Vector Mechanics for Engineers, Statics and Dynamics, 12th Edition Solution Manual by TBtutors

Type:

Solution Manual

Resource:

Vector Mechanics for Engineers, Statics and Dynamics

Edition:

12th Edition

Author(s):

Ferdinand P. Beer E. Russell Johnston David Mazurek Brian Self

Instructor’s and Solutions Manual to accompany

Vector Mechanics for Engineers, Dynamics Twelfth Edition Ferdinand P. Beer Late of Lehigh University

E. Russell Johnston, Jr. University of Connecticut

Phillip J. Cornwell Rose-Hulman Institute of Technology

Brian P. Self California Polytechnic State University – San Luis Obispo

Prepared by Charles B. Birdsong California Polytechnic State University – San Luis Obispo

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw-Hill Education and protected by copyright and other state and federal laws. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

TO THE INSTRUCTOR As indicated in its preface, Vector Mechanics for Engineers: Dynamics is designed for a first course in dynamics. 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 that have been included, this text can also be used to teach a course that 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. To assist instructors in making up schedules 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 maximum number of periods have been suggested, and the topics which form the standard basic course in dynamics have been separated from those that are optional. The total number of periods required to teach the basic material varies from 27 to 48, while covering the entire text would require from 40 to 67 periods. In most instances, of course, the instructor will want to include some, but not all, of the additional material presented in the text. If allowance is made for the time spent for review and exams, it is seen that this text is equally suitable for teaching the standard basic dynamics course in 40 to 45 periods and for teaching a more complete dynamics course to advanced students. In addition, it should be noted that Statics and Dynamics can be used together to teach a combined 4- or 5-credit-hour course covering all the essential topics in dynamics as well as those sections of statics that are prerequisites to the study of dynamics. 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 they may have brought up in discussing a problem assigned for homework. Educational research has shown that students can often choose appropriate equations and solve algorithmic problems without having a strong conceptual understanding of mechanics principles. To help assess and develop student conceptual understanding, we have included Concept Questions, which are multiple choice problems that require few, if any, calculations. Each possible incorrect answer typically represents a common misconception (e.g., students often think that a vehicle moving in a curved path at constant speed has zero acceleration). Students are encouraged to solve these problems using the principles and techniques discussed in the text and to use these principles to help them develop their intuition. Mastery and discussion of these Concept Questions will deepen students’ conceptual understanding and help them to solve dynamics problems. Drawing diagrams correctly is a critical step in solving kinetics problems in dynamics. A new type of problem has been added to the text to emphasize the importance of drawing these diagrams. In Chapters 12 and 16 the Free-Body Practice Problems require students to draw a

free-body diagram (FBD) showing the applied forces and an equivalent diagram called a “kinetic diagram” (KD) showing ma (or its components) and I a. These diagrams provide students with a pictorial representation of Newton’s second law and are critical in helping students to correctly solve kinetic problems. In Chapters 13 and 17 the Impulse-Momentum Practice problems require students to draw diagrams showing the momenta of the bodies before impact, the impulses exerted on the body during impact, and the final momenta of the bodies after the impact. A group of problems designed to be solved with computational software, analyses of the problems, problem solutions and output for the most widely used computational programs can be found through the instructor resources available within Connect. 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 desirable. To illustrate this point, sample lesson schedules are shown in Tables III and IV, together with various alternative lists of assigned problems. Half of the problems in each of the six lists suggested in Table III 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 IV are stated in SI units and 25% in U.S. customary units. Because the approach used in this text differs in a number of respects from the approach used in other books, instructors will be well advised 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 Professor Charles B. Birdsong of California Polytechnic State University for his careful preparation of the solutions contained in this manual. Phillip J. Cornwell Brian P. Self

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DESCRIPTION OF THE MATERIAL CONTAINED IN VECTOR MECHANICS FOR ENGINEERS: DYNAMICS, Twelfth Edition Chapter 11 Kinematics of Particles In this chapter, the motion of bodies is studied without regard to their size; all bodies are assumed to reduce to single particles. The analysis of the effect of the size of a body and the study of the relative motion of the various particles forming a given body are postponed until Chap. 15. In order to present the simpler topics first, Chap. 11 has been divided into two parts: rectilinear motion of particles, and curvilinear motion of particles. In Sec. 11.1A, position, velocity, and acceleration are defined for a particle in rectilinear motion. They are defined as quantities which may be either positive or negative and students should be warned not to confuse the position coordinate and distance traveled, or velocity and speed. The significance of positive and negative acceleration should be stressed. Negative acceleration may indicate a loss in speed in the positive direction or a gain in speed in the negative direction. As they begin the study of dynamics, many students are under the belief that the motion of a particle must be either uniform or uniformly accelerated. To destroy this misconception, the motion of a particle is first described under very general conditions, assuming a variable acceleration which may depend upon the time, the position, or the velocity of the particle (Sec. 11.1B). To facilitate the handling of the initial conditions, definite integrals, rather than indefinite integrals, are used in the integration of the equations of motion. The special equations related to uniform and uniformly accelerated motion are derived in Secs. 11.2A and 11.2B. Before using these equations, students should be warned to check carefully that the motion under consideration is actually a uniform or a uniformly accelerated motion. Two important concepts are introduced in Sec. 11.2C: (1) the concept of relative motion, which will be developed further in Secs. 11.4D, 15.2B and (2) the concept of dependent motions and degrees of freedom. The first part of Chap. 11 ends with the presentation of several graphical methods of solution of rectilinear-motion problems (Sec. 11.3). This material is optional and may be omitted. Several problems in which the data are given in graphical form have been included. The second part of the chapter begins with the introduction of the vectors defining the position, velocity and acceleration of a particle in curvilinear motion. The derivative of a vector function is defined and introduced at this point (Sec. 11.4B). The motion of a particle is first studied in terms of rectangular components (Sec. 11.4C); it is shown that in many cases (for example, projectiles) the study of curvilinear motion can be reduced to that of two independent rectilinear motions. The concept of fixed and moving frames of reference is introduced in Sec. 11.4D and is immediately used to treat the relative motion of particles.

The use of tangential and normal components, and of radial and transverse components is discussed in Secs. 11.5. Each system of components is first introduced in two dimensions and then extended to include three-dimensional space.

Chapter 12 Kinetics of Particles: Newton’s Second Law As indicated earlier, this chapter and the following two are concerned only with the kinetics of particles and systems of particles. They neglect the effect of the size of the bodies considered and ignore the rotation of the bodies about their mass center. The effect of size will be taken into account in Chaps. 16 through 18, which deal with the kinetics of rigid bodies. Sec. 12.1 presents Newton’s second law of motion and introduces the concept of a newtonian frame of reference. In Sec. 12.1B the concept of linear momentum of a particle is introduced, and Newton’s second law is expressed in its alternative form, which states that the resultant of the forces acting on a particle is equal to the rate of change of the linear momentum of the particle. Section 12.1C reviews the two systems of units used in this text, the SI metric units and the U.S. customary units which were previously discussed in Sec. 1.3. This section also emphasizes the difference between an absolute and a gravitational system of units. A number of problems with two degrees of freedom have been included (Problems 12.31 through 12.35), some of which require a careful analysis of the accelerations involved (see Sample Problem 12.5). Section 12.1D applies Newton’s second law to the study of the motion of a particle in terms of rectangular components, tangential and normal components, and radial and transverse components. The term inertia vector is used in preference to inertia force or effective force to avoid any possible confusion with actual forces. In Sec. 12.2 the concept of angular momentum of a particle is introduced, and Newton’s second law is used to show that the sum of the moments about a point O of the forces acting on a particle is equal to the rate of change of the angular momentum of the particle about O. Section 12.2B considers the particular case of the motion of a particle under a central force. The early introduction of the concept of angular momentum greatly facilitates the discussion of this motion. Section 12.2C presents Newton’s law of gravitation and its application to the study of the motion of earth satellites. Section 12.3 is optional. Section 12.3A derives the differential equation of the trajectory of a particle under a central force, while Sec. 12.3B discusses the trajectories of satellites and other space vehicles under the gravitational attraction of the earth. While the general equation of orbital motion is derived (Eq. 12.36), its application is restricted to launchings in which the velocity at burnout is parallel to the surface of the earth. (Oblique launchings are considered in Sec. 13.2D.) The periodic time is found directly from the fundamental definition of areal velocity rather than by formulas requiring a previous knowledge of the properties of conic sections.

Instructors may omit Section 12.3 and yet assign a number of interesting space mechanics problems to their students after they have reached Sec. 13.2D. Chapter 13 Kinetics of Particles: Energy and Momentum Methods After a brief introduction designed to give to students some motivation for the study of this chapter, the concept of work of a force is introduced in Sec. 13.1A. The term work is always used in connection with a well-defined force. Three examples considered are the work of a weight (i.e., the work of the force exerted by the earth on a given body), the work of the force exerted by a spring on a given body, and the work of a gravitational force. Confusing statements such as the work done on a body, or the work done on a spring are avoided. The concept of kinetic energy is introduced in Sec. 13.1B and the principle of work and energy is derived by integration of Newton’s equation of motion. In applying the principle of work and energy, students should be encouraged to draw separate sketches representing the initial and final positions of the body (Sec. 13.1C). Section 13.1D introduces the concepts of power and efficiency. Section 13.2 is devoted to the concepts of conservative forces and potential energy and to the principle of conservation of energy. Potential energy should always be associated with a given conservative force acting on a body. By avoiding statements such as “the energy contained in a spring” a clearer presentation of the subject is obtained, which will not conflict with the more advanced concepts that students may encounter in later courses. In applying the principle of conservation of energy, students should again be encouraged to draw separate sketches representing the initial and final positions of the body considered. In Sec. 13.2D, the principles of conservation of energy and conservation of angular momentum are applied jointly to the solution of problems involving conservative central forces. A large number of problems of this type, dealing with the motion of satellites and other space vehicles, are available for homework assignment. As noted earlier, these problems (except the last two, Probs. 13.117 and 13.118) can be solved even if Section 12.3 has been omitted. The second part of Chap. 13 is devoted to the principle of impulse and momentum and to its application to the study of the motion of a particle. Section 13.3 introduces the concept of linear impulse and derives the principle of impulse and momentum from Newton’s second law. The instructor should emphasize the fact that impulses and momenta are vector quantities. Students should be encouraged to draw three separate sketches when applying the principle of impulse and momentum and to show clearly the vectors representing the initial momentum, the impulses, and the final momentum. It is only after the concept of impulsive force has been presented that students will begin to appreciate the effectiveness of the method of impulse and momentum (Sec. 13.3B). Direct central impact and oblique central impact are studied in Section 13.4. Note that the coefficient of restitution is defined as the ratio of the impulses during the period of restitution and the period of deformation. This more basic approach will make it possible in Chapter 17 to extend the results obtained here for central impact to the case of eccentric impact. Emphasis vi

should be placed on the fact that, except for perfectly elastic impact, energy is not conserved. Note that the discussion of oblique central impact in Sec. 13.4B has been expanded to cover the case when one or both of the colliding bodies are constrained in their motions. Section 13.4C shows how to select from the three fundamental methods studied in Chaps. 12 and 13 the one best suited for the solution of a given problem. It also shows how several methods can be combined to solve a given problem. Note that problems have been included (Probs. 13.177 through 13.189) which require the use of both the method of energy and the method of momentum in their solutions.

Chapter 14 Systems of Particles Chapter 14 is devoted to the study of the motion of systems of particles. Sections 14.1A and 14.1B derive the fundamental equations (14.10) and (14.11) relating, respectively, the resultant and the moment resultant of the external forces to the rate of change of the linear and angular momentum of a system of particles. Sections 14.1C and 14.1D are devoted, respectively, to the motion of the mass center of a system and to the motion of the system about its mass center. Section 14.1E discusses the conditions under which the linear momentum and the angular momentum of a system of particles are conserved. Sections 14.2A and 14.2B deal with the application of the workenergy principle to a system of particles, and in Sec. 14.2C the application of the impulsemomentum principle is discussed. A number of challenging problems have been provided to illustrate the application of the principles discussed in this chapter. The first group of problems (Probs. 14.1 through 14.30) deal chiefly with the conservation of the linear momentum of a system of particles and with the motion of the mass center of the system, while the second group of problems (Probs. 14.31 through 14.58) involve the combined use of the principles of conservation of energy, linear momentum, and angular momentum. However, the main purpose of these sections is to lay the proper foundation for the later study of the kinetics of rigid bodies (Chaps. 16 through 18). Depending upon the preparation and interest of the students, a greater or lesser emphasis may be placed on this part of the course. It is essential, however, that the significance of Eqs. (14.16) and (14.23) be pointed out to students, in view of the role played by these equations in the study of the motion of rigid bodies. The instructor should note the distinction made in Sec. 14.1A between equivalent systems of forces (i.e., systems of forces which have the same effect) and equipollent systems of forces (i.e., systems of forces which have the same resultant and the same moment resultant). The equivalence of two systems of forces has been indicated in diagrams by red equals signs, and their equipollence by blue equals signs. Section 14.3 is optional. It is devoted to the study of variable systems of particles, with applications to the determination of the forces exerted by deflected streams and the thrust of propellers, jet engines, and rockets. Since Newton’s second law F = ma was stated for a particle with a constant mass and does not apply, in general, to a system with a variable mass, the

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derivations given in Sec. 14.3A for a steady stream of particles and in Sec. 14.3B for a system gaining or losing mass are based on the consideration of an auxiliary system consisting of unchanging particles. This approach should give students a basic understanding of the subject and lead them to more advanced courses in mechanics of fluids. Chapter 15 Kinematics of Rigid Bodies With this chapter we start the study of the dynamics of rigid bodies. After an introduction in which the fundamental types of plane motion are defined, the relations defining the velocity and the acceleration of any given particle of a rigid body are established for two particular cases: translation (Sec. 15.1A) and rotation about a fixed axis (Secs. 15.1B and 15.1C). In Sec. 15.2 it is shown that the most general plane motion can always be considered as the sum of a translation and a rotation. This property is established by considering the relative motion of two particles of the rigid body and is immediately applied in Sec. 15.2B to the determination of velocities. Section 15.3 introduces the concept of the instantaneous center of rotation. The instructor should stress the fact that, while the instantaneous center method simplifies the solution of many problems involving velocities, it cannot be used to determine accelerations. In Sec. 15.4 the concept of relative motion is used again, this time to determine accelerations in plane motion. Students should be warned against any unwarranted assumptions concerning the direction of unknown accelerations. Section 15.4B is optional. It presents an analytical method for the determination of velocities and accelerations based on the use of a parameter. Section 15.5A discusses the rate of change of a vector with respect to a rotating frame, and Sec. 15.5B applies the results obtained to the determination of Coriolis acceleration in plane motion. It should be kept in mind, however, that Coriolis acceleration does not depend upon the existence of an actual slab or rod on which the particle moves. It can appear whenever rotating axes are used. There lies the fundamental difference between the approach of Sec. 15.5 and that of Secs. 11.4D and 15.4A where moving axes of fixed direction were used. The remaining sections of Chap. 15 are devoted to the kinematics of rigid bodies in three dimensions and are optional. However, they should be included in a course covering the kinetics of rigid bodies in three dimensions (Chap. 18) and should be taught either at this point or immediately before Chap. 18. In Sec. 15.6A the motion of a rigid body about a fixed point is presented. Students should note that while the finite rotations of a rigid body have magnitude and direction, the rotations are not vectors; on the other hand, both angular velocity and angular acceleration are vector quantities. In Sec. 15.6B, the general motion of a rigid body is analyzed. It should be emphasized that in this section the moving frame of reference is of fixed orientation and does not rotate. Section 15.7 considers the motion of particles and rigid bodies with respect to rotating frames of reference and extend the concept of Coriolis acceleration to three-dimensional motion. viii

Chapter 16 Plane Motion of Rigid Bodies: Forces and Accelerations This chapter is devoted to the plane motion of rigid bodies which consist of plane slabs or which are symmetrical with respect to the reference plane. Cases involving the plane motion of nonsymmetrical bodies and, more generally, the motion of rigid bodies in three dimensions are considered in Chap. 18. If the determination of mass moments of inertia has not been covered in the previous statics course, the instructor should include material from Appendix B at this point. In Sec. 16.1 the fundamental relations derived in Chap. 14 for a system of particles are used to show that the external forces acting on a rigid body are equipollent to the vector ma attached at  . This result, which is illustrated in the mass center G of the body and the couple of moment H G Fig. 16.3, is valid in the most general case of motion of a rigid body (three-dimensional as well as plane motion). It is shown in Sec. 16.1B that in the case of the plane motion of a slab or symmetrical body, the angular momentum H G reduces to I w and its rate of change to I a . It is shown that the external forces acting on a rigid body are actually equivalent to the inertial terms represented by the vector ma and the couple I a . As noted in Sec. 16.1D, this result is obtained independently of the principle of transmissibility and can be used to derive this principle from the other axioms of mechanics. At this point students will have reached the climax of their study of rigid-body motion in two dimensions. Indeed, they can solve any problem by drawing two sketches—one showing the external forces, the other the vector ma and the couple I a —and then by expressing that the two systems of vectors shown are equivalent. The various types of plane motion problems have been grouped according to their kinematic characteristics. Translation, centroidal rotation, and plane motion consisting of a translation and an unrelated centroidal rotation are considered first, since they are the simplest ones to analyze. They are followed by plane motions with various kinematic constraints: non-centroidal rotation, rolling motion, and other types of plane motion. Problems involving systems of rigid bodies have been included at the end of this chapter (Probs. 16.126 through 16.141). The instructor should stress the fact that, in spite of the different kinematic characteristics of these various motions, the approach to the kinetics of the motion is consistently the same: all problems are solved by drawing two sketches—one showing the external forces, the other the vector ma and the couple I a —and then expressing that the two systems of vectors are equivalent. Since the approach used in this text differs from others in the emphasis placed on the direct application of Newton’s second Law, rather than on specialized formulas, it might be appropriate at this point to summarize the advantages derived from this approach. (1) A single method is emphasized, which applies to all cases of plane motion, regardless of their kinematic characteristics, and which can be used safely under any conditions. This is in contrast to using the equation SM = Ia which is limited in its applications.

(2) By stressing the use of the free-body diagram and kinetic diagram, this method provides a better understanding of the kinetics of the motion. There will be little danger, for example, in the solution of a problem of non-centroidal rotation, that students will forget the effect of the acceleration of the mass center on the reaction at the fixed point, a mistake which occurs frequently when the specialized formula SMO = IO a is used. (3) The method used divides the solution of a problem into two main parts, one in which the kinematic and kinetic characteristics of the problem are considered (separately if necessary), and the other in which the methods of statics are used. In this way the techniques of each separate field can be used most efficiently. For example, moment equations can be written to eliminate unwanted reactions, just as it was done in statics; this can be done independently of the kinetic characteristics of the problem. (4) By resolving every plane motion (even a non-centroidal rotation) into a translation and a centroidal rotation, a unified approach is obtained, which will also be used in Chap. 17 with the method of work and energy and with the method of impulse and momentum, and which will be extended in Chap. 18 to the study of the three-dimensional motion of a rigid body. This approach is a basic one, which can be applied effectively throughout the study of mechanics in advanced courses as well as in elementary ones. Chapter 17 Plane Motion of Rigid Bodies: Energy and Momentum Methods The first portion of the chapter extends the method of work and energy, already used in Chap. 13, to the study of the plane motion of rigid bodies. The expressions for the work of a couple and for the kinetic energy of a rigid body are derived in Secs. 17.1B and 17.1C. Using the results obtained in Sec. 14.2A, the kinetic energy of a rigid body is separated into a translational part and a rotational part (about the mass center). The authors believe that, while it may lead to slightly longer solutions, this method is more fundamental and should be used in preference to special formulas. Indeed, it follows the basic idea of resolving every plane motion into a translation and a centroidal rotation. It is shown in Sec. 17.1D that the method of work and energy is especially effective in the case of systems of rigid bodies connected by pins, inextensible cords, and meshed gears. In Sec. 17.1E the principle of conservation of energy is used to analyze the plane motion of rigid bodies. In the second part of Chap. 17, the method of impulse and momentum is extended to the study of the motion of rigid bodies. The approach used is different from that of most elementary textbooks. Ready-to-use formulas are avoided; instead, students are taught to express the general principle of impulse and momentum by means of impulse-momentum diagrams and to write the equations most appropriate to the solution of the problem considered. The results obtained in Sec. 14.2C for a system of particles are directly applicable to the system of particles forming a rigid body and can be used to analyze the plane motion of rigid bodies. It is shown in Sec. 17.2A that the momenta of the various particles forming a rigid body reduce to a vector mv and a couple I w in the most general case of plane motion.

While the principle of conservation of angular momentum is discussed in Sec. 17.2C because of its physical and historical significance, it is not actually used in the solution of problems. To solve any problem, regardless of the type of motion, and whether it involves constant forces of finite magnitude applied for a finite time or impulsive forces applied for a very short time interval, students are told to draw three separate sketches showing, respectively, the initial momenta, the impulses of the external forces, and the final momenta. The momenta of a rigid body are represented in the most general case by a momentum vector mv attached at the mass center and a momentum couple I w . If students then consider the components of the vectors involved, they obtain relations between linear impulses and linear momenta. If they consider the moments of the same vectors, they obtain angular impulses and angular momenta. If, by equating moments about a point such as a pivot, they obtain an equation which does not involve any of the external forces, they will have automatically established conservation of angular momentum about that point. The advantages derived from this approach can be summarized as follows: (1) Students learn only one method of solution, a method based directly on a fundamental principle and which can be used safely under any conditions. This is in contrast with the equation å M0Dt = I0 (w2 - w1 ) , which is limited in its applications. (2) The method stresses the use of impulse and momentum diagrams and thus provides a better understanding of the kinetics of the motion. It is unlikely, for example, that students will forget an impulsive reaction at a fixed support. (3) Students use the basic tools they learned in statics: reduction of a system of vectors to a vector and a couple and equations relating the components or the moments of these vectors. (4) Again, the same unified approach is used: every plane motion is resolved into a translation and a centroidal rotation. In Chap. 18 this approach will be extended to the solution of problems involving the three-dimensional motion of rigid bodies. Some teachers may fear that the inclusion of momentum vectors and momentum couples in the same diagrams may lead to confusion. This will not be the case, however, if students are instructed to write separate equations involving either components or moments, as they did in statics. The first equations will contain linear impulses and linear momenta expressed in N⋅s or lb⋅s, and the latter angular impulses and angular momenta expressed in N⋅m⋅s or lb⋅ft⋅s. Section 17.3 is devoted to the eccentric impact of two rigid bodies, a topic seldom included in an elementary text. No special difficulty will be encountered, however, if separate sketches are used as indicated above. Chapter 18 Kinetics of Rigid Bodies in Three Dimensions In this chapter the restrictions imposed in preceding chapters (e.g., plane motion, symmetrical bodies) are lifted and students proceed to the analysis of more general (and more difficult)

problems, such as the rotation of nonsymmetrical bodies about fixed axes and the motion of gyroscopes. In Sec. 18.1, the general result obtained in Sec. 16.1 is recalled, namely, that the external forces acting on a rigid body are equipollent to the vector ma attached at the mass center G of the body  . It is also pointed out that the main feature of the impulseand the couple of moment H G momentum method, namely, the reduction of the momenta of the particles of a rigid body to a linear momentum vector mv attached at G and an angular momentum, remains valid and that the work-energy principle and the principle of conservation of energy still apply in the case of the motion of a rigid body in three dimensions. The difficulties encountered in the study of the three dimensional motion of a rigid body are related to the determination of the angular  , and of the kinetic energy of the body. momentum H G , of its rate of change H G The determination of the angular momentum H G of a rigid body from its angular velocity w is discussed in Sec. 18.1A. Since this requires the use of mass products of inertia, as well as the use of mass moments of inertia, the instructor should cover the material from Appendix B (or from the second part of Chap. 9) if this material has not been included in the previous statics course. Section 18.1B is devoted to the application of the impulse-momentum principle to the threedimensional motion of a rigid body, and Sec. 18.1C to the determination of its kinetic energy. In Secs. 18.2A and 18.2B, the rate of change of the angular momentum H G is computed and the equations of motion for a rigid body in three dimensions are derived. Newton’s second law is extended to the case of three-dimensional motion by showing that the external forces are actually  . Sections 18.2C equivalent to the inertial terms represented by the vector ma and the couple H G and 18.2D are devoted to the particular cases of the motion of a rigid body about a fixed point and the rotation of a rigid body about a fixed axis, with applications to the balancing of rotating shafts. While Euler’s equations of motion have been derived, it should be noted that the more fundamental vector relations represented by Equations (18.22), (18.23), and (18.28) are used in the solution of problems. The remaining portion of this chapter (Secs. 18.2D through 18.3C) is designed for advanced students and, in general, should be omitted for ordinary classes. In Secs. 18.3A and 18.3B the motion of a gyroscope is considered. At this point Eulerian angles are introduced. It should be carefully noted that the rotating system of axes Oxyz is attached to the inner gimbal; these axes are principal axes of inertia and they follow the precession and nutation of the gyroscope; they do not, however, spin with the gyroscope. The special case of steady precession is considered in Sec. 18.3B. Several problems dealing with the steady precession of a top and other axisymmetrical bodies have been included and, in one of them (Prob. 18.110), it is shown that the formula usually given in introductory texts is only approximate.

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The motion of an axisymmetrical body under no force (Sec. 18.3C) introduces students to one of the most interesting aspects of classical dynamics—an aspect which has gained widespread attention in recent years due to the interest in space vehicles and artificial satellites. In this connection, it should be pointed out that Poinsot’s theory of the motion of a nonsymmetrical body under no force may be covered by assigning Probs. 18.143 through 18.145. An additional problem (Prob. 18.146) relates to the stability of the rotation of a nonsymmetrical body about a principal axis. Chapter 19 Mechanical Vibrations This chapter provides an introduction to the study of mechanical vibrations. While only onedegree-of-freedom systems are included, all the basic principles are presented. The various topics covered are as follows:

(a) Free, undamped vibrations of a particle (Sec. 19.1A). The differential equation characterizing simple harmonic motion is derived and all basic terms, such as period, natural frequency, and amplitude, are defined. Both the analytical and the geometrical methods of solution are described. It is shown in Sec. 19.1B that the motion of a simple pendulum can be approximated by a simple harmonic motion. Section 19.1C, which is optional, shows how an exact solution can be obtained for the period of oscillations of a simple pendulum. (b) Free, undamped vibrations of a rigid body. The principle of equivalence of the systems of applied forces and inertial terms is first used to determine the natural frequency and the period of oscillations of a rigid body (Sec. 19.2). Note that the same positive sense is assumed for the angular acceleration and displacement. The principle of conservation of energy is then used to solve the same type of problems (Sec. 19.3). (c) Forced, undamped vibrations of a particle (Sec. 19.4). This section introduces students to the concepts of forced frequency, transient and steady-state vibrations, and resonance. While all students will be able to understand this section, those with a knowledge of elementary differential equations will derive a greater benefit from it since it provides a direct application of the solution of linear nonhomogeneous equations with constant coefficients.

(d) Free, damped vibrations of a particle (Sec. 19.5A), and (e) Forced, damped vibrations of a particle (Sec. 19.5B). These two sections take into account the effect of friction and thus provide a more rigorous analysis of the vibrations of a particle. They are not recommended, however, to students who do not possess a basic knowledge of elementary differential equations. In Sec. 19.5C the electrical analogue for a vibrating mechanical system is discussed; this section is optional and should not be assigned unless Secs. 19.5A and 19.5B have been covered.

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TABLE I. LIST OF TOPICS COVERED IN VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Suggested Number of Periods Additional Sections

Topics

Basic Course

Topics

11. KINEMATICS OF PARTICLES 11.1 11.2 *11.3 11.4 11.5 11.5B

Rectilinear Motion Special Cases and Relative Motion Graphical Methods Curvilinear Motion (Rectangular Components) Tangential and Normal Components Radial and Transverse Components

1 1 1 1-2 0.5-1 0.5-1

12. KINETICS OF PARTICLES: NEWTON'S SECOND LAW 12.1

Newton’s Second Law and Linear Momentum

12.2

Angular Momentum and Orbital Mechanics

*12.3

Applications of Central-Force Motion

1-2 1-2 1-2

13. KINETICS OF PARTICLES: ENERGY AND MOMENTUM METHODS 13.1 13.2 13.3 13.4

Work and Energy Conservation of Energy; Central Forces Impulse and Momentum; Impulsive Motion Impact; Mixed Problems

1-2 1-2 1-2 1-2

14. SYSTEMS OF PARTICLES 14.1

Motion of Systems of Particles

0.5-1

14.2

Energy and Momentum Methods for Systems of Particles

0.5-1

*14.3

Variable Systems of Particles

1-2

15. KINEMATICS OF RIGID BODIES 15.1 15.2 15.3 15.4 15.5 *15.6 *15.7

Translation; Rotation About a Fixed Axis Velocities in Plane Motion Instantaneous Center Accelerations in Plane Motion Motion With Respect to a Rotating Frame Motion of a Rigid Body in Space Motion Relative to a Moving Reference Frame

1-1.5 1-1.5 1 1-2 1-2

1-2 1-2

16. PLANE MOTION OF RIGID BODIES: FORCES AND ACCELERATIONS 16.1

Kinetics of a Rigid Body in Plane Motion

2-3

16.2

Constrained Plane Motion

2-4

17. PLANE MOTION OF RIGID BODIES: ENERGY AND MOMENTUM METHODS 17.1

Energy Methods for a Rigid Body

1-2

17.2

Momentum Methods for a Rigid Body

1-2

17.3

Eccentric Impact

1-2

xiv

0.5

18. KINETICS OF RIGID BODIES IN THREE DIMENSIONS *18.1

Energy and Momentum in Three Dimensions

*18.2

Motion of a Rigid Body in Three Dimensions

1-2

*18.3

Motion of a Gyroscope

19. MECHANICAL VIBRATIONS 19.1

Vibrations of Particles without Damping

1-2

*19.1C

Exact Solution of a Simple Pendulum

19.2

Free Vibrations of Rigid Bodies

0.5

19.3

Energy Methods

19.4

Forced Vibrations

*19.5

Damped Vibrations

1-2

*19.5C

Electrical Analogues

1-2 1-1.5 1-1.5

Total Number of Periods

_____

27-48

13-19

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS

Problem Number* SI Units

U.S. Units

Problem Description

CHAPTER 11: KINEMATICS OF PARTICLES RECTILINEAR MOTION OF PARTICLES Analyze the motion of a particle given 11.3,4

11.1,2

position = f(t)

11.5,7 11.6,8 11.11,12

also find total distance traveled

11.9,10

acceleration = f(t)

11.16,17

acceleration = f(x)

11.22,23

acceleration = f(v)

11.28

velocity = f(x)

11.29,30

acceleration µ 1 / r

11.13,14 11.15,18 11.19,20 11.21,24 11.25,26 11.27

11.31

11.32

Simple harmonic motion

11.33,34

11.35,36

motion of one particle

11.39,40

11.41,42

motion of two particles

11.44,45

11.43,46

11.47,49

11.48,50

11.51,52

11.53,54

11.57,58

11.55,56

Uniformly accelerated motion 11.37,38

Relative motion of particles systems with one degree of freedom systems with two degrees of freedom

11.59,60

Graphical methods 11.61,62

analyze motion using a given motion curve

11.66

11.65

motion of one particle

11.69,70

11.67,68

11.63,64

xvi

11.71,75

11.72,73

11.76

11.74

11.78

11.77,79

motion of two particles problems involving rate of change of acceleration

80 11.81,82

11.83

problems solved by numerical integration

11.84

problem solved using a v-x curve

11.87,88

11.85,86

problems solved using the moment-area method

CURVILINEAR MOTION OF PARTICLES Analyze the motion of a particle given 11.89,90

11.91,92

11.93

11.94

11.95

*11.96

x and y as functions of t the position vector: two dimensions the position vector: three dimensions

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xvii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

11.99,99

11.97,100

with horizontal initial velocity

11.102,103

11.101,104

with initial velocity of a given magnitude and direction

11.106,108

11.105,107

for a given direction, find magnitude of initial velocity

Projectiles

11.109,110 11.111,114

11.112,113

11.115,*116

for a given magnitude, find direction of initial velocity (Solution of a quadratic equation is required.) Relative motion of particles with

11.118,120

11.117,119

constant velocities

11.121,122 11.123, 124 11.125,126

given velocities and accelerations given relative acceleration

11.127,128

constant accelerations

11.129,131

11.130,132

two given observations of relative motion

11. 133,134,135

11. 136

zero tangential acceleration

11.138,139

11.137,141

constant tangential acceleration

11.144,145

11.143,146

find radii of curvature of trajectory

Normal and tangential components

11.140,142 11.147,149

11.148,150

*11.151

*11.152

find radius of curvature of a curve in three dimensions

11.153,154

11.156,157

problems involving a satellite in a circular orbit

11.158,159

11.157,160 Radial and transverse components 11.161,162

given r and q as functions of time

11.167,168

11.169,*170

find relations among v, a, q , and q

11.171

11.172

11.163,164 11.165,166

find approximate velocity from two observations of r and q for motion along a spiral

11.174

11.173

11.175

11.176

11.177

11.179

motion of a particle in three dimensions: cylindrical coordinates

11.178

motion of a particle along a hyperbola or a circle

*11.181

find direction of the binormal

*11.180

find velocity find acceleration

xviii

11.182,183

11.186,187

11.184,185

11.189,190

11.188,192

11.191,193

Review problems

CHAPTER 12: KINETICS OF PARTICLES NEWTON'S SECOND LAW 12.3,4

12.1,2

Definition of mass and weight Rectilinear motion

12.7,8

12.5,6

single particles

12.9,10

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xix

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

12.11,12

12.13,14

several connected bodies

12. 17,18

12. 15,16

12.19 12.20,21

two bodies with impending relative motion

12.22,23 12. 24,25

12.26,27

12.28,30

12.29,31

problems with variable acceleration systems with two degrees of freedom

12.32,33 12.34,35 Curvilinear motion: normal and tangential components 12.36,37

circular paths contained in a horizontal plane

12.38,39 12.40,41

*12.42,*43

motion defined by two criteria

12.46,47

12.44,45

circular paths contained in a vertical plane

12.52

cars traveling on banked curves

12.48,49 12.50,51 12,53,54 12.55,56

impending relative motion of a particle with respect to a moving support

12.57,58

12.60

12,59

12.62,63

12.64

12.65

cathode-ray tubes

Curvilinear motion: radial and transverse components 12.68,69

12.66,67

12.71,72

12.70,73

find acceleration of and/or forces acting on a particle for a given motion

12.74,75

12.76,77

motion under a central force: theoretical problems

12.79,81

12.78,80

motion defined by Newton's Law of Gravitation

12.82,85

12.83,84

12.86,87

12.88,89

12.90,93

12.91,92

12.96,97

12.94,95

determination of the central force for a given trajectory

12.98,100

12.99

analysis of spacecraft orbits

Conservation of angular momentum orbital motion of spacecraft mechanical devices

Trajectory of a particle under a central force

12.101

12.102,104

12.103,105

analysis of more advanced orbital problems

12.106,107 12.108

12.109

determination of the time in orbit

12.110,111 12.112,113 12.115,116

12.114,117

determination of the point of impact of a descending spacecraft

12.119,120

12.118,121

derivation of classical formulas

12.123,126

12.122,124

12.127,128

12.125,130

12.129,131

12.131,132

Review problems

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxi

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

CHAPTER 13: KINETICS OF PARTICLES ENERGY AND MOMENTUM METHODS 13.1,4

13.2,3

Determine kinetic energy

13.7,8

13.5,6

Method of work and energy applied to problems involving

13.9,10 13.11,12

a single body 13.13,14

13.15,16

13.17,18

13.21,22

13.19,20

two connected bodies

13.23 13.24

13.25

a body which drops out of the system

13.26,27

13.28.29

work of springs

13.30,31 13.32,35

13.33,34

work of non-linear forces

13.37,38

13.36

work of a gravitational force

13.39,40

13.41,42

use of F = ma

13.44

13.43,45

13.46,47

Introductory problems involving power and efficiency

13.48,51

13.49,50

Power in uniform and uniformly accelerated motion

13.52,53

13.54

13.55,56

13.58,59

springs in a horizontal plane

13.64,65

13.62,63

springs and the effect of gravity

Method of conservation of energy applied to problems involving 13.57 13.60,61 13.68,69

13.66,67

13.70,71

13.72,73

13.76,77

13.74,75

use of F = ma

*13.78 13.*79,80

13.*81,*82

Theory problems on conservative forces

13.85,86

13.87,88

Space mechanics problems involving gravitational potential energy

13.89,90

13.91,92

13.*83,*84

13.93,94

13.95,96

13.98,99

13.97

13.100,103

13.101,102

Conservation of energy and conservation of angular momentum Applications to space mechanics velocity perpendicular to the radius vector

13.104,105

xxii

13.106,107

13.110,*111

13.108,109

13.112,113

13.*114,115

13.116,*117

velocity oblique to the radius vector derivations and theoretical problems

13.*118 Method of impulse and momentum applied to problems involving 13.119,120

13.121

13.122,123

13.124,125

13.126,128

13.127

a single body

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxiii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

13.131,132

13.129,130

two bodies

13.135

13.134

a force defined by a graph

13.136

13.137,138

13.141,142

13.139,140

impulsive motion of a single body

13.145,148

impulsive motion of two bodies

13.133

13.143,144 13.146,147

13.149 13.151,152

13.150

impulsive motion with different supports

13.153,154

13.155,156

13.157,158

13.159,160

13,161

Direct central impact

13.162,163 13.166,167

13.164,165

Oblique central impact

13.170,171

Impact against fixed surfaces

13.175,177

13.174,176

Problems involving energy and momentum in direct impact

13.179,180

13.178,181

13.168 13.169 13.172,173

13.182,183 13.185,186

13.184,188

Problems involving energy and momentum in oblique impact

13.193,195

13.190,191

Review problems

13.197,198

13.192,194

13.199,200

13.196,201

13.187,189

CHAPTER 14: SYSTEMS OF PARTICLES 14.1,2

14.3,4

14.7,8

14.5,6

14.9,10

14.11,12

Conservation of linear momentum (one dimension) Conservation of angular momentum

14.13,14 14.15,16

14.19,20

Motion of mass center

14.17,18

xxiv

14.24,25

14.21,22

14.26

14.23

Conservation of linear momentum (2 or 3 dimensions)

14.27,28

14.29,30

Derivations and proofs

14.31,34

14.32,33

Conservation of linear momentum and computation of change in kinetic energy

14.34,36

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxv

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

14.37,38

14.39,40

Conservation of linear momentum and conservation of energy

14.43,44

14.41 ,42

14.45,48

14.46,47

Conservation of linear and angular momentum

14.49,50 14.51,52

14.55,56

Conservation of linear momentum, angular momentum and energy

14.59,60

Thrust caused by a diverted flow

14.63,65

Reactions at supports of vanes or conveyor belts

14.69,71

14.70,72

Thrust developed by a jet engine, a propeller, or a fan

14.73,75

14.74

14.53,54

14.57,68 14.61,62 14.64,66 14.67,68

14.76 14.77,78

Power and efficiency

14.79

14.80

14. 81,83

14.82,*84

Theory problems from fluid mechanics

14.88

Motion of chains

14.*85 14.86,87 14.91,94

14.90,91

Motion of a toy car

14.92,93

Thrust of rockets

14.95,96

14.97,98

14.101

14.99,100 14.102 14.103,104

Efficiency formulas for jet engine and rocket

14.106,107

14.105,108

Review problems

14.109,110

14.112,113

14.111,115

14.114,116

CHAPTER 15: KINEMATICS OF RIGID BODIES Rotation about a fixed axis 15.1,2

15.3,4

15.5,6

15.7,8

solutions using equations

15.9

xxvi

15.10,11

15.12,13

rotation about a skew axis

15.14,15 15.17

15.16

15.18,19 15.20,21

motion of the earth motion in two dimensions

15.22,23 15.24

15.26,27

15.25

rolling contact with no slipping

15.29,30

15.28,31

linear and angular motion

15.34,35

15.32,33

disks brought into contact

*15.36

*15.37

special problems

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxvii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

15.40,41

15.38,39

Problem Description

General plane motion: velocities motion of s single rigid body

15.42,43 15.44,45

application of vector algebra to motion of a plate

15.46,47 15.49,50

15.48,52

15.51

15.53,54

15.55,56

15.57,58

15.61,62

15. 59,60

planetary gear systems motion of rod connected to

15.63,64

a crank and a sliding block two cranks

15.65

15.66,67

15.68,69

15.70,72

motion involving a rolling wheel

15.71

Instantaneous center of rotation 15.76,77

15.73,74

15.78,79

15.75,81

problems involving parallel velocities

15.80 15.84,85

15.82,83

single rod: angle between controlling velocities = 90

15.89

single rod angle between controlling velocities ¹ 90

15.86,87 15.88,90 15.91,92 15.93

15.94

15.96,97

15.95,98

problems involving two instantaneous centers

15.100

15.99

space and body centrodes

15.102,103

15.101,104

previous problems solved using instantaneous center

General plane motion: accelerations find relations among accelerations ( w = 0)

15.105,106 15.107,108

15.109,110

15.111

15.112,113

15.115,116

15.114

15.117,119

15.118

rolling motion

motion of a body controlled by 15.120,121

15.123,124

15.122

15.125,126

15.127,128

15.129,130

one crank and a sliding block two cranks

xxviii

15.132,134

15.131,133

15.136

15.135

15.137

special problems

*15.138,*139

*15.140,*141

*15.142,*143

*15.144,147

*15.145,*146

*15.148,*149

analysis of plane motion in terms of a parameter

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxix

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

15.150,151

15.152,155

15.153,154

15.156,157

Problem Description

Plane motion of a particle relative to a rotating frame velocity of a particle with respect to a rotating frame Coriolis acceleration 15.160,162

15.158,159

15.163,164

15.161

acceleration of a particle moving at constant speed with respect to frame rotating with constant velocity

15.165,166 15.167,168 15.169

15.170,171

15.172,173 15.174,175

acceleration of a particle with respect to a frame rotating with an

15.176,177

angular acceleration

15.178

15.179,180

*15.183

15.*181,*182

Motion about a fixed point. General Motion 15.184,185

relation between linear and angular velocity for a rigid body with

15.186

15.187

15.190,191

15.188,189

15.192,193

15.194,195

15.197,198

15.196

15.199,200

15.201,202

15.205

15.203,204

15.208,209

15.206,207

a fixed point angular acceleration of a body rotating simultaneously about two axes velocity and/or acceleration of a body rotating about a fixed point rod attached to moving collars: find velocities of collars (angular velocity is indeterminate)

15.210,211

universal joints

15.213

15.112

rod with clevis attached at one end

15.214

15.215,216

rod attached to moving collars: find acceleration of collars (angular velocities

15.218,219

15.117

and angular accelerations are indeterminate)

Three dimensional motion of a particle relative to a rotating frame. Coriolis 15.222,223

15.220,221

15.224

15.225,226

Acceleration W = 0 and u = 0

15.227,228 15.229

15.230

W and u not constant

15.231,233

15.232

previous problems to be solved using rotating frames

15.234,235

15.236

axis of W in plane of mechanism

xxx

15.237,238

14,239

15.244,245

15.240,241

15.246,247

15.242,243

15.248,251

15.249,250

15.252,254

15.253,257

15.255,256

15.258,259

frame of reference in general motion

Review problems

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxxi

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

CHAPTER 16: PLANE MOTION OF RIGID BODIES FORCES AND ACCELERATIONS Rectilinear translation 16.4,5

16.1,2

16.6

16.3

16.9,10

16.7,8

find reactions or accelerations determine whether body tips or slides

16.11,12 Curvilinear translation bodies with zero velocity ( an = 0 )

16.13,14 16.15,16 16.17 16.18

bodies with velocities ( an ¹ 0 )

16.19,20 16.*22

16.*21

draw shear and bending-moment diagrams for accelerated rod Centroidal rotation

16.23

16.24

proofs that effective forces reduce to ma and I a

16.25,29

16.26,27

single body

16.30,31

16.28

16.32,35

16.33,34

16.36

16.37,38

16.41,42

16.39,40

16.43,44

16.45

connected pulleys and blocks friction disks General plane motion

16.46

16.47

proofs concerning system of effective forces

16.50,51

16.48,49

find accelerations for given loading

16.53,54

16.52

16.55,56

16.57,60

16.58,59

16.61,62

16.63

16.64,65

16.66,67

16.68

16.69,70

16.71,72

vertical motion controlled by applied forces and gravity bars suspended from springs and suddenly released spheres, disks, or hoops rotating and sliding on horizontal surface

16.73,74

Non-centroidal rotation 16.75 16.76,77

derivation involving center of percussion 16.78,79

find acceleration and reactions for a suspended body find reactions for a body rotating with

xxxii

16.80,83

16.81,82

constant angular velocity

16.84

16.85,86

angular acceleration and velocity

16.87,90

16.88,89

two connected bodies

16.91 Rolling motion 16.92

16.93

derivations involving use of SMC = IC a

16.96,97

16.94,95

single body rolling on an incline under gravity

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxxiii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

16.98,99

16.102,103

single body rolling, or rotating and sliding

16.100,101

16.104,105

16.108,109

16.106.107

connected bodies

16.110 16.113,114

16.111,112

unbalanced bodies

16.115,116 General plane motion single rigid body, find accelerations and reactions 16.119,120

16.117,118

problems in which the velocity is zero

16.125,126

problems in which the velocity is not zero

16.121,122 16.123 16.124,127 16.128,129 16.130,131 16.132,133

problems involving an integration connected rigid bodies

16.135,136

16.134

16.137,138

16.139,140

16.142

16.141

16.*143,*145

16.*144

16.*146,

16.*147,*148

systems with one degree of freedom needs rotating axis systems with two degrees of freedom

16.*149,*150 16.*150

16.*151,*152

16.155,156

16.153,154

16.159,160

16.157,158

16.163,164

16.161,164

draw shear and bending-moment diagrams for accelerated rod

Review problems

CHAPTER 17: PLANE MOTION OF RIGID BODIES ENERGY AND MOMENTUM METHODS ENERGY METHODS Centroidal rotation 17.1,4

17.2,3

17.6

17.5

17.11,12

17.7,8

17.13,14

17.9,10

single body with applied couple connected bodies

17.15

xxxiv

Noncentroidal rotation 17.17

17.16,18

17.21,22

17.19,20

17.23,24

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxxv

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

17.25,26

17.27

17.28,29

17.30

Problem Description

Rolling motion single body

17.31 17.34,35

17.32,33

connected bodies General plane motion

17.36,37

17.39,40

17.38

17.41

single body

17.42,43

connected bodies

17.44,45 17.46,47 17.49,50

17.48,51

Power

MOMENTUM METHODS Centroidal rotation 17.52,54

17.53,55

17.56,57

17.58,59

17.60,61

17.63,64

single body connected bodies

17.62 17.65,66

17.67,68

Derivations involving noncentroidal rotation (center of percussion) Rolling motion

17.70,71

17.69

single body

17.74,75

17.72,73

connected bodies

17.76 17.77

17.78

sliding and rotating sphere Conservation of angular momentum

17.79

17.80

problems involving changes in the configuration of a rotating assembly

17.82,83

17.81

problems involving relative motion between components of a rotating assembly

17.86,87

17.84,85

17.88,89

17.90,91

17.92,93

17.95

problems involving conservation of energy

17.94

Eccentric impact 17.96

17.97,98

17.101,102

17.99,100

17.102,103

17. 104,105

17.106,107

17.108,109

projectile becomes embedded in a body suspended from a fixed point single body, rotation about a fixed point

xxxvi

17.110,111

17.112

17.113 17.114,115

17.118,119

17.116,117

17.120

17.121,122

17.123,124

17.126

17.125

17.127,128

17.130

problems involving method of work and energy two bodies, general plane motion

17.129 17.131,*134

17.132,133

special problems

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxxvii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

17.138,139

17.135,136

17.140,141

17.137,142

17.143,145

17.144,146

Review problems

CHAPTER 18: KINETICS OF RIGID BODIES IN THREE DIMENSIONS 18.1,5

18.2,3

18.7,8

18.4,6

Computation of angular momentum (no product of inertia)

18.10,12

18.9,11

Use of equation HO = r ´ mv + H G

18.17,18

Computation of angular momentum (using products of inertia)

18.23,24

Impulse and momentum in three dimensions

18.13,14 18.15,16 18.19,20 18.21,22 18.27,28

18.25,26

18.29,30

18.31,32

18.35,36

18.33,34

18.37

18.38

18.39,43

18.40,41

18.44,45

18.42,46

18.47,49

18.48

impact problems

Computation of kinetic energy

18.50 18.51,52

18.53,54

18.55,59

18.56,57

18.60

18.58,61

18.63,64

18.62

Computation of derivative of angular momentum

Dynamic reactions on rotating shaft 18.65,68

18.66,67

for w = constant for w = 0 and a ¹ 0

18.69,70 18.71,74

18.72,73

18.77,78

18.75,76

18.79,81

18.80,83

Simple problems on gyroscopic effect

18.86,87

18.85,88

Relative equilibrium under rotation about a fixed axis

18.90,91

18.89,92

18.82,84

xxxviii

Dynamic reactions for combined rotations 18.95,96

18.93,94

with constant rates

18.99,100

18.97,98

18.103,104

18.101,102

with variable rates

18.107,108

Steady precession. Simple problems

18.105,*106

18.109,110

18.111,112

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xxxix

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

18.115,116

18,113.114

Steady precession. More advanced problems

18.118

18.117

18.119,120

18.121,122

Precession under no force (theory)

18.123,124

18.125,128

Precession under no force (applications)

18.126,127

18.129,130

18.131,134

18.132,133

for bodies with precession and nutation

18.*137,*138

for bodies with precession, nutation, and spin

Conservation of energy and angular momentum 18.135,136 18.*139,*140

18.*141,*142 18.*143,144

18.*145,*146

Poinsot analysis

18.148,151

18.147,149

Review problems

18.152,153

18.150,154

18.155,157

18.156,158

CHAPTER 19: MECHANICAL VIBRATIONS VIBRATIONS WITHOUT DAMPING Free vibrations of a particle 19.1,4

19.2,3

problems involving amplitude, frequency, maximum velocity and acceleration

19.5,7

19.6,9

19.8

19.10

19.14,15

19.11,12

19.16

19.13

19.17,20

19.18,18

springs in parallel or in series

19.23,25

change mass or stiffness of the system special problems

problems involving velocity and acceleration at a given instant

19.21,22 19.24,26 19.28,30

19.27,29

19.31

19.32

19.*34,*36

19.*33,*35

19.37,39

19.38,41

19.40,43

19.42

simple pendulum with large amplitude

Free vibrations of rigid bodies systems with elastic restoring forces

19.44,45

19.46,47

19.48,49

19.50,51

systems with restoring forces due to gravity

19.52,53

19.54

derivations for motion of compound pendulum

19.55,56

19.57,58

systems with elastic and gravity restoring forces

19.60,62

19.59,61

determine velocity and acceleration

19.63,64

19.67,68

torsional vibrations

19.65.66

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xli

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

19.69,72

19.70,71

particles

19.75,76

19.73,74

single rigid body; pendulum

19.77,80

19.78,79

single rigid body; elastic restoring forces

19.85,86

connected rigid bodies

19.89,92

19.90,91

special problems

19.95,96

19.93,94

19.*98

19.*97

19.99,100

19.101,102

problems involving periodic applied force

19.106,107

19.105,108

problems involving periodic support motion

19.109,112

19.110,111

Conservation of energy used to analyze the motion of

19.81,82 19.83,84 19.87,88

Forced vibrations 19.103,104

19.113,114

19.115,116

19.117,119

19.118,120

19.121,122

19.123,125

problems involving rotating machinery special problems

19.124,126

DAMPED VIBRATIONS Damped free vibrations 19.127

19.128

proofs concerning damped free vibration

19. 129,130

19.131

proofs involving the logarithmic decrement

19.133

19.132

problems that use logarithmic decrement

19.136

19.134,135

maximum deflection of damped free vibration

19.137

19.138

damped free vibration of a rigid body

19.141,142

19.139,140

19.144

19.143,145

Damped forced vibrations problems involving rotating machinery

19.146 19.147,148

19.149,

forced transmission with damping

19.*150

energy dissipation

19.151*,152* 19.153,154

write the differential equation 19.155,156

Electrical analogues

19.157,158

xlii

19.159,162

19. 160,161

19.166,167

19. 163,164

19.168,170

19. 165,169

Review problems

* Problems which do not involve any specific system of units have been indicated by underlining their numbers. Answers are not given to problems with the numbers set in italic type.

xliii

xliv

15.3

15.4

15.5

15.6

15.7

16.1

17.1

Work and Energy

Constrained Plane Motion

16.2

EXAM NUMBER THREE

Plane Motion of Rigid Bodies

Moving Reference Frames in Three Dimensions

Motion of a Rigid Body in Space

Motion With Respect to a Rotating Frame

Acceleration in Plane Motion

Instantaneous Center

General Plane Motion

Translation, Rotation

15.2

Variable Systems of Particles

15.1

14.3

Systems of Particles

14.2

Systems of Particles

14.1

Impact

EXAM NUMBER TWO

13.4

Application to Space Mechanics

Impulse and Momentum

13.2D

13.3

Conservation of Energy

Work and Energy, Power

13.2

13.1

Review

Angular Momentum and Orbital Mechanics

15.77,82,90,95

17.1,19,25,40

16.117,125,134,143

16.76,86,98,116

16.48,56,64,70

16.4,19,25,40

15.222,232,234,243

15.187,198,203,213

15.150,163,168,178

17.6,18,26,39

16.118,124,139,144

16.77,85,99,107

16.48,55,65,69

16.5,18,30,39

15.223,230,235,242

15.188,197,204,213

15.151,162,169,177

15.113,120,128,138

15.76,83,88,98 15.112,121,127,139

15.39,49,58,64

15.2,10,19,28

14.62,70,78,86

14.33,37,47,51

14.2,11,16,21

13.158,166,176,185

13.120,129,142,148

13.86,95,103,110

13.59,64,74,79

13.8,17,27,41

12.58,6880,87

12.8,15,30,46

11.141,143,156,165

11.99,105,120,127

11.36,39,50,57

11.4,9,18,22

List 2

15.38,50,57,70

15.1,11,18,29

14.61,72,77,87

14.32,38,46,52

14.1,12,15,22

13.157,168,174,186

13.119,130,141,150

13.85,96,100,111

13.58,65,72,80

13.7,18,26,45

12.57,69,78,88

12.7,16,28,47

Newton’s Second Law and Linear Momentum

EXAM NUMBER ONE

12.2

11.137,146,155,166

11.98,106,118,128

11.35,40,48,55

11.3,10,15,23

List 1

Curvilinear Motion (Other Comp.)

Curvilinear Motion (Rect. Comps)

Uniformly Accelerated Motion

12.1

Topics

Rectilinear Motion

Introduction

11.4

11.5

11.2

11.1

Sections

11.1

Period

17.2,23,27,37

16.121,126,135,147

16.78,87,102,109

16.50,58,63,72

16.1,15,27,42

15.220,233,236,241

15.184,196,205,211

15.152,161,166,175

15.111,125,129,141

15.74,87,89,97

15.40,53,55,69

15.3,13,23,31

14.59,71,80,94

14.31,40,45,56

14.3,10,19,25

13.155,165,175,188

13.121,132,139,146

13.87,94,101,107

13.55,67,70,82

13.5,16,28,40

12.55,67,79,91

12.5,23,29,45

11.138,145,153,170

11.97,108,117,126

11.33,42,47,56

11.1,12,16,24

List 3

17.7,17,30,36

16.121,125,136,146

16.79,84,103,108

16.51,57,63,71

16.2,14,28,41

15.221,231,239,240

15.185,195,205,210

15.155,160,167,174

15.115,124,130,140

15.75,86,94,96

15.41,48,56,68

15.4,12,24,30

14.60,69,83,94

14.35,39,48,55

14.4,9,20,24

13.156,164,177,184

13.124,131,140,145

13.88,93,102,106

13.57,62,71,81

13.4,15,29,39

12.56,66,81,90

12.6,20,31,44

11.139,144,154,169

11.100,107,119,123

11.34,41,49,55

11.2,11,17,21

List 4

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK

17.14,22,28,42

16.122,133,141,150

16.81,91,100,110

16.53,62,68,74

16.7,17,32,45

15.225,229,237,247

15.191,202,208,215

15.156,159,172,180

15.116,126,131,146

15.73,85,92,104

15.42,54,59,67

15.7,15,20,35

14.63,75,92,98

14.34,42,49,54

14.5,14,17,27

13.160,171,178,189

13.125,137,143,154

13.91,99,104,114

13.60,63,73,84

13.13,22,30,50

12.59,73,83,93

12.13,22,32,54

11.140,150,157,168

11.101,110,121,132

11.37,46,53,60

11.6,14,19,30

List 5

17.15,21,34,41

16.123,132,142,149

16.82,90,101,106

16.54,61,68,73

15.8,16,35,45

15.226,229,238,246

15.192,201,209,214

15.157,158,173,179

15.117,123,135,145

15.81,84,93,101

15.43,52,60,66

15.8,14,21,34

14.65,73,93,97

14.36,41,50,53

14.6,13,18,23

13.162,170,181,187

13.127,133,144,153

13.92,98,105,113

13.61,66,75,83

13.14,21,35,49

12.60,70,84,92,

12.14,21,33,52

11.142,148,160,167

11.104,109,122,130

11.38,43,54,59

11.8,13,20,29

List 6

FOR ANY OF THESE PROBLEMS

NO ANSWERS ARE GIVEN IN THE BOOK

50% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 50% U.S. CUSTOMARY UNITS

This schedule includes all of the material of VECTOR MECHANICS FOR ENGINEERS: DYNAMICS with the exception of Sections 11.7, 11.8, 12.11–12.13, 18.9–18.11 and 19.8–19.10.

TABLE III: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN DYNAMICS

xlv

Motion in Three Dimensions

19.2

19.3

19.4

Forced Vibrations

Energy Methods

Free Vibrations of Rigid Bodies

Free Vibrations of Particles

19.1

Motion in Three Dimensions

18.2

Momentum and Energy in Three Dimensions

18.1

EXAM NUMBER FOUR

18.1

Impulse and Momentum

Eccentric Impact

17.2

17.3

17.55,71,80,89

19.101,107,115,122

19.69,79,83,91

19.38,45,57,64

19.1,14,17,25

18.86,95,102

18.55,68,71,78

18.30,41,50

18.1,11,15,26

17.96,108,116,124

17.59,70,81,88

19.102,106,116,121

19.72,78,87,90

19.41,44,59,63

19.4,8,20,23

18.85,96.101

18.56,67,72,77

18.29,42,49

18.5,9,16,25

17.101,105,117,123

17.52,72,79,95

19.99,110,113,125

19.70,80,85,92

19.37,51,55,68

19.3,12,18,26

18.88,93,104

18.57,66,75,80

18.32,39,54

18.2,12,17,22

17.97, 107,119,122

17.54,69,82,90

17.56,75,84,93

19.100,105,114,123

19.71,77,86,89

19.41,50,56,67

19.103,111,118126

19.73,2,84,94

19.40,47,58,66

19.9,16,21,32

18.92,97,106

19.5,11,19,24

18.60,70,73,82

18.87,94,103

18.36,45,52

18.4,14,19,24

17.99,111,113,130

18.58,65,76,79

18.31,40,53

18.3,10,18,21

17.98,109,120,121

17.57,73,85,92

19.104,108,120,124

19.74,81,88,93

19.43,46,61,65

19.10,15,22,27

18.91,98,105

18.64,69,74,81

18.35,46,51

18.6,13,20,23

17.100,110,114,125

xlvi

15.4

15.5

15.6

15.7

16.1

15.3

EXAM NUMBER THREE

Plane Motion of Rigid Bodies

Moving Reference Frames in Three Dimensions

Motion of a Rigid Body in Space

Motion With Respect to a Rotating Frame

Acceleration in Plane Motion

Instantaneous Center

General Plane Motion

Translation, Rotation

Variable Systems of Particles

15.2

14.3

Systems of Particles

15.1

14.2

Systems of Particles

14.1

Impact

13.4

Impulse and Momentum

EXAM NUMBER TWO

13.3

Application to Space Mechanics

13.2D

Conservation of Energy

Work and Energy, Power

13.1

13.2

EXAM NUMBER ONE

Angular Momentum and Orbital Mechanics

Newton’s Second Law and Linear Momentum

Curvilinear Motion (Other Comp.)

Curvilinear Motion (Rect. Comps)

Uniformly Accelerated Motion

Review

12.2

Topics

Rectilinear Motion

Introduction

11.5

12.1

11.4

11.2

11.1

Sections

11.1

Period

16.48,56,66,70

16.4,19,25,44

15.222,232,234,245

15.187,198,206,213

15.150,163,168,176

16.48,55,67,69

16.5,18,30,43

15.223,230,235,244

15.188,197,207,213

15.151,162,169,181

15.113,120,133,138

15.77,82,90,102

15.76,83,88,103 15.112,121,132,139

15.39,49,62,64

16.50,60,63,72

16.9,15,27,42

15.224,233,236,241

15.184,199,205,211

15.153,161,166,175

15.111,122,129,141

15.78,87,89,97

15.40,51,55,69

15.5,13,23,31

14.64,71,80,94

14.62,70,78,95

15.2,10,19,32

14.31,44,45,56

14.7,10,19,25

13.155,172,175,188

13.126,132,139,146

13.89,94,101,107

13.55,69,70,82

13.11,16,28,40

12.55,72,79,91

12.9,23,29,45

11.138,149,153,170

11.102,108,117,126

11.33,45,47,56

11.5,12,16,24

List 3a

14.33,37,51,52

14.2,11,16,26

13.158,166,180,185

13.120,129,142,151

13.86,95,103,108

13.59,64,77,79

13.8,17,27,51

12.58,68,85,87

12.8,15,30,48

11.141,143,159,165

11.99,105,120,129

11.36,39,52,57

11.4,9,18,26

List 2a

15.38,50,61,70

15.1,11,18,33

14.61,72,77,96

14.32,38,51,52

14.1,12,15,28

13.157,168,179,186

13.119,130,141,152

13.85,96,100,109

13.58,65,76,80

13.7,18,26,52

12.57,69,82,88

12.7,16,28,49

11.137,146,158,166

11.98,106,118,131

11.35,40,51,55

11.3,10,15,27

List 1a

16.51,59,63,71

16.10,14,28,41

15.227,231,239,240

15.185,200,205,210

15.154,160,167,174

15.115,122,130,140

15.79,86,94,96

15.41,50,56,68

15.6,12,24,30

14.66,69,83,94

14.35,43,48,55

14.8,9,20,24

13.156,169,177,184

13.128,131,140,145

13.90,93,102,106

13.57,68,71,81

13.12,15,29,39

12.56,71,81,90

12.10,20,31,44

11.139,147,154,169

11.103,107,119,123

11.34,44,49,55

11.7,11,17,21

List 4a

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK

75% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 25% U.S. CUSTOMARY UNITS

This schedule includes all of the material of VECTOR MECHANICS FOR ENGINEERS: DYNAMICS with the exception of Sections 11.7, 11.8, 12.11–12.13, 18.9–18.11 and 19.8–19.10.

TABLE IV: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN DYNAMICS

xlvii

19.3

19.4

Forced Vibrations

Energy Methods

Free Vibrations of Rigid Bodies 19.101,107,117,122

19.69,79,83,96

19.38,45,60,64

19.1,14,17,31

18.86,99,102

18.55,68,71,84

Motion in Three Dimensions

18.2

18.1,11,15,28

17.96,108,116,128

17.55,71,83,89

17.1,19,25,46

18.30,44,50

Free Vibrations of Particles

18.2

16.76,86,98,114 16.117,125,137,143

Momentum and Energy in Three Dimensions

19.2

18.1

19.1

18.1

Eccentric Impact

17.3

Impulse and Momentum

17.3

Work and Energy

EXAM NUMBER FOUR

17.1

Constrained Plane Motion

16.2

16.77,85,99,113

19.102,106,119,121

19.72,78,87,95

19.41,44,62,63

19.4,8,20,28

18.85,100.101

18.56,67,72,82

18.29,45,49

18.5,9,16,27

17.101,105,117,127

17.59,70,86,88

17.6,18,26,45

16.118,124,138,144

19.99,112,113,125

19.75,80,85,92

19.37,49,55,68

19.5,12,18,26

18.91,93,104

18.60,66,75,80

18.36,39,54

18.7,12,17,22

17.102,107,119,122

17.52,75,79,95

17.11,23,27,37

16.121,129,135,147

16.80,87,102,109

19.100,109,114,123

19.76,77,86,89

19.41,48,56,67

19.7,11,19,24

18.90,94,103

18.64,65,76,79

18.35,40,53

18.8,10,18,21

17.103,109,120,121

17.54,74,82,90

17.12,17,30,36

16.121,128,136,146

16.83,84,103,108

Instructor’s and Solutions Manual to accompany

Vector Mechanics for Engineers, Statics Twelfth Edition Ferdinand P. Beer Late of Lehigh University

E. Russell Johnston, Jr. Late of University of Connecticut

David F. Mazurek United States Coast Guard Academy

Prepared by

Amy Mazurek

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. To assist 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 maximum number 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 I). 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. Accessible through Connect are problem sets for each chapter that are designed to be solved with computational software. Solutions for these problems, including analyses of the problems and problem solutions and output for the most widely used computational programs, are also available through Connect. 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 III 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 IV 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 advised 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. Acknowledgement and thanks are given to Amy Mazurek for her careful preparation of the solutions contained in this manual. David Mazurek DESCRIPTION OF THE MATERIAL CONTAINED IN VECTOR MECHANICS FOR ENGINEERS: STATICS, Twelfth 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. l.4, it is shown how numerical 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, iii

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 25 when printed or as F = 280 lb 25 when handwritten. Unit vectors i and j are introduced in Sec. 2.7, where the rectangular components of forces are considered. 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 ΣFx = 0, ΣFy = 0 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 = Fx i + Fy j + Fz k = Fλ, where i, j, and k are the unit vectors directed respectively along the x, y, and z axes, and λ 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 iv

Sec. 2.3C 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∙s2/ft (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. 16.1D 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. 3.41 and 3.44) 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 I). In Sec. 3.3 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. In Sec. 3.4, the concept of equivalent systems of forces is carefully presented. This concept is made more intuitive through the extensive use of free-body-diagram equations (see Figs. 3.36 through 3.41). Note that the moment of a force is either not shown or is represented by a green vector (Figs. 3.10 and 3.22). A red vector with the symbol is used only to represent a couple, that is, an actual system consisting of two forces (Figs. 3.33 through 3.41). Section 3.4D 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 may occasionally 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 v

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 freebody 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. 4.1B) 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 4.1C introduces the concepts of statical indeterminacy and partial constraints. Section 4.2 is 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 5.1C 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 5.2A explains the use of differential elements in the determination of centroids by integration. The theorems of Pappus-Guldinus are given in Sec. 5.2B. Sections 5.3A and 5.3B are optional; they show how the resultant of a distributed load can be determined by evaluating an area and by locating its centroid. Section 5.4 deals 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/m3), rather than by its specific 3 weight (weight per unit volume, expressed in N/m ). The specific weight of the material can then be obtained by multiplying its density by g = 9.81 m/s2 (see the footnote of Sec. 5.1B). Chapter 6 Analysis of Structures In this chapter students learn to determine the internal 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 Sec. 6.1A, the method of joints and the method of sections are explained in detail in Sec. 6.1B and Sec. 6.2A, 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 6.1C and 6.1D are optional. It is shown in Sec. 6.1C how the analysis of certain trusses can be expedited by recognizing joints under special loading conditions, while in Sec. 6.1D the method of joints is applied to the solution of three-dimensional trusses. It is pointed out in Sec. 6.1B 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. 6.2A) 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 6.2B is optional; it deals with the trusses vii

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 been detached from their supports, others will collapse (Sec. 6.3B). 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. 6.4). Students should be urged to resolve a force of unknown magnitude and direction into two components but to represent a force of known direction 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 7.1 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 Section 7.2 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 freebody diagrams of the various portions of the beams. The relations among load, shear, and bending moment are introduced in Sec. 7.3. Problems in this section should be solved by evaluating areas under load and shear curves or by formal integration (as in Probs. 7.85 through 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. 7.3. Others will feel that the study of shear and bending-moment diagrams is incomplete without this 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.

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Section 7.4 is 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 7.5 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. Section 8.1 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 Secs. 8.2 through 8.4. They are divided into the following groups: wedges and screws (Sec. 8.2); axle and disk friction, rolling resistance (Sec. 8.3); belt friction (Sec. 8.4). 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. 9.1A 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. Section 9.2 introduces 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 9.3 and 9.4 are optional; they are devoted to products of inertia and to the determination of principal axes of inertia. ix

Sections 9.5 and 9.6 deal with the moments of inertia of masses. Particular emphasis is placed on the moments of inertia of thin plates (Sec. 9.5C) and on the use of these plates as differential elements in the computation of moments of inertia of symmetrical threedimensional bodies (Sec. 9.5D). Section 9.6 is optional but should be used whenever the following dynamics course includes the motion of rigid bodies in three dimensions. Sections 9.6A and 9.6B 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 9.6C 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 material 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∙m2 directly from the dimensions of the body in meters and from its density in kg/m3. 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∙s2/ft (or slugs). The mass moment of inertia of the body is then obtained in lb∙ft∙s2 (or slug∙ft2). Sample Problem 9.12 provides an example of such a computation. Attention is also called to the footnote of Sec. 9.5A 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 inclusion 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. Section 10.1 is devoted to the derivation of the principle of virtual work and to its direct application to the solution of equilibrium problems. Section 10.2 introduces the concept of potential energy and shows that equilibrium requires that the derivative of the potential energy be zero. Section 10.1D defines the mechanical efficiency of a machine and Sec. 10.2D discusses the stability of equilibrium. The first groups of problems in each assignment 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).

TABLE I: LIST OF THE TOPICS COVERED IN VECTOR MECHANICS FOR ENGINEERS: STATICS Suggested Number of Periods

Sections

Topics

Basic Course

Additional

Abridged Course to be

Topics

used as an introduction to dynamics+

1. INTRODUCTION 1.1–6 This material may be used for the first assignment or for later reference 2. STATICS OF PARTICLES 2.1 Addition and Resolution of Forces 2.2 Rectangular Components 2.3 Equilibrium of a Particle 2.4 Forces in Space 2.5 Equilibrium in Space 3. RIGID BODIES: EQUIVALENT SYSTEMS OF FORCES 3.1 Vector Product; Moment of a Force about a Point 3.2 Scalar Product; Moment of a Force about an Axis 3.3 Couples 3.4A–4C Equivalent Systems of Forces * 3.4D Reduction of a Wrench 4. EQUILIBRIUM OF RIGID BODIES 4.1A–1B Equilibrium in Two Dimensions 4.1C Indeterminate Reactions; Partial Constraints 4.2 Two- and Three-Force Bodies 4.3 Equilibrium in Three Dimensions 5. CENTROIDS AND CENTERS OF GRAVITY 5.1 Centroids and First Moments of Areas and Lines 5.2 Centroids by Integration 5.3 Beams and Submerged Surfaces 5.4 Centroids of Volumes 6. ANALYSIS OF STRUCTURES 6.1A–1B Trusses by Method of Joints * 6.1C Joints under Special Loading Conditions * 6.1D Space Trusses 6.2A Trusses by Method of Sections 6.2B Combined Trusses 6.3 Frames 6.4 Machines

0.5–1 0.5–1 1 1 1

1–2 1–2 1 1–1.5

1–2 1–2 1 1–1.5 0.5–1

1.5–2 0.5–1 1 2

1–2 1–2 1–1.5 1–2

1–1.5 0.25–0.5 0.5–1 1–2 0.25–0.5 2–3 1–2

7. INTERNAL FORCES AND MOMENTS 7.1 Internal Forces in Members 7.2 Shear and Moment Diagrams by FB Diagram 7.3 Shear and Moment Diagrams by Integration * 7.4 Cables with Concentrated Loads; Parabolic Cable * 7.5 Catenary 8. FRICTION 8.1 Laws of Friction and Applications 8.2 Wedges and Screws * 8.3 Axle and Disk Friction; Rolling Resistance 8.4 Belt Friction

1.5–2

1–2 0.5–1.5

1 1–2 1–2 1–2 1

1–2 1

1–2 1–2

9. MOMENTS OF INERTIA 9.1 Moments of Inertial of Areas 9.2 Composite Areas * 9.3 Products of Inertia; Principal Axes * 9.4 Mohr’s Circle 9.5 Moments of Inertia of Masses# * 9.6 Mass Products of Inertia; Principal Axes and Principal Moments of Inertia

1 1–2 1–2 1 1–2 1–2

10. METHOD OF VIRTUAL WORK 10.1A–1C Principle of Virtual Work 10.1D Mechanical Efficiency 10.2 Potential Energy; Stability Total Number of Periods

1–2 0.5–1 1–1.5 26–39

15–26

14–21

+ A sample assignment schedule for a course in dynamics including this minimum amount of introductory material in statics is given Table V. It 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 of mechanics of materials. # Mass moments of inertia have not been included in the basic statics course since this material is often taught in dynamics.

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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 of concurrent forces graphical method law of sines

2.1, 3 2.5, 8 2.9, 10 2.14 2.16, 18 2.19, 20

2.2, 4 2.6, 7 2.11, 12 2.13 2.15, 17

2.22, 23 2.25, 27 2.29, 30 2.32, 34 2.35, 36 2.39, 40

2.21, 24 2.26, 28 2.31, 33 2.37, 38 2.41, 42

Resultant by ΣFx = 0, ΣFy = 0

2.F1, F4

2.F2, F3

Free Body Practice Problems

2.44, 45 2.46, 47 2.49, 50 2.53, 54 2.57, 58 2.61, 62 2.65, 66 2.69, 70

2.43, 48

Equilibrium. Free-Body Diagram equilibrium of 3 forces

2.51, 52 2.55, 56 2.59, 60 2.63, 64 2.67, 68

special problems laws of cosines and sines

Rectangular components of force simple problems more advanced problems

Select force so that resultant has a given direction

equilibrium of 4 forces find parameter to satisfy specified conditions special problems

FORCES IN SPACE Rectangular components of a force in space given F,q, and f, find components and direction angles

2.71, 72 2.73, 74 2.80, 82 2.83, 84 2.85, 86 2.89, 90 2.91, 92 2.95, 96

2.75, 76 2.77, 78 2.81, 82

2.F5, F6

2.F7, F8

Free Body Practice Problems

2.99, 100 2.101, 102

2.103, 104

Equilibrium of a particle in space load applied to three cables, introductory problems

relations between components and direction angles

2.87, 88

direction of force defined by two points on its line of action

2.93, 94 2.97, 98

resultant of two or three forces

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

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TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

2.107, 108 2.109, 110

2.105, 106 2.111, 112

intermediate problems

2.115, 116 2.117, 118

2.113, 114 2.119, 120

advanced problems

2.121, 122 2.125, 126

2.123, 124

problems involving cable through ring special problems

2.127, 131 2.132, 134 2.135, 136

2.128, 129 2.130, 133 2.137, 138

Review problems

2.C1, C3 2.C4

2.C2 2.C5

Computer problems

CHAPTER 3: RIGID BODIES; EQUIVALENT SYSTEMS OF FORCES 3.1, 2 3.4, 5 3.9, 10 3.13, 14 3.15 3.17, 18

3.3, 6 3.7, 8 3.11, 12

3.19 3.21, 23 3.24, 26 3.28, 29 3.32, 33

3.20 3.22, 25

3.35 3.37, 38 3.41, 42 3.45 3.47, 48 3.51, 52 3.57, 58 3.59, 60

3.36 3.39, 40 3.43, 44 3.46 3.49, 50 3.53, 54 3.55, 56 3.61, 62 3.63 *3.64, *65

*3.66, *67 *3.68, *69

3.70, 71 3.74 3.78, 79

3.16

3.27, 30 3.31, 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 of a formula applications of the vector product Moment of a force about a point: Three dimensions computing M = r ´ F, introductory problems computing M = r ´ F, more involved problems using M to find the perpendicular distance from a point to a line

Scalar Product Finding the angle between two lines Mixed triple product Moment of a force about the coordinate axes Moment of a force about an oblique axis Finding the perpendicular distance between two lines

3.72, 73

Couples in two dimensions

3.75, 76 3.77, 80

Couples in three dimensions

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

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TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

3.81, 83 3.85, 86 3.87, 88 3.90, 92 3.93, 95 3.96, 98

3.82, 84

Replacing a force by an equivalent force-couple system: two dimensions

3.89, 91

Replacing a force-couple system by an equivalent force or forces

3.94, 97 3.99, 100

Replacing a force by an equivalent force-couple system: three dimensions

3.104

Equivalent force-couple systems

3.105, 106 3.108, 109 3.112, 113 3.114, 115 3.121, 122 3.123, 126 3.129, 130

Finding the resultant of parallel forces: two dimensions Finding the resultant and its line of action: two dimensions

3.101, 102 3.103 3.107 3.110, 111 3.116, 117 3.118 3.119, 120 3.124, 125 3.127, 128 *3.131, *132

Problem Description

Reducing a three-dimensional system of forces to a single force-couple system Finding the resultant of parallel forces: three dimensions Reducing three-dimensional systems of forces or forces and couples to a wrench axis of wrench is parallel to a coordinate axis or passes through O force-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

*3.133, *135 *3.137 *3.139, 140 *3.141 *3.143, *144

*3.134, *136

3.147, 148 3.151, 153 3.156, 157

3.149, 150 3.152, 154 3.155, 158

Review problems

3.C1, C4 3.C5

3.C2, C3 3.C6

Computer problems

*3.138 *3.142 *3.145, *146

CHAPTER 4: EQUILIBRIUM OF RIGID BODIES EQUILIBRIUM IN TWO DIMENSIONS 4.F1, F4

4.F2, F3

Free Body Practice Problems

4.1, 2 4.5, 6 4.9, 10 4.11, 13 4.15, 16 4.19, 20 4.22, 25 4.26, 28 4.29, 30 4.31, 32 4.35, 36 4.39, 40 4.45, 46

4.3, 4 4.7, 8 4.12, 14

Parallel forces

4.17, 18

Rigid bodies with one reaction of unknown direction and one of known direction

Parallel forces, find range of values of loads to satisfy multiple criteria

4.21, 23 4.24, 27 4.33, 34 4.37, 38 4.41, 42 4.43, 44

Rigid bodies with three reactions of known direction Rigid bodies with a couple included in the reactions

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

4.47, 50 4.51, 52 4.56, 58 4.59

4.48, 49 4.53, 54 4.55, 57 4.60

Partial constraints, statical indeterminacy

4.62, 63 4.65, 67 4.69, 70 4.73, 74 4.75, 76 4.78, 81 4.82 4.83, 84 4.85, 88

4.61, 64 4.66, 68 4.71, 72

Three-force bodies simple geometry, solution of a right triangle required simple geometry, frame includes a two-force member more involved geometry

Find position of rigid body in equilibrium

4.77, 79 4.80 4.86, 87 4.89, 90

find position of equilibrium

EQUILIBRIUM IN THREE DIMENSIONS 4.F6, F7

4.F5

Free Body Practice Problems

4.91, 92 4.95, 96 4.99, 100 4.101, 102 4.106, 107 4.108, 109 4.113, 114 4.117, 118 4.119, 122 4.123, 124 4.129, 130 4.131, 132 4.133, 134 4.138, 139

4.93, 94

Rigid bodies with two hinges along a coordinate axis and an additional reaction parallel to another coordinate axis Rigid bodies supported by three vertical wires or by vertical reactions

4.97, 98 4.103, 104 4.105, 110 4.111, 112 4.115, 116 4.120, 121 4.125, 126 4.127, 128

Derrick and boom problems involving unknown tension in two cables Rigid bodies with two hinges along a coordinate axis and an additional reaction not parallel to a coordinate axis Problems involving couples as part of the reaction at a hinge Advanced problems

4.135, 136 4.137, 140 4.141

Problems involving taking moments about an oblique line passing through two supports

4.144, 145 4.146, 148 4.150, 153

4.142, 143 4.147, 149 4.151, 152

Review problems

4.C2, C5 4.C6

4.C1, C3 4.C4

Computer problems

CHAPTER 5: DISTRIBUTED FORCES: CENTROIDS AND CENTERS OF GRAVITY 5.1, 3 5.4 5.6, 9

5.2 5.5, 7 5.8

Centroid of an area formed by combining rectangles and triangles rectangles, triangles, and portions of circular areas

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xvi

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

5.11, 12 5.14, 15 5.16, 17 5.20, 22 5.24, 25 5.29, 30 5.31, 32

5.10, 13

5.36 5.37, 38 5.40

5.34, 35 5.39 5.41 5.42 5.44 *5.47

5.43 5.45, 46 *5.48, *49

5.18, 19 5.21, 23 5.26, 27 5.28, 33

5.50, 51 5.52, 54 5.55, 57 5.60, 61 5.64, *65

5.66, 67 5.71 5.73 5.74, 75 5.78, 79 5.81, 82 5.83, 86 5.87 5.88, 89 5.92, 93

5.53, 56 5.58, 59 5.62, 63

5.68, 69 5.70 5.72 5.76, 77 5.80, 84 5.85 5.90, 91 5.94, 95

5.98, 99 5.100, 101 5.104, 105 5.106, 107 5.110, 113 5.114, 116

5.96, 97 5.102, 103 5.108, 109 5.111, 112 5.115, 117

5.118, 121

5.119, 120

5.122 5.125, 126 5.127

5.123, 124 *5.128, *129

5.132

*5.130, 131

Problem Description triangles, portions of circular or elliptical areas, and areas of analytical functions Derive expression for location of centroid First moment of an area Center of gravity of a wire figure Equilibrium of wire figures Find dimension to maximize distance to centroid 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 parabolic loadings special problems Forces on submerged surfaces reactions on dams or vertical gates reactions on non-vertical gates

Centroids and centers of gravity of three-dimensional bodies composite bodies formed from two common shapes composite bodies formed from three or more elements composite bodies formed from a material of uniform thickness composite bodies formed from a wire or structural shape of uniform cross section composite bodies made of two different materials use integration to locate the centroid of standard shapes: single integration bodies of revolution: single integration special applications: single integration

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xvii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

5.133, *134 5.135, 136

Problem Description bodies formed by cutting a standard shape with an oblique plane: single integration special applications: double integration

5.138, 141 5.142, 144 5.145, 148

5.137, 139 5.140, 143 5.146, 147

Review problems

5.C2, C3 5.C5, C6

5.C1, C4 *5.C7

Computer problems

CHAPTER 6: ANALYSIS OF STRUCTURES TRUSSES 6.2, 4 6.6, 8 6.9, 10 6.13, 14 6.17, 18 6.21, 24 6.25, 26 6.29, 30 6.32, 34 *6.36, *37 *6.39, *40

6.43, 44 6.47, 48 6.49, 50 6.53, 54 6.55, 56 6.61, 62 6.65, 66 6.69, 70 6.73, 74

6.1, 3 6.5, 7 6.11, 12 6.15, 16 6.19, 20 6.22, 23 6.27, 28 6.31, 33 *6.35, *38 *6.41, *42

6.45, 46 6.51, 52 6.57, 58 6.59, 60 6.63, 64 6.67, 68 6.71, 72

Method of joints simple problems problems of average difficulty more advanced problems

designate simple trusses find zero-force members space trusses

Method of sections two of the 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 6.F1, F4

6.F2, F3

6.75, 77 6.79, 80 6.83, 84 6.85, 86

6.76, 78 6.81, 82 6.87, 88 6.89

Free Body Practice Problems – Frames Analysis of Frames easy problems problems where internal forces are changed by repositioning a couple or by moving a force along its line of action

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xviii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units 6.90 6.91, 92 6.97, 98 6.101, 102 6.105, 106 6.107, 108 6.112, 113 6.115 6.117 6.119, 120

U.S. Units

6.93, 94 6.95, 96 6.99, 100 6.103, 104 6.109, 110 6.111, 114 6.116 6.118 6.121

Problem Description replacement of pulleys by equivalent loadings analysis of frames supporting pulleys or pipes analysis of highway vehicles analysis of frames consisting of multiforce members problems involving the solution of simultaneous equations unusual floor systems rigid and non-rigid frames

6.F6, F8

6.F5, F7

Free Body Practice Problems – Machines

6.122, 125 6.126, 127 6.128 6.131, 132 6.133, 134 6.137, 138 6.139, 140 6.143, 144 6.147, 148 6.149, 150 6.155, 156

6.123, 124

Analysis of Machines toggle-type machines

6.159, 160 *6.163

6.129, 130

machines involving cranks

6.135, 136

machines involving a crank with a collar robotic machines tongs pliers, boltcutters, pruning shears wrench, find force to maintain position of toggle force in hydraulic cylinder

6.141, 142 6.145, 146 6.151, 152 6.153, 154 6.157, 158 *6.161, *162

gears and universal joints special tongs

6.164, 165 6.169, 170 6.173, 174

6.166, 167 6.168, 171 6.172, 175

Review problems

6.C2, C4 6.C6

6.C1, C3 6.C5

Computer problems

CHAPTER 7: FORCES IN BEAMS AND CABLES * Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xix

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description CHAPTER 7: FORCES IN BEAMS AND CABLES

7.3, 4 7.5, 6 7.9, 10 7.11, 12 7.15, 16 7.19, 20 7.23, 24 7.27, 28

7.1, 2

Internal forces in members simple frames

7.7, 8 7.13, 14 7.17, 18

curved members frames with pulleys or pipes

7.21, 22 7.25, 26

effect of supports bending moment in circular rods due to their own weight

BEAMS 7.29, 30 7.35, 36 7.39, 40 7.43, 44 7.47, 48 7.49, 50 7.53, 54 7.55, 56 7.57, 58

7.63, 64 7.67, 68 7.69, 70 7.71, 72 7.77, 78 7.79, 80 7.85, 88 7.89, 90

7.31, 32 7.33, 34 7.37, 38 7.41, 42 7.45, 46

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

7.51, 52

beams subjected to forces and couples

7.59, 60 7.61, *62

find value of parameter to minimize absolute value of bending moment

7.65, 66 7.73, 74 7.75, 76 7.81, 82 7.83, 84 7.86, 87 *7.91, *92

Shear and bending-moment diagrams using relations among w, V, and M problems involving no numerical values problems involving numerical values find magnitude and location of maximum bending moment Determine V and M by integrating w , twice Find values of loads given bending moment at two points CABLES

7.93, 94 7.97, 98 7.101, 102 7.105, 106

7.95, 96 7.99, 100 7.103, 104

7.107, 108 7.112, 113 7.115, 116

7.109, 110 7.111, 114 7.117, 118

*7.119 7.120, 121

7.122, 123

Cables with concentrated loads vertical loads horizontal and vertical loads Parabolic cables supports at the same elevation supports at different elevations Derive analogy between a beam and a cable use analogy to solve previous problems

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

*7.124

*7.125, *126

Derive or use d2y/dx2 = w(x) T0

7.129, 130 7.131, 132 7.133, 136 7.139, 140 7.141, 142 7.145, 146 *7.151, *152 *7.153

7.127, 128

Catenary given length of cable and sag or Tm, find span of cable

7.134, 135 7.137, 138 7.143, 144 *7.147, *148 7.149, *150

Problem Description

given span and length of cable, find sag and/or weight given span, Tm, and w, find sag given T0, w, and sag or slope, find span or sag special problems

7.154, 155 7.157, 159 7.163, 164

7.156, 158 7.160, 161 7.162, 165

Review problems

7.C2, C4 7.C5, C6

7.C1, C3

Computer problems

CHAPTER 8: FRICTION 8.F1, F4

8.F2, F3

Free Body Practice Problems – Frames

8.1, 2 8.7, 9 8.10 8.13, 14 8.15, 16 8.21, 22 8.25, 26 8.27, 28 8.32, 33

8.3, 4 8.5, 6 8.8 8.11, 12 8.17, 18 8.19, 20 8.23, 24 8.29, 30 8.31, 34 8.35 8.36, 38 8.41, 42 8.46, 47

For given loading, determine whether block is in equilibrium and find friction force Find minimum force required to start, maintain, or prevent motion

8.37, 39 8.40 8.43, 44 8.45

8.48, 49 8.50, 51 8.54, 55 8.56 8.59, 64 8.65 *8.66, *67 8.68 8.72, 73 8.75, 76

Analyze motion of several blocks Sliding and/or tipping of a rigid body Problems involving wheels and cylinders Problems involving rods Analysis of mechanisms with friction Analysis of more advanced rod and beam problems Analysis of systems with possibility of slippage for various loadings

8.52, 53 8.57

Wedges, introductory problems

8.58, 60 8.61, 62 8.63

Wedges, more advanced problems

8.69, 70 8.71, 74

Square-threaded screws

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xxi

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

Problem Description

8.77, 79 8.80, 81 8.82, 85 8.86, 91 8.92, *94 *8.95, *96 8.100, 101 8.102

8.78, 83 8.84, 87 8.88, 89 8.90 8.93, 97

Disk friction

8.98, 99

Rolling resistance

8.105, 106 8.107, 108 8.109, 112 8.113, 114

8.103, 104

Belt friction belt passing over fixed drum

Axle friction

8.118, 119 8.124, 125 8.126, 127 8.132, 133

8.110, 111 8.115, 116 8.117 8.120, 121 8.122, 123 8.128, 129 8.130, 131

transmission belts and band brakes

8.134, 135 8.138, 139 8.143, 144

8.136, 137 8.140, 141 8.142, 145

Review problems

8.C1, C3 8.C4, C6 8.C8

8.C2, C5 8.C7

Computer problems

advanced problems derivations, V belts

CHAPTER 9: DISTRIBUTED FORCES: MOMENTS OF INERTIA MOMENTS OF INERTIA OF AREAS 9.1, 3 9.6, 8 9.9, 12 9.13, 14 9.15, 17 9.19, 20 9.21, 22 9.25, 26 *9.29

9.2, 4 9.5, 7 9.10, 11

Find by direct integration moments of inertia of an area

9.16, 18

moments of inertia and radii of gyration of an area

9.23, 24 9.27, 28 *9.30

polar moments of inertia and polar radii of gyration of an area Special problems

9.31, 33 9.35, 36

9.32, 34

Parallel-axis theorem applied to composite areas to find moment of inertia and radius of gyration

9.39, 40 9.41, 42

9.37, 38 9.43, 44

centroidal moment of inertia, given I or J centroidal moments of inertia

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xxii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

9.47, 48

9.45, 46

9.49, 52 9.54, 55

9.50, 51 9.53, 56

9.58, *60 9.61, 62 *9.64 *9.65

9.57, 59 *9.63 *9.66

9.67, 68 9.71, 72 9.75, 78

9.69, 70 9.73, 74 9.76, 77

9.79, 80 9.82, 84 9.85, 86 9.88, 90

9.81, 83

9.91, 92 9.94, 96 9.97, 98

9.93, 95

9.104, 105 9.107, 108

9.87, 89

9.99, 100 9.101, 102 9.103, *106 9.109, 110

Problem Description centroidal polar moment of inertia centroidal moments of inertia of composite areas consisting of rolled-steel shapes: doubly-symmetrical composite areas singly-symmetrical composite areas (first locate centroid of area) Center of pressure for end panel of a trough for a submerged vertical gate or cover used to locate centroid of a volume special problems

Products of inertia of areas found by direct integration parallel-axis theorem Using the equations defining the moments and products of inertia with respect to rotated axes to find Ix', Iy', Ix'y' for a given angle of rotation principal axes and principal moments of inertia

Using Mohr's circle to find Ix', Iy', Ix'y' for a given angle of rotation principal axes and principal moments of inertia Special problems

MOMENTS OF INERTIA OF MASSES 9.111, 112 9.117, 118 9.119, 120 9.123, 125 9.128 9.129, 130 9.132, 133 9.135, 136 9.137, *140 9.141, 142 9.145, 148 9.149, 150 9.153, 154 9.155, 156 9.157, 158 9.162

9.113, 114 9.115, 116 9.121, 122 9.124, *126 9.127 9.131, 134

Mass moment of inertia of thin plates: two-dimensions of simple geometric shapes by direct single integration and radius of gyration of composite bodies special problems using the parallel-axis theorem

9.138, 139

of bodies formed of sheet metal or of thin plates: three-dimensions

9.143, 144 9.146, 147

of machine elements and of bodies formed of homogeneous wire

9.151, 152

Mass products of inertia of machine elements of bodies formed of sheet metal or of thin plates

9.159, 160 9.161

of bodies formed of homogeneous wire Derivation and special problem

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xxiii

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

9.165, 166 9.169, 170

9.163, 164 9.167, 168 9.171, 172 9.173, 174

Mass moment of inertia of a body with respect to a skew axis

*9.182, *183

Principal mass moments of inertia and principal axes of inertia

9.187, 188 9.189, 193 9.194, 195

9.185, 186 9.190, 191 9.192, 196

Review problems

9.C3, C5 *9.C7, *C8

9.C1, C2 9.C4, C6

Computer problems

9.175, 176 9.177, 178 *9.179, *180 *9.181, *184

Problem Description

Ellipsoid of inertia and special problems

CHAPTER 10: METHOD OF VIRTUAL WORK 10.1, 3 10.5, 6 10.11, 12 10.13, 14 10.17, 18 10.21, 22 10.23, 24 10.27, 28 10.29, 32 10.33, 34 10.35, 37 10.39, 40 10.43, 44 10.48, 50 10.51, 52 10.53, 54 10.55

10.2, 4 10.7, 8 10.9, 10 10.15, 16 10.19, 20 10.25, 26 10.30, 31 10.36, 38 10.41, 42 10.45, 46 10.47, 49

10.56 10.57, 58

10.62, 63 10.64, 65 10.67

10.59, 60 10.61, 66 10.68

10.69, 72 10.73, 74 10.77, 78 10.79, 82 10.85, 86

10.70, 71 10.75, 76 10.80, 81 10.87, 88

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 of loads linear springs included in mechanism torsional spring included in mechanism Problems requiring the use of the law of cosines Problems involving the effect of friction Use method of virtual work to find: reactions of a beam internal forces in a mechanism movement of a truss joint

Potential energy method used to solve problems from Sec. 10.4 establish that equilibrium is neutral find position of equilibrium and determine its stability for a system involving: gears and drums torsional springs linear springs

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xxiv

TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS (CONTINUED)

Problem Number* SI Units

U.S. Units

10.83, 84

Problem Description two applied forces determine stability of a known position of equilibrium for a system with one degree of freedom

10.89, 93 10.94, 96 *10.97, *98

10.90, 91 10.92, 95 *10.99, *100

10.103, 104 10.105, 107 10.111, 112

10.101, 102 10.106, 108 10.109, 110

Review problems

10.C2, C3 10.C5, C6 10.C7

10.C1, C4

Computer problems

two degrees of freedom

* Problems which do not involve any specific system of units have been indicated by underlining their number. Answers are not given to problems with a number set in italic type.

xxv

xxvi

4.1 4.1 4.2 4.3 4.3 5.1 5.2 5.3 5.4

6.1 6.2 6.3 6.4 6.1–4 7.1 7.2 7.3

8.1 8.2 8.4 9.1 9.2 9.3 9.4 9.5 9.6

10.1 10.1 10.2

21 22 23 24 25 26 27 28 29

30 31 32 33 34 35 36 37 38 39

40 41 42

Sections 1.1–6 2.1–2 2.3 2.4 2.5 3.1 3.2 3.3 3.4

11 12 13 14 15 16 17 18 19 20

Period 1 2 3 4 5 6 7 8 9 10

Virtual Work Virtual Work Potential Energy

Laws of Friction Wedges and Screws Belt Friction Moments of Inertia of Areas Composite Areas Product of Inertia Mohr’s Circle Moments of Inertia of Masses Principal Axes EXAM NUMBER FOUR

Trusses: Method of Joints Trusses: Method of Sections Analysis of Frames Analysis of Machines Review of Chapter 6 Internal Forces in Members Beams Beams EXAM NUMBER THREE

Equilibrium in Two Dimensions Equilibrium in Two Dimensions Two- and Three-Force Bodies Equilibrium in Three Dimensions Equilibrium in Three Dimensions Centroids and First Moments Centroids by Integration Beams and Submerged Surfaces Centroids of Volumes EXAM NUMBER TWO

Topics Introduction Addition and Resolution of Forces Equilibrium of a Particle Forces in Space Equilibrium in Space Vector Product, Moment of a Force about a Point Scalar Product, Moment of a Force about an Axis Couples Equivalent Systems of Forces EXAM NUMBER ONE

10.1, 9, 18, 20 10.31, 35, 46, 53 10.69, 80, 86, 91

8.1, 18, 32, 36 8.52, 54, 63, 73 8.104, 109, 121, 126 9.1, 10, 17, 23 9.34, 42, 51, 58 9.72, 83, 86 9.93, 97, 103 9.112, 121, 135, 147 9.152, 158, 167, 175

6.2, 11, 21, 30 6.45, 54, 68, 69 6.75, 88, 102, 109 6.123, 134, 145, 149 6.15, 52, 91, 151 7.1, 12, 18, 23 7.36, 46, 50, 59 7.65, 70, 81, *89

10.3, 10, 17, 19 10.30, 37, 45, 54 10.72, 81, 83, 92

8.2, 17, 27, 41 8.53, 55, 62, 72 8.103, 112, 120, 133 9.3, 11, 15, 28 9.32, 41, 50, *60 9.71, 81, 85 9.95, 98, *106 9.111, 122, 136, 143 9.151, 155, 168, 176

6.4, 12, 24, 27 6.46, 53, 67, 70 6.77, 89, 101, 110 6.124, 133, 141, 155 6.19, 51, 92, 152 7.2, 11, 17, 24 7.35, 45, 49, 59 7.66, 69, 82, *90

4.2, 12, 15, 24 4.33, 35, 48, 51 4.62, 71, 81, 90 4.94, 99, 105, 113 4.119, 128, 138 5.3, 13, 20, 26 5.35, 40, 53, 60 5.67, 77, 82, 90 5.103, 106, 117, 125

2.10, 17, 29, 37 2.48, 49, 59, 69 2.72, 82, 89, 94 2.104, 108, 113, 121 3.5, 11, 26, 30 3.39, 47, 55, *67 3.70, 82, 90, 97 3.105, 110, 122, 128

2.5, 15, 27, 31 2.43, 50, 63, 65 2.71, 81, 85, 97 2.103, 107, 119, 125 3.4, 12, 23, 27 3.43, 48, 61, *68 3.71, 84, 87, 99 3.106, 111, 126, 127

4.1, 14, 19, 23 4.34, 36, 43, 52 4.63, 72, 75, 86 4.93, 100, 110, 117 4.122, 127, 133 5.1, 10, 24, 33 5.34, 43, 58, 64 5.66, 76, 81, 94 5.102, 107, 119, 122

List 2

List 1

10.2, 12, 16, 24 10.32, 36, 43, 57 10.71, 77, 88, 89

8.3, 14, 30, 40 8.48, 57, 65, 74 8.105, 110, 119, 128 9.4, 12, 16, 21 9.31, 43, 52, 57 9.76, 79, 87 9.92, 100, 107 9.113, 119, 138, 142 9.149, 159, 165, 174

6.1, 13, 23, 29 6.43, 59, 62, 71 6.76, 83, 100, 107 6.127, 129, 144, 153 6.17, 50, 94, 140 7.3, 8, 19, 26 7.42, 48, 52, 55 7.64, 73, 78, 86

4.4, 13, 17, 25 4.31, 42, 46, 57 4.66, 73, 77, 88 4.91, 103, 107, 116 4.121, 131, 135 5.5, 14, 23, 30 5.37, 44, 54, 63 5.70, 74, 80, 89 5.100, 111, 114, 124

2.11, 16, 26, 36 2.46, 55, 61, 67 2.75, 84, 88, 92 2.101, 106, 116, 123 3.7, 13, 25, 28 3.41, 53, 57, *64 3.73, 86, 89, 96 3.102, 108, 120, 129

List 3

10.4, 14, 15, 23 10.33, 38, 44, 58 10.70, 78, 87, 93

8.4, 13, 29, 39 8.49, 60, 64, 71 8.106, 111, 118, 129 9.2, 9, 18, 22 9.33, 44, 49, 59 9.74, 80, 89 9.91, 99, 108 9.114, 120, 139, 141 9.150, 160, 166, 173

6.3, 9, 22, 28 6.44, 60, 61, 72 6.81, 85, 99, 108 6.122, 130, 143, 154 6.18, 49, 93, 137 7.4, 7, 20, 25 7.41, 47, 51, 58 7.63, 74, 77, 87

4.7, 9, 18, 22 4.32, 41, 47, 55 4.61, 69, 79, 83 4.95, 97, 106, 115 4.120, 129, 140 5.2, 11, 21, 31 5.39, 41, 52, 62 5.69, 78, 84, 88 5.104, 109, 116, 123

2.6, 19, 28, 35 2.44, 51, 62, 68 2.77, 79, 87, 91 2.99, 112, 115, 124 3.6, 9, 22, 29 3.37, 49, 58, *65 3.80, 83, 91, 95 3.101, 114, 119, 130

List 4

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK

10.8, 13, 21, 25 10.34, 42, 47, 55 10.76, 79, 84, 95

8.8, 16, 33, 38 8.50, 58, 70, 76 8.108, 115, 123, 127 9.5, 14, 19, 24 9.35, 45, 56, 61 9.77, 84, 88 9.94, 102, 110 9.116, 123, *140, 146 9.154, 163, 171, 178

6.7, 14, 25, 33 6.47, 57, 64, 73 6.78, 84, 105, 111 6.126, 135, 142, 150 6.16, 48, 98, 157 7.5, 14, 22, 27 7.38, 43, 54, 60 7.68, 75, 83, 85

4.3, 10, 16, 27 4.30, 37, 44, 58 4.68, 74, 76, 87 4.92, 104, 111, 118 4.126, 132, 134 5.8, 15, 25, 28 5.38, 42, 59, *65 5.72, 75, 83, 95 5.101, 112, 115, 126

2.12, 18, 30, 33 2.45, 56, 64, 66 2.76, 80, 86, 93 2.102, 105, 120, 126 3.3, 14, 21, 31 3.42, 50, 56, *69 3.75, 85, 88, 100 3.107, 112, 121, 131

List 5

10.7, 11, 22, 26 10.29, 41, 51, 56 10.75, 82, 85, 90

8.6, 15, 28, 42 8.51, 61, 69, 75 8.107, 116, 122, 132 9.7, 13, 20, 27 9.36, 46, 53, 62 9.73, 82, 90 9.96, 101, 109 9.115, 125, 137, 144 9.153, 164, 172, 177

6.5, 10, 26, 31 6.48, 58, 63, 74 6.82, 86, 106, 114 6.128, 136, 146, 156 6.20, 47, 97, 158 7.6, 13, 21, 28 7.37, 44, 53, 61 7.67, 76, 84, 88

4.8, 11, 20, 21 4.29, 38, 49, 56 4.64, 70, 82, 89 4.96, 98, 112, 114 4.125, 130, 139 5.7, 12, 22, 27 5.36, *47, 56, 61 5.68, 79, 86, 91 5.105, 108, 120, 127

2.7, 20, 25, 38 2.47, 52, 60, 70 2.78, 83, 90, 98 2.100, 111, 114, 122 3.8, 10, 24, 34 3.38, 54, 62, *66 3.72, 81, 92, 94 3.103, 109, 123, 132

List 6

NO ANSWERS ARE GIVEN IN THE BOOK FOR ANY OF THESE PROBLEMS

50% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 50% U.S. CUSTOMARY UNITS

This schedule includes all of the material of VECTOR MECHANICS FOR ENGINEERS: STATICS with the exception of Sections 3.4D, 6.1D, 7.4–7.5, 8.3, and 9.6C.

TABLE III: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN STATICS

xxvii

4.1 4.1 4.2 4.3 4.3 5.1 5.2 5.3 5.4

6.1 6.2 6.3 6.4 6.1–4 7.1 7.2 7.3

8.1 8.2 8.4 9.1 9.2 9.3 9.4 9.5 9.6

10.1 10.1 10.2

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29

30 31 32 33 34 35 36 37 38 39

40 41 42

Sections

1.1–6 2.1–2 2.3 2.4 2.5 3.1 3.2 3.3 3.4

1 2 3 4 5 6 7 8 9 10

Period

Topics

Virtual Work Virtual Work Potential Energy

Laws of Friction Wedges and Screws Belt Friction Moments of Inertia of Areas Composite Areas Product of Inertia Mohr’s Circle Moments of Inertia of Masses Principal Axes EXAM NUMBER FOUR

Trusses: Method of Joints Trusses: Method of Sections Analysis of Frames Analysis of Machines Review of Chapter 6 Internal Forces in Members Beams Beams EXAM NUMBER THREE

Introduction Addition and Resolution of Forces Equilibrium of a Particle Forces in Space Equilibrium in Space Vector Product, Moment of a Force about a Point Scalar Product, Moment of a Force about an Axis Couples Equivalent Systems of Forces EXAM NUMBER ONE

10.1, 9, 18, 28 10.31, 35, 52, 53 10.69, 80, 86, 96

8.1, 18, 32, 37 8.52, 54, *67, 73 8.104, 109, 125, 126 9.1, 10, 17, 26 9.34, 42, 55, 58 9.72, 83, 86 9.93, 97, 103 9.112, 121, 135, 148 9.152, 158, 170, 175

6.2, 11, 21, 32 6.45, 54, 66, 69 6.75, 88, 102, 113 6.123, 134, 147, 149 6.15, 52, 91, 125 7.1, 12, 15, 23 7.36, 46, 50, 57 7.65, 70, 80, *89

4.1, 14, 19, 28 4.34, 36, 45, 52 4.63, 72, 75, 84 4.93, 100, 108, 117 4.122, 127, 133 5.1, 10, 24, 29 5.34, 43, 57, 64 5.66, 76, 81, 93 5.102, 107, 118, 122

2.5, 15, 27, 34 2.43, 50, 57, 65 2.71, 81, 85, 95 2.103, 107, 118, 125 3.4, 12, 23, 32 3.43, 48, 59, *68 3.71, 84, 87, 98 3.106, 111, 125, 127

List 1a

10.3, 10, 17, 27 10.30, 37, 48, 54 10.72, 81, 83, 94

8.2, 17, 27, 43 8.53, 55, *66, 72 8.103, 112, 124, 133 9.3, 11, 15, 25 9.32, 41, 54, *60 9.71, 81, 85 9.95, 98, *106 9.111, 122, 136, 145 9.151, 155, 169, 176

6.4, 12, 24, 34 6.46, 53, 65, 70 6.77, 89, 101, 112 6.124, 133, 148, 155 6.19, 51, 92, 139 7.2, 11, 16, 24 7.35, 45, 49, 56 7.66, 69, 79, *90

4.2, 12, 15, 26 4.33, 35, 50, 51 4.62, 71, 81, 85 4.94, 99, 109, 113 4.119, 128, 138 5.3, 13, 20, 32 5.35, 40, 55, 60 5.67, 77, 82, 92 5.103, 106, 121, 125

2.10, 17, 29, 32 2.48, 49, 58, 69 2.72, 82, 89, 96 2.104, 108, 117, 121 3.5, 11, 26, 33 3.39, 47, 60, *67 3.70, 82, 90, 93 3.105, 110, 124, 128

List 2a

10.5, 12, 16, 24 10.32, 40, 43, 57 10.74, 77, 88, 89

8.7, 14, 30, 40 8.48, 56, 65, 74 8.105, 113, 119, 128 9.8, 12, 16, 21 9.31, 47, 52, 57 9.75, 79, 87 9.92, 104, 107 9.117, 119, 138, 142 9.149, 157, 165, 174

6.6, 13, 23, 29 6.43, 56, 62, 71 6.79, 83, 100, 107 6.127, 132, 144, 153 6.50, 94, 140, 164 7.3, 10, 19, 26 7.40, 48, 52, 55 7.64, 71, 78, 86

4.5, 13, 17, 25 4.31, 40, 46, 57 4.67, 73, 77, 88 4.91, 102, 107, 116 4.124, 131, 135 5.6, 14, 23, 30 5.37, 46, 54, 63 5.73, 74, 80, 89 5.100, 113, 114, 124

2.8, 16, 26, 36 2.46, 54, 61, 67 2.74, 84, 88, 92 2.101, 109, 116, 123 3.1, 13, 25, 28 3.41, 52, 57, *64 3.74, 86, 89, 96 3.102, 116, 120, 129

List 3a

10.6, 14, 15, 23 10.33, 39, 44, 58 10.73, 78, 87, 93

8.9, 13, 29, 39 8.49, 59, 64, 71 8.106, 114, 118, 129 9.6, 9, 18, 22 9.33, 48, 49, 59 9.78, 80, 89 9.91, 105, 108 9.118, 120, 139, 141 9.150, 156, 166, 173

6.8, 9, 22, 28 6.44, 55, 61, 72 6.80, 85, 99, 108 6.122, 131, 143, 154 6.49, 93, 137, 165 7.4, 9, 20, 25 7.39, 47, 51, 58 7.63, 72, 77, 87

4.6, 9, 18, 22 4.32, 39, 47, 55 4.65, 69, 79, 83 4.95, 101, 106, 115 4.123, 129, 140 5.9, 11, 21, 31 5.39, 45, 52, 62 5.71, 78, 84, 88 5.104, 110, 116, 123

2.9, 19, 28, 35 2.44, 53, 62, 68 2.73, 79, 87, 91 2.99, 110, 115, 124 3.2, 9, 22, 29 3.37, 51, 58, *65 3.78, 83, 91, 95 3.101, 117, 119, 130

List 4a

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK

75% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 25% U.S. CUSTOMARY UNITS

This schedule includes all of the material of VECTOR MECHANICS FOR ENGINEERS: STATICS with the exception of Sections 3.4D, 6.1D, 7.4–7.5, 8.3, and 9.6C.

TABLE IV: SAMPLE ASSIGNMENT SCHEDULE FOR A COURSE IN STATICS

xxviii

4.1 4.1 6.3 6.4 4.3 4.3 8.1 8.2

11.1 11.2 11.4 11.5 12.1 12.2

13.1 13.2 13.2D 13.3 13.4 14.1 14.2

15.1 15.2 15.3 15.4 15.5 9.5

16.1 16.1 16.2 16.2 17.1 17.2 17.3

19.1 19.2 19.3 19.4

20 21 22 23 24 25 26 27

28 29 30 31 32 33 34 35

36 37 38 39 40 41 42 43

44 45 46 47 48 49 50 51 52

53 54 55 56

Sections 1.1–6 2.1–2 2.3 2.4 2.5 3.1 3.2 3.3 3.4

11 12 13 14 15 16 17 18 19

Period 1 2 3 4 5 6 7 8 9 10

TABLE V: SAMPLE ASSIGNMENT SCHEDULE FOR A COMBINED COURSE IN STATICS AND DYNAMICS

Free Vibrations of Particles Free Vibrations of Rigid Bodies Energy Methods Forced Vibrations

Plane Motion of Rigid Bodies Plane Motion of Rigid Bodies Constrained Plane Motion Constrained Plane Motion Work and Energy Impulse and Momentum Eccentric Impact Review EXAM NUMBER SIX

Translation, Rotation General Plane Motion Instantaneous Center Acceleration in Plane Motion Coriolis Acceleration in Plane Motion Moments of Inertia of Masses Review EXAM NUMBER FIVE

Work and Energy, Power Conservation of Energy Application to Space Mechanics Impulse and Momentum Impact Systems of Particles Systems of Particles EXAM NUMBER FOUR

Rectilinear Motion Uniformly Accelerated Motion Curvilinear Motion (Rect. Comps) Curvilinear Motion (Other Comp.) Equations of Motion Angular Momentum Review EXAM NUMBER THREE

Equilibrium in Two Dimensions Equilibrium in Two Dimensions Analysis of Frames Analysis of Machines Equilibrium in Three Dimensions Equilibrium in Three Dimensions Laws of Friction Wedges and Screws EXAM NUMBER TWO

List 1

9.112, 121, 135, 147

4.1, 14, 19, 23 4.34, 36, 43, 52 6.75, 88, 102, 109 6.123, 134, 145, 149 4.93, 100, 110, 117 4.122, 127, 133 8.1, 18, 32, 36 8.52, 54, 63, 73

2.5, 15, 27, 31 2.43, 50, 63, 65 2.71, 81, 85, 97 2.103, 107, 119, 125 3.4, 12, 23, 27 3.43, 48, 61, *68 3.71, 84, 87, 99 3.106, 111, 126, 127

9.111, 122, 136, 143

4.2, 12, 15, 24 4.33, 35, 48, 51 6.77, 89, 101, 110 6.124, 133, 141, 155 4.94, 99, 105, 113 4.119, 128, 138 8.2, 17, 27, 41 8.53, 55, 62, 72

2.10, 17, 29, 37 2.48, 49, 59, 69 2.72, 82, 89, 94 2.104, 108, 113, 121 3.5, 11, 26, 30 3.39, 47, 55, *67 3.70, 82, 90, 97 3.105, 110, 122, 128

List 2

9.113, 119, 138, 142

4.4, 13, 17, 25 4.31, 42, 46, 57 6.76, 83, 100, 107 6.127, 129, 144, 153 4.91, 103, 107, 116 4.121, 131, 135 8.3, 14, 30, 40 8.48, 57, 65, 74

2.11, 16, 26, 36 2.46, 55, 61, 67 2.75, 84, 88, 92 2.101, 106, 116, 123 3.7, 13, 25, 28 3.41, 53, 57, *64 3.73, 86, 89, 96 3.102, 108, 120, 129

List 3

9.114, 120, 139, 141

4.7, 9, 18, 22 4.32, 41, 47, 55 6.81, 85, 99, 108 6.122, 130, 143, 154 4.95, 97, 106, 115 4.120, 129, 140 8.4, 13, 29, 39 8.49, 60, 64, 71

2.6, 19, 28, 35 2.44, 51, 62, 68 2.77, 79, 87, 91 2.99, 112, 115, 124 3.6, 9, 22, 29 3.37, 49, 58, *65 3.80, 83, 91, 95 3.101, 114, 119, 130

List 4

ANSWERS TO ALL OF THESE PROBLEMS ARE GIVEN IN THE BACK OF THE BOOK

9.116, 123, *140, 146

4.3, 10, 16, 27 4.30, 37, 44, 58 6.78, 84, 105, 111 6.126, 135, 142, 150 4.92, 104, 111, 118 4.126, 132, 134 8.8, 16, 33, 38 8.50, 58, 70, 76

2.12, 18, 30, 33 2.45, 56, 64, 66 2.76, 80, 86, 93 2.102, 105, 120, 126 3.3, 14, 21, 31 3.42, 50, 56, *69 3.75, 85, 88, 100 3.107, 112, 121, 131

List 5

9.115, 125, 137, 144

4.8, 11, 20, 21 4.29, 38, 49, 56 6.82, 86, 106, 114 6.128, 136, 146, 156 4.96, 98, 112, 114 4.125, 130, 139 8.6, 15, 28, 42 8.51, 61, 69, 75

2.7, 20, 25, 38 2.47, 52, 60, 70 2.78, 83, 90, 98 2.100, 111, 114, 122 3.8, 10, 24, 34 3.38, 54, 62, *66 3.72, 81, 92, 94 3.103, 109, 123, 132

List 6

NO ANSWERS ARE GIVEN IN THE BOOK FOR ANY OF THESE PROBLEMS

50% OF THE PROBLEMS IN EACH OF THE FOLLOWING LISTS USE SI UNITS AND 50% U.S. CUSTOMARY UNITS

This schedule is intended for a 4-semester-credit-hour course consisting of (1) the abridged Statics course defined in Table I of this manual and (2) the standard Dynamics course defined in Table I of the Dynamics Instructor’s Manual. Note that the Statics portion of the course is designed to serve only as an introduction to Dynamics; a more complete coverage of Statics is recommended for curricula including courses in Mechanics of Materials (see Tables III and ff. of the Statics Manual).

CHAPTER 2

600 N 45°

900 N

30°

PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 1391 kN, a = 47.8

R = 1391 N

47.8 ◀

PROBLEM 2.2

800 lb

Two forces are applied as shown to a bracket support. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

60° 35° 500 lb

SOLUTION

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 906 lb, a = 26.6

R = 906 lb

26.6 ◀

PROBLEM 2.3 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:

(b)

Triangle rule:

We measure:

R = 179 N, a = 75.1

R = 179 N

75.1

PROBLEM 2.4 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.

SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 77.1 lb, a = 85.4

R = 77.1 lb

85.4

120 N 25°

PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

Using the triangle rule and the law of sines:

120 N

P sin 25

(a)

sin30

(b)

30 + b + 25 = 180

P = 101.4 N ◀

b = 180 - 25 - 30 = 125

120 N sin 30

R sin125

R = 196.6 N ◀

A 15°

PROBLEM 2.6

25°

A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1= 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines: (a)

75 + 40 + a = 180 a = 180 - 75 - 40 = 65 800 lb sin 65

(b)

800 lb sin 65

T2 sin 75

T 2 = 853 lb ◀

R sin 40

R = 567 lb ◀

A 15°

PROBLEM 2.7

25°

A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines: (a)

75 + 40 + b = 180 b = 180 - 75 - 40 = 65

(b)

T1 1000 lb = sin 75° sin 65

T1 = 938 lb ◀

1000 lb R = sin 75° sin 40

R = 665 lb ◀

PROBLEM 2.8 B A

30° α C

A disabled automobile is pulled by means of two ropes as shown. The tension in rope AB is 2.2 kN, and the angle α is 25. Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A.

SOLUTION

Using the law of sines: TA C R 2.2 kN = = sin 30° sin125 sin 25

TAC = 2.603 kN R = 4.264 kN

(a)

T A C = 2.60 kN ◀

(b)

R = 4.26 kN ◀

PROBLEM 2.9 B A

30° α

A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigonometry the tension in rope AC and the value of α so that the resultant force exerted at A is a 4.8-kN force directed along the axis of the automobile.

SOLUTION

Using the law of cosines:

TAC 2 = (3 kN)2 + (4.8 kN)2 - 2(3 kN)(4.8 kN)cos 30° TAC = 2.6643 kN

Using the law of sines:

sin a sin 30 = 3 kN 2.6643 kN a = 34.3 TA C = 2.66 kN

34.3 ◀

PROBLEM 2.10 50 N 25° α P

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 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 triangle rule and law of sines: (a)

sin a

sin 25

50 N 35 N sin a = 0.60374 a = 37.138 

(b)

a = 37.1  ◀

a + b + 25 = 180 b = 180 - 25 - 37.138 = 117.862 R 35 N = sin117.862 sin 25

R = 73.2 N ◀

30°

PROBLEM 2.11

425 lb α

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 at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines: (a)

b + 50 + 60 = 180 b = 180 - 50 - 60 = 70

(b)

425 lb P = sin 70 sin 60

P = 392 lb ◀

425 lb R = sin 70 sin 50

R = 346 lb ◀

30°

PROBLEM 2.12

425 lb α

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 at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the law of sines: (a)

(a + 30) + 60 + b = 180 b = 180 - (a + 30) - 60 b = 90 - a sin (90 - a ) 425 lb

(b)

sin 60 500 lb

90 - a = 47.402

a = 42.6 ◀

R 500 lb = sin (42.598 + 30) sin 60

R = 551 lb ◀

425 lb 30°

PROBLEM 2.13 A steel tank is to be positioned in an excavation. 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, (b) the corresponding magnitude of R.

SOLUTION

The smallest force P will be perpendicular to R. (a)

P = (425 lb) cos30

(b)

R = (425 lb)sin 30

P = 368 lb

◀

R = 213 lb ◀

PROBLEM 2.14 50 N 25° α P

For the hook support of Prob. 2.10, 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 of R.

SOLUTION

The smallest force P will be perpendicular to R. (a)

P = (50 N)sin 25

P = 21.1 N ◀

(b)

R = (50 N) cos 25

R = 45.3 N ◀

PROBLEM 2.15 The barge B is pulled by two tugboats A and C. At a given instant the tension in cable AB is 4500 lb and the tension in cable BC is 2000 lb. Determine by trigonometry the magnitude and direction of the resultant of the two forces applied at B at that instant.

SOLUTION

Using the law of cosines:

b = 180 - 30 - 45 b = 105 R 2 = (4500 lb) + (2000 lb) - 2 (4500 lb)(2000 lb) cos 105 2

R = 5380 lb

Using the law of sines:

2000 lb R = sin b sin (30 - a) 5380 lb 2000 lb = sin105 sin (30 - a) a = 8.94  R = 5380 lb

8.94 ◀

PROBLEM 2.16 600 N 45°

900 N

30°

Solve Prob. 2.1 by trigonometry.

PROBLEM 2.1 Two forces are applied as shown to a hook. 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 = (900 N)2 + (600 N )2 - 2(900 N )(600 N)cos(135) R = 1390.57N

Using the law of sines:

sin(a - 30) sin (135) = 600N 1390.57N a - 30 = 17.7642 a = 47.764 R = 1391N

47.8 ◀

PROBLEM 2.17 Solve Problem 2.4 by trigonometry.

SOLUTION Using the triangle rule and the law of cosines:

20 + 35 + a = 180 a = 125 R 2 = P 2 + Q 2 - 2 PQ cos a R 2 = (60 lb)2 + (25 lb)2 - 2(60 lb)(25 lb)cos125 R 2 = 3600 + 625 + 3000(0.5736) R = 77.108 lb Using the law of sines:

sin125 sin b = 25 lb 77.108 lb b = 15.402

70 + b = 85.402

R = 77.1 lb

85.4 ◀

120 N 25°

PROBLEM 2.18 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N.

SOLUTION

Using the laws of cosines and sines:

P 2 = (120 N)2 + (160 N)2 - 2(120 N)(160 N)cos25 P = 72.096 N And

sin a sin 25 = 120 N 72.096 N sin a = 0.70343 a = 44.703 P = 72.1 N

44.7 ◀

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 force is 10 kN in member A and 15 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 triangle and the laws of cosines and sines We have

g = 180 - (40 + 20) = 120

Then

R 2 = (10 kN)2 + (15 kN)2 - 2(10 kN)(15 kN)cos120 = 475 kN2 R = 21.794 kN 15 kN 21.794 kN = sin a sin120 æ 15 kN ö÷ sin a = çç ÷ sin120 çè 21.794 kN ÷÷ø = 0.59605 a = 36.588

and

Hence:

f = a + 50 = 86.588

R = 21.8 kN

86.6°

PROBLEM 2.20 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 15 kN in member A and 10 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 triangle and the laws of cosines and sines: We have

g = 180 - (40 + 20) = 120

Then

R 2 = (15 kN)2 + (10 kN)2 - 2(15 kN)(10 kN)cos120 = 475 kN2 R = 21.794 kN 10 kN 21.794 kN = sin a sin120 æ 10 kN ö÷ sin a = çç ÷ sin120 çè 21.794 kN ø÷÷ = 0.39737 a = 23.414

and

Hence:

f = a + 50 = 73.414

R = 21.8 kN

73.4°

PROBLEM 2.21

y 28 in.

96 in.

Determine the x and y components of each of the forces shown.

84 in.

50 lb

80 in. 29 lb

51 lb 90 in.

48 in.

SOLUTION Compute the following distances: OA = (84)2 + (80)2 = 116 in. OB = (28)2 + (96)2 = 100 in. OC = (48)2 + (90)2 = 102 in. 29-lb Force:

50-lb Force:

Fx = +(29 lb)

84 116

Fx = + 21.0 lb ◀

Fy = +(29 lb)

80 116

Fy = +20.0 lb ◀

Fx = -(50 lb)

28 100

Fx = - 14.00 lb ◀

96 100

Fy = + 48.0 lb ◀

Fx = +(51 lb)

48 102

Fx = + 24.0 lb ◀

Fy = -(51 lb)

90 102

Fy = -45.0 lb ◀

Fy = +(50 lb) 51-lb Force:

PROBLEM 2.22

Determine the x and y components of each of the forces shown.

800 Dimensions in mm

800 N 600 O

424 N

408 N

560

480

900

SOLUTION Compute the following distances: OA = (600)2 + (800)2 = 1000 mm OB = (560)2 + (900)2 = 1060 mm OC = (480)2 + (900)2 = 1020 mm 800-N Force:

424-N Force:

408-N Force:

Fx = +(800 N)

800 1000

Fx = + 640 N ◀

Fy = +(800 N)

600 1000

Fy = +480 N ◀

Fx = -(424 N)

560 1060

Fx = - 224 N ◀

Fy = -(424 N)

900 1060

Fy = -360 N ◀

Fx = +(408 N)

480 1020

Fx = + 192.0 N ◀

Fy = -(408 N)

900 1020

Fy = -360 N ◀

PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION

350-N Force:

800-N Force:

600-N Force:

Fx = +(350 N)cos25

Fx = + 317 N ◀

Fy = +(350 N)sin 25

Fy = +147.9 N ◀

Fx = +(800 N)cos70

Fx = + 274 N ◀

Fy = +(800 N)sin 70

Fy = +752 N ◀

Fx = -(600 N)cos60

Fx = - 300 N ◀

Fy = +(600 N)sin 60

Fy = +520 N ◀

PROBLEM 2.24 Determine the x and y components of each of the forces shown.

SOLUTION

80-lb Force:

120-lb Force:

150-lb Force:

Fx = +(80 lb)cos30

Fx = + 69.3 lb ◀

Fy = -(80 lb)sin30

Fy = -40.0 lb ◀

Fx = +(120 lb)cos75

Fx = + 31.1 lb ◀

Fy = -(120 lb)sin 75

Fy = -115.9 lb ◀

Fx = -(150 lb)cos 40

Fx = - 114.9 lb ◀

Fy = -(150 lb)sin 40

Fy = -96.4 lb ◀

PROBLEM 2.25 A

Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

720 mm

650 mm

SOLUTION BC = (650 mm)2 + (720 mm)2 = 970 mm æ 650 ö÷ Px = P çç ÷ çè 970 ÷÷ø

(a)

æ 970 ö÷ ÷ P = Px çç çè 650 ÷÷ø æ 970 ö÷ = 325 N çç ÷ çè 650 ÷÷ø = 485 N P = 485 N ◀

(b)

æ 720 ö÷ Py = P çç ÷ èç 970 ÷ø÷ æ 720 ÷ö = 485 N çç ÷ çè 970 ÷÷ø = 360 N Py = 970 N ◀

C 35°

PROBLEM 2.26 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

B A Q

SOLUTION

P sin 35 = 300 lb

(a)

P= (b)

Vertical component

300 lb sin 35

P = 523 lb ◀

Pv = P cos 35 

= (523 lb)cos35

Pv = 428 lb ◀

PROBLEM 2.27

The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB.

M 45° A

30°

SOLUTION 180 = 45 + a + 90 + 30 a = 180 - 45 - 90 - 30 = 15 (a)

Px P P P= x cos a 600 N = cos15 = 621.17 N

cos a =

P = 621 N ◀

(b)

tan a =

Px Py = Px tan a = (600 N) tan15 = 160.770 N Py = 160.8 N ◀

PROBLEM 2.28 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

55° A B Q

SOLUTION

(a)

Py cos 55

350 lb cos 55

= 610.21 lb (b)

P = 610 lb ◀

Px = P sin 55

= (610.21 lb)sin 55 = 499.85 lb

Px = 500 lb ◀

PROBLEM 2.29

A B Q

60° C

The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC.

50° D

SOLUTION

(a)

750 N = P sin 20 P = 2192.9 N

(b)

P = 2190 N ◀

PA B C = P cos 20 

= (2192.9 N)cos20

PA B C = 2060 N ◀

PROBLEM 2.30

The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

2.4 m

SOLUTION (a)

37 P 12 x 37 = (720 N) 12 = 2220 N

P = 2.22 kN ◀

(b)

35 P 12 x 35 = (720 N) 12 = 2100 N

Py =

Py = 2.10 kN ◀

PROBLEM 2.31

y 28 in.

96 in.

Determine the resultant of the three forces of Problem 2.21.

84 in.

PROBLEM 2.21 Determine the x and y components of each of the forces shown. 50 lb

80 in. 29 lb

51 lb 90 in.

48 in.

SOLUTION Components of the forces were determined in Problem 2.21: Force

x Comp. (lb)

y Comp. (lb)

29 lb

+21.0

+20.0

50 lb

-14.00

+48.0

51 lb

+24.0

-45.0

R x = + 31.0

R y = + 23.0

R = Rx i + Ry j = (31.0 lb)i + (23.0 lb) j Ry tan a = Rx 23.0 = 31.0 a = 36.573 R=

23.0 lb sin (36.573) R = 38.6 lb

= 38.601 lb

36.6 ◀

PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.23.

PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.23: Force

x Comp. (N)

y Comp. (N)

350 N

+317

+147.9

800 N

+274

+752

600 N

-300

+520

R x = + 291

R y = +1419.9

R = R x i + Ry j = (291 N)i + (1419.9 N) j tan a =

Rx 1419.9 N = 291 N a = 78.418 R=

1419.9 N sin (78.418) R = 1449 N

= 1449 N

78.4 ◀

PROBLEM 2.33 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. (lb)

y Comp. (lb)

80 lb

+69.3

-40.0

120 lb

+31.1

-115.9

150 lb

-114.9

-96.4

R x = - 14.50

Ry = -252.3

R = Rx i + Ry j = -(14.50 lb)i - (252.3 lb) j Rx Ry -14.50 = -252.3 b = 3.289

tan b =

252.3 lb cos(3.289) R = 253 lb

= 253 lb

86.7 ◀

PROBLEM 2.34

Determine the resultant of the three forces of Problem 2.22.

800 Dimensions in mm

PROBLEM 2.22 Determine the x and y components of each of the forces shown.

800 N 600 O

424 N

408 N

560

480

900

SOLUTION Components of the forces were determined in Problem 2.22: Force

x Comp. (N)

y Comp. (N)

800 lb

+640

+480

424 lb

-224

-360

408 lb

+192

-360

R x = + 608

R y = - 240

R = Rx i + R y j = (608 lb)i + (-240 lb) j tan a =

Rx 240 = 608

a = 21.541 R=

240 N sin(21.541°)

= 653.65 N

R = 654 N

21.5 ◀

PROBLEM 2.35 Knowing that a = 35°, determine the resultant of the three forces shown. α

α 30°

100 N

200 N 150 N

SOLUTION 100-N Force:

Fx = +(100 N) cos35 = +81.915 N Fy = -(100 N)sin 35 = -57.358 N

150-N Force:

Fx = +(150 N) cos65 = +63.393 N Fy = -(150 N)sin 65 = -135.946 N

200-N Force:

Fx = -(200 N) cos35 = -163.830 N Fy = -(200 N)sin 35 = -114.715 N

Force

x Comp. (N)

y Comp. (N)

100 N

+81.915

-57.358

150 N

+63.393

-135.946

200 N

-163.830

-114.715

R x = - 18.522

R y = -308.02

R = Rx i + R y j = (-18.522 N)i + (-308.02 N) j Ry tan a = Rx 308.02 = 18.522 a = 86.559

308.02 N sin 86.559

R = 309 N

86.6 ◀

PROBLEM 2.36 Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at Point B of beam AB.

SOLUTION Cable BC Force:

500-N Force:

840 = -525 N 1160 840 Fy = (725 N) = 500 N 1160

Fx = -(725 N)

3 Fx = -(500 N) = -300 N 5 4 Fy = -(500 N) = -400 N 5

12 = 720 N 13 5 Fy = -(780 N) = -300 N 13

780-N Force:

Fx = (780 N)

and

Rx = SFx = -105 N Ry = SFy = -200 N R = (-105 N)2 + (-200 N)2 = 225.89 N

Further:

tan a =

200 105

a = tan-1 = 62.3

200 105

R = 226 N

Thus:

62.3

PROBLEM 2.37

80 lb 120 lb

60 lb a'

Knowing that a = 40°, determine the resultant of the three forces shown.

20° a

SOLUTION 60-lb Force:

Fx = (60 lb) cos 20 = 56.382 lb Fy = (60 lb)sin 20 = 20.521 lb

80-lb Force:

Fx = (80 lb) cos60 = 40.000 lb Fy = (80 lb)sin 60 = 69.282 lb

120-lb Force:

Fx = (120 lb) cos30 = 103.923 lb Fy = -(120 lb)sin 30 = -60.000 lb

and

R x = SFx = 200.305 lb R y = SFy = 29.803 lb R = (200.305 lb)2 + (29.803 lb)2 = 202.510 lb

Further:

tan a =

29.803 200.305

a = tan-1 = 8.46

29.803 200.305

R = 203 lb

8.46 ◀

PROBLEM 2.38

80 lb 120 lb

Knowing that a = 75°, determine the resultant of the three forces shown.

60 lb a'

20° a

SOLUTION 60-lb Force:

Fx = (60 lb) cos 20 = 56.382 lb Fy = (60 lb)sin 20 = 20.521 lb

80-lb Force:

Fx = (80 lb) cos 95  = -6.9725 lb Fy = (80 lb)sin 95 = 79.696 lb

120-lb Force:

Fx = (120 lb) cos 5  = 119.543 lb Fy = (120 lb)sin 5 = 10.459 lb

Then

R x = SFx = 168.953 lb R y = SFy = 110.676 lb

and

R = (168.953 lb)2 + (110.676 lb)2 = 201.976 lb

110.676 168.953 tan a = 0.65507 tan a =

a = 33.228

R = 202 lb

33.2 ◀

PROBLEM 2.39 A collar that can slide on a vertical rod is subjected to the three forces shown. Determine (a) the required value of α if the resultant of the three forces is to be horizontal, (b) the corresponding magnitude of the resultant.

SOLUTION

(a)

Rx = SFx = (85 N) cos a + (170 N)sin (a )

(1)

Ry = SFy = +(110 N) + (85 N)sin (a ) - (170 N) cos a

(2)

For R to be horizontal, we must have R y = 0. We make R y = 0 in Eq. (2):

110 + 85sin a - 170 cos a = 0 22 + 17sin a - 34 cos a = 0 17sin a + 22 = -34 1 - sin 2 a

(

289sin 2 a + 748sin a + 484 = 1156 1 - sin 2 a

)

1445sin a + 748sin a - 672 = 0 2

Solving by use of the quadratic formula: sin a = 0.47059 a = 28.072

(b)

a = 28.1 ◀

Since R = Rx using Eq. (1):

R = 85cos28.072 + 170sin 28.072 = 155.0 N R = 155.0 N ◀

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 Rx = SFx = Rx = -

21 T + 420 N 29 BC

Ry = SFy = Ry =

(a)

(1)

800 5 4 TBC - (780 N) - (500 N) 1160 13 5

20 T - 700 N 29 BC

(2)

For R to be vertical, we must have Rx = 0 Set Rx = 0 in Eq. (1)

(b)

840 12 3 TBC + (780 N) - (500 N) 1160 13 5

21 T + 420 N = 0 29 BC

TBC = 580 N

Substituting for TBC in Eq. (2):

20 (580 N) - 700 N 29 Ry = -300 N Ry =

R = | Ry | = 300 N

R = 300 N

PROBLEM 2.41

65°

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.

25° 35°

75 lb 50 lb

SOLUTION

Using the x and y axes shown:

(a)

R x = SFx = T A C sin10 + (50 lb) cos35 + (75 lb) cos60 = T A C sin10 + 78.458 lb

(1)

R y = SFy = (50 lb)sin 35 + (75 lb)sin 60 - T A C cos10 R y = 93.631 lb - T A C cos10

(2)

Set R y = 0 in Eq. (2): 93.631 lb - TA C cos10 = 0 T A C = 95.075 lb

(b)

T A C = 95.1 lb ◀

Substituting for T A C in Eq. (1):

R x = (95.075 lb)sin10 + 78.458 lb = 94.968 lb R = Rx

R = 95.0 lb ◀

120 lb

80 lb

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.

60 lb a'

20° a

SOLUTION

Select the x axis to be along a a¢. Then R x = S Fx = (60 lb) + (80 lb) cos a + (120 lb) sin a

(1)

R y = SFy = (80 lb)sin a - (120 lb) cos a

(2)

and

(a)

Set R y = 0 in Eq. (2). (80 lb)sin a - (120 lb) cos a = 0

Dividing each term by cosa gives:

(80 lb) tan a = 120 lb 120 lb tana = 80 lb a = 56.310 (b)

a = 56.3 ◀

Substituting for a in Eq. (1) gives: R x = 60 lb + (80 lb) cos 56.31 + (120 lb)sin 56.31 = 204.22 lb

R x = 204 lb ◀

PROBLEM 2.43 A

50°

30°

Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

400 lb

SOLUTION Free-Body Diagram

Force Triangle

Law of sines: TAC T 400 lb = BC = sin 60 sin 40 sin 80

(a)

TAC =

400 lb (sin 60) sin 80

TAC = 352 lb ◀

(b)

TBC =

400 lb (sin 40) sin80

T B C = 261 lb ◀

PROBLEM 2.44 A

Two cables are tied together at C and are loaded as shown. Knowing that α = 30°, determine the tension (a) in cable AC, (b) in cable BC. 6 kN

55° C

α B

SOLUTION Free-Body Diagram

Force Triangle

Law of sines: TA C T 6 kN = BC = sin 60 sin 35 sin 85

(a)

TA C =

6 kN (sin 60) sin85

T A C = 5.22 kN ◀

(b)

TBC =

6 kN (sin 35) sin85

T BC = 3.45 kN ◀

PROBLEM 2.45 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

(

Force Triangle

)

W = mg = (120 kg) 9.81 m/s2 = 1177 N Law of sines: TAC T 1.177 kN = BC = sin130 sin 30 sin 20

(a)

TAC =

1.177 kN (sin130) sin 20

T AC = 2.64 kN ◀

(b)

TBC =

1.177 kN (sin 30) sin 20

TBC = 1.721 kN ◀

A 25°

45°

PROBLEM 2.46 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:

Force Triangle

TA C T 500 N = BC = sin 35 sin 75 sin 70°

(a)

TA C =

(b)

TBC =

500 N sin 70

sin 35

T A C = 305 N ◀

sin 75

T B C = 514 N ◀

PROBLEM 2.47 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram

(

Force Triangle

)

W = mg = (50 kg) 9.81 m/s2 = 490 N

Law of sines:

TAC T 490 N = BC = sin110 sin 30 sin 40°

(a)

TAC =

(b)

TBC =

490 N sin 40 490 N sin 40

sin110

T AC = 716 N ◀

sin 30

TBC = 381 N ◀

PROBLEM 2.48

5°

Knowing that a = 20, determine the tension (a) in cable AC, (b) in rope BC.

C 1200 lb

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

TA C T 1200 lb = BC = sin 110 sin 5 sin 65

(a)

TA C =

1200 lb sin 110 sin 65

T A C = 1244 lb ◀

(b)

T BC =

1200 lb sin 5 sin 65

T B C = 115.4 lb ◀

Two cables are tied together at C and are loaded as shown. Knowing that P = 300 N, determine the tension in cables AC and BC.

30° 30°

200 N

PROBLEM 2.49

45° P

SOLUTION Free-Body Diagram

SFx = 0

- TCA sin 30 + TCB sin 30 - P cos 45 - 200N = 0

For P = 200N we have, - 0.5TCA + 0.5TCB + 212.13 - 200 = 0 (1)

SFy = 0

TCA cos30 - TCB cos30 - P sin 45 = 0 0.86603TCA + 0.86603TCB - 212.13 = 0

(2) TCA = 134.6 N ◀

Solving equations (1) and (2) simultaneously gives,

TCB = 110.4 N ◀

Two cables are tied together at C and are loaded as shown. Determine the range of values of P for which both cables remain taut.

30° 30°

200 N

PROBLEM 2.50

45° P

SOLUTION Free-Body Diagram

SFx = 0

- TCA sin 30 + TCB sin 30 - P cos 45 - 200N = 0

For TCA = 0 we have, 0.5TCB + 0.70711P - 200 = 0 (1)

SFy = 0

TCA cos30 - TCB cos30 - P sin 45 = 0 ; again setting TCA = 0 yields, 0.86603TCB - 0.70711P = 0

Adding equations (1) and (2) gives,

(2)

1.36603TCB = 200 hence T CB = 146.410N and P = 179.315N

Substituting for TCB = 0 into the equilibrium equations and solving simultaneously gives, - 0.5TCA + 0.70711P - 200 = 0 0.86603TCA - 0.70711P = 0

And TCA = 546.40N, P = 669.20N Thus for both cables to remain taut, load P must be within the range of 179.315 N and 669.20 N. 179.3 N < P < 669 N ◀

PROBLEM 2.51 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 600 lb and Q = 800 lb, determine the tension in rods A and B.

SOLUTION Free-Body Diagram

Resolving the forces into x- and y-directions: R = P + Q + TA + TB = 0

Substituting components:

R = -(600 lb) j - [(800 lb)cos30]i + [(800 lb)sin 30]j + TB i + (TA cos60)i + (TA sin 60) j = 0

Summing forces in the y-direction: -600 lb + (800 lb)sin 30 + TA sin 60 = 0 TA = 230.94 lb

TA = 231 lb ◀

Summing forces in the x-direction: -(800 lb)cos30 + TB + TA cos60 = 0

Thus,

TB = -(230.94 lb) cos60 + (800 lb) cos30

= 577.35 lb

TB = 577 lb ◀

PROBLEM 2.52 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the tensions in rods A and B are TA = 240 lb and TB = 500 lb, determine the magnitudes of P and Q.

SOLUTION Free-Body Diagram

Resolving the forces into x- and y-directions: R = P + Q + TA + TB = 0

Substituting components:

R = -Pj - Q cos 30i + Q sin 30 j + [(240 lb)cos 60]i + [(240 lb)sin 60]j + (500 lb)i

Summing forces in the x-direction: -Q cos 30 + (240 lb)cos 60 + 500 lb = 0

Q = 715.91 lb Summing forces in the y-direction: -P + Q sin 30 + (240 lb)sin 60 = 0 P = Q sin 30 + (240 lb)sin 60 = (715.91 lb)sin 30 + (240 lb)sin 60 = 565.80 lb

P = 566 lb; Q = 716 lb ◀

PROBLEM 2.53

FB B

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.

C FC

3 4

FA D FD

SOLUTION Free-Body Diagram of Connection

SFx = 0:

3 3 FB - FC - FA = 0 5 5

FA = 8 kN FB = 16 kN

With

FC =

4 4 (16 kN) - (8 kN) 5 5

FC = 6.40 kN ◀

3 3 S Fy = 0: - FD + FB - FA = 0 5 5 With FA and FB as above:

3 3 FD = (16 kN) - (8 kN) 5 5

FD = 4.80 kN ◀

PROBLEM 2.54

FB B C FC

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.

4 FA D FD

SOLUTION Free-Body Diagram of Connection

3 3 SFy = 0: -FD - FA + FB = 0 5 5 or

3 FB = FD + FA 5

With

FA = 5 kN, FD = 8 kN FB =

ù 5é 3 ê6 kN + (5 kN)ú ê úû 3ë 5

FB = 15.00 kN ◀

4 4 SFx = 0: -FC + FB - FA = 0 5 5

4 FC = (FB - FA ) 5 4 = (15 kN - 5 kN) 5

FC = 8.00 kN ◀

PROBLEM 2.55

D A α

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 = 30° and b = 10° and that the combined weight of the boatswain’s chair and the sailor is 200 lb, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

SOLUTION Free-Body Diagram

SFx = 0: T A CB cos 10 - T A CB cos 30 - TCD cos 30 = 0 TCD = 0.137158T A CB

(1)

SFy = 0: T A CB sin 10 + T A CB sin 30 + TCD sin 30 - 200 = 0 0.67365T A CB + 0.5T CD = 200

(a)

(b)

Substitute (1) into (2):

From (1):

(2)

0.67365T A CB + 0.5(0.137158T A CB ) = 200 T A CB = 269.46 lb

T A CB = 269 lb ◀

TCD = 0.137158(269.46 lb)

TCD = 37.0 lb ◀

PROBLEM 2.56

D A α

B β

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 α = 25° and b = 15° and that the tension in cable CD is 20 lb, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) the tension in the support cable ACB.

SOLUTION Free-Body Diagram

SFx = 0: T A CB cos 15 - T A CB cos 25 - (20 lb) cos 25 = 0 T A CB = 304.04 lb

SFy = 0: (304.04 lb)sin 15 + (304.04 lb)sin 25

+ (20 lb)sin 25 -W = 0 W = 215.64 lb (a)

W = 216 lb ◀

(b) T A CB = 304 lb ◀

PROBLEM 2.57 A 6 kN

55° C

For the cables of prob. 2.44, find the value of α for which the tension is as small as possible (a) in cable bc, (b) in both cables simultaneously. In each case determine the tension in each cable.

α B

SOLUTION Free-Body Diagram

Force Triangle

(a) For a minimum tension in cable BC, set angle between cables to 90 degrees. a = 35.0 ◀

By inspection, T A C = (6 kN) cos 35 

T A C = 4.91 kN ◀

T B C = (6 kN) sin 35 

T B C = 3.44 kN ◀

(b) For equal tension in both cables, the force triangle will be an isosceles. a = 55.0 ◀

Therefore, by inspection,

TA C = TBC = (1 / 2)

6 kN cos35

T A C = T BC = 3.66 kN ◀

A 25°

45°

PROBLEM 2.58 For the cables of Problem 2.46, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of a.

SOLUTION Free-Body Diagram

(a)

Law of cosines

Force Triangle

P 2 = (600)2 + (750)2 - 2(600)(750)cos(25 + 45) P = 784.02 N

(b)

Law of sines

P = 784 N ◀

sin b sin(25 + 45) = 600 N 784.02 N

b = 46.0

\ a = 46.0 + 25

a = 71.0 ◀

PROBLEM 2.59

For the situation described in Figure P2.48, 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.

5° C 1200 lb

SOLUTION Free-Body Diagram

Force Triangle

To be smallest, T B C must be perpendicular to the direction of T A C . (a) (b)

Thus,

a = 5.00

T BC = (1200 lb)sin 5

◀

T B C = 104.6 lb ◀

PROBLEM 2.60 Two cables tied together at C are loaded as shown. Determine the range of values of P for which both cables remain taut.

SOLUTION Free-Body Diagram

SFx = 0:

3 P -TBC - TAC sin 30 = 0 5 P=

SFy = 0:

(

5 T + TAC sin 30 3 BC

)

(1)

4 P + TAC cos30 - 120 = 0 5 P=

(

5 120 - TAC cos30 4

Requirement:

TAC = 0:

From Eq. (2):

P = 150.0 lb

Requirement:

TBC = 0:

(

(2)

)

From Eq. (1): P =

5 T sin 30 3 AC

From Eq. (2): P =

5 120 - TAC cos 30 4

(

)

Solving simultaneously yields: TAC = 78.294 lb and

P = 65.2 lb 65.2 lb< P < 150.0 lb ◀

PROBLEM 2.61

C A

A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.

0.7 m

1.2 m

SOLUTION Free-Body Diagram

tan a =

h 0.6 m

(1)

Isosceles Force Triangle

Law of sines:

sin a =

1 (2.8 kN) 2

TA C TA C = 5 kN 1

sin a = 2

(2.8 kN)

5 kN a = 16.2602

From Eq. (1): tan16.2602 =

h 0.6 m

\ h = 0.175000 m

Half-length of chain = AC = (0.6 m)2 + (0.175 m)2

= 0.625 m Total length:

= 2 ´ 0.625 m

1.250 m ◀

PROBLEM 2.62

For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium.

SOLUTION Free-Body Diagram

T A C = T B C = 800 N

(h + d )

A C = BC =

SFy = 0: 2(800 N)

h h +d2 2

-P = 0

æ d ö÷2 P 800 = 1 + çç ÷÷ çè h ÷ø 2 Data:

P = 200 N, d = 600 mm and solving for h 800 N =

æ 600 mm ö÷2 200 N 1 + çç ÷ çè h ÷÷ø 2

h = 75.6 mm ◀

PROBLEM 2.63

x B

20 in.

Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in.

50 lb P A

SOLUTION (a)

Free Body: Collar A

Force Triangle

P 50 lb = 4.5 20.5

(b)

Free Body: Collar A

P = 10.98 lb ◀

Force Triangle

P 50 lb = 15 25

P = 30.0 lb ◀

PROBLEM 2.64

x B

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. 20 in.

50 lb P A

SOLUTION Free Body: Collar A

Force Triangle

N 2 = (50)2 - (48)2 = 196 N = 14.00 lb

Similar Triangles

x 48 lb = 20 in. 14 lb x = 68.6 in. ◀

PROBLEM 2.65 A cable loop of length 1.5 m is placed around a crate. Knowing that the mass of the crate is 300 kg, determine the tension in the cable for each of the arrangements shown.

SOLUTION Free-Body Diagram

(

Isosceles Force Triangle

)

W = (300 kg) 9.81 m/s2 = 2943.0 N 1 EB = (1500 mm - 400 mm - 300 mm - 300 mm) 2 EB = 250 mm æ 200 mm ö÷ a = cos-1 çç ÷ = 36.87 çè 250 mm ø÷÷

TAE = TBE 1 TAE sin a = (2943.0 N) 2 1 TAE sin 36.87 = (2943.0 N) 2 TAE = 2452.5 N Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 66

TAE = 2450 N ◀

(a)

Problem 2.65 (Continued)

Isosceles Force Triangle Free-Body Diagram

1 (1500 mm - 300 mm - 400 mm - 400 mm) 2 EB = 250 mm æ 150 mm ÷ö ÷ = 41.41 a = cos-1 çç èç 200 mm ÷÷ø EB =

TAE = TBE 1 TAE sin a = (2943.0 N) 2 1 TAE sin 41.41 = (2943.0 N) 2 TAE = 2224.7 N TAE = 2220 N ◀

(b)

PROBLEM 2.66

0.75 m B

A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. 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 a simple pulley. This can be proved by the methods of Ch. 4.)

2.4 m P α A

200 kg

SOLUTION Free-Body Diagram: Pulley A

æ 5 ö÷ ÷ + P cos a = 0 SFx = 0: - 2P ççç çè 281 ÷÷ø cos a = 0.59655 a = 53.377

For a = + 53.377: æ 16 ö÷ ÷ + P sin 53.377 - 1962 N = 0 SFy = 0: 2 P ççç çè 281 ÷÷ø P = 724 N

53.4 ◀

For a = - 53.377: æ 16 ö÷ ÷ + P sin(-53.377) - 1962 N = 0 SFy = 0: 2 P ççç çè 281 ÷÷ø P = 1773

53.4 ◀

PROBLEM 2.67 T

A 600-lb crate is supported by several ropeand-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.)

(a)

(b)

(c)

(d)

(e)

SOLUTION Free-Body Diagram of Pulley (a)

SFy = 0: 2T - (600 lb) = 0 1 T = (600 lb) 2 T = 300 lb ◀

(b)

SFy = 0: 2T - (600 lb) = 0 1 T = (600 lb) 2 T = 300 lb ◀

(c)

SFy = 0: 3T - (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb ◀

(d )

SFy = 0: 3T - (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb ◀

(e)

SFy = 0: 4T - (600 lb) = 0 1 T = (600 lb) 4 T = 150.0 lb ◀

PROBLEM 2.68 T

Solve Parts b and d of Problem 2.67, assuming that the free end of the 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 2.66.) (a)

(b)

(c)

(d)

(e)

SOLUTION Free-Body Diagram of Pulley and Crate

(b)

SFy = 0: 3T - (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb ◀

(d )

SFy = 0: 4T - (600 lb) = 0 1 T = (600 lb) 4 T = 150.0 lb ◀

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 and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.

25° D 55° P

SOLUTION Free-Body Diagram: Pulley C

(a)

SFx = 0: T A CB (cos 25 - cos 55) - (750 N) cos 55° = 0 T A CB = 1292.88 N

Hence:

T A CB = 1293 N ◀

(b)

SFy = 0: T A CB (sin 25 + sin 55) + (750 N)sin 55 - Q = 0 (1292.88 N)(sin 25 + sin 55) + (750 N)sin 55 - Q = 0

Q = 2219.8 N

Q = 2220 N ◀

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, (b) the magnitude of load P.

25° D 55° P

PROBLEM 2.70

SOLUTION Free-Body Diagram: Pulley C SFx = 0: T A CB (cos 25 - cos 55) - P cos 55 = 0 P = 0.58010 TACB (1)

SFy = 0: T A CB (sin 25 + sin 55) + P sin 55 - 1800 N = 0 1.24177T A CB + 0.81915P = 1800 N (2)

or (a)

Substitute Equation (1) into Equation (2): 1.24177T A CB + 0.81915(0.58010T A CB ) = 1800 N T A CB = 1048.37 N

Hence:

T A CB = 1048 N ◀

(b)

Using (1),

P = 0.58010(1048.37 N) = 608.16 N P = 608 N ◀

PROBLEM 2.71 Determine (a) the x, y, and z components of the 500-N force, (b) the angles qx, qy, and qz that the force forms with the coordinate axes.

SOLUTION

(a)

Fx = (500 N)sin 40 cos30 Fx = 278 N ◀

Fx = 278.34 N

Fy = (500 N)cos 40° Fy = 383.02 N

Fy = 383 N ◀

Fz = (500 N)sin 40 sin30 Fz = 160.697 N (b)

cos qx = cos qy =

cos qz =

Fz = 160.7 N ◀

Fx 278.34 N = 500 N F Fy F

Fz F

q x = 56.2  ◀

383.02 N 500 N

qy = 40.0 ◀

160.697 N 500 N

q z = 71.3  ◀

PROBLEM 2.72 Determine (a) the x, y, and z components of the 800-N force, (b) the angles qx, qy, and qz that the force forms with the coordinate axes.

SOLUTION

(a)

Fx = -(800 N)cos 70 sin 25 Fx = - 115.635 N

Fx = -115.6 N ◀

Fy = (800 N)sin 70°

(b)

Fy = 751.75 N

Fy = 752 N ◀

Fz = (800 N)cos 70 cos 25 Fz = 247.98 N

Fz = 248 N ◀

Fx -115.635 N = 800 N F

q x = 98.3  ◀

cos qx = cos qy =

cos qz =

Fy F

Fz F

751.75 N 800 N

qy = 20.0 ◀

247.98 N 800 N

q z = 71.9  ◀

Note: From the given data, we could have computed directly

qy = 90 - 35 = 55, which checks with the answer obtained.

PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION Recoil force

F = 400 N \ FH = (400 N) cos 40 = 306.42 N

(a)

Fx = -FH sin35 = -(306.42 N)sin35 = -175.755 N Fy = -F sin 40 = -(400 N)sin 40 = -257.12 N

Fx = - 175.8 N ◀

Fy = -257 N ◀

Fz = +FH cos35 = +(306.42 N)cos35

Fz = +251 N ◀

= +251.00 N

(b)

cos qx =

cos qy =

cos qz =

Fx -175.755 N = 400 N F Fy F

-257.12 N 400 N

Fz 251.00 N = 400 N F

qx = 116.1 ◀

qy = 130.0 ◀

qz = 51.1  ◀

PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION Recoil force

F = 400 N

\ FH = (400 N) cos 25 = 362.52 N

(a)

Fx = +FH cos15 = +(362.52 N)cos15 = +350.17 N

Fx = + 350 N ◀

Fy = -F sin 25 = -(400 N)sin 25

Fy = -169.0 N ◀

= -169.047 N Fz = +FH sin15 = +(362.52 N)sin15

Fz = +93.8 N ◀

= +93.827 N

(b)

cos qx =

cos qy =

cos qz =

Fx +350.17 N = F 400 N Fy F

qx = 28.9  ◀

-169.047 N 400 N

Fz +93.827 N = 400 N F

qy = 115.0 ◀

qz = 76.4  ◀

PROBLEM 2.75 The angle between the guy wire AB and the mast is 20°. Knowing that the tension in AB is 300 lb, determine (a) the x, y, and z components of the force exerted on the boat at B, (b) the angles qx, qy, and qz defining the direction of the force exerted at B.

SOLUTION

Fh = FAB sin 20 = (300 lb)sin 20 Fh = 102.606 lb Fx = -Fh cos 40

Fy = FAB cos 20

Fz = -Fh sin 40

Fx = (-102.606 lb) cos 40

Fy = (300 lb)cos20

Fz = (-102.606 lb)sin 40

Fx = -78.601 lb

Fy = 281.91 lb

Fz = -65.954 lb

(a) Fx = - 78.6 lb ◀

Fy = 282 lb ◀ Fz = - 66.0 lb ◀

(b)

Fx -78.601 lb = F 300 lb Fy 281.91 lb cos qy = = 300 lb F Fz -65.954 lb = cos qz = F 300 lb cos qx =

q x = 105.2 ◀

qy = 20.0 ◀ q z = 102.7  ◀

PROBLEM 2.76 The angle between the guy wire AC and the mast is 20°. Knowing that the tension in AC is 300 lb, determine (a) the x, y, and z components of the force exerted on the boat at C, (b) the angles qx, qy, and qz defining the direction of the force exerted at C.

SOLUTION

Fh = FAC sin 20 = (300 lb)sin 20 Fh = 102.606 lb Fx = Fh cos 40

Fy = FAC cos 20

Fx = (102.606 lb) cos 40

Fy = (300 lb)cos20

Fz = (-102.606 lb)sin 40

Fz = -Fh sin 40

Fx = 78.601 lb

Fy = 281.91 lb

Fz = -65.954 lb

(a) Fx = 78.6 lb ◀

Fy = 282 lb ◀ Fz = - 66.0 lb ◀

(b)

Fx 78.601 lb = F 300 lb Fy 281.91 lb cos qy = = 300 lb F F -65.954 lb cos qz = z = F 300 lb cos qx =

q x = 74.8  ◀

qy = 20.0 ◀ q z = 102.7  ◀

PROBLEM 2.77

y A

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 qx , qy , and qz defining the direction of that force.

56 ft

50° z

B 20° x

SOLUTION 56 ft 65 ft = 0.86154

cos qy =

From triangle AOB:

qy = 30.51

Fx = -F sin qy cos20

(a)

= -(3900 lb)sin 30.51 cos20 Fx = - 1861 lb ◀

Fy = +F cos qy = (3900 lb)(0.86154) Fy = +3360 lb ◀

(b)

Fz = +(3900 lb)sin 30.51° sin 20°

Fz = + 677 lb ◀

Fx 1861 lb == - 0.4771 3900 lb F

qx = 118.5 ◀

qy = 30.51

qy = 30.5 ◀

cos qx = From above:

cos qz =

Fz 677 lb =+ = + 0.1736 F 3900 lb

qz = 80.0 ◀

PROBLEM 2.78 A

Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles qx, qy, and qz defining the direction of that force.

56 ft

50° z

B 20° C

SOLUTION AC = 70 ft

In triangle AOB:

OA = 56 ft F = 5250 lb 56 ft 70 ft qy = 36.870

cos qy =

FH = F sin qy = (5250 lb)sin 36.870 = 3150.0 lb (a)

Fx = -FH sin 50 = -(3150.0 lb)sin 50 = -2413.0 lb Fx = - 2410 lb ◀

Fy = +F cos qy = +(5250 lb)cos36.870 = +4200.0 lb

Fy = +4200 lb ◀ Fz = -FH cos50 = -3150 cos50 = -2024.8 lb (b)

cos qx =

From above:

Fz = - 2025 lb ◀

Fx -2413.0 lb = 5250 lb F

qx = 117.4 ◀

qy = 36.870

qy = 36.9 ◀

qz =

Fz -2024.8 lb = 5250 lb F

qz = 112.7 ◀

PROBLEM 2.79 Determine the magnitude and direction of the force F = (260 N)i – (320 N)j + (800 N)k.

SOLUTION F = Fx2 + Fy2 + Fz2 F = (260 N)2 + (-320 N)2 + (800 N)2 cos qx = cos qy =

cos qy =

Fx 260 N = 900 N F Fy F

Fz F

F = 900 N ◀ q x = 73.2  ◀

- 320 N 900 N

qy = 110.8 ◀

800 N 900 N

q z = 27.3  ◀

PROBLEM 2.80 Determine the magnitude and direction of the force F = (700 N)i – (820 N)j + (960 N)k.

SOLUTION F = Fx2 + Fy2 + Fz2 F = (700 N)2 - (820 N)2 + (960 N)2 F = 1443.61 N cos qx = cos qy =

cos qy =

F = 1444 N ◀

Fx 700 N = F 1443.61 N

q x = 61.0  ◀

-820 N 1443.61 N

qy = 124.6 ◀

960 N 1443.61 N

qz = 48.3 ◀

Fy F

Fz F

PROBLEM 2.81 A force F of magnitude 250 lb acts at the origin of a coordinate system. Knowing that qx = 65, qy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle qz.

SOLUTION cos2 qx + cos2 qy + cos2 qz = 1 cos2 (65) + cos2 (40) + cos2 qz = 1 cos qz = 0.48432 (b)

\ qz = 61.0 ◀

Since Fz > 0, we choose cos qz = 0.48432

(a) Fx = F cos qx = (250 lb)cos65 Fy = F cos qy = (250 lb)cos 40

Fx = 105.7 lb ◀ Fy = 191.5 lb ◀

Fz = F cos qz = (250 lb)cos 61

Fz = 121.2 lb ◀

PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles qx = 70.9 and qy = 144.9. Knowing that the z component of the force is s-52.0 lb, determine (a) the angle qz, (b) the other components and the magnitude of the force.

SOLUTION cos2 qx + cos2 qy + cos2 qz = 1 cos2 70.9 + cos2 144.9 + cos2 qz = 1 cos qz = 0.47282 (a)

Since Fz < 0, we choose cos qz = - 0.47282

(b)

Fz = F cos qz -52.0 lb = F (-0.47282)

\ qz = 118.2 ◀

F = 110.0 lb

F = 110.0 lb ◀

Fx = F cos qx = (110.0 lb) cos 70.9

Fx = 36.0 lb ◀

Fy = F cos qy = (110.0 lb)cos144.9

Fy = -90.0 lb ◀

PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, qz= 151.2, and Fy< 0, determine (a) the components Fy and Fz, (b) the angles qx and qy.

SOLUTION Fz = F cos qz = (210 N)cos151.2

(a)

= -184.024 N Then: So: Hence:

F 2 = Fx2 + Fy2 + Fz2

(210 N)2 = (80 N)2 + (Fy )2 + (184.024 N)2 Fy = - (210 N)2 - (80 N)2 - (184.024 N)2

= -61.929 N (b)

Fz = -184.0 N ◀

cos qx =

cos qy =

Fy = -62.0 lb ◀

Fx 80 N = = 0.38095 210 N F

Fy F

qx = 67.6  ◀

61.929 N = -0.29490 210 N qy = 107.2 ◀

PROBLEM 2.84 A force acts at the origin of a coordinate system in a direction defined by the angles qy = 120 and qz = 75. Knowing that the x component of the force is +40 N, determine (a) the angle qx, (b) the magnitude of the force.

SOLUTION cos2 qx + cos2 qy + cos2 qz = 1 cos2 qx + cos2 120 + cos2 75  = 1 cos qx = 0.82644 (b)

\ qx = 34.3 ◀

Since Fx > 0, we choose cos q x = 0.82644

(a) Fx = F cos q x 40 N = F cos34.3

F = 48.4 N ◀

PROBLEM 2.85 Two cables BG and BH are attached to frame ACD as shown. Knowing that the tension in cable BG is 540 N, determine the components of the force exerted by cable BG on the frame at B.

SOLUTION

 BG = -(0.8 m)i + (1.48 m) j - (0.64 m)k BG = (-0.8 m)2 + (1.48 m 2 ) + (-0.64 m)2 = 1.8 m TBG = TBG λ BG  BG = TBG BG 540 N = [(-0.8 m)i + (1.48 m)j + (-0.64 m)k ] 1.8 m TBG = (-240 N)i + (444 N) j - (192.0 N)k Fx = -240 N, Fy = +444 N, Fz = +192.0 N ◀

PROBLEM 2.86 Two cables BG and BH are attached to frame ACD as shown. Knowing that the tension in cable BH is 750 N, determine the components of the force exerted by cable BH on the frame at B.

SOLUTION

 BH = (0.6 m)i + (1.2 m) j - (1.2 m)k BH = (0.6 m)2 + (1.2 m)2 - (1.2 m)2 = 1.8 m TBH = TBH λ BH  BH = TBH BH 750 N = [i + 2 j - 2 k ] (m) 3m TBH = (250 N)i + (500 N) j - (500 N)k Fx = +250 N, Fy = +500 N, Fz = -500 N ◀ Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 88

PROBLEM 2.87

C 28.8 ft

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 in cable AB is 2 kips, determine the components of the force exerted at A by the cable.

45 ft

18 ft

36 ft

54 ft A

30°

SOLUTION

Cable AB:

 AB (-46.765 ft)i + (45 ft) j + (36 ft)k λAB = = AB 74.216 ft -46.765i + 45 j + 36 k TAB = TAB λAB = 74.216

(TA B )x = -1.260 kips ◀ (TA B )y = +1.213 kips ◀ (TA B )z = +0.970 kips ◀

PROBLEM 2.88

C 28.8 ft

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 in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable.

45 ft

18 ft

36 ft

54 ft A

30°

SOLUTION

Cable AB:

 AC (-46.765 ft)i + (55.8 ft) j + (-45 ft)k λAC = = AC 85.590 ft -46.765i + 55.8 j - 45k TAC = TAC λAC = (1.5 kips) 85.590

(TA C )x = -0.820 kips ◀ (TAC )y = +0.978 kips ◀ (TA C )z = -0.789 kips ◀

PROBLEM 2.89

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force exerted on the plate at B.

480

250 O

D 360

130

320 z

450

360 C

Dimensions in mm

SOLUTION

We have:

 BA = +(320 mm)i + (480 mm)j - (360 mm)k Thus:

BA = 680 mm

 æ8 12 9 ö BA FB = TBA λBA = TBA = TBA çç i + j - k ÷÷÷ ç 17 17 ÷ø BA è17

æ8 ö æ ö æ ö çç T ÷÷ i + çç12 T ÷÷ j- çç 9 T ÷÷ k = 0 BA BA BA ÷ ÷ èç17 ø÷ èç17 ø÷ èç17 ø÷÷ Setting T BA = 408 N yields,

Fx = +192.0 N, Fy = +288 N, Fz = -216 N ◀

PROBLEM 2.90

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force exerted on the plate at D.

480

250 O

D 360

130

320 z

450

360 C

Dimensions in mm

SOLUTION

We have:

 DA = -(250 mm)i + (480 mm)j + (360 mm)k Thus:

DA = 650 mm

 æ 5 48 36 ö÷ DA FD = TDA λDA = TDA j+ k÷ = TDA çç- i + 65 65 ÷ø÷ DA èç 13

æ5 ö æ 48 ö æ 36 ö -çç TDA ÷÷÷ i + çç TDA ÷÷÷ j + çç TDA ÷÷÷ k = 0 çè13 ÷ø çè 65 ÷ø èç 65 ø÷ Setting T DA = 429 N yields,

Fx = -165.0 N, Fy = +317 N, Fz = +238 N ◀

PROBLEM 2.91

y Q

Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N.

50° 20° 30°

15° z

SOLUTION P = (300 N)[- cos30 sin15i + sin 30 j + cos30 cos15k ] = - (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos50 cos20i + sin 50 j - cos50 sin 20k ] = (400 N)[0.60402i + 0.76604 j - 0.21985] = (241.61 N)i + (306.42 N) j - (87.939 N)k R= P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N)2 + (456.42 N)2 + (163.011 N)2

R = 515 N ◀

= 515.07 N

cos qx =

cos qy = cos qz =

Rx 174.367 N = = 0.33853 515.07 N R

qx = 70.2  ◀

456.42 N = 0.88613 515.07 N

qy = 27.6 ◀

Rz 163.011 N = = 0.31648 515.07 N R

qz = 71.5  ◀

PROBLEM 2.92

y Q

Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N.

50° 20° 30°

15° z

SOLUTION P = (400 N)[- cos30 sin15i + sin 30 j + cos30 cos15k ] = - (89.678 N)i + (200 N) j + (334.61 N)k Q = (300 N)[cos 50 cos 20i + sin 50 j - cos 50 sin 20k ] = (181.21 N)i + (229.81 N)j - (65.954 N)k R= P+Q = (91.532 N)i + (429.81 N) j + (268.66 N)k R = (91.532 N)2 + (429.81 N)2 + (268.66 N)2

R = 515 N ◀

= 515.07 N

cos qx =

cos qy = cos qz =

Rx 91.532 N = = 0.177708 515.07 N R

qx = 79.8  ◀

429.81 N = 0.83447 515.07 N

qy = 33.4 ◀

Rz 268.66 N = = 0.52160 515.07 N R

qz = 58.6  ◀

PROBLEM 2.93

40 in.

Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

45 in.

60 in.

B z

60 in.

SOLUTION  AB = (40 in.)i - (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.)2 = 85 in.  AC = (100 in.)i - (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.  é (40 in.)i - (45 in.) j + (60 in.)k ù AB ú TAB = TAB λAB = TAB = (425 lb) êê ú AB 85 in. êë úû TAB = (200 lb)i - (225 lb) j + (300 lb)k  é (100 in.)i - (45 in.) j + (60 in.)k ù AC ú TAC = TAC λAC = TAC = (510 lb) êê ú 125 in. AC êë úû TAC = (408 lb)i - (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i - (408.6 lb) j + (544.8 lb)k

Then: and

R = 912.92 lb

R = 913 lb ◀

cos qx =

608 lb = 0.66599 912.92 lb

qx = 48.2  ◀

cos qy =

408.6 lb = -0.44757 912.92 lb

qy = 116.6 ◀

cos qz =

544.8 lb = 0.59677 912.92 lb

qz = 53.4  ◀

PROBLEM 2.94

40 in.

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.

45 in.

60 in.

B z

60 in.

SOLUTION  AB = (40 in.)i - (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.)2 = 85 in.  AC = (100 in.)i - (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.  é (40 in.)i - (45 in.) j + (60 in.)k ù AB ú TAB = TAB λAB = TAB = (510 lb) êê ú AB 85 in. êë úû TAB = (240 lb)i - (270 lb) j + (360 lb)k  é (100 in.)i - (45 in.) j + (60 in.)k ù AC ú TAC = TAC λAC = TAC = (425 lb) êê ú 125 in. AC êë úû TAC = (340 lb)i - (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i - (423 lb) j + (564 lb)k

Then: and

R = 912.92 lb

R = 913 lb ◀

cos qx =

580 lb = 0.63532 912.92 lb

qx = 50.6  ◀

cos qy =

-423 lb = -0.46335 912.92 lb

qy = 117.6 ◀

cos qz =

564 lb = 0.61780 912.92 lb

qz = 51.8  ◀

PROBLEM 2.95 For the frame of Prob. 2.85, determine the magnitude and direction of the resultant of the forces exerted by the cables at B knowing that the tension is 540 N in cable BG and 750 N in cable BH. PROBLEM 2.85 Two cables BG and BH are attached to frame ACD as shown. Knowing that the tension in cable BG is 540 N, determine the components of the force exerted by cable BG on the frame at B.

SOLUTIO ON JJJG BG = −(0.8 m)i + (1.448 m) j − (0.644 m)k BG = (−0.8 m)2 + ((1.48 m 2 ) + (− 0.64 m)2 = 11.8 m JJJG BH = (00.6 m)i + (1.2 m) j − (1.2 m))k BH = (0.6 m)2 + (1..2 m)2 − (1.2 m m)2 = 1.8 m TBG = TBG B λ BG JJJG BG 540 N = TBG = [(−0.8 m)i + (1.48 m)j + (−0.64 m)k ] B BG 1.8 m = (−240 N)i + (4444 N) j − (1922.0 N)k TBH = TBH B λ BH JJJG BH 750 N [i + 2 j − 2 k ] (m) = TBH = B BH 3m TBH = (2250 N)i + (5000 N) j − (500 N N)k R = TBG + i N) j − (692 N )k + T = (1 10 N) (944 B BH

Then:

and

(10 + 944 + (−692) ) = 1170.51 N 2

R = 1171 N ◀

cos θx =

10 N 1170.51 N

θ x = 89.5 ° ◀

cos θy =

944 N 1170.51 N

θy = 36.2° ◀

−692 N 1170.51 1 N

θ z = 126.2 ° ◀

cos θz =

PROBLEM 2.96

For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable AC is 54 N and that the resultant of the forces exerted by the three cables at A must be vertical.

480

250 O

D 360

130

320 z

450

360 C

Dimensions in mm

SOLUTION

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 mm  AD = (250 mm)i - (480 mm) j - (360 mm) k AD = 650 mm Thus:

 AB TAB TAB = TAB λAB = TAB = (-320i - 480 j + 360k) AB  680 54 AC TAC = TAC λAC = TAC = (450i - 480 j + 360k) AC 750  AD TAD TAD = TAD λAD = TAD = (250i - 480 j - 360k) AD 650

Substituting into the Eq. R = SF and factoring i, j, k :

æ 320 ö 250 TAB + 32.40 + TAD ÷÷÷ i R = çççè 680 ÷ø 650 æ 480 ö 480 + ççT - 34.560 TAD ÷÷÷ j çè 680 AB ÷ø 650 æ 360 ö 360 + çç TAB + 25.920 TAD ÷÷÷ k 650 èç 680 ø÷

PROBLEM 2.96 (Continued) Since R is vertical, the coefficients of i and k are zero:

320 250 TAB + 32.40 + T =0 680 650 AD

(1)

360 360 TAB + 25.920 T =0 680 650 AD

(2)

Multiply (1) by 3.6 and (2) by 2.5 then add:

252 T + 181.440 = 0 680 AB

T A B = 489.60 N

T A B = 490 N ◀

Substitute into (2) and solve for TAD :

360 360 (489.60 N) + 25.920 T =0 680 650 AD T A D = 514.80 N T A D = 515 N ◀

PROBLEM 2.97

y 36 in.

The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

C 25 in.

24 in. B

29 in.

O 48 in.

A x P

SOLUTION

Cable AB:

Cable AC:

Load P:

TA B = 183 lb  (-48 in.)i + (29 in.) j + (24 in.)k AB = (183 lb) TAB = TAB λAB = TAB 61 in. AB TAB = -(144 lb)i + (87 lb) j + (72 lb)k  (-48 in.)i + (25 in.) j + (-36 in.)k AC = TAC TAC = TAC λAC = TAC AC 65 in. 48 25 36 TAC = - TAC i + TAC j - TAC k 65 65 65

P=P j

For resultant to be directed along OA, i.e., x-axis Rz = 0: SFz = (72 lb) -

36 ¢ TAC = 0 65

T A C = 130.0 lb ◀

PROBLEM 2.98

y 36 in.

For the boom and loading of Problem. 2.97, determine the magnitude of the load P.

25 in. 24 in.

PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

29 in.

O 48 in.

A x P

SOLUTION See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write

Ry = 0: SFy = (87 lb) +

25 T -P = 0 65 AC

T A C = 130.0 lb from Problem 2.97. Then

(87 lb) +

25 (130.0 lb) - P = 0 65

P = 137.0 lb ◀

PROBLEM 2.99

450 mm 500 mm D 360 mm B

320 mm C

z A

A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN.

600 mm x

SOLUTION Free-Body Diagram at A:

The forces applied at A are:

TA B , TA C , TA D , and W

where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write  AB = - (450 mm)i + (600 mm) j AB = 750 mm  AC = + (600 mm) j - (320 mm)k AC = 680 mm  AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm  (-450 mm)i + (600 mm) j AB and TAB = λAB TAB = TAB = TAB 750 mm AB æ 45 60 ÷ö = çç- i + j÷T çè 75 75 ÷÷ø A B

 (600 mm)i - (320 mm) j AC TAC = λAC TAC = TAC = TAC 680 mm AC æ 60 ö÷ 32 = çç j - k÷÷TAC çè 68 68 ÷ø  AD (500 mm)i + (600 mm) j + (360 mm)k TAD = λADTAD = TAD = TAD AD 860 mm æ 50 ö÷ 60 36 j + k÷÷TAD = çç i + çè 86 86 86 ÷ø Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 102

PROBLEM 2.99 (Continued) Equilibrium condition:

SF = 0: \ TA B + TA C + TA D + W = 0

Substituting the expressions obtained for TA B , TA C , and TA D ; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: From j: From k:

45 50 TA B + TA D = 0 75 86

(1)

60 60 60 T + T + T -W = 0 75 A B 68 A C 86 A D

(2)

32 36 TA C + TA D = 0 68 86

(3)

Setting T A B = 6 kN in (1) and (2), and solving the resulting set of equations gives

TAC = 6.1920 kN TAC = 5.5080 kN

W = 13.98 kN ◀

PROBLEM 2.100

450 mm 500 mm D 360 mm B

320 mm C

z A

A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN.

600 mm x

SOLUTION See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: 45 50 TA B + TA D = 0 75 86

(1)

60 60 60 TA B + TA C + TA D - W = 0 75 68 86

(2)

32 36 T + T =0 68 A C 86 A D

(3)

Setting TA D = 4.3 kN into the above equations gives

TAB = 4.1667 kN TAC = 3.8250 kN

W = 9.71 kN ◀

PROBLEM 2.101

Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N.

5.60 m

B D 4.20 m

O C

3.30 m 4.20 m

2.40 m

SOLUTION

The forces applied at A are:

FREE-BODY DIAGRAM AT A

TA B , TA C , TA D , and P

where P = Pj . 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 m  AC = (2.40 m)i - (5.60 m) j + (4.20 m)k AC = 7.40 m  AD = - (5.60 m) j - (3.30 m)k AD = 6.50 m  AB and = (-0.6 i - 0.8 j)TAB TAB = TAB λAB = TAB AB  AC = (0.32432 i - 0.75676 j + 0.56757k )TAC TAC = TAC λAC = TAC AC  AD = (-0.86154 j - 0.50769k )TAD TAD = TAD λAD = TAD AD Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 105

PROBLEM 2.101 (Continued) Equilibrium condition:

SF = 0: TA B + TA C + TA D + Pj = 0

Substituting the expressions obtained for TA B , TA C , and TA D and factoring i, j, and k:

(- 0.6TAB + 0.32432TAC )i + (-0.8TAB - 0.75676TAC - 0.86154TAD + P ) j + (0.56757TAC - 0.50769TAD )k = 0 Equating to zero the coefficients of i, j, k: - 0.6T A B + 0.32432T A C = 0

(1)

- 0.8T A B - 0.75676T A C - 0.86154T A D + P = 0

(2)

0.56757T A C - 0.50769T A D = 0

(3)

Setting T A D = 481 N in (2) and (3), and solving the resulting set of equations gives

TAC = 430.26 N TAD = 232.57 N

P = 926 N ◀

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, determine the tension in each cable.

5.60 m

B D 4.20 m

O C

2.40 m

3.30 m 4.20 m x

SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). -0.6T A B + 0.32432TA C = 0

(1)

-0.8T A B - 0.75676T A C - 0.86154T A D + P = 0

(2)

0.56757T A C - 0.50769T A D = 0

(3)

From Eq. (1):

TA B = 0.54053TA C

From Eq. (3):

TA D = 1.11795TA C

Substituting for TA B and TAD in terms of TAC into Eq. (2) gives - 0.8(0.54053T A C ) - 0.75676TA C - 0.86154(1.11795T A C ) + P = 0 2.1523T A C = P ; P = 800 N 800 N 2.1523 = 371.69 N Substituting into expressions for TA B and TAD gives TA C =

T A B = 0.54053(371.69 N) T A D = 1.11795(371.69 N) T A B = 201 N, T A C = 372 N, T A D = 416 N ◀

PROBLEM 2.103 A crate is supported by three cables as shown. Determine the weight W of the crate, knowing that the tension in cable AD is 924 lb.

SOLUTION The forces applied at A are:

TAB , TAC , TAD and W where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write

 AB = (28 in.)i + (45 in.) j AB = 53 in.  AC = (45 in.) j - (24 in.)k AC = 51 in.  AD = -(26 in.)i + (45 in.) j + (18 in.)k AD = 55 in.

 AB TAB = TAB λAB = TAB AB = (0.5283i + 0.84906 j)TAB  AC TAC = TAC λAC = TAC AC = (0.88235 j - 0.47059k )TAC  AD TAD = TAD λAD = TAD AD = (-0.47273i + 0.81818 j + 0.32727k )TAD

and

Equilibrium Condition with

W = - Wj

SF = 0: TAB + TAC + TAD - Wj = 0

PROBLEM 2.103 (Continued)

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (0.5283TAB - 0.47273TAD )i + (0.84906TAB + 0.88235TAC + 0.81818TAD - W ) j + (-0.47059TAC + 0.32727TAD )k = 0

Equating to zero the coefficients of i, j, k: 0.5283TAB - 0.47273TAD = 0 (1) 0.84906TAB + 0.88235TAC + 0.81818TAD - W = 0 (2) -0.47059TAC + 0.32727TAD = 0 (3)

Substituting TAD = 924 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: TAB = 826.81 lb TAC = 642.59 lb

W = 2030 lb ◀

PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight W of the crate, knowing that the tension in cable AB is 1378 lb.

SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

0.5283TAB - 0.47273TAD = 0 (1) 0.84906TAB + 0.88235TAC + 0.81818TAD - W = 0 (2) -0.47059TAC + 0.32727TAD = 0 (3)

Substituting TAB = 1378 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAD = 1539.99 lb TAC = 1070.98 lb

W = 3380 lb ◀

PROBLEM 2.105 A 12-lb circular plate of 7-in. radius is supported as shown by three wires, each of 25-in. length. Determine the tension in each wire, knowing that α = 30°.

SOLUTION Let q be angle between the vertical and any wire. OA =

(25 - 7 ) = 24 in. thus cos q = 24 25 2

By symmetry TAB = TAC

SFx = 0: -2(TAB sin q )(cos a ) + TAD (sin q ) = 0

For a = 30 :

TAD = 2 (cos30)TAB = 1.73205TAB SFy = 0: 12 lb - 2TAB (cos q) - TAD (cos q) = 0 or

12 lb = (2TAB + TAD ) cos q

æ 24 ö 12 lb = (2TAB + 1.73205TAB )çç ÷÷÷ èç 25 ÷ø

PROBLEM 2.106 Solve Prob. 2.105, knowing that α = 45°. PROBLEM 2.105 A 12-lb circular plate of 7-in. radius is supported as shown by three wires, each of 25-in. length. Determine the tension in each wire, knowing that α = 30°.

SOLUTION Let q be angle between the vertical and any wire. OA =

(25 - 7 ) = 24 in. thus cos q = 24 25 2

By symmetry TAB = TAC

SFx = 0: -2(TAB sin q )(cos a ) + TAD (sin q ) = 0

For a = 45 :

TAD = 2 (cos 45)TAB = 1.41421TAB

SFy = 0: 12 lb - 2TAB (cos q) - TAD (cos q) = 0 or

12 lb = (2TAB + TAD ) cos q

æ 24 ö 12 lb = (2TAB + 1.41421TAB )çç ÷÷÷ çè 25 ÷ø

TAB = TAC = 3.66 lb ◀ TAD = 5.18 lb ◀

PROBLEM 2.107

220 mm

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N.

960 mm

320 mm C

380 mm

240 mm

960 mm

P x

SOLUTION SFA = 0: TAB + TAC + TAD + P = 0 where P = P 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 = 1040 mm  AD = -(960 mm)i + (720 mm) j - (220 mm)k AD = 1220 mm  æ 48 AB 12 19 ö TAB = TAB λAB = TAB = TAB çç- i - j + k÷÷÷ ç AB 53 53 ÷ø è 53  æ 12 AC 3 4 ö TAC = TAC λAC = TAC = TAC çç- i - j - k÷÷÷ ç AC 13 13 ø÷ è 13

305 N [(-960 mm)i + (720 mm) j - (220 mm)k] 1220 mm = -(240 N)i + (180 N) j - (55 N)k

TAD = TAD λAD =

Substituting into SFA = 0, factoring i, j, k, and setting each coefficient equal to f gives: i: P= j:

48 12 TAB + TAC + 240 N 53 13 12 3 T + T = 180 N 53 A B 13 A C 19 4 T A B - TA C = 55 N 53 13

(1) (2) (3)

Solving the system of linear equations using conventional algorithms gives: T A B = 446.71 N

P = 960 N ◀

T A C = 341.71 N

PROBLEM 2.108

220 mm

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut.

960 mm

320 mm C

380 mm

240 mm

960 mm

P x

SOLUTION We assume that T A D = 0 and write SFA = 0: TAB + TAC + Qj + (1200 N)i = 0  AB = -(960 mm)i - (240 mm)j + (380 mm)k AB = 1060 mm  AC = -(960 mm)i - (240 mm) j - (320 mm)k AC = 1040 mm  12 19 ö AB æç 48 = ç- i - j + k ÷÷÷TAB TAB = TAB λAB = TAB ç AB è 53 53 53 ÷ø  AC æç 12 3 4 ö TAC = TAC λAC = TAC = ç- i - j - k ÷÷÷TAC ç AC è 13 13 13 ø÷

Substituting into SFA = 0, factoring i, j, k, and setting each coefficient equal to f gives: 48 12 TAB - TAC + 1200 N = 0 53 13 12 3 j : - TAB - TAC + Q = 0 53 13 19 4 k: TAB - TAC = 0 53 13 Solving the resulting system of linear equations using conventional algorithms gives:

i: -

TA B = 605.71 N TA C = 705.71 N Q = 300.00 N

(1) (2) (3)

0 £ Q < 300 N ◀

Note: This solution assumes that Q is directed upward as shown (Q ³ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = -460 N. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 114

PROBLEM 2.109

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate.

480

250 O

D 360

130

320 z

450

360 C

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 : SF = 0: TAB + TAC + TAD + Pj = 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 mm  AD = (250 mm)i - (480 mm) j - (360 mm) k AD = 650 mm Thus:

 12 9 ö AB æç 8 = ç- i - j + k ÷÷÷TAB Dimensions in mm TAB = TAB λAB = TAB 17 17 ÷ø AB çè 17  AC TAC = TAC λAC = TAC = (0.6 i - 0.64 j + 0.48k )TAC AC  9.6 7.2 ö÷ AD æç 5 =ç iTAD = TAD λAD = TAD jk÷T 13 13 ÷÷ø AD AD çè13

Substituting into the Eq. SF = 0 and factoring i, j, k : æ 8 ö çç- T + 0.6T + 5 T ÷÷ i AB AC AD ÷ø÷ 13 èç 17 æ 12 ö 9.6 + çç- TAB - 0.64TAC TAD + P ÷÷÷ j çè 17 ÷ø 13 æ9 ö 7.2 + çç TAB + 0.48TAC TAD ÷÷÷ k = 0 çè17 ÷ 13 ø Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 115

PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i:

8 5 T + 0.6TA C + TA D = 0 17 A B 13

(1)

12 9.6 TA B - 0.64TA C T +P=0 7 13 A D

(2)

9 7.2 TA B + 0.48T AC T =0 17 13 A D

(3)

8 5 T + 36 N + TA D = 0 17 A B 13

(1¢)

9 7.2 TA B + 28.8 N T =0 17 13 A D

(3¢)

Making T A C = 60 N in (1) and (3): -

Multiply (1¢) by 9, (3¢) by 8, and add: 554.4 N -

12.6 T = 0 TAD = 572.0 N 13 A D

Substitute into (1¢) and solve for TA B : TA B =

ö 17 æç 5 çç36 + ´ 572÷÷÷÷ 8è 13 ø

T A B = 544.0 N

Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N

Weight of plate = P = 845 N ◀

PROBLEM 2.110

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.

480

250 O

D 360

130

320 z

360

450

Dimensions in mm

SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 TA B + 0.6TA C + TA D = 0 17 13

(1)

12 9.6 TAB + 0.64 TAC T +P =0 17 13 AD

(2)

9 7.2 TA B + 0.48T AC T =0 17 13 A D

(3)

Making T A D = 520 N in Eqs. (1) and (3): 8 T + 0.6TA C + 200 N = 0 17 A B

(1¢)

9 T + 0.48TA C - 288 N = 0 17 A B

(3¢)

Multiply (1¢) by 9, (3¢) by 8, and add: 9.24T A C - 504 N = 0 T A C = 54.5455 N

Substitute into (1¢) and solve for TA B : TA B =

17 (0.6 ´ 54.5455 + 200) T A B = 494.545 N 8

Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 = 768.00 N

Weight of plate = P = 768 N ◀

PROBLEM 2.111

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 840 lb, determine the vertical force P exerted by the tower on the pin at A.

A 100 ft

20 ft

25 ft B

O 74 ft

20 ft 60 ft

18 ft

SOLUTION SF = 0: TAB + TAC + TAD + Pj = 0

Free-Body Diagram at A:

 AB = -20 i - 100 j + 25k AB = 105 ft  AC = 60 i - 100 j + 18k AC = 118 ft  AD = -20i - 100 j - 74 k AD = 126 ft We write

 AB TAB = TAB λ AB = TAB AB æ 4 20 5 ö = çç- i - j + k÷÷÷TAB çè 21 21 21 ÷ø  AC TAC = TAC λAC = TAC AC æ 30 50 9 ö = çç i j + k÷÷÷TAC 59 59 ø÷ èç 59  AD TAD = TAD λAD = TAD AD æ 10 50 37 ö ç = ç- i - j - k÷÷÷TAD 63 63 ø÷ èç 63

Substituting into the Eq. SF = 0 and factoring i, j, k : Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 118

PROBLEM 2.111 (Continued) æ 4 ö çç- T + 30 T - 10 T ÷÷ i AB AC AD ÷ ÷ø 59 63 èç 21 æ 20 ö 50 50 + çç- TAB - TAC - TAD + P ÷÷÷ j çè 21 59 63 ø÷ æ5 ö 9 37 + çç TAB + TAC - TAD ÷÷÷ k = 0 çè 21 59 63 ø÷

Setting the coefficients of i, j, k, equal to zero: i:

4 30 10 T + T - T =0 21 A B 59 A C 63 A D

(1)

20 50 50 TA B - TA C - TA D + P = 0 21 59 63

(2)

5 9 37 TA B + TA C - TA D = 0 21 59 63

(3)

Set TA B = 840 lb in Eqs. (1) - (3): 30 10 TA C - TA D = 0 59 63

(1¢)

50 50 T - T +P =0 59 A C 63 A D

(2¢)

9 37 TA C - TA D = 0 59 63

(3¢)

-160 lb + -800 lb -

200 lb +

Solving,

T A C = 458.12 lb T A D = 459.53 lb P = 1552.94 lb

P = 1553 lb ◀

PROBLEM 2.112

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 590 lb, determine the vertical force P exerted by the tower on the pin at A.

A 100 ft

20 ft

25 ft B

O 74 ft

20 ft 60 ft

18 ft

SOLUTION SF = 0: TAB + TAC + TAD + Pj = 0

We write

Free-Body Diagram at A:

 AB = -20 i - 100 j + 25k AB = 105 ft  AC = 60 i - 100 j + 18k AC = 118 ft  AD = -20i - 100 j - 74 k AD = 126 ft  AB TAB = TAB λ AB = TAB AB æ 4 20 5 ö = çç- i - j + k÷÷÷TAB çè 21 21 21 ÷ø  AC TAC = TAC λAC = TAC AC æ 30 50 9 ö = çç i j + k÷÷÷TAC çè 59 59 59 ÷ø  AD TAD = TAD λAD = TAD AD æ 10 50 37 ö = çç- i - j - k÷÷÷TAD 63 63 ø÷ èç 63

PROBLEM 2.112 (Continued) æ 4 ö çç- T + 30 T - 10 T ÷÷ i AB AC AD ÷ø÷ 59 63 èç 21 æ 20 ö 50 50 + çç- TAB - TAC - TAD + P ÷÷÷ j çè 21 ÷ø 59 63 æ5 ö 9 37 + çç TAB + TAC - TAD ÷÷÷ k = 0 çè 21 59 63 ø÷

Setting the coefficients of i, j, k, equal to zero: i:

4 30 10 TA B + TA C - TA D = 0 21 59 63

(1)

20 50 50 TA B - TA C - TA D + P = 0 21 59 63

(2)

5 9 37 T + T - T =0 21 A B 59 A C 63 A D

(3)

Set T A C = 590 lb in Eqs. (1) – (3): 4 10 T A B + 300 lb - TA D = 0 21 63

(1¢)

20 50 TA B - 500 lb - TA D + P = 0 21 63

(2¢)

5 37 T + 90 lb - T A D = 0 21 A B 63

(3¢)

Solving,

TA B = 1081.82 lb TA D = 591.82 lb

P = 2000 lb ◀

PROBLEM 2.113

y O

4 ft x

8 ft z 16 ft

12 ft

32 ft

In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

30 ft

SOLUTION Free-Body Diagram at A

æ 16 30 ö÷ N = N çç i + j÷ çè 34 34 ÷÷ø

and W = W j = -(175 lb) j

 (-30 ft)i + (20 ft) j - (12 ft)k AC TAC = TAC λ AC = TAC = TAC 38 ft AC æ 15 10 6 ö = TAC çç- i + j - k÷÷÷ çè 19 19 19 ø÷  (-30 ft)i + (24 ft) j + (32 ft)k AB = TAB TAB = TAB λ AB = TAB 50 ft AB æ 15 12 16 ö = TAB çç- i + j + k÷÷÷ çè 25 25 25 ÷ø Equilibrium condition: SF = 0 TA B + TA C + N + W = 0

PROBLEM 2.113 (Continued) Substituting the expressions obtained for TAB , TAC , N , and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: From j: From k:

15 15 16 TA B - TA C + N = 0 25 19 34

(1)

12 10 30 TA B + TA C + N - (175 lb) = 0 25 19 34

(2)

16 6 T - T =0 25 A B 19 A C

(3)

Solving the resulting set of equations gives: T A B = 30.8 lb; T A C = 62.5 lb ◀

PROBLEM 2.114

y O

4 ft x

8 ft z 16 ft

12 ft

32 ft

Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P =-(45 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

30 ft

SOLUTION Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force P = (- 45 lb) k. 15 15 16 TA B - TA C + N = 0 25 19 34

(1)

12 10 30 T + T + N - (175 lb) = 0 25 A B 19 A C 34

(2)

16 6 TA B - TA C - (45 lb) = 0 25 19

(3)

Solving the resulting set of equations simultaneously gives: T A B = 81.3 lb ◀ TA C = 22.2 lb ◀

PROBLEM 2.115

For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

480

250 O

D 360

130

320 z

450

360 C

Dimensions in mm

SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TA B + 0.6TA C + TA D = 0 17 13

(1)

12 9.6 TA B - 0.64TA C T + 792 N = 0 17 13 A D

(2)

9 7.2 TA B + 0.48TA C T =0 17 13 A D

(3)

Solving Equations (1), (2), and (3) by conventional algorithms gives TA B = 510.00 N

TA B = 510 N ◀

T A C = 56.250 N

TA C = 56.2 N ◀

T A D = 536.25 N

T A D = 536 N ◀

PROBLEM 2.116

220 mm

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.

960 mm

320 mm C

380 mm

240 mm

960 mm

SOLUTION SFA = 0: TAB + TAC + TAD + P + Q = 0 P = Pi and Q = Qj  AB = -(960 mm)i - (240 mm) j + (380 mm)k AB = 1060 mm  AC = -(960 mm)i - (240 mm) j - (320 mm)k AC = 1040 mm  AD = -(960 mm)i + (720 mm) j - (220 mm)k AD = 1220 mm  æ 48 12 19 ö AB TAB = TAB λAB = TAB = TAB çç- i - j + k÷÷÷ ç 53 53 ÷ø AB è 53  æ 12 3 4 ö AC TAC = TAC λAC = TAC = TAC çç- i - j - k÷÷÷ ç 13 13 ÷ø AC è 13  æ 48 36 11 ö AD TAD = TAD λAD = TAD j - k÷÷÷ = TAD çç- i + çè 61 61 61 ø÷ AD Substituting into SFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations:

Where

i: -

48 12 48 T - T - T + 2880 N = 0 53 AB 13 AC 61 AD

(1)

j: -

12 3 36 TAB - TAC + TAD = 0 53 13 61

(2)

19 4 11 TAB - TAC - TAD = 0 53 13 61

(3)

PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: T A B = 1340.14 N TA C = 1025.12 N TA B = 1340 N ◀

TA D = 915.03 N

TA C = 1025 N ◀ TA D = 915 N ◀

PROBLEM 2.117

220 mm

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.

960 mm

320 mm C

380 mm

240 mm

960 mm

SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: -

48 12 48 TA B - TA C - TA D + P = 0 53 13 61

12 3 36 TA B - TA C + TA D + Q = 0 53 13 61 19 4 11 TA B - TA C - TA D = 0 53 13 61

(1) (2) (3)

Setting P = 2880 N and Q = 576 N gives: -

48 12 48 TA B - TA C - TA D + 2880 N = 0 53 13 61

(1¢)

12 3 36 TA B - T A C + T A D + 576 N = 0 53 13 61 19 4 11 T - T - T =0 53 A B 13 A C 61 A D

(2¢)

(3¢)

Solving the resulting set of equations using conventional algorithms gives: T A B = 1431.00 N TA C = 1560.00 N TA D = 183.010 N

TA B = 1431 N ◀ TA C = 1560 N ◀ T A D = 183.0 N ◀

PROBLEM 2.118 Three cables are connected at D, where an upward force of 30 kN is applied. Determine the tension in each cable.

SOLUTION

Free-Body Diagram of Point D:

SFD = 0: TDA + TDB + TDC + P = 0

Where

P = Pj

 DA = -(1 m)i DA = 1 m  DB = +(0.75 m)i - (3 m) j - (1 m)k DB = 3.25 m  DC = +(1 m)i - (3 m) j + (1.5 m)k DC = 3.5 m  DA = TDA (-i) TDA = TDA λDA = TDA DA  æ3 DB 12 4 ö = TDB çç i - j - k÷÷÷ TDB = TDB λDB = TDB ç DB 13 13 ø÷ è13  æ2 DC 6 3 ö = TDC çç i - j + k÷÷÷ TDC = TDC λDC = TDC ç DC 7 7 ÷ø è7

Substituting into SFD = 0, setting P = (30 kN) j and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i : - TDA +

3 2 TDB + TDC = 0 13 7

(1)

PROBLEM 2.118 (Continued)

j: -

12 6 TDB - TDC + 30 kN = 0 (2) 13 7

k: -

4 3 T + T =0 13 DB 7 DC

(3)

Solving the system of linear equations using conventional algorithms gives: TDA = 8.50 kN ◀ TDB = 19.50 kN ◀ TDC = 14.00 kN ◀

PROBLEM 2.119

For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 1800 lb.

A 100 ft

20 ft

PROBLEM 2.111 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 840 lb, determine the vertical force P exerted by the tower on the pin at A.

25 ft B

O 74 ft

20 ft 60 ft

18 ft

SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: i:

4 30 10 TA B + TA C - TA D = 0 21 59 63

(1)

20 50 50 TA B - TA C - TA D + P = 0 21 59 63

(2)

5 9 37 TA B + TA C - TA D = 0 21 59 63

(3)

Substituting for P = 1800 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 4 30 10 TA B + TA C - TA D = 0 21 59 63

(1¢)

20 50 50 T A B - T A C - T A D + 1800 lb = 0 21 59 63

(2¢)

5 9 37 TA B + TA C - TA D = 0 21 59 63

(3¢)

T A B = 973.64 lb T A C = 531.00 lb T A D = 532.64 lb

T A B = 974 lb ◀ TA C = 531 lb ◀ T A D = 533 lb ◀

PROBLEM 2.120

24 in.

B 16 in.

24 in.

Three wires are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-lb cylinder is suspended from point D as shown.

22 in. C A 18 in.

D 180 lb

SOLUTION Free-Body Diagram of Point D:

The forces applied at D are: TDA , TDB , TDC and W

where W = - 180.0 lbj . To express the other forces in terms of the unit vectors i, j, k, we write  DA = (18 in.) j + (22 in.)k

DA = 28.425 in.  DB = -(24 in.)i + (18 in.) j - (16 in.)k DB = 34.0 in.  DC = (24 in.)i + (18 in.) j - (16 in.)k DC = 34.0 in.

PROBLEM 2.120 (Continued)  DA TDA = TDa λDA = TDa DA = (0.63324 j + 0.77397k )TDA  DB TDB = TDB λDB = TDB DB = (-0.70588i + 0.52941j - 0.47059k )TDB  DC TDC = TDC λDC = TDC DC = (0.70588i + 0.52941j - 0.47059k )TDC

and

Equilibrium Condition with

W = -W j

SF = 0: TDA + TDB + TDC - W j = 0

Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:

(-0.70588TDB + 0.70588TDC )i (0.63324TDA + 0.52941TDB + 0.52941TDC - W ) j (0.77397TDA - 0.47059TDB - 0.47059TDC )k Equating to zero the coefficients of i, j, k: -0.70588T DB + 0.70588T DC = 0

(1)

0.63324TDA + 0.52941T DB + 0.52941TDC - W = 0

(2)

0.77397T DA - 0.47059T DB - 0.47059T DC = 0

(3)

Substituting W = 180 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, T DA = 119.7 lb ◀ T DB = 98.4 lb ◀ T DC = 98.4 lb ◀

PROBLEM 2.121

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.)

0.40 m

0.86 m 1.20 m

1.30 m

O 0.78 m

F D P

z 1.60 m

0.40 m

A W

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

and

 AB TAB = T λAB = TAB AB TAB [-(0.78 m)i + (1.6 m) j + (0 m)k ] = 1.78 m TAB = TAB (-0.4382i + 0.8989 j + 0 k )  AC = (0)i + (1.6 m) j + (1.2 m)k

and

AC = (0 m)2 + (1.6 m)2 + (1.2 m)2 = 2 m  AC TAC TAC = T λAC = TAC = [(0)i + (1.6 m) j + (1.2 m)k ] AC 2 m TAC = TAC (0.8 j + 0.6 k )  AD = (1.3 m)i + (1.6 m) j + (0.4 m)k AD = (1.3 m)2 + (1.6 m)2 + (0.4 m)2 = 2.1 m  T AD TAD = T λAD = TAD = AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ] AD 2.1 m TAD = TAD (0.6190i + 0.7619 j + 0.1905k )

PROBLEM 2.121 (Continued)  AE = -(0.4 m)i + (1.6 m) j - (0.86 m)k

Finally,

AE = (-0.4 m)2 + (1.6 m)2 + (-0.86 m)2 = 1.86 m  AE TAE = T λAE = TAE AE TAE = [-(0.4 m)i + (1.6 m) j - (0.86 m)k ] 1.86 m TAE = TAE (-0.2151i + 0.8602 j - 0.4624 k ) With the weight of the container

W = -W j, at A we have: SF = 0: TA B + TA C + TA D - W j = 0

Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: -0.4382TA B + 0.6190T A D - 0.2151TA E = 0

(1)

0.8989TA B + 0.8T A C + 0.7619TA D + 0.8602T A E - W = 0

(2)

0.6T A C + 0.1905T A D - 0.4624TA E = 0

(3)

Knowing that W = 1000 N and that because of the pulley system at B TA B = TAD = P, where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P.

P = 378 N ◀

PROBLEM 2.122

Knowing that the tension in cable AC of the system described in Problem 2.121 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container.

0.40 m

0.86 m 1.20 m

1.30 m

O 0.78 m

F D P

z 1.60 m

0.40 m A W

PROBLEM 2.121 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 Here, as in Problem 2.121, 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 TA B = TA D = P

and using the linear algebraic equations of Problem 2.131 with T A C = 150 N, we obtain (a)

P = 454 N ◀

(b)

W = 1202 N ◀

PROBLEM 2.123 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 W = 270 lb, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLUTION Free-Body Diagram of Point A:

TAB = T λAB  AB =T AB (-48 in.)i + (72 in.) j - (16 in.)k =T 88 in. æ 6 9 2 ö = T çç- i + j - k÷÷÷ 11 11 ø÷ èç 11

TAC = T λAC  AC =T AC (24 in.)i + (72 in.) j + (-13 in.)k =T 77 in. æ 24 72 13 ö j - k÷÷÷ = T çç i + çè 77 77 77 ÷ø SF = 0: TAB + TAC + Q + P + W = 0 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 137

PROBLEM 2.123 (continued) Setting coefficients of i, j, k equal to zero:

Data:

i: -

6 24 T + T +P=0 11 77

æ 18 ö - çç ÷÷÷T + P = 0 çè 77 ø÷

(1)

j: +

9 72 T + T -W = 0 11 77

æ135 ö÷ çç ÷T - W = 0 èç 77 ø÷÷

(2)

k: -

2 13 T - T +Q = 0 11 77

æ 27 ö - çç ÷÷÷T + Q = 0 çè 77 ÷ø

(3)

æ 77 ÷ö ÷ 270 lb T = 154.0 lb W = 270 lb T = çç çè135 ÷÷ø

Substituting for T in Eqn . (1) and (2):

P = 36.0 lb ◀ Q = 54.0 lb ◀

PROBLEM 2.124 For the system of Prob. 2.123, determine W and P knowing that Q = 36 lb. PROBLEM 2.123 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 W = 270 lb, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLUTION Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3). Setting Q = 36 lb we have: æ 77 ö T = çç ÷÷÷ 36 lb èç 27 ø÷

Eq. (3):

18 (102.667 lb) + P = 0 77

Eq. (1):

Eq. (2):

135 (102.667 lb) - W = 0 77

T = 102.667 lb P = 24.0 lb ◀ W = 180.0 lb ◀

PROBLEM 2.125

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.

P A y

200 mm O

SOLUTION Free-Body Diagrams of Collars:

For both Problems 2.125 and 2.126: ( A B )2 = x 2 + y 2 + z 2

Here

(0.525 m)2 = (0.20 m)2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2

or Thus, when y given, z is determined, Now

 AB λAB = AB 1 (0.20 i - yj + zk )m 0.525 m = 0.38095i - 1.90476 yj + 1.90476 zk =

Where y and z are in units of meters, m. From the F.B. Diagram of collar A:

SF = 0: N x i + N z k + Pj + T A B lA B = 0

Setting the j coefficient to zero gives

P - (1.90476 y )T A B = 0

P = 341 N

With

TAB =

341 N

Now, from the free body diagram of collar B:

1.90476 y SF = 0: N x i + N y j + Qk - T A B λA B = 0

Setting the k coefficient to zero gives

Q - TA B (1.90476z ) = 0

And using the above result for TAB , we have

Q = TA B z =

341 N (1.90476) y

(1.90476z ) =

(341 N)(z ) y

PROBLEM 2.125 (Continued) Then from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 - (0.155 m)2 z = 0.46 m

and

341 N 0.155(1.90476) = 1155.00 N

TA B =

(a)

TA B = 1155 N ◀

or and

(b)

341 N(0.46 m)(0.866)

(0.155 m) = (1012.00 N) Q = 1012 N ◀

PROBLEM 2.126

Solve Problem 2.125 assuming that y = 275 mm.

PROBLEM 2.125 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.

A y

200 mm O

SOLUTION From the analysis of Problem 2.125, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N 1.90476 y 341 N Q= z y

TAB =

With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 - (0.275 m)2 z = 0.40 m

and TAB =

(a)

341 N = 651.00 (1.90476)(0.275 m) T A B = 651 N ◀

or and Q=

(b)

341 N(0.40 m) (0.275 m) Q = 496 N ◀

PROBLEM 2.127

85°

Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

P 55° 25°

SOLUTION Using the force triangle and the laws of cosines and sines: We have

g = 180 - (20 + 10) = 150 R 2 = (48 N)2 + (60 N)2 - 2(48 N)(60 N) cos150 R = 104.366 N

Then

48 N 104.366 N = sin a sin150 sin a = 0.22996

and

a = 13.2947 Hence:

f = 180 - a - 80 = 180 - 13.2947 - 80 = 86.705 R = 104.4 N

86.7° ◀

PROBLEM 2.128

y 120 N

Determine the x and y components of each of the forces shown. 80 N 150 N

30° 35°

40° x

SOLUTION 80-N Force:

120-N Force:

150-N Force:

Fx = +(80 N) cos 40 

Fx = 61.3 N ◀

Fy = +(80 N)sin 40

Fy = 51.4 N ◀

Fx = +(120 N) cos 70 

Fx = 41.0 N ◀

Fy = +(120 N)sin 70

Fy = 112.8 N ◀

Fx = -(150 N) cos 35 

Fx = -122.9 N ◀

Fy = +(150 N)sin 35

Fy = 86.0 N ◀

PROBLEM 2.129

α 400 lb

A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.

α 200 lb

SOLUTION Rx =

S Fx = P + (200 lb) sin 40  - (400 lb) cos 40 

R x = P - 177.860 lb

Ry =

SFy = (200 lb) cos 40 + (400 lb)sin 40

R y = 410.32 lb

(a)

(1)

(2)

For R to be vertical, we must have R x = 0. Set

R x = 0 in Eq. (1)

0 = P - 177.860 lb

P = 177.9 lb ◀

P = 177.860 lb

(b)

Since R is to be vertical: R = R y = 410 lb

R = 410 lb ◀

PROBLEM 2.130

30°

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.

20° 300 lb

SOLUTION Free-Body Diagram

Law of sines:

Force Triangle

FAC T 300 lb = BC = sin 35 sin 50 sin 95

(a)

FAC =

300 lb sin 35 sin 95

FA C = 172.7 lb ◀

(b)

TBC =

300 lb sin 50 sin 95

T B C = 231 lb ◀

PROBLEM 2.131

600 mm

Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable AC, (b) in cable BC.

250 mm

P C

Q = 480 N

SOLUTION Free Body: C

(a)

SFx = 0: -

(b)

SFy = 0:

12 4 TAC + (360 N) = 0 13 5

TA C = 312 N ◀

5 3 (312 N) + TBC + (360 N) - 480 N = 0 13 5 T B C = 480 N - 120 N - 216 N

TBC = 144.0 N ◀

35°

Two cables tied together at C are loaded as shown. Knowing that the 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.

50°

PROBLEM 2.132

α P

SOLUTION Free-Body Diagram: C

Force Triangle

Force triangle is isosceles with

2b = 180 - 85 b = 47.5 P = 2(800 N)cos 47.5° = 1081 N

(a) Since P > 0, the solution is correct. (b)

P = 1081 N ◀ a = 180 - 50 - 47.5 = 82.5

a = 82.5 ◀

PROBLEM 2.133 A

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 qx, qy, and qz that the force forms with the coordinate axes.

36° B

60° 20°

48°

C x

SOLUTION (a)

(b)

Fx = (120 lb)cos 60 cos 20 Fx = 56.382 lb

Fx = + 56.4 lb ◀

Fy = -(120 lb)sin 60 Fy = -103.923 lb

Fy = -103.9 lb ◀

Fz = -(120 lb) cos 60 sin 20 Fz = -20.521 lb

Fz = - 20.5 lb ◀

cos qx = cos qy =

cos qz =

Fx 56.382 lb = 120 lb F Fy F

-103.923 lb 120 lb

Fz -20.52 lb = 120 lb F

qx = 62.0  ◀

qy = 150.0 ◀ qz = 99.8  ◀

PROBLEM 2.134

y 360 mm

Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C.

920 mm A O B

600 mm D

900 mm

SOLUTION  CA = -(900 mm)i + (600 mm) j - (920 mm)k CA = (900 mm)2 + (600 mm)2 + (920 mm)2 = 1420 mm TCA = TCA λCA  CA = TCA CA 2130 N TCA = [-(900 mm)i + (600 mm) j - (920 mm)k ] 1420 mm = -(1350 N)i + (900 N) j - (1380 N)k (TCA )x = -1350 N, (TCA )y = 900 N, (TCA )z = -1380 N ◀

PROBLEM 2.135

Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N.

40° 55° 30° O

z 25°

SOLUTION P = (600 N)[sin 40 sin 25i + cos 40 j + sin 40 cos 25k ] = (162.992 N)i + (459.63 N) j + (349.54 N)k Q = (450 N)[cos 55 cos30i + sin 55 j - cos 55 sin 30k ] = (223.53 N)i + (368.62 N) j - (129.055 N)k R= P+Q = (386.52 N)i + (828.25 N) j + (220.49 N)k R = (386.52 N)2 + (828.25 N)2 + (220.49 N)2 = 940.22 N

cos qx =

cos qy = cos qz =

R = 940 N ◀

Rx 386.52 N = R 940.22 N

qx = 65.7  ◀

828.25 N 940.22 N

qy = 28.2 ◀

Rz 220.49 N = 940.22 N R

qz = 76.4  ◀

PROBLEM 2.136

y 17.5 in.

25 in.

45 in. E

Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

60 in.

A 80 in. P

SOLUTION Free Body Diagram at A:

Since T B A C = tension in cable BAC, it follows that TA B = TA C = TBA C

TAB = TBAC λAB = TBAC TAC = TBAC λAC = TBAC TAD = TAD λAD = TAD TAE = TAE λAE = TAE

(-17.5 in.)i + (60 in.) j 62.5 in. (60 in.)i + (25 in.)k

65 in. (80 in.)i + (60 in.) j

100 in. (60 in.) j - (45 in.)k 75 in.

æ -17.5 60 ö÷ i+ j÷ = TBAC çç çè 62.5 62.5 ø÷÷

æ 60 25 ö = TBAC çç j + k÷÷÷ çè 65 65 ÷ø

æ4 3 ö = TAD çç i + j÷÷÷ çè 5 5 ø÷ æ4 3 ö = TAE çç j - k÷÷÷ çè 5 5 ÷ø

PROBLEM 2.136 (Continued) Substituting into S FA = 0, setting P = (-200 lb) j, and setting the coefficients of i, j, k equal to f, we obtain the following three equilibrium equations:

17.5 4 TBAC + TAD = 0 62.5 5

From

i: -

From

æ 60 60 ö 3 4 j : çç + ÷÷÷TBAC + TAD + TAE - 200 lb = 0 çè 62.5 65 ÷ø 5 5

From

(1)

25 3 TBAC - TAE = 0 65 5

(2)

(3)

Solving the system of linear equations using conventional algorithms gives: TBA C = 76.7 lb; TA D = 26.9 lb; TA E = 49.2 lb ◀

PROBLEM 2.137

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.

x P A

O 20 in. Q

SOLUTION Free-Body Diagrams of Collars: A:

 AB -xi - (20 in.) j + zk λAB = = AB 25 in. SF = 0: Pi + N y j + N z k + TAB λ AB = 0

Collar A:

Substitute for λA B and set coefficient of i equal to zero:

TAB x =0 25 in. SF = 0: (60 lb)k + N x¢ i + N y¢ j - TAB λ AB = 0 P-

Collar B:

(1)

Substitute for λA B and set coefficient of k equal to zero:

60 lb x = 9 in.

(a)

TAB z =0 25 in.

(2)

(9 in.)2 + (20 in.)2 + z 2 = (25 in.)2 z = 12 in.

(b)

From Eq. (2):

60 lb - TAB (12 in.) 25 in.

From Eq. (1):

(125.0 lb)(9 in.) 25 in.

T A B = 125.0 lb ◀

P = 45.0 lb ◀

PROBLEM 2.138

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 lb.

x P A

O 20 in. Q

SOLUTION See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below:

TAB x =0 25 in.

(1)

60 lb -

TAB z =0 25 in.

(2)

For P = 120 lb, Eq. (1) yields

T A B x = (25 in.)(20 lb)

(1¢)

From Eq. (2):

T A B z = (25 in.)(60 lb)

(2¢)

x =2 z

Dividing Eq. (1¢) by (2¢), Now write

(3)

x 2 + z 2 + (20 in.)2 = (25 in.)2

(4)

Solving (3) and (4) simultaneously,

4z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.7082 in. From Eq. (3):

x = 2z = 2(6.7082 in.) = 13.4164 in. x = 13.42 in., z = 6.71 in. ◀

1100 mm

PROBLEM 2F1

400 mm

Two cables are tied together at C and loaded as shown. Draw the free-body diagram needed to determine the tension in AC and BC.

A 960 mm

1600 kg

SOLUTION Free-Body Diagram of Point C:

W = (1600 kg)(9.81 m/s2 ) W = 15.6960(103 ) N W = 15.696 kN

40°

PROBLEM 2.F2 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-body diagram needed to determine the magnitudes of the forces TC and TD.

SOLUTION Free-Body Diagram of Point E:

PROBLEM 2.F3

15 in. B

The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. C 65 lb

60 lb

SOLUTION Free-Body Diagram of Point A:

14 m

24 m

PROBLEM 2.F4

8.25 m B E 10 m

A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F.

C F

1.10 m

SOLUTION Free-Body Diagram of Point B:

W E = 250 N + 765 N = 1015 N 8.25 qA B = tan-1 = 30.510 14 10 qBC = tan-1 = 22.620 24

Use this free body to determine TAB and TBC.

Free-Body Diagram of Point C:

qCD = tan-1

1.1 = 10.3889 6

Use this free body to determine TCD and WF. Then weight of skier WS is found by

W S = W F - 250 N ◀

PROBLEM 2.F5

Three cables are used to tether a balloon as shown. Knowing that the tension in cable AC is 444 N, draw the free-body diagram needed to determine the vertical force P exerted by the balloon at A.

5.60 m

B D 4.20 m

O C

2.40 m

3.30 m 4.20 m x

SOLUTION Free-Body Diagram of Point A:

PROBLEM 2.F6

y 360 mm

C D

450 mm

500 mm

A container of mass m = 120 kg is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each cable.

320 mm

x A

600 mm

SOLUTION Free-Body Diagram of Point A:

W = (120 kg)(9.81 m/s2 ) = 1177.2 N

PROBLEM 2.F7

y 10.8 ft

A 150-lb cylinder is supported by two cables AC and BC that are attached to the top of vertical posts. A horizontal force P, which is perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable.

10.8 ft B

3.6 ft

C P 7.2 ft

15 ft

SOLUTION Free-Body Diagram of Point C:

PROBLEM 2.F8

A 90 ft

20 ft

30 ft B

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tension in wire AB is 630 lb, draw the free-body diagram needed to determine the vertical force P exerted by the tower on the pin at A.

60 ft 45 ft z

30 ft

65 ft

SOLUTION Free-Body Diagram of point A:

CHAPTER 3

PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28, determine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components.

SOLUTION Note that

q = a - 20 = 28 - 20 = 8

and

Fx = (16 N) cos8 = 15.8443 N Fy = (16 N)sin 8 = 2.2268 N

Also

x = (0.17 m)cos 20 = 0.159748 m y = (0.17 m)sin 20 = 0.058143 m

Noting that the direction of the moment of each force component about B is counterclockwise,

MB = xFy + yFx = (0.159748 m)(2.2268 N) + (0.058143 m)(15.8443 N) = 1.277 N ⋅ m

or M B = 1.277 N ⋅ m

◀

PROBLEM 3.2 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28, determine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC.

SOLUTION First resolve the 4-lb force into components P and Q, where Q = (16 N)sin 28 = 7.5115 N

Then

M B = rA /B Q = (0.17 m)(7.5115 N)

= 1.277 N ⋅ m

or M B = 1.277 N ⋅ m

◀

PROBLEM 3.3 A

It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α = 10°, (c) the smallest force P that creates the same moment about B.

P 18 in. 70°

4 in.

SOLUTION

(a)

M B = rC/B FN

We have

= (4 in.)(200 lb) = 800 lb ⋅ in.

or MB = 800 lb ⋅ in. (b)

By definition,

◀

M B = rA/B P sin q q = 10 + (180 - 70) = 120

Then

800 lb ⋅ in. = (18 in.) ´ P sin120 or P = 51.3 lb ◀

(c)

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus

M B = dPmin d = rA/B

800 lb ⋅ in. = (18 in.)Pmin

Pmin = 44.4 lb

20° ◀

PROBLEM 3.4

200 mm

100 mm

D 125 mm 200 mm

A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D.

25°

300 N

SOLUTION

(a)

Fx = (300 N)cos 25 = 271.89 N Fy = (300 N)sin 25 = 126.785 N F = (271.89 N)i + (126.785 N) j

 r = DA = -(0.1 m)i - (0.2 m) j MD = r ´ F M D = [-(0.1 m)i - (0.2 m) j] ´ [(271.89 N)i + (126.785 N) j] = -(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k = (41.700 N ⋅ m)k

MD = 41.7 N ⋅ m ◀ (b)

The smallest force Q at B must be perpendicular to  DB at 45°  M D = Q( DB ) 41.700 N ⋅ m = Q(0.28284 m)

Q =147.4 N

45.0° ◀

PROBLEM 3.5

200 mm

100 mm

D 125 mm 200 mm C B

25°

A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.

300 N

SOLUTION (a)

See Problem 3.3 for the figure and analysis leading to the determination of MD

MD = 41.7 N ⋅ m ◀

(b)

Since C is horizontal C = C i  r = DC = (0.2 m)i - (0.125 m) j M D = r ´ C i = C (0.125 m)k 41.7 N ⋅ m = (0.125 m)(C )

C = 334 N ◀

C = 333.60 N

(c)

The smallest force C must be perpendicular to DC; thus, it forms a with the vertical

0.125 m 0.2 m a = 32.0

tan a =

M D = C ( DC ); DC = (0.2 m)2 + (0.125 m)2 = 0.23585 m

41.70 N ⋅ m = C(0.23585 m)

C = 176.8 N

58.0° ◀

PROBLEM 3.6 An 8-lb force P is applied to a shift lever. Determine the moment of P about B when a is equal to 25°.

SOLUTION First note

Px = (8 lb) cos 25 = 7.2505 lb Py = (8 lb)sin 25 = 3.3809 lb

Noting that the direction of the moment of each force component about B is clockwise, have M B = -xPy - yPx = -(8 in.)(3.3809 lb) - (22 in.)(7.2505 lb) = -186.6 lb ⋅ in.

or M B = 186.6 lb ⋅ in.

◀

PROBLEM 3.7 For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 210-lb ⋅ in. clockwise moment about B.

SOLUTION For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,

a=q 8 in. 22 in. = 19.98 = tan-1

and

M B = dPmin

Where

d = rA /B = (8 in.)2 + (22 in.)2 = 23.409 in.

Then

210 lb ⋅ in. 23.409 in. = 8.97 lb

Pmin =

Pmin = 8.97 lb

19.98°◀

PROBLEM 3.8 An 11-lb force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of 250 lb ⋅ in. Determine the value of a.

SOLUTION By definition

M B = rA /B P sin q

where

q = a + (90 - f )

and

f = tan-1

also

8 in. = 19.9831 22 in.

rA /B = (8 in.)2 + (22 in.)2 = 23.409 in.

Then

250 lb ⋅ in = (23.409 in.)(11 lb) x sin(a + 90 - 19.9831)

sin(a + 70.0169) = 0.97088

a + 70.0169 = 76.1391

and

a + 70.0169 = 103.861

a = 6.12 33.8 ◀

PROBLEM 3.9 Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c = 360 mm, determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C.

SOLUTION Free-Body Diagram of Rod AB: (a) F = 1350 N

cos a =

AC = (450)2 + (600)2 = 750 mm

450 = 0.6 750

sin a =

600 = 0.8 750

 F = F cos ai + F sin a j = (1350 N)0.6i + (1350 N)0.8 j

  MB = r

A/ B

 ´ F = (-0.45i - 0.24 j) ´ (810 i + 1080 j)

= (810 N)i + (1080N)j  r = -(0.45 m)i - (0.24 m)j A/ B

= -486 k + 194.4 k

=-(291.6 N ⋅ m)k (b)

MB = 292 N ⋅ m

◀

MB = 292 N ⋅ m

◀

F = (810 N)i + (1080 N)j

rC / B = (0.36 m) j

M B = rC / B ´ F = 0.36 j´ (810i + 1080 j) = -(291.6 N ⋅ m)k

PROBLEM 3.10 Rod AB is held in place by the cord AC. Knowing that c = 840 mm and that the moment about B of the force exerted by the cord at point A is 756 N⋅m, determine the tension in the cord.

SOLUTION Free-Body Diagram of Rod AB: M B = -(756 N ⋅ m )k

cos a =

450 1170

AC = (450)2 + (1080)2 = 1170 mm

sin a =

1080 1170

 F = F cos ai + F sin a j 450 1080 = Fi + Fj 1170 1170  r A/ B = -(0.45 m)i - (0.24 m)j

  MB = r

 F ´ F = (-0.45i - 0.24 j) ´ (450 i + 1080 j) A/ B 1170 F = (-486k + 108k ) 1170 æ 378 ÷ö F k Substituting for MB : = ççè 1170 ÷÷ø 378 -756 = F 1170

F = 2340 N ◀

PROBLEM 3.11

15.3 in. A 12.0 in. C

2.33 in.

The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

12.0 in.

SOLUTION First note

dCB = (12.0 in.)2 + (2.33 in.)2 = 12.2241 in.

Then

cos q = sin q =

and

12.0 in. 12.2241 in. 2.33 in. 12.2241 in.

FCB = FCB cos q i - FCB sin q j =

125 lb 12.2241 in.

[(12.0 in.) i - (2.33 in.) j]

Now

MA = rB/A ´ FCB

where

rB/A = (15.3 in.) i - (12.0 in. + 2.33 in.) j = (15.3 in.) i - (14.33 in.) j

Then

M A = [(15.3 in.)i - (14.33 in.) j]´

125 lb 12.2241 in.

(12.0i - 2.33 j)

= (1393.87 lb ⋅ in.)k

= (116.156 lb ⋅ ft)k

or M A = 116.2 lb ⋅ ft

◀

PROBLEM 3.12

20.5 in. 4.38 in.

7.62 in.

The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

17.2 in.

SOLUTION First note

dCB = (17.2 in.)2 + (7.62 in.)2 = 18.8123 in.

Then

cos q = sin q =

and

17.2 in. 18.8123 in. 7.62 in. 18.8123 in.

FCB = (FCB cos q )i - (FCB sin q ) j =

125 lb 18.8123 in.

[(17.2 in.)i + (7.62 in.) j]

Now

MA = rB/A ´FCB

where

rB/A = (20.5 in.)i - (4.38 in.) j

Then

MA = [(20.5 in.)i - (4.38 in.) j] ´

125 lb 18.8123 in.

(17.2i - 7.62 j)

= (1538.53 lb ⋅ in.)k or M A = 128.2 lb ⋅ ft

= (128.2 lb ⋅ ft)k

◀

PROBLEM 3.13 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

A 144 mm B 56 mm C 42 mm

SOLUTION Using (a):

MC = y1 (FAB ) x + x1 (FAB ) y æ7 ö æ 24 ö = (0.056 m) çç ´ 2500 N÷÷÷ + (0.042 m) çç ´ 2500 N÷÷÷ è 25 ø è 25 ø = 140.0 N ⋅ m

(a)

MC = 140.0 N ⋅ m

◀

Using (b):

MC = y2 (FAB ) x æ7 ö = (0.2 m) çç ´ 2500 N÷÷÷ è 25 ø = 140.0 N ⋅ m

(b)

MC = 140.0 N ⋅ m

◀

PROBLEM 3.14 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

A 88 mm C

56 mm

B 42 mm

SOLUTION Using (a):

MC = -y1 (FAB ) x + x1 (FAB ) y æ 24 ö æ7 ö = -(0.056 m) çç ´ 2500 N÷÷÷ + (0.042 m) çç ´ 2500 N÷÷÷ è 25 ø è 25 ø = +61.6 N ⋅ m

(a)

MC = 61.6 N ⋅ m ◀

Using (b):

MC = y2 (FAB ) x æ7 ö = (0.088 m) çç ´ 2500 N÷÷÷ è 25 ø = +61.6 N ⋅ m

(b)

MC = 61.6 N ⋅ m

◀

PROBLEM 3.15 Form the vector product P1 × P2 and use the result obtained to prove the identity sin (θ1 – θ2) = sin θ1 cos θ2 – cos θ1 sin θ2.

SOLUTION P1 = P1 (cos q1 i + sin q1 j) P2 = P2 (cos q2 i + sin q2 j)

Note:

(1) (2)

By definition,

P1 ´ P2 = -P1P2 sin(q1 - q2 )k

From Eqns. (1) and (2)

P1 ´ P2 = (P1 cos q1i + P1 sin q1 j) ´(P2 cos q2 i + P2 sin q2 j)

= P1P2 (cos q1 sin q2 - sin q1 cos q2 )k

(3)

(4)

Equating the right-hand members of Equations (3) and (4) yields:

-P1P2 sin(q1 - q2 )k = P1P2 (cos q1 sin q2 - sin q1 cos q2 )k or sin (q1 - q2 ) = sin q1 cos q2 - cos q1 sin q2 ◀

PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = –2i + 2j – 6k and Q = 8i + 2j + 4k, (b) P = 3i – 9j – 7k and Q = –4i + 2j – 5k.

SOLUTION (a)

We have

A = |P ´Q|

where

P =-2i + 2 j - 6k Q = 8i + 2 j + 4k

Then

i j k P ´ Q = -2 2 -6 8 2 4 = [(8 + 12)i + (-48 + 8) j + (-4 - 16)k ] = (20)i + (-40) j + (-20)k A = (20)2 + (-40)2 + (-20)2

(b)

We have

A = |P ´Q|

where

P = 3i - 9 j - 7k

or A = 49.0 ◀

Q = -4i + 2 j - 5k

Then

i j k P ´ Q = 3 -9 -7 -4 2 -5 = [(45 + 14)i + (28 + 15) j + (6 - 36)k ] = (59)i + (43) j + (-30)k A = (59)2 + (43)2 + (-30)2

or A = 78.9 ◀

PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 7i + 8j – 2k and 9i – 4j – 5k, (b) 6i – 3j + 9k and –5i + 4j – 3k.

SOLUTION (a)

A´ B A´ B

We have

where

A = 7i + 8 j - 2k

Then

i j k A ´ B = 7 8 -2 9 -4 -5

B = 9i - 4 j - 5k

and

= (-40 - 8)i + (-18 + 35) j + (-28 - 72)k = (-48)i + (17) j + (-100)k

A´ B = l=

(-48) + (17) + (-100) = 12593 2

(-48)i + (17) j + (-100)k 12593

or l = -0.428i + 0.1515 j - 0.891k ◀ (b)

A´B A´B

We have

where

A = 6 i - 3 j + 9k

Then

i A´ B = 6

j -3

k 9

-5

-3

and

B = -5i + 4 j - 3k

= (9 - 36)i + (-45 + 18) j + (24 - 15)k = -27i - 27 j + 9k = 9 (-3i - 3 j + k)

A ´ B = 9 (-3) + (-3) + (1) = 9 19 2

9(-3i - 3 j + 1)k 9 19

or l =

(-3i - 3 j + k)

◀

PROBLEM 3.18 A line passes through the points (–4 m, –3 m) and (2 m, 7 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLUTION d AB =

We have

(2 - (-4)) + (7 - (-3)) = 136m 2

Assume that a force F, of magnitude F(N), acts at point A and is directed from A to B:

F = FlAB

Where

r -r l AB = B A d AB 1 = (6 i + 10 j) 136

By definition

M 0 = |rA ´ F |= - Fd k

where

rA = (-4 m) i + (-3 m) j M0 = (-4 i - 3j)´

136

(6i -10 j)

(-40k +18k) 136 22 M0 = Fk (N ⋅ m) 136 M0 =

then

Finally

Fk = -Fdk

136 d=

(m)

or d = 1.886 m ◀

136

PROBLEM 3.19 Determine the moment about the origin O of the force F = –2i – 3j + 5k that acts at a point A. Assume that the position vector of A is (a) r = i + j + k, (b) r = 4i + 6j – 10k, (c) r = 4i + 3j – 5k.

SOLUTION

MO = r ´ F

(a)

MO = 1

-2 -3 5

= (5 + 3)i + (-2 - 5) j + (-3 + 2)k

(b)

MO = 4

-10

-2 -3

= (30 - 30)i + (20 - 20) j + (-12 +12)k

(c)

MO = 4

-5

-2 -3

= (15 -15)i + (10 - 20) j + (-12 + 6)k

MO = 8i - 7j - k ◀

MO = 0 ◀

MO =-10j -6k ◀

Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the

determinant are proportional.

PROBLEM 3.20 Determine the moment about the origin O of the force F = 4i + 10j + 6k that acts at a point A. Assume that the position vector of A is (a) r = 2i – 3j + 4k, (b) r = 2i + 6j + 3k, (c) r = 2i + 5j + 6k.

SOLUTION

MO = r ´ F i (a)

M O = 2 -3 4 4 10

= (-18 - 40)i + (16 -12) j + (20 +12)k

(b)

MO = 2

MO =-58i + 4 j + 32k ◀

4 10 6

= (36 - 30)i + (12 -12) j + (20 - 24)k

(c)

MO = 2

MO = 6i - 4k ◀

4 10 6

= (30 - 60)i + (24 -12) j + (20 - 20)k

MO =-30i +12j ◀

PROBLEM 3.21

Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

B 7m

4.25 m

6m O A 0.75 m

1m x

SOLUTION

d BA = (-0.75 m)2 + (-7 m)2 + (6 m)2 = 9.25 m d BC = (4.25 m)2 + (-7 m)2 + (1 m)2 = 8.25 m

Have

 BA 555 N = TB A = TB A (-0.75 m i - 7 m j + 6 m k ) dB A 9.25 m

TBA = -(45 N)i - (420 N)j + (360 N)k  660 N BC = (4.25 m i - 7 m j + k ) TBC = TBC dBC 8.25 m

TBC = (340 N)i - (540 N) j + (80 N)k R = TBA + TBC

(

) (

)

\ R = 295 N i - 980 N j + 440 N k

MO = rB / O ´R where rB / O = (7 m)j i MO = 0

j 7

k 0 N⋅m

295 -980 440

(

) (

)

= 3080 N ⋅ m i - 2065 N ⋅ m k

(

) (

)

MO = 3080 N ⋅ m i - 2070 N ⋅ m k ◀

y 8 ft

PROBLEM 3.22

C 4.8 ft

The 12-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 380 lb, determine the moment about A of the force exerted by the cable at B.

z 12 ft

B x

SOLUTION First note

dBC = (-12)2 + (4.8)2 + (-8)2 = 15.2 ft 380 lb (-12i + 4.8 j - 8k) 15.2

Then

TBC =

We have

MA = rB/A ´ TBC

where

rB/A = (12 ft)i

Then

M A = (12 ft)i ´

380 lb (-12i + 4.8 j - 8k ) 15.2 or M A = (2400 lb⋅ ft) j + (1440 lb⋅ ft)k ◀

PROBLEM 3.23

y 200 N

A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.

30°

60 mm

60° B

x 25 mm

50 mm

SOLUTION We have

MA = rC/A ´ FC

where

rC/A = (0.06 m)i + (0.075 m) j FC = -(200 N)cos 30 j + (200 N)sin 30k i

Then

MA = 200 0.06 0

0.075

- cos30 sin 30

= 200[(0.075sin 30)i - (0.06 sin 30) j - (0.06 cos 30)k ] or MA = (7.50 N⋅ m)i - (6.00 N⋅ m) j - (10.39 N⋅ m)k ◀

PROBLEM 3.24 A force P of magnitude 200 N acts along the diagonal BC of the bent plate shown. Determine the moment of P about point E.

SOLUTION First determine the components of the force P:  BC 200 N = P = P l BC = P ((-0.3 m) j + (0.225 m) k ) BC 0.375 m P = -(160 N) j + (120 N)k

We have

 ME = EB ´ P

where

 EB = (-0.2 m)i + (0.3 m) j

i Then

ME = 200 -0.2 0

0.3

-160 120 or ME = (36.0 N⋅ m)i + (24.0 N⋅ m) j + (32.0 N⋅ m)k ◀

PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. y

B 8°

30°

45° D

SOLUTION We have

Txz = (6 lb)cos 8 = 5.9416 lb

Then

Tx = Txz sin 30 = 2.9708 lb Ty = TBC sin 8 = - 0.83504 lb Tz = Txz cos 30 = -5.1456 lb

Now

M A = rB/A ´TBC

where

rB/A = (6 sin 45) j - (6 cos 45)k =

Then

6 MA = 2 =

6 ft ( j - k) 2

i 0

j 1

k -1

2.9708 -0.83504 -5.1456

6 6 6 (-5.1456 - 0.83504)i (2.9708) j (2.9708)k 2 2 2

MA =-(25.4 lb ⋅ ft)i - (12.60 lb ⋅ ft) j - (12.60 lb ⋅ ft)k

◀

PROBLEM 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.

SOLUTION  BD F=F BD

where F = 900 N

 BD = -(1 m)i - (2 m) j + (2 m)k BD = (-1 m)2 + (-2 m)2 + (2 m)2 =3 m

-i - 2 j + 2 k 3 = -(300 N)i - (600 N) j + (600 N)k

F = (900 N)

rB/O = (2.5 m)i + (2 m) j

MO = rB/O ´ F i

= 2.5 2 0 -300 -600 600 = 1200 i - 1500 j + (-1500 + 600)k

MO = (1200 N⋅ m)i - (1500 N⋅ m) j - (900 N⋅ m)k ◀

y 8 ft

PROBLEM 3.27

C 4.8 ft

In Prob. 3.22, determine the perpendicular distance from point A to cable BC.

PROBLEM 3.22 The 12-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 380 lb, determine the moment about A of the force exerted by the cable at B.

z 12 ft

B x

SOLUTION M A = (2400)2 + (1440)2

From the solution to problem 3.22

= 2798.9 lb ⋅ ft

But

MA = FB ´ d d=

MA FB

2798.9 lb ⋅ ft 380 lb or d = 7.37 ft ◀

PROBLEM 3.28 In Prob. 3.23, determine the perpendicular distance from point A to the line of action of the 200-N force. PROBLEM 3.23 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.

SOLUTION From the solution to Prob. 3.23

F = 200 N M A = (7.50 N ⋅ m)i - (6.00 N ⋅ m) j - (10.3923 N ⋅ m)k M A = (7.50)2 + (6.00)2 + (10.3923)2 = 14.1510 N ⋅ m

But

M A = Fd

MA F 14.1510 N ⋅ m d= 200 N = 0.0708 m

d = 70.8 mm ◀

PROBLEM 3.29 In Prob. 3.24, determine the perpendicular distance from point E to the line of action of force P. PROBLEM 3.24 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.

SOLUTION From the solution to Prob. 3.24 P = 200 N M E = 36.0 i + 24.0 j + 32.0 k

ME = (36.0)2 + (24.0)2 + (32.0)2 = 53.814 N ⋅ m M E = Pd

ME P 53.814 N ⋅ m d= 200 N = 0.26907 m

d = 269 mm ◀

PROBLEM 3.30 In Prob. 3.25, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. y

B 8°

30°

45° D

SOLUTION From the solution to Prob. 3.25:

M A = -(25.4 lb ⋅ ft)i - (12.60 lb ⋅ ft) j - (12.60 lb ⋅ ft)k

M A = (-25.4)2 + (-12.60)2 + (-12.60)2 = 31.027 lb ⋅ ft M A = TBC d

MA TBC 31.027 lb ⋅ ft = 6 lb = 5.1712 ft

d = 5.17 ft ◀

PROBLEM 3.31 In Prob. 3.25, determine the perpendicular distance from point D to a line drawn through points B and C. PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. y

B 8°

30°

45° D

SOLUTION

AB = 6 ft TBC = 6 lb

We have

| MD | = TBC d

where d = perpendicular distance from D to line BC. MD = rB/D ´TBC rB/D = (6sin 45 ft) j = (4.2426 ft)

TBC : (TBC )x = (6 lb)cos8 sin30 = 2.9708 lb (TBC ) y = -(6 lb)sin8 = -0.83504 lb (TBC )z = -(6 lb)cos8 cos30 = -5.1456 lb TBC = (2.9708 lb)i - (0.83504 lb) j - (5.1456 lb)k MD =

i 0

j 4.2426

k 0

2.9708 -0.83504 -5.1456 = -(21.831 lb ⋅ ft)i - (12.6039 lb ⋅ ft)

| MD | = (-21.831)2 + (-12.6039)2 = 25.208 lb ⋅ ft

25.208 lb ⋅ ft = (6 lb)d

d = 4.20 ft ◀

PROBLEM 3.32 In Prob. 3.26, determine the perpendicular distance from point O to cable BD. PROBLEM 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.

SOLUTION From the solution to Prob. 3.26 we have MO = (1200 N ⋅ m)i - (1500 N ⋅ m) j - (900 N ⋅ m)k MO = (1200)2 + (-1500)2 + (-900)2 = 2121.3 N ⋅ m MO = Fd

MO F 2121.3 N ⋅ m

900 N

d = 2.36 m ◀

PROBLEM 3.33 In Prob. 3.26, determine the perpendicular distance from point C to cable BD. PROBLEM 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.

SOLUTION From the solution to Prob. 3.26 we have F = -(300 N)i - (600 N) j + (600 N)k rB/C = (2 m) j MC = rB/C ´ F = (2 m) j ´ (-300 Ni - 600 Nj + 600 Nk )

= (600 N ⋅ m)k + (1200 N⋅ m)i MC = (600)2 + (1200)2 = 1341.64 N ⋅ m MC = Fd

d= =

MC F 1341.64 N ⋅ m 900 N

d = 1.491 m ◀

PROBLEM 3.34

Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B.

a 16 ft

C 2 ft 3 ft

18 ft 8 ft

A 10 ft

24 ft

SOLUTION Assuming a force F acts along AB,

|MC | = |rA / C ´ F| = F (d ) d = perpendicular distance from C to line AB

where

F = λAB F =

(24 ft) i + (24 ft) j - (28 ft) k (24)2 + (24)2 + (28)2 ft

F (6) i + (6) j - (7) k 11 rA/C = (3 ft)i - (10 ft) j - (a - 10 ft)k =

i j k F MC = 3 -10 10 a 11 6 6 -7 = [(10 + 6 a)i + (81 - 6 a) j + 78 k ]

|MC | = |rA/C ´ F2 |

Since

F 11

|rA/C ´ F2 | = (dF )2

1 (10 + 6 a)2 + (81 - 6 a)2 + (78)2 = d 2 121 d (d 2 ) = 0 to find a to minimize d: Setting da

1 [2(6)(10 + 6a ) + 2(-6)(81 - 6 a)] = 0 121 Solving

a = 5.92 ft

or a = 5.92 ft ◀

PROBLEM 3.35 Given the vectors P = 2i + j + 2k, Q = 3i + 4j – 5k, and S = –4i + j – 2k, compute the scalar products P · Q, P · S, and Q · S.

SOLUTION P ⋅ Q = (2 i + j + 2 k ) ⋅ (3i + 4 j - 5k ) = (2)(3) + (1)(4) + (2)(-5) = 6 + 4 - 10

P⋅Q = 0 ◀

P ⋅ S = (2i + j + 2k) ⋅ (-4i + j - 2k) = (2)(-4) + (1)(1) + (2)(-2) = -8 + 1 - 4 P ⋅ S = -11 ◀ Q ⋅ S = (3i + 4 j - 5k ) ⋅ (-4 i + j - 2 k ) = (3)(-4) + (4)(1) + (-5)(-2) = -12 + 4 + 10

Q⋅ S = 2 ◀

PROBLEM 3.36

Form the scalar product B · C and use the result obtained to prove the identity

C α β x

cos (α - β) = cos α cos β + sin α sin β .

SOLUTION

B = B cos ai + B sin a j

(1)

C = C cos b i + C sin b j

(2)

By definition:

B ⋅ C = BC cos(a - b )

(3)

From (1) and (2):

B ⋅ C = ( B cos ai + B sin a j) ⋅ (C cos b i + C sin b j) = BC (cos a cos b + sin a sin b )

(4)

Equating the right-hand members of (3) and (4),

cos(a - b ) = cos a cos b + sin a sin b ◀

PROBLEM 3.37 Three cables are attached to the top of the tower at A. Determine the angle formed by cables AB and AC.

SOLUTION First note:

 AB = (16 m)i - (48 m) j + (12 m)k  AC = (16 m)i - (48 m) j - (24 m)k

AB = (16 m)2 + (-48 m)2 + (12 m)2 = 52.0 m AC = (16 m)2 + (-48 m)2 + (-24 m)2 = 56.0 m

By definition,

  AB ⋅ AC = ( AB)( AC )cos BAC   AB ⋅ AC cos BAC = AB ( )( AC )

((16 m)i - (48 m) j + (12 m)k) ⋅ ((16 m)i - (48 m) j - (24 m)k) (52.0 m)(56.0 m) 2 2 (16) + (48) - (12)(24) cos BAC = (52.0)(56.0) cos BAC =

cos BAC = 0.78022

PROBLEM 3.38 Three cables are attached to the top of the tower at A. Determine the angle formed by cables AD and AB.

SOLUTION First note:  AB = (16 m)i - (48 m) j + (12 m)k  AD = -(14 m)i - (48 m) j

AB = (16 m)2 + (-48 m)2 + (12 m)2 = 52.0 m AD = (-14 m)2 + (-48 m)2 = 50.0 m

By definition,

  AB ⋅ AD = ( AB)( AD)cos BAD   AB ⋅ AD cos BAD = ( AB)( AD)

((16 m)i - (48 m) j + (12 m)k) ⋅ ((-14 m)i - (48 m) j) (52.0 m)(50.0 m) 2 (16)(-14) + (48) cos BAD = (52.0)(50.0) cos BAD =

cos BAD = 0.800

PROBLEM 3.39

y 6 ft 3 ft

Knowing that the tension in cable AC is 280 lb, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at point A.

D 7.5 ft

C A

6.5 ft

B 6 ft

4.5 ft

SOLUTION (a)

First note

AC = (-6)2 + (2)2 + (3)2 = 7.00 ft AB = (-6)2 + (-4.5)2 + (0)2

and

By definition

= 7.50 ft  AC = -(6 ft)i + (2 ft) j + (3 ft)k  AB = -(6 ft)i - (4.5 ft) j   AC ⋅ AB = ( AC )( AB)cos q

(-6i + 2 j + 3k) ⋅ (-6i - 4.5 j) = (7.00)(7.50)´ cosq

(-6)(-6) + (2)(-4.5) + (3)(0) = 52.50cosq

cos q = 0.514 29

or (b)

We have

or q = 59.0 ◀

(TAC ) AB = TAC ⋅ λ AB = TAC cos q = (280 lb)(0.51429)

or (TAC ) AB = 144.0 lb ◀

PROBLEM 3.40

y 6 ft 3 ft

Knowing that the tension in cable AD is 180 lb, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exerted by cable AD at point A.

D 7.5 ft

C A

6.5 ft

B 6 ft

4.5 ft

SOLUTION (a)

First note

AD = (-6)2 + (3)2 + (-6)2 = 9.00 ft AB = (-6)2 + (-4.5)2 + (0)2

and

By definition,

= 7.50 ft  AD = -(6 ft)i + (3 ft) j - (6 ft)k  AB = -(6 ft)i - (4.5 ft) j   AD ⋅ AB = ( AD)( AB)cos q

(-6i + 3 j - 6k ) ⋅ (-6 i - 4.5 j) = (9.00)(7.50)cos q (-6)(-6) + (3)(-4.5) + (-6)(0) = 67.50 cos q cos q =

(b)

1 3

q = 70.5 ◀

(TAD ) AB = TAD ⋅ λ AB = TAD cos q

æ1ö = (180 lb)çç ÷÷÷ è 3ø

(TAD ) AB = 60.0 lb ◀

PROBLEM 3.41 y

3m A

Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B.

0.38 m

1m z

1.5 m

0.08 m 0.16 m Detail of the stake at B

SOLUTION First note:

BA = (-3)2 + (3)2 + (-1.5)2 = 4.5 m BD = (-0.08)2 + (0.38)2 + (0.16)2 = 0.42 m

TBA (-3i + 3 j - 1.5k ) 4.5 T = BA (-2i + 2 j - k ) 3  1 BD = (-0.08i + 0.38 j + 0.16k) λ BD = BD 0.42 1 = (-4 i + 19 j + 8k) 21 TBA =

Then

(a)

We have or

TBA ⋅ λBD = TBA cos q TBA 1 (-2 i + 2 j - k ) ⋅ (-4 i + 19 j + 8k ) = TBA cos q 3 21

1 [(-2)(-4) + (2)(19) + (-1)(8)] 63 = 0.60317

cos q =

or q = 52.9 ◀ (b)

We have

(TBA ) BD = TBA ⋅ λ BD = TBA cos q = (540 N)(0.60317) or (TBA )BD = 326 N ◀

PROBLEM 3.42 y

3m A

C 3m

Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at Point B.

0.38 m

1m z

1.5 m

0.08 m 0.16 m Detail of the stake at B

SOLUTION First note:

BC = (1)2 + (3)2 + (-1.5)2 = 3.5 m BD = (-0.08)2 + (0.38)2 + (0.16)2 = 0.42 m

TBC (i + 3 j -1.5k ) 3.5 T = BC (2i + 6 j - 3k ) 7  1 BD (-0.08i + 0.38 j + 0.16k) = λ BD = BD 0.42 1 = (-4 i + 19 j + 8k) 21 TBC =

(a)

TBC ⋅ λ BD = TBC cos q TBC 1 (2 i + 6 j - 3k ) ⋅ (-4 i + 19 j + 8k ) = TBC cos q 7 21

1 [(2)(-4) + (6)(19) + (-3)(8)] 147 = 0.55782

cos q =

q = 56.1 ◀ (b)

(TBC )BD = TBC ⋅ λ BD = TBC cos q = (490 N)(0.55782)

(TBC )BD = 273 N ◀

PROBLEM 3.43

y 24 in.

C 12 in.

O z

The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC.

x 20 in.

B x

SOLUTION (a)

Using Eq. (3.32):  CA = 11i - 12 j + 24 k CA = (11)2 + (-12)2 + (24)2 = 29 in.  CB = 31i - 12 j + 24 k CB = (31)2 + (-12)2 + (24)2 = 41 in.   CA ⋅ CB cos q = (CA)(CB) (11i - 12 j + 24 k ) ⋅ (31i - 12 j + 24 k ) = (29)(41) (11)(31) + (-12)(-12) + (24)(24) = (29)(41) = 0.89235

(b)

q = 26.8 ◀

Law of cosines:

( AB)2 = (CA)2 + (CB)2 - 2(CA)(CB)cos q (20)2 = (29)2 + (41)2 - 2(29)(41)cos q cos q = 0.89235

q = 26.8 ◀

PROBLEM 3.44

y 24 in.

Solve Prob. 3.43 for the position corresponding to x = 4 in.

12 in.

O z

x 20 in.

B x

PROBLEM 3.43 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC.

SOLUTION (a)

Using Eq. (3.32):  CA = 4 i - 12 j + 24 k CA = (4)2 + (-12)2 + (24)2 = 27.129 in.  CB = 24 i - 12 j + 24 k CB = (24)2 + (-12)2 + (24)2 = 36 in.

  CA ⋅ CB cos q = (CA)(CB) (4 i - 12 j + 24 k ) ⋅ (24 i - 12 j + 24 k ) = (27.129)(36) = 0.83551 (b)

q = 33.3 ◀

Law of cosines:

( AB)2 = (CA)2 + (CB)2 - 2(CA)(CB)cos q (20)2 = (27.129)2 + (36)2 - 2(27.129)(36)cos q cos q = 0.83551

q = 33.3 ◀

PROBLEM 3.45 Determine the volume of the parallelepiped of Fig. 3.20b when (a) P = 2i – 4j + 3k, Q = –5i – j + 2k, and S = i – 7j – k, (b) P = 6i – 5 j – 2k, Q = i + 2j + 3k, and S = –4i – 3j + 2k.

SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a)

Vol. = P ⋅ (Q ´ S) 2 -4

= -5 -1

1 -7 -1 = (2 - 8 + 105 + 3 + 28 + 20) = 150 or Volume = 150.0 ◀ (b)

Vol. = P ⋅ (Q ´ S) 6 -5 -2 =

-4 -3

= (24 + 60 + 6 - 16 + 54 + 10) = 138 or Volume = 138.0 ◀

PROBLEM 3.46 Given the vectors P = 3i + 2j + k, Q = 5i + j – 2k, and S = i + 3j + Sz k, determine the value of Sz for which the three vectors are coplanar.

SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q ´ S).

P ⋅ (Q ´ S) = 0 (or, the volume of a parallelepiped defined by P, Q, and S is zero).

3 2 Then

5 1 -2 = 0 1 3 Sz 3S z - 4 + 15 - 1 - 10 Sz + 18 = 0

Qy = 4. ◀

PROBLEM 3.47

y A

O 15 m C z

A crane is oriented so that the end of the 25-m boom AO lies in the yz plane. At the instant shown, the tension in cable AB is 4 kN. Determine the moment about each of the coordinate axes of the force exerted on A by cable AB. x

2.5 m

SOLUTION OC = (OA)2 - ( AC )2 = (25 m)2 - (15 m)2 = 20 m

rA = (15 m)j + (20 m)k AB = (2.5 m)2 + (15 m)2 = 15.2069 m  AB P =P AB 2.5i - 15 j = (4 kN) 15.2069 = (0.65760 kN)i - (3.9456 kN) j

Using Eq. (3.11):

MO = rA ´ P =

15 20

0.65760 3.9456

k 0

MO = (78.912 kN ⋅ m)i + (13.1520 kN ⋅ m) j - (9.8640 kN ⋅ m)k But Therefore,

MO = M x i + M y j + M z k

M x = 78.9 kN⋅ m, M y = 13.15 kN⋅ m, Mz = -9.86 kN⋅ m ◀

PROBLEM 3.48

y A

O 15 m

The 25-m crane boom AO lies in the yz plane. Determine the maximum permissible tension in cable AB if the absolute value of moments about the coordinate axes of the force exerted on A by cable AB must be x

£ 60 kN⋅m, M y £ 12 kN⋅m, M z £ 8 kN⋅m

Mx C z

2.5 m

SOLUTION OC = (OA)2 - ( AC )2 = (25 m)2 - (15 m)2 = 20 m

rA = (15 m)j + (20 m)k AB = (2.5 m)2 + (15 m)2 = 15.2069 m  AB P =P AB 2.5i - 15 j = ( P) 15.2069 = (0.164399P kN)i - (0.98639 P kN) j

Using Eq. (3.11):

i MO = rA ´ P =

0 15 20 0.163499P -0.98639P 0

MO = (19.7278P )i + (3.2700 P ) j - (2.4525P )k But

M x < 60 kN ⋅ m:

19.7278P < 60

P £ 3.04 kN

M y < 12 kN ⋅ m:

3.270P < 12

P £ 3.67 kN

M z < 8 kN ⋅ m:

2.4525P < 8

P £ 3.15 kN

P = 3.04 kN ◀

Therefore, the maximum permissible tension is

PROBLEM 3.49

F ϕ

θ A

4 in.

To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that q = 25, Mx = -61 lb ⋅ ft, and M z = -43 lb ⋅ ft, determine f and d.

11 in.

y B

x z

SOLUTION We have

SMO : rA/O ´ F = MO

where

rA/O = -(4 in.)i + (11 in.) j - ( d )k F = F (cos q cos f i - sin q j + cos q sin f k )

F = 70 lb, q = 25

For

F = (70 lb)[(0.90631cos f )i - 0.42262 j + (0.90631sin f)k ]

MO = (70 lb)

i -4

j 11

k -d

in.

-0.90631cos f -0.42262 0.90631sin f = (70 lb)[(9.9694 sin f - 0.42262d ) i + (-0.90631d cos f + 3.6252 sin f ) j + (1.69048 - 9.9694 cos f )k ] in.

and

M x = (70 lb)(9.9694 sin f - 0.42262d ) in. = -(61 lb ⋅ ft)(12 in./ft)

(1)

M y = (70 lb)(-0.90631d cos f + 3.6252sin f) in.

(2)

Mz = (70 lb)(1.69048 - 9.9694 cos f) in. = -43 lb ⋅ ft(12 in./ft)

(3)

From Equation (3):

æ 634.33 ö÷ = 24.636 f = cos-1 çç è 697.86 ø÷÷

From Equation (1):

æ1022.90 ÷ö d = çç = 34.577 in. è 29.583 ø÷÷

or d = 34.6 in. ◀

f = 24.6 ◀

PROBLEM 3.50

F ϕ

θ A

When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, M x = -77 lb ⋅ ft and M z = -81 lb ⋅ ft. For d = 27 in., determine the moment My of F about the y axis.

4 in. 11 in.

y B

x z

SOLUTION We have

SMO : rA/O ´ F = MO

Where

rA/O = -(4 in.)i + (11 in.) j - (27 in.)k F = F (cos q cos fi - sin q j + cos q sin fk )

MO = F

i -4

j 11

k -27

lb ⋅ in.

cos q cos f - sin q cos q sin f = F [(11cos q sin f - 27sin q )i + (-27 cos q cos f + 4 cos q sin f) j + (4 sin q - 11cos q cos f)k ](lb ⋅ in.)

and

Mx = F(11cos q sin f - 27sin q)(lb ⋅ in.)

(1)

M y = F (-27cos q cos f + 4 cos q sin f) (lb ⋅ in.)

(2)

Mz = F (4 sin q -11cos q cos f) (lb ⋅ in.)

(3)

ö 1 æ Mx ç + 27sin q÷÷÷ ç ø 11 è F M ö 1æ cos q cos f = ççç4 sin q - z ÷÷÷ and Equation (3) F ø 11 è Substituting Equations (4) and (5) into Equation (2), ìï é1æ é 1 æ Mx M z ö÷ù öùüï My = F ï + 27sin q÷÷÷úï ÷ú + 4 ê çç í-27 êê ççç4 sin q ý ÷ ú øúûþ êë 11 è F F øû ë11 è ïï ïï î 1 or M y = (27 M z + 4 M x ) 11 Now, Equation (1)

cos q sin f =

(4) (5)

PROBLEM 3.50 (Continued) 27 4 Noting that the ratios 11 and 11 are the ratios of lengths, have

27 4 (-81 lb ⋅ ft) + (-77 lb ⋅ ft) 11 11 = 226.82 lb ⋅ ft

My =

or M y = -227 lb ⋅ ft ◀

PROBLEM 3.51 The 0.61´1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

SOLUTION First note

z = (0.61)2 - (0.11)2 = 0.60 m

Then

dDE = (0.3)2 + (0.6)2 + (-0.6)2 = 0.9 m 66 N (0.3i + 0.6 j - 0.6 k ) 0.9 = 22[(1 N)i + (2 N) j - (2 N)k ]

and

TDE =

Now

M A = rD /A ´ TDE

where

rD /A = (0.11 m) j + (0.60 m)k

i Then

M A = 22 0 0.11 0.60 1 2 -2 = 22[(-0.22 - 1.20)i + 0.60 j - 0.11k ] = - (31.24 N ⋅ m)i + (13.20 N ⋅ m) j - (2.42 N ⋅ m)k M x = -31.2 N ⋅ m, M y = 13.20 N ⋅ m, M z = -2.42 N ⋅ m ◀

PROBLEM 3.52 The 0.61´1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C.

SOLUTION First note

z = (0.61)2 - (0.11)2 = 0.60 m

Then

dCE = (-0.7)2 + (0.6)2 + (-0.6)2 = 1.1 m 66 N (-0.7i + 0.6 j - 0.6k ) 1.1 = 6[-(7 N)i + (6 N) j - (6 N)k ]

and

TCE =

Now

M A = rE /A ´ TCE

where

rE /A = (0.3 m)i + (0.71 m) j

Then

i j k M A = 6 0.3 0.71 0 -7 6 -6 = 6[-4.26i + 1.8 j + (1.8 + 4.97)k ] = - (25.56 N ⋅ m)i + (10.80 N ⋅ m) j + (40.62 N ⋅ m)k M x = -25.6 N ⋅ m, M y = 10.80 N ⋅ m, M z = 40.6 N ⋅ m ◀

PROBLEM 3.53

A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb⋅ft, determine the magnitude of TDE when TAB = 255 lb.

A B C

12 ft z

14 ft

1 ft

F 1.5 ft

12 ft

SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have

SM x : (TAB )z ( yA ) + (TDE )z ( yD ) = M x

where

(TAB ) z = k ⋅ TAB = k ⋅ (TAB λAB ) é æ - i - 12 j + 12k ö÷ù ç ÷÷ú = k ⋅ êê 255 lb çç ÷ú 17 ç êë è ø÷úû = 180 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE ) é æ1.5i -14 j + 12k ö÷ù = k ⋅ êTDE çç ÷ú ê çè ÷÷øú 18.5 ë û = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft

(180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and

TDE = 282.79 lb

or TDE = 283 lb ◀

PROBLEM 3.54

Solve Problem 3.53 when the tension in cable AB is 306 lb.

A B

12 ft z

14 ft

PROBLEM 3.53 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

1 ft

F 1.5 ft

12 ft

SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have

SM x : (TAB )z ( yA ) + (TDE )z ( yD ) = M x

Where

(TAB ) z = k ⋅ TAB = k ⋅ (TAB λAB ) é æ - i - 12 j + 12 k ö÷ù ú = k ⋅ êê306 lb ççç ÷÷÷ú ç 17 è ø ëê ûú = 216 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE ) é æ1.5i -14 j + 12k ÷öù = k ⋅ êTDE çç ÷÷øúú êë è 18.5 û = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft

(216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and

TDE = 235.21 lb

or TDE = 235 lb ◀

PROBLEM 3.55 A force P of magnitude 520 lb acts on the frame shown at point E. Determine the moment of P about a line joining points O and D.

SOLUTION First develop expression for P in terms of its unit vector:  EH = (30 in.)i - (7.5 in.) j + (10 in.)k EH = (30)2 + (- 7.5)2 + (10)2 = 32.50 in.  EH 30i - 7.5 j + 10 k λEH = = 32.50 EH λEH = 0.92308i - 0.23077 j + 0.30769k

P = PλEH = 520 lb (0.92308i - 0.23077 j + 0.30770 k )

P = (480.0 lb)i - (120.0 lb) j + (160.0 lb)k The moment of P about axis OD is given by the mixed triple product Eq. (3.43): MOD = λOD ⋅ (rE / D ´ P )

rE / D =-(30 in.)i - (7.5 in.) j

 OD 30 i + 15 j + 10 k λOD = = 35.0 OD λOD = 0.85714 i + 0.42857 j + 0.28571k

0.85714 0.42857 0.28571 0 MOD = λOD ⋅ (rE / D ´ P ) = -30.0 in. -7.50 in. 480 lb

-120 lb

160 lb

MOD = 0.85714[(-7.50 in.)(160 lb) - 0] + 0.42857[0 - (-30.0 in.)(160 lb)] + 0.28571[(-30.0 in.)(-120 lb) - (-7.50)(480] = +3085.7 lb ⋅ in.

PROBLEM 3.56 A force P acts on the frame shown at point E. Knowing that the absolute value of the moment of P about a line joining points F and B is 300 lb∙ft, determine the magnitude of the force P.

P = PλEH = P (0.92308i - 0.23077 j + 0.30770 k )

The moment of P about axis FB is given by the mixed triple product Eq. (3.43): MFB = λFB ⋅ (rE / F ´ P )

rE / F = (7.5 in.) j

 FB 30 i + 15 j - 10 k λFB = = 35.0 FB λFB = 0.85714 i + 0.42857 j - 0.28571k

0.85714 MFB = λFB ⋅ (rE / F ´ P ) = 0

0.42857 7.50 in.

-0.28571 0

0.92308P -0.23077P 0.30769P

MFB = 0.85714[(7.50 in.)(0.30769 P) - 0] + 0.42857(0) - 0.28571[0 - (7.50)(0.92308P)]

Thus

= +3.9560 P MFB 3600 lb ⋅ in. P= = 3.9560 3.9560 in.

PROBLEM 3.57

y 0.35 m

The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

0.875 m

H O

0.925 m

0.75 m D

A z

0.5 m

0.75 m

0.5 m P

SOLUTION MAD = λ AD ⋅ (rB/A ´ TBH ) Where

1 λ AD = (4 i - 3k ) 5 rB/A = (0.5 m)i

dBH = (0.375)2 + (0.75)2 + (-0.75)2

and

= 1.125 m 450 N (0.375i + 0.75 j - 0.75k ) 1.125 = (150 N)i + (300 N) j - (300 N)k

Then

TBH =

Finally,

-3 4 0 1 0 MAD = 0.5 0 5 150 300 -300 1 = [(-3)(0.5)(300)] 5 or M AD = - 90.0 N ⋅ m ◀

PROBLEM 3.58

y 0.35 m

In Problem 3.57, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

0.875 m

H O

0.925 m

0.75 m D

A z

0.5 m

0.75 m

PROBLEM 3.57 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

0.5 m P

SOLUTION

MAD = λ AD ⋅ (rB/A ´TBG ) Where

1 λ AD = (4 i - 3k ) 5 rB/A = (0.5 m) j

BG = (-0.5)2 + (0.925)2 + (-0.4)2

and

= 1.125 m 450 N (-0.5i + 0.925 j - 0.4 k ) 1.125 = -(200 N)i + (370 N) j - (160 N)k

Then

TBG =

Finally,

-3 4 0 1 0.5 0 0 MAD = 5 -200 370 -160 1 = [(-3)(0.5)(370)] 5

M AD = -111.0 N⋅ m ◀

PROBLEM 3.59

y 0.4 m 0.2 m

The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

0.9 m

0.7 m A

0.6 m F

0.6 m z

0.9 m

B 0.35 m 0.6 m

0.3 m

0.4 m x

SOLUTION First note:

 AE TAE = TAE AE AE = (0.9)2 + (-0.6)2 + (0.2)2 = 1.1 m

55 N (0.9i - 0.6 j + 0.2 k ) 1.1 = 5[(9 N)i - (6 N) j + (2 N)k ]

Then

TAE =

Also,

DB = (1.2)2 + (-0.35)2 + (0)2

Now

= 1.25 m  DB λ DB = DB 1 (1.2i - 0.35 j) = 1.25 1 = (24 i - 7 j) 25 M DB = λ DB ⋅ (rA/D ´ TAE )

where

rA/D =-(0.1 m) j + (0.2 m)k

Then

24 -7 0 1 MDB = (5) 0 -0.1 0.2 25 9 -6 2

Then

1 = (-4.8 - 12.6 + 28.8) 5

or MDB = 2.28 N ⋅ m ◀

PROBLEM 3.60

y 0.4 m 0.2 m

The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

0.9 m

0.7 m A

0.6 m F

0.6 m z

0.9 m

B 0.35 m 0.6 m

0.3 m

0.4 m x

SOLUTION First note:

 CF TCF = TCF CF CF = (0.6)2 + (-0.9)2 + (-0.2)2 = 1.1 m

33 N (0.6 i - 0.9 j + 0.2k ) 1.1 = 3[(6 N)i - (9 N) j - (2 N)k ]

Then

TCF =

Also,

DB = (1.2)2 + (-0.35)2 + (0)2

Then

= 1.25 m  DB λ DB = DB 1 (1.2i - 0.35 j) = 1.25 1 = (24 i - 7 j) 25

Now

MDB = λ DB ⋅ (rC/D ´TCF )

where

rC/D = (0.2 m) j - (0.4 m)k

Then

24 -7 0 1 MDB = (3) 0 0.2 -0.4 25 6 -9 -2 =

3 (-9.6 + 16.8 - 86.4) 25

PROBLEM 3.61

y A

A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.

C z

SOLUTION We have

MOA = λOA ⋅ (rC/O ´ P)

From triangle OBC:

(OA) x =

a 2

(OA) z = (OA) x tan 30 =

Since or

a æç 1 ö÷ a ç ÷= 2 çè 3 ÷ø 2 3

(OA)2 = (OA)2x + (OA)2y + (OAz )2

æ a ÷ö2 æ a ö2 a 2 = çç ÷÷÷ + (OA)2y + çç ÷ è2ø èç 2 3 ÷ø (OA) y = a 2 -

a2 a2 2 =a 4 12 3

Then

rA/O =

a 2 a i+a j+ k 2 3 2 3

and

λOA =

1 2 1 i+ j+ k 2 3 2 3

P = λBC P =

( a sin 30)i - ( a cos 30)k P ( P ) = ( i - 3k ) a 2

rC/O = ai

1 2 MOA = 1

2 3 0

2 3 æPö 0 (a)ççè 2 ÷÷÷ø

- 3

aP æç 2 ö÷ aP çç- ÷÷ (1)(- 3) = ÷ 2 è 3ø 2

MOA =

aP ◀ 2

PROBLEM 3.62

y A

A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.61 to determine the perpendicular distance between edges OA and BC.

C z

SOLUTION (a)

For edge OA to be perpendicular to edge BC,   OA ⋅ BC = 0 From triangle OBC:

(OA) x =

a 2

aæ 1 ö a (OA) z = (OA) x tan 30 = ççç ÷÷÷ = ÷ ç 2è 3ø 2 3  æ a ö æ a ö÷ ÷k OA = çç ÷÷÷ i + (OA) y j + ççç çè 2 3 ÷÷ø èç 2 ø÷

and

 BC = ( a sin 30) i - ( a cos30) k =

Then

a a 3 a ik = (i - 3 k ) 2 2 2

éa æ a ö÷ ù a ê i + (OA) y j + çç k ú ⋅ (i - 3k ) = 0 çè 2 3 ÷÷ø ú ê2 2 ë û a2 a2 + (OA) y (0) =0 4 4   OA ⋅ BC = 0

  OA is perpendicular to BC. ◀   We have MOA = Pd, with P acting along BC and d the perpendicular distance from OA to BC.

so that (b)

From the results of Problem 3.57,

MOA =

Pa 2

Pa = Pd 2

or d =

a ◀ 2

PROBLEM 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 .

SOLUTION

First note that

F1 = F1 λ1 and F2 = F2 λ 2

Let M1 = moment of F2 about the line of action of F1 and M2 = moment of F1 about the line of action of F2 . Now, by definition,

M1 = λ1 ⋅ (rB/A ´ F2 ) = λ1 ⋅ (rB/A ´ λ 2 )F2 M 2 = λ 2 ⋅ (rA/B ´ F1 ) = λ 2 ⋅ (rA/B ´ λ1 )F1

F1 = F2 = F and rA/B = -rB/A

Since

M1 = λ1 ⋅ (rB/A ´ λ 2 )F M2 = λ 2 ⋅ (-rB/A ´ λ1 )F Using Equation (3.39): so that

λ1 ⋅ (rB/A ´ λ2 ) = λ2 ⋅ (-rB/A ´ λ1 ) M2 = λ 1⋅ (rB/A ´ λ 2 )F

M12 = M21 ◀

PROBLEM 3.64 In Prob. 3.55, determine the perpendicular distance between a line joining points O and D and the line of action of P. PROBLEM 3.55 A force P of magnitude 520 lb acts on the frame shown at point E. Determine the moment of P about a line joining points O and D.

SOLUTION From the solution to Prob. 3.55: P = (480.0 lb)i - (120.0 lb) j + (160.0 lb)k and P = 520 lb  OD 30 i + 15 j + 10k MOD = 3090 lb ⋅ in. λOD = = and OD 35.0 Based on the discussion of section 3.2C it follows that only the perpendicular component of P will contribute to the moment of P about line OD. Now

( P )parallel = P ⋅ λOD 1 (30i + 15j + 10k) 35 1 = éê(480)(30) + (-120)(15) + (160)(10)ùú û 35 ë = 405.71 lb

= (480i -120 j + 160k) ⋅

Also P = ( P )parallel + ( P )perpendicular So that

( P )perpendicular =

(520) - (405.71) = 325.27 lb 2

since λOD and (P)perpendicular are perpendicular, it follows that MOD = d ( P )perpendicular

3090 lb ⋅ in. = d (325.27 lb) d = 9.4998 in.

PROBLEM 3.65 In Prob. 3.56, determine the perpendicular distance between a line joining points F and B and the line of action of P. PROBLEM 3.56 A force P acts on the frame shown at point E. Knowing that the absolute value of the moment of P about a line joining points F and B is 300 lb∙ft, determine the magnitude of the force P.

SOLUTION From the solution to Prob. 3.56:

P = PλEH = P(0.92308i - 0.23077 j + 0.30769k )

and

P = 910 lb

P = 910(0.92308i - 0.23077 j + 0.30769k) P = (840.0 lb)i - (210.0 lb) j + (280.0 lb)k)  FB 30 i + 15 j - 10 k λFB = = and M FB = 3600 lb ⋅ in. FB 35.0 Based on the discussion of section 3.2C it follows that only the perpendicular component of P will contribute to the moment of P about line FB. Now

( P )parallel = P ⋅ λFB 1 (30i + 15j -10k) 35 1 = éê(840)(30) + (-210)(15) + (280)(10)ùú û 35 ë = 550.0 lb

= (840i - 210 j + 280k) ⋅

Also P = ( P )parallel + ( P )perpendicular

So that

( P )perpendicular =

(910) - (550.0) = 724.98 lb 2

since λOD and (P)perpendicular are perpendicular, it follows that MFB = d ( P )perpendicular

3600 lb ⋅ in. = d (724.98 lb) d = 4.9657 in.

PROBLEM 3.66

y 0.35 m

In Problem 3.57, determine the perpendicular distance between portion BH of the cable and the diagonal AD. 0.875 m

H O

0.925 m

0.75 m D

A z

0.5 m

0.75 m

0.5 m P

SOLUTION From the solution to Problem 3.57:

TBH = 450 N TBH = (150 N)i + (300 N) j - (300 N)k

MAD = 90.0 N ⋅ m 1 λ AD = (4 i - 3k ) 5

Based on the discussion of Section 3.11, itfollows that only the perpendicular component of TBH will  contribute to the moment of TBH about line AD.

(TBH )parallel = TBH ⋅ λ AD

Now

1 = (150i + 300 j - 300k) ⋅ (4 i - 3k ) 5 1 = [(150)(4) + (-300)(-3)] 5 = 300 N Also,

TBH = (TBH )parallel + (TBH )perpendicular

so that

(TBH ) perpendicular = (450)2 - (300) 2 = 335.41 N

Since λ AD and (TBH )perpendicular are perpendicular, it follows that M AD = d (TBH )perpendicular

90.0 N ⋅ m = d (335.41 N)

d = 0.26833 m

d = 0.268 m ◀

PROBLEM 3.67

y 0.35 m

In Problem 3.58, determine the perpendicular distance between portion BG of the cable and the diagonal AD. 0.875 m

PROBLEM 3.58 In Problem 3.57, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

H O

0.925 m

0.75 m D

A z

0.5 m

0.75 m C

0.5 m

SOLUTION From the solution to Problem 3.58:

Τ BG = 450 N TBG = -(200 N)i + (370 N) j - (160 N)k

MAD = 111 N ⋅ m 1 λ AD = (4 i - 3k ) 5

Based on the discussion of Section 3.11, itfollows that only the perpendicular component of TBG will  contribute to the moment of TBG about line AD. (TBG )parallel = TBG ⋅ λ AD

Now

1 = (-200i + 370 j - 160k ) ⋅ (4 i - 3k ) 5 1 = [(-200)(4) + (-160)(-3)] 5 = -64 N TBG = (TBG )parallel + (TBG )perpendicular

Also, so that

( TBG ) perpendicular = (450) 2 - (- 64) 2 = 445.43 N

Since λ AD and (TBG )perpendicular are perpendicular, it follows that M AD = d (TBG )perpendicular

111 N ⋅ m = d (445.43 N)

d = 0.24920 m

d = 0.249 m ◀

PROBLEM 3.68

y 0.4 m 0.2 m

In Problem 3.59, determine the perpendicular distance between cable AE and the line joining Points D and B.

0.9 m

PROBLEM 3.59 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

0.7 m A

0.6 m F

0.6 m z

0.9 m

B 0.35 m 0.6 m

0.3 m

0.4 m x

SOLUTION From the solution to Problem 3.59:

Τ AE = 55 N TAE = 5[(9 N)i - (6 N) j + (2 N)k]

MDB = 2.28 N ⋅ m λ DB =

1 (24 i - 7 j) 25

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. (TAE )parallel = TAE ⋅ λ DB

Now

= 5(9i - 6 j + 2k ) ⋅

1 (24 i - 7 j) 25

1 = [(9)(24) + (-6)(-7)] 5 = 51.6 N TAE = (TAE )parallel + (TAE )perpendicular

Also, so that

( TAE ) perpendicular =

(55) 2 + (51.6) 2 = 19.0379 N

Since λDB and (TAE )perpendicular are perpendicular, it follows that M DB = d (TAE )perpendicular

2.28 N ⋅ m = d (19.0379 N)

d = 0.119761

d = 0.1198 m ◀

PROBLEM 3.69

y 0.4 m 0.2 m

In Problem 3.60, determine the perpendicular distance between cable CF and the line joining Points D and B.

0.9 m

PROBLEM 3.60 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

0.7 m A

0.6 m F

0.6 m z

0.9 m

B 0.35 m 0.6 m

0.3 m

0.4 m x

SOLUTION From the solution to Problem 3.60:

ΤCF = 33 N TCF = 3[(6 N)i - (9 N) j - (2 N)k ]

MDB = 9.50 N ⋅ m λ DB =

1 (24 i - 7 j) 25

Based on the discussion of Section 3.11, itfollows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. (TCF )parallel = TCF ⋅ λ DB

Now

= 3(6i - 9 j - 2k ) ⋅

1 (24 i - 7 j) 25

3 [(6)(24) + (-9)(-7)] 25 = 24.84 N =

TCF = (TCF )parallel + (TCF )perpendicular

Also, so that

(TCF )perpendicular = (33)2 - (24.84)2 = 21.725 N

Since λDB and (TCF )perpendicular are perpendicular, it follows that

MDB = d (TCF )perpendicular or

9.50 N ⋅ m = d ´ 21.725 N or d = 0.437 m ◀.

PROBLEM 3.70

50°

Two 80-N forces are applied as shown to the corners B and D of a rectangular plate. (a) Determine the moment of the couple formed by the two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples. (b) Use the result obtained to determine the perpendicular distance between lines BE and DF.

80 N

300 mm 80 N 50° A

F 500 mm

SOLUTION (a)

Resolving forces into components:

P = (80 N)sin 50 = 61.284 N Q = (80 N)cos50 = 51.423 N M = (51.423 N)(0.5 m) - (61.284 N)(0.3 m) = 7.3263 N ⋅ m

M = 7.33 N⋅ m (b)

◀

Distance between lines BE and DF M = Fd

7.3263 N ⋅ m=(80 N)d d = 0.091579 m

d = 91.6 mm ◀

PROBLEM 3.71 Two parallel 60-N forces are applied as shown to the corners A and C of a 200-mm square plate. Determine the moment of the couple formed by the two forces (a) by multiplying their magnitude by their perpendicular distance, (b) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples.

SOLUTION

(a)

Using the perpendicular distance between the force couple: d = AE = ( AB) cos60 + ( BC ) sin 60 = 0.2 m (cos60 + sin 60) = 0.27321 m

M = (60 N)(0.27321 m) (b)

M =16.39 N ⋅ m ◀

= 16.3926 N ⋅ m Resolving into components: P = (60 N) sin 60 = 51.962 N Q = (60 N) cos60 = 30.0 N

M = Pa + Qa M = a(P + Q) M = (0.2 m)(51.962 N + 30.0 N)

M = 16.39 N⋅ m

◀

PROBLEM 3.72 A multiple-drilling machine is used to drill simultaneously six holes in the steel plate shown. Each drill exerts a clockwise couple of magnitude 40 lb∙in. on the plate. Determine an equivalent couple formed by the smallest possible forces acting (a) at A and C, (b) at A and D, (c) on the plate.

SOLUTION Equivalent couple must have moment:

M = 6(40 lb ⋅ in.) = 240 lb ⋅ in. (a)

Smallest forces at A and C are perpendicular to AC.

M = Fd 240 lb ⋅ in. = F (12 in.) F = 20.0 lb ◀

(b)

Smallest forces at A and D are perpendicular to AD.

M = Fd 240 lb ⋅ in. = F (15 in.) F = 16.00 lb ◀

M = Fd 240 lb ⋅ in. = F (20 in.) F = 12.00 lb ◀

PROBLEM 3.73 40 lb 60 lb

Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 1132.5 lb·in. counterclockwise.

9 in.

40 lb

60 lb

12 in.

SOLUTION M = d AD FAD + dBC FBC 1132.5 lb ⋅ in. = [(9 + d ) in.](60 lb) + [(12 + d ) in.](40 lb)

d = 1.125 in. ◀

PROBLEM 3.74 A piece of plywood in which several holes are being drilled successively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N⋅m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C. A

450 mm

C B

240 mm

SOLUTION (a)

M = Fd 12 N ⋅ m = F (0.45 m) F = 26.7 N ◀

(b)

M = Fd 12 N ⋅ m = F (0.24 m) F = 50.0 N ◀

(c)

M = Fd

d = (0.45 m)2 + (0.24 m)2 = 0.510 m

12 N ⋅ m = F (0.510 m) F = 23.5 N ◀

PROBLEM 3.75 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION Based on

M = M1 + M2

where

M1 = -(8 lb ⋅ ft)j M 2 = -(6 lb ⋅ ft)k M = -(8 lb ⋅ ft)j - (6 lb ⋅ ft)k

and

|M| = (8)2 + (6)2 = 10 lb ⋅ ft λ= =

or M = 10.00 lb⋅ ft ◀

M |M| -(8 lb ⋅ ft)j - (6 lb ⋅ ft)k

10 lb ⋅ ft = -0.8 j - 0.6k

M = |M|λ = (10 lb⋅ ft)(-0.8j - 0.6k) cos qx = 0 qx = 90 cos qy = -0.8 qy = 143.130 cos qz = -0.6 qz = 126.870 or qx = 90.0 qy = 143.1 qz = 126.9 ◀

PROBLEM 3.76 15 in.

16 lb

40 lb

16 lb

A 10 in. P

If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

15 in.

40 lb –P C 10 in. x

10 in.

SOLUTION M = M1 + M2 ; F1 = 16 lb, F2 = 40 lb M1 = rC ´ F1 = (30 in.)i ´[-(16 lb) j] = -(480 lb ⋅ in.)k M 2 = rE/B ´ F2 ; rE/B = (15 in.)i - (5 in.) j dDE = (0)2 + (5)2 + (10)2 = 5 5 in. F2 =

40 lb (5 j - 10k ) 5 5

= 8 5[(1 lb) j - (2 lb)k ] i

M2 = 8 5 15 -5 0 0 1 -2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] M = -(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] = (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j - (211.67 lb ⋅ in.)k

M = (178.885)2 + (536.66)2 + (-211.67)2 = 603.99 lb ⋅ in

M = 604 lb ⋅ in. ◀

M = 0.29617i + 0.88852 j - 0.35045k M cos qx = 0.29617 λ axis =

cos qy = 0.88852 cos qz = -0.35045

qx = 72.8 qy = 27.3 qz = 110.5 ◀

PROBLEM 3.77 15 in.

16 lb

40 lb

16 lb

A 10 in. P

If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

15 in.

40 lb –P C 10 in. x

10 in.

SOLUTION From the solution to Problem. 3.76: 16-lb force:

M1 = -(480 lb ⋅ in.)k

40-lb force:

M2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]

P = 20 lb

M3 = rC ´ P = (30 in.)i ´ (20 lb)k = (600 lb ⋅ in.) j

M = M1 + M2 + M3 = -(480)k + 8 5 (10i + 30 j + 15k ) + 600 j = (178.885 lb ⋅ in.)i + (1136.66 lb ⋅ in.) j - (211.67 lb ⋅ in.)k M = (178.885)2 + (113.66)2 + (211.67)2 = 1169.96 lb ⋅ in.

M = 1170 lb ⋅ in. ◀

M = 0.152898i + 0.97154 j - 0.180921k M cos qx = 0.152898 λ axis =

cos qy = 0.97154 cos qz = -0.180921

qx = 81.2 qy = 13.70 qz = 100.4 ◀

PROBLEM 3.78 The two couples shown are to be replaced with a single equivalent couple. Determine (a) the couple vector representing the equivalent couple, (b) the two forces acting at B and C that can be used to form that couple.

SOLUTION (a) Represent each couple by a couple vector in rectangular coordinates :

M1 = (25 N)(0.75 m) = 18.75 N ⋅ m M2 = (20 N)(0.4 m) = 8.0 N ⋅ m M1 = (18.75 sin 40)i + (18.75 cos 40) j = (12.0523 N ⋅ m)i + (14.3633 N ⋅ m)j M2 = -(8.0 N ⋅ m) Resultant couple:

M1 = M1 + M2 = (12.052 N ⋅ m)i + (6.3633 N ⋅ m)j M=

(12.0523) + (6.3633) = 13.6290 N⋅ m 2

12.0523 qx = 27.8 6.3633 qy = 90 - 27.8 = 62.2 tan qx =

qz = 90

M = 13.63 N⋅ m; qx = 27.8, qy = 62.2, qz = 90.0 ◀ (b)

Forces at B and C used to form a couple M

M = Fd 13.6290 = F (0.75 m) 18.17 N

62.2 at B; 18.17 N

62.2 at C ◀

PROBLEM 3.79 Solve part a of Prob. 3.78, assuming that two 15-N vertical forces have been added, one acting upward at A and the other downward at C. PROBLEM 3.78 The two couples shown are to be replaced with a single equivalent couple. Determine (a) the couple vector representing the equivalent couple, (b) the two forces acting at B and C that can be used to form that couple.

SOLUTION

The moment M3 of the additional couple is equal to the moment about A of the force applied at C. M 3 = (0.4 i - 0.75k ) ´ (-15 j) = -(11.25 N ⋅ m)i - (6 N ⋅ m)k Adding this couple to the couples M1 and M3 obtained in Prob. 3.78:

M = M1 + M2 + M3 = 12.05i + 14.36 j - 8 j -11.25i - 6k = (0.80 N ⋅ m)i + (6.36 N ⋅ m) j - (6 N ⋅ m)k M = (0.80)2 + (6.36)2 + (6)2 = 8.7801 N ⋅ m 0.80 8.7801 6.36 cos qy = 8.7801 -6 cos qz = 8.7801

M = 8.78 N⋅ m ◀

cos qx =

qx = 84.8 qy = 43.6 qz = 133.1 ◀

PROBLEM 3.80

y 900 lb·ft B

20°

840 lb·ft 20°

C x

Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts by a single equivalent couple, specifying its magnitude and the direction of its axis.

1200 lb·ft

SOLUTION M A = -1200sin 20 j + 1200 cos 20k = -410.42 j + 1127.63k M B = 900sin 20 j + 900 cos 20k = 307.82 j + 845.72k MC = -840i The single equivalent couple is the sum of the three moments M = -(840 lb ⋅ ft)i - (102.60 lb ⋅ ft) j + (1973.35 lb ⋅ ft )k

M = (-840)2 + (-102.60)2 + (1973.35)2 = 2147.3 lb ⋅ ft

M = 2150 lb ⋅ ft ◀

M = -0.39119i - 0.047921j + 0.91899k M cos qx = -0.39119 λ axis =

cos qy = -0.047921 cos qz = 0.91899

qx = 113.0 qy = 92.7 qz = 23.2 ◀

PROBLEM 3.81 A

75 mm 30°

B 500 N

175 mm

A 500-N force is applied to a bent plate as shown. Determine (a) an equivalent force-couple system at B, (b) an equivalent system formed by a vertical force at A and a force at B.

300 mm 125 mm

SOLUTION

(a)

Force-couple system at B

F = (500 N)sin 30 i - (500 N)cos30 k F = (250.0 N)i - (433.01 N)k M B = (0.3i - 0.175 j) ´ (250 i - 433.01j) = -86.153k N ⋅ m The equivalent force-couple system at B is

FB = 500 N (b)

MB = 86.2 N ⋅ m

60.0

◀

Require

Equivalence requires

å M B = SM B 86.153 N ⋅ m = A(0.125 m); A = (689.22 N)j F = A + B or B = F - A;

B = (250 N)i - (1122.23 N)j = 1149.74 N

77.4

B = (250 N)i - (433.01 N)j - (689.22 N)j B = 1150 N

77.4 ◀

PROBLEM 3.82 A crane column supports a 16-kip load as shown. Replace the load with an equivalent system consisting of an axial force along AB and a couple.

SOLUTION

F = 16.00 kips  M A = (16.00 kips)(12.00 in.) = 192.0 kip ⋅ in.

F = 16.00 kips  along AB, MA = 192.0 kip ⋅ in. ◀

PROBLEM 3.83

A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.

6.7 m 4 m

60°

SOLUTION

Require the equivalent forces acting at A and C be parallel and at an angle of a with the vertical. Then for equivalence,

SFx : (1040 N)sin30 = FA sin a + FB sin a

(1)

SFy : -(1040 N)cos30 =-FA cos a - FB cos a

(2)

Dividing Equation (1) by Equation (2),

(1040 N)sin 30 -(1040 N)cos30

(FA + FB )sin a -(FA + FB )cos a

Simplifying yields a = 30. Based on

SMC : [(1040 N)cos30](4 m) = (FA cos30)(10.7 m) FA = 388.79 N FA = 389 N

60.0° ◀

Based on

SMA : -[(1040 N)cos30](6.7 m) = (FC cos30)(10.7 m) FC = 651.21 N

FC = 651 N

60.0° ◀

PROBLEM 3.84

5 in. A

2 in. B 3 in.

A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a.

SOLUTION (a)

M B = (30 lb)(5 in.) = 150.0 lb ⋅ in.

(b)

B=C =

F = 30.0 lb , MB = 150.0 lb ⋅ in.

◀

B = 50.0 lb

◀

150 lb ⋅ in. = 50.0 lb 3.0 in. ; C = 50.0 lb

PROBLEM 3.85

40° 360 N

A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a.

B 0.4 m

C 0.35 m D 0.3 m 30° 2.4 m

SOLUTION (a)

(a)

We have

(b)

SF : 360 N(- sin 40°i - cos40 j) = -(231.40 N)i - (275.78 N) j = F or F = 360N

We have where

50° ◀

SMD : rB/D ´ R = M rB/D = -[(0.65 m)cos30]i + [(0.65 m)sin 30]j = -(0.56292 m)i + (0.32500 m) j i

M = -0.56292 -231.40

0.32500

k 0 N⋅m

-275.78 0

= [155.240 + 75.206)N ⋅ m]k

or M = 230 N ⋅ m

= (230.45 N ⋅ m)k (b)

We have where

◀

SMD : M = rA/D ´ FA rB/D = -[(1.05 m)cos30]i + [(1.05 m)sin 30]j = -(0.90933 m)i + (0.52500 m) j i

FA = -0.90933 0.52500 0 N ⋅ m -1 0 0 = [230.45 N ⋅ m]k Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 250

PROBLEM 3.85 (Continued) (0.90933FA )k = 230.45k

FA = 253.42 N We have

or FA = 253 N ◀

SF: F = FA + FD

-(231.40 N)i - (275.78 N) j =-(253.42 N) j + FD (-cos q i - sin q j) From

i : 231.40 N = FD cos q

(1)

22.36 N = FD sin q

(2)

Equation (2) divided by Equation (1)

tan q = 0.096629 q = 5.5193 or q = 5.52 Substitution into Equation (1)

FD =

231.40 = 232.48 N cos5.5193

or FD = 232 N

5.52° ◀

PROBLEM 3.86

40° 360 N

A worker tries to move a rock by applying a 360-N force to a steel bar as shown. If two workers attempt to move the same rock by applying a force at A and a parallel force at C, determine these two forces so that they will be equivalent to the single 360-N force shown in the figure.

B 0.4 m

C 0.35 m D 0.3 m 30° 2.4 m

SOLUTION SFX : - 360sin40= - FA sin a - FC sina (1) SFY : - 360cos40 = FA cos a - FC cos a (2)

Dividing (1) by (2) yields a = 40 and then FA + FC = 360 N (3)

SM B : 0 = (0.4 m)FA cos10 - (0.35 m)FC cos10 7 FC (4) 8 Substituting into (3) ....

FA =

7 FC + FC = 360 8 Finally

FA = 168.0 N

50.0°; FC = 192.0 N

50.0°; ◀

120 mm

PROBLEM 3.87

x B

A 250 N 90 mm

C 900 N

90 mm E

250 N

The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.)

SOLUTION Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple

M H = (0.18)(250 N) = 45 N ⋅ m Then or

MH = x(900 N) 45 N ⋅ m = x (900 N) x = 0.05 m F = 900 N

x = 50.0 mm ◀

PROBLEM 3.88 Knowing that α = 60°, replace the force and couple shown with a single force applied at a point located (a) on line AB, (b) on line CD. In each case determine the distance from the center O to the point of application of the force.

SOLUTION We first reduce system to a force and couple at O: Since the 120-N forces cancel out we have:

F = 960 N

60.0°

MO = (120 N)(0.2 m) = 24.0 N ⋅ m

(a)

We move force through x so that

24 N ⋅ m = (960 N)(cos30) x x = 0.028868 m Single force 960 N

(b)

60.0°, 28.9 mm to the right of O◀

We move force through y so that

24 N ⋅ m = (960 N)(sin30) y y = 0.05 m Single force 960 N

60.0°, 50.0 mm below O ◀

PROBLEM 3.89

48 lb 55° 30 in. 20 lb

40 in.

20°

C 30°

20°

20 lb

Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in Part a, and specify its point of application on the lever.

SOLUTION (a )

First note that the two 20-lb forces form A couple. Then F = 48 lb

where

q = 180 - (60 + 55) = 65

and

M = SM B = (30 in.)(48 lb)cos55 - (70 in.)(20 lb)cos20 = -489.62 lb ⋅ in

The equivalent force-couple system at B is F = 48.0 lb

(b)

65.0°;

M = 490 lb ⋅ in.

◀

The single equivalent force F ¢ is equal to F. Further, since the sense of M is clockwise, F ¢ must be applied between A and B. For equivalence.

SM B : M = -aF ¢ cos 55 where a is the distance from B to the point of application of F¢. Then

-489.62 lb ⋅ in. =-a(48.0 lb)cos 55 or

a = 17.78 in.

F ¢ = 48.0 lb

and is applied to the lever 17.78 in. to the left of pin B

65.0° ◀ ◀

PROBLEM 3.90

α 48 N

15 N

B α

240 mm

15 N

A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For a = 40, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of a if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D.

400 mm

SOLUTION (a )

The given force-couple system (F, M ) at B is F = 48 N

and

M = SM B = (0.4 m)(15 N)cos 40 + (0.24 m)(15 N)sin 40

M = 6.9103 N ⋅ m

The single equivalent force F ¢ is equal to F. Further for equivalence

SM B : M = dF ¢ 6.9103 N ⋅ m = d ´ 48 N

d = 0.14396 m

and the line of action of F ¢ intersects line AB 144.0 mm to the right of A. (b)

F ¢ = 48.0 N ◀ ◀

Following the solution to Part a but with d = 0.1 m and a unknown, have

SMB : (0.4 m)(15 N)cos a + (0.24 m)(15 N)sin a = (0.1 m)(48 N) or

5cos a + 3sin a = 4

Rearranging and squaring

25 cos2 a = (4 - 3 sin a )2

Using cos2 a = 1 - sin2 a and expanding 25(1 - sin 2 a ) = 16 - 24 sin a + 9 sin 2 a

or Then

34 sin 2 a - 24 sin a - 9 = 0

24  (-24)2 - 4(34)(-9) 2(34) sin a = 0.97686 or sin a = -0.27098 sin a =

a = 77.7 or a =-15.72

◀

25°

PROBLEM 3.91

2.9 lb B

2.8 in.

While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.

C 3.2 in. D

A 2.65 lb

25° x

SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a forcecouple. We have FB = 2.9 lb - 2.65 lb = 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two parallel forces, Mcouple = (2.65 lb)[(3.2 in. + 2.8 in.)cos25]

= 14.4103 lb ⋅ in.

Mcouple = 14.4103 lb ⋅ in.

and

A single equivalent force will be located in the negative z direction. Based on

SMB : -14.4103 lb ⋅ in. = [(0.25 lb)cos 25](a)

a = 63.600 in. F¢ = (0.25 lb)(cos25i + sin 25k) F¢ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B.

◀

PROBLEM 3.92

P α C

A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.

300 N

F 0.2 m

SOLUTION From the statement of the problem, it follows that SME = 0 for the given force-couple system. Further, for Pmin, we must require that P be perpendicular to rB/E . Then

SM E : (0.2 sin 30 + 0.2)m ´ 300 N + (0.2 m)sin 30´ 300 N - (0.4 m)Pmin = 0 or

Pmin = 300 N Pmin = 300 N

30.0° ◀

PROBLEM 3.93

An eccentric, compressive 250-kN force P is applied to the end of a column. Replace P with an equivalent force-couple system at G.

30 mm

G x 60 mm

SOLUTION

SF :

Have

(

)

- 250 kN j = F

(

)

or F = - 250 kN j ◀ Also have

SMG :

rP ´ P = M

i - 0.030

j 0

- 250

k 0.060 kN ⋅ m = M 0

(

) (

)

\ M = 15 kN ⋅ m i + 7.5 kN ⋅ m k

(

) (

)

or M = 15.00 kN ⋅ m i + 7.50 kN ⋅ m k ◀

PROBLEM 3.94

A 2.6-kip force is applied at Point D of the cast iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section.

12 in.

2.6 kips

A z

6 in.

E 5 in. x

SOLUTION  DE = -(12 in.) j - (5 in.)k; DE = 13.00 in.  DE F = (2.6 kips) DE

F = (2.6 kips)

-12 j - 5k 13

F = -(2.40 kips) j - (1.000 kip)k ◀ M A = rD/A ´ F where

rD/A = (6 in.)i + (12 in.) j i M A = 6 in. 0

12 in.

-2.4 kips -1.0 kips

M A = -(12.00 kip ⋅ in.)i + (6.00 kip ⋅ in.) j - (14.40 kip ⋅ in.)k ◀

PROBLEM 3.95

Replace the 150-N force with an equivalent force-couple system at A.

200 mm A B x

120 mm 35° 150 N

C D

20 mm

40 mm 60 mm

SOLUTION Equivalence requires

SF : F = (150 N)(- cos 35 j - sin 35k) = -(122.873 N) j - (86.036 N)k

SMA : M = rD/A ´ F where

rD/A = (0.18 m)i - (0.12 m) j + (0.1 m)k

Then

i j k M = 0.18 -0.12 0.1 N ⋅ m 0 -122.873 -86.036 = [(-0.12)(-86.036) - (0.1)(-122.873)]i + [-(0.18)(-86.036)]j + [(0.18)(-122.873)]k = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j - (22.1 N ⋅ m)k

The equivalent force-couple system at A is

F = -(122.9 N) j - (86.0 N)k ◀ M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j - (22.1 N ⋅ m)k ◀

PROBLEM 3.96 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.

SOLUTION We have

SF: PAB = FC where

PAB = λAB PAB =

(33 mm)i + (990 mm) j - (594 mm)k 1155.00 mm

(175 N)

or FC = (5.00 N)i + (150.0 N) j - (90.0 N)k ◀ We have

SMC : rB/C ´ PAB = MC i j k MC = 5 0.683 -0.860 0 N⋅m 1 30 -18 = (5){(- 0.860)(-18)i - (0.683)(-18) j + [(0.683)(30) - (0.860)(1)]k} or MC = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k ◀

PROBLEM 3.97 A 46-lb force F and a 2120-lb⋅in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H.

SOLUTION

Then

d AJ = (18)2 + (-14)2 + (-3)2 = 23 in. 46 lb F= (18i -14 j - 3k) 23 = (36 lb)i - (28 lb) j - (6 lb)k

Also

d AC = (-45)2 + (0)2 + (-28)2 = 53 in.

Then

Now

M ¢ = M + rA/H ´ F

where

rA/H = (45 in.)i + (14 in.) j

We have

2120 lb ⋅ in. (-45i - 28k ) 53 = -(1800 lb ⋅ in.)i - (1120 lb ⋅ in.)k

i Then

M ¢ = (-1800i -1120k) + 45 14 0 36 -28 -6 = (-1800 i - 1120 k ) + {[(14)(-6)]i + [-(45)(-6)]j + [(45)(-28) - (14)(36)]k} = (-1800 - 84)i + (270) j + (-1120 - 1764)k = -(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j - (2884 lb ⋅ in.)k = -(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j - (240 lb ⋅ ft)k F ¢ = (36.0 lb)i - (28.0 lb) j - (6.00 lb)k ◀

The equivalent force-couple system at H is

M ¢ = -(157.0 lb ⋅ ft)i + (22.5 lb ⋅ ft) j - (240 lb ⋅ ft)k ◀

PROBLEM 3.98 110 N

15° B

35°

A 110-N force acting in a vertical plane parallel to the yz-plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent force-couple system at the origin O of the coordinate system.

150 mm

O z x

SOLUTION We have

SF : PB = F

where

PB = 110 N[- (sin15) j + (cos15)k] = -(28.470 N) j + (106.252 N)k

or F = -(28.5 N) j + (106.3 N)k ◀ We have where

SMO : rB/O ´ PB = MO rB/O = [(0.22 cos35)i + (0.15) j - (0.22sin 35)k ] m = (0.180213 m)i + (0.15 m) j - (0.126187 m)k i

0.180213

0.15

-28.5

k 0.126187 N ⋅ m = MO 106.3

MO = [(12.3487)i - (19.1566) j - (5.1361)k] N ⋅ m or MO = (12.35 N ⋅ m)i - (19.16 N ⋅ m)j - (5.13 N ⋅ m)k ◀

PROBLEM 3.99 The 12-ft boom AB has a fixed end A, and the tension in cable BC is 570 lb. Replace the force that the cable exerts at B with an equivalent force-couple system at A.

SOLUTION We have Then

Now

 BC = -(12 ft)i + (4.8 ft) j - (8 ft)k; BC = 15.20 ft  BC F = (570 lb) BC (-12 i + 4.8 j - 8k ) = (570 lb) 15.20 = -(450 lb)i + (180 lb) j - (300 lb)k

M A = rB / A ´ F = (12 ft)i ´ éë-(450 lb)i + (180 lb) j - (300 lb)k ùû = (2160 lb ⋅ ft)k + (3600 lb ⋅ ft) j

The equivalent force-couple system at A is

F = -(450 lb)i + (180 lb) j - (300 lb)k ◀ M = (3600 lb ⋅ ft) j + (2160 lb ⋅ ft)k ◀

PROBLEM 3.100 The jib crane shown is oriented so that its boom AD is parallel to the x axis and is used to move a heavy crate. Knowing that the tension in cable AB is 2.6 kips, replace the force exerted by the cable at A with an equivalent force-couple system at the center O of the base of the crane.

SOLUTION We have Then

Now

 AB = -(15 ft) j + (6.25 ft)k; AB = 16.25 ft  AB F = (2.6 kips) AB (-15 j + 6.25k ) = (2.6 kips) 16.25 = -(2.4kips) j + (1.0 kips)k MO = rA /O ´ F i

= 10 15 0 kip ⋅ ft 0 -2.4 1.0

MO = [(15.00)i - (10.00) j - (24.0)k] kip ⋅ ft The equivalent force-couple system at O is

F = -(2.40kips) j + (1.000 kips)k ◀ M = (15.00 kip ⋅ ft)i - (10.00 kip ⋅ ft) j - (24.0 kip ⋅ ft)k ◀

PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION (a)

(a)

We have

SFy : -400 N - 200 N = Ra

or and

R a = 600 N ◀

SM A : 1800 N ⋅ m - (200 N)(4 m) = M a

or M a = 1000 N ⋅ m (b)

We have

SFy : - 600 N = Rb

or and

We have

M b = 900 N ⋅ m

◀

SFy : 300 N - 900 N = Rc

or and

R b = 600 N ◀

SM A : - 900 N ⋅ m = M b

or (c)

◀

R c = 600 N ◀

SM A : 4500 N ⋅ m - (900 N)(4 m) = M c

M c = 900 N ⋅ m

◀