 10.5,6
10.11,12 10.9,10
10.73, .14
10.15, 18 10.16,17
10.79,20 10.21,22
10.24,25 10.23,27
10.26, 28
10.29, 32 10.30,31
10.33, 34 10.36,37
10.35,38
10.39,40
10.43,44 10.41,42
10.45,46
10.49,50 10.47,42
10.51,52
10.53,54
10.55 10.56
10,57, 58
TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)
Problem Number*SI Units U.S. Units Problem Description ___„
10.80, &? 10.78,81
10.07,88 10.85,86
1 0.83, 84 two applied forces
determine stability of a known position ofequilibrium for a system with
10.92,93 10.32,90 one degree of freedom
10.94,96 10.9I,«5* 10.97, *98 * 1 0.99, * 100 two degrees offreedom
1 0. 1 04, 105 10.101, 102 Review problems
10.106,130 10.703,107
10.7/7,1.12 10.108,709
10.C2.C3 1 0.C 1 , C4 Computer problems
10.C5.C6
10.C7
Problems which do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
xxvii
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XXX
CHAPTER 2
PROBLEM 2.1
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P = 75 N and Q - 125 N, determine graphically the magnitude and direction of their
resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a) Parallelogram law:
ISNi
IZ5"NJ
(/;) Triangle rule:
~15-M
We measure:
1Z5"N
tf = 179N, a^JS.V R = 179N ^ 75.1° <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it without permission.
wmm PROBLEM 2.2
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P - 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their
resultant using (a) the parallelogram law, (b) the triangle rule.
m* 35° \%Q
SOLUTION
(a) Parallelogram law:
6o\b
zs\b
(b) Triangle rule:
CO lb
We measure:
25 \b
/? = 77.1 lb, ar = 85.4c Rs 77.1 lb 7" 85.4° ^
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
you are using it without permission.
0B
10 ft
oh—
Sfl (i ri
PROBLEM 2.3
The cable stays AB and /ID help support pole AC. Knowingthat the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of theforces exerted by the stays at A using (a) the parallelogram
law, (/;) the triangle rule.
SOLUTION
We measure:
(a) Parallelogram law:
C £.,ffriTrnrrnTt i /
7J,rnrrr-ry-t,^,
t— 8 |t'—***— i ft-»l
a- 5 1.3°
/? - 59.0°
s^\Jys^.o°7f > MO lb
/ /
I /
I /
(h) Triangle rule:
s<\o
We measure: /? = 139.11b, f=67.0c
.^ = 139.11b^67.0° ^
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PROBLEM 2.4
Two forces are applied at Point B of beam AB. Determine
graphically the magnitude and direction of their resultant using
(a) the parallelogram law, (b) the triangle rule.
3kN
2lcN
SOLUTION
(a) Parallelogram law:
2k/M3kN
(b) Triangle rule:
We measure:
3kN
J? = 3.30kN, « = 66.6° R = 3.30k.N ^T 66.6° <
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;X«) Jl PROBLEM 2.5
The 300-1b force is to be resolved into components along lines a-d and h-b'
.
(a) Determine the angle a by trigonometry knowing that the componentalong line a-d is to be 240 lb. (h) What is the corresponding value of the
component along /;>-//?
SOLUTION
(a) Using the triangle rale and law of sines:
sin/5 sin 60c
(b) Law of sines:
240 lb 3001b
sin/? == 0.69282
P == 43.854°
a + /? + 60° == 180°
a == 1.80°- 60
= 76,146°
F* 300 lb
sin 76. 146° sin 60°
F =30O\b
F • ^Z40\b
43.854 c
a = 76.1° <
Fbl/ ^336\b<
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior writ/en permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
300 lh PROBLEM 2.6
The 300-lb force is to be resolved into components along lines a-d and h~ti
.
(a) Determine the angle a by trigonometry knowing that the component along
line b-h' is to be 120 lb. {b) What is the corresponding value ofthe component
along a~a"l
SOLUTION
Using the triangle rule and law of sines:
sintf sin60c
(b)
F.* 120 Ik 3oo %
120 lb 300 lb
sin a = 0.34641
a = 20.268°
tf + />' + 60° = 180°
/? = 180° -60° -20.268°
= 99.732°
or = 20.3° M
F. 300 1b/<;..., = 341 lb 4
sin 99.732° sin60c
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PROBLEM 2,7
Two forces are applied as shown to a hook support. Knowing that the
magnitude of P is 35 N, determine by trigonometry (a) the required
angle or if the resultant R of the two forces applied to the support is to
be horizontal, (b) the corresponding magnitude of JR.
SOLUTION
Using the triangle rule and law of sines:? = 35rJy^
(, sin a sin 25°
35 N*A^ «T^
(a) :
50 N Ps
sin a --0.60374
a-= 37.138° «r = 37.1° A
(b) a + fi + 25° == 180°
f3~~-180°-25°-
= 117.86°
-37.138°
R 35 Nsin 25°
R = 73.2 N 4sin 117.86
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed,
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PROBLEM 2,8
For the hook support of Problem 2.1, knowing that the magnitude of P is 75 N,
determine by trigonometry (a) the required magnitude of the force Q if the
resultant R of the two forces applied at A is to be vertical, (b) the corresponding
magnitude of R.
PROBLEM 2.1 Two forces P and Q are applied as shown at Point /f of a hook
support. Knowing that .P-75N and <2 = 125N, determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law, (/;) the
triangle rule.
SOLUTION
ZO°
P~>Sn /
RA b)L_
3S°q %
Using the triangle rule and law of sines:
M C - 75N{a)
sin 20° sin 35°= 44.7N <
(b) a + 20° + 35° -180°
a = 180° -20° -35°
= 125°
R 75 N R = 107.1 N ^sin 125° sin 35°
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10
ifiOON"*"*
PROBLEM 2.9
A trolley that moves along a horizontal beam is acted upon by two
forces as shown, (a) Knowing that a - 25°, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical, (b) What is the corresponding magnitude of the
resultant?
SOLUTION
t
iSVj.\COON^^-J \ tS^
\UR
p 'o-t. ™ Z5°
Using the triangle rule and the law of sines:
1600 N PP = 3660'N <v" y
sin. 25° sin 75°
(b) 25° + /? + 75° -180°
/? = 180° -25°~-75°
= 80°
1600N R
sin 25° sin 80°.fi = 3730N 4
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11
KiOO N"^\
PROBLEM 2.10
A trolley that moves along a horizontal beam is acted upon by two forces
as shown. Determine by trigonometry the magnitude and direction of the
force P so that the resultant is a vertical force of 2500 N.
SOLUTION
I
\U>0MJ5^^
F* = zsoo N
\ ©*-
£ \ I
Using the law of cosines: P2 = (1 600 N)2 + (2500 N)
2 - 2(1600 N)(2500 N)cos 75°
P = 2596N
Using the law of sines:sina sin 75°
1600 N 2596 Na = 36.5°
Pis directed 90°-36.5°oi 53.5° below the horizontal. P - 2600 N ^x 53,5° <
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12
i 251
r-
1i
PROBLEM 2.11
'
A steel tank is to be positioned in an excavation. Knowing that
a = 20°, determine by trigonometry {a) the required magnitude
of the force P if the resultant R of the two forces applied sXA is
to be vertical, (b) the corresponding magnitude ofR.
SOLUTION
?m' E
/^v"*-/T ^c ~ 2o"
Mzslb^\^s. LO°
3o° r^c
Using the triangle rule and the law of sines:
(a) /? + 50° + 60° = l80°
/? = 1 80° -50° -60°
-70°
425 lb Pj ^ 392 lb <
sin 70° sin 60°
425 lb R^ = 346 lb <
sin 70° sin 50°
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13
IW 111
30°
,
PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing that
the magnitude of P is 500 lb, determine by trigonometry (a) the
required angle a if the resultant R of the two forces applied atAis to be vertical, (b) the corresponding magnitude ofR.
SOLUTION
P^SOGtb/^'
/-<gN R
/AcA
M7S Lb\\. 6°°
3°°J\Using the triangle rule and the law of sines:
(a) (a + 30°) + 60° + /? = 180°
/? = 180°-(a+ 30°)--60°
/? = 90°~a
sin(90°"«)_sin60°
425 lb 500 lb
90°-a = 47.40° or = 42.6° ^
R _ 500 lbtf = 5511b ^(;)
sin (42.6° + 30°) sin 60°
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l<
PROBLEM 2.13
For the hook support of Problem 2.7, determine by trigonometry (a) the
magnitude and direction of the smallest force P for which the resultant R of
the two forces applied to the support is horizontal, (b) the corresponding
magnitude ofR.
SOLUTION
€50 m
R
The smallest force P will be perpendicular to R.
(a) P = (50 N)sin 25° P = 2l...lN| M
(/>) # = (50 N)eos 25° 7? = 45.3N <
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15
A 251
i
j i
.1\
PROBLEM 2.14
For the steel tank, of Problem 2.11, determine by trigonometry
(a) the magnitude and direction of the smallest force P for
which the resultant R of the two forces applied at A is
'. vertical, (/?) the corresponding magnitude ofR.
PROBLEM 2.11 A steel tank is to be positioned in an
excavation. Knowing that a = 20°, determine by trigonometry
(a) the required magnitude of the force P if the resultant Rof the two forces applied at A is to be vertical, (/;) the
corresponding magnitude of R.
SOLUTION
14zs\b^ Lo
a""~C '*
The smallest force P will be perpendicular to R.
(a) P = (425 lb) cos 30° P«3681b— M
(/;) R = (425 lb) sin 30° R = 2)3\b A
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16
"Vtnr PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hooksupport. Knowing that P = 60lb and g = 25 1b, determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law, (b) the
triangle rule.
SOLUTION
Using the triangle rule and the law of cosines:
Using the law of sines:
35° + a--= 180° 10°\Ji
a~= 125°
R 2--=P2 +Q2 ~ 2PQ cosa
R 2~---(60 lb)
2+(25 lb)
2
-2(60 lb)(25 lb) cos 125°zc?
7 \-
R 2== 3600 + 625 + 3000(0.5736)
r\"LI
7? == 77.108 lb\ I
sin/? sin 125° Q^- zS lb ^
25 lb 77.108 1b
P~-= 15.402°
70° + /? == 85.402° R = 77.llb 7^ 85.4° <
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17
AH-
10 ft
sfl
1£ !LU~
i,fi ft
J
PROBLEM 2.16
Solve Problem 2.3 by trigonometry.
PROBLEM 2.3 The cable stays AB and ^D help support
pole AC. Knowing that the tension is 120 lb in AB and
40 lb in AD, determine graphically the magnitude and
direction of the resultant of the forces exerted by the stays
at A using (a) the parallelogram law, (/;) the triangle rule.
SOLUTION
/B i C X>\1
H~ S fI 1— 6 ft —
lott
tana8
tan j3
Using the triangle rule:
Using the law of cosines:
Using the law of sines:
10
a - 38.66°
A10
[3 = 30.96°
flf+ y? +^ = 180°
38.66° + 30.96° +^ = 180°
^ = 110.38° 40lb
R 2 =(120 lb)2+(40 lb) -2(120 lb)(40 1b)cosl 10.38
c
y? = 139.081b
amy sin 11 0.38°
401b 139.081b
7=15.64°
^ = (90°~- a) + y
(f,= (90°- 38.66°) + 15.64
c
<p - 66.98° R = 139.1 lb f 67.0° A
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18
2kN
PROBLEM 2.17
Solve Problem 2.4 by trigonometry,
PROBLEM 2.4 Two forces are applied at Point B of
beam AB. Determine graphically the magnitude and
direction of their resultant using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
Using the law of cosines: R 2 = (2 kN)2 + (3 kN)
2
4-2(2kN)(3kN)cos80°
MOVR = 3.304 kN
Using the law of sines:sin y _ sin 80° jneo* 1
2kN 3.304 kN
y= 36.59° "\^1 R
fi+ /+80° = l80°
7 = 180°-80°-36.59°
y = 63A\°3kN
= 18O°~/? + 5O°
^-66.59° R = 3.30 kN X 66.6° 4
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19
.4.0°-205 PROBLEM 2.18
Two structural members A and B are bolted to a bracket as shown.
Knowing that both members are in compression and that the force
is 1 5 klM in member A. and 1 kN in member B, determine by
trigonometry the magnitude and direction of the resultant of the
forces applied to the bracket by members A and B.
SOLUTION
Using the force triang le and the laws of cosines and sines: Y*°%We have y = .
180° -(40° + 20°) \py 40°= 120° IbW
Then #2 =(15kN)2 +(10kN)2
- VM-2(15 kN)(10kN) cos 120°
-475 kN 2\ /2o
°
R = 21.794 kN ^iOkM
and10 kN 21.794 kN
sin a sin 120°
sina -f 10kN
VinP0°i, 21.794 kN )
-0.39737
# = 23.414
Hence: = tf + 50° = 73.414 R = 21.8 klM^; 73.4° 4
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211
-10° -20'i PROBLEM 2.19
Two structural members A and B are bolted to a bracket as shown.Knowing that both members are in compression and that the forceis 10 kN in member A and 15 kN in member B, determine bytrigonometry the magnitude and direction of the resultant of theforces applied to the bracket by members A. and B.
SOLUTION
Using the force triangle and the laws of cosines and sines
^180° -(40°+ 20°)We have
Then
and
= 120°
J?2 =(10kN)2
+(15kN)2
~2(10k:N)(15kN)cosl20 c
- 475 kN2
J? = 2 1.794 kN
15 kN 21.794 kN
Ok.M
sin a
since
sin 120°
15k.N "\.
sinl20c
Hence:
2\.794kNJ
= 0.59605
cr- 36.588°
= a+ 50° = 86.588° R = 21.8kNX86.6 ^
PROHUhTARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual mav be displayedreproduced or distributed in anyjorm or by any means, without the prior mitten permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manualyou are using it without permission.
21
SON
PROBLEM 2.20
For the hook support of Problem 2.7, knowing that P = 75 N and a - 50°,
determine by trigonometry the magnitude and direction of the resultant of
the two forces applied to the support.
PROBLEM 2.7 Two forces are applied as shown to a hook support.
Knowing that the magnitude ofP is 35 N, determine by trigonometry (a) the
required angle a if the resultant R of the two forces applied to the support is
to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the force triangle and the laws of cosines and sines: SO^We have P-= 180° -(50° + 25°)
^--^°\cy^c-105° A^n>
Then R l-~= (75 N)
2 + (50 N)2
/ /^R 2
--
-2(75 N)(50N)cos 105°
= 10066.1 N 2
^sn//^R~= 100.330 N
andsin y75 N"
sin 1 05°
100.330 Nsin. y z= 0.72206
y.= 46.225°
Hence: y-25°--= 46.225° -25° = 23.225° R = 100.3 N ~F 21.2° A
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22
Dimensions
in iiini
-800-
MH) N
900
PROBLEM 2.21
Determine the x andy components of each of the forces shown.
SOLUTION
Compute the following distances:
800-N Force:
424-N Force:
408-N Force:
o^ = 7(60°)2+
(800)2
= 1000 mm
0£ = V(56O)2+(9OO)
2
= 1060 mm
oc - V(480)
2+ (90°)
2
-1020 mm
FV=+(800N)
8°°
Fy= +(800N)
Fx = -(424 N)
F3,= -(424 N)
FV =+(408N)
Fj, = -(408N)
1000
600
1000
560
1060
900
1060
480
1020
900
1020
y
A
o /"424-^J \4oaA
/
(
\
\
B c
Fv =+640N <
/;, =4480N <
FX =-224H <
Fy^~36QN A
Fx=+192.0 N 4
F,, --360N A
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23
28 in
96 in.
-84 in.-
48 in.-
SO in.
90 in.
PROBLEM 2.22
Determine thex andy components of each of the forces shown.
SOLUTION6
y
Compute the following distances:A,
/
a4 = 7(84)2+
(80)2 TSo)b ^
= 116 in. Xz^ tb
OB = yl{2S?+(96)2
\siibX
= 100 in.
OC = V(48)2+(90)
2
= 102 in. c
29-lb Force:84
F = +(29 lb)— F. = +2 1 .0 lb <116
80F = +(29 lb) F. = +20.0 lb ^> 116
J'
50-Ib Force: Fv =-(50 lb)— F.L =-14.00 lb <100
96F =+(50 lb)— F = +48.0 lb M
100
5 1 -lb Force: Fr= +(5 1 lb)— FT
= +24.0 lb <102
90F, = ™(5 1 lb)— F = -45 .0 lb A
102
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24
Si^Sfe^
fiOlb
25°
60°
50'J
50 lb4.0 ib
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
40-fb Force: K = +(40 lb) cos 60° F*--= 20.0 lb A
Fy= -(40 lb) sin 60° F
y= -34.6 lb 4
50-lb Force: F* = -(50 lb) sin 50° Fx = -38.3 lb M
Fy
= -(50 lb) cos 50° Fy = -32.1 lb 4
60-lb Force: K == +(60 lb) cos 25° Fx = 54.4 lb <
Fj-= +(60lb)sin25° Fy= 25.4 lb <
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25
PROBLEM 2.24
Determine the a: andy components of each of the forces shown.
SOLUTION
80-N Force: Fx =+(80 N) cos 40°
Fv=+(80N)sin40°
120-N Force: Fx =+(120 N) cos 70°
Fv=+(120N)sin70°
150-N Force: Fv =-(150N)cos35° 1
Fy=+(150 N) sin 35°
Fx =613*1 <
/; = 51.4N <
FV=41.0N <
Fr= I.12.8N <
Fv =-122.9 N <
F.=86.0N <
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26
7,Hd
35° AW
PROBLEM 2.25
Member BD exerts on member ABC a force P directed along line BD.Knowing that P must have a 300-ib horizontal component, determine
(a) the magnitude of the force P, (/?) its vertical component.
SOLUTION
3ooib
p.
(fl)
(b) Vertical component
P sin 35° = 300 lb
_ 300 lb
"sin 35°
^, = Pcos35°
= (523 1b)cos35(
P = 523lb A
/^,=428lb M
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it without permission.
27
1. PROBLEM 2.26
^%t^ The hydraulic cylinder BD exerts on member ABC a force P directed along
\; -;.'\ mo line BD. Knowing that P must have a 750-N component perpendicular to
^'\^-^lf-. member ABC, determine (a) the magnitude of the force P, (b) its component
^%K '*S$t- parallel to ABC.
50&f
SOLUTION
60-fC^3&°
li'OH -J \jA^P
ifl) 750N = />sin20°
F = 2193N /> = 2190N <
(b) PABC~ ^ cos 20°
- (2193 N) cos 20° PABC = 2060 N ^
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28
PROBLEM 2.27
The guy wire BD exerts on the telephone pole AC a force P directed along
BD. Knowing that P must have a 120-N component perpendicular to the
pole AC, determine (a) the magnitude of the force P, (b) its componentalong WmAC.
3H"
SOLUTION
Ar
Jg» tzo N
\ * j
38° \ i
P \ >
(a)
sin 38°
120 Nsin 38°
= 194.91 N or P= 194.9 N A
(t>) P, _ ^tan 38°
120N
tan 3 8°
= 153.59 N or ^ = 153.6 N A
PROPRIETARY MATERIAL. O 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,
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29
A3« PROBLEM 2.28
t.Jj The guy wire BD exerts on the telephone pole AC a force P directed
B U along BD. Knowing that P has a 180-N component along line AC,
ij\ determine (a) the magnitude of the force P, (b) its component in a
m \ direction perpendicular to AC.
jr 38°\is \pjj '%
ill \
C;
# , _Jt„
SOLUTION
A r4
£ H mma^tp,.
I80N
C
(«)cos38°
180 Ncos 38°
= 228.4 N P = 228N <
(*) Px z=Py tm3$°
- (1.80 N) tan 38°
= 140.63 N Px = 140.6 N ^
PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. All rtghls reserved. Wo />«rt o///«s Manual may be displayed,
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30
PROBLEM 2.29
Member CB of the vise shown exerts on block B a force Pdirected along line CB. Knowing that P must have a 1200-N
horizontal component, determine (a) the magnitude of the
force P, (b) its vertical component.
SOLUTION
We note:
CB exerts force P on B along CB, and the horizontal component ifPis Px
-= .1200N:
Then
Px = .Psin55
1
°/«s
P-._ Px
i
(
>^ r
1200 Nsin 55°
1200N
sin 55°
= 1464.9 N P = 1465N 4
(*) = /;,tan55
__Px
tan 55°
1200N
tan 55°
= 840.2 N Pv =840nJ <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AM rights reserved. No part of this Manual may be displayed,
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31
PROBLEM 2.30
Cable AC exerts on beam AB a force P directed along line AC. Knowing thai Pmust have a 350-lb vertical component, determine (a) the magnitude of the
force P, (6) its horizontal component.
SOLUTION
?y
& A?
A\f/ 1
fa ^X
ifl) P— P>
cos 55°
350 lb
cos 55°
-610.2 1b P = 610lb M
ib) Px = /'sin 55°
= (610.2 lb) sin 55°
= 499.8 lb P, =500 lb <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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M
I!
2S in
96 in.
•S4in.-
O51 lh
- 48 in.
PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.22.
PROBLEM 2.22 Determine the x and y components of each of
the forces shown.80 in.
90 in.
SOLUTION
Components of the forces were determined in Problem 2.22:
Force x Comp. (lb) y Comp. (lb)
291b
50 lb
51 lb
+21.0
-14.00
+24.0
+20.0
+48.0
-45.0
7?V =+3U) #,,=+23.0
tan a
= (3 1.0 lb)i + (23.0 lb)|
^ 23.0
31.0
a = 36.573°
23.0 lbR
sin (36.573°)
38.601 lb
f? *2B,oJ
/'~A
R
R^ - 31.0 I
R = 38.6 lb ^C 36.6£
PROPRIETARY MATERIAL. O 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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33
PROBLEM 2.32
Determine the resultant of the three forces of Problem 2.24.
PROBLEM 2.24 Determine the x and y components of each of the
forces shown.
SOLUTION
Components of the forces were determined in Problem 2.24:
Force x Comp. (N) y Comp. (N)
SON
120 N
150 N
+61.3
+41.0
-122.9
+51.4
+ 112.8
+86.0
Rx = -20.6 Rv= +250.2
-(-20.6 N)i+ (250.2 N)j R
tan a
tan a •
5l
250.2 N20.6 N
tana = 12.1456
a = 85.293°
R250.2 N
sin 85.293°
K =ZS0.2J
R =-zo.fr^
R = 251N ib. 85.3° <
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34
1PROBLEM 2.33
Determine the resultant of the three forces ofProblem 2.23.
PROBLEM 2.23 Determine the x and y components of each ol
the forces shown.
SOLUTION
Force x Comp. (lb) y Comp. (lb)
40 1b
501b
601b
+20.00
-38.30
+54.38
-34.64
-32.14
+25.36
Rx = +36.08 Ry= -41.42
R
tana
tan a •
tana =
a-
W + Xyl
(+36.08 lb)i + (~4 1.42 lb)
j
41.421b
36.08 lb
1.14800
48.942°
41.421b
sin 48.942°
Rv =. 30-O&0
R^-mkhz)--*R
R = 54.9 lb X 48.9° <4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it without permission.
35
Dimensions
in mm
-.800-
900
600
PROBLEM 2,34
Determine the resultant of the three forces ofProblem 2.21
.
PROBLEM 2.21 Determine the x and y components of each of
the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.21
:
Force x Comp. (N) y Comp. (N)
8001b
424 1b
408 lb
+640
-224
+192
+480
-360
-360
Rx = +608 RY= -240
R^Rxi+ RJ
= (608ib)i + (~240lb)j
/?,
tan a
_240~608
or = 21.541°
240 N
p.* - (oO$> L
Rsin(21.541°)
= 653.65 N
•>£o<
£*« Z40J
R
R-654N ^21.5° <
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36
200 N
150 N
100 N
PROBLEM 2.35
Knowing that a- 35°, determine the resultant of the three forces
shown.
SOLUTION
100-N Force:
150-N Force:
200-N Force:
£„ - -I6.5*Z2 j
Fx = +(1 00 N) cos 35° = +8 1 .9 1 5 NFy= -(100 N)sin35° - -57.358 N
Fx = +(150 N) cos 65° = +63.393 N
Fy= -(150 N)sin 65° = -135.946 N
Fx = -(200 N) cos35° = -163.830 NFv= -(200 N)sin35° = -114.71 5 N
•5o& ~oz I
Force x Comp. (N) y Comp. (N)
100N
150N
200 N
+8.1.915
+63.393
-163.830
-57.358
-135.946
-114.715
Rx =-18.522 ^ -308.02
tana?
R=*,i + /g= (»1 8.522 N)i + (-308.02 N)j
5l
308.02
18.522
a = 86,559°
A'
308.02 Nsin 86.559
R = 309 N ZF 86.6° <4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, hie. Al! rights reserved. No part of this Manual may be displayedreproduced or distributed in anyJam or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedbyMcGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
37
800 nun
840 mm
% k = 1160 mm
500 N
13
B
4°
12
PROBLEM 2.36
Knowing that the tension in cable BC is 725 N, determine the
resultant of the three forces exerted at Point B ofbeam A.B.
imn
SOLUTION
Cable BC Force: Fx =-
F,=<
500-N Force: Fv=-
Fy=-
780-N Force: K =
F,=
and Rx -
Ry=
R-r-
Further: tan a =
a =
Thus:
-(725 N)840
1.160
-525 N
(725N)-^- = 500N1160
-(500N)- = -300N
-(500N)- = -400N
(780N)— = 720 N
-(780 N)— = -300 N
1FS -105 N-200 N
7(-105N)2 +(-200N)2
225.89 N
200
105
tan"
62.3<
200
105
^= -(£oo*}j
R = 226N ^62.3° <
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38
120 11
a*\fiO
PROBLEM 2.37
Knowing that a~ 40°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force: F* = (601b)cos20° = 56.381b
?y = (601b)sin20° = 20.521b
80-lb Force: ^v = (80lb)cos60° = 40.001b %* CM.6t"b)j <* £
120-Ib Force:
Fy= (801b)sin60° = 69.281b
= (120 lb) cos 30° = 103 .92 lb
f j—-_^^ir^i=—^jj? • Cioo.iollJ^
Fy
= -(1 20 lb) sin 30° = -60.00 lb
and K = XFX = 200.30 lb
R
= £FV= 29.80 lb
= 7(200.30 lb)2 + (29.80 lb)
2
= 202.50 lb
Further: tana -
29.80
200.30
a-, _, 29.80
= tan200.30
= 8.46° R = 203 lb ^ 8.46° ^
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39
! 2.0 ib
SO !b
A,
PROBLEM 2.38
Knowing that a— 75°, determine the resultant of the three forces
shown.60 lb
a
J.20°
SOLUTION
60-lb Force:
80-lb Force:
120-lb Force:
Then
and
Fx = (60 lb) cos 20° = 56.38 lb
Fv= (60 lb) sin 20° = 20.52 lb
Fx = (80 lb) cos 95 °= -6.97 lb
F =(801b)sin95° = 79.701b
Fv=(1201b)cos5°= 119.541b Rv r()io,<,8-»lO^
R,
F,*:(120Ib)sin5° = IO 46 lb
*x =--ZFX =168.95 lb
V= SFy=110.68 lb
./? == 7(168.95 lb)2 + (1
-201.981b
10.68 lb)2
tan«
=
1 10.68
168.95
tan« == 0.655
a = 33.23°
H*- (\ue.fiS\bNjc
R = 202 1b ^33.2° A
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40
200 N
iollX
100 N
PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value ofa if the resultant of the three forces shown is to be vertical, (b) the
corresponding magnitude ofthe resultant.
SOLUTION
RX =ZFX= (lOON)cosa + (150N)cos(tf + 30°)-(200N)cosar
Rx - -(1 00 N) cosa+ (150 N) cos(a + 30°)
R,=XF,
= -00ON)sina-(15ON)sin(«+ 30°)-(2O0N)sina
Ry= -(300 N)sina -(ISO N)sin (a+ 30°)
(a) For R to be vertical, we must have Rx - 0. We make /?v- in Eq. (I ):
-100cos a + 150cos(a + 30°) =
-100cos a+ 1 50 (cos a cos 30° -sin a sin 30°) =
29.904cosflr = 75sinor
(1)
(2)
tan a29.904
(b) Substituting for a in Eq. (2):
R.
75
= 0.3988
ar= 21.74°
-300 sin 2 1. .74°- 1 50 sin 5 1 .74c
-228.9 N
R = |» 1 = 228.9 N
ar = 21.7° ^
7? = 229N ^
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. M> par/ o/rto Manual may be displayed,
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41
800 mm
840 nun
L = 1 160 mm
500 N
'so\
PROBLEM 2.40
For the beam of Problem 2.36, determine (a) the required
tension in cable BC if the resultant of the three forces exerted at
Point B is to be vertical, (b) the corresponding magnitude of the
resultant.
SOLUTION
&:•= SF840
f'2 3
1160liC
13 5
V=-—7V+420N29
u'(1)
V= ZF =80° TBC -—(780 N) --(500 N)
y 1160*c
13 5
Ry
=
20=—V-700N29
Bt (2)
(a) For R to be vertical, we must haveRx—
Set/?, =0 inEq.(l) ~rRr +420N=029
BC 7},C = 580N ^
(b) Substituting for TBC in Eq. (2):
20ff =—(580N)-700N} 29
tfv=-300N
/J = |/fv |
= 300N /? = 300 IS! A
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42
^65°
35°.
PROBLEM 2.41
Determine (a) the required tension in cable AC, knowing that the resultant of
the three forces exerted at Point C of boom BC must be directed along BC,
(b) the corresponding magnitude of the resultant.
m ib
SOLUTION
r^
///is
r _J ISlbSo ! b
X
Using the x andy axes shown:
Rs= XFX = TAC sin 10° + (50 lb) cos35°+ (75 lb) cos 60°
= TAC sin 10° + 78.46 lb (0
Ry= E/?
y= (50 lb)sin35° + (75 lb)sin 60° ~~ TAC cos 1
0°
^=93 .63 lb-7^ cos 10° (2)
(a) SetRy=0 in Eq, (2):
93.63 lb -TAC cos 1.0° =
7^c =95.07 lb TAC = 95.1 lb 4
(b) Substituting for TAC in Eq. ( 1):
Rx =(95.07 lb) sin 10° + 78.46 lb
= 94.97 lb
R = Rx R --= 95.0 lb ^
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43
SO lb
120 lh
m n>
PROBLEM 2.42
For the block of Problems 2.37 and 2.38, determine (a) the
required value of a if the resultant of the three forces shown is to
be parallel to the incline, (b) the corresponding magnitude of the
resultant.
\ 20°
SOLUTION
80b
IZ0&o(
Select the x axis to be along a a.
Then
and
Rx=ZFX =(60 lb) + (80 lb) cos or + (120 lb) sina
Ry= ZF
y= (80 lb) sina- (120 lb) cosa
(I)
(2)
{a) Set Ry= in Eq. (2).
(80 lb) sin a - (1 20 lb) cos a =
Dividing each term by cosa gives:
(80 lb) tan a = 120 lb
120 lbtana :
801b
ar = 56.310°
(b) Substituting for a in Eq. (1) gives:
Rx= 60 lb + (80 lb) cos 56.3 1 ° + (120 lb)sin 56.3 1
° = 204.22 lb
a = 56.3° ^
/? = 204 lb ^
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44
J40°"Zi
L-ll^^J?
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Knowing that a = 20°, determine the tension (a) in cable AC,(h) in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
TAC"lac
4o° >\ X^=zo°
^N
6o°r>
IH^
i— (2oo^(H,6l m(i^
\<^Z N
Law of sines:^c _ 5V _1962N
sin 70° sin 50° sin 60°
(«)1 96? N
= '^ sin 70° = 2 1 28.9 Nsin 60°
7^;=2.13kN «
(6)1962N
7- =iyo
sin 50° = 1 735.49 Nsin 60°
TBC =; 1.735 kN ^
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45
APROBLEM2.44
m Two cables are tied together at C and are loaded as shown. Determine the
JK-tension (a) in cable AC, (b) in cable BC.
"* 50° ^%. SIX) N
-f
SOLUTION
Free-Body Diagram Force Triangle
Tacl V 50°
5<90aJ
•5£0aj
N/ Xe><i
fa*
TSc /
Tac _Law of sines:
sin6()0
-TBC _ 500 N
sin 40° sin 80°
(«) TAC~-=
50° Nsin 60° ~ 439.69 N TAC - 440 N <
sin 80°
(b) TBC =:J^l s jn 40°== 326.35 N V = 326N <sin 80°
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46
AW i
PROBLEM 2.45
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and a~ 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
Law of sines:?ac TBC 500 N
sin 35° sin 75° sin70c
(a)
(b)
T 500 N . __<TAC — sin 35
T -—^—sin75 c
iLsin 70°
ry(C =305N
TBC = 5.1.4 N -4
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47
PROBLEM 2.46
Two cables are tied together at C and are loaded as shown. Determine the tension
(a) in cable ,4C, (/;) in cable BC.
111200 kg
SOLUTION
Free-Body Diagram
IfflH
W'— mg
= (200kg)(9.81m/s2)
-1962N
Law of sines:
(«)
1962 Nsin 15° sin 105° sin 60°
(1962 N) sin 15°
sin 60°
(1.962 N)sin 105°
TAc
Tbcsin 60°
7^=586>H
rBC =2190N <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed',
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48
PROBLEM 2.47
Knowing that a = 20°, determine the tension (a) in cable AC,(b) in rope BC.
'I"n200 lb B<J
SOLUTION
Free-Body Diagram Force Triangle
5T<l
^ = lO'
-7; c,
8CIZOOlb
(b<-
lioolb
Law of sines:
(«)
(6)
^c ^ ^?c ^ 1200 lb
sin 110° sin 5° sin 65°
T1200 1b .
ACsin 65 f
sinll0 c
T,1200 1b .
«csin 65°
sin 5(
7^=1244 lb A
TBC -115.41b ^
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displaved,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
49
PROBLEM 2.48
Knowing that a = 55° and that boom AC exerts on pin C a force directed
along line AC, determine (a) the magnitude of that force, (b) the tension in
cable BC.
300 lb
SOLUTION
Free-Body Diagram
Law of sines:
300 lb
Pac
Pac
Tbc
T,BC 300 1b
sin 35° sin 50° sin 95°
3001b .
sin 95
300 1b .
sin 35°
sin 95c
sin 50°
7^=172.7 lb <
7^ =231 lb <
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50
40°\Q
PROBLEM 2.49
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium andthat P= 500ib and = 650 lb, determine the magnitudes ofthe forces exerted on the rods A and B.
Free-Body Diagram
if
SOLUTION
Resolving the forces into x- and ^-directions:
R = P +Q + F4+F
B=0
Substituting components: R = -(500 lb)j + [(650 lb)cos50°]i
-[(650 lb) sin 50°]j
+FBi- (FA cos 50°)i -f (FA sin 50°)j =
In thejMlirection (one unknown force)
-500 lb - (650 lb) sin 50°+ FA sin 50° =
500 lb+ (650 lb) sin 50°Thus,
In thex-direction:
Thus,
- - -* (*5o ib
Sooib
fAsin 50°
= 1302.70 lb
(650 lb) cos 50°+FB ~FA cos 50° =
FB -FA cos 50°- (650 lb) cos 50°
= (1302.70 lb) cos 50° -(650 lb)cos50c
= 419.55 lb
F, =1303 lb <
Ffi =4201b 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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51
PROBLEM 2.50
.^\,
w\ ^\
Two forces P and Q are applied as shown to an aircraft,
connection. Knowing that the connection is in equilibrium
and that the magnitudes of the forces exerted on rods A and Bare FA = 750 lb and FB = 400 lb, determine the magnitudes of
P and Q.
SOLUTION
Resolving the forces into x- and ^-directions:
R = P +Q + F_4+FB =0
Substituting components: R = ~P\ +Q cos 50°i -Q sin 50°
j
-[(750 1b)cos50°]i
+ [(750 lb)sin 50°]j + (400 lb)i
In the x-direction (one unknown force)
Q cos 50°- [(750 lb) cos 50°]+ 400 lb =
(750 lb) cos 50° -400 lb
cos 50°
= 127.710 lb
-P~Q sin 50° + (750 lb) sin 50° =
P = -Q sin 50°+ (750 lb) sin 50°
= -(127.710 lb)sin 50° + (750 lb)sin 50c
= 476.70 lb
Tnthej-direction:
Free-Body Diagram
ISO \b
P = 477 1b; £ = 127.7 lb A
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52
PROBLEM 2.51
A welded connection is in equilibrium under the action of the
four forces shown. Knowing that FA - 8 kN and FB = 16 kN,determine the magnitudes of the other two forces.
SOLUTION
1?ree-Body Diagram of Connection
J
^,5
£ *
5
IFX -0: 1fb -*-j*-
With = 8kN
= 16kN
Fc = -(16 kN)- 1(8 kN) ^c = 6.40kN <
*Fy= 0: -FD + 1 F -If
5 " 5*" =
With FA and FB as above: FD = -(16kN)- |(8 kN)*i>
= 4.80kN -4
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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53
PROBLEM 2.52
A welded connection is in equilibrium under the action of the four
forces shown. Knowing that FA - 5 kN and FD = 6 kN, determine
the magnitudes of the other two forces.
\D
SOLUTION
Free-Body Diagram of Connection
n •5
5tt
5fo
*Fy
3 3= 0: ~FD—FA +-FB ~-
D g A5
B=
or FB - p +—P
With Fa = 5kN, FD =8kN
Fs_5~3 6kN + -(5kN) F
7i=15.00kN ^
XFX
4 4= 0: -FC+1F.-FA
.=
Fc •±(F.-V
=-(15kN-5kN) Fc =8.00kN A
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54
PROBLEM 2.53
Two cables tied together at C are loaded as shown. Knowing that
Q — 60 lb, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
\
VIca
'P
TC>^/^V9.
2^=0: 7^ -gcos30°=:0
With g = 601b
(a) TCA =(60\b)(0M6) rCVf= 52.0 lb A
(6) XFX =0: P~-7^-0 sin 30° =
With /> = 751b
7c»=751b- (60 lb)(0.50) or 7^ =45 .01b -«
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55
PROBLEM 2.54
Two cables tied together at C are loaded as shown. Determine the
range of values of Q for which the tension will not exceed 60 lb in
either cable.
SOLUTION
Free-Body Diagram XFX = 0: -7^ - cos 60°+ 75 lb =
V5c \
f\l5lb IF, = 0:
TBC =75 lb- Q cos 60°
r,c -0sin6O° = O
(1)
\ \3&°TAC ^Qsm60° (2)
Requirement TAC < 60 lb:
\ ' n From Eq. (2): gsin60°<60]b
Q< 69.3 lb
Requirement TBC < 60 lb:
From Eq, (1): 75 1b- gsin60°<601b
£>>30.0Ib 30.01b<gfs 69.3 lb ^
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56
PROBLEM 2.55
A sailor is being rescued using a boatswain's chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that # = 30° and /?= 10° and that the combined
weight of the boatswain's chair and the sailor is 900 N,
determine the tension {a) in the support cable ACB, (b) in the
traction cable CD.
SOLUTION
Free-Body Diagram
t & Y
\ I^ca3o«?^>y^^
- '
V.(r>o« *
T ^oo M
±~ £FV= ; TACB cos 1 0° - TACB cos 30°-TCD cos 30° =
7^=0.1371587^(1)
+t ^,=0: 7^ sin 10° +7^ sin 30° + 7a> sin 30°--900 =
0.67365TACB + Q.5TCD =900 (2)
(a) Substitute (1) into (2): 0.67365.7^ + 0.5(0. 1371 58 7*,JC/i )
= 900
TACB ~ 1212.56 N 'ACB = 1213N A
(b) From ( 1 ): TCD =0.1371 58 (1 2
1
2.56 N) Ta.r~= 166.3 N A
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57
PROBLEM 2.56
A sailor is being rescued using a boatswain's chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that a = 25° and /? = 15° and that the tension in
cable CD is 80 N, determine (a) the combined weight of the
boatswain's chair and the sailor, (b) in tension in the support
cable ACB.
SOLUTION
Free-Body Diagram
Y
£<* g°M T„ACB2S°fX>
f >
W '
f
-±.SFv-0: r,c/J cosl5° - TACB cos 25° - (80 N) cos 25° -0
TACB =1216.15 N
+J ZFy =0: (1216.15 N)sin 15° + (1216. 15 N)sin 25°
+(80 N)sin 25°-W =
FT = 862.54 N
(«) ^ = 863 N ^
(/') 7^= 1.216 N <
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58
PROBLEM 2.57
For the cables of Problem 2.45, it is known that the maximumallowable tension is 600 N in cable AC and 750 N in cable BC.Determine {a) the maximum force P that can be applied at C, (b) the
corresponding value of a.
SOLUTION
Free-Body Diagram Force Triangle
7 c*6eoN
T&.1SON
(a) Law of cosines
(b) Law of sines
P2 = (600)2+ (750)
2 - 2(600)(750)cos(25° + 45°)
P = 784,02 N
sin/? __ sin (25°+ 45°)
600 N 784.02 N
p * 46.0°
p = 784N <
a^ 46.0°+ 25° = 71.0° <4
PROPRIETARY MATERIAL. © 20)0 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual way be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
59
GO D
PROBLEM 2.58
For the situation described in Figure P2.47, determine (a) the
value of a for which the tension in rope BC is as small as
possible, (b) the corresponding value of the tension.
PROBLEM 2.47 Knowing that a = 20°, determine the tension
(a) in cable AC, (b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Tc
To be smallest, 7^c must be perpendicular to the direction of TAC.
(a) Thus, a = 5°
(6) r/fC
=(1200 lb) sin 5°
IZOO lb
a = 5.00° <dL <
TBC = 104.6 lb <«
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60
PROBLEM 2.59
For the structure and loading of Problem 2.48, determine (a) the value of a for
which the tension in cable BC is as small as possible, (/?) the corresponding
value ofthe tension.
300 lb
SOLUTION
TliC must be perpendicular to FAC to be as small as possible.
Force Triangle is a right triangle
lit, ^Free-Body Diagram: C
30o(t>
k
3oolt>
To be a minimum, r0C must be perpendicular to FAC .
(a) We observe: a = 90° -30°
(b) rBC =(3001b)sin50°
or JBC =229.81 lb
a = 60.0° A
7^.= 230 lb <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
61
2.1 m 2.1 in
m*\ PROBLEM 2.60
fjft~ Knowing that portions AC and BC of cable ACB must be equal,
determine the shortest length of cable that can be used to support the
load shown if the tension in the cable is not to exceed 870 N.
* 1200 N
SOLUTION
Free-Body Diagram: C
(ForF = 725N)
\1QO N
670« sy 7[
B
x!
!
630 U I
- 2Jm *]
+|SFv-0: 27; -1200 N =
V= 600N
7;2+:r,
2-= T2
r2 +(600N)2== (870 N)
2
r,== 630N
. ., . , BCy similar triangles:
870 N2.1m
630 N
BC:= 2.90 m
L = 2(BC)
= 5.80 m L- 5,80m <
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62
PROBLEM 2.61
Two cables tied together at C are loaded as shown. Knowing thatthe maximum allowable tension in each cable is 800 N, determine(a) the magnitude of the largest force P that can be applied at C,(b) the corresponding value of a.
SOLUTION
Free-Body Diagram: C
"(^z &OON
$oc
Force triangle is isosceles with
2/? = 180° -85°
/J = 47.5°
^) P = 2(800 N)cos 47.5° = 1 08 1
N
Since P> 0, the solution is correct.
(b) tf = 180°- 50° -47.5° = 82.5°
P = 1081N <
a = 82.5° <
ZoZToZ^lifiL-
nVm) T,
;
e McGraw-HiU crpafs' lnc- An righ,s rescrved-
'v° '-' *** *"* «* *•-mnZ, f , , , , , /^ /W7M °'' ^ my
' ""*""'' """""" "?e "*»' "**"" /*™«fc*fo» <>/^ pwWisfer. or used beyond the limited
63
PROBLEM 2.62
Two cables tied together at C are loaded as shown. Knowing that the
maximum allowable tension is 1200 N in cable AC and 600 N in
cable BC, determine (a) the magnitude of the largest force P that can
be applied at C, (b) the corresponding value ofa
SOLUTION
Free-Body Diagram Force Triangle
Te,C =r c^>oN
tZOON
P2 = (1200 N)2 + (600 N)
2 - 2(1200 N)(600 N)cos 85c
P = 1 294 N
Since P> 1200 N, the solution is correct.
(a) Law of cosines:
P = 1294N <
(b) Law of sines:
sin/? sin 85°
1200N 1294 N/? = 67.5°
a = 180°"50°-67.5 < a -62.5° A
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64
/
"WX
/M:l
50 lb
2(1 in.
PROBLEM 2.63
Collar^ is connected as shown to a 50-lb load and can slide ona frictionless horizontal rod. Determine the magnitude of theforce P required to maintain the equilibrium of the collar when(a) j = 4.5 in., (/>) x = 15 in.
SOLUTION
(a) Free Body: Collar A
-«(f_ -f~A"zy
Force Triangle
50 1bP
4.5 20.5P = 10.98 lb <
N
(b) Free Body: Collar A
So 0> 5b lb
Force Triangle
P = 30.0 lb <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of Urn Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used bevond the limiteddistribution to teachers and educatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student mine this Manualyou are using it withoutpermission.
65
PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the
collar is in equilibrium when P = 48 lb.
SOLUTION
Free Body: Collar A
50 &
hSfo
Force Triangle
U \ ^'b
Similar Triangles
7V2=(50)
2-(48)
2 =:196
N = 14.00 lb
x 48 lb
20 in. 141b|*
—
11— X >|
x = 68.6 in. -^
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66
160 kg
PROBLEM 2.65
A 160-kg load is supported by the rope-and-pulley arrangement shown.Knowing that ft = 20°, determine the magnitude and direction of the
force P that must be exerted on the free end of the rope to maintain
equilibrium. {Hint: The tension in the rope is the same on each side of asimple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley^
W - ov|
- sS^^O N
±* £Ft= 0: IP sin 20° - Pcosa =
and cosa= 0.8452 or a = ±46.840°
a =, +46.840
For +t ZFy = 0: 2Pcos 20° + Fsin 46.840° - 1 569.60 N =
or P = 602 N^ 46.8° <For a = -46.840
+t ZFy=Q: 2Pcos20° + Psm(-46.840°)~l569.60N =
or P = 1365 N ^46.8° <
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
67
\ V
PROBLEM 2.66
A 160-kg load is supported by the rope-and-pulley arrangement shown.
Knowing that a = 40°, determine (a) the angle /?, (b) the magnitude of
the force P that must be exerted on the free end of the rope to maintain
equilibrium. (See the hint for Problem 2.65.)
> An
160
!
SOLUTION
Free-Body Diagram: Pulley .d
y?
?
o4*H0*
GO £/<;. = : 2Fsin sin /?- Pcos 40° =
y-
sm5 =—cos40c
' 2
,0 = 22.52°
/? = 22.5° <
(b) IFy= 0: Psin 40° + 2Pcos 22.52° - 1 569.60 N =
W~ 'TO3
p = 630N -4
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68
PROBLEM 2.67
A 600-lb crate is supported by several rope-and-
pulley arrangements as shown. Determine for each
arrangement the tension in the rope. (See the hint
for Problem 2.65.)
SOLUTION
Free-Body Diagram of Pulley
(«) T "X
<oOO\b
(b) T T
r kooib
+tsF=(): 2f- (6001b) =
7 = -(600 lb)
+t SF =0: 27" -(600 lb) =
T = -(600 lb)
7 = 300 lb <
T^ 300 lb <
(c)
I'COOlb
(rf) T T X
Goo lb
(c)
GOO )b
+txF„=0: 37 - (600 lb) =
7 = -(600 lb)
+1zF=0: 37 - (600 lb) =
7 = -(600 lb)
+t SF =0: 47 - (600 lb) =
7 = -(600 lb)
T = 200 lb <
7 = 200 lb <
T = 150.0 lb <
PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
69
PROBLEM 2,68
Solve Parts b and d ofProblem 2.67, assuming that
the free end ofthe rope is attached to the crate.
PROBLEM 2.67 A 600-lb crate is supported by
several rope-and-pulley arrangements as shown.
Determine for each arrangement the tension in the
rope. (See the hint for Problem 2.65.)
SOLUTION
Free-Body Diagram of Pulley and Crate
(b) T r t- +1 SF =0: 37* - (600 lb) =
r = -(6001b)
r = 200 lb <
QsOOib
(d) r tt t-
i i i
+\ZFy=0: 47
,
-(600ib) =
T = -(600 lb)
T = 150.0 lb <4
GOO lb
PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, inc. All rights reserved. No pari of this Manual may be displayed,
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70
m"25°
PROBLEM 2.69
A load Q is applied to the pulley C, which can roll on the
cable ACB. The pulley is held in the position shown by a
second cable CAD, which passes over the pulley A andsupports a load P. Knowing that P = 750N, determine
{a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
T»AC6
35
(a) -+*£/<; -0: r,,0i (cos25 ~cos55 )~(750N)cos55° =
Hence: TACB = 1292.88 N
7^CT = 1293N <
(b) +1 ZFy = 0: 7^CB (sin25o+ sin55o) + (750N)sin55°-e =
(1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° - Q =
= 2219.8 N g = 2220N <or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afa /*«/ of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
71
IE MJT
D pr/SSQ^5
Q
PROBLEM 2.70
An 1800-N load Q is applied to the pulley C, which can roll
on the cable ACB. The pulley is held in the position shown
by a second cable CAD, which passes over the pulley A and
supports a load P. Determine (a) the tension in cable ACB,
(/>) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
~±~ LPX= 0: TACIi {cos 25° - cos55°) - Pcos55° =
or P = 0.580107^ (1)
+| ZFy=0: r,
(C/i(sin25
o + sm55o) + .Psin55°-1800 N =
sCtS or 1 .24177^ + 0.81 915^ = 1800 N (2)
(a) Substitute Equation (1) into Equation (2):
1 .24
1
77TACB +0.8191 5(0.580
1
QTACB ) = 1 800 N
Hence: TACB =1048.37 N
7^= 1048 N <
Ta<*
iboom
(b) Using (1), P = 0.58010(1048.37 N) = 608.16 N
P = 608N <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed
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72
900 N
7.10 N
PROBLEM 2.71
Determine (a) the x, y, and 2 components of the 750-N force, (b) the
angles 0„ V> and #z that the force forms with the coordinate axes.
SOLUTION
Fh
Fh
= fsin35°
= (750 N) sin 35°
- 430.2 Ni ^
(a)
Fx =Fh cos 25°
= (430.2 N) cos 25°
F,= Fcos35°
= (750N)cos35°
Fz = /}(sin25°
= (430.2 N)sh
Fx = +390N, Fy = +614 N, F2
'-= +181.8N
(t>) cos
cos
cos 0,
_ Fx _ +390 NF 750 N
_Fy _+614NF 750 N
_FZ _ +181,8 NF 750 N
v=58.7° A
6>v=35.0° <
6, = 76.0° <
PROPRIETARY MATERIAL. © 20] The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
73
900 N
750 N
PROBLEM 2.72
Determine (a) the x, y, and z components of the 900-N force, (/>) the
angles 9X, &y, and Z that the force forms with the coordinate axes.
SOLUTION
{a)
Fx = Fh sin 20°
= (380.4 N)sin 20c
Fx =-130.1. N,
/^=Fcos65°
(900N)cos65<
FA = 380.4 N
F^F sin 65*
- (900 N) sin 65°
F. =+816 N,
fZ*q00N *\
FZ =FA cos 20°
= (380.4 N) cos 20°
F.-+357N
(/') cos 6?
cos 6?
cos 0,
Fx _ -130.1 N
F "900
N
Fy _ +816M.,
T ~900
N
Fz _ +357 NF ~
900 N
0„-98.3° <
d =25.0° ^
0, =: 66.6° ^
PROPRIETARY MATERIAL. © 2050 The McGraw-Hill Companies, Inc. All rights reserved. No /x/rf o/r/rw Manual may be displayed,
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74
PROBLEM 2.73
A horizontal circular plate is suspended as shown from three wires that
are attached to a support at D and form 30° angles with the vertical.
Knowing that the x component of the force exerted by wire AD on the
plate is 1 10.3 N, determine (a) the tension, in wire AD, (b) the angles
X3 6y, and ft that the force exerted atA forms with the coordinate axes.
SOLUTION
(«)
(b)
F
cos ft-
cosV
-
F_ =
Fx = Fsin 30°sin 50° = 1 1 0.3 N (Given)
1I0.3N
sin 30° sin 50c
K. 110.3 N
287.97 N
F 287.97 N
F cos 30° = 249.39
0.38303
K 249.39 N0.86603
cos tf, =
F 287.97 N
~F sin 30° cos 50°
-(287.97 N) sin 30°cos 50c
-92.552 N
F. -92.552 NF 287.97 N
-0.32139
F = 288N A
ft= 67.5° <
9 =30.0° ^
ft =108.7° •*
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75
o%.
PROBLEM 2.74
A horizontal circular plate is suspended as shown from three wires that
are attached to a support at D and form 30° angles with the vertical.
Knowing that the z component of the force exerted by wire BD on the
plate is -32.14 N, determine (a) the tension in wire BD, (b) the angles
X, 6yi and Z that the force exerted at B forms with the coordinate axes.
SOLUTION
(«) ^ =~-F sin 30°sin40° = 32.14 N (Given)
F-32.14
00.0 N F = 100.0 N ^sin 30° sin 40°
(b) Fx '-= -Fsin30°cos40c
= -(100.0 N)sin 30
= -38.302 N
"cos 40°
cos 9X
-_ Fx __ 38.302 N __
F 100.0 N-0.38302
V=112.5° -4
py-= Fcos30° = 86,603 N
COSOy '
_ Fy _ 86.603 N _
F 100 N0.86603 ev
= 3o.o° <
F**= -32.14 N
cos#z
=_F2 _ -32.14 N_F 100 N
-0.32140 ez ^\m.i°<
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76
DX
it %1 \
L %
PROBLEM 2.75
A horizontal circular plate is suspended as shown from three
wires that are attached to a support at D and form 30° angles
with the vertical. Knowing that the tension in wire CD is 60 lb,
determine (a) the components of the force exerted by this wire
on the plate, (b) the angles X , V , and 6Z that the force forms
with the coordinate axes.
SOLUTION
(a) F* - ~(60 lb) sin 30°cos 60° = -1 5 lb Fx =-15.00 lb <
Fy
= (60 lb) cos 30° = 5 1.96 lb Fy= +52.0 lb <
F*-'= (60 lb) sin 30° sin 60° = 25.98 lb Fz=+26.0 lb <
(b) cos#v
=-5lF 60 1b
V^.104.5° <
cos 6 -
F=5L96lb
= 0.866601b
6V= 30.0° <
cos#, -
_F2~ F
=25 '981b
=0.433601b
6^=64.3° <
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77
DM.
PROBLEM 2.76
A horizontal circular plate is suspended as shown from three wires that
are attached to a support at D and form 30° angles with the vertical.
Knowing that the x component of the force exerted by wire CD on the
plate is -20.0 lh, determine (a) the tension in wire CD, (b) the angles
Xl Vi and Z that the force exerted at C forms with the coordinate axes.
SOLUTION
(a) Fx = -F sin 30° cos 60° - -20 lb (Given)
201bF =
sin 30° cos 60°801b
(/>)a F
x -201bcos r
——^ = = -0.25x F 80 lb
Fy= (80 lb) cos 30° = 69.282 lb
cos 0.,1 F 801b
Fz = (80 Ib)sin30°sin60° = 34.641 lb
a Fz 34.641
cos ft = —£- = -> = 0.43301
2 F 80
F = 80.0 lb ^
0, =104.5° ^
6?v= 30.0° ««
6> = 64.3° <
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78
PROBLEM 2.77
The end of the coaxial cable AE is attached to the pole AB, which is
strengthened by the guy wires AC and AD. Knowing that the tension
in wire AC is 120 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles 0„ y, and $z that the
force forms with the coordinate axes.
SOLUTION
(fl) K--(120 lb) cos 60°cos 20°
K--= 56.382 lb Fx = +56.4 lb <
K = -(120 lb) sin.60°
Fy= -103.923 lb Fy=-103.9 lb <
fz
-~= -(1201b)cos60°sin20°
Fz~-= -20.521 lb F
z=-20.5 lb «
(b) cos 6X
-
_FX _ 56.382 lb
F 1201b6»r= 62.0° A
cos Ov
s_ Fy _ -103.923 lb
F 120 lbBy=150.0° ^
cos 6Z_ Fz _ -20.52 lb
F 1201b&=99.8° «
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo /«»*/ o/"/Aw Mmwi/ /««>' te displayed,
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79
PROBLEM 2.78
The end of the coaxial cable AE is attached to the pole AB, which is
strengthened by the guy wires AC and AD. Knowing that the tension
in wire AD is 85 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles $n y, and 6Z that the
force forms with the coordinate axes.
SOLUTION
(fl) ^; = (85 1b)sin36°sin 48°
-37.1291b /;;=37..i lb <
Fv=-(85 1b)cos36°
= -68.766 lb Fv=~68.8Ib <
F, =(85 lb)sin36°cos 48°
= 33.431 lb F2=33.4 lb A
(b)rt
Fx 37.129 1bcos #v
- —^ -—1 F 85 lb
X=64.1° A
„ Fy -68.766 lb
cos = —— =' F 85 lb
0, =144.0° A
n F. 33.431. lbcos 6/_ = —^ =
* F 85 lb
6^=66.8° «
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> ptn-f £>///»* Manual may be displayed,
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80
PROBLEM 2.79
Determine the magnitude and direction of the force F = (320 N)i + (400 N)j - (250 N)k.
SOLUTION
F = JF:+F: + F?
F = 7(320 N)2 + (400 N)2 + (-250 N)2 F = 570N <
F ^20 Ncase ^tJL = ¥CiSL #.=55.8° <
.
x F 570 N
*»52i *„=45.4°«570 N '
250 Nft=116.0°«
570 N
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81
PROBLEM 2.80
Determine the magnitude and direction of the force F = (240 N)i - (270 N)j + (680 N)k.
SOLUTION
F = 7(240 N)2 + (-270 N)2 + (680 N) F = 770N <
cos*,-*-*™ ex=n.r<
x F 770N
1 F 770 N
z F 770
N
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82
PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles ox = 70.9° and
By = 144.9°. Knowin g that the z component of the force is -52.0 lb, determine (a) the angle ft, {b) the other
components and the magnitude of the force.
SOLUTION
(a) We have
(cos#Y)2+(cos eyf + (cosez )
2 = i => (cos eyf = \--(cos0
v )2
--(cos0z )2
Since Fz< we must have cos#2 <
Thus, taking the negative s
cos6Z
>quare root, from above, we have:
= 0.47282 6Z =118.2° <= -j\ - (cos 70.9 )2 - (cos 1 44.9°)
2
(b) Then:
F =F
' =52 '0lb
=109.978 lb
cos6>z 0.47282
and Fx = Fcos0x =(1.09.978 lb) cos 70.9° Fv=36.01b <
Fy= Fco$O
y=(109.978 lb) cos 144.9° F
v=-90.0 lb <
F = l 10.01b <
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S3
PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles 8Y = 55° and Z = 45°.
Knowing that the x component of the force is - 500 lb, determine (a) the angle X, (/?) the other components
and the magnitude of the force.
SOLUTION
{a) We have
(cos Xf + (cosyf + (cos Zf = .1 => (cos
yf = 1 - (cosVK-(cos0z )2
Since Fx< we must have cos<9
v<
Thus, tak ng the negative square root, from above, we have:
ox -114.4° 4cos6X = -yj\ - (cos 55)2 - (cos45)
2 = 0.41 353
(b) Then:
F _ Fx _ 5001b _ p0OI01b F-~= 1209Jb <cos0
v0.41353
and Fv= F cose
v= (1 209. 1 lb) cos 55° K = 694 lb A
Fz = F cos 6Z= (1 209. 3 lb) cos 45° K -8551b ««
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84
PROBLEM 2.83
A force F ofmagnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, Z ~ 1 51 .2°,
and Fy< 0, determine {a) the componentsFv and Fz, (b) the angles 6X and 6y.
SOLUTION
(a) F2 = Fcos^=(210N)cosl51.2°
= -184.024 N Fr=-184.0 N ^
Then: F2= F?+F$+F?
So:
Hence:
(2 1 ON)2
.
Fy
= (80 N)2 + (F,,)
2 + (1 84.024 N)2
= -^(2 1 N)2 - (80 N)2 - (184.024 N)
2
= -61.929 N Fy= -62.0 lb ^
(b) cos dx = ^= 80N=0.38095
F 210 N6>
v=67.6° *
cos 6 = ^= 6L929N=-0.29490
F 210 N6^=107.2 ^
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85
PROBLEM 2.84
A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that 6X = 32.5°, Fy ~ ~ 60 N,
and Fz > 0, determine (a) the components Fx and Fz,(h) the angles y and Z.
SOLUTION
(a) We have
Fx - F cos $x = (230 N) cos 32.5° Fv= -194.0 N ^
Then: Fx = 193.980 N
F2-.
So:
Hence:
(230 N)2
=
F2
--
- (1 93.980 N)2 + (-60 N)
2 + F*
/?= 108.0 N ^= +a/(230 N)2 - (193.980 N)2 - (-60 N)2
(b) Fz
--= 108.036 N
cos =/; = -6on = _ 026087F 230 N
6^, =105.1° ^
cosOz -
__F,=108.036 N
F 230 N^=62.0° «
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86
PROBLEM 2.85
A transmission tower is held by three guy wires anchored by bolts
at B, C, and D. If the tension in wire AB is 525 lb, determine the
components of the force exerted by the wire on the bolt at B.
SOLUTION
BA = (20 ft)i + (1 00 ft)j ~ (25 ft)k
BA = 7(20 ft)2 + (100 ft)
2 + (-25 ft)2
= 105 ft
F = FXW
BA
- 1||^[(20 ft)i + (1 00 ft)j - (25 ft)k]
F = (1 00.0 lb)i + (500 lb)j - (1 25.0 lb)k
Fx =+100.0 lb, Fy= +500 lb, F
z = -125.01b <
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87
PROBLEM 2.86
A transmission tower is held by three guy wires anchored by bolts
at B, C, and D. If the tension in wire AD is 3 1 5 lb, determine the
components ofthe force exerted by the wire on the bolt at D.
SOLUTION
DA = (20 ft)i+ (100 ft)j + (70 ft)k
DA = 7(20 ft)2 + (100 ft)
2 + (+70 ft)2
= 126 ft
F = FXDA
= F9±DA
=~|~[(20 ft)i + (1 00 ft)j + (74 ft)k]
F = (50 lb)i + (250 lb)j + (1 85 lb)k
Fv =+501b, F
y= +250 lb, Fz
= +185.0 lb <
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88
PROBLEM 2.87
A frame ABC is supported in part by cable DEE that passes
through a fricti.on.less ring at B. Knowing that the tension in the
cable is 385 N, determine the components of the force exerted by
the cable on the support at D.
480 mm
SOLUTION
DB = (480 mm)i -(510 mm)j + (320 mm)k
DB = 7(480 mm)2 + (510 mm2
) + (320 mm)2
= 770 mm
F
Dli
DBDB385 N
[(480 mm)i - (5 1 mm)j + (320 mm)k]
+ (160 N)k
/^.=+240N; /^V =-255N, Fz ^ +160.0 N <
770 mm(240 N)i - (255 N)j + (1 60 N)k
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89
PROBLEM 2.88
For the frame and cable of Problem 2.87, determine the components
of the force exerted by the cable on the support at E.
PROBLEM 2.87 A frame ABC is supported in part by cable DBEthat passes through a frictionless ring at B. Knowing that the tension
in the cable is 385 N, determine the components of the force exerted
by the cable on. the support at D.
SOLUTION
EB = (270 mm)i - (400 mm)j + (600 mm)k
EB ~ 7(270 mm)2+ (400 mm)
2+ (600 mm>
2
= 770 mmF = n.M
EB
=385N
[(270 mm)i - (400 mm)j + (600 mm)k]770 mm
F = (1 35 N)i - (200 N)j + (300 N)k
Fx = + 135.0 N, Fv= -200 N, F
z= +300N <
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90
IJ
360 nun— *PROBLEM 2.89
. / Knowing that the tension in cable AB is 1425 N, determine the920 mm components of the force exerted on the plate at B.
.1 n600 mm u
X/ 900 X <•'
1)111 /
£r
SOLUTION
BA = -(900 mm)i + (600 mm)j + (360 mm)k
BA = ^(900 mm)2 + (600 mm)2
4- (360 mm)2
= 1140 mm
T BA' **BA
1425 NX = _ —[-(900 mm)i + (600 mm)j + (360 mm)k]
1140 mm= -(1 1 25 N)i + (750 N)j + (450 N)k
(7^),=-11.25N, (3V)„=750N, {TBA )Z= 450N A
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91
PROBLEM 2.90
Knowing that the tension in cable ^C is 2130 N, determine the5)20 min
J}
components of the force exerted on the plate at C.
/ 9 ~"*^\
/ M . :~&$—
_
t „600 mm u
X'-'Sf
/ 900 mm X' C
SOLUTION
CA:= ~(900mm)i + (600 um)j--(920mm)k
CA--= 7(900 mm)2
= 1420 mm4(600 mm)2 + (920 mm)2
T -
CACA
900 mr ili + f -SOOmm'ii-fC)1420 mm~(1 350 N)i + (900 M)j - (1 380 N)k
(7^), =-1350 N, (.r^)y=900N, (TCA )z
~-mQH<
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92
PROBLEM 2.91
Find the magnitude and direction of the resultant of the two forces
shown knowing that P - 300 N and Q = 400 N.
SOLUTION
Q
R
R
cos 6>
cos 6„
cost/.
- (300 N)[-cos30°sin 1 5°i + sin 30°j+ cos 30° cos 1 5°k]
: - (67.243 N)i + (1 50 N)j + (250.95 N)k
= (400 N)[cos 50°cos 20°i + sin 50°j - cos 50°sin 20°k]
= (400 N)[0.60402i + 0.76604j - 0.2 1 985]
= (241 .61 N)i + (306.42 N)j -(87.939 N)k
P +Q= (1 74.367 N)i + (456.42 N) j + (1 63.0 1 1 N)k
7(174.367 N)2 + (456.42 N)2+(163.01 1 N)2
515.07 N
R
R
*LR
1 74.367 N515.07N
456.42 N_
515.07 N"
163.01 IN
515.07 N
0.33853
0.88613
0.3 1 648
# = 5.15 N A
6X = 70.2° <
y= 27.6° <
6 =71.5° <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No pari of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.
93
.'/ PROBLEM 2.92
1'
30"/ >/
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 400 N and Q = 300 N.
X
;>^5°7
SOLUTION
P =
Q =
R--
= (400N)[-eos30 sinl5 i + sin30 j + cos30 cosl5°kj
= -(89.678 N)i + (200 N)j+ (334.6.1 N)k
= (300 N)[cos50°cos20°i + sin 50°j - cos50°sin20°k]
= (1 8 1 .21 N)i + (229.8 1 N)j - (65.954 N)k
= P +Q= (91 .532 N)i + (429.8 1 N)j + (268.66 N)k
= 7(9 1 -532 N)24- (429.81 N)2 + (268.66 N)
2
= 515.07 N /? == 515N <
cos#r := ^= 9L532N=0.177708
R 51 5.07 N0.v
= 79.8° <
cos V
K. 429 8 1 N= y =^ZA5IIN
-0.83447R 5 15.07 N y
= 33.4° <
cos Oz -= ^ =m66N
=0.52.60R 515.07 N
ez= 58.6° <
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94
I D
60 in.
PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb in
cable >4C, determine the magnitude and direction ofthe resultant
of the forces exerted atA by the two cables.
SOLUTION
Then:
and
AB
AB
AC :
AC--
TV
(40in.)i-(45in.)j + (60in.)k
-J(40 in.)2+ (45 in.)
2 + (60 in.)2 = 85 in.
(1 00 i.n.)i - (45 in.)j + (60 in.)k
^(1 00 in.)2 + (45 in.)
2 + (60 in.)2 = 125 in.
TAI)KB -TA/i^ = (425\b)
AB
(40in.)i-(45in.)j + (60in.)k
85 in.
TAfi
= (200 lb)i - (225 lb)j + (300 lb)k
AC'AC
Vac
R
R
cos#.
cos tfy
cos #.
TAC XAC =TAC ~^(510\b)(100 in.)i - (45 in.)j + (60 in.)k
125 in.
(408 lb)i - (1 83.6 lb)j + (244.8 lb)k
tab + tac = (608)1 - (408.6 lb)j+ (544.8 lb)k
912.921b
608 lb
912.921b
408.61b
91.2.92 lb
544.8 lb
9 J 2.92 lb
0.66599
-0.44757
= 0.59677
# = 913 lb A
6>v=48.2° <
9y=116.6° <
'=53.4° <
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
95
50 in.
m o
60 in.
PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb in cable
AC, determine the magnitude and direction of the resultant of the
forces exerted at A by the two cables.
SOLUTION
AB-
AB~~
= (40 in.)i - (45 in.)j + (60 in.)k
= 7(40 in.)2 + (45 in.)
2 + (60 in.)2 = 85 in.
AC-~
AC--
= (100 in.)i - (45 in.)j + (60 in.)k
= 7(1 00 in.)2 + (45 in.)
2 + (60 in.)2 = 1 25 in.
X»
=
AR=^^=7-^— = (510 lb)
AB
(40 in.)i - (45 in.)j + (60 in.)k
85 in.
T« == (240 lb)i - (270 lb)j + (360 Ib)k
X,c =
AC~-TAC\ic-TAC~- -(425 lb)
AC
(100 in.)i - (45 in.)j+ (60 in.)*
125 in.
^ac -= (340 lb)i - (1 53 lb)j + (204 lb)k
R = TAB + TAC = (580 lb)i - (423 lb)j + (564 lb)k
Then: R-= 912.92Ib tf = 9131b <
and cos#v
==580ib
-0.63532 t =5O.6° <912.921b
cos =~423 lb
= -0.46335 V- 1
1
7.6° M912.921b
}
cos $z
-=5641b
-0.61780 ft -51.8° 4912.921b
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96
PROBLEM 2.95
For the frame of Problem 2.87, determine the magnitude and
direction of the resultant of the forces exerted by the cable at Bknowing that the tension in the cable is 385 N.
PROBLEM 2.87 A frame ABC is supported in part by cable
DBE that passes through a frictionless ring at B. Knowing that
the tension in the cable is 385 N, determine the components of
the force exerted by the cable on the support at D.
SOLUTION
BD-
BD--
be-
BE--
= -(480 mm)i + (510 mm)j - (320 mm)k
- 7(480 mm)2 + (51.0 mm)2 + (320 mm)2 = 770 mm
~ lBDKI!D ~ L HO ~~ZX
=— ^-[-(480 mm)i + (510 mm)j - (320 mm)k](770 mm)
= -(240 N)i + (255 N)j- (1 60 N)k
- -(270 mm)i + (400 mm)j - (600 mm)k
= 7(270 mm)2 + (400 mm)2 + (600 mm)2 = 770 mm
~ lmKtiE~~ i be
Rf7
= —[-(270 mm)i + (400 mm)j - (600 mm)k](770 mm)
= -(135 N)i + (200 N)j - (300 N)k
R
R
= fbd + Fw = ~(375 N )' + (455 N )J " (460 N)k
7? = 748 N <= 7(375 N)2 + (455 N)2 + (460 N)2 = 747.83 N
cos Bx -
-375 N747.83 N
ex = 120.1° <
cosV
-455 N
747.83 N9
y= 52.5° <
cos $z -_ -460
N
747.83 N9Z =128.0° 4
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97
360 mmPROBLEM 2.96
For the cables of Problem 2.89, knowing that the tension is 1425 Nin cable AB and 2130 N in cable AC, determine the magnitude and
direction of the resultant of the forces exerted at A by the two
cables.
SOLUTION
Tab := -TBA (use results of Problem 2.89) V
(Tab),-.+1125 N (r
fB ) v=~750N (7^),== -450 N
A -Jab
Tac = ~TCA (use results of Problem 2.90) ^
(Tac )x'= +1350 N (TAC )y
=-900 N (TAC )Z--= +1380N
Resultant: *, = XFX - +1 125 + 1 350 = +2475 N z ^*, = 2F
y- -750 - 900 = - 1 650 N c
R
= XFZ = -450 + 1380 = +930 N
-^R2
x +Rl+R2
z
= 7(+2475)2 + (-1650)
2+(+930)
2
= 3116.6N # = 3120N ^
cos X-_ Rx _ +2475
R 3116.6<?v
=37.4° <
cosy
-
_ Ry _ -1650
7? 311 6.60,, =122.0° ^
cos Oz
•
_ i?2 _ +930
7? 3116.6r- 72.6° 4
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98
PROBLEM 2.97
The end of the coaxial cable AE is attached to the pole AB, whichis strengthened by the guy wires AC and AD. Knowing that the
tension in AC is 150 lb and that the resultant of the forces exerted
at A by wires AC and AD must be contained in the xy plane,
determine (a) the tension in AD, (b) the magnitude and direction
of the resultant ofthe two forces,
SOLUTION
R-T^+T^= (150 lb)(cos 60° cos 20°i-sin 60°j-cos 60° sin 20°k)
+ 7^D (sin 36°sin 48°i-cos 36°j + sin 36° cos 48°k)
(a) Since Rz- 0, The coefficient of k must be zero.
(3 50 lb)(-cos 60° sin 20°)+ TAD (sin. 36° cos 48°) -
TiD= 65.220 lb
(b) Substituting for TAD into Eq. (1) gives:
R = [(150 lb) cos 60°cos 20°+ (65.220 lb)sin 36° sin 48°)]i
- [(1 50 lb) sin 60°+ (65.220 lb) cos 36°]j +
R = (98.966 lb)i - (1 82.668 lb)j
(1)
TAD =65.2 lb A
/? =^(98.966 lb)2+(182.668 lb)
2
= 207.76 lb
cos ft.
cos ft
98.966 lb
207.76 lb
182.6681b
207.76 lb
cos ft -
/e = 208 lb A
<9V=61.6° <
ey= 151.6° <
ft =90.0° A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
99
PROBLEM 2.98
The end of the coaxial cable AE is attached to the pole AB, which
is strengthened by the guy wires AC and AD. Knowing that the
tension in AD is 125 lb and that the resultant of the forces exerted
at A by wires AC and AD must be contained in the xy plane,
determine (a) the tension in AC, (b) the magnitude and direction
ofthe resultant of the two forces.
SOLUTION
R = T^C +T^= TAC (cos 60° cos 20°i-sin 60°j-cos 6()°sin 20°k)
+ (125 lb)(sin 36°sin 48°i-cos 36°j + sin 36° cos 48°k)
(a) Since R r= 0, The coefficient ofk must be zero.
TAC (- cos 60° sin 20°) + (3 25 lb)(sin 36° cos 48°) =
TM .- 287.49 lb
(1)
TAC = 287 lb <
(h) Substituting for TAC into Eq. (1) gives:
R = [(287.49 lb) cos 60° cos 20° + (125 lb) sin 36° sin 48°]i
-[(287.49 lb)sin 60° + (125 lb)cos 36°]j +
R~ (189.677 lb)i-(350.10 lb)J
cos 0..~
cos #
R = 7(1 89.677 lb)2 + (350. 1 lb)
2
= 398.18 lb
189.677 lb
398.18 lb
350.101b
398.181b
cos 0„ =
#-398 lb 4
6,, =151.6° <
0. = 90.0° <
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100
5.60 m
PROBLEM 2.99
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension
in cable AB is 259 N.
SOLUTION
The forces applied at A are: TAB , T,(C , TM) , and P
where P = P\. To express the other forces in terms of the unit vectors i, j, k, we write
AB - -(4.20 m)i - (5.60 m)j AB - 7.00 mAC = (2.40 m)i - (5.60 m)j + (4.20 m)k AC = 7.40 mAD = -(5.60 m)j - (3.30 m)k AD = 6.50 m
ABand %w = TabKb = TAB ~~ = (-0.61 - 0.8j)7^fl
Trfc =TA<\AC =TAC~L = (0.32432 -0.75676] + 0.56757k)^c
ADTi/> ^/A/; =^D^ = (-0.861 54j-0.50769k)7^
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101
PROBLEM 2.99 (Continued)
Equilibrium condition UP" = 0: TAB + TAC + TAD + P$~0
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(-0.67^ + 0.324327^)1 + (-0.87^ -0.756767^.. -0M\54T/iD + P)i
+(0.567577^ -0.507697*^ )k =
Equating to zero the coefficients of i, j, k:
-0.67^ + 0,32432rfC =0 (1)
-0.87^-0.756767^ -0.86.1.54F/fO+P = (2)
0.56757f/K . - 0.50769TAD = (3)
Setting TAB = 259 N in (1) and (2), and solving the resulting set of equations gives
TAC =479.15 N
7^ = 535.66 N P = 103 I.NH
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102
PROBLEM 2.100
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension in
cable AC is 444 N.
m&
SOLUTION
See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2),
and (3) below:
-0.67^+0.324327^=0 (1)
-0.STAB - 0.756767^. - 0.861547^ +P= (2)
0.56757TAC -0.507697^ =0 (3)
Substituting TAC - A4A N in Equations (1), (2), and (3) above, and solving the resulting set of
equations using conventional algorithms gives
7^=240NTAD = 496.36 N P == 956 N J <
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103
PROBLEM 2.101
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension
in cable AD is 481 N.
SOLUTION
See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3).
-0.6r }g+0.32432F,lc =0
-0.87^ - 0.756767^. - 0.86.1 54TAD + P =
0.567577V ~ 0.507697V, =AC
0)
(2)
(3)
Substituting TAD = 481 M in Equations (I), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives
TAC = 430.26 N
TAS = 232.57 N P = 926 N I 4
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104
5.60 in
PROBLEM 2.102
Three cables are used to tether a balloon as shown. Knowing that the
balloon exerts an 800-N vertical force at A t determine the tension in
each cable.
SOLUTION
See Problem 2.99 for the figure and analysis leading to the linear algebraic Equations (l), (2), and (3).
-0.67^+0.324327^=0 (1)
-0.87\(jS- 0.75676TAC - 0.86 J 547^ + P = (2)
(3)0.56757TAC -Q.50769TAD =0
From Eq. (J)
From Eq. (3)
7^=: 0.540537V/in AC
TAn =\.M95TAD AC
Substituting for TAB and TAD in terms of TAC into Eq. (2) gives:
-0.8(0.54053^) - 0.75676TAC - 0.861 54(1 . 1 1 795?fC ) + P =
2.I5237^C «P; /> = 800N
Tac800 N2.1523
371.69 N
Substituting into expressions for TAB and TAD gives:
TAB -0.54053(371.69 N)
TAD = 1.1.1795(371.69 N)
TAB =201N, 7\r = 372 N, T\D 4J6N
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105
PROBLEM 2.103
A crate is supported by three cables as shown. Determine
the weight of the crate knowing that the tension in cable ABis 750 ib.
SOLUTION
The forces applied at A are:
T^/I^;,T
/(/JandW
where P - P\. To express the other forces in terms of the unit vectors i, j, k we write
"AB = -(36 in.)i + (60 in.)j - (27 in.)k y
AB = 75 in.
ZC = (60m.)j + (32in.)k
/iC = 68in.
32'^X. 1
,3t>»<>.
Mom.
AD = (40 in.)i + (60 in.)j - (27 in.)k ;<fAD = 77 in. /\3»l / V 1? ,r\.
A Rand TAB = TAB XAB — TAB—
—
AB= (-0.48I + 0.8j - 0.36k)TAB
A
/So
AC
./4C
f tf \s,,
f
= (0.88235j + 0.47059k)7,
/JC
1AD ~ *AD KAD ~~ ' AD .^= (0.5 1 948i + 0.77922j - 0.35065k)7^D
Equilibrium Condition with W = -W\
ZF = 0: TAIi +TAC +TAD ~rV^Q
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106
PROBLEM 2.103 (Continued)
Substituting the expressions obtained for TA8 , TAC> and TAO and factoring i, j, and k:
(-0.48r,,s + 0.5 1 9487^)1 + (O.HTAB + 0.882357^ + 0J1922TAD - W)\
+(-0.367^ + 0.470597^. - 0.350657^ )k =
Equating to zero the coefficients of i, j, k:
-0.487^ + 0.5
1
948TAD =
0.STAli + 0.88235FiC + ().V922TAD -W =
-0.367^+0.470597^, -0.350657^ =0
Substituting TAH = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using
conventional algorithms for solving linear algebraic equations, gives:
TAC =1090.1 lb
7^ =693 lb W = 2100 lb <
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107
m in.
PROBLEM 2.104
A crate is supported by three cables as shown. Determine
the weight of the crate knowing that the tension in cable ADis 616 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear
below:
algebraic Equations (1), (2), and (3)
-0.487^+0.519487^ ==
i)MAB + 0.882357;^. + 0.17922TAD -W -=
-0.367^ + 0.470597^. -0.350657^ ==
Substituting TAO == 616 lb in Equations (1), (2), and (3) above, and solvii g the resulting set of equations using
conventional algorithms, gives:
TAB = 667.67 lb
TAC = 969.00 lb W = 1868 lb <
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108
PROBLEM 2.105
«)i„. A crate is supported by three cables as shown. Determine the weight
of the crate knowing that the tension in cable AC is 544 lb.
60 in.
SOLUTION
See Problem 2.103 for
below:
the figure and the analysis leading to the linear algebraic Equiitions (1), (2), and (3)
-0.487^+0.519487^ ==
0.87^ + 0.88235r/fC
+ i).11922TAD - W -=
-0.36TIB +0A7059TAC -035065TAD ==
Substituting TAC =544 lb in Equations (1), (2), and (3) above, and solving the resulting
conventional algorithms, gives:
set of'equ itions using
TAB = 374.27 lb
TAD ~ 345.82 lb W---10491b <
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109
PROBLEM 2.106
A 1600-lb crate is supported by three cables as shown. Determine
the tension in each cable.
60 in.
SOLUTION
See Problem 2.103 for the
below:
figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
-0.487^+0.519487^ ==
0.8.7^ + 0.882357^ + 0.779227^ - W ==
~i)MTAB + ().47059r/fC
-0.350657^-, ==
Substituting W - .1600 lb in Equations (1), (2), and (3) above, and solving the resulting set
conventional algorithms, gives;
of equal ions using
Tab*= 571 lb <
TAC = 830 lb A
Tad =-. 528 lb A
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1 10
320 nun -
380 nun •
960 1n it i
c>\iu
960 mm
PROBLEM 2.107
Three eables are connected at A, where the forces P and Q are
applied as shown. Knowing that £? = 0, find the value of P for
which the tension in cable AD is 305 N.
SOLUTION
2*^=0: T^+T^+T^+P-0 where P = Pi
AB - -(960 mra)i - (240 mm)j + (380 mm)k AB == 1060 mmAC = -(960 mm)i - (240 mm)j - (320 mm)k AC = 1040 mmAD = -(960 mm)i + (720 mm)j - (220 mm)k AD = 1220 mm
1/f/i ~ * //# ^i/J ~ 1 AB jn~ AB
f 48. 12 . 19, >
^ 53 53 53 ;
T r 1 r ^C _ f 12. 3 . 4,^,«. AC AC AC AC Aty
j3 1;J
J]3 j
T„> ^/>^z> =1<i°n
5N[(-960 mm)i + (720 mm)j-
1 220 mm-(220mm)k]
= -(240 N)i + (1 80 N)j - (55 N)k
Substituting into X>FA== 0, factoring i, j, k, and setting each coefficient equal to <j> gives:
AQj J
i: P=—J;„,+—7>.+240N53
AB13
AL (1)
r- f3 rAB+^c=m " (2)
19 4
53Ali
13/JC (3)
Solving the system of 1 near equations using conventional algorithms gives:
7^ =446.71 NTAC = 341.71 N P == 960N <
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Ill
320 mm -
380 nun
9G0 mm
PROBLEM 2.108
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that P - 1 200 N, determine the values
ofQ for which cable AD is taut.
SOLUTION
We assume that TAD and write 2^=0: JAB + T (C + 0j + (1 200 N)i =
AB = -(960 mm)i - (240 mm)j + (380 mm)k AB = 1060 mm
AC = -(960 mm)i - (240 mm)j - (320 mm)k AC = 1 040 mm
AB _
AB~
~AC
' ABT 1 — T' A If *"AB ~ l AB
48. 12. 19, ,„.— ij +—k 7'
53 53 53(/i
'AC Tac Ktc TAC12. 3 .— i
j13 13
4k \T4CAC V 13 13" 13
Substituting into ZJ?A= 0, factoring i, j, k, and setting each coefficient equal to <j> gives:
All
X
k:
48
53
|2
53
53AB
13AC
ill
12
13
A13
r/lc +1200N =
tac + Q = o
(1)
(2)
(3)
Solving the resulting system of linear equations using conventional algorithms gives:
TAB
T
605.71 N
705.71 NAC
Q = 300.00 N 0<g<300N^
Note: This solution assumes that Q is directed upward as shown (£>^ 0), if negative values ofQare considered, cable AD remains taut, but AC becomes slack for Q = -460 N.
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112
PROBLEM 2.109
A transmission tower is held by three guy wires attached to a
pin at A and anchored by bolts at B, C, and D. If the tension in
wire AB is 630 lb, determine the vertical force P exerted by the
tower on the pin at A.
SOLUTION
We write
SF-0: TAB +TAC +TAD + Pj =
AB = -45i - 90j + 30k AB = 1 05 ft
AC = 30i - 90j + 65k AC = .1 1 5 ft
^75 = 20i - 90j - 60k AD = 1 1 ft
VJfi ~ *Air-AB ~ 1ABAB
fl-fj+fk|^
T ~ 7" ) ~ T ACAC '"AC -!C
' 6 . 18. 13. ,,r
23 23 23
7/3*/(D ~ * AD**AD ~~ l Al> AD
2 , 9 . 6,_, j~—
k
II/i/.)
Substituting into the Eq. SF = and factoring i, j, k
:
"
j * AB + 7AC + TA0
+ —T18 9 m
- yj/i ~^ -TjC , «
'
1AD + ^ ! J
13+^TAB+—TAc~-riD \k^o
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.
113
PROBLEM 2.109 (Continued)
Setting the coefficients of i, j, k, equal to zero:
3 T a.6 T m 2 T
7AB
23A<
11.AD = 0)
6 T 18 9J- „ J AB j*
1AC , . * AD fP -0 (2)
i
2 r J 37>
6 r = (3)
Set 7^ = 630 lb in Eqs. (!) -- (3):
~270}b +—TAC +^T,D --
23 11
= (10
18 9-540 lb TAC—TAD + P =
23 '" 11= (20
1 ^ /"'
1801b +~7>—
T
/(/>=
23AC
11 "= (3')
Solving, 7^c =467.42 lb TA0 -814.35 lb P = 1572.10 lb P = 1572 lb ^
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114
PROBLEM 2.110
A transmission tower is held by three guy wires attached to a
pin at A and anchored by bolts at B, C, and D. If the tension in
wire AC is 920 lb, determine the vertical force P exerted by the
tower on the pin at A,
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the 1 near algebrtlie Equations (1), (2), and (3)
below:
At +Ar- ' All ^ y~ 'AC
AT _i8 r _!7
AB23
AC11
2 13
7* All
T *~ *AC
11 ^
t^ + p=o
-—T =011
"D
(0
(2)
(3)
Substituting for TAC - 920 lb in Equations (1), (2), and (3) above and solving the resulting set of e filiations
using conventional algorithms gives:
~f^B+2401b +~7^-0 CO
6 9_r^-720 1b-^ TAD + P = (20
1^ +520 lb--—T =011 ^
(3')
Solving, = 1240.00 lb
= 1602.86 lb
P = 3094.3 lb P = 3090 lb <
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115
'J PROBLEM 2.111
A rectangular plate is supported by three cables as shown.|~~~\^ ~y~ Knowing that the tension in cable AC is 60 N, determine the
480 y ST weight of the plate.
Ky^'4. itr>
X --. "" M£ X360 *
Dimensions in mm
SOLUTION
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.
Free Body A :
£F = 0: T/W +TAC +TAD + P^0
We have:
AB = -(320 mm)i - (480 mm)j + (360 mm)k AB = 680 mm
AC - (450 mm)i - (480 mm)j + (360 mm)k AC- 750 mmAD = (250 mm)i - (480 mm)j - (360 mm)k AD = 650 mm
Thus:
*AB ~~'AB*"AB ~ *AU
AB { 8 . 12. 9
/*£ , 17 17 17 '
T,c =^c^c = TAC~ - (0.61 - 0.64J + 0.48k)TA
*AD ~ 'Ati *"AD ~ 'AD
ADAD
9.6 . 7.2
13 13.1
13TID
Substituting into the Eq. XF = and factoring i, j, k:
' 8
]7Lw+0WAC +-TAD \i
-^TM -QMTAC -^-TAD +P\}
9 7 2—TAB + 0AST4C—-rm1.7
AB 'K1.3
-w k =
Dimensions in mm
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116
PROBLEM 2.111 (Continued)
Setting the coefficient of i, j, k equal to zero:
1.7'"J /u
" 13
12 9
7 '"' /K.13
9 7
'
—7^ +0.487*,--—17
Als A <-
13
i: -y^ -0.64?;,,. -^r,D +.p=o (2)
k: 1^+0.48^-^7^=0 (3)
Making 7^c = 60 N in (1 ) and (3):
17 ~13
Multiply (]') by 9, (3') by 8, and add:
97^+28.8N-^r^=0 (3')
554.4 N-i^r/1D
=0 TAD =572.0 N
Substitute into (I') and solve for Tin :
?ab =yf 36 + -—X572|
TAB = 544.0 N
Substitute for the tensions in Eq. (2) and solve for/':
p = ~(544 N) + 0.64(60 N) +— (572 N)
= 844.8 M Weight of plate = P = 845 N ^
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117
J 30320 ;><.
450^•c -
Dimensions in mm
PROBLEM 2.112
A rectangular plate is supported by three cables as shown. Knowing
that the tension in cable AD is 520 N, determine the weight of the plate.
SOLUTION
See Problem 2. Ill for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
-^+0.67-^+^=0 (1)
~TAB + 0.647>~TAD + P = (2)
^+0.48^-^^=0 (3)
Making TAD = 520 N in Eqs. (1) and (3):
~TAB + Q.6TAC + 200 N = 00
—TAB + 0ASTAC - 288 N = (3')
Multiply (10 by 9, (30 by 8, and add:
9.24TAC - 504 N = TAC = 54.5455 N
Substitute into (V) and solve for TAB \
TA8 = ~(0.6 x 54.5455 + 200) TAB = 494.545 No
Substitute for the tensions in Eq. (2) and solve for P:
P = 11(494.545 N) + 0.64(54.5455 N)+^(520 N)
= 768.00 N Weight ofplate = P = 768 N A
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118
PROBLEM 2.113
For the transmission tower of Problems 2.109 and 2.1 10, determine
the tension in eaeh guy wire knowing that the tower exerts on the
pin at A an upward vertical force of 2 1 00 lb.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
_±T +-2-T 4-—T -A(1)
~~Z *ab
~~ZZ
*
AC -—TAD + P = (2)
- 1 AB T ~~ *AC . . 'AD u (3)
Substituting for P- 2 100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
-Z.T 4-_-Lt +J-~t -07
' AB r y~ lAC^ ., lAD~ K} (10
~1Tab ~x Tac "u Tad +2I0° lb = (2')
1T +—T ~—T -o- ' AB + y~ 'AC ,. JAD~ V (3')
TAB -841.551b
TAC =624.38 lb
TAD =1087.81 lb T,w -8421b <
Tac = 624 lb A
'AD~ 1088 lb <
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119
D,
PROBLEM 2.114
A horizontal circular plate weighing 60 lb is suspended as shown from
three wires that are attached to a support at D and form 30° angles with
the vertical. Determine the tension in each wire.
SOLUTION
~LFX = 0:
-TAD (sin 30°)(sin 50°) + TBD (sin 30°)(cos 40°) + TCD (sin 30°)(cos 60°) =
Dividing through by sin 30° and evaluating:
-Q.76604TAD -f 0.766047^ + 0.5TCD = ( 1
)
ZFy= 0: -TAD (cos 30°)-TBD (cos 30°) - TCD (cos 30°) + 60 lb =
or TAD + TBD + TCD = 69.282 lb (2)
ZFZ= 0: TAD sin 30° cos 50°+ Tm sin 30° sin 40° - TCD sin 30° sin 60° =
or 0.642797^ + 0.642797^ - 0.86603rCD = (3)
Solving Equations (1), (2), and (3) simultaneously:
7\fD =29.5lb A
TBD =10.25 lb <
rc/)
=29.5 1b <
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120
Dimensions in mm
PROBLEM 2,115
For the rectangular plate of Problems 2. 1 1 1 and 2.1 12, determine
the tension in each, of the three cables knowing that the weight of
the plate is 792 N.
SOLUTION
See Problem 2.11 1. for the figure and the analysis leading to the linear algebraic Equations (0,(2), and
(3) below. Setting P = 792 N gives:
~TAB +QMAC +^TAD = (1)
~r^-0.64^c-^D +792N = (2)
^TAS +0AXTAC -^TAD = (3)
Solving Equations (1), (2), and (3) by conventional algorithms gives
7^ =510.00 N ^=510N«
TAC = 56.250 N 7^C =56.2N <
TAD = 536.25 N 7 iD =536N <
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121
320 nm i-
..580 mm -
220 mm
060 mm
OfiOitmi
•240 mm
PROBLEM 2.116
For the cable system of Problems 2.107 and 2.108, determine the
tension in each cable knowing that P - 2880 N and Q = 0.
SOLUTION
£F, = 0: T/W +TAC +TAD+ V +Q = <)
Where P
AB
AC
AD
Tad
= />i and Q = fiJ
- -(960 mm)i - (240 mm)j + (380 mm)k AB = 1 060 mm
= -(960 mm)i - (240 mm)j - (320 mm)k AC -1040 mm= -(960 mm)i + (720 mm)j - (220 mm)k AD = 1 220 mm
T 1 T A® T (48
«12 -J 9 !^-IabKb-IabJj-'ab^
53*
53J +
53kJ
T 1 T A^ T (12
'3
'4 I^= 7\r x4C - r,r— - 7\r — i— j—
k
,1C A ,«. AC AC,y|3 j 3
J u J
^D f 48. 36. 11 /j
Substituting into £F. = 0, setting /> = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0,
we obtain the fol owing three equi ibrium equations:
i
:
48 12 48~TAB~TAC -^TAD +2880 N = (1)53 1.3 61
j: ~TAB -2-TAC +%TAD =0 (2)53 13 61
k:19 4 11—TAB -—T4C -—TAD ^() (3)53 ^ 13
AC61
/,D
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122
PROBLEM 2.116 (Continued)
Solving the system of linear equations using conventional algorithms gives:
TAB =1340.14 N
TAC = 1025.12 N
7\n =915.03N 7',, =1340 N AAD --"•>•"-> '•< i AB
TAC =1()25N <
r jD =915N ^
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123
320 hiiii -
380 mm -
960 mm
PROBLEM 2.117
For the cable system of Problems 2.107 and 2.108, determine the
tension in each cable knowing that P - 2880 N and Q = 576 N.
SOLUTION
See Problem 2. 1 1 6 for the analysis leading to the linear algebraic Equations <1).<2>. and (3) below
53M
13AC
6.1M> = (0
53Ali
13Ai
61AD Q = (2)
53A,i
13AC
6}AD = (3)
Setting P = 2880N and £2 = 576 N gives:
48 12 48~TAB~tac—r^ +2880 N == (10
53 13 61
53 13 61= (2')
19 4 11
c~ ' AB ,^ 'AC < , * AD53 13 6
1
= (3')
Solving the resulting set of equations using conventional algorithms gives:
7^= 1431.00 NTAC =1560.00 NTAD =183.010 N Tab
'-
TAC =
ho =
=1431 N A
= 1560N ^
183.0 N ^
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124
320 nun •
380 mm •
\\ 960 mm
960 mm
PROBLEM 2.118
For the cable system of Problems 2.107 and 2.108, determine thetension in each cable knowing that /» = 2880N and 0--576N.(Q is directed downward).
SOLUTION
See Problem 2. 1 16 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
53M n AC
T\Tao+p - q
I^-^c +f^ +^019^, II
5?>* AB n AC6\
Tad ~°
Setting /> = 2880N and g = -576N gives:
48 m 12 48
13 6153
J±T53 -l|^ +f r,o -576N =
19_/ -±T -lir53
Ali n AC61
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1249.29 N
r^. =490.31 NTAd =1646.97 N
(0
(2)
(3)
(10
(20
(30
^ == 1.249 N <
Tac = 490N <«
Tad ~-= 1.647 N ^
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125
PROBLEM 2.119
Using two ropes and a roller chute, two workers are unloading a 200-ib
cast-iron counterweight from a truck. Knowing that at the instant
shown the counterweight is kept from moving and that the positions of
Points A, fl, and C are, respectively, A(0, -20 in., 40 in.), B(-~40 in.,
50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists
between the counterweight and the chute, determine the tension in each
rope. (Hint: Since there is no friction, the force exerted by the chute on
the counterweight must be perpendicular to the chute.)
SOLUTION
From the geometry of the chute:
NN
(2J + k)4~5
Af(0.8944j + 0.4472k)
The force in each rope can be written as the product ofthe magnitude of the
force and the unit vector along the cable. Thus, with
AB = (40 in.)i + (70 in.)j - (40 in.)k
AB = /(40 in.)2 + (70 in.)
2 + (40 in.)2
= 90 in.
AB
and
Then;
*AB ~^\iS ~~ *AB ABTAH
90 in.f-[(~40 in.)i + (70 in.)j-(40 in.)k]
l AB
4. 7. 4TAB "T^J
9 9 9
AC = (45 i.n.)i + (60 in.)j - (40 in,)k
AC = V(45 in.)' + (60 in.)2 + (40 in.) = 85 in.
ACTAC - TXAC ~ TAC AC
--^[(45 in.)i + (60 in.)j -(40 in.)k]
85 in.
8
AC '< I , 7 3?J
l?
£F = 0: N +T^+T^ +W^O
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126
PROBLEM 2.119 (Continued)
With W = 200 lb, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
* ~~TAB+-~TAC ={)(1)
J- |^fl
+l|^c +-^-200lb = (2)
9AB
\1AC +S
Using conventional methods for solving linear algebraic equations we obtain:
*•• ~~Tas ~~~Tac+—N =(3)
TAB = 65.6 lb «
r^c =55.1 lb ^
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127
PROBLEM 2.120
Solve Problem 2.1 1 9 assuming that a third worker is exerting a force
p = -(40 lb)i on the counterweight.
PROBLEM 2.119 Using two ropes and a roller chute, two workers are
unloading a 200-lb cast-iron counterweight from a truck. Knowing that
at the instant shown the counterweight is kept from moving and that the
positions of Points A, B, and C are, respectively, A(Q, - -20 in., 40 in.),
#(-40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no
friction exists between the counterweight and the chute, determine the
tension in each rope. (Hint: Since there is no friction, the force exerted
by the chute on the counterweight must be perpendicular to the chute.)
SOLUTION
See Problem 2.1 1 9 for the analysis leading to the vectors describing the tension in each rope.
4. 7. 4.TAB-TAB \--l +-l~*.
T —TlAC ' AC
r 9 . 12. 8,
17 17 17
Then:
Where
and
£F,,-0: N +T^+TjC. + P +W =
p = -(40 lb)i
W = (200 1b)j
S
Equating the factors of i, j, and k to zero, we obtain the linear equations:
i: -*TAB +—r^c -401b =
j :
^ 7 ]200lb==0
S 9AB
17AC
1 4 8
S 9AB
17AC
Using conventional methods for solving linear algebraic equations we obtain
7^ =24.8 lb <
TAC =9(>A\h<
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128
PROBLEM 2.121
A container of weight W is suspended from ring A.Cable BAC passes through the ring and is attached to
fixed supports at B and C. Two forces P = Pi and
Q = Qk are applied to the ring to maintain the
container in the position shown. Knowing that
}'F = 376N, determine P and Q. {Hint: The tension is
the same in both portions of cable BAC.)
SOLUTION
Free Body A: T„=7VabTAft --T T*£~-T1 Aft
= TAB_
AB
- 7-(~ { 30 mm)i + (400 mm)j + (1 60 mm)k
450 mmQ--QW
1
/k:
r
n.
13. 40. 16,— 1 +— j +—
k
45 45 45
AC
fAC_
AC
T(-1 50 mm)i + (400 mm)j + (-240 mm)k
490 mm
~T 15. 40. 24,— , +—
,
^49 49 49
IF = 0: TAB + TAC +Q + P +W =
Setting coefficients of i, j, k equal to zero:
,: -HT -11t +P =45 49
j: +i£r+i£ r _r =45 49
k: + i6 r _24 r +e =45 49
0.595017 = P
1 .70521 r = jp
0.1 342407' =
W s -(37414^0
(1)
(2)
(3)
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' *
129
PROBLEM 2.121 (Continued)
Data: W = 376N 1 .70521^ = 376 N 7* = 220.50 N
0.59501(220.50 N) = P P = 13i.2N <
0.134240(220.50 N) =£ g = 29.6N M
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13<)
150 nun
PROBLEM 2.122
For the system of Problem 2.121, determine W and Qknowing that P - 164 N.
PROBLEM 2.121 A container of weight W is suspended
from ring A. Cable BAC passes through the ring and is
attached to fixed supports at B and C. Two forces P ~ P\
and Q = Qk are applied to the ring to maintain the container
in the position shown. Knowing that W7 = 376N >determine
P and Q. {Hint: The tension is the same in both portions of
cable BAC.)
SOLUTION
Refer to Problem 2.121 for the figure and analysis resulting in Equations (1), (2), and (3) for P, Wy and Q in
terms ofT below. Setting P = 164 N we have:
Eq.(l):
Eq. (2):
Eq. (3):
0.59501T = 164 N
1 .70521(275.63 N) =W
0.1 34240(275.63 N) = £
T = 275.63 N
W = 470 N <
g = 37.0N A
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1.31
PROBLEM 2.123
A container of weight W is suspended from ring A, to
which cables AC and AE are attached. A. force P is
applied to the end F of a third cable that passes over a
pulley at B and through ring A. and that is attached to a
support at D. Knowing that W ~ 1000 N, determine the
magnitude of P. (Hint: The tension is the same in all
portions of cable FBAD.)
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along
the cable. That is, with
AB - -(0.78 m)i + (1 .6 m)j + (0 m)k
AB = ^(-0.78 m)2 + (1 .6 m)2 + (0)
2
-1.78 m
AB*AB "^^AB ~ ' AB AB
THi
1.78 m[-(0.78 m)i + (1 .6 m)j + (0 m)k]
l Ali TAB (-0.43821 + 0.8989J + 0k)
and
and
"AC = (0)i + (1 .6 m)j+ (1 .2 m)k
AC = V(° m)2 + (1 .6 m)2 + (1 .2 m)
2 = 2 m
Xtc = TXAC = 7>i£ - ^£-[(0)1 + (1 .6 m)j + (1 .2 m)k]' ,4C 2 m
^c=^c(0-8j + 0.6k)
y*D - (1 .3 m)i + (1 .6 m)j + (0.4 m)k
AD - y[(] .3 m)2 + (1 .6 m)2 + (0.4 m)2 =2.1m
*AD ~ ' *"AD TAD Tin
ID AD 2.1 m[(1 .3 m)i + (1 .6 m)j + (0.4 m)k]
AD T4D (0.61901 + 0.76 1 9 j + 0, 1 905k)
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132
PROBLEM 2.123 (Continued)
Finally, AE = -(0.4 m)i + (1 .6 m)j - (0.86 m)k
1 .86 mAE = yji-OA m)2 + (1.6 m)2 + (-0.86 m)2 =
AE*AE ~ l **AE
~~ 1AE ,„. . Ah
=—^-[-(0.4 m)i + (1 .6 m)j - (0.86 it
1 .86 m)fc]
tae = tae (~ -2 l5li + 0.8602j- 0.4624k)
With the weight of the container W = —W\ , at A we have:
^ = 0: TAB +TAC +TAD -W1 = Q
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
-0.43827^ +0.6190.7^ -0.21517^ -0 0)
0.89897^ + Q.$TAC + 0.76\9TJD + 0.86027^- -W = (2)
0.6TAC + 0.
1
905TM) - 0A624TAI, = (3)
Knowing that W = 1000 N and that because of the pulley system at BTAB = TAD - P,
applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3)
where Puniquely
s the externally
for P.
P = 378 N <
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133
PROBLEM 2.124
Knowing that the tension in cable AC of the system
described in Problem 2.123 is 150 N, determine (a) the
magnitude of the force P, (b) the weight W of the
container.
PROBLEM 2.123 A container of weight W is
suspended from ring A, to whieh cables AC and AE are
attached. A force P is applied to the end F of a third
cable that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing that
W - 1000 N, determine the magnitude ofP. (Hint: The
tension is the same in all portions of cable FBAD.)
SOLUTION
Here, as in Problem 2.123, the support of the container consists of the four cables AE, AC, AD, and AB, with
the condition that, the force in cables AB and AD is equal to the externally applied force P. Thus, with the
condition
T =T = P' ab l ad '
and using the linear algebraic equations ofProblem 2.131 with TAC - 1 50 N, we obtain
(a) P = 454N <
(b) JF = 1202N <
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134
PROBLEM 2.125
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. If a 60-lb force Q is applied to collar B as
shown, determine (a) the tension in the wire when x = 9 in., (b) the
corresponding magnitude of the force P required to maintain the
equilibrium of the system.
SOLUTION
A:
Free Body Diagrams of Collars:
B:
Collar A:
_ AB _ ~jA- (20 in.)j + zkAB " AB~ 25 in.
"EF = 0: Pi +Ay + A^k +r^M)
Substitute for "kAB and set coefficient of i equal to zero:
25 in.
Collar B: ZF = : (60 lb)k + N'xl + N'
yj~ TMZAB =
Substitute forX^ and set coefficient of k equal to zero:
60Ib—2ffiL = o
(a)
From Eq. (2):
(/;) FromEq.(l):
x = 9 in.
25 in.
(9in.)2+(20in.)
2 +z2=(25in.y
z = 12 in.
60 lb ~TAB (12 in.)
25 in.
,y _ (1.25.0 tb)(9 in.)
25 in.
jAe>'d A£>
(I)
(2)
7\„ =125.0 lb ^All
F = 45.0 lb <
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135
PROBLEM 2.126
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. Determine the distances x and z for
which the equilibrium of the system is maintained when P - 120 lb
and Q~60\b.
SOLUTION
See Problem 2. 1.25 for the diagrams and analysis leading to Equations (1) and (2) below:
60 lb
25 in.
TAazAB'
For P = 120 lb, Eq. (J) yields
From Eq. (2)
Dividing Eq.(l') by (20:
Now write
25 in.
r^v = (25in.)(20 1b)
r^sz = (25 in.) (60 lb)
* = 2z
x2 +z2
+(20in.)2=(25in.)
2
Solving (3) and (4) simultaneously
4r+z2 +400 = 625
From Eq. (3)
zz =45
z = 6.708 in.
x = 2z = 2(6.708 in.)
= 13.416 in.
0)
(2)
(10
(2')
(3)
(4)
x = 13.42 in., z = 6J\ in. <
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136
2'JO H
7511
PROBLEM 2,127
The direction of the 75-lb forces may vary, but the angle between the forces is
always 50°. Determine the value ofa for which the resultant of the forces acting at ,4
is directed horizontally to the left.
SOLUTION
We must first replace the two 75-lb forces by their resultant R, using the triangle rule.
IS lb
7$Jfb
Ri= 2(75 lb) cos 25°
= 135.946 lb
Rj =135.946 lb ~p- a + 25 c
Next we consider the resultant R2 of R, and the 240-lb force where R2must be horizontal and directed to
the left. Using the triangle rule and law of sines,
"R?. sin (a + 25°) sin (30°)
^4olb
I^S.^C lb
2401b 135.946
sin (or+ 25°) = 0.88270
a + 25° = 6 1.970°
a = 36.970° « = 37.0° <
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137
30 lb PROBLEM 2.128
A stake is being pulled out of the ground by means of two ropes as shown.
Knowing the magnitude and direction of the force exerted on one rope,
determine the magnitude and direction of the force P that should be exerted on
the other rope if the resultant ofthese two forces is to be a 40-lb vertical force.
SOLUTION
Triangle rule:PyS~f of '
Law of cosines: P2--= (30)
2+(40)
2--2(30)(40)cos 25° V 1 R= 4o \\>
P == 18.0239 lb 3o iv>Vs ,,
Law of sines:sin a
30 lb"
sin 25°
\18.0239 1b
a-= 44.703°
90°~tf == 45.297° P = 18.02 lb A. 45.3° <
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138
D
/ :
A40°
PROBLEM 2,129
MemberBD exerts on memberABC a force P directed along line BD.Knowing that P must have a 240-lb vertical component, determine(a) the magnitude of the force P, (b) its horizontal component.
SOLUTION
(a) P
2HQ\h
zMo
1
3p
P. 240 1b
sin 35° sin 40c
Pv 240 lb
tan 40° tan 40c
or ^ = 373 lb 4
or />. = 286 lb <
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139
S.5 ft
12 it
PROBLEM 2.130
Two cables are tied together at C and loaded as shown. Determine
the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free Body Diagram at C:
_ A 12 ft 7.5 ft T12.5 ft
AC8.5 ft
/iC -^V T^
Tgc = 1 .088007^ 3,5'
p-^W-.S' b&S'm
(
3 5 ft 4 ft
XFV- 0: 7^ + rBC - 396 lb -
•' 12 ftAC
8.5 ftBC
»*• 1-5'
(a)3 '5 ft
TAC +4 ft
(1.08800^0-396 lb -v 7
12.5 ftAC
8.5 ft
i, 3%>\b
(0.28000+ 0.5 1200)7'/1C
= 396 lb
7^, =500.0 lb r/iC
= 500 lb <
(ft) 7^.= (1.08800)(500.0 lb) r/iC =544 1b <
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140
600 mm
250 inni
Q -- 480 N
PROBLEM 2.131
Two cables are tied together at C and loaded as shown. Knowingthat P = 360N, determine the tension (a) in cable AC, (b) incable BC.
SOLUTION
(a) £F =0:
Free Body: C
Tfle T6cf f-jJpN
[J
12 4^C +|(360N) = 7\r =312N <«
XF^O: ~(312N) +^C +|(360 N)-480N =
rfit = 480N-120N-216N r,.r = 144N <
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72t7*™f*?^MfVJ<»m or by any means, without the prior written permission of the publisher, or used beyond the United
141
600 mm PROBLEM 2.132
250mm
Q = im hi
Two cables are tied together at C and loaded as shown.
Determine the range of values ofP for which both cables
remain taut.
SOLUTION
Free Body: C
12 4
13AC
5
AC15
2Fy= 0: ^TAC + 7),c +|p - 480 N =
CD
Substitute for TiC from(l): — if— \P + Tm.+~P-4&QN =13 A 15 J 5
71 =480N-~P*c
15(2)
From (1), TAC > requires P > 0.
From (2), TBC > requires —/>< 480 N, P< 514.29 N
Allowable range: 0<P<514N<«
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142
PROBLEM 2.133
A force acts at the origin of a coordinate system in a direction defined by the anglesX= 69.3° and
62 -57.9°. Knowing that the^ component of the force is -MAX) lb, determine (a) the angle"'
(b) the othercomponents and the magnitude of the force,
"'"
SOLUTION
(a) To determine 8y) use the relation
cos2
+ cos2
V + cos2 B7
~I
^ cos2
y= l- cos
2
X- cos
2<9,
Since Fy < 0, we must have cos
y<
cos6y= -VI - cos
269.3° - cos
257.9°
- -0.76985
(« F =-^-=- 174 -0lb-226.02 lb
cos#(
, -0.76985
Fx = Feos#r = (226.02 lb) cos 69.3° Fx = 79.9 ]D -4
Fs =F cos^ = (226.02 lb) cos 57.9° Fz = 1 20. lib ^
>,, = 140.3° ^
F = 226 Jb <«
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' *
143
PROBLEM 2.134
Cable AB is 65 ft long, and the tension in that cable is 3900 lb.
Determine (a) the x, y, and z components of the force exerted by
the cable on the anchor B, (b) the angles x9 6V , and Z defining
the direction of that force.
SOLUTION
From triangle AOB: cos 6>
(a)
(b)
56 ft
65 ft
0.86 1 54
6V=30.51°
Fx. =-Fsin<?
vcos20°
= -(3900 Ib)sin30.51°cos20°
Fy= +Fcos0
y= (3900 lb)(0.86154)
F, =+(3900 lb)sin 30.51° sin 20°
cos V
F. 1861 lb
From above:
F 3900 lb
=30.51°
-0.477.
Fx =-1861 lb <
Fy=+3360 lb <
/<;=+677 1b <
6X =118,5° <
6„ = 30.5° A
F2 677 lbcos 0, = —— = H = +0. 1736
z F 3900 lb
51=80.0° <
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144
SOLUTION
PROBLEM 2.135
In order to move a wrecked truck, two cables are attached at
A and pulled by winches B and C as shown. Knowing that the
tension is 10 kN in cable AB and 7.5 kN In cable AC,determine the magnitude and direction of the resultant of the
forces exerted at A by the two cables.
AB
AB-
AC-
AC--
;-15.588i + 15j + 12k
24.739 m
-15.588i + 18.60j~15k
28.530 m
TAB
w AB '"Aft AB
T,fl=(10kN)
AB•15.588i + 15j + 12k
VAB
T,c
%1C
R
R
cos 6^
cosV
cos ft.
24.739
(6.301 kN)i + (6.063 kN)j + (4.851 k.N)k
t i t ^ncu, ,-15.588i + 18.60j-15kTAcKc^c—0.5 kN)2g 53()
-(4.098 kN)i + (4.890 k.N)j- (3.943 kN)k
tab +Tjc =~(I0.399 kN)i + (10,953 kN)j + (0.908 kN)k
V(10.399)2+(10.953)
2+(0.908)
2
15.130 kN
-10.399 kN
R
5lR
Ik
15.130 kN
10.953 kN
15.130 kN"
0.908 kN
15.130 kN"
-0.6873
0.7239
0.0600
/? = 15.13 kN -4
Oz -133.4° A
#,..=43.6° <
ft =86.6° <
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145
360 mm
500 imn
PROBLEM 2.136
4^q iimi^ container of weight W = 1 165 N is supported by three cables as
shown. Determine the tension in each cable.
SOLUTION
Free Body: A
We have:
XF = 0: T^+TJ|C +T^+W =
^B _ (45() mm)i+ (600 mm)j
45 = 750 mm
AC = (600 mm)j - (320 mm)k
4C = 680mm
AD = (500 mm)i + (600 mm)j+ (360 mm)k
AD = $60 mm
^Ali ~ ^AB^Ali ~" *A8AB
AB
(0.6i + 0.8j)7^
450. 600.1 + 1 \TAl,
750 750 '
3fco
£-00
J -T X -T *£-(*™i-™k« -J/" — / Aff^Af — * AS' —I j ».MC ~-*^C /l
'/IC '/it:
*AD ~ -
' AD^AD ~~ 'AD
AC 1680 680
'ad
TA
15.
17k Vac
AD500. 600. 360,
860 860 860w
T,)/;
25. 30. 187
1
^j/>
43 43 43
Substitute into £F = 0, factor i, j, k, and set their coefficient equal to zero:
0.67'25__T -0
us4o ^*>
~ 7^=0.968997^
0.87^ +~^C+^D -1.165 N=0
]8
17^c + 43^° r,JC = 0.88953^
(I)
(2)
(3)
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146
PROBLEM 2.136 (Continued)
Substitute for TAB and TAC from (1) and (3) into (2):
{0.8x0.96899 +—x 0.88953 +~|7\ -
V 17 53 JAi)
-1165 N =
2.25787^-1 165 N = TAD == 516N ^
From (1 ): TAB = 0.96899(5 J 6 N) ^ == 500N ^
From (3): r„c - 0.88953 (5 16 N) ^c = 459 N ^
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147
PROBLEM 2.137
Collars A and B are connected by a 525-mm-long wire and can
slide freely on frictionless rods. If a force P = (341 N)j is
applied to collar A, determine (a) the tension in the wire when
y = 155 mm, (b) the magnitude of the force Q required to
maintain the equilibrium of the system.
SOLUTION
For both Problems 2. 1 37 and 2. 3 38: Free Body Diagrams of Collars:
(AB)2 = x
2 + y2 + z'
Here
or
(0.525 m)2 = (.20 m)2 +y2 + z
2
y2 +z2 =0.23563 nV
Thus, why j> given, z is determined,
Now "AB
AB
AB
-(0.20i->a + 2k)m0.525 m
= 0.38095i - 1 .90476j>j + 1 .90476zk
Where y and z are in units ofmeters, m.
From the F.B . Diagram of collar A: IF = : Nx \ + Nzk + P\ + TABXAB
Setting the j coefficient to zero gives: P - (1 .9Q476y)TA/) —
With P = 34IN
341 NTAB
.90476^
Now, from the free body diagram of collar B; IF - : Nx\ + Nv j + Qk -~ TAB XAB =
Setting the k coefficient to zero gives
:
Q - TAli (1 .90476z) =
And using the above result for TAB we have Q - TABz341 N
(l.90476z)^M1NXz>
(1 .90476) v
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148
PROBLEM 2.137 (Continued)
Then, from the specifications of the problem, y - 155 mm = 0.155 m
z2=0.23563 m 2
-(0.155 m)2
z = 0.46 mand
(a) T.34i N
All
0.155(1.90476)
1155.00 N
or
and
r,,, =1.1.55 kN <
(b) ?_ 34rN(0.46m)(0.866)
(0.155 m)
= (1012.00 N)
or = I.O12kN <4
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149
PROBLEM 2,138
Solve Problem 2.137 assuming that y — 275 mm.
PROBLEM 2.137 Collars A and B are connected by a
525-mm-long wire and can slide freely on frictionless
rods. If a force P = (341 N)j is applied to collar A,
determine (a) the tension in the wire when y — 1 55 mm,(b) the magnitude of the force Q required to maintain
the equilibrium of the system.
SOLUTION
From the analysis jfProblem 2.137, par iculaiiy I
2 >y +z~ -
TAB =
Q =
he results:
= 0.23563 m2
341 N1.90476;/
341 N= z
y
With y = 275 mm = 0.275 m, we obtain
z2
'.= 0.23563 m2 -(0.275 m)2
z -= 0.40 m
and
(«) TAB '
341
N
-65100(1.90476)(0.275 m)
or r^=651N <
and
(b) Q341 N(0.40 m)
(0.275 m)
or <2 = 496N <
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