solucionario berr

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

Boston Burr Ridge, !L Dubuque, IA New York San Francisco St. Louis

Bangkok Bogota Caracas Kuala Lumpur Lisbon London Madrid Mexico City

Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto

The McGraw-Hill Companies

Higher Education

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,

New York, NY 10020. Copyright© 2010, 2007, 2004, and 1997 by The McGraw-Hill Companies, Inc. All rights reserved.

The contents, or parts thereof, may be reproduced in print form solely for classroom use with VECTOR MECHANICS FOR ENGINEERS,

STATICS, NINTH EDITION provided such reproductions bear copyright notice, but may not be reproduced in any other form or for any other

purpose without the prior written consent ofThe McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic

storage or transmission, or broadcast for distance learning.

This book is printed on acid-free paper.

1234567 8 90 CCW/CCW

ISBN: 978-0-07-724918-2

MHID: 0-07-724918-6

www.mhhe.com

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)


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.


xvi

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


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.


xvn



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



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



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



xix



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


Answers are not given to problems with a number set in italic type,

xx



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





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



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)


*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



xxm



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



xxiv



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



XXV

TABLE II; CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)


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



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


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



xxvii

ill

'J.

oocc

£§Z c-w w> c/>

< Or/> vOJ £UJ 4

i~; *l' 1 ---" i -1

<N >4J <•> ^ •-"- •*

"" ^^f- r~ >.o ^ r-' —

r-l r-l Ol OI tl n ft f*l

— £.01* r~ , Os•— r-l © -... , v£! f»

m r-i r-' ..

•<# *cf "& <& vf Vi Vl Vi

c. „ vc- — . r-l

no vo — in ca r-l

•sj- 'i-i oo _ <f- m * r—

.. r- o" cm oo . r-' ad<» •* OO r- ^" * f'l «vci vri ^ <> 6 r- r- s~-

O .

(-4 ^>

•£> a-

1' i -

,

o O o

r-i

v 6A o

^

7--J

mUJ

iioo

u.iyi m

a;t_

2 o^

u

o CO

It]

LU2:r-

O o'J

^< r •= o

o /:

u-£/)

w % ^-^

q 5 oa.

o

t/3

si.

'0,

XHft o

u.1

X

? o" O

6 s d

^

fc

(X.

-> r/)

< /.

oc >n \f%

vO t r- ^j "* (-•

<M r-l in ri "; CI rl CI

rl >C o(1 vo Oi — CI * C* ol

o ©' >£>" _ . >o" © .m ic •» in o « a »-OM .. . O—, — en —

<

.. O '/"I OO ..

<> CT>

tO r-

O r>i ,. .. —OO .. (N -^f OO ,

- .. .. vo,. O M O•• 3:

o\ >* r- ™ rn * c

oi <~i c-> c-'i t<S r<i

«S\ , o

,'-.- HI VI M (fj

1^ O mr-'"-* "* •* *t * I/I /-> in in

OO r C**l O Oin in .—

—~

ol (N NMMWffin "t ^f •* •* * n in

n * r~ —• t-~ a «n oo— , .. , 0\ <N - ,.

. M r- <3\ —* .^Ci<0

•=t \-~ .. -* — ' vo id oofi .. in — , <N .. ., «T O - r- .• O -—

in vi> »—< >o ^ — i/i oo

r-l .. ..en .. c*l ...„ c ""> — > r^ ,oo-o

o in oo . w*> o •* t-

r vi

fsftt ii '"

™ 'JJ T

1 - ^»- 'i i

••£> vo \o 'C o r- r- (~~

o ..— —• .. , r-i (N„oo— .i^'to^-£ m . rn * oo — ^.

,. o" O .. vi' r- tt —

*

1/

1

tr r-~ rl t 'O t-i •sj- _ r-l f- 1H~- —

•o <o >o o -o r^ r~ r-- '-Xi oo oo a. o !3l O «• O

» * M N 'n o <•"

.. rl'' _ _ r-i «n *oo vo o r- in oo .. ©r-l .. © — .. , •* ol

-* ** oo" c< in'

i/^ c^» r-i

O-. © r- ^<

_ 1 - -^

L-J oc oc o O. o-. t— .-.

.— o --. i^ »-* ^-^ r- r-l n on

•i

c^ - o\' r-" - •* r- « k_'™. cl O , i— c-i tt O r

'\ ©Jf; ^ oC

'"' Tf oo

!'..' Hi in"* "' ' J "• ....

—T —''^" © —

i-• -s- P-. r~ rt^ n r-l v£> c?-. in >j> T~ ^^ I - <X\ —

••—

•

f> *—*

--^

r-l r-l n r-l r'i fO rl ei •^- tt ^j- -i xf m m in in so •O «5 * « M-- r- crj VI oo o O C* a> O-. O

«*<

4i ,

a H ^ *J- S

§ If 111I E -! £ t ^

^ o* o cr,J

u. w*+-.

up OS

U.I

a;*-'^

/.

O O O f> "U tj

.S -S ^ .5 -S T3

e w o2 "o

OT ^a oo Si H

-C- CO

<t0u,iij>(yiuujuj

.3 .2 •« .2 .3 " 49 c "> ST.

^ rS «» rg & 'g '5 g o y.

% % g '3 % S e I § <

42 ss

h- &- < < cd

sixt~

atwCQ

5;

?

illCQ 03 UJ

j m<* D

'

Hi u.

$ .'-': ,;J

1^ cii

ly_£

« CQ^

mi

i* P'' *- < J

.. CO li.O oD. 3

U

'is

cs'->

F

a. w

^

^

t.'M

JJ ^ m S S s I>

rl r-i r-l r-l

7 *? 7 T T~ n. l(J » a•tj- -^- **t ^r ^t

II!©Cf> C^ -3n t>

ie r- oo O •3- in \0 t~-c-i r-i r-i n ol r-i r-l r-l r-l

XXV11I

goN_;

H ^1

00 ol> D oCO u CQ

O 00 wr^ P X

H1

a? U-

Q5

o

CQrl

rooo

C g HO 3 s'J GO z

(/! UJ tu<*_ OO >O D

(W K oo oO O H w<

o,"J(J

00

3 <H K O oooo

5•u

J3 ££ J

w J= O CQtrt

S_1 o

c£ J odD ft o ft.

OO £

(A,UJ00

< 7r- K LUS-, H X

Oa.

W_J-J

<-0

Oasu<

O

<Q ^ OJ

a_ < s o

u 00 C/3

00 S 2 Hif_ p w wz t*i ~j £w ft

CQ 00

:>*?

O

UJ

§OoooO

<1

1a? O

a. O 5?? K i/i

< O r^

oo £> O

w 3-J 'Cm Si

< 03

H S

n « o>

© o" vo

* On r-i-<r

O <* 00 ..

*-• . .. r-l, r- © ©

r-l . .CN -=r

.*3- 00 .' n© ©

CO r -. CN r~«« K .

r-l .. . © r-i, C-- r-l T—

-i «* 00 o-1 , .. no- O <N O

y? 'O ff- -< r^ * O* «CO . . *— r-l . . >—

.. r-i <n" ~ .©".-".—ki via) .n «i 00 .

. r-" *3-" © . o>'d O

c-l. r'i

. ..© ON >0 00(N r. *

Ol {S"- O »*

00 •* 00 .(>

-oo"oCo™ -inooMmiC" „ "

.. <>. -O "

<) rO lO CJ- —• On K} r~

r~ CO 00 CO Ovc O NO CO

00 O "1r-i id rT

tt r- O cfr~ f'i r-l

r~ O r-i' ** r~-0--

0" -0 r-l r-l

*/i ro •0 .7. »—

1

•q f*1 1---

-*' •t 1-" •J1 * vi VI "ri

.. ^r — >& l~H ro-O 00 , CI n£>

r-l fc . O r~< m -Ov On —

'

- r-- toIT) rr r- — 1^ O -r> i»

r\l (S .r-l —• On .. NO-* [-- 0. — -^J- r^

r-ir-i -=t r-l - "*' -»

ci M O On '™f — ro ND

n- >» tj- * *

1^ -/'t 00 O fT] rn '^) 0\ ^_O ror-' NO

ON ri *'r

S (N r-l<*' O'' r-

"*- "* t~- * ^tOn *

On" r-" r-'fl r-i NO 0. "- f-1 fO NO —"* •* ^ 't 't vi '^•! /> vl

M T—

(

O .MM CM i>3

NT> C^ — r r^M co , r-c-l

"*' ™ -ef —

.

*'

o-''

Ul _ CJN „•0 _ r-

* r~

N£> r-i M Oi - NC" •J-"f-i ^ ~Xj ~ 1* en c^ -oNO NC- •6 no no r-^ t> r^

On O •^ co r^ NOo-. NC Y- *r * '-'!

C-l

iri On COrl

COr— NO On r^ 0- no" y-k tr~

*0 co 0"co r-iO '/I 0O

r- 2 * r-

in —

"

r> On' 0"NO •* 00 •-t "* ô-D NO NO NO NC- r^ r-^ r>

— . oO „ no -<— r~- - ** , coCO ,Mi-OM-r-IO -<* ,.Oa-*

C-J On

00 f^->r>ao .. On "«r >* VO

- * no" f-| r-T , in" no*

no NO* ,ovONOr--"r--: r-:

— .On-^ co m

XI

no"©"r-> .. ..

- 1- Oncn x-i 00— - . r-l„ CO Vi CM -—'ifr^'—r-l'-icONOr-l O

-t r-

,. oT

#

r-i

— .COON . . ,•*©* <?> o . — 00 —. —• m

OOCOOOOnOnO-OnOnO^

Mr- -t"* .. — rs . <rt.O . .On r~ O

r. 00 o . co r-l r-lfO-tf—.NOCOI-;0-—;—

•

ooooooOnOnOnOnOnOn.

ri Tj- I--

ON . ", ",-0 r- vO«Oi

«> . -Hr-l

"\ O CO NO

.NO .CO * On r- CO . r-CN „ O J C-IVl. ^r — O — —

1—* *t 00 *-•«-.r. r^ O ^ -* — *n •-< 'nOntt—; r; mrÔN -; «

COOOOOOnOÔnOnOnOn

O -r~ co r-

-» „ r-l 00 «->

O Tt —' cn .> NO - .COo » cn »o »n

. r-" <— . «n"o m . r-i -<t

, -o" o . r-f

in" o

r>On «5•<t CO

_TOn Onco r-

C-"CO COco rÔ O

NO "3- CO

. in'o

o -* 00

. o -"

Ha.

Pd

3 I

4> W _-" c yO <U

J 5 § <» 3 .5 S "5 -S -g Q-3 .2 Ji « -o i-o .*• ^ o -2 o

£ < u u. w >

IS

isa -* ^> *. J

vlOlUlU

3 ^

Q (5 p

EH

B E ga i5 °a u ^

IIIS S xi

.0 x> m Sj

£ -S •? ffl

ZXI « 13o a '5 s

u Ni- u B si a XWWOOUJOW•3 "3

T .0*

.s 1"5Cfl Ml 1

00 .£NO2

-0 XI £ U.3

S S-'-j £ 2 -J

22 <1HON-

Oli.

S s ill- 11 S

# CD

a »<« oo ft.

.seac g « £0

N« .fa O < Joc_> <2-ZS -

w § .S-51

%* B g<o o o .is «0- 5 2 a, wj

J r-l <nO- —• —

-

c-i r-i c-i

t .i/> f> ON On

•a- •* -Nj- -Q- -*t in

tr» 00 —• —NOoiS ^-1 f^ On — —«i-l to »i^ NO NO VO >0 no

t ;? o 7 7 3 o J- nA

odododoÔNONONONON

I 1 I

r~. ~~ lOOOO

NO C^ CO On no r^ 00 On o "«nfiniOr-MO>NPINHMNNNrl O'-Mn'ruiNOMBOi o — r-l

XXIX

Z5

IS a-* ts •* t

O * — -r, „

. o-7 10 ~ o :

o^-ta

3- oo , tr .

> Ch 1> t>

p <o

o3

2^

M £ £

u S^-5 ° B

Ui

< "' V

e hS f-

Q ^Q t3

z oa;

UJ-j SJ

u sSH O -E

Ii.

£ s g UJ*'

r . , o —

1««l > i>-c^O * O

flci

M-K3~&

'.

-or . *" i

i^'OO'- «' »' N I

^-1 -, .*

. o r-' .

M t oo - __-

-*" * r^ ~- —

.. r*> o r-J *t «/-» © .. — Tt o

o- ~ j; is o

HI O .

at;_>

OuQwr*i

< >

o<ai

Ou<

«j

U

s o

-o >/, ^ 3"

>v.

=5 e

ti 3o

< \i

-i ?,

>

">-

G

,

C

'-

ca

-:«?-• 3

."1-

t^ r

'

, o

— -* -fi

OO ~ <* ~.

" r-(s, * On <A 1"

o' * —, ,._- ^J

ci ri C\ — c~

®

o"O ,3 . .«->"

r-^ — £• oo *r, -*

vf «' ~' —. . o

3,

13,21.3

40.

49,

57,

73,

83,

92.

112.

122,

1

i;;2™ (M mnmmitT £12 £ £

S ..- CI —

co ~ oC " " t'

. ci , rM —< . "". + OO -=f

<r vt r t vi <o •?

5n •>* — O .OO «

***£>*© «d- *J-

r-( , _ •«-

— r oo •<* <N -„ o o — . r-

r-1 * — . <* >£i

Z> .ri t--'"*

2 « -oo'

S - 9 p „ «", w « & (S S * — i^ ™ (?i o> oo

5 *-. ~ « S 2 £ s^.-as".o-as-.-.a

.<^^> - '*v> ,!^, .. 00 O O* .. C7>o i« oo _ r- r-i _ <^* '1 <~; -: -: ~ <-j >r r • •- v, in

-1 T~ r fl o

« . ,.eiN .

. i~ oo — . 't .

— .. .. o— . - o

si

o

o

u. oc c § o Bi-g

ii> & 1

-S 3 |

c.

6 S

3 ia

I- 'r-

S .£

s e O|63 w

U4

3 c

;;

°6IS 11^

£ 2 -s's

c.g 3

"8^6c

I ill

fill

X

5 oi

8 9

b

3 5=

•1 —1 2 8

c.11

E E,3 .a

.-S £o o § §,a .S3 "c 'f-5 Jfi _.£ o

6 <.S

~ < w li. u > v> U Ui EM wia < <iy w s- aj K300U

s |g |

? 1

11[/> o-.5

.a .a tul-sli|SIl^il

c ""2

;d « 2 °^-i5

'!. ^ £E? ~ &%*>'§. o6* Q- c

t/S t/> UJS § g y o oH o £ < u S

5>S 5

I ^ .2 .2 £ G

.a ^ ||l>s g§ > > &;"S

sSx g a g 2

Tf AAf tt

T *»**' '

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


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,


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:


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° ^

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 educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

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


2k/M3kN

(b) Triangle rule:

We measure:

3kN

J? = 3.30kN, « = 66.6° R = 3.30k.N ^T 66.6° <

PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill .Companies, Inc. All rights reserved. No part of this Manual may be displayed,




;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


reproduced or distributed in any form 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 coursepreparation. Ifyou are a student using this Manual,


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,

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.

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°


reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the. limited

distribution to teachers and educatorspermittedby McGraw-Hillfar their individual coursepreparation. Ifyou are a student using this Manual,


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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Ato /«ri-/ o///»& Manual may be displayed,




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° <

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,




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°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,


distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyouarea student using this Manual,you are using it without permission.

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°



distribution lo teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,

you are using it withoutpermission.

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 <

PROPRIETARY MATERIAL. © 20 1 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

reproduced or distributed hi any form or by any means, without the prior written permission of the publisher, or used beyond the limited

distribution to teachers and educators permitted by McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,

you are using if without permission.

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





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° <

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 limiteddistribution to teachers ami educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

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

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,


distribution to teachers andeducators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


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



distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


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

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




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



distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manna I,


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





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



distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


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 <





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 <

PROPRIETARY MATERIAL. £) 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

reproduced or distributed in anyfonn 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 coursepreparation. Ifyou are a student using this Manual',


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 ^



distribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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,


distribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

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,




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,

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 educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

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 <



distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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,

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 permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

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° <

PROPRIETARY MATERIAL. © 20 10 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,


distribution to teachers andeducators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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

:


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° <





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


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° <





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° ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o /*//•/ <?/*//?« Manual may he 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 individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed


distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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,

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 educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

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

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,


distribution to teachers and educatorspermittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


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 ^

PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. Ail rights reserved. No pan of this Manual may be displayed,


distribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,


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 ^





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 ^



distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,

you are using il without permission.

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


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°

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All lights reserved. A'o />«// of this Manual may he displayed,




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



distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


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',




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


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 <



distribution to teachers and educatorspermittedby McGraw-Hillfor their individualcourse preparation. Ifyou are a student using this Manual,


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


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.

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

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,




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


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 coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

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





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 -«

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may he. displayed,




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 ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVo ,part «////«• jViamia/ may fee displayed,


distribution to teachers and educatorspermitted by McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


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

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed,




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 <



distribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


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


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


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 <«





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 <



distribution to teachers and educatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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


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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, hie. AH rights reserved. No pari of this Manual may be displayed,


distribution to teachers and educators permittedby McGraw-Hillfor their individualcourse preparation. Ifyou are a student using this Manual,


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. -^





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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No pari of (his Manual may be displayed,




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,


distribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


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




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,




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° •*

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayedreproduced 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.

75

o%.

PROBLEM 2.74



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°<





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° <





77

DM.

PROBLEM 2.76



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° <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'<> />«// o/rtw Manual may be displayed,




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,




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,




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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVo /wrt o/rfiw A/ai»«rf ///«>• Ae displayed,



yon are using it without permission.

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

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of/his Manual may be displayed,




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 <

PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




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 ««


reproduced or distributed in any form or by any means, without lite prior written permission of (he publisher, or used beyond (he limited

distribution to teachers and educators permitted by McGraw-Hillfor (heir individual course preparation. Ifyou are a student using this Manual,


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 ^





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° «





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 <





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 <



distribution to teachers and educators permittedby McGraw-Hillfor their individualcourse preparation. Ifyou are a student using this Manual,


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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,




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 <

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of (his Manual may be displayed,


distribution to teachers and educators'permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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

PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




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<



distribution to teachers and educatorspermitted by McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


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° <





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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No pari of this Manual may be displayed,




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

PROPRIETARY MATERIAL. «5 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,


distribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed,


distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


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° <

PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All righls reserved. No part of litis 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. limited



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^

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. Afa par/ o//A* Minna/ way fe <%,/«»•«/,reproduced or distributed m anyform or by any means, without the prior written permission of the publisher, or used bevond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

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

PROPRIETARY MATERIAL. €) 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




102

PROBLEM 2.100


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 <


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 individualcourse preparation. Ifyou are a student using this Manual,you are using it without permission.

103

PROBLEM 2.101


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

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed


distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coarsepreparation. Ifyou are a student using this Manual,


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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, foe. 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 permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o part of this Manual may be displayed,




106


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 <

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 (he prior written permission of the publisher, or used beyond the limited

distribution to teachers and educatorspermittedby McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual,


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 <



distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student fixing this Manual,


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 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manila} may be displayed,


distribution to teachers andeducatorspermittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

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





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 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,


distribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are. using it without permission.

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.

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No pari of this Manual may be displayed,

reproduced or distributed in anyform or by any means, without the prior whiten permission of the publisher, or used beyond the limited

distribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyon are a student using this Manual,


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


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 ^

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




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 <





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

PROPRIETARY MATERIAL © 2010 the McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




116


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 ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> par/ o//Aw 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 educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manualyou are using it without permission.

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





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 <

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,




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 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of (his Manual may be displayed,




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 <





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

PROPRIETARY MATERIAL CO 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,




122


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 ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. Afo /ww-f o//A& it^mtia/ »«iy fa? displayed,




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 ^

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of (his Manual may be displayed




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 ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o par/ tf/Afc MaW »«p Ae <%,AyWreproduced or distributed m 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 individual coursepreparation. Ifyou are a student mine this Manualyou are using it without permission.

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



distribution to teachers andeducators permittedby McGraw-Hillfar their individual course preparation. Ifyou are a student using this Manual,


126


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 ^

PROPRIETARY NUIkRIAL ©2010 The McGraw-Hill Companies, Inc. Ail rights reserved. M, /*»/ «///,*M^^ * tf^wXr^TflS

V rryMm °'' h}'T 'mam

-wUh0Ut 'hepri°r Wrt"e" /*"***» of the publisher, or used beyond the limiteddilution to teachers and^mtors permittedby McCmiw-Hillfor their individual an,rs^

you are using it without permission.' '

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<

PROPRIETARY MATERIAL. £> 2010 The McGraw-Hill Companies, inc. Ail rights reserved. No part of this Manual may be displayed,




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)

PROPUhlARY MATERIAL. C 2010 The McGraw-Hill Companies. Inc. All rights served. Ato^rf «///,/, AAaiwa/w fc dtofa,*/reproduced or distributed in any jorm or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfar their individual course preparation. Ifyou are a student mine this Manualyou are using it without permission.

' *

129


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





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

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of litis Manual may be displayed,




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)





132


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 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayedreproduced 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 individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

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 <





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 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All lights reserved. No part of this Manual may be displayed,


distribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,yon are using it withoutpermission.

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. <





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° <



distribution to teachers and educators permittedby McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

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° <





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 <

PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual mav 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 educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manualyou are using it without permission.

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 <



distribution to teachers and educatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


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 <


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<«

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. M> /)«// of this Manual may be displayed,




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 <«

PROPRIETARY MATERIAL. (0 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayedreproduced or distributed in anyjorm or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-imifor their individual course preparation. Ifyou are a studentyou are using it without permission.

' *

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° <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo pcw7 o/r/iw Manual may be displayed,




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° <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ait rights reserved. No pari of this Manual may be displayed,reproduced or distributed in anyJarm 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 Manualyou are using it without permission.

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

Âli ~ ÂBÂli ~" *A8AB

AB

(0.6i + 0.8j)7^

450. 600.1 + 1 \TAl,

750 750 '

3fco

£-00

J -T X -T *£-(*™i-™k« -J/" — / AffÂf — * AS' —I j ».MC ~-*^C /l

'/IC '/it:

*AD ~ -

' ADÂD ~~ '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)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. /Vo /«w/ <>///?« Manual may be displayed,

reproduced or distributed in anyform or by any means, without (he prior written permission of the publisher, or used beyond the limited



146


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 ^





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



distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using (his Manual,


148


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

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 limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using ii without permission.

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 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed,




ISO

solucionario berr

Documents