Engineering Mechanics PDF - Chapter 1

Chapter 1 (© Andrew Peacock/Lonely Planet Images/Getty Images) Large cranes such as this one are required to lift extremely large loads. Their design is based on the basic principles of statics and dynamics, which form the subject matter of engin...

Chapter 1 (© Andrew Peacock/Lonely Planet Images/Getty Images) Large cranes such as this one are required to lift extremely large loads. Their design is based on the basic principles of statics and dynamics, which form the subject matter of engineering mechanics. General Principles CHAPTER OBJECTIVES n To provide an introduction to the basic quantities and idealizations of mechanics. n To give a statement of Newton’s Laws of Motion and Gravitation. n To review the principles for applying the SI system of units. n To examine the standard procedures for performing numerical calculations. n To present a general guide for solving problems. 1.1 Mechanics Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies that are subjected to the action of forces. In general, this subject can be subdivided into three branches: rigid-body mechanics, deformable-body mechanics, and fluid mechanics. In this book we will study rigid-body mechanics since it is a basic requirement for the study of the mechanics of deformable bodies and the mechanics of fluids. Furthermore, rigid-body mechanics is essential for the design and analysis of many types of structural members, mechanical components, or electrical devices encountered in engineering. Rigid-body mechanics is divided into two areas: statics and dynamics. Statics deals with the equilibrium of bodies, that is, those that are either at rest or move with a constant velocity; whereas dynamics is concerned with the accelerated motion of bodies. We can consider statics as a special case of dynamics, in which the acceleration is zero; however, statics deserves separate treatment in engineering education since many objects are designed with the intention that they remain in equilibrium.  4 C h a p t e r 1    G e n e r a l P r i n c i p l e s Historical Development. The subject of statics developed very 1 early in history because its principles can be formulated simply from measurements of geometry and force. For example, the writings of Archimedes (287–212 B.C.) deal with the principle of the lever. Studies of the pulley, inclined plane, and wrench are also recorded in ancient writings—at times when the requirements for engineering were limited primarily to building construction. Since the principles of dynamics depend on an accurate measurement of time, this subject developed much later. Galileo Galilei (1564–1642) was one of the first major contributors to this field. His work consisted of experiments using pendulums and falling bodies. The most significant contributions in dynamics, however, were made by Isaac Newton (1642–1727), who is noted for his formulation of the three fundamental laws of motion and the law of universal gravitational attraction. Shortly after these laws were postulated, important techniques for their application were developed by other scientists and engineers, some of whom will be mentioned throughout the text. 1.2 Fundamental Concepts Before we begin our study of engineering mechanics, it is important to understand the meaning of certain fundamental concepts and principles. Basic Quantities. The following four quantities are used throughout mechanics. Length. Length is used to locate the position of a point in space and thereby describe the size of a physical system. Once a standard unit of length is defined, one can then use it to define distances and geometric properties of a body as multiples of this unit. Time. Time is conceived as a succession of events. Although the principles of statics are time independent, this quantity plays an important role in the study of dynamics. Mass. Mass is a measure of a quantity of matter that is used to compare the action of one body with that of another. This property manifests itself as a gravitational attraction between two bodies and provides a measure of the resistance of matter to a change in velocity. Force. In general, force is considered as a “push” or “pull” exerted by one body on another. This interaction can occur when there is direct contact between the bodies, such as a person pushing on a wall, or it can occur through a distance when the bodies are physically separated. Examples of the latter type include gravitational, electrical, and magnetic forces. In any case, a force is completely characterized by its magnitude, direction, and point of application.  1.2 Fundamental Concepts 5 Idealizations. Models or idealizations are used in mechanics in order to simplify application of the theory. Here we will consider three 1 important idealizations. Particle. A particle has a mass, but a size that can be neglected. For example, the size of the earth is insignificant compared to the size of its orbit, and therefore the earth can be modeled as a particle when studying its orbital motion. When a body is idealized as a particle, the principles of mechanics reduce to a rather simplified form since the geometry of the body will not be involved in the analysis of the problem. Rigid Body. A rigid body can be considered as a combination of a Three forces act on the ring. Since these forces all meet at a point, then for any large number of particles in which all the particles remain at a fixed force analysis, we can assume the ring to distance from one another, both before and after applying a load. This be represented as a particle. (© Russell model is important because the body’s shape does not change when a C. Hibbeler) load is applied, and so we do not have to consider the type of material from which the body is made. In most cases the actual deformations occurring in structures, machines, mechanisms, and the like are relatively small, and the rigid-body assumption is suitable for analysis. Concentrated Force. A concentrated force represents the effect of a loading which is assumed to act at a point on a body. We can represent a load by a concentrated force, provided the area over which the load is applied is very small compared to the overall size of the body. An example would be the contact force between a wheel and the ground. Steel is a common engineering material that does not deform very much under load. Therefore, we can consider this railroad wheel to be a rigid body acted upon by the concentrated force of the rail. (© Russell C. Hibbeler)  6 C h a p t e r 1    G e n e r a l P r i n c i p l e s Newton’s Three Laws of Motion. Engineering mechanics is 1 formulated on the basis of Newton’s three laws of motion, the validity of which is based on experimental observation. These laws apply to the motion of a particle as measured from a nonaccelerating reference frame. They may be briefly stated as follows. First Law. A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force, Fig. 1–1a. F1 F2 v F3 Equilibrium (a) Second Law. A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force, Fig. 1–1b.* If F is applied to a particle of mass m, this law may be expressed mathematically as F = ma(1–1) a F Accelerated motion (b) Third Law. The mutual forces of action and reaction between two particles are equal, opposite, and collinear, Fig. 1–1c. force of A on B F F A B force of B on A Action – reaction (c) Fig. 1–1 *Stated another way, the unbalanced force acting on the particle is proportional to the time rate of change of the particle’s linear momentum.  1.3 Units of Measurement 7 Newton’s Law of Gravitational Attraction. Shortly after formulating his three laws of motion, Newton postulated a law governing 1 the gravitational attraction between any two particles. Stated mathematically, m 1m 2 F= G (1–2) r2 where F = force of gravitation between the two particles G = universal constant of gravitation; according to experimental evidence, G = 66.73 ( 10 - 12 ) m3 > ( kg # s2 ) m1, m2 = mass of each of the two particles r = distance between the two particles Weight. According to Eq. 1–2, any two particles or bodies have a mutual attractive (gravitational) force acting between them. In the case of a particle located at or near the surface of the earth, however, the only gravitational force having any sizable magnitude is that between the earth and the particle. Consequently, this force, termed the weight, will be the only gravitational force considered in our study of mechanics. From Eq. 1–2, we can develop an approximate expression for finding the weight W of a particle having a mass m1 = m. If we assume the earth to be a nonrotating sphere of constant density and having a mass m2 = Me, then if r is the distance between the earth’s center and the particle, we have mM e W = G r2 The astronaut’s weight is diminished since 2 she is far removed from the gravitational Letting g = GM e >r yields field of the earth. (© NikoNomad/ Shutterstock) W = mg (1–3) By comparison with F = ma, we can see that g is the acceleration due to gravity. Since it depends on r, then the weight of a body is not an absolute quantity. Instead, its magnitude is determined from where the measurement was made. For most engineering calculations, however, g is determined at sea level and at a latitude of 45°, which is considered the “standard location.” 1.3 Units of Measurement The four basic quantities—length, time, mass, and force—are not all independent from one another; in fact, they are related by Newton’s second law of motion, F = ma. Because of this, the units used to measure these quantities cannot all be selected arbitrarily. The equality F = ma is maintained only if three of the four units, called base units, are defined and the fourth unit is then derived from the equation.  8 C h a p t e r 1    G e n e r a l P r i n c i p l e s SI Units. The International System of units, abbreviated SI after the 1 French “Système International d’Unités,” is a modern version of the 1 kg metric system which has received worldwide recognition. As shown in Table 1–1, the SI system defines length in meters (m), time in seconds (s), and mass in kilograms (kg). The unit of force, called a newton (N), is 9.81 N derived from F = ma. Thus, 1 newton is equal to a force required to give (a) 1 kilogram of mass an acceleration of 1 m>s2 ( N = kg # m>s2 ). If the weight of a body located at the “standard location” is to be determined in newtons, then Eq. 1–3 must be applied. Here measurements give g = 9.806 65 m>s2; however, for calculations, the value g = 9.81 m>s2 will be used. Thus, W = mg ( g = 9.81 m>s2 ) (1–4) Therefore, a body of mass 1 kg has a weight of 9.81 N, a 2-kg body weighs 19.62 N, and so on, Fig. 1–2a. U.S. Customary. In the U.S. Customary system of units (FPS) length is measured in feet (ft), time in seconds (s), and force in pounds (lb), Table 1–1. The unit of mass, called a slug, is derived from F = ma. Hence, 1 slug is equal to the amount of matter accelerated at 1 ft>s2 when acted upon by a force of 1 lb ( slug = lb # s2 >ft ). Therefore, if the measurements are made at the “standard location,” 1 slug where g = 32.2 ft>s2, then from Eq. 1–3, W m = ( g = 32.2 ft>s2 ) (1–5) 32.2 lb g (b) And so a body weighing 32.2 lb has a mass of 1 slug, a 64.4-lb body has a Fig. 1–2 mass of 2 slugs, and so on, Fig. 1–2b. Table 1–1 Systems of Units Name Length Time Mass Force International meter second kilogram newton* System of Units N SI m s kg kg # m ¢ ≤ s2 U.S. Customary foot second slug* pound FPS lb # s2 ft s ¢ ≤ lb ft *Derived unit.  1.4 The International System of Units 9 Conversion of Units. Table 1–2 provides a set of direct conversion factors between FPS and SI units for the basic quantities. Also, in the 1 FPS system, recall that 1 ft = 12 in. (inches), 5280 ft = 1 mi (mile), 1000 lb = 1 kip (kilo-pound), and 2000 lb = 1 ton. Table 1–2 Conversion Factors Unit of Unit of Quantity Measurement (FPS) Equals Measurement (SI) Force lb 4.448 N Mass slug 14.59 kg Length ft 0.3048 m 1.4 The International System of Units The SI system of units is used extensively in this book since it is intended to become the worldwide standard for measurement. Therefore, we will now present some of the rules for its use and some of its terminology relevant to engineering mechanics. Prefixes. When a numerical quantity is either very large or very small, the units used to define its size may be modified by using a prefix. Some of the prefixes used in the SI system are shown in Table 1–3. Each represents a multiple or submultiple of a unit which, if applied successively, moves the decimal point of a numerical quantity to every third place.* For example, 4 000 000 N = 4 000 kN (kilo-newton) = 4 MN (mega-newton), or 0.005 m = 5 mm (milli-meter). Notice that the SI system does not include the multiple deca (10) or the submultiple centi (0.01), which form part of the metric system. Except for some volume and area measurements, the use of these prefixes is to be avoided in science and engineering. Table 1–3 Prefixes Exponential Form Prefix SI Symbol Multiple 1 000 000 000 109 giga G 1 000 000 106 mega M 1 000 103 kilo k Submultiple 0.001 10–3 milli m 0.000 001 10–6 micro m 0.000 000 001 10–9 nano n *The kilogram is the only base unit that is defined with a prefix.  10 C h a p t e r 1    G e n e r a l P r i n c i p l e s Rules for Use. Here are a few of the important rules that describe 1 the proper use of the various SI symbols: Quantities defined by several units which are multiples of one another are separated by a dot to avoid confusion with prefix notation, as indicated by N = kg # m>s2 = kg # m # s - 2. Also, m # s (meter-second), whereas ms (milli-second). The exponential power on a unit having a prefix refers to both the unit and its prefix. For example, mN2 = (mN) 2 = mN # mN. Likewise, mm2 represents (mm) 2 = mm # mm. With the exception of the base unit the kilogram, in general avoid the use of a prefix in the denominator of composite units. For example, do not write N>mm, but rather kN>m; also, m>mg should be written as Mm>kg. When performing calculations, represent the numbers in terms of their base or derived units by converting all prefixes to powers of 10. The final result should then be expressed using a single prefix. Also, after calculation, it is best to keep numerical values between 0.1 and 1000; otherwise, a suitable prefix should be chosen. For example, (50 kN)(60 nm) = 3 50 ( 103 ) N 4 3 60 ( 10 - 9 ) m 4 = 3000 ( 10 - 6 ) N # m = 3 ( 10 - 3 ) N # m = 3 mN # m 1.5 Numerical Calculations Numerical work in engineering practice is most often performed by using handheld calculators and computers. It is important, however, that the answers to any problem be reported with justifiable accuracy using appropriate significant figures. In this section we will discuss these topics together with some other important aspects involved in all engineering calculations. Dimensional Homogeneity. The terms of any equation used to describe a physical process must be dimensionally homogeneous; that is, each term must be expressed in the same units. Provided this is the case, Computers are often used in engineering for all the terms of an equation can then be combined if numerical values advanced design and analysis. (© Blaize are substituted for the variables. Consider, for example, the equation Pascall/Alamy) s = vt + 12 at 2 , where, in SI units, s is the position in meters, m, t is time in seconds, s, v is velocity in m>s and a is acceleration in m>s2. Regardless of how this equation is evaluated, it maintains its dimensional homogeneity. In the form stated, each of the three terms is expressed in meters 3 m, ( m>s ) s, ( m>s2 ) s2 4 or solving for a, a = 2s>t2 - 2v>t, the terms are each expressed in units of m>s2 3 m>s2, m>s2, ( m>s ) >s 4. Keep in mind that problems in mechanics always involve the solution of dimensionally homogeneous equations, and so this fact can then be used as a partial check for algebraic manipulations of an equation.  1.5 Numerical Calculations 11 Significant Figures. The number of significant figures contained in any number determines the accuracy of the number. For instance, the 1 number 4981 contains four significant figures. However, if zeros occur at the end of a whole number, it may be unclear as to how many significant figures the number represents. For example, 23 400 might have three (234), four (2340), or five (23 400) significant figures. To avoid these ambiguities, we will use engineering notation to report a result. This requires that numbers be rounded off to the appropriate number of significant digits and then expressed in multiples of (103), such as (103), (106), or (10–9). For instance, if 23 400 has five significant figures, it is written as 23.400(103), but if it has only three significant figures, it is written as 23.4(103). If zeros occur at the beginning of a number that is less than one, then the zeros are not significant. For example, 0.008 21 has three significant figures. Using engineering notation, this number is expressed as 8.21(10–3). Likewise, 0.000 582 can be expressed as 0.582(10–3) or 582(10–6). Rounding Off Numbers. Rounding off a number is necessary so that the accuracy of the result will be the same as that of the problem data. As a general rule, any numerical figure ending in a number greater than five is rounded up and a number less than five is not rounded up. The rules for rounding off numbers are best illustrated by examples. Suppose the number 3.5587 is to be rounded off to three significant figures. Because the fourth digit (8) is greater than 5, the third number is rounded up to 3.56. Likewise 0.5896 becomes 0.590 and 9.3866 becomes 9.39. If we round off 1.341 to three significant figures, because the fourth digit (1) is less than 5, then we get 1.34. Likewise 0.3762 becomes 0.376 and 9.871 becomes 9.87. There is a special case for any number that ends in a 5. As a general rule, if the digit preceding the 5 is an even number, then this digit is not rounded up. If the digit preceding the 5 is an odd number, then it is rounded up. For example, 75.25 rounded off to three significant digits becomes 75.2, 0.1275 becomes 0.128, and 0.2555 becomes 0.256. Calculations. When a sequence of calculations is performed, it is best to store the intermediate results in the calculator. In other words, do not round off calculations until expressing the final result. This procedure maintains precision throughout the series of steps to the final solution. In this text we will generally round off the answers to three significant figures since most of the data in engineering mechanics, such as geometry and loads, may be reliably measured to this accuracy.  12 C h a p t e r 1    G e n e r a l P r i n c i p l e s 1 1.6 General Procedure for Analysis Attending a lecture, reading this book, and studying the example problems helps, but the most effective way of learning the principles of engineering mechanics is to solve problems. To be successful at this, it is important to always present the work in a logical and orderly manner, as suggested by the following sequence of steps: Read the problem carefully and try to correlate the actual physical situation with the theory studied. Tabulate the problem data and draw to a large scale any necessary diagrams. When solving problems, do the work as Apply the relevant principles, generally in mathematical form. When neatly as possible. Being neat will writing any equations, be sure they are dimensionally homogeneous. stimulate clear and orderly thinking, and vice versa. (© Russell C. Hibbeler) Solve the necessary equations, and report the answer with no more than three significant figures. Study the answer with technical judgment and common sense to determine whether or not it seems reasonable. Important Points Statics is the study of bodies that are at rest or move with constant velocity. A particle has a mass but a size that can be neglected, and a rigid body does not deform under load. A force is considered as a “push” or “pull” of one body on another. Concentrated forces are assumed to act at a point on a body. Newton’s three laws of motion should be memorized. Mass is measure of a quantity of matter that does not change from one location to another. Weight refers to the gravitational attraction of the earth on a body or quantity of mass. Its magnitude depends upon the elevation at which the mass is located. In the SI system the unit of force, the newton, is a derived unit. The meter, second, and kilogram are base units. Prefixes G, M, k, m, m, and n are used to represent large and small numerical quantities. Their exponential size should be known, along with the rules for using the SI units. Perform numerical calculations with several significant figures, and then report the final answer to three significant figures. Algebraic manipulations of an equation can be checked in part by verifying that the equation remains dimensionally homogeneous. Know the rules for rounding off numbers.  1.6 General Procedure for Analysis 13 Example 1.1 1 Convert 2 km>h to m>s How many ft>s is this? Solution Since 1 km = 1000 m and 1 h = 3600 s, the factors of conversion are arranged in the following order, so that a cancellation of the units can be applied: 2 km 1000 m 1h 2 km>h = ¢ ≤¢ ≤ h km 3600 s 2000 m = = 0.556 m>s Ans. 3600 s From Table 1–2, 1 ft = 0.3048 m. Thus, 0.556 m 1 ft 0.556 m>s = a ba b s 0.3048 m = 1.82 ft>s Ans. NOTE: Remember to round off the final answer to three significant figures. Example 1.2 Convert the quantities 300 lb # s and 52 slug>ft3 to appropriate SI units. Solution Using Table 1–2, 1 lb = 4.448 N. 300 lb # s = 300 lb # s a 4.448 N b 1 lb = 1334.5 N # s = 1.33 kN # s Ans. Since 1 slug = 14.59 kg and 1 ft = 0.3048 m, then 52 slug 14.59 kg 3 1 ft 52 slug>ft3 = a b a b ft3 1 slug 0.3048 m      = 26.8 ( 103 ) kg>m3 = 26.8 Mg>m3  Ans.  14 C h a p t e r 1    G e n e r a l P r i n c i p l e s 1 Example 1.3 Evaluate each of the following and express with SI units having an appropriate prefix: (a) (50 mN)(6 GN), (b) (400 mm)(0.6 MN)2, (c) 45 MN3 >900 Gg. Solution First convert each number to base units, perform the indicated operations, then choose an appropriate prefix. Part (a) (50 mN)(6 GN) = 3 50 ( 10-3 ) N 4 3 6 ( 109 ) N 4 = 300 ( 106 ) N2 1 kN 1 kN = 300 ( 106 ) N2 a 3 ba 3 b 10 N 10 N = 300 kN2  Ans. NOTE: Keep in mind the convention kN2 = (kN) 2 = 106 N2. Part (b)    (400 mm)(0.6 MN)2 = 3 400 ( 10-3 ) m 4 3 0.6 ( 106 ) N 4 2     = 3 400 ( 10-3 ) m 4 3 0.36 ( 1012 ) N2 4     = 144 ( 109 ) m # N2     = 144 Gm # N2  Ans. We can also write 144 ( 109 ) m # N2 = 144 ( 109 ) m # N2 a 1 MN 1 MN ba 6 b 106 N 10 N = 0.144 m # MN2  Ans. Part (c) 45 MN3 45 ( 106 N ) 3 = 900 Gg 900 ( 106 ) kg = 50 ( 109 ) N3 >kg 1 kN 3 1 = 50 ( 109 ) N3 a b 103 N kg = 50 kN3 >kg  Ans.  Problems 15 Problems 1 The answers to all but every fourth problem (asterisk) are given in the back of the book. 1–1. What is the weight in newtons of an object that 1–13. The density (mass > volume) of aluminum is has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg? 5.26 slug>ft3. Determine its density in SI units. Use an appropriate prefix. 1–2. Represent each of the following combinations of units in the correct SI form: (a) kN>ms, (b) Mg>mN, and 1–14. Evaluate each of the following to three significant (c) MN>(kg · ms). figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52 800 ms)2, and 1–3. Represent each of the following combinations of (c) [548(106)]1>2 ms. units in the correct SI form: (a) Mg>ms, (b) N>mm, (c) mN>(kg · ms). 1–15. Using the SI system of units, show that Eq. 1–2 is a dimen­sionally homogeneous equation which gives F in 350 lb>ft3 to kN>m3, *1–4. Convert: (a) 200 lb · ft to N · m, (b) newtons. Determine to three significant figures the (c) 8 ft>h to mm>s. Express the result to three significant gravitational force acting between two spheres that are figures. Use an appropriate prefix. touching each other. The mass of each sphere is 200 kg and 1–5. Represent each of the following as a number between the radius is 300 mm. 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, *1–16. The pascal (Pa) is actually a very small unit of (b) 568(105) mm, and (c) 0.00563 mg. pressure. To show this, convert 1 Pa = 1 N>m2 to lb>ft2. 1–6. Round off the following numbers to three significant Atmosphere pressure at sea level is 14.7 lb>in2. How many figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg. pascals is this? 1–7. Represent each of the following quantities in the 1–17. Water has a density of 1.94 slug>ft3. What is the correct SI form using an appropriate prefix: (a) 0.000 431 kg, density expressed in SI units? Express the answer to three (b) 35.3 ( 103 ) N, (c) 0.005 32 km. significant figures. *1–8. Represent each of the following combinations of units 1–18. Evaluate each of the following to three significant in the correct SI form using an appropriate prefix: (a) Mg>mm, figures and express each answer in SI units using an (b) mN>ms, (c) mm # Mg. appropriate prefix: (a) 354 mg(45 km)>(0.0356 kN), (b) (0.004 53 Mg)(201 ms), (c) 435 MN>23.2 mm. 1–9. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: 1–19. A concrete column has a diameter of 350 mm and (a) m>ms, (b) mkm, (c) ks>mg, and (d) km # mN. a length of 2 m. If the density (mass>volume) of concrete is 2.45 Mg>m3, determine the weight of the column in pounds. 1–10. Represent each of the following combinations of units in the correct SI form: (a) GN # mm, (b) kg>mm, (c) N>ks2, *1–20. If a man weighs 155 lb on earth, specify (a) his and (d) kN>ms. mass in slugs, (b) his mass in kilograms, and (c) his weight in newtons. If the man is on the moon, where the acceleration 1–11. Represent each of the following with SI units having due to gravity is gm = 5.30 ft>s2, determine (d) his weight an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg. in pounds, and (e) his mass in kilograms. *1–12. Evaluate each of the following to three significant 1–21. Two particles have a mass of 8 kg and 12 kg, figures and express each answer in SI units using respectively. If they are 800 mm apart, determine the force an appropriate prefix: (a) (684 mm)>(43 ms), of gravity acting between them. Compare this result with (b) (28 ms)(0.0458 Mm)>(348 mg), (c) (2.68 mm)(426 Mg). the weight of each particle.  Chapter 2 (© Vasiliy Koval/Fotolia) This electric transmission tower is stabilized by cables that exert forces on the tower at their points of connection. In this chapter we will show how to express these forces as Cartesian vectors, and then determine their resultant. Force Vectors CHAPTER OBJECTIVES n To show how to add forces and resolve them into components using the Parallelogram Law. n To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction. n To introduce the dot product in order to use it to find the angle between two vectors or the projection of one vector onto another. 2.1 Scalars and Vectors Many physical quantities in engineering mechanics are measured using either scalars or vectors. Scalar. A scalar is any positive or negative physical quantity that can be completely specified by its magnitude. Examples of scalar quantities include length, mass, and time. Vector. A vector is any physical quantity that requires both a magnitude and a direction for its complete description. Examples of vectors encountered in statics are force, position, and moment. A vector is shown graphically by an arrow. The length of the arrow represents the Line of action magnitude of the vector, and the angle u between the vector and a fixed 1 Head axis defines the direction of its line of action. The head or tip of the arrow P A indicates the sense of direction of the vector, Fig. 2–1. In print, vector quantities are represented by boldface letters such as Tail 20 A, and the magnitude of a vector is italicized, A. For handwritten work, it O is often convenient to denote a vector quantity by simply drawing an S arrow above it, A. Fig. 2–1  18 C h a p t e r 2    F o r c e V e c t o r s 2.2 Vector Operations Multiplication and Division of a Vector by a Scalar. If a vector is multiplied by a positive scalar, its magnitude is increased by that 2 2A amount. Multiplying by a negative scalar will also change the directional sense of the vector. Graphic examples of these operations are shown A A in Fig. 2–2.  0.5 A Vector Addition. When adding two vectors together it is important to account for both their magnitudes and their directions. To do this we Scalar multiplication and division must use the parallelogram law of addition. To illustrate, the two Fig. 2–2 component vectors A and B in Fig. 2–3a are added to form a resultant vector R = A + B using the following procedure: First join the tails of the components at a point to make them concurrent, Fig. 2–3b. From the head of B, draw a line parallel to A. Draw another line from the head of A that is parallel to B. These two lines intersect at point P to form the adjacent sides of a parallelogram. The diagonal of this parallelogram that extends to P forms R, which then represents the resultant vector R = A + B, Fig. 2–3c. A A A R P B B B RAB Parallelogram law (a) (b) (c) Fig. 2–3 We can also add B to A, Fig. 2–4a, using the triangle rule, which is a special case of the parallelogram law, whereby vector B is added to vector A in a “head-to-tail” fashion, i.e., by connecting the head of A to the tail of B, Fig. 2–4b. The resultant R extends from the tail of A to the head of B. In a similar manner, R can also be obtained by adding A to B, Fig. 2–4c. By comparison, it is seen that vector addition is commutative; in other words, the vectors can be added in either order, i.e., R = A + B = B + A.  2.2 Vector Operations 19 A A B R R B A B 2 RAB RBA Triangle rule Triangle rule (a) (b) (c) Fig. 2–4 As a special case, if the two vectors A and B are collinear, i.e., both have the same line of action, the parallelogram law reduces to an algebraic or scalar addition R = A + B, as shown in Fig. 2–5. R A B RAB Addition of collinear vectors Fig. 2–5 Vector Subtraction. The resultant of the difference between two vectors A and B of the same type may be expressed as R = A - B = A + (-B) This vector sum is shown graphically in Fig. 2–6. Subtraction is therefore defined as a special case of addition, so the rules of vector addition also apply to vector subtraction. B A A or R¿ R¿ A B B Parallelogram law Triangle construction Vector subtraction Fig. 2–6  20 C h a p t e r 2    F o r c e V e c t o r s 2.3 Vector Addition of Forces Experimental evidence has shown that a force is a vector quantity since it has a specified magnitude, direction, and sense and it adds according to 2 the parallelogram law. Two common problems in statics involve either finding the resultant force, knowing its components, or resolving a known force into two components. We will now describe how each of these F2 problems is solved using the parallelogram law. F1 FR Finding a Resultant Force. The two component forces F1 and F2 acting on the pin in Fig. 2–7a can be added together to form the resultant force FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or using The parallelogram law must be used the triangle rule, Fig. 2–7c, we can apply the law of cosines or the law of to determine the resultant of the sines to the triangle in order to obtain the magnitude of the resultant two forces acting on the hook. force and its direction. (© Russell C. Hibbeler) F1 F1 F1 F2 FR FR F2 F2 FR  F1  F2 F v u (a) (b) (c) Fu Fig. 2–7 Fv Finding the Components of a Force. Sometimes it is necessary to resolve a force into two components in order to study its pulling or pushing effect in two specific directions. For example, in Fig. 2–8a, F is to be resolved into two components along the two members, defined by the u and v axes. In order to determine the magnitude of each component, a parallelogram is constructed first, by drawing lines starting from the tip of F, one line parallel to u, and the other line parallel to v. These lines then intersect with the v and u axes, forming a parallelogram. The force components Fu and Fv are then established by simply joining the tail of F to the intersection points on the u and v axes, Fig. 2–8b. This parallelogram Using the parallelogram law the supporting force F can be resolved into can then be reduced to a triangle, which represents the triangle rule, components acting along the u and v axes. Fig. 2–8c. From this, the law of sines can then be applied to determine the (© Russell C. Hibbeler) unknown magnitudes of the components.  2.3 Vector Addition of Forces 21 v v F F Fv F Fv u u 2 Fu Fu (a) (b) (c) Fig. 2–8 Addition of Several Forces. If more than two forces are to be F1  F2 FR added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force. For example, if three forces F2 F1, F2, F3 act at a point O, Fig. 2–9, the resultant of any two of the forces is found, say, F1 + F2—and then this resultant is added to the third force, yielding the resultant of all three forces; i.e., FR = (F1 + F2) + F3. Using F1 the parallelogram law to add more than two forces, as shown here, often F3 requires extensive geometric and trigonometric calculation to determine O the numerical values for the magnitude and direction of the resultant. Fig. 2–9 Instead, problems of this type are easily solved by using the “rectangular- component method,” which is explained in Sec. 2.4. FR F1  F2 F2 F1 F3 The resultant force FR on the hook requires the addition of F1 + F2, then this resultant is added to F3. (© Russell C. Hibbeler)  22 C h a p t e r 2    F o r c e V e c t o r s Important Points A scalar is a positive or negative number. 2 A vector is a quantity that has a magnitude, direction, and sense. Multiplication or division of a vector by a scalar will change the magnitude of the vector. The sense of the vector will change if the scalar is negative. As a special case, if the vectors are collinear, the resultant is formed by an algebraic or scalar addition. F1 FR Procedure for Analysis F2 Problems that involve the addition of two forces can be solved as (a) follows: Parallelogram Law. v Two “component” forces F1 and F2 in Fig. 2–10a add according to the parallelogram law, yielding a resultant force FR that forms the F diagonal of the parallelogram. u Fv If a force F is to be resolved into components along two axes Fu u and v, Fig. 2–10b, then start at the head of force F and construct lines parallel to the axes, thereby forming the parallelogram. The (b) sides of the parallelogram represent the components, Fu and Fv. Label all the known and unknown force magnitudes and the angles c on the sketch and identify the two unknowns as the magnitude and A B direction of FR, or the magnitudes of its components. b a C Trigonometry. Cosine law: Redraw a half portion of the parallelogram to illustrate the triangular head-to-tail addition of the components. C  A2  B2  2AB cos c Sine law: A  B  C From this triangle, the magnitude of the resultant force can be sin a sin b sin c determined using the law of cosines, and its direction is (c) determined from the law of sines. The magnitudes of two force components are determined from the law of sines. The formulas Fig. 2–10 are given in Fig. 2–10c.  2.3 Vector Addition of Forces 23 Example 2.1 The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force. 10 2 F2  150 N A 150 N F1  100 N 115 65 15 10 FR 360  2(65)  115 2 u 100 N 15 90  25  65 (a) (b) SOLUTION Parallelogram Law. The parallelogram is formed by drawing a line from the head of F1 that is parallel to F2, and another line from the head of F2 that is parallel to F1. The resultant force FR extends to where these lines intersect at point A, Fig. 2–11b. The two unknowns are the magnitude of FR and the angle u (theta). FR Trigonometry. From the parallelogram, the vector triangle is 150 N constructed, Fig. 2–11c. Using the law of cosines 115 FR = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115 u = 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N f 15 100 N = 213 N Ans. (c) Applying the law of sines to determine u, Fig. 2–11 150 N 212.6 N 150 N = sin u = (sin 115) sin u sin 115 212.6 N u = 39.8 Thus, the direction f (phi) of FR, measured from the horizontal, is f = 39.8 + 15.0 = 54.8 Ans. NOTE: The results seem reasonable, since Fig. 2–11b shows FR to have a magnitude larger than its components and a direction that is between them.  24 C h a p t e r 2   F o r c e V e c t o r s Example 2.2 Resolve the horizontal 600-lb force in Fig. 2–12a into components acting along the u and v axes and determine the magnitudes of these components. 2 u u B Fu Fu 30 30 30 30 Fv 120 120 30 30 30 A 600 lb 600 lb 600 lb 120 Fv C v v (a) (b) (c) Fig. 2–12 SOLUTION The parallelogram is constructed by extending a line from the head of the 600-lb force parallel to the v axis until it intersects the u axis at point B, Fig. 2–12b. The arrow from A to B represents Fu. Similarly, the line extended from the head of the 600-lb force drawn parallel to the u axis intersects the v axis at point C, which gives Fv. The vector addition using the triangle rule is shown in Fig. 2–12c. The two unknowns are the magnitudes of Fu and Fv. Applying the law of sines, Fu 600 lb = sin 120 sin 30 Fu = 1039 lb Ans. Fv 600 lb = sin 30 sin 30 Fv = 600 lb Ans. NOTE: The result for Fu shows that sometimes a component can have a greater magnitude than the resultant.  2.3 Vector Addition of Forces 25 Example 2.3 Determine the magnitude of the component force F in Fig. 2–13a and the magnitude of the resultant force FR if FR is directed along the positive y axis. 2 y y F 45 45 F FR FR 200 lb 45 45 75 F 45 30 60 60 30 200 lb 200 lb 30 (a) (b) (c) Fig. 2–13 SOLUTION The parallelogram law of addition is shown in Fig. 2–13b, and the triangle rule is shown in Fig. 2–13c. The magnitudes of FR and F are the two unknowns. They can be determined by applying the law of sines. F 200 lb = sin 60 sin 45 F = 245 lb Ans. FR 200 lb = sin 75 sin 45 FR = 273 lb Ans.  26 C h a p t e r 2   F o r c e V e c t o r s Example 2.4 It is required that the resultant force acting on the eyebolt in Fig. 2–14a be directed along the positive x axis and that F2 have a minimum magnitude. Determine this magnitude, the angle u, and the corresponding 2 resultant force. F1  800 N F1  800 N F2 F1  800 N F2 60 60 u 60 x x x FR FR u u  90 (b) (c) F2 (a) Fig. 2–14 SOLUTION The triangle rule for FR = F1 + F2 is shown in Fig. 2–14b. Since the magnitudes (lengths) of FR and F2 are not specified, then F2 can actually be any vector that has its head touching the line of action of FR, Fig. 2–14c. However, as shown, the magnitude of F2 is a minimum or the shortest length when its line of action is perpendicular to the line of action of FR, that is, when u = 90 Ans. Since the vector addition now forms the shaded right triangle, the two unknown magnitudes can be obtained by trigonometry. FR = (800 N)cos 60 = 400 N Ans. F2 = (800 N)sin 60 = 693 N Ans. It is strongly suggested that you test yourself on the solutions to these examples, by covering them over and then trying to draw the parallelogram law, and thinking about how the sine and cosine laws are used to determine the unknowns. Then before solving any of the problems, try to solve the Preliminary Problems and some of the Fundamental Problems given on the next pages. The solutions and answers to these are given in the back of the book. Doing this throughout the book will help immensely in developing your problem-solving skills.  2.3 Vector Addition of Forces 27 Preliminary Problems Partial solutions and answers to all Preliminary Problems are given in the back of the book. P2–1. In each case, construct the parallelogram law to P2–2. In each case, show how to resolve the force F into show FR = F1 + F2. Then establish the triangle rule, where 2 components acting along the u and v axes using the FR = F1 + F2. Label all known and unknown sides and parallelogram law. Then establish the triangle rule to show internal angles. FR = Fu + Fv. Label all known and unknown sides and interior angles. F  200 N v F1  200 N u F2  100 N 15 30 70 45 45 (a) (a) F  400 N 70 v F1  400 N 130 F2  500 N 120 (b) u (b) F1  450 N 30 20 v 40 F2  300 N u F  600 N (c) (c) Prob. P2–1 Prob. P2–2  28 C h a p t e r 2   F o r c e V e c t o r s FUNDAMENTAL PROBLEMS Partial solutions and answers to all Fundamental Problems are given in the back of the book. F2–1. Determine the magnitude of the resultant force F2–4. Resolve the 30-lb force into components along the 2 acting on the screw eye and its direction measured clockwise u and v axes, and determine the magnitude of each of these from the x axis. components. v 30 lb 15 x 45 30 60 u 2 kN 6 kN Prob. F2–1  Prob. F2–4 F2–2. Two forces act on the hook. Determine the magnitude of the resultant force. F2–5. The force F = 450 lb acts on the frame. Resolve this force into components acting along members AB and AC, and determine the magnitude of each component. 30 A C 45 30 450 lb 200 N 40 500 N Prob. F2–2 B F2–3. Determine the magnitude of the resultant force and Prob. F2–5 its direction measured counterclockwise from the positive F2–6. If force F is to have a component along the u axis of x axis. Fu = 6 kN, determine the magnitude of F and the magnitude of its component Fv along the v axis. y u 800 N F 45 105 x 30 v 600 N Prob. F2–3  Prob. F2–6  2.3 Vector Addition of Forces 29 PROBLEMS 2–1. If u = 60 and F = 450 N, determine the magnitude *2–4. The vertical force F acts downward at A on the two- of the resultant force and its direction, measured membered frame. Determine the magnitudes of the two counterclockwise from the positive x axis. components of F directed along the axes of AB and AC. 2 Set F = 500 N. 2–2. If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude 2–5. Solve Prob. 2–4 with F = 350 lb. of force F and its direction u. B y F 45 u x 15 A 700 N F 30 Probs. 2–1/2 C Probs. 2–4/5 2–3. Determine the magnitude of the resultant force 2–6. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise FR = F1 + F2 and its direction, measured clockwise from from the positive x axis. the positive u axis. 2–7. Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the y components. *2–8. Resolve the force F2 into components acting along F1  250 lb the u and v axes and determine the magnitudes of the components. 30 v 30 x 75 F1  4 kN 45 30 u F2  375 lb F2  6 kN Prob. 2–3 Probs. 2–6/7/8  30 C h a p t e r 2    F o r c e V e c t o r s 2–9. If the resultant force acting on the support is to be 2–13. The force acting on the gear tooth is F = 20 lb. 1200 lb, directed horizontally to the right, determine the Resolve this force into two components acting along the force F in rope A and the corresponding angle u. lines aa and bb. F 2–14. The component of force F acting along line aa is A required to be 30 lb. Determine the magnitude of F and its 2 component along line bb. u B 60 900 lb b F a Prob. 2–9 80 2–10. Determine the magnitude of the resultant force and its 60 direction, measured counterclockwise from the positive x axis. a y b 800 lb 40 Probs. 2–13/14 x 35 2–15. Force F acts on the frame such that its component acting along member AB is 650 lb, directed from B 500 lb towards A, and the component acting along member BC is Prob. 2–10 500 lb, directed from B towards C. Determine the magnitude of F and its direction u. Set f = 60. 2–11. The plate is subjected to the two forces at A and B as shown. If u = 60, determine the magnitude of the resultant *2–16. Force F acts on the frame such that its component of these two forces and its direction measured clockwise acting along member AB is 650 lb, directed from B from the horizontal. towards A. Determine the required angle f (0 … f … 45) and the component acting along member BC. Set F = 850 lb *2–12. Determine the angle u for connecting member A to and u = 30. the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force? FA  8 kN B u A u F f 45 40 A C B FB  6 kN Probs. 2–11/12 Probs. 2–15/16  2.3 Vector Addition of Forces 31 2–17. Determine the magnitude and direction of the 2–21. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first resultant force, FR measured counterclockwise from finding the resultant F = F1 + F2 and then forming the positive x axis. Solve the problem by first finding the FR = F + F3. resultant F = F1 + F2 and then forming FR = F + F3. 2–18. Determine the magnitude and direction of the 2–22. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first resultant force, measured counterclockwise from the positive 2 finding the resultant F = F2 + F3 and then forming x axis. Solve l by first finding the resultant F = F2 + F3 and FR = F + F1. then forming FR = F + F1. y y F1  30 N 5 F1  400 N 3 F2  200 N 4 F3  50 N 90º x x 150º 20 F3  300 N F2  20 N Probs. 2–21/22 Probs. 2–17/18 2–23. Two forces act on the screw eye. If F1 = 400 N and F2 = 600 N, determine the angle u (0 … u … 180) between 2–19. Determine the design angle u (0 … u … 90) for them, so that the resultant force has a magnitude of strut AB so that the 400-lb horizontal force has a component FR = 800 N. of 500 lb directed from A towards C. What is the component of force acting along member AB? Take f = 40. *2–24. Two forces F1 and F2 act on the screw eye. If their lines of action are at an angle u apart and the magnitude of *2–20. Determine the design angle f (0 … f … 90) each force is F1 = F2 = F, determine the magnitude of the between struts AB and AC so that the 400-lb horizontal resultant force FR and the angle between FR and F1. force has a component of 600 lb which acts up to the left, in the same direction as from B towards A. Take u = 30. F1 400 lb A u f B u C F2 Probs. 2–19/20 Probs. 2–23/24  32 C h a p t e r 2    F o r c e V e c t o r s 2–25. If F1 = 30 lb and F2 = 40 lb, determine the angles u *2–28. Determine the magnitude of force F so that the and f so that the resultant force is directed along the resultant FR of the three forces is as small as possible. What positive x axis and has a magnitude of FR = 60 lb. is the minimum magnitude of FR? 2 8 kN y F 30 F1 6 kN θ