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Study Unit
Engineering Mechanics, Part 1 By
Andrew Pytel, Ph.D. Associate Professor, Engineering Mechanics The Pennsylvania State University
About the Author
Dr. Andrew Pytel is well qualified to write about engineering mechanics. He obtained his M.S. and Ph.D. degrees in engineering mechanics from The Pennsylvania State University. Since 1957 he has taught engineering mechanics at Rochester Institute of Technology, Northeastern University, and The Pennsylvania State University. He has authored and coauthored several articles on engineering mechanics.
Dr. Pytel is a member of the American Academy of Mechanics, American Society of Engineering Education, American Society of Mechanical Engineers, and Society of Sigma Chi.
Copyright © 1995 by Penn Foster, Inc.
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w When you complete this study unit, you’ll be able to
• Explain some of the methods, laws, and procedures used in engineering mechanics
• Describe the characteristics and classifications of forces and their effects on bodies
• Apply the rules and formulas that pertain to collinear forces
• Apply the analytic methods used to determine the resultant of concurrent forces
• Calculate the resultant of a system of nonconcurrent forces as they relate to couples and components
• Describe the principles pertaining to the center of gravity of a body and the methods used to calculate the distance to the centroid
INTRODUCTORY EXPLANATIONS 1
Scope of Engineering Mechanics 1 Applications of Engineering Mechanics 2 Branches of Engineering Mechanics 2 Mass and Weight of Body 3 Historical Introduction to Newtonian Mechanics 4 Statement of Newton’s Laws 5 Scalar Quantities and Vector Quantities 6 Graphic and Analytic Methods of Solving Problems 6 General Procedure in Solving a Problem 7 Suggestions for Studying Engineering Mechanics 7
FORCES 11
Representation of Forces 11 Combining Collinear Forces 20 Combining Concurrent Forces 26 Combining Nonconcurrent Forces 47
CENTER OF GRAVITY 65
Simple Body 65 Composite Body 78
SELF-CHECK ANSWERS 85
PRACTICE PROBLEMS ANSWERS 87
EXAMINATION 89
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1
INTRODUCTORY EXPLANATIONS
Scope of Engineering Mechanics
1 Engineering mechanics is that branch of engineering that deals with problems involving the effects of forces on bodies. There are many different kinds of actions that are called forces. For example, a person exerts a force on a book when he or she moves the book from one place to another. Also, the earth exerts a force on every object, and this force tends to keep the object in contact with the earth. Numerous other types of forces could be mentioned. It is therefore impos- sible to give a simple definition of the term force in the usual way. The best general definition of a force is as follows: A “force” is any action that causes or tends to cause a change in the motion of a body. Various kinds of forces must be con- sidered in engineering. Many general types will be discussed in these texts on engineering mechanics.
A body may be composed entirely of one material, or it may consist of parts made of two or more different materials. In these texts, we shall assume that every body is composed of material in the solid state (rather than in the liquid state or the gaseous state) and that each body has a specific shape and size. In a certain problem, it may be desirable to con- sider the entire earth as a single body. In another problem, an automobile may be treated as a single body. In still another problem, the body under consideration may be a small block having the shape of a rectangular prism.
Engineering Mechanics, Part 1
Engineering Mechanics, Part 12
Applications of Engineering Mechanics
2 The principles of engineering mechanics are applied inall kinds of engineering. A few examples will be men- tioned here. A civil engineer is interested in the forces exerted on the parts of a building frame by objects on the floors. A mechanical engineer must consider the forces acting upon and produced by rotating machinery. An electrical engineer may have to consider the forces acting on cables for trans- mission lines. An aeronautics engineer must deal with the forces acting on aircraft. A nuclear engineer must protect a reactor against the forces that would be produced by an earthquake.
Branches of Engineering Mechanics
3 Every actual body is always under the influence of atleast two forces. The effects of the forces acting on a par- ticular body at any instant depend on the conditions at that instant. Let us consider the simple case of a book resting on the top of a desk. If you push vertically downward on the book, the book will not move noticeably, because movement will be prevented by the support provided by the desk. If you push the book sideways and apply enough force, the book will slide along the top of the desk in the direction in which you push it. In either case, the shape and size of the book will be changed, even though the change may be so small that you will not be able to see any difference in the appear- ance of the book. Any change in the dimensions of an object is called a deformation.
Three types of problems involving systems of forces acting on bodies are common in engineering. In one type of problem, it is desired to keep a body at rest. In such a problem, the body is subjected to the actions of certain forces that tend to cause the body to move, but movement is prevented by the applica- tion of other forces. In another type of problem, the body moves under the actions of a system of forces, and it is nec- essary to analyze the movement. In either of these cases, the deformation of the body is ignored. In a problem of the third type, the deformation of a body under the actions of the forces is considered.
Engineering Mechanics, Part 1 3
Problems of the first type are studied in a branch of engineering mechanics called statics. The branch that includes problems of the second type is called dynamics. The branch that pertains to problems of the third type is called mechanics of materials or strength of materials. In these texts on engineering mechanics, only the basic principles of statics and dynamics will be covered. The deformation of a body will be ignored, and it will be assumed that every body is rigid. In other words, it will be assumed that the shape and size of the body will not be changed by the forces acting on it.
Mass and Weight of Body
4 Every material of which bodies are composed may beincluded in the general term matter. What is called the mass of a body indicates the amount of matter contained in the body. However, it is not practicable to attempt to deter- mine the mass of a body by direct measurement. The usual practice is to measure what is called the weight of the body and to compute the mass from the weight.
It is a well-established fact that every body exerts a force of attraction on every other body. At this time, we shall refer only to the force of attraction between the earth and any other body. Such a force is called the force of gravity, and the mag- nitude of this force for a particular body is known as the weight of the body. The weight of a body that is of reasonable size can be measured directly by means of a suitable type of scale.
In the English system of measures, the unit most commonly used for expressing weights is the pound. Small weights may be expressed in ounces, and large weights may be expressed in tons. In the metric system, the basic unit of weight is the gram, but the unit most commonly used is the kilogram, which is equal to 1000 grams. Since a weight is a force, the units used for measuring weights are also used for forces of all kinds.
Engineering Mechanics, Part 14
The relation between the weight of a body and its mass need not be considered at this time. In Engineering Mechanics, Parts 1, 2, and 3, we shall discuss only the effects of forces on bodies, and the mass of a body is not a factor in the types of problems covered in these texts.
Historical Introduction to Newtonian Mechanics
5 In the year 1687, Sir Isaac Newton published a bookentitled Mathematical Principles of Natural Philosophy, in which he presented his now famous three laws of motion. These laws form the basis of the principles of engineering mechanics that are covered in these texts. However, before we discuss these laws themselves, we shall mention some of the outstanding scientific developments that led up to Newton’s conclusions.
Even in very ancient times, human beings were interested in the motions of bodies. In particular, the movements of the heavenly bodies across the sky have had an extraordinary influence on the philosophy, religions, and daily life of people. We can readily understand why some of the first scientific studies included attempts to explain and describe the motions of the planets with respect to the sun and the stars. Early philosophers, among them Aristotle and Ptolemy, based their theories on the concept that the earth was a stationary body at the center of the universe, and that the sun and the other planets revolved around the earth along complex orbits. This concept was accepted for many centuries. However, in 1543, the Polish astronomer Copernicus published his theory that the planets moved in circular orbits around the sun and that each planet also rotated on its own axis. This theory stimu- lated astronomers in the sixteenth century to keep more careful records pertaining to the motions of the planets. The most significant information was obtained by Brahe, and his records were studied extensively by one of his assistants, named Kepler.
In addition to stating his three laws of motion for mechanical systems, Newton was able to apply these laws to the motions of the planets and to derive results that agreed with the data
Engineering Mechanics, Part 1 5
obtained by Brahe and Kepler. For more than 200 years, scientists believed that Newton’s principles could be used to solve all problems pertaining to the motions of mechanical systems. However, exceptions to the universal validity of Newton’s laws were discovered by Einstein in 1905 and by Schroedinger and others in 1925. Now scientists realize that Einstein’s theory of relativity must replace Newton’s laws in the analysis of a system that moves with a speed approaching that of light (186,000 miles per second). Also, Schroedinger’s wave equation, or quantum mechanics, must be used in the analysis of the motion of a body that moves through only an extremely small distance. Although the theories of Einstein and Schroedinger replace Newton’s laws in certain special cases, Newtonian mechanics has a very high degree of accu- racy for the speeds and distances involved in ordinary engineering problems. Since Newton’s theory can be applied to such problems with relatively little difficulty, Newton’s laws are generally used in the analysis of these problems.
Statement of Newton’s Laws
6 Newton’s three laws of motion are usually stated in thefollowing way: First Law. A particle continues in a state of rest, or of uniform motion along a straight line, unless acted upon by a force.
Second Law. If a resultant force acts upon a particle, the particle will be accelerated in the direction of the force. Furthermore, the magnitude of the acceleration will be directly proportional to the magnitude of the resultant force and inversely proportional to the mass of the particle.
Third Law. For every action, there is an equal and opposite reaction.
The meaning of each of these laws will be clarified when applications of that law are discussed.
Engineering Mechanics, Part 16
Scalar Quantities and Vector Quantities
7 Some quantities can be described completely by a num-ber and a unit of measurement. For example, it may be said that a person’s age is 40 years or that the cost of an automobile is 4,200 dollars. No additional information is needed to clarify such a measurement. In these texts on engi- neering mechanics, however, it is often necessary to consider a quantity that involves not only a number of units but also a direction. Three of the quantities that are included in this class are forces, distances through which bodies are moved, and rates of movement of bodies.
A quantity that can be described completely by its magnitude, or by a number of units, is called a scalar quantity, or scalar. A quantity that involves both magnitude and direction is called a vector quantity, or vector. The characteristics of scalars and the methods of performing mathematical calculations with scalars should require no additional explanations. However, the characteristics of vectors and the methods of combining vectors will be explained in detail.
In Engineering Mechanics, Parts 1 and 2, the only vectors that will be considered are forces. Hence, the general principles relating to vectors will be explained in the discussion of forces.
Graphic and Analytic Methods of Solving Problems
8 In some cases, it is convenient to solve a problem completely by constructing a suitable diagram. Such a solution is called a graphic solution. In other cases, it is advisable to solve a problem by performing the proper calcu- lations. Such a solution is called an analytic solution.
An advantage of a graphic solution is that the use of a diagram makes it easier for the person to visualize the condi- tions and to understand the solution. However, the accuracy obtainable by using a graphic solution is limited by the degree of precision possible with the instruments used for
Engineering Mechanics, Part 1 7
measuring distances and angles. When a problem is solved by an analytic method, it is possible to determine results to any desirable degree of precision. On the other hand, the mathematical procedure that is applied does not always indi- cate clearly the meaning of the calculations. For example, it may be easy for a person to sum numbers in a formula and compute a result without understanding the significance of the result.
In these texts on engineering mechanics, computed values will usually be expressed to three significant figures when the first significant figure is 2 or more, and to four figures when the first figure is 1. The number of figures used in a result based on a graphic solution will depend on the conditions in the particular problem.
General Procedure in Solving a Problem
9 Since it is usually easier for a person to visualize theconditions in a problem when a graphic solution is used, the procedure in this text for presenting a general principle will be to explain the graphic method before the analytic method. Also, when a problem is to be solved by an analytic method, the first step will be to prepare a suitable diagram to indicate the conditions. This diagram need not picture the conditions accurately, but it should be complete enough to show all the known information and also the relation of the required quantities to the known quantities.
Suggestions for Studying Engineering Mechanics
10 In the texts on Engineering Mechanics, the primaryemphasis will be on the techniques for solving prob- lems. Nevertheless, we shall include a discussion of the principle applied in the solution of each problem and enough of the theory underlying the principle to enable you to under- stand the significance of the principle and its limitations.
Engineering Mechanics, Part 18
To solve some of the problems, you will have to apply princi- ples of arithmetic, algebra, geometry, and trigonometry. Therefore, if you have forgotten some of the mathematical procedures that will be used, you should review the portions of previous texts in which these procedures are explained. Also, many of the principles described in the texts on engi- neering mechanics are applied in other texts in your course. Hence, you should study each principle in these texts very carefully and be sure that you understand it completely before you go on to the next one. Moreover, if you come to some procedure that requires a knowledge of a principle that was explained previously and you find that you cannot recall that principle, you are expected to review the appropriate text material. Where we have reason to believe that you may encounter difficulty because of insufficient knowledge of a preceding explanation, we shall try to help you by including a reference to the place in which the necessary basic informa- tion is given. Remember that every explanation in these texts is important.
Engineering Mechanics, Part 1 9
Self-Check 1 At the end of each section of Engineering Mechanics, Part 1, you’ll be asked to pause and check your understanding of what you’ve just read by completing a “Self-Check” exercise. Answering these questions will help you review what you’ve studied so far. Please complete Self-Check 1 now.
1. What general types of problems are dealt with in engineering mechanics?
__________________________________________________________
2. How would you define the term force in a general way?
__________________________________________________________
3. What are the names commonly given to the three main branches of engineering mechanics?
__________________________________________________________
4. When a rigid body is acted upon by a number of forces, is it necessary to consider the deformation of the body?
__________________________________________________________
5. What is meant by the “weight” of a body?
__________________________________________________________
6. Name the unit commonly used to measure force in the a) English system and b) metric system.
__________________________________________________________
7. How many separate laws are included in what are called Newton's laws of motion?
__________________________________________________________
(Continued)
Engineering Mechanics, Part 110
Self-Check 1 8. What are the two essential characteristics of a vector quantity?
__________________________________________________________
9. What is the main advantage of a graphic solution of a problem over an analytic solution?
__________________________________________________________
10. In a problem in engineering mechanics, when a value is computed, how many significant figures are generally used to express the result?
__________________________________________________________
Check your answers with those on page 85.
Engineering Mechanics, Part 1 11
FORCES
Representation of Forces
Characteristics of a Force
11 When any force is applied to a body, the force actsalong a certain straight line, which is called the line of action of the force. The line of action of a force may be hor- izontal or vertical, or it may be inclined at any angle to the horizontal or vertical. Also, a force may act in either of two directions along the line of action. A force really involves two distinct directions. One is the direction of the line of action in space; the other is the direction of the force along its line of action. To avoid confusion in regard to the meaning of the direction of a force, we shall use the expressions “inclination of the line of action” and “direction of the force along the line of action.” Of course, the magnitude of a force must be considered. Therefore, to describe a force completely, it is necessary to specify the inclination of the line of action of the force, the direction of the force along that line of action, and the magnitude of the force. In these texts on engineering mechanics, the magnitude of every force will be expressed in pounds. The inclination of the line of action of a force and the direction of the force along its line of action may be described or indicated in more than one way.
The usual method of indicating the inclination of the line of action of a force is by means of the angle between that line of action and either a horizontal or a vertical reference line. This method will be used in all problems in this text. The direction of a force along its line of action is indicated in most cases by placing an arrowhead on the line of action. Other methods will be described where they are used.
In the analysis of some problems in engineering mechanics, it is necessary to assume that a force is applied to a body at some particular point on the body. In other problems, it may be assumed that a force is applied to a body at any point on the line of action of the force.
Engineering Mechanics, Part 112
Graphic Representation of a Force
12 Although it is possible to describe a force by usingwords alone, the easiest and clearest method of describing a force is by means of a diagram, as indicated in Figure 1. For each of the forces represented, the straight line 1 indicates the position of the line of action of the force with respect to the body 2 on which the force acts, and the arrow- head 3 indicates the direction of the force along its line of action. The arrowhead may be placed at one end of the line of action or at some convenient point on the line of action. The magnitude of each force is the number of pounds shown, and the length of the line of action is chosen arbitrarily. In some diagrams constructed for solving problems, the relative mag- nitudes of two or more forces are indicated by the lengths of specific segments of lines that are drawn parallel to the lines of action of the forces. The use of such segments will be dis- cussed a little later in this text. For simplicity, each body in this illustration is represented by a small rectangle, but the body in a practical problem may have any shape and size.
FIGURE 1—Graphic Representation of Forces
1. Line of action of force
2. Body acted upon by force
3. Arrowhead showing direction of force along its line of action
Engineering Mechanics, Part 1 13
In Figure 1A, the line of action of the force is horizontal and the force acts toward the right along the line of action. Because of the relative positions of the force and the body, the force acts toward the body and, therefore, the force tends to push the body horizontally toward the right. In Figure 1B, the line of action of the force is vertical, and the force acts downward along its line of action. This force acts away from the body and tends to pull the body vertically downward. The line of action of the force in Figure 1C is inclined to the hori- zontal at an angle equal to 30° (degrees) and the force tends to push the body. In Figure 1D, the line of action of the force is inclined at the indicated angle with the horizontal, and the force tends to pull the body. For the conditions in Figures 1C or 1D, the actual effect of the force on the body will depend on the effects of other forces that will also act on the body.
For the problems in these texts on engineering mechanics, it is not necessary to consider the manner in which a force is applied to a body. In other words, we shall only consider the direction in which a force acts along its line of action, and it will not be necessary to specify whether the force tends to push a body or tends to pull the body.
Description of a Force
13 Now that you have seen how a force can be repre-sented graphically, you should be able to understand a description of a force in words. Typical descriptions of the four forces represented in Figure 1 are as follows: Figure 1A represents a force acting horizontally toward the right and having a magnitude equal to 200 lb (pounds); Figure 1B represents a force acting vertically downward and having a magnitude equal to 50 lb; Figure 1C represents a force acting downward and toward the right at an angle with the horizon- tal equal to 30° and having a magnitude equal to 115 lb; Figure 1D represents a force acting upward and toward the right at an angle with the horizontal equal to 45° and having a magnitude equal to 78 lb.
Engineering Mechanics, Part 114
In the solution of a problem or in an explanation of a proce- dure, the common practice is to use a symbolic notation for describing a force. The notations for the forces in Figure 1 would be as shown in Figure 2. In every case, the magnitude of the force is given as if no other information were needed, and the inclination of the line of action and the direction of the force along its line of action are indicated diagrammati- cally. This symbolic method of describing a force will be used frequently in these texts.
Vector Representing Force
14 When a problem involving forces is to be solved com-pletely by a graphic method, each force must be represented by a line that has both the correct inclination and the correct length. Such a line is, of course, a vector. A vector must usually be parallel to the actual line of action of the force represented by the vector, and the length of the vec- tor must represent the magnitude of the force to some suitable scale. For example, if the magnitudes of two forces that act on a single body at the same time are 200 lb and 50 lb, the relative lengths of the vectors representing the forces may be determined as follows: It is assumed that each inch of length of a vector corresponds to some convenient number of pounds of force, such as 100 lb. Then the length of the vector repre- senting the 200-lb force should be 200/100 or 2.00 in. (inches) and the length of the vector representing the 50-lb force should be 50/100 = 0.50 in. The number of pounds correspon- ding to 1 in. of length of a line in a diagram is called the scale of the diagram. In the example just mentioned, the scale would be 1 in. = 100 lb.
The person graphically solving a problem involving forces must also consider which end of the vector representing a force is the termination of the vector, or its tail, and which
FIGURE 2—Representation of Forces by Symbolic Notation
Engineering Mechanics, Part 1 15
end is the beginning of the vector, or its tip, or head. In general, the direction of a vector from its tail to its tip must correspond to the direction of the arrowhead on the line of action of the force represented by the vector. In Figure 3 are shown vectors representing each of the forces in Figure 1 to a scale of 1 in. = 100 lb. In each case, the end of the vector marked A is its tail, and the end marked B is the tip of the vector. The arrowhead is here placed on each vector at its tip, but the arrowhead may be placed anywhere on the vector. It is often convenient to describe a vector representing a force by means of two letters that are placed at the ends of the vector. For instance, each vector in Figure 3 may be called the vector AB. When such a method of describing a vector is used, the first letter should always be the letter at the tail. The order of the letters then indicates the direction of the force along its line of action.
Classification of Systems of Forces
15 In a practical problem involving forces, it is oftennecessary to consider the combined effects of two or more forces. A group of forces that act on a single body at the same time is called a system of forces. Systems of forces can be divided into four general classes:
1. Collinear forces
2. Concurrent forces
3. Parallel forces
4. Nonconcurrent, non-parallel forces
FIGURE 3—Vectors Representing Forces
A. Tail of vector
B. Tip of vector
SCALE: 1 in. = 100 lb
Engineering Mechanics, Part 116
Bodies or groups of bodies that are acted upon by each of the four types of systems of forces will be mentioned in prob- lems that are considered in these texts. The bodies to which references are made are only a few of those that could be mentioned.
In a system of collinear forces, or a collinear system, the lines of action of all the forces coincide. That is, all the forces have a common line of action. For example, in Figure 4A the forces F1 and F2 are collinear. The two forces F1 and F2 in Figure 4B are also collinear. In each case, it is assumed that the two forces of the system act on a small body (rectangular prism). As indicated by the arrowheads, the two forces in Figure 4A act in the same direction, and the two forces in Figure 4B act in opposite directions. A system of collinear forces may con- sist of more than two forces. All the forces in such a system may act in the same direction, or some forces may act in one direction while the other forces act in the opposite direction.
In a system of concurrent forces, or a concurrent system, the lines of action of all the forces pass through a common point. A collinear system is a special type of concurrent system. Typical systems of concurrent forces that are not collinear are represented in Figure 5. Here the body acted upon by the forces is represented by a small circle so that you can easily see that the forces of each system pass through a single point. There may be any number of forces in a concurrent system; the lines of action of the forces may have any inclina- tions; and each force may act in either direction along its line of action.
FIGURE 4—Systems of Two Collinear Forces
A. Forces acting in same direction
B. Forces acting in opposite directions
Engineering Mechanics, Part 1 17
In a system of parallel forces, or parallel system, the lines of action of all the forces are parallel, as indicated in Figure 6. In each of these cases, it is assumed that the body on which the forces act is a long, slender bar in a horizontal position. There may be any number of forces; the lines of action of the forces may be located in any positions; and each force may act in either direction along the line of action.
In a system of nonconcurrent, nonparallel forces, the lines of action of the forces do not pass through a common point and are not parallel. A typical system of forces of this type is rep- resented in Figure 7, in which the body acted upon by the forces is shown with an irregular shape. There may be any number of forces in the system; the lines of action of the forces may have any positions and any inclinations; and each force may act in either direction along its line of action.
FIGURE 5—Typical Systems of Concurrent Forces
FIGURE 6—Typical Systems of Parallel Forces
Engineering Mechanics, Part 118
In most problems in ordinary engineering work, the lines of action of all the forces in a system lie in one plane. In other words, the forces are coplanar. However, in some problems the lines of action of all the forces of a system do not lie in the same plane. We shall assume that all the forces in any system considered in Engineering Mechanics, Part 1, are coplanar.
Resultant of System of Forces
16 To solve a problem in engineering, it is often desir-able to replace two or more forces acting on a body by a single force that would have the same effect on the body as the given forces it replaces. A single force that can replace two or more given forces is called the resultant of the given forces. For example, the motion of either body in Figure 4 theoretically would not be affected if the two forces F1 and F2 were replaced by a single horizontal force having the proper magnitude and the proper direction along its line of action. The method of determining the magnitude and the direction of the resultant will be explained in the next article. Also, the motion of either body in Figure 5 theoretically would not be affected if the three given forces acting on the body were replaced by an equivalent single force whose line of action passes through the common point of intersection of the lines of action of the given forces. Procedures for determining the inclination of the line of action of this resultant force, the magnitude of the resultant, and the direction of the resultant along its line of action will be described a little later in this text. The characteristics of the resultant of a system of paral- lel forces and of a system of nonconcurrent, nonparallel forces will also be discussed in this text.
FIGURE 7—System of Nonconcurrent, Nonparallel Forces
Engineering Mechanics, Part 1 19
Self-Check 2 1. A force is described by the following notation: . What is a) the magnitude of the force,
b) the angle between a horizontal reference line and the line of action of the force, and c) the direction of the force along its line of action?
__________________________________________________________
2. A force acts upward and toward the left at an angle with the horizontal equal to 20°, and its magnitude is 120 lb. How would you show the characteristics of this force by the symbolic method?
__________________________________________________________
3. Draw a vector representing the force in question 2 so that 1 in. corresponds to 40 lb.
__________________________________________________________
4. Place the words tail and tip at the proper ends of the vector in question 3.
__________________________________________________________
5. Describe the relative positions of the lines of action of a group of forces that form a collinear system.
__________________________________________________________
6. If the forces forming a system are concurrent, what requirement must be satisfied by the lines of action of the forces?
__________________________________________________________
7. The lines of action of the forces of a system are not parallel and do not pass through a single point. How would you describe such a system of forces in the simplest terms?
__________________________________________________________
8. A certain body is acted upon by two concurrent forces. If these two forces are replaced by a single force that would have the same effect on the body as the given forces, what is this single force commonly called?
__________________________________________________________
Check your answers with those on page 85.
200
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Engineering Mechanics, Part 120
Combining Collinear Forces
Resultant of Collinear Forces
17 To determine the magnitude of the resultant of twocollinear forces that act in the same direction, it is simply necessary to add the magnitudes of the two given forces. Obviously, the direction of the resultant along its line of action is the same as the direction of each given force. For instance, let us consider a freight train that is pulled along a railroad track by one locomotive at the front of the train with a force whose magnitude is 20,000 lb, and is also pushed by a second locomotive at the rear of the train with a force whose magnitude is 15,000 lb. The conditions are really similar to those represented in Figure 4A, where the body represents the train, the force F1 represents the pull exerted by the front locomotive, and the force F2 represents the push exerted by the rear locomotive. In this case, the pull and the push tend to move the train along the track in the same direction. Hence, the magnitude of the resultant of these forces—the total force tending to move the train—is 20,000 + 15,000 = 35,000 lb. Also, the direction of the resultant, or the direction of move- ment, would be the same as the direction of each actual force applied to the train.
If two collinear forces act in opposite directions, the magni- tude of the resultant of the given forces may be found by subtracting the magnitude of the smaller given force from the magnitude of the larger given force; and the direction of the resultant along its line of action is the same as the direction of the larger force. For example, let us suppose that two men are having a tug-of-war. The conditions are similar to those represented in Figure 4B, where the body represents the rope, the force F1 represents the pull exerted by the man at the right, and the force F2 represents the pull exerted by the man at the left. If the magnitude of the first force F1 is 150 lb and the magnitude of the second force F2 is 120 lb, then the magnitude of the resultant of these forces, or the force tend- ing to move the rope, would be 150 – 120 = 30 lb; and the direction of the resultant, or the direction of movement, would be the same as that of the larger force F1, toward the right.
Engineering Mechanics, Part 1 21
When a system of collinear forces consists of more than two forces and all the given forces act in the same direction, the magnitude of the resultant of the system is equal to the sum of the magnitudes of all the given forces, and the direction of the resultant along its line of action is the same as the direc- tion of each given force.
If a system of collinear forces consists of more than two forces and some of the given forces act in one direction while the other given forces act in the opposite direction, the proce- dure for determining the magnitude and the direction of the resultant of the system may be outlined as follows: First, the magnitude and direction of the partial resultant of all the forces acting in one direction are determined by applying the method described in the preceding paragraph, and the mag- nitude and direction of the partial resultant of all the forces acting in the other direction are also determined in a similar manner. Then, the magnitude and direction of the resultant of the entire system are determined by combining the two partial resultants, which are treated as two collinear forces acting in opposite directions.
Example Problem
At intervals throughout this text you will find one or more example problems solved to illustrate clearly the application of a principle, rule, or formula. Read each problem carefully, and study the solution until you understand it thoroughly.
Problem: In Figure 8 is represented a system consisting of six forces, all of which have a common horizontal line of action. The forces are shown separately for easier identifica- tion of the magnitude and direction of each. Determine the magnitude and the direction of the resultant of the entire system.
FIGURE 8—Forces in Example Problem, Article 17
Engineering Mechanics, Part 122
Solution: The magnitude of the partial resultant of the three forces acting toward the right is
60 + 110 + 80 = 250 lb
and the magnitude of the partial resultant of the three forces acting toward the left is
90 + 100 + 40 = 230 lb
Therefore, the magnitude of the resultant of the entire system is 250 – 230 = 20 lb, and the direction of this resultant along the common line of action of the given forces is toward the right.
Symbolic Notation for Combination of Collinear Forces
18 Since forces are vectors, a convenient symbolic wayto indicate the procedure for determining the result- ant of a system of collinear forces is to represent each force in the manner indicated in Figure 2 and to place the symbol
between each two forces. For instance, the resultant of the 20,000-lb force and the 15,000-lb force referred to in the first paragraph of Article 17 may be indicated as follows:
20,000 lb 15,000 lb = 35,000 lb
Likewise, the resultant of the 120-lb force and the 150-lb force referred to in the second paragraph of Article 17 would
be
120 lb 150 lb = 30 lb
Similarly, the steps in the solution of the preceding example problem could be indicated as follows:
60 lb 110 lb 80 lb = 250 lb
90 lb 100 lb 40 lb = 230 lb
250 lb 230 lb = 20 lb
Engineering Mechanics, Part 1 23
The symbol is intended to be a combination of the symbol + and an arrow. It is used to distinguish between the addition of scalar quantities and the “addition” of vector quantities. You should note that the symbol is used between two forces regardless of whether the forces act in the same direction or in opposite directions, and regardless of whether a force acts toward the right or toward the left. However, if two forces act in the same direction, their magnitudes are added; whereas, if two forces act in opposite directions, the magnitude of one force is subtracted from the magnitude of the other force.
Simplified Formula for Resultant of Collinear Forces
19 Since it is so easy to determine analytically the magnitude and the direction of the resultant of any system of collinear forces, the graphic method is rarely, if ever, used for solving a problem of collinear forces. The sev- eral rules given in Article 17 for determining the magnitude of the resultant may be replaced by a single simple formula, which may be stated as follows:
R = �F
In this formula, R represents the resultant of a system, and the combination of the Greek letter � (sigma) and F is used to denote the vector sum of the magnitudes of all the given forces in a system.
To determine the vector sum �F for any system of collinear forces, it is simply necessary to consider as positive quantities all forces acting in one direction along the common line of action, and to consider as negative quantities all forces acting in the opposite direction. The usual practice is to assume that horizontal forces acting toward the right are positive and that horizontal forces acting toward the left are negative. Also, vertical forces acting upward are usually considered as positive and vertical forces acting downward are considered as negative. The magnitude of the resultant is the same as the magnitude of the vector sum. If the vector sum for a sys- tem of horizontal forces is a positive quantity, the resultant of the system acts toward the right; and if the vector sum is negative, the resultant acts toward the left. Similarly, if the
Engineering Mechanics, Part 124
vector sum for a system of vertical forces is positive, the resultant acts upward; and if the vector sum is negative, the resultant acts downward.
One method of applying the formula R = �F to a system of forces that act along a common inclined line is to imagine that the inclined line of action is horizontal and to treat the system as if it were a system of horizontal forces.
When the preceding formula is applied to solve the example problem in Article 17, the vector sum may be found as follows:
The sum of the positive forces is
+(60 + 110 + 80) = +250 lb
The sum of the negative forces is
–(90 + 100 + 40) = –230 lb
The vector sum of the forces for the entire system is
+250 + (–230) = +20 lb
Thus, the magnitude of the resultant is 20 lb. Also, since the vector sum is positive, the resultant acts toward the right.
Engineering Mechanics, Part 1 25
Practice Problems 1 Practice problems are included in this text to test your ability to apply a rule or a formula. Work each problem carefully and check your answer against the answer given after the statement of the problem.
In each of the following problems, assume that the forces are collinear. Determine the magnitude of the resultant and the direction of the resultant along the common line of action of the given forces.
1. F1 = 25 lb toward the right and F2 = 75 lb toward the right
__________________________________________________________
2. F1 = 150 lb upward and F2 = 60 lb upward
__________________________________________________________
3. F1 = 80 lb toward the left and F2 = 60 lb toward the right
__________________________________________________________
4. F1 = 1500 lb upward and F2 = 800 lb downward
__________________________________________________________
5. F1 = 1200 lb toward the left, F2 = 1000 lb toward the right, F3 = 2000 lb toward the right, and F4 = 2400 lb toward the left
__________________________________________________________
Check your answers with those on page 87.
Engineering Mechanics, Part 126
Combining Concurrent Forces
Parallelogram of Forces
20 So that you may clearly understand the relationshipbetween two given concurrent forces and their resultant, we shall present first a graphic method for deter- mining the characteristics of the resultant, and later an analytic method. Figure 9A shows a body that is acted upon by two forces F1 and F2. For example, one of these forces may represent the effect of the propeller in moving a small airplane, while the other force represents the effect of the wind on the movement of the airplane. Numerous other bodies are acted upon by two concurrent forces at the same time. The diagram in Figure 9A shows the relative positions of the lines of action of the forces, the direction of each force along its line of action, and the magnitude of each force. Since the lines of action of these two forces must intersect at some point, the forces are concurrent. Figure 9B shows a graphic method for determining directly the magnitude of the resultant, the incli- nation of its line of action, and its direction along its line of action.
The first step in constructing the diagram in Figure 9B is to select a point O in some convenient position and to draw through this point the vectors OA and OB, which represent the given forces. It is assumed that the point O is the tail of each of these vectors. To draw the vector OA, an inclined line of unlimited length is drawn through the point O so that the angle between a horizontal reference line OH through O and this inclined line is 18°. Then the distance OA along this line is laid off to represent the magnitude of the force F1 to some convenient scale. In this case, the selected scale is 1 in. = 50 lb, and the distance OA represents 100 lb to that scale. The vector OB is drawn in a similar manner. An inclined line of unlimited length is drawn through the point O at the proper angle with the horizontal reference line OH, and the distance OB along this inclined line is laid off to represent the magnitude of the force F2 to the selected scale. Thus, the point A is the tip of the vector OA and the point B is the tip of the vector OB. The distances OA and OB must be measured
Engineering Mechanics, Part 1 27
in the proper directions from the point O. These directions are indicated by the arrows on the lines of action of the forces F1 and F2 in Figure 9A.
After the lines OA and OB have been drawn in Figure 9B, the next step is to complete the parallelogram OACB in the fol- lowing manner: From the point A, a line of unlimited length is drawn parallel to the vector OB, and from the point B a line of unlimited length is drawn parallel to the vector OA.
FIGURE 9—Resultant of Two Concurrent Forces by Parallelogram
A. Representation of forces
B. Parallelogram of forces
Engineering Mechanics, Part 128
These two new lines, which are the broken lines in the illustration, intersect at the point C. Finally, the vector repre- senting the resultant R of the given forces F1 and F2 is drawn from the initial point O to the point C. The magnitude of the resultant is indicated by the length of the line OC expressed in pounds by the relation 1 in. = 50 lb. The inclination of the line of action of the resultant with respect to the horizontal can be determined by measuring the angle between the lines OH and OC. The direction of the resultant along its line of action is determined by assuming that the tail of the vector OC is at O and the tip of this vector is at C. For the given characteristics of the forces F1 and F2 in Figure 9A, the mag- nitude of the resultant would be about 225 lb, and the angle between the horizontal and the line of action of the resultant would be about 48°.
The actual line of action of the resultant of two concurrent forces must pass through the point at which the lines of action of the two given forces intersect.
A parallelogram like OACB in Figure 9B is called a parallelogram of forces. Such a parallelogram can be con- structed for determining the characteristics of the resultant of any two concurrent forces. The procedure may be outlined as follows:
Step 1. Select any convenient initial point, and through this point draw vectors that represent the two given forces. Be careful to draw these vectors in the proper directions and to lay off their lengths correctly to a selected scale.
Step 2. Complete the parallelogram by drawing a line that passes through the tip of each vector located in Step 1 and is parallel to the other vector.
Step 3. Draw the vector representing the required resultant by connecting the initial point and the point of intersection of the two lines drawn in Step 2.
Engineering Mechanics, Part 1 29
Step 4. Determine the magnitude of the resultant by meas- uring the length of the vector representing the resultant; determine the inclination of the line of action of the resultant by measuring the angle between the vector representing it and some con- venient reference line; and determine the direction of the resultant along its line of action by consider- ing the direction from the tail to the tip of the vector representing the resultant.
In Figure 9B, the angle between the vectors representing the given forces F1 and F2 is less than 90°. In the following exam- ple problem, the angle between the vectors representing the given forces in the parallelogram of forces is greater than 90°.
Example Problem
Problem: The characteristics of two forces acting on a body are indicated in Figure 10A. Determine the characteristics of the resultant of the given forces by constructing a parallelo- gram of forces.
Solution: The parallelogram of forces in which the scale is 1 in. = 100 lb is shown in Figure 10B. The vector OA repre- sents the force F1, and the vector OB represents the force F2. The parallelogram is completed by drawing a line from A par- allel to OB and drawing a line from B parallel to OA. These lines intersect at the point C, and the vector OC represents the required resultant.
The magnitude of the resultant is about 80 lb, the angle between a horizontal line and the line of action of the result- ant is a little greater than 48°, and the direction of the resultant along its line of action is upward and toward the right.
In this case, the magnitude of the resultant is less than the magnitude of the smaller given force because of the effect of the other given force.
Engineering Mechanics, Part 130
Triangle of Forces
21 Actually drawing a parallelogram to determine thecharacteristics of the resultant of two concurrent forces requires some unnecessary work. For example, the parallelogram in Figure 9B consists of the two similar triangles OAC and OBC, and the characteristics of the resultant can be determined just as well by constructing either of these trian- gles instead of the complete parallelogram. This procedure is illustrated in Figure 11. The diagram in Figure 11A, which shows the characteristics of the given forces F1 and F2, is the same as the diagram in Figure 9A. The two alternative trian- gles of forces, in which 1 in. = 50 lb, are shown in Figures 11B and 11C.
To construct the triangle of forces in Figure 11B, the first step is to draw the vector OA representing the given force F1 in the manner described in Figure 9B. The next step is to treat the tip A of this vector in Figure 11B as the tail of the vector rep- resenting the given force F2, and to draw the vector AC so that it has the same inclination as the line of action of the force F2 in Figure 11A and its length represents the magni- tude of the force F2 to the specified scale. The last step is to draw the vector from O to C to represent the resultant of the forces F1 and F2.
FIGURE 10—Parallelogram of Forces for Example Problem in Article 20
A. Representation of forces
B. Parallelogram of forces
Engineering Mechanics, Part 1 31
A. Representation of forces
B. and C. Triangles of forces
FIGURE 11—Resultant of Two Concurrent Forces by Triangle
Engineering Mechanics, Part 132
The procedure for constructing the triangle of forces in Figure 11C is basically similar to that just described for the triangle in Figure 11B. However, the vector OB representing the force F2 is drawn first, and the tail of the vector BC repre- senting the force F1 coincides with the tip of the vector OB.
The general procedure for constructing a triangle of forces for determining the characteristics of the resultant of any two concurrent forces may be outlined as follows:
Step 1. Select any convenient initial point, and draw a vector representing either of the two given forces.
Step 2. Starting at the tip of this first vector, draw a second vector representing the other given force.
Step 3. Draw the vector representing the required resultant of the given forces by connecting the initial point and the tip of the vector drawn in Step 2.
Step 4. Determine the characteristics of the resultant in the manner described in Step 4, Article 20.
You should note that the same result is obtained regardless of which vector representing a given force is drawn first. However, it is important to draw the second vector through the proper end of the first vector and in the proper direction from that end. Thus, in Figure 11B, the vector AC must be started at the end A of the vector OA, and it must be laid off upward and to the right from the point A.
The symbolic notation for indicating the resultant R of any two concurrent forces F1 and F2 is either R = F1 F2 or R = F2 F1. Such a notation is used on the vector repre- senting the resultant in Figures 11B or 11C, and both notations are shown in Figure 9B.
Example Problem
Problem: By constructing a triangle of forces, determine the magnitude of the resultant of the forces F1 = 100 lb and F2 = 150 lb in Figure 10A
20° 10° and the angle between a horizontal line and the line of action of the resultant.
Engineering Mechanics, Part 1 33
Solution: The characteristics of the forces are shown in Figure 12A, which is the same as Figure 10A, and the triangle of forces is shown in Figure 12B. The scale used for this tri- angle is 1 in. = 100 lb. The vector OA represents the force F1. The vector AC represents the force F2, and its tail coincides with the tip of the vector OA. The vector OC representing the resultant is drawn by connecting the initial point O and the tip of the vector AC. The results are the same as those obtained in the example problem in Article 20. Thus, the notation for the resultant is 80 lb.
Analytic Determination of Resultant of Two Concurrent Forces
22 In a triangle of forces for determining the character-istics of the resultant of two concurrent forces, the lengths of the two sides representing the magnitudes of the two given forces will always be known, and the angle between these two sides either will be given or can be easily determined from other given information. For example, in Figure 9B it will be known that the length of the side OA of the triangle OAC is 100 lb, the length of the side AC is 150 lb, and the angle OAC is 180° – 50° = 130°. In other words, the known parts of the triangle will be two sides and the angle included between them, and it will be required to compute the length of the third side and either of the other angles by applying principles of trigonometry.
If F1 and F2 denote the magnitudes of the given forces and if, as indicated in Figures 11B or 11C, Z denotes the angle between the two known sides F1 and F2 of the triangle, Z1
FIGURE 12—Triangle of Forces in Example Problem Article 21
A. Representation of forces
B. Triangle of forces
Engineering Mechanics, Part 134
denotes the angle opposite the side F1, and Z2 denotes the angle opposite the side F2, then the trigonometric formulas for determining these unknown values are as follows:
(1)
(2)
When the angle Z is less than 90°, both cos Z and sin Z are positive numbers. When the angle Z is between 90° and 180°, cos Z is a negative number and sin Z is positive. In this text, a computed angle will be expressed as the nearest multiple of 10� (minutes).
The direction of the resultant along its line of action is not determined by calculation. If the direction cannot be deter- mined by inspection of the directions of the given forces along their lines of action, the usual practice is to sketch a triangle of forces (in which the lengths of the sides and the angles need not be laid off accurately to scale) and to apply the principle described in the last sentence in Step 4, Article 20. Drawing such a triangle of forces will also help you visualize the rela- tive positions of the vectors in the triangle of forces and in choosing the proper value of the angle Z in formula 1.
Example Problems
Problem 1: Compute the magnitude of the resultant of forces F1 = 100 lb and F2 = 150 lb
18° 68° in either Figure 9A or Figure 11A; also compute the angle between the line of action of the force F1 and the line of action of the resultant.
Solution: In this problem, the magnitudes of F1 and F2 are, respectively, 100 lb and 150 lb. Also, as will be seen from the triangle of forces, Figures 11B and C, angle Z must be taken as 180° – 50° = 130°. Since cos 130° = cos 50° = –0.643, the magnitude of the resultant is, by formula 1,
Engineering Mechanics, Part 1 35
Also, it will be seen from a triangle of forces that the required angle is Z2. By formula 2, in which sin 130° = sin 50° = 0.766,
Since the angle between a horizontal line and the line of action of the force F1 is 18°, the angle between the horizontal line and the line of action of the resultant would be 18° + 30° = 48°.
Problem 2: Compute the magnitude of the resultant of the forces in the example problem in Article 21; also compute the angle between the line of action of the force F1 and the line of action of the resultant.
Solution: The characteristics of the forces are shown in Figure 12A and the triangle of forces is shown in Figure 12B. The magnitude of the resultant may be computed by applying formula 1, in which the magnitudes of F1 and F2 are, respec- tively, 100 lb and 150 lb and the angle Z must be taken as 20° + 10° = 30°. Since cos 30° = + 0.866,
The angle between the line of action of the force F1 and the line of action of the resultant, or the angle Z2 in Figure 12B, can be found by formula 2 as follows:
As will be seen from the triangle of forces, the angle Z2 must be greater than 90°. Since the acute angle for which the sine is 0.929 is 68°20�, the angle Z2 must be equal to 180° – 68°20� = 111°40�.
sin sin
. .Z 2
150 130 80 7
0 929= °
=
sin sin
.Z Z2 2 150 130
228 0 500 30 00=
° = = °and �
R = + =100 150 2 100 150 0 866 80 72 2 � � � � . . lb
R = +
= + + =
100 150 2 100 150 0 643
10 000 22 500 19 290 228
2 2 � � � �� .
, , , lb
Engineering Mechanics, Part 136
Resultant of Two Perpendicular Forces
23 Quite often it is desired to determine the character-istics of the resultant of two forces whose lines of action are at right angles to each other. Typical conditions are represented in Figure 13A. For example, the vertical force F1 may represent the effect of the weight of a body, and the horizontal force F2 may represent the effect of a machine pushing against the body. In such a case, the triangle of forces is a right triangle, as shown in Figure 13B. The vector OA in Figure 13B, which represents the force F1 in A, is laid off along a vertical line, and the length of the vector represents 1000 lb when the scale is 1 in. = 400 lb. The vector AC, which represents the force F2, is laid off along a horizontal line, and its length represents 400 lb to the same scale. The vector OC representing the resultant is drawn from the tail of the vector for F1 to the tip of the vector for F2. The magnitude of the resultant, which is indicated by the length of the vector OC, is 1080 lb. The inclination of the line of action of the resultant with respect to the horizontal, or the angle Z1, is slightly greater than 68°. The direction of the resultant along its line of action is downward and toward the left.
When the lines of action of two forces are at right angles to each other, the formulas for determining the magnitude of the resultant and the inclination of its line of action are rela- tively simple. By application of the principles of trigonometry, it can be seen that
(1)
(2)
In most problems in which it is required to determine the characteristics of the resultant of two forces whose lines of action are at right angles to each other, one of these lines of action is horizontal and the other is vertical. However, the lines of action of the given forces may be inclined in any per- pendicular directions.
tan tanZ Z1 1
2 2
2
1
= = F F
F F
and
Engineering Mechanics, Part 1 37
Rectangular Components of a Force
24 It is often convenient to imagine that a single force isreplaced by two concurrent forces. These imaginary forces are called the components of the given force. The lines of action of the two concurrent components of a given force must intersect at some point on the line of action of the given force.
In almost every problem in which two concurrent components are substituted for a single force, the lines of action of the components are at right angles to each other. Such perpendi- cular components are called rectangular components. Although the lines of action of rectangular components may have any two perpendicular inclinations, it is usually most convenient to assume that the line of action of one component is hori- zontal and that of the other component is vertical. Such components are called the horizontal component and the vertical component of the given force. Figure 14A shows an inclined force F, its horizontal component Fx, and its vertical component Fy.
FIGURE 13—Resultant of Forces at Right Angles
A. Representation of forces
B. Triangle of forces
Engineering Mechanics, Part 138
To determine the characteristics of the horizontal and vertical components of a force by the graphic method, the procedure is as follows:
Step 1. Draw a vector representing the given force. This vector must be parallel to the line of action of the given force, and its length must represent the mag- nitude of the force to a selected scale.
In Figures 14B and 14C, the vector OA represents force F in Figure 14A to a chosen scale, which is 1 in. = 200 lb.
Step 2. Complete a right triangle by drawing a horizontal line through one end of the vector in Step 1 and drawing a vertical line through the other end of that vector. These two lines must be drawn in such positions that they will intersect.
FIGURE 14—Horizontal and Vertical Components of a Force
A. Representation of forces
B. and C. Triangles of forces
Engineering Mechanics, Part 1 39
In Figure 14B, the horizontal line is drawn through the end A of the vector OA and the vertical line is drawn through O. In Figure 14C, the horizontal line is drawn through O and the vertical line is drawn through A. Either procedure may be used.
Step 3. Consider the tail of the vector representing the given force as the tail of a vector forming the leg of the right triangle that passes through this tail. Also consider the tip of the vector representing the given force as the tip of the vector forming the third side of the right triangle. Place an arrowhead on each leg of the triangle to conform to this procedure.
In Figure 14B, the tail of the vector OA is also the tail of the vector OB, and the tip of the vector OA is also the tip of the vector BA. In Figure 14C, the tail of the vector OA is also the tail of the vector OC, and the tip of the vector OA is also the tip of the vector CA. From either triangle of forces, it is seen that the horizontal component of the given force acts toward the left and the vertical component acts downward.
Step 4. Determine the magnitude of each component of the given force by measuring the length of the vector representing that component in the triangle of forces to the chosen scale.
In Figure 14, the magnitude of the horizontal com- ponent of the force would be 250 lb, and the magnitude of the vertical component would be about 430 lb.
You should check the direction of each component along its line of action by making certain that the given force is the resultant of the two components of that force.
The magnitudes of the horizontal and vertical components of a given inclined force can be computed very easily by apply- ing the following formulas:
Fx = F cos H (1) Fy = F sin H (2)
Engineering Mechanics, Part 140
in which Fx and Fy denote the magnitudes of the components, F denotes the magnitude of the given force, and H denotes the angle between a horizontal line and the line of action of the given force.
When the horizontal and vertical components of an inclined force are computed, you should be able to determine the direction of each component along its line of action by noting the direction of the given force along its line of action. If you are in doubt, you should sketch a triangle of forces like the one in Figure 14.
Example Problem
Problem: Compute the magnitudes of the horizontal and vertical components of the force F = 500 lb represented in Figure 14A. 60°
Solution: By formulas 1 and 2, in which F = 500 lb and H = 60°,
Fx = 500 cos 60° = 250 lb Fy = 500 sin 60° = 433 lb
Polygon of Forces
25 It is seldom necessary in a practical problem to con-sider the resultant of more than two concurrent forces. However, the characteristics of such a resultant can be determined graphically by extending the principle of the triangle of forces. The procedure is illustrated in Figure 15.
Figure 15A shows the characteristics of four concurrent forces, F1, F2, F3, and F4, which act on a body. In Figure 15B, vectors representing the given forces are drawn in sequence in such a manner that each vector is parallel to the line of action of the corresponding force, the length of each vector indicates the magnitude of the force to a selected scale, and the tail of each vector after the first one coincides with the tip of the preceding vector. Thus, the vector OA represents the force F1, the vector AB represents the force F2, the vector BC represents the force F3, and the vector CD represents the force F4. The vector representing the resultant R of the given forces is drawn from the initial point O to the tip D of the
Engineering Mechanics, Part 1 41
vector for the last given force F4. The characteristics of the resultant can be determined in the manner described in Article 20, Step 4. For the forces in Figure 15, the magnitude of the resultant is about 170 lb, the angle between the hori- zontal reference line and the line of action of the resultant is about 18°, and the direction of the resultant along its line of action is toward the right and downward.
A diagram like that shown in Figure 15B is called a polygon of forces. The lines OA, AB, BC, CD, and DO in this diagram actually form a polygon. However, in some cases, the vector representing the resultant may cross a vector representing one of the given forces, as illustrated in the following example problem. Even in such a case, it is convenient to use the term “polygon of forces.”
Example Problem
Problem: A body is acted upon by four concurrent forces, F1, F2, F3, and F4, whose characteristics are shown in Figure 16C. Determine the characteristics of the resultant graphically.
FIGURE 15—Resultant of Concurrent Forces by Polygon
A. Representation of forces
B. Polygon of forces
Engineering Mechanics, Part 142
Solution: In Figure 16B, vectors representing the forces are drawn to a scale of 1 in. = 100 lb in the following sequence: The vector OA represents force F1; the vector AB represents F2; BC represents F3; and CD represents F4. The vector repre- senting the resultant is then drawn from the initial point O to the tip D of the vector representing the last given force. Even though the vector OD crosses the vector AB, the principle of the polygon of forces is still applicable. The magnitude of the resultant, which is indicated by the length of the vector OD, is about 240 lb. The angle between a horizontal reference line and the line of action of the resultant is about 32°. The direc- tion of the resultant along its line of action is toward the left and downward.
Analytic Method for More Than Two Concurrent Forces
26 The characteristics of the resultant of any number ofconcurrent forces can be determined analytically by performing the following operations.
FIGURE 16—Polygon of Forces in Example Problem, Article 25
A. Representation of forces
B. Polygon of forces
Engineering Mechanics, Part 1 43
Step 1. Compute the magnitudes of the horizontal compo- nent and the vertical component of each of the given forces by using the formulas in Article 24. Also, indicate the direction of each component along its line of action by placing the sign + (plus) or – (minus) in front of its magnitude.
It is usually convenient to use the plus sign for a horizontal component that acts toward the right or for a vertical component that acts upward, and to use the minus sign for a horizontal component that acts toward the left or for a vertical component that acts downward.
Step 2. Since the horizontal components of the given forces form a collinear system of horizontal forces and the vertical components of the given forces form a collinear system of vertical forces, compute the hor- izontal component and the vertical component of the resultant by applying the formula in Article 19 to each of these systems. In other words, apply the following relations:
Rx = �Fx (1) Ry = �Fy (2)
in which Rx and Ry denote the horizontal and verti- cal components of the resultant and �Fx and �Fy denote the vector sums of the components of the given forces.
Step 3. Determine the characteristics of the resultant by treating the components Rx and Ry as two perpen- dicular forces and applying the formulas in Article 23.
As shown in the following example problems, it is often con- venient to tabulate the horizontal and vertical components of the given forces.
Engineering Mechanics, Part 144
Example Problems
Problem 1: Determine analytically the characteristics of the forces F1, F2, F3, and F4 shown in Figure 15A.
Solution: The horizontal and vertical components of the given forces are shown in the accompanying tabulation. Since the line of action of the force F2 is horizontal, the magnitude of its horizontal component is the same as the magnitude of the force itself and its vertical component is zero. The computa- tions for the magnitudes of the components of the other forces and the direction of each component along its line of action are indicated in the tabulation.
The magnitudes and directions of the components of the resultant, which are determined by formulas 1 and 2, are as follows: Rx = �Fx = + 41.0 + 100.0 + 96.4 – 73.7 = + 163.7, or 163.7
"
Ry = �Fy = + 112.8 – 114.9 – 51.6 = – 53.7, or 53.7 $
The magnitude of the resultant, which is computed by for- mula 1, Article 23, is
The angle H between a horizontal reference line and the line of action of the resultant is found from formula 2, Article 23, as follows:
tan . .
.H R R
Hy x
= = = = ° 53 7
163 7 0 328 18 10and �
R = + =163 7 53 7 172 32 2. . . lb
Force Horizontal Component, Fx Vertical Component, Fy
F1 120 cos 70° = 41.0, or + 41.0
" 120 sin 70° = 112.8, or + 112.8
#
F2 100.0, or + 100.0 "
0.0
F3 150 cos 50° = 964, or + 96.4
" 150 sin 50° = 114.9, or – 114.9
$
F4 90 cos 35° = 73.7, or – 73.7
! 90 sin 35° = 51.6, or – 51.6
$
Engineering Mechanics, Part 1 45
Since the horizontal component of the resultant acts toward the right and its vertical component acts downward, the direction of the resultant along its line of action is toward the right and downward.
Problem 2: Determine by the analytic method the character- istics of the resultant of the forces F1, F2, F3, and F4 in Figure 16A.
Solution: The horizontal and vertical components of the given forces are shown in the accompanying tabulation.
The components of the resultant are found as follows: Rx = – 77.3 – 86.6 – 42.4 = – 206.3, or 206.3
!
Rx = – 20.7 – 200.0 + 50.0 + 42.4 = – 128.3, or 128.3 $
The magnitude of the resultant is
Also,
The resultant acts toward the left and downward along its line of action.
tan . .
.H H= = = ° 128 3 206 3
0 622 31 50and �
R = + =206 3 128 3 2422 2. . lb
Force Horizontal Component, Fx Vertical Component, Fy
F1 80 cos 15° = 77.3, or – 77.3
! 80 sin 15° = 20.7, or – 20.7
$
F2 0.0 200.0, or – 200.0 $
F3 100 cos 30° = 86.6, or – 86.6
! 100 sin 30° = 50.0, or + 50.0
#
F4 60 cos 45° = 42.4, or – 42.4
! 60 sin 45° = 42.4, or + 42.4
#
Engineering Mechanics, Part 146
Practice Problems 2 Solve each of the following problems first by using the specified graphic method and then by using an analytic method. The given answers are those obtained by calculation. The answers obtained graphically should be very nearly the same but rounded off somewhat.
1. Determine the characteristics of the resultant of the concurrent forces F1 = 200 lb and !
F2 = 250 lb. In the graphic solution draw a parallelogram of forces for which the scale is 30°
1 in. = 100 lb.
2. Determine the characteristics of the resultant of the two concurrent forces F1 = 150 lb and 45°
F2 = 125 lb. In the graphic solution draw a triangle of forces for which the scale is 25°
1 in. = 100 lb. Describe the inclination of the line of action of the resultant by specifying the angle between that line of action and a vertical reference line.
3. Determine the characteristics of the resultant of the following two concurrent forces: F1 = 75 lb acting horizontally toward the right and F2 = 125 lb acting vertically downward. In the graphic solution draw a triangle of forces for which the scale is 1 in. = 50 lb.
4. Determine the magnitudes and the directions along the lines of action of the horizontal and vertical components of a force described by the notation 300 lb.
65°. In the graphic solution draw a triangle of forces for which the scale is 1 in. = 200 lb.
5. Determine the characteristics of the resultant of the four concurrent forces F1, F2, F3, and F4 which are represented in Figure 17. In the graphic solu- tion draw a polygon of forces for which the scale is 1 in. = 200 lb.
________________________________
Check your answers with those on page 87.
FIGURE 17—Representation of Forces in Practice Problem 5, Article 26
Engineering Mechanics, Part 1 47
Combining Nonconcurrent Forces
Translation and Rotation of Body
27 When a system of concurrent forces acts on the center of gravity (explained in Article 33) of a body at rest, each force simply tends to accelerate the body along a straight line that coincides with the line of action of the force. The direction of motion along the line of action of the force is the same as the direction of the force along the line of action. Such motion is called translation. The effect of the entire sys- tem of forces is to tend to cause translation of the body along the line of action of the resultant of the given forces. For example, the body in Figure 9 would tend to move in the direction of the line of action of the resultant force R.
When a body is acted on by a system of nonconcurrent forces, or by a system of concurrent forces not acting on the center of gravity of the body, there are two distinct effects produced by each force on the body. One effect of each force is to tend to cause translation of the body, just as if the forces were concurrent. The other effect of each force is to tend to cause the body to turn around a certain point, which serves as a pivot. This kind of motion is called rotation. For example, the effects of the forces in Figure 7 on the body would be to cause both translation of the body along the line of action of the resultant of the given forces and rotation of the body about some point within the body.
The effect of the force in producing translation and rotation of a body is illustrated in Figure 18. In Figure 18A a rectan- gular block is placed on a horizontal surface in the position 1, and is pushed by a horizontal force F. Other forces that act on the block are a downward vertical force W representing the weight of the block and a vertical force N acting upward and representing the effect of the horizontal surface on the block. If it may be assumed that the surface is smooth enough to permit the block to slide along the surface without any resistance, and the force, F, acts parallel to the surface and toward the body’s center of gravity, then the lines of action of the forces F, W, and N pass through a common point; that is,
Engineering Mechanics, Part 148
they are concurrent. The only motion of the block will be translation toward position 2, in which the block is indicated by broken lines.
In Figure 18B a block is placed on a horizontal surface and is pushed by a horizontal force F. In addition, the block is acted upon by the forces W and N, representing the weight of the block and the effect of the horizontal surface on the block. These three forces F, W, and N are concurrent. However, there is an irregularity in the horizontal surface that tends to prevent the block from sliding along the surface under the action of the force F. Because of this obstruction, the block is acted upon by a fourth force, which is represented by the force P, and the line of action of this force does not pass through the common point of intersection of the lines of action of the forces F, W, and N. When the block reaches the position 1 and comes in contact with the irregularity in the horizontal surface, the effect of the forces acting on the block is to tend to cause the block to turn or rotate about the point a until the block occupies the position 2 indicated by the bro- ken lines.
Translation of a body is the same under the action of both nonconcurrent and concurrent forces. The body tends to move along a straight line parallel to the line of action of the resultant of the given forces, and the direction of this move-
FIGURE 18—Translation and Rotation of Body
1. Original position of body
2. Final position of body
A. Translation
B. Rotation
Engineering Mechanics, Part 1 49
ment corresponds to the direction of the resultant force along its line of action. Also, the magnitude of the resultant is a measure of the ability of the given system of forces to cause translation of the body.
To measure the ability of a force to cause rotation of a body about a point, it is necessary to consider what is called the moment of the force with respect to the point. When a body is acted upon by a system of nonconcurrent forces, the ability of the system to cause rotation of the body is measured by the resultant moment of the system.
Moment of a Force
28 The magnitude of the moment of a force with respectto a point is determined by multiplying the magni- tude of the force by the perpendicular distance from the point to the line of action of the force. Figure 19A, for example, shows the line of action of a force whose magnitude is F and some point O; and the distance from this point to this force, measured along a line that is at right angles to the line of action of the force, is denoted by d. Then the magnitude of the moment of the force F with respect to the point O is F � d. Likewise, in Figure 19B, the perpendicular distance from the point O� to the line of action of the force F� is d�, and the magnitude of the moment of the force F� with respect to the point O� is F� � d�.
The point from which the distance to a force is measured for the purpose of determining the moment of the force is called the center of moments; and the perpendicular distance from the center of moments to the line of action of the force is called the moment arm. In Figure 19A, the moment arm of the force F with respect to the center O is the distance d. If the line of action of a force passes through the center of moments, the moment arm of the force is zero and the moment of the force is also zero.
Engineering Mechanics, Part 150
Since the magnitude of the moment of a force is the product of the magnitude of the force and a distance, the unit used for expressing a moment is a combination of a unit of force and a unit of distance. The unit of moment most commonly used in engineering is called either the pound-foot or the foot- pound. Sometimes, it is convenient to use the unit pound-inch or inch-pound. In all problems in these texts on engineering mechanics, we shall express moments in foot-pounds.
To describe the moment of a force completely, it is necessary to include both the magnitude of the moment and the direc- tion in which the force tends to rotate the body with respect to the center of moments. The usual practice in describing the direction is by comparison with the movement of the hands of a clock. If the force tends to rotate the body in the direction in which the hands of a clock move, it is said that the moment is clockwise. If the rotation is in the opposite direction, it is said that the moment is counterclockwise. In Figure 19A, the force F tends to cause rotation about the point O in the direction in which the hands of a clock move, and so the moment of the force F with respect to the point O is clockwise. In Figure 19B, the moment of the force F� with respect to the point O� is counterclockwise.
It is a common practice to use the notation MO to indicate a moment that tends to rotate a body about some point O. Also, the direction of a moment may be indicated by placing a
FIGURE 19—Moment of a Force
A. Clockwise moment
B. Counterclockwise moment
Engineering Mechanics, Part 1 51
short arc with an arrowhead above the magnitude of the moment as in the following notations: MO, MO�, Fd, and F�d�. When it is necessary to write an equation involving moments of forces, it is usually convenient to assume that a clockwise moment is positive and that a counterclockwise moment is negative. Thus, the moment in Figure 19A would be shown as +Fd, and the moment in Figure 19B would be –F�d�.
To illustrate the method of computing the moment of a force, we shall assume that the magnitude of the force F in Figure 19A is 250 lb and that the perpendicular distance d from the point O to the line of action of the force is 6 ft (feet). Then the magnitude of the moment with respect to the point O is 250 � 6 = 1500 ft-lb (foot-pounds) and the direction of the moment is clockwise. Hence, the moment can be indi- cated as MO = 1500 ft-lb. Likewise, if the magnitude of the force F� in Figure 19B is 3000 lb and the distance d� is 4 ft, the magnitude of the moment with respect to the point O� is 3000 � 4 = 12,000 ft-lb and MO� = 12,000 ft-lb.
It may sometimes be necessary to compute the moment arm of a force or to measure the moment arm on an accurate dia- gram, as in the following example problem.
Example Problem
Problem: Figure 20A shows the characteristics and locations of three forces F1, F2, and F3. Determine the moment of each force with respect to the point O.
Solution: It is known that the angle between a horizontal ref- erence line through the point O and the line of action of the force F1 is 30°, and that the distance along this horizontal line from the point O to the line of action of the force F1 is 4 ft. The moment arm of the force F1 with respect to the point O, or the perpendicular distance d1 from the point O to the line of action of the force F1, can be determined by consider- ing the right triangle in Oa1b1 in Figure 20B. Thus, d1 = 4 sin 30° = 2.00 ft; the magnitude of the moment of the force F1 is 100 � 2 = 200 ft; and the moment is 200 ft-lb.
� � � �
�
�
�
Engineering Mechanics, Part 152
It is also known that the angle between the horizontal line through O and the line of action of the force F2 is 60° and that the horizontal distance from O to the line of action of F2 is 5 ft. The moment arm d2 of this force can be determined by considering the right triangle Oa2b2 in Figure 20C. Here, d2 = 5 sin 60° = 43.3 ft. So the magnitude of the moment of F2 is 500 � 4.33 = 2170 ft-lb and the moment is 2170 ft-lb.
As indicated in Figure 20A, the angle between a vertical refer- ence line through O and the line of action of F3 is 45°, and the vertical distance from O to the line of action of F3 is 3 ft.
FIGURE 20—Forces in Example Problem in Article 28
A. Location of forces
B. Triangle for computing d1
C. Triangle for computing d2
D. Triangle for computing d3
�
Engineering Mechanics, Part 1 53
The moment arm d3 of this force can be determined by con- sidering the right triangle Oa3b3 in Figure 20D. From this triangle, d3 = 3 sin 45° = 2.12 ft. Hence, the magnitude of the moment of F3 is 300 � 2.12 = 636 ft-lb and the moment is 636 ft-lb.
Resultant Moment of System of Forces
29 When a system of nonconcurrent forces acts on abody, it is always possible to determine the magni- tude and direction of a single moment that will produce the same rotation of the body about a certain point as would the combined moments of the given forces. This single moment is called the resultant moment of the system of forces with respect to the selected point. The procedure for determining the magnitude and direction of the resultant moment of a system of nonconcurrent forces is basically the same as that for determining the resultant of a system of collinear forces. The formula for moments may be written in the form
MR = �MF in which MR denotes the resultant moment and �MF denotes the vector sum of the moments of the given individual forces.
The vector sum of the moments of the given forces is deter- mined by first finding the sum of the magnitudes of all clockwise, or positive, moments and the sum of the magni- tudes of all counterclockwise, or negative, moments. Then the magnitude of the resultant moment is computed by subtract- ing the magnitude of the smaller of these two sums from the magnitude of the larger sum. The direction of the resultant moment is the same as the direction of the larger sum.
Example Problem
Problem: Figure 21 shows a body that is acted upon by four forces, F1, F2, F3, and F4, which are located as shown. Determine the resultant moment of the forces with respect to the point O.
�
Engineering Mechanics, Part 154
Solution: If clockwise moments are considered to be positive, the moments of the individual forces, in foot-pounds, are as follows:
For F1, + 400 � 2.5 = +1000 For F2, 0 For F3, – 800 � 5.0 = – 4000 For F4, + 500 � 1.2 = + 600
The moment of the force F2 is zero because the line of action of that force passes through the center of moments O.
The magnitude of the sum of the positive moments is 1000 + 600 = 1600, and the magnitude of the sum of the negative moments is 4000. Hence, the magnitude of the vector sum, or the magnitude of the resultant moment, is 4000 – 1600 = 2400 and the direction of the resultant moment is negative. The resultant moment can be expressed by the notation 2400 ft-lb.
Characteristics of Resultant of Nonconcurrent Forces
30 The magnitude and the direction along its line ofaction of the resultant of a system of nonconcurrent parallel forces can be determined by treating the forces as if they were collinear. Thus, it is simply necessary to apply the formula R = �F. In the case of a system of nonconcurrent, nonparallel forces, it is possible to determine graphically the magnitude of the resultant, the inclination of the line of
FIGURE 21—Forces in Example Problem in Article 29
�
Engineering Mechanics, Part 1 55
action of the resultant, and the direction of the resultant along its line of action by drawing a polygon of forces as if the forces were concurrent. The same characteristics of this resultant can also be determined analytically by applying the procedure described in Article 26.
For a system of nonconcurrent forces, it is necessary also to determine the position of some point on the line of action of the resultant. It is possible to determine the location of some point on the line of action of the resultant of a system of non- concurrent forces by a graphic method, but it is usually more convenient to compute the distance from some chosen refer- ence point to a point on the line of action of the resultant.
Such a distance can be computed by applying the following principle:
The moment of the resultant of a system of forces with respect to any point is equal to the resultant moment of the system with respect to that point.
Procedures for determining the characteristics of the result- ant of a system of nonconcurrent forces are illustrated in the following example problems.
Example Problems
Problem 1: Figure 22 shows a horizontal bar that is acted upon by the three vertical forces F1, F2, and F3. The magni- tudes of the forces and the horizontal distances between their lines of action are as indicated. Determine a) the magnitude of the resultant of the given forces and the direction of the resultant along its line of action, and b) the horizontal dis- tance from the line of action of the force F1 to the line of action of the resultant.
Solution: a) The magnitude of the resultant and its direction along its line of action can be determined, just as if the forces were collinear, by applying the relation R = �F. If upward forces are considered positive, R = �F = + 200 – 300 + 450 = 350 lb. or 350 lb.
#
Engineering Mechanics, Part 156
b) For the center of moments at the point O on the line of action of the force F1, the moment of that force is zero. If clockwise moments are considered positive, the resultant moment of all the given forces with respect to the point O is
MR = �MF = 0 + 300 � 4 – 450 � (4 + 5) = +1200 – 4050
= – 2850 ft-lb, or 2850 ft-lb.
Since the resultant force acts upward along its line of action and since the moment of the resultant must be counterclock- wise, the line of action of the resultant must lie to the right of the force F1. If the horizontal distance from the point O to the line of action of the resultant is denoted by d, the moment of the resultant is –350 d. Hence,
–350 d = – 2850
and
To make the conditions clearer to you, the resultant force R is included in Figure 22, and its line of action is represented by a broken line.
You should note that the magnitude of the force F1 must be included in the vector sum �F for determining the magnitude of the resultant, even though the moment of that force with respect to the point O is zero.
FIGURE 22—Conditions in Example Problem 1, Article 30
�
Engineering Mechanics, Part 1 57
Problem 2: A horizontal bar represented in Figure 23 by the line AB is acted upon by the three forces F1, F2, and F3 whose characteristics are as indicated. a) Determine the magnitude of the resultant of the given forces, the angle between the bar and the line of action of the resultant, and the direction of the resultant along its line of action.
b) Compute the horizontal distance from the point A to the point at which the line of action of the resultant intersects the bar AB.
Solution: a) In this part of the problem, it is convenient to consider the given forces as concurrent. The first step is to replace each of the forces F1 and F3 by its horizontal and vertical components at the point at which the line of action of the given force intersects the bar. The magnitudes and direc- tions of these components may be computed by applying the formulas in Article 24.
For F1, F1x = 1500 cos 60° = 750 and F1y = 1500 sin 60° = 1299
" $
For F3, F3x = 4000 cos 45° = 2830 and F3y = 4000 sin 45° = 2830
! $
If horizontal components acting toward the right and vertical components acting upward are considered to be positive, the vector sums of the horizontal and vertical components of the three given forces are as follows: Fx = + 750 + 0 – 2830 = – 2080 lb, or 2080 lb
!
H
FIGURE 23—Conditions in Example Problem 2, Article 30
Engineering Mechanics, Part 158
Fy = – 1299 – 3000 – 2830 = – 7130 lb, or 7130 lb $
By formula 1, Article 23, the magnitude of the resultant is
Also, the acute angle H between the horizontal bar and the line of action of the resultant can be determined by applying formula 2, Article 23, as follows:
The direction of the resultant along its line of action must be toward the left and downward.
The resultant may be described by the notation 7430 lb. 73°45�
b) If each of the forces F1 and F3 is replaced by its components at the point at which the line of action of the force intersects the bar, the lines of action of the horizontal components will pass through the point A. Hence, the moment of each of these horizontal components with respect to the point A will be zero. If clockwise moments are considered positive, the expression for the resultant moment of the three given forces with respect to the point A may be written as follows:
MR = �MF = + 1299 � 6 + 3000 � (6 + 6) + 2830 � (6 + 6 + 6)
= + 94,700 ft-lb, or 94,700 ft-lb
If the resultant is replaced by horizontal and vertical compo- nents at the point at which the line of action of the resultant intersects the bar, the moment of the horizontal component will be zero, and the magnitude of the moment of the vertical component with respect to the point A will be equal to the product of the magnitude of the vertical component and the horizontal distance from the point A to the line of action of the resultant. If this horizontal distance is denoted by d, the magnitude of the moment of the vertical component will be 7130 d. Since this moment must equal the resultant moment of the given forces, which is 94,700 ft-lb, and since the
tan .H F F
Hy x
= ∑ ∑
= = = ° 7130 2080
3 43 73 45and �
R F Fx y= + = + = 2 2 2 22080 7130 7430 lb
�
�
Engineering Mechanics, Part 1 59
vertical component acts downward, the resultant must lie to the right of the point A and
+ 7130 d = + 94,700
Hence,
In Figure 23 the line of action of the resultant R is represented by a broken line.
Use of Components of Inclined Forces
31 When the lines of action of nonconcurrent forces ina system are parallel, it is usually easy to determine the moment arm of each given force with respect to the point selected as the center of moments. For instance, in example problem 1, Article 30, the moment arms of the force F2 and F3 with respect to the point O can be determined by inspec- tion of the diagram in Figure 22.
When the lines of action of nonconcurrent forces are not parallel, it would be possible to compute the moment arm of each force with respect to the center of moments by either applying principles of trigonometry or measuring each arm on an accurate diagram. For example, the moment arm of the force F1 in Figure 22 with respect to the point A would be equal to 6 sin 60°, and the moment arm of the force F3 would be (6 + 6 + 6) � sin 45°. However, since it is necessary to compute the vertical component of each of these forces in order to determine the vertical component of the resultant, it is much more convenient to compute the moment of such a force with respect to the center of moments by multiplying the vertical component of the force by a horizontal distance that is determined by inspection of the diagram.
Furthermore, after you have determined the angle between a horizontal reference line and the line of action of the result- ant of a system of nonconcurrent forces, it would be possible to express the moment arm of the resultant with respect to the center of moments in terms of a horizontal distance and a trigonometric function of that angle. For example, if the horizontal distance from the point A in Figure 23 to the line
d = = 94 700 7130
13 28 ,
. ft
Engineering Mechanics, Part 160
of action of the resultant is denoted by d, as in example prob- lem 2, Article 30, the moment arm of the resultant would be equal to d sin 73°45�. This calculation can be avoided by using the moment of the vertical component of the resultant instead of the moment of the resultant itself.
Couples
32 A kind of system of forces that must be consideredin many engineering problems consists of only two forces having the following characteristics: 1) The magnitudes of the two forces are the same; 2) the lines of action of the two forces are parallel; and 3) the forces act in opposite direc- tions along their lines of action. Such a system is called a couple. Four typical couples are represented in Figure 24. Every couple has several important properties.
Since the two forces forming a couple act along parallel lines in opposite directions, and since the magnitudes of the two forces are equal, the vector sum of the forces must be zero. In other words, the magnitude of a force corresponding to the resultant of the given forces would be zero.
FIGURE 24—Typical Couples
Engineering Mechanics, Part 1 61
Furthermore, the resultant moment of the two forces forming a couple is always the same, regardless of the location of the center of moments. The magnitude of this moment is equal to the product of the magnitude of either of the given forces and the perpendicular distance between the lines of action of the forces. The direction of this moment is always clockwise or always counterclockwise. For example, if the center of moments in Figure 24A is at the point O, the moment arm of the force F1 would be 2 ft and the moment arm of F2 would be 1 ft. Since the moment of each force would be clockwise, the resultant moment would be + 60 � 2 + 60 � 1 = + 60 � (2 + 1) = + 60 � 3, or 60 � 3. Likewise, if the center of moments is at the point O′, the moment arm of the force F1 would be 2.5 ft, and the moment of that force would be counterclockwise and its magnitude would be equal to 60 � 2.5. Also, the moment arm of F2 would be 2.5 + 3 = 5.5 ft, and the moment of F2 would be clockwise and its magnitude would be equal to 60 � 5.5. Therefore, the resultant moment would be –60 � 2.5 + 60 � 5.5 = + 60 � (5.5 – 2.5) = + 60 � 3, or 60 � 3. Similarly, the resultant moment of the two forces in Figure 24B would be 90 � 2, the resultant moment of the two forces in Figure 24C would be 80 � 5, and the resultant moment of the two forces in Figure 24D would be 100 � 4.
It follows from the preceding explanations that the resultant of any couple is a moment, and that the magnitude and direction of this moment are the same for every possible center of moments. It is said that two couples are equivalent if the magnitudes of their moments are the same and their directions are also the same. For instance, the couple in Figure 24A would be equivalent to the couple in Figure 24B because the resultant moment of each is 180 ft-lb. However, although both the couple in Figure 24C and the couple in Figure 24D have the same magnitude, 400 ft-lb, the moment in Figure 24C is clockwise and the moment in Figure 24D is counterclockwise. Therefore, these couples are not equivalent.
�
�
� �
�
�
Engineering Mechanics, Part 162
In some problems, the moment of a certain couple is known and it is desired to determine the characteristics of an equiv- alent couple. For example, the magnitude of each force of the equivalent couple may be specified, and it may be required to determine the perpendicular distance between the lines of action of the forces by dividing the known moment of the couple by the magnitude of either force. On the other hand, the perpendicular distance between the lines of action of the forces of the equivalent couple may be specified, and it may be required to determine the magnitude of each force of the couple by dividing the known moment of the couple by the distance between the forces.
Engineering Mechanics, Part 1 63
Practice Problems 3 1. If the magnitude of force F in Figure 19A is 500 lb and the distance d is 8 ft, what is the
moment of the force with respect to the point O?
__________________________________________________________
2. Figure 25 shows the characteristics of the three forces F1, F2, and F3. Determine the moment, in foot-pounds, of each of these forces with respect to the point O.
__________________________________________________________
3. What is the resultant moment of the forces F1, F2, and F3 in problem 2 with respect to the point O?
__________________________________________________________
(Continued)
FIGURE 25—Forces in Practice Problems 2 and 3, Article 32
Engineering Mechanics, Part 164
Practice Problems 3 4. The horizontal bar in Figure 26 is acted upon
by the four vertical forces F1, F2, F3, and F4. Determine a) the magnitude and the direc- tion of the resultant of the given forces, and b) the horizontal distance d from the line of action of the force F1 to the line of action of the resultant.
____________________________
5. A horizontal bar is acted upon by three forces whose characteristics and locations are indicated in Figure 27. Compute a) the magnitude of the resultant of the given forces, b) the acute angle between the bar and the line of action of the resultant, c) the direction of the resultant along its line of action, and d) the horizontal distance d from the point O on the bar to the line of action of the resultant.
____________________________
6. The magnitude of each of the two forces forming a couple is 300 lb, and the perpen- dicular distance between the parallel lines of action of the forces is 4 ft. What is the mag- nitude of the moment of the couple?
____________________________
7. The couple in problem 6 is to be replaced by an equivalent couple in which the magnitude of each force is to be 800 lb. What should be the perpendicular distance between the lines of action of the forces forming the new couple?
__________________________________________________________
Check your answers with those on page 87.
FIGURE 26—Forces in Practice Problem 4, Article 32
FIGURE 27—Forces in Practice Problem 5, Article 32
Engineering Mechanics, Part 1 65
CENTER OF GRAVITY
Simple Body
General Location of Center of Gravity
33 Any body is composed of a number of particles ofmaterial. The total weight of a body is the sum of the weights of all the particles contained in the body. Since the weight of each particle acts vertically downward, the weights of the particles form a system of parallel forces. Therefore, the weight of a body is really the resultant of a system of par- allel forces, each of which acts vertically downward. When a body is held in different positions, the line of action of the weight of the entire body will pass through different points in the body. However, for every possible position of the body, there will be one point through which the line of action of the weight always passes.
The conditions for a simple body are indicated in Figure 28. In Figure 28A, the edge ab of the body is horizontal, and the vertical line of action of the force R representing the weight has the position 1. In Figure 28B, the edge ab of the body is vertical, and the vertical line of action of the force R repre- senting the weight has the position 2. In Figure 28C, the edge ab of the body is in some inclined position, and the vertical line of action of the force R representing the weight has the position 3. For all three positions of the body, as well as for any other position, the line of action of the force R will always pass through a certain point designated as G. In Figure 28, the point G lies within the body, but no attempt is made to indicate the distance from a point on the exposed surface of the body to the point G.
A point, such as the point G, through which the line of action of the force representing the weight of the body always passes, is called the center of gravity of the body. Regardless of the position of a body in space, the center of gravity of the body has a certain fixed position with respect to other points in the body.
Engineering Mechanics, Part 166
In some problems in engineering mechanics, it is permissible to treat an entire body as a single unit and to assume that the weight of the entire body passes through the center of gravity of the body. In other problems, it is convenient to imagine that a body is composed of two or more elementary portions and to consider the weight of each such portion as a separate force. In such a case, it is assumed that the weight of each portion passes through the center of gravity of the corresponding portion. For example, the body represented in Figure 29 consists of a spherical portion 1 and a cylindrical portion 2, and it may be convenient to consider the weights of the two parts as separate forces. The resultant of these two forces is equivalent to the single force representing the weight of the entire body.
FIGURE 28—Center of Gravity of a Body
FIGURE 29—Elementary Portions of a Body
1. Spherical portion
2. Cylindrical portion
Engineering Mechanics, Part 1 67
Principles of Center of Gravity of a Body
34 The general principles involving the center of gravityof a body can be stated as follows: 1. The weight of a body or of a part of a body may be
treated as a single force acting vertically downward through the center of gravity of the body or of the part. This principle is often expressed as follows: All the material in a body or a part is concentrated at the center of gravity of the body or particle.
2. The position of the center of gravity of a body composed of the same material throughout depends only on the shape and size of the body. It does not depend on the nature of the material.
3. If a body is suspended from a point in such a manner that the body is free to rotate, it will finally come to rest in some position in which the center of gravity of the body is vertically below the point of suspension.
4. The center of gravity of a body may be outside the body. For example, if a body has the shape shown in Figure 30, the center of gravity would be located outside the body, as at the point G.
Center of Gravity of Symmetrical Body
35 When a body is so shaped that a certain plane bisectsevery line that passes through the body perpendicu- lar to the plane, it is said that the body is symmetrical with respect to the plane. The plane is called a plane of symmetry. For instance, a sphere is symmetrical with respect to any plane that passes through the geometric center of the sphere. A rectangular solid is symmetrical with respect to any plane that passes through the geometric center of the body and is perpendicular to any pair of opposite faces of the body. A right circular cylinder or a right cone is symmetrical with respect to any plane that passes through the axis of the body. A hemisphere is symmetrical with respect to a plane that passes through the geometric center of the sphere and is perpendicular to the plane surface of the hemisphere. Various other bodies are symmetrical with respect to certain planes.
FIGURE 30—Center of Gravity Outside the Body
Engineering Mechanics, Part 168
If a body is symmetrical with respect to a certain plane, the center of gravity of the body must lie in that plane. If a body is symmetrical with respect to each of two planes that are perpendicular to each other, the center of gravity of the body must lie on the line in which these planes of symmetry inter- sect. If a body is symmetrical with respect to each of three planes that are mutually perpendicular, the center of gravity of the body must be at the point of intersection of these three planes.
The center of gravity of a sphere, of a right circular cylinder, or of a rectangular solid must be at the geometric center of the body. Thus, the distance from any point on the surface of a sphere to the center of gravity of the sphere is equal to the radius of the sphere. The distance from any surface of a rectangular solid to the center of gravity, measured perpendi- cular to that surface, is equal to one-half the perpendicular distance between that surface and the opposite, or parallel, surface or one-half the length of an edge connecting these parallel surfaces. The distance from any face of a cube to the center of gravity of the cube is equal to one-half the length of any edge of the cube.
The center of gravity of a hemisphere is on the line of symme- try, which passes through the center of the entire sphere and is perpendicular to the plane surface of the hemisphere. The distance along the line of symmetry from the plane surface of the hemisphere to the center of gravity is equal to three- eighths of the radius of the sphere.
The center of gravity of a right cone or of a right pyramid is on the axis of the body. The distance along the axis from the base of the body to its center of gravity is equal to one-fourth of the altitude of the body.
In Figure 29, the center of gravity of the spherical portion 1 of the body would be at the geometric center of the sphere. The position of this center is indicated by the point G1, which lies within the sphere. The center of gravity of the cylindrical portion 2 of the body is on the axis of the cylinder and mid- way between the ends of the cylinder. The position of this center of gravity is indicated by the point G2, which lies within the cylinder.
Engineering Mechanics, Part 1 69
Centroid of Plane Surface
36 In an engineering problem, it is often desirable todeal with a plane surface of a body instead of the entire body. Theoretically, a surface has no thickness and consequently cannot have weight. It is possible to imagine that a surface is a very thin body with two parallel faces, each of which has the same shape and size as the given sur- face. The term “center of gravity” can then be used to apply to a surface. Another concept is to neglect the thickness of the surface and to use the term centroid instead of center of gravity. When the centroid of a plane surface is considered, a line of symmetry may be used instead of a plane of symme- try. The positions of the centroids of several common plane figures are discussed in the next five articles.
Centroid of Circle or Parallelogram
37 A circle is symmetrical with respect to any diameter,and the centroid of a circle obviously is at the geo- metric center of the circle. In other words, the distance from any point on a circle to the centroid of the circle is equal to the radius of the circle.
A square or a rectangle is symmetrical with respect to two lines, each of which passes through the geometric center of the figure and is perpendicular to a pair of opposite sides of the figure. In the rectangle in Figure 31A, the two lines of symmetry 1 and 2 are represented by broken lines. Similar lines can be drawn for a square. The centroid of a square or a rectangle, as the point G in Figure 31, is at the geometric center of the figure. It will be seen that the diagonal lines 3 and 4 in Figure 31A also pass through the geometric center of the figure. So the centroid of a square or a rectangle is also at the intersection of the diagonals of the figure.
The centroid of any parallelogram (a square or a rectangle is a special kind of parallelogram) is located at the geometric center of the parallelogram, which is at the intersection of the two diagonals. Figure 31B shows the outlines of a parallelo- gram, its diagonals 3 and 4, and its centroid G. The distance from any side of a parallelogram to the centroid, measured perpendicular to that side, is equal to one-half the perpendi- cular distance between that side and the parallel side. For a
Engineering Mechanics, Part 170
square, the perpendicular distance from any side to the cen- troid is one-half the length of any side. For a rectangle, the perpendicular distance from either long side to the centroid is one-half the short dimension; and the perpendicular distance from either short side to the centroid is one-half the long dimension.
Centroid of Triangle
38 In any triangle, there are three lines, called medians,each of which runs from a vertex to the central point on the opposite side. In Figure 32, for example, the three ver- tices of the triangle are marked A, B, and C; the point A� is at the middle of the side BC; the point B� is at the middle of the side AC; and the point C� is at the middle of the side AB. The medians are the three broken lines 1, 2, and 3. The three medians in any triangle will intersect at a common point, which is the centroid of the triangle. In this illustration, the centroid is marked G.
The distance along any median from the vertex at the end of that median to the centroid of the triangle is equal to two- thirds of the length of the entire median. Thus, in Figure 32, AG = 2/3AA�, BG =
2/3BB�, and CG = 2/3CC�. However, when it
is desired to consider the position of the centroid of a triangle in a practical problem, it is usually convenient to use the dis- tance from some side of the triangle to the centroid measured perpendicular to that side. This perpendicular distance can be easily expressed in terms of the altitude of the triangle when the selected side is taken as the base of the triangle.
FIGURE 31—Centroid of Parallelogram
1, 2.Lines through center and perpendicular to sides
3, 4.Diagonals
A. Rectangle
B. Parallelogram
Engineering Mechanics, Part 1 71
For instance, if the side AB is taken as the reference side, the perpendicular distance from that side to the centroid is rep- resented by the line GH, and the corresponding altitude of the triangle is represented by the line CK. Then, since the distance GC� along the median CC� is equal to 1/3CC� and since the right triangles C�GH and C�CK are similar, it follows that the distance GH is equal to 1/3CK. In general, if one side of a triangle is taken as the base of the triangle, the perpen- dicular distance from the base to the centroid is equal to one-third of the altitude. If the side AB is considered as the base and the altitude CK is 1.5 ft, the perpendicular distance HG from the side AB to the centroid of the triangle would be 1/3 � 1.5 = 0.5 ft.
Centroid of Trapezoid
39 Figure 33 shows a trapezoid in which the length ofthe longer base is denoted by b1, the length of the shorter base is denoted by b2, and the altitude is denoted by h. The centroid of the trapezoid must lie on the broken line connecting the central points of the two bases, as at the point G. The distance y from the longer base to the centroid, meas- ured perpendicular to the bases, can be computed by the relation
y h b b b b
= + +
⎛
⎝ ⎜
⎞
⎠ ⎟
1 3
21 2 1 2
FIGURE 32—Centroid of Triangle
1, 2, 3.Medians
Engineering Mechanics, Part 172
Example Problem
Problem: In a trapezoid like the one in Figure 33, the lengths of the bases are b1 = 3 ft and b2 = 2 ft, and the alti- tude is h = 1.5 ft. What is the distance y from the longer base to the centroid of the trapezoid?
Solution: By the formula,
Centroid of Sector of Circle
40 In Figure 34 is shown a sector of a circle. The centerof the circle is at the point O, and the boundaries of the sector are the circular arc 1 and the radii 2 and 3. The length of each radius of the circle is denoted by r, and the central angle between the radii 2 and 3 is denoted by L. The sector is symmetrical with respect to the radius 4, which bisects the angle L. The centroid of the sector must lie on the radius 4, and the distance y along that radius from the cen- ter of the circle to the centroid G can be computed by the following formula:
(1)
The value of the angle L in the denominator must be expressed in degrees and a decimal part of a degree.
For a semicircle, L = 180° and formula 1 becomes
y = 0.424r (2)
y r L
L =
76 4 1 2
. sin
FIGURE 33—Centroid of Trapezoid
Engineering Mechanics, Part 1 73
For a quadrant, L = 90° and the formula is
y = 0.600r (3)
For a sextant, L = 60° and the formula is
y = 0.637r (4)
Example Problem
Problem: In a sector of a circle whose radius is 4 ft, the cen- tral angle is 40°30�. What is the distance from the center of the circle to the centroid of the sector?
Solution: In formula 1, r = 4 ft, L = 40°30� = 40.5°, and sin 1/2L = sin 20°15� = 0.346. Hence, the required distance is
Centroid of Segment of Circle
41 In Figure 35 is shown a segment of a circle. The cen-ter of the circle is at the point O, and the boundaries of the segment are the circular arc 1 and the chord 2. The length of each radius of the circle is denoted by r, the angle at the center of the circle between the radii 3 and 4 through the ends of the arc 1 is denoted by L, and the length of the chord 2 is denoted by c. The segment is symmetrical with respect to the radius 5, which bisects the angle L. The cen-
y = = 76 4 4 0 346
40 5 2 61
. . .
. � �
ft
FIGURE 34—Centroid of Sector of Circle
1. Circular arc
2, 3, 4. Radii
Engineering Mechanics, Part 174
troid G of the segment lies on the radius 5 and the distance y from the center of the circle to the centroid can be computed by the formula
In the first term in the denominator, the value of L must be expressed in degrees and a decimal part of a degree.
Example Problem
Problem: In a segment of a circle whose radius is 8 in., the length of the chord bounding the segment is 10 in. What is the distance from the center of the circle to the centroid of the segment?
Solution: First, determine the angle 1/2L in the following manner:
sin . 1 2
1 2 5
8 0 625L
c
r = = =
y c
r L cr L =
−
3
20 1047 6 1 2
. cos
FIGURE 35—Centroid of Segment of Circle
1. Circular arc
2. Chord
3, 4, 5. Radii
Engineering Mechanics, Part 1 75
L = 2 � 38°40� = 77°20� = 77.3°
and
cos1/2L = 0.781
When these values are substituted in the formula, the required distance from the center of the circle to the centroid of the segment is found to be
1 2
38 40L = ° �
Engineering Mechanics, Part 176
Self-Check 3 1. When the weight of an entire body is represented by a single force, what term is used to refer
to the point through which the line of action of this force always passes?
__________________________________________________________
2. For a body that is composed of only one kind of material, what characteristics of the body influence the position of the center of gravity of the body?
__________________________________________________________
3. Must the center of gravity of a body always be inside the body?
__________________________________________________________
4. When a body is suspended from a point in such a way that the body can rotate freely, finally coming to rest, what will be the direction of a line from the point of suspension to the center of gravity of the body?
__________________________________________________________
5. If the diameter of a sphere is 4 ft, what is the distance from any point on the surface of a sphere to the center of gravity of the sphere?
__________________________________________________________
6. If a hemisphere is formed by cutting the sphere in question 5 into two equal parts, what is the distance from the plane surface of the hemisphere to the center of gravity, measured perpendicular to that plane surface?
__________________________________________________________
7. The length of each side of a cube is 8 in. What is the distance from any face of the cube to the center of gravity, measured perpendicular to that face?
__________________________________________________________
8. The altitude of a right cone is 6 ft. What is the distance from the base of the cone to the center of gravity of the cone, measured perpendicular to the base?
__________________________________________________________
(Continued)
Engineering Mechanics, Part 1 77
Self-Check 3 9. The length of each short side of a rectangle is 5 ft. What is the distance from either long side
of the rectangle to the centroid, measured perpendicular to that long side?
__________________________________________________________
10. When a certain side of a triangle is considered to be the base, the altitude of the triangle is 4.5 ft. What is the distance from that side to the centroid, measured perpendicular to that side?
__________________________________________________________
Check your answers with those on page 86.
Practice Problems 4 1. The lengths of the bases of a trapezoid are 6 ft and 4 ft, and the altitude is 3 ft. What is the
distance from the longer base to the centroid, measured perpendicular to that base?
__________________________________________________________
2. The radius of the circle from which a sector is formed is 10 ft. If the central angle of the sec- tor is 45°, what is the distance from the center of the circle to the centroid of the sector?
__________________________________________________________
3. The radius of a circle from which a segment is formed is 5 in. If the length of the chord that bounds the segment is 8 in., what is the distance from the center of the circle to the centroid of the segment?
__________________________________________________________
Check your answers with those on page 88.
Engineering Mechanics, Part 178
Composite Body
General Explanations
42 It is often desirable in an engineering problem todetermine the position of the center of gravity of a body that consists of two or more parts. For example, it may be required to locate the center of gravity of the entire body represented in Figure 29, which consists of a spherical por- tion 1 and a cylindrical portion 2. For this purpose, it would be convenient to compute the horizontal distance from the left-hand end of the cylindrical portion to the center of gravity of the entire body. This distance could be determined in the following way:
Step 1. Represent the weight of the spherical portion 1 by a vertical force whose line of action passes through the center of gravity G1 of that portion.
Step 2. Represent the weight of the cylindrical portion 2 by a vertical force whose line of action passes through the center of gravity G2 of that portion.
Step 3. Compute the horizontal distance from the left-hand end of the cylindrical portion to the line of action of the resultant of these two parallel forces.
If an entire body is composed of the same material, the weights of the several portions of the body are proportional to the volumes of the portions. It is therefore convenient to deter- mine the magnitudes of the parallel forces by using volumes instead of weights. Moreover, the lines of action of the forces do not have to be vertical, but may be horizontal or inclined at any angle. However, all volumes must be expressed in the same units, as cubic feet or cubic inches; it must be imag- ined that the lines of action of all the forces are parallel; and the distances between the lines of action must be correct and expressed in suitable units.
If it is desirable to determine the position of the centroid of a plane surface that consists of two or more parts, it is conven- ient to determine the magnitudes of the parallel forces of a system by using the areas of the parts.
Engineering Mechanics, Part 1 79
Procedure for Body Composed of Parts
43 The general formula for computing the distance fromsome reference plane to the center of gravity of a body may be expressed as follows:
in which p = distance, measured perpendicular to a reference plane, from that plane to the center of gravity of a body
�Mv = vector sum of the moments of the volumes of the portions of the body with respect to the reference plane
V = total volume of the body
The reference plane may have any direction, but the distance from the reference plane to the center of gravity of each por- tion of the body and the distance from the reference plane to the center of gravity of the entire body must be measured perpendicular to the reference plane.
Example Problem
Problem: A body having the shape indicated in Figure 29 has the following dimensions:diameter d1 of the spherical portion, 9 in.; diameter d2 of the cylindrical portion, 1
1/2 in.; length l2 of the cylindrical portion, 20 in. Determine the hori- zontal distance from the left-hand end of the cylindrical portion to the center of gravity of the entire body.
Solution: In this problem, the volumes of the parts of the body will be expressed in cubic inches and the moment arms will be expressed in inches.
The volume of the cylindrical portion is
and the volume of the spherical portion is
1 6
9 3823� �π = cu in.
p M V
v= ∑
Engineering Mechanics, Part 180
When the center of moments is at the left-hand end of the cylindrical portion, the moment arm for the cylindrical portion is 1/2 � 20 = 10 in. and the moment arm for the spherical portion is 20 + 1/2 � 9 = 24.5 in. Hence,
Since the volume of the entire body is V = 35.3 + 382 = 417 cu in., the horizontal distance from the left-hand end of the cylindrical portion to the center of gravity of the body is
The calculations can be arranged conveniently as shown in the accompanying tabulation.
Therefore,
Procedure for Area Composed of Parts
44 When it is desired to compute the distance fromsome reference line to the centroid of a plane surface, the formula to be used is similar to the formula in Article 43. However, since areas may be used instead of volumes, the formula may be written as follows:
p M A
a= ∑
p = = 9713 417
23 3. .in
p M V
v= ∑
= = 9710 417
23 3. .in
∑ = + =Mv 35 3 10 382 24 5 9710. .� �
Part Volume Moment Arm Moment
Cylinder π 4
1.5 20 = 35.3 2
� � 1 2
20 = 10� 353
Sphere 1 6
� �π 9 =3 382 20 1 2
9 = 24.5+ � 9360
Sum = 417 Sum = 9713
Engineering Mechanics, Part 1 81
in which p = distance, measured perpendicular to a reference line, from that line to the centroid of a plane surface
�Ma = vector sum of the moments of the areas of the portions of the surface with respect to the refer- ence line
A = total area of the surface
Example Problem
Problem: In Figure 36 is represented an L-shaped plane sur- face OABCDE having the dimensions shown. Determine the perpendicular distance to the centroid of the surface a) from the edge OA and b) from the edge OE.
Solution: a) In this problem, assume that the given surface consists of two rectangles. The division line between these two parts may be located in either of two ways. One way is to prolong the edge DC to the point F on the edge OA as indi- cated in Figure 36A, and thus form the rectangles OEDF and ABCF. The other way is to prolong the edge BC to the point H
FIGURE 36—Surface in Example Problem in Article 44
Engineering Mechanics, Part 182
on the edge OE, as indicated in Figure 36B, and thus form the rectangles OABH and CDEH. The procedure in which the rectangles OEDF and ABCF in Figure 36A are used is as follows:
The centroid of the rectangle OEDF is at G1, and the centroid of the rectangle ABCF is at G2. The area of the rectangle OEDF is equal to the product of the lengths of the edges OE and ED, or 31/2 �
3/4 = 2.63 sq in. (square inches); the area of the rectangle ABCF is equal to the product of the lengths of the edges BC and AB (6 – 3/4) �
3/4 = 3.94 sq in.; and the area of the entire surface is A = 2.63 + 3.94 = 6.97 sq in. The perpendicular distance from the reference line OA to the centroid G1 is JG1 =
1/2 � 3 1/2 = 1.75 in., and the perpendi-
cular distance from the reference line OA to the centroid G2 is KG2 =
1/2 � 3/4 = 0.375 in. Therefore, when the center of
moments is on the edge OA,
�Ma = 26.3 � 1.75 + 3.94 � 0.375 = 6.08
The perpendicular distance NG from the edge OA to the cen- troid G of the entire surface is
b) When the reference line is the edge OE, the perpendicular distance to G1 is LG1 =
1/2 � 3/4 = 0.375 in., and the perpen-
dicular distance to G2 is MG2 = 3/4 +
1/2 � (6 – 3/4) = 3.38 in.
In this case,
�Ma = 2.63 � 0.375 + 3.94 � 3.38 = 14.30
and the perpendicular distance PG from OE to the centroid G is
p = = 14 30 6 57
2 18 . .
. .in
p M A
a= ∑
= = 6 08 6 57
0 93 . .
. .in
Engineering Mechanics, Part 1 83
Practice Problems 5 1. Figure 37 represents a steel bolt. Its head is a prism, each base of which is a square 1 1/4 in.
wide and the height of which is 5/8 in. It may be assumed that the shank is a circular cylinder, the diameter of which is 3/4 in. and the length of which is 3 in. Compute the distance from the left-hand edge of the head to the center of gravity of the entire bolt.
__________________________________________________________
2. Determine the perpendicular distance from the edges OA and OE of the L-shaped surface in Figure 36 to the centroid G of the entire surface by dividing the given surface in the two rec- tangles OABH and CDEH, as indicated in Figure 36B. The positions of the centroids G3 and G4 of these rectangles are indicated in Figure 36B.
__________________________________________________________
Check your answers with those on page 88.
FIGURE 37—Dimensions of Bolt in Practice Problem 1, Article 44
Engineering Mechanics, Part 184
NOTES
Self-Check 1 1. Engineering mechanics deals with
problems involving the effects of forces on bodies. Article 1
2. A force is any action that causes or tends to cause a change in the motion of a body. Article 1
3. Statics Dynamics Mechanics of materials or strength of materials Article 3
4. No Article 3
5. The weight of a body is the magnitude of the force of attraction between the earth and the body. Article 4
6. (a) The pound (b) The kilogram Article 4
7. Three Article 5
8. Magnitude and direction Article 7
9. With a graphic solution, it is easier for the person to visualize the conditions and to understand the solution. Article 8
10. Three (unless the first figure is 1) Article 8
Self-Check 2 1. a) 200 lb
b) 60° c) Downward and toward the left Article 13
2. 120 lb Article 13 20°
85
A n s w e r s
A n s w e r s
3. and 4. Article 14
5. The lines of action of all the forces must coincide, or all the forces must have a common line of action. Article 15
6. The lines of action of all the forces must pass through a common point. Article 15
7. A nonconcurrent, nonparallel system Article 15
8. The resultant Article 16
Self-Check 3 1. Center of gravity of the body Article 33
2. The shape and size of the body Article 34
3. No Article 34
4. Vertical Article 35
5. 1/2 � 4 = 2 ft Article 35
6. 3/8 � 2 = 0.75 ft Article 35
7. 1/2 � 8 = 4 in. Article 35
8. 1/4 � 6 = 1.5 ft Article 35
9. 1/2 � 5 = 2.5 ft Article 37
10. 1/3 � 4.5 = 1.5 ft Article 38
Self-Check Answers86
FIGURE 38
Self-Check Answers 87
Practice Problems 1 1. 100 lb
"
2. 210 lb #
3. 20 lb !
4. 700 lb #
5. 600 lb !
Practice Problems 2 1.
2. 159 lb 2°40�
3.
4. Fx = 127 lb; Fy + 272 lb " #
5.
Practice Problems 3 1. 4000 ft-lb
2. for F1, 334; for F2, 717; for F3, 507
3. 124 ft-lb 4. a) 600 lb;
#
b) 6.17 ft
�
�
� ��
5. a), b), and c)
d) 8.79 ft
6. 1200 ft-lb
7. 1.50 ft
Practice Problems 4 1. 1.4 ft
2. 6.50 ft
3. 3.82 in.
Practice Problems 5 1. 1.36 in.
2. NG = 0.93 in.; PG = 2.18 in.
14.130 lb
83�00�
Practice Problems Answers88
89
1. Each of the following general principles apply to the center of gravity of a body except
A. all the material in a body or a part is concentrated at the center of gravity of the body part.
B. the position of the center of gravity of a body composed of the same material throughout depends only on the shape and size of the body.
C. if a body is suspended from a point in such a manner that the body is free to rotate, it will finally come to rest in some position in which the center of gravity of the body is vertically below the point of suspension.
D. the center of gravity of a body may not be outside the body.
2. The lines of action of the forces of a system are not parallel and do not pass through a single point. The system is said to be
A. collinear. B. concurrent, nonparallel.
E x a m in a t io n
E x a m in a t io n
Engineering Mechanics, Part 1
When you feel confident that you have mastered the material in this study unit, go to http://www.takeexamsonline.com and submit your answers online. If you don’t have access to the Internet, you can phone in or mail in your exam. Submit your answers for this examination as soon as you complete it. Do not wait until another examination is ready.
Questions 1–20: Select the one best answer to each question.
EXAMINATION NUMBER
28603602 Whichever method you use in submitting your exam
answers to the school, you must use the number above.
For the quickest test results, go to http://www.takeexamsonline.com
Examination90
C. nonconcurrent, nonparallel. D. parallel.
3. Assuming the forces are collinear, if F1 = 25 lb toward the right and F2 = 75 lb toward the right, the magnitude of the resultant and the direction of the resultant along the common line of action of the given forces would be
A. 50 #. C. 100 $. B. 100 ". D. 50 !.
4. In Examination Figure 1 are shown the location of the line of action of a force F and the characteristics of the force. The magnitude of the moment of this force with respect to the point O is
A. 360 ft-lb. C. 540 ft-lb. B. 480 ft-lb. D. 600 ft-lb.
5. An action that causes a change in the motion of a body is known as
A. vector. C. force. B. acceleration. D. velocity.
6. A speed of 60 miles/hour converted to units of feet/second is
A. 88 ft/s. C. 880 ft/s. B. 1220 ft/s. D. 3.67 ft/s.
7. A body is acted upon by two forces F1 and F2. F1 acts due west with a 100 lb magni- tude and F2 acts due south with a 150 lb magnitude. The magnitude of the resultant of these forces is
Examination Figure 1
Examination 91
A. 175 lb. C. 180 lb. B. 200 lb. D. 250 lb.
8. The characteristics of two concurrent forces F1 and F2 are indicated in Examination Figure 2. The magnitude of the resultant of these forces is
A. 100 lb. C. 134 lb. B. 118 lb. D. 140 lb.
9. The dimensions of a trapezoid ABCD are shown in Examination Figure 3. The perpendi- cular distance from the side AB to the centroid G of the trapezoid is
A. 4.13 in. C. 4.75 in. B. 4.50 in. D. 4.87 in.
10. When two concurrent forces acting upon a body are replaced by a single force that has the same effect, this single force is usually called the
A. coplanar. C. resultant. B. magnitude. D. vector.
11. When F1 = 1500 lb upward and F2 = 800 lb downward, the magnitude of the resultant and its direction would be
Examination Figure 2
Examination Figure 3
Examination92
A. 600 lb $. C. 700 lb #. B. 700 lb $. D. 2300 lb #.
12. Examination Figure 4 shows two concurrent forces F1 and F2. The angle (measured counterclockwise) between the horizontal reference line and the line of action of the resultant of the two given forces is
A. 61°20�. C. 102°40�. B. 95°30�. D. 134°50�.
13. As indicated in Examination Figure 5, a body is acted upon by the four forces F1, F2, F3, and F4. The locations of the lines of action and the characteristics of these forces are shown. The resultant moment of the forces with respect to the point O is
A. 700 ft-lb. C. 700 ft-lb.
Examination Figure 4
Examination Figure 5
� �
� �
Examination 93
B. 600 ft-lb. D. 800 ft-lb.
14. The horizontal bar represented in Examination Figure 6 is acted upon by the four vertical forces F1, F2, F3, and F4, whose locations and characteristics are shown. The horizontal distance from the point O to the line of action of the resultant of the given forces is
A. 13.25 ft. C. 26.5 ft. B. 16.75 ft. D. 33.5 ft.
15. A system of forces can be divided into all of the following general classes except
A. collinear forces. C. concurrent forces. B. parallel forces. D. circular forces.
16. As indicated in Examination Figure 7, a body is acted upon by three concurrent forces F1, F2, and F3, whose characteristics are given. The horizontal component of the result- ant of the specified forces is
A. 181.4 lb. C. 247.7 lb.
Examination Figure 6
Examination Figure 7
Examination94
" " B. 191.3 lb. D. 304.0 lb.
" "
17. The vertical component of the resultant of the forces F1, F2, and F3 in Examination Figure 7 is
A. 64.6 lb. C. 85.4 lb. $ $
B. 75.0 lb. D. 95.3 lb. $ $
18. Examination Figure 8 shows two forces that form a couple. It is desired to replace this couple by an equivalent couple. If the magnitude of each force of this new couple is to be 80 lb, the perpendicular distance between the lines of action of these forces
should be
A. 4 ft. C. 8 ft. B. 6 ft. D. 10 ft.
19. In a trapezoid, the lengths of the bases are b1 equals 4 ft and b2 equals 2 ft, and the altitude h is 3 ft. What is the vertical distance y from the longer base to the centroid of the trapezoid?
A. 0.75 ft C. 1.33 ft B. 0.92 ft D. 1.67 ft
20. If F1 = 50 lb upward and F2 = 60 lb upward, the resultant and the direction of the resultant along the common line of action would be
A. 10 lb $. C. 110 lb #. B. 90 lb #. D. 210 lb #.
Examination Figure 8
