Chapter 2. ( Vasiliy Koval/Fotolia)

Transcription

1 hapter ( Vasili Koval/otolia) This electric transmission tower is stabilied b cables that eert forces on the tower at their points of connection. In this chapter we will show how to epress these forces as artesian vectors, and then determine their resltant.

2 orce Vectors HPTER OJETIVES To show how to add forces and resolve them into components sing the Parallelogram Law. To epress force and position in artesian vector form and eplain how to determine the vector s magnitde and direction. To introdce the dot prodct in order to se it to find the angle between two vectors or the projection of one vector onto another..1 Scalars and Vectors Man phsical qantities in engineering mechanics are measred sing either scalars or vectors. Scalar. scalar is an positive or negative phsical qantit that can be completel specified b its magnitde. Eamples of scalar qantities inclde length, mass, and time. Vector. vector is an phsical qantit that reqires both a magnitde and a direction for its complete description. Eamples of vectors encontered in statics are force, position, and moment. vector is shown graphicall b an arrow. The length of the arrow represents the magnitde of the vector, and the angle between the vector and a fied ais defines the direction of its line of action. The head or tip of the arrow indicates the sense of direction of the vector, ig. 1. In print, vector qantities are represented b boldface letters sch as, and the magnitde of a vector is italicied,. or handwritten work, it is often convenient to denote a vector qantit b simpl drawing an S arrow above it,. 1 Tail O Line of action Head P 0 ig. 1

3 18 H P T E R O R E VETORS. Vector Operations 0. Scalar mltiplication and division ig. Mltiplication and Division of a Vector b a Scalar. If a vector is mltiplied b a positive scalar, its magnitde is increased b that amont. Mltipling b a negative scalar will also change the directional sense of the vector. Graphic eamples of these operations are shown in ig.. Vector ddition. When adding two vectors together it is important to accont for both their magnitdes and their directions. To do this we mst se the parallelogram law of addition. To illstrate, the two component vectors and in ig. 3a are added to form a resltant vector R = + sing the following procedre: irst join the tails of the components at a point to make them concrrent, ig. 3b. rom the head of, draw a line parallel to. Draw another line from the head of that is parallel to. These two lines intersect at point P to form the adjacent sides of a parallelogram. The diagonal of this parallelogram that etends to P forms R, which then represents the resltant vector R = +, ig. 3c. P R R Parallelogram law (c) (a) (b) ig. 3 We can also add to, ig. 4a, sing the triangle rle, which is a special case of the parallelogram law, whereb vector is added to vector in a head-to-tail fashion, i.e., b connecting the head of to the tail of, ig. 4b. The resltant R etends from the tail of to the head of. In a similar manner, R can also be obtained b adding to, ig. 4c. comparison, it is seen that vector addition is commtative; in other words, the vectors can be added in either order, i.e., R = + = +.

4 . VETOR OPERTIONS 19 R R R R (a) Triangle rle (b) Triangle rle (c) ig. 4 s a special case, if the two vectors and are collinear, i.e., both have the same line of action, the parallelogram law redces to an algebraic or scalar addition R = +, as shown in ig.. R R ddition of collinear vectors ig. Vector Sbtraction. The resltant of the difference between two vectors and of the same tpe ma be epressed as R = - = + (-) This vector sm is shown graphicall in ig. 6. Sbtraction is therefore defined as a special case of addition, so the rles of vector addition also appl to vector sbtraction. R or R Parallelogram law Triangle constrction Vector sbtraction ig. 6

5 0 H P T E R O R E VETORS.3 Vector ddition of orces 1 R The parallelogram law mst be sed to determine the resltant of the two forces acting on the hook. ( Rssell. Hibbeler) Eperimental evidence has shown that a force is a vector qantit since it has a specified magnitde, direction, and sense and it adds according to the parallelogram law. Two common problems in statics involve either finding the resltant force, knowing its components, or resolving a known force into two components. We will now describe how each of these problems is solved sing the parallelogram law. inding a Resltant orce. The two component forces 1 and acting on the pin in ig. 7a can be added together to form the resltant force R = 1 +, as shown in ig. 7b. rom this constrction, or sing the triangle rle, ig. 7c, we can appl the law of cosines or the law of sines to the triangle in order to obtain the magnitde of the resltant force and its direction R R v (a) (b) R 1 (c) v ig. 7 Using the parallelogram law the spporting force can be resolved into components acting along the and v aes. ( Rssell. Hibbeler) inding the omponents of a orce. Sometimes it is necessar to resolve a force into two components in order to std its plling or pshing effect in two specific directions. or eample, in ig. 8a, is to be resolved into two components along the two members, defined b the and v aes. In order to determine the magnitde of each component, a parallelogram is constrcted first, b drawing lines starting from the tip of, one line parallel to, and the other line parallel to v. These lines then intersect with the v and aes, forming a parallelogram. The force components and v are then established b simpl joining the tail of to the intersection points on the and v aes, ig. 8b. This parallelogram can then be redced to a triangle, which represents the triangle rle, ig. 8c. rom this, the law of sines can then be applied to determine the nknown magnitdes of the components.

6 1 3.3 VETOR DDITION O ORES 1 v v v v (a) (b) (c) ig. 8 ddition of Several orces. If more than two forces are to be added, sccessive applications of the parallelogram law can be carried ot in order to obtain the resltant force. or eample, if three forces 1,, 3 act at a point O, ig. 9, the resltant of an two of the forces is fond, sa, 1 + and then this resltant is added to the third force, ielding the resltant of all three forces; i.e., R = ( 1 + ) + 3. Using the parallelogram law to add more than two forces, as shown here, often reqires etensive geometric and trigonometric calclation to determine the nmerical vales for the magnitde and direction of the resltant. Instead, problems of this tpe are easil solved b sing the rectanglarcomponent method, which is eplained in Sec..4. O 1 1 ig. 9 3 R R 1 The resltant force R on the hook reqires the addition of 1 +, then this resltant is added to 3. ( Rssell. Hibbeler)

7 H P T E R O R E VETORS Important Points scalar is a positive or negative nmber. vector is a qantit that has a magnitde, direction, and sense. Mltiplication or division of a vector b a scalar will change the magnitde of the vector. The sense of the vector will change if the scalar is negative. s a special case, if the vectors are collinear, the resltant is formed b an algebraic or scalar addition. 1 R Procedre for nalsis (a) Problems that involve the addition of two forces can be solved as follows: v v (b) Parallelogram Law. Two component forces 1 and in ig. 10a add according to the parallelogram law, ielding a resltant force R that forms the diagonal of the parallelogram. If a force is to be resolved into components along two aes and v, ig. 10b, then start at the head of force and constrct lines parallel to the aes, thereb forming the parallelogram. The sides of the parallelogram represent the components, and v. b c (c) osine law: cos c Sine law: sin a sin b sin c ig. 10 a Label all the known and nknown force magnitdes and the angles on the sketch and identif the two nknowns as the magnitde and direction of R, or the magnitdes of its components. Trigonometr. Redraw a half portion of the parallelogram to illstrate the trianglar head-to-tail addition of the components. rom this triangle, the magnitde of the resltant force can be determined sing the law of cosines, and its direction is determined from the law of sines. The magnitdes of two force components are determined from the law of sines. The formlas are given in ig. 10c.

8 .3 VETOR DDITION O ORES 3 EXMPLE.1 The screw ee in ig. 11a is sbjected to two forces, 1 and. Determine the magnitde and direction of the resltant force N 10 N N R 360 (6 ) N 1 (a) 90 6 (b) SOLUTION Parallelogram Law. The parallelogram is formed b drawing a line from the head of 1 that is parallel to, and another line from the head of that is parallel to 1. The resltant force R etends to where these lines intersect at point, ig. 11b. The two nknowns are the magnitde of R and the angle (theta). Trigonometr. rom the parallelogram, the vector triangle is constrcted, ig. 11c. Using the law of cosines R 10 N R = (100 N) + (10 N) - (100 N)(10 N) cos 11 = (-0.46) = 1.6 N = 13 N ns. ppling the law of sines to determine, 10 N sin = 1.6 N sin = 10 N (sin 11 ) sin N = 39.8 Ths, the direction f (phi) of R, measred from the horiontal, is 11 f N (c) ig. 11 f = = 4.8 ns. NOTE: The reslts seem reasonable, since ig. 11b shows R to have a magnitde larger than its components and a direction that is between them.

9 4 H P T E R O R E VETORS EXMPLE. Resolve the horiontal 600-lb force in ig. 1a into components acting along the and v aes and determine the magnitdes of these components. 600 lb v lb lb v v (a) v (b) (c) ig. 1 SOLUTION The parallelogram is constrcted b etending a line from the head of the 600-lb force parallel to the v ais ntil it intersects the ais at point, ig. 1b. The arrow from to represents. Similarl, the line etended from the head of the 600-lb force drawn parallel to the ais intersects the v ais at point, which gives v. The vector addition sing the triangle rle is shown in ig. 1c. The two nknowns are the magnitdes of and v. ppling the law of sines, sin 10 v sin = 600 lb sin = 1039 lb = 600 lb sin v = 600 lb ns. ns. NOTE: The reslt for shows that sometimes a component can have a greater magnitde than the resltant.

10 .3 VETOR DDITION O ORES EXMPLE.3 Determine the magnitde of the component force in ig. 13a and the magnitde of the resltant force R if R is directed along the positive ais. 00 lb R lb R lb (a) (b) ig. 13 (c) SOLUTION The parallelogram law of addition is shown in ig. 13b, and the triangle rle is shown in ig. 13c. The magnitdes of R and are the two nknowns. The can be determined b appling the law of sines. sin 60 = 00 lb sin = 4 lb ns. R sin 7 = 00 lb sin R = 73 lb ns.

11 6 H P T E R O R E VETORS EXMPLE.4 It is reqired that the resltant force acting on the eebolt in ig. 14a be directed along the positive ais and that have a minimm magnitde. Determine this magnitde, the angle, and the corresponding resltant force N N N R R 90 (b) (c) (a) ig. 14 SOLUTION The triangle rle for R = 1 + is shown in ig. 14b. Since the magnitdes (lengths) of R and are not specified, then can actall be an vector that has its head toching the line of action of R, ig. 14c. However, as shown, the magnitde of is a minimm or the shortest length when its line of action is perpendiclar to the line of action of R, that is, when = 90 ns. Since the vector addition now forms the shaded right triangle, the two nknown magnitdes can be obtained b trigonometr. R = (800 N)cos 60 = 400 N = (800 N)sin 60 = 693 N ns. ns. It is strongl sggested that o test orself on the soltions to these eamples, b covering them over and then tring to draw the parallelogram law, and thinking abot how the sine and cosine laws are sed to determine the nknowns. Then before solving an of the problems, tr to solve the Preliminar Problems and some of the ndamental Problems given on the net pages. The soltions and answers to these are given in the back of the book. Doing this throghot the book will help immensel in developing or problem-solving skills.

12 .3 VETOR DDITION O ORES 7 PRELIMINRY PROLEMS Partial soltions and answers to all Preliminar Problems are given in the back of the book. P 1. In each case, constrct the parallelogram law to show R = 1 +. Then establish the triangle rle, where R = 1 +. Label all known and nknown sides and internal angles. P. In each case, show how to resolve the force into components acting along the and v aes sing the parallelogram law. Then establish the triangle rle to show R = + v. Label all known and nknown sides and interior angles N 100 N v 00 N (a) (a) 400 N N v 00 N 10 (b) (b) 1 40 N 0 40 v 300 N 600 N (c) Prob. P 1 (c) Prob. P

13 8 H P T E R O R E VETORS UNDMENTL PROLEMS Partial soltions and answers to all ndamental Problems are given in the back of the book. 1. Determine the magnitde of the resltant force acting on the screw ee and its direction measred clockwise from the ais. 4. Resolve the 30-lb force into components along the and v aes, and determine the magnitde of each of these components. v 1 30 lb 60 kn 6 kn Prob. 1. Two forces act on the hook. Determine the magnitde of the resltant force. Prob. 4. The force = 40 lb acts on the frame. Resolve this force into components acting along members and, and determine the magnitde of each component. 40 lb N 00 N Prob. 3. Determine the magnitde of the resltant force and its direction measred conterclockwise from the positive ais. Prob. 6. If force is to have a component along the ais of = 6 kn, determine the magnitde of and the magnitde of its component v along the v ais. 800 N N v Prob. 3 Prob. 6

14 .3 VETOR DDITION O ORES 9 PROLEMS 1. If = 60 and = 40 N, determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais.. If the magnitde of the resltant force is to be 00 N, directed along the positive ais, determine the magnitde of force and its direction. * 4. The vertical force acts downward at on the twomembered frame. Determine the magnitdes of the two components of directed along the aes of and. Set = 00 N.. Solve Prob. 4 with = 30 lb N Probs. 1/ Probs. 4/ 3. Determine the magnitde of the resltant force R = 1 + and its direction, measred conterclockwise from the positive ais. 6. Determine the magnitde of the resltant force R = 1 + and its direction, measred clockwise from the positive ais. 1 0 lb 7. Resolve the force 1 into components acting along the and v aes and determine the magnitdes of the components. * 8. Resolve the force into components acting along the and v aes and determine the magnitdes of the components. v kn 37 lb 6 kn Prob. 3 Probs. 6/7/8

15 30 H P T E R O R E VETORS 9. If the resltant force acting on the spport is to be 100 lb, directed horiontall to the right, determine the force in rope and the corresponding angle. 13. The force acting on the gear tooth is = 0 lb. Resolve this force into two components acting along the lines aa and bb. 14. The component of force acting along line aa is reqired to be 30 lb. Determine the magnitde of and its component along line bb. 60 Prob lb 10. Determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais. 800 lb 60 a 80 b b a 40 Probs. 13/14 3 Prob lb 11. The plate is sbjected to the two forces at and as shown. If = 60, determine the magnitde of the resltant of these two forces and its direction measred clockwise from the horiontal. * 1. Determine the angle for connecting member to the plate so that the resltant force of and is directed horiontall to the right. lso, what is the magnitde of the resltant force? 1. orce acts on the frame sch that its component acting along member is 60 lb, directed from towards, and the component acting along member is 00 lb, directed from towards. Determine the magnitde of and its direction. Set f = 60. * 16. orce acts on the frame sch that its component acting along member is 60 lb, directed from towards. Determine the reqired angle f (0 f ) and the component acting along member. Set = 80 lb and =. 8 kn 40 6 kn f Probs. 11/1 Probs. 1/16

16 .3 VETOR DDITION O ORES Determine the magnitde and direction of the resltant R = of the three forces b first finding the resltant = 1 + and then forming R = Determine the magnitde and direction of the resltant R = of the three forces b first finding the resltant = + 3 and then forming R = Determine the magnitde and direction of the resltant force, R measred conterclockwise from the positive ais. Solve the problem b first finding the resltant = 1 + and then forming R = Determine the magnitde and direction of the resltant force, measred conterclockwise from the positive ais. Solve l b first finding the resltant = + 3 and then forming R = N N N 00 N 90º 10º 0 N N Probs. 17/18 Probs. 1/ 19. Determine the design angle (0 90 ) for strt so that the 400-lb horiontal force has a component of 00 lb directed from towards. What is the component of force acting along member? Take f = 40. * 0. Determine the design angle f (0 f 90 ) between strts and so that the 400-lb horiontal force has a component of 600 lb which acts p to the left, in the same direction as from towards. Take =. 3. Two forces act on the screw ee. If 1 = 400 N and = 600 N, determine the angle (0 180 ) between them, so that the resltant force has a magnitde of R = 800 N. * 4. Two forces 1 and act on the screw ee. If their lines of action are at an angle apart and the magnitde of each force is 1 = =, determine the magnitde of the resltant force R and the angle between R and lb f Probs. 19/0 Probs. 3/4

17 3 H P T E R O R E VETORS. If 1 = 30 lb and = 40 lb, determine the angles and f so that the resltant force is directed along the positive ais and has a magnitde of R = 60 lb. * 8. Determine the magnitde of force so that the resltant R of the three forces is as small as possible. What is the minimm magnitde of R? 8 kn 1 6 kn θ φ Prob. 8 Prob. 6. Determine the magnitde and direction of so that the resltant force is directed along the positive ais and has a magnitde of 10 N. 7. Determine the magnitde and direction, measred conterclockwise from the positive ais, of the resltant force acting on the ring at O, if = 70 N and =. 9. If the resltant force of the two tgboats is 3 kn, directed along the positive ais, determine the reqired magnitde of force and its direction. 30. If = 3 kn and =, determine the magnitde of the resltant force of the two tgboats and its direction measred clockwise form the positive ais. 31. If the resltant force of the two tgboats is reqired to be directed towards the positive ais, and is to be a minimm, determine the magnitde of R and and the angle. kn O = 800 N Probs. 6/7 Probs. 9/30/31

18 .4 DDITION O SYSTEM O OPLNR ORES 33.4 ddition of a Sstem of oplanar orces When a force is resolved into two components along the and aes, the components are then called rectanglar components. or analtical work we can represent these components in one of two was, sing either scalar or artesian vector notation. Scalar Notation. The rectanglar components of force shown in ig. 1a are fond sing the parallelogram law, so that = +. ecase these components form a right triangle, the can be determined from = cos and = sin Instead of sing the angle, however, the direction of can also be defined sing a small slope triangle, as in the eample shown in ig. 1b. Since this triangle and the larger shaded triangle are similar, the proportional length of the sides gives (a) or = a c and or = a a c b = b c b c a (b) ig. 1 = - a b c b Here the component is a negative scalar since is directed along the negative ais. It is important to keep in mind that this positive and negative scalar notation is to be sed onl for comptational prposes, not for graphical representations in figres. Throghot the book, the head of a vector arrow in an figre indicates the sense of the vector graphicall; algebraic signs are not sed for this prpose. Ths, the vectors in igs. 1a and 1b are designated b sing boldface (vector) notation.* Whenever italic smbols are written near vector arrows in figres, the indicate the magnitde of the vector, which is alwas a positive qantit. *Negative signs are sed onl in figres with boldface notation when showing eqal bt opposite pairs of vectors, as in ig..

19 34 H P T E R O R E VETORS j ig. 16 (a) 3 i 1 artesian Vector Notation. It is also possible to represent the and components of a force in terms of artesian nit vectors i and j. The are called nit vectors becase the have a dimensionless magnitde of 1, and so the can be sed to designate the directions of the and aes, respectivel, ig. 16.* Since the magnitde of each component of is alwas a positive qantit, which is represented b the (positive) scalars and, then we can epress as a artesian vector, = i + j oplanar orce Resltants. We can se either of the two methods jst described to determine the resltant of several coplanar forces, i.e., forces that all lie in the same plane. To do this, each force is first resolved into its and components, and then the respective components are added sing scalar algebra since the are collinear. The resltant force is then formed b adding the resltant components sing the parallelogram law. or eample, consider the three concrrent forces in ig. 17a, which have and components shown in ig. 17b. Using artesian vector notation, each force is first represented as a artesian vector, i.e., 1 1 = 1 i + 1 j = - i + j 3 = 3 i - 3 j 3 (b) ig The vector resltant is therefore R = = 1 i + 1 j - i + j + 3 i - 3 j = ( ) i + ( ) j = ( R )i + ( R )j If scalar notation is sed, then indicating the positive directions of components along the and aes with smbolic arrows, we have 1 +h ( R) = c ( R ) = These are the same reslts as the i and j components of R determined above. The resltant force of the for cable forces acting on the post can be determined b adding algebraicall the separate and components of each cable force. This resltant R prodces the same plling effect on the post as all for cables. ( Rssell. Hibbeler) *or handwritten work, nit vectors are sall indicated sing a circmfle, e.g., î and ĵ. lso, realie that and in ig. 16 represent the magnitdes of the components, which are alwas positive scalars. The directions are defined b i and j. If instead we sed scalar notation, then and cold be positive or negative scalars, since the wold accont for both the magnitde and direction of the components.

20 .4 DDITION O SYSTEM O OPLNR ORES 3 We can represent the components of the resltant force of an nmber of coplanar forces smbolicall b the algebraic sm of the and components of all the forces, i.e., ( R ) = ( R ) = ( 1) Once these components are determined, the ma be sketched along the and aes with their proper sense of direction, and the resltant force can be determined from vector addition, as shown in ig. 17c. rom this sketch, the magnitde of R is then fond from the Pthagorean theorem; that is, ( R ) R R = ( R ) + ( R ) ( R ) lso, the angle, which specifies the direction of the resltant force, is determined from trigonometr: (c) ( = tan -1 R ) ( R ) ig. 17 (cont.) The above concepts are illstrated nmericall in the eamples which follow. Important Points The resltant of several coplanar forces can easil be determined if an, coordinate sstem is established and the forces are resolved along the aes. The direction of each force is specified b the angle its line of action makes with one of the aes, or b a slope triangle. The orientation of the and aes is arbitrar, and their positive direction can be specified b the artesian nit vectors i and j. The and components of the resltant force are simpl the algebraic addition of the components of all the coplanar forces. The magnitde of the resltant force is determined from the Pthagorean theorem, and when the resltant components are sketched on the and aes, ig. 17c, the direction can be determined from trigonometr.

21 ( 36 H P T E R O R E VETORS EXMPLE N 13 1 Determine the and components of 1 and acting on the boom shown in ig. 18a. Epress each force as a artesian vector. SOLUTION Scalar Notation. the parallelogram law, 1 is resolved into and components, ig. 18b. Since 1 acts in the - direction, and 1 acts in the + direction, we have 1 = -00 sin N = -100 N = 100 N d ns. 60 N 1 = 00 cos N = 173 N = 173 Nc ns N (a) 1 00 cos N The force is resolved into its and components, as shown in ig. 18c. Here the slope of the line of action for the force is indicated. rom this slope triangle we cold obtain the angle, e.g., = tan , and then proceed to determine the magnitdes of the components in the same manner as for 1. The easier method, however, consists of sing proportional parts of similar triangles, i.e., 1 00 sin N (b) 60 N = 1 13 Similarl, = 60 Na 1 13 b = 40 N = 60 Na 13 b = 100 N ( N 60 ( 1 13 N N (c) ( Notice how the magnitde of the horiontal component,, was obtained b mltipling the force magnitde b the ratio of the horiontal leg of the slope triangle divided b the hpotense; whereas the magnitde of the vertical component,, was obtained b mltipling the force magnitde b the ratio of the vertical leg divided b the hpotense. Hence, sing scalar notation to represent these components, we have = 40 N = 40 N S = -100 N = 100 NT ns. ns. ig. 18 artesian Vector Notation. Having determined the magnitdes and directions of the components of each force, we can epress each force as a artesian vector. 1 = -100i + 173j6N = 40i - 100j6N ns. ns.

22 .4 DDITION O SYSTEM O OPLNR ORES 37 EXMPLE.6 The link in ig. 19a is sbjected to two forces 1 and. Determine the magnitde and direction of the resltant force. SOLUTION I Scalar Notation. irst we resolve each force into its and components, ig. 19b, then we sm these components algebraicall. 400 N N S + ( R ) = ; ( R ) = 600 cos N sin N = 36.8 N S (a) + c( R ) = ; ( R ) = 600 sin N cos N = 8.8 Nc The resltant force, shown in ig. 19c, has a magnitde of R = (36.8 N) + (8.8 N) = 69 N ns. 400 N N rom the vector addition, = tan -1 a 8.8 N b = 67.9 ns N (b) SOLUTION II artesian Vector Notation. rom ig. 19b, each force is first epressed as a artesian vector. Then, 1 = 600 cos i sin j6n = -400 sin i cos j6n R = 1 + = (600 cos N sin N)i + (600 sin N cos N)j 8.8 N R 36.8 N = 36.8i + 8.8j6N (c) The magnitde and direction of R are determined in the same manner as before. ig. 19 NOTE: omparing the two methods of soltion, notice that the se of scalar notation is more efficient since the components can be fond directl, withot first having to epress each force as a artesian vector before adding the components. Later, however, we will show that artesian vector analsis is ver beneficial for solving three-dimensional problems.

23 38 H P T E R O R E VETORS EXMPLE.7 The end of the boom O in ig. 0a is sbjected to three concrrent and coplanar forces. Determine the magnitde and direction of the resltant force N N N (a) SOLUTION Each force is resolved into its and components, ig. 0b. Smming the components, we have 00 N 3 4 (b) 0 N 400 N S + ( R ) = ; ( R ) = -400 N + 0 sin N N = N = 383. N d The negative sign indicates that R acts to the left, i.e., in the negative direction, as noted b the small arrow. Obviosl, this occrs becase 1 and 3 in ig. 0b contribte a greater pll to the left than which plls to the right. Smming the components ields + c( R ) = ; ( R ) = 0 cos N N = 96.8 Nc R 96.8 N The resltant force, shown in ig. 0c, has a magnitde of R = (-383. N) + (96.8 N) = 48 N ns N rom the vector addition in ig. 0c, the direction angle is = tan -1 a 96.8 b = 37.8 ns (c) ig. 0 NOTE: pplication of this method is more convenient, compared to sing two applications of the parallelogram law, first to add 1 and then adding 3 to this resltant.

24 .4 DDITION O SYSTEM O OPLNR ORES 39 UNDMENTL PROLEMS 7. Resolve each force acting on the post into its and components N 40 N N If the resltant force acting on the bracket is to be 70 N directed along the positive ais, determine the magnitde of and its direction N Prob Determine the magnitde and direction of the resltant force. 0 N N 300 N Prob N 11. If the magnitde of the resltant force acting on the bracket is to be 80 lb directed along the ais, determine the magnitde of and its direction. 0 lb Prob Determine the magnitde of the resltant force acting on the corbel and its direction measred conterclockwise from the ais. 90 lb 3 4 Prob Determine the magnitde of the resltant force and its direction measred conterclockwise from the positive ais lb lb lb 1 1 kn 0 kn kn 3 Prob. 9 Prob. 1

25 40 H P T E R O R E VETORS PROLEMS * 3. Determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais. 34. Resolve 1 and into their and components. 3. Determine the magnitde of the resltant force and its direction measred conterclockwise from the positive ais N N 10 N Prob. 3 0 N Probs. 34/3 33. Determine the magnitde of the resltant force and its direction, measred clockwise from the positive ais. * 36. Resolve each force acting on the gsset plate into its and components, and epress each force as a artesian vector. 37. Determine the magnitde of the resltant force acting on the plate and its direction, measred conterclockwise from the positive ais. 400 N 800 N 3 60 N N N Prob. 33 Probs. 36/37

26 .4 DDITION O SYSTEM O OPLNR ORES Epress each of the three forces acting on the spport in artesian vector form and determine the magnitde of the resltant force and its direction, measred clockwise from positive ais. 1 0 N 4 80 N N Prob Determine the and components of 1 and. * 40. Determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais Epress 1,, and 3 as artesian vectors. 43. Determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais N 6 N 3 4 Probs. 4/ N * 44. Determine the magnitde of the resltant force and its direction, measred clockwise from the positive ais. 40 lb 1 00 N N 30 lb Probs. 39/ Determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais. 3 8 kn 60 kn lb Prob Determine the magnitde and direction of the resltant force R. Epress the reslt in terms of the magnitdes of the components 1 and and the angle f. 1 4 kn 1 R f Prob. 41 Prob. 4

27 4 H P T E R O R E VETORS 46. Determine the magnitde and orientation of so that the resltant force is directed along the positive ais and has a magnitde of 100 N. 47. Determine the magnitde and orientation, measred conterclockwise from the positive ais, of the resltant force acting on the bracket, if = 600 N and = Epress 1,, and 3 as artesian vectors. 1. Determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais. 700 N 6 kn kn 3 36 kn Probs. 46/47 Probs. 0/1 * 48. Three forces act on the bracket. Determine the magnitde and direction of 1 so that the resltant force is directed along the positive ais and has a magnitde of 800 N. 49. If 1 = 300 N and = 10, determine the magnitde and direction, measred conterclockwise from the positive ais, of the resltant force acting on the bracket. *. Determine the and components of each force acting on the gsset plate of a bridge trss. Show that the resltant force is ero. 00 N 6 kn N kn kn kn Probs. 48/49 Prob.

28 .4 DDITION O SYSTEM O OPLNR ORES Epress 1 and as artesian vectors. 4. Determine the magnitde of the resltant force and its direction measred conterclockwise from the positive ais. * 6. If the magnitde of the resltant force acting on the bracket is to be 40 N directed along the positive ais, determine the magnitde of 1 and its direction f. 7. If the resltant force acting on the bracket is reqired to be a minimm, determine the magnitdes of 1 and the resltant force. Set f = kn f N N Probs. 6/ kn Probs. 3/4. Determine the magnitde of force so that the resltant force of the three forces is as small as possible. What is the magnitde of the resltant force? 8. Three forces act on the bracket. Determine the magnitde and direction of so that the resltant force is directed along the positive ais and has a magnitde of 8 kn. 9. If = kn and = 30, determine the magnitde of the resltant force and its direction, measred conterclockwise from the positive ais. 4 kn 14 kn 1 ' 8 kn 6 kn Prob. Probs. 8/9

29 44 H P T E R O R E VETORS. artesian Vectors The operations of vector algebra, when applied to solving problems in three dimensions, are greatl simplified if the vectors are first represented in artesian vector form. In this section we will present a general method for doing this; then in the net section we will se this method for finding the resltant force of a sstem of concrrent forces. ig. 1 ( Rssell. Hibbeler) Right-Handed oordinate Sstem. We will se a righthanded coordinate sstem to develop the theor of vector algebra that follows. rectanglar coordinate sstem is said to be right-handed if the thmb of the right hand points in the direction of the positive ais when the right-hand fingers are crled abot this ais and directed from the positive towards the positive ais, ig. 1. Rectanglar omponents of a Vector. vector ma have one, two, or three rectanglar components along the,, coordinate aes, depending on how the vector is oriented relative to the aes. In general, thogh, when is directed within an octant of the,, frame, ig., then b two sccessive applications of the parallelogram law, we ma resolve the vector into components as = + and then = +. ombining these eqations, to eliminate, is represented b the vector sm of its three rectanglar components, = + + ( ) ig. artesian Unit Vectors. In three dimensions, the set of artesian nit vectors, i, j, k, is sed to designate the directions of the,, aes, respectivel. s stated in Sec. 4, the sense (or arrowhead) of these vectors will be represented analticall b a pls or mins sign, depending on whether the are directed along the positive or negative,, or aes. The positive artesian nit vectors are shown in ig. 3. k i j ig. 3

30 . RTESIN VETORS 4 artesian Vector Representation. Since the three components of in Eq. act in the positive i, j, and k directions, ig. 4, we can write in artesian vector form as k = i + j + k ( 3) There is a distinct advantage to writing vectors in this manner. Separating the magnitde and direction of each component vector will simplif the operations of vector algebra, particlarl in three dimensions. i i k j j Magnitde of a artesian Vector. It is alwas possible to obtain the magnitde of provided it is epressed in artesian vector form. s shown in ig., from the ble right triangle, = +, and from the gra right triangle, = +. ombining these eqations to eliminate ields = + + ( 4) ig. 4 k Hence, the magnitde of is eqal to the positive sqare root of the sm of the sqares of its components. i j oordinate Direction ngles. We will define the direction of b the coordinate direction angles a (alpha), b (beta), and g (gamma), measred between the tail of and the positive,, aes provided the are located at the tail of, ig. 6. Note that regardless of where is directed, each of these angles will be between 0 and 180. To determine a, b, and g, consider the projection of onto the,, aes, ig. 7. Referring to the colored right triangles shown in the figre, we have cos a = cos b = cos g = ( ) ig. k g b a j These nmbers are known as the direction cosines of. Once the have been obtained, the coordinate direction angles a, b, g can then be determined from the inverse cosines. i ig. 6

31 46 H P T E R O R E VETORS n eas wa of obtaining these direction cosines is to form a nit vector in the direction of, ig. 6. If is epressed in artesian vector form, = i + j + k, then will have a magnitde of one and be dimensionless provided is divided b its magnitde, i.e., 90 a g b = = i + j + k ( 6) where = + +. comparison with Eqs., it is seen that the i, j, k components of represent the direction cosines of, i.e., = cos a i + cos b j + cos g k ( 7) ig. 7 Since the magnitde of a vector is eqal to the positive sqare root of the sm of the sqares of the magnitdes of its components, and has a magnitde of one, then from the above eqation an important relation among the direction cosines can be formlated as cos a + cos b + cos g = 1 ( 8) Here we can see that if onl two of the coordinate angles are known, the third angle can be fond sing this eqation. inall, if the magnitde and coordinate direction angles of are known, then ma be epressed in artesian vector form as = = cos a i + cos b j + cos g k ( 9) = i + j + k f Transverse and mth ngles. Sometimes, the direction of can be specified sing two angles, namel, a transverse angle and an amth angle f (phi), sch as shown in ig. 8. The components of can then be determined b appling trigonometr first to the light ble right triangle, which ields O = cos f and = sin f ig. 8 Now appling trigonometr to the dark ble right triangle, = cos = sin f cos = sin = sin f sin

32 .6 DDITION O RTESIN VETORS 47 Therefore written in artesian vector form becomes = sin f cos i + sin f sin j + cos f k Yo shold not memorie this eqation, rather it is important to nderstand how the components were determined sing trigonometr..6 ddition of artesian Vectors The addition (or sbtraction) of two or more vectors is greatl simplified if the vectors are epressed in terms of their artesian components. or eample, if = i + j + k and = i + j + k, ig. 9, then the resltant vector, R, has components which are the scalar sms of the i, j, k components of and, i.e., R = + = ( + )i + ( + )j + ( + )k ( )k R ( )j If this is generalied and applied to a sstem of several concrrent forces, then the force resltant is the vector sm of all the forces in the sstem and can be written as R = = i + j + k ( 10) ( )i ig. 9 Here,, and represent the algebraic sms of the respective,, or i, j, k components of each force in the sstem. Important Points artesian vector has i, j, k components along the,, aes. If is known, its magnitde is defined b = + +. The direction of a artesian vector can be defined b the three angles a, b, g, measred from the positive,, aes to the tail of the vector. To find these angles formlate a nit vector in the direction of, i.e., = >, and determine the inverse cosines of its components. Onl two of these angles are independent of one another; the third angle is fond from cos a + cos b + cos g = 1. The direction of a artesian vector can also be specified sing a transverse angle and aimth angle f. artesian vector analsis provides a convenient method for finding both the resltant force and its components in three dimensions. ( Rssell. Hibbeler)

33 48 H P T E R O R E VETORS EXMPLE.8 Epress the force shown in ig. 30a as a artesian vector. 100 lb 60 SOLUTION The angles of 60 and 4 defining the direction of are not coordinate direction angles. Two sccessive applications of the parallelogram law are needed to resolve into its,, components. irst = +, then = +, ig. 30b. trigonometr, the magnitdes of the components are (a) = 100 sin 60 lb = 86.6 lb = 100 cos 60 lb = 0 lb = cos = 0 cos lb = 3.4 lb = sin = 0 sin lb = 3.4 lb Realiing that has a direction defined b j, we have 100 lb = 3.4i - 3.4j k6 lb ns. 60 To show that the magnitde of this vector is indeed 100 lb, appl Eq. 4, = + + = (3.4) + (3.4) + (86.6) = 100 lb (b) If needed, the coordinate direction angles of can be determined from the components of the nit vector acting in the direction of. Hence, = = i + j + k 100 lb = i j k so that = 0.34i j k a = cos -1 (0.34) = 69.3 b = cos -1 (-0.34) = 111 (c) g = cos -1 (0.866) = 30.0 ig. 30 These reslts are shown in ig. 30c.

34 .6 DDITION O RTESIN VETORS 49 EXMPLE.9 Two forces act on the hook shown in ig. 31a. Specif the magnitde of and its coordinate direction angles so that the resltant force R acts along the positive ais and has a magnitde of 800 N. SOLUTION To solve this problem, the resltant force R and its two components, 1 and, will each be epressed in artesian vector form. Then, as shown in ig. 31b, it is necessar that R = 1 +. ppling Eq. 9, 1 = 1 cos a 1 i + 1 cos b 1 j + 1 cos g 1 k N (a) = 300 cos i cos 60 j cos 10 k = 1.1i + 10j - 10k6N = i + j + k Since R has a magnitde of 800 N and acts in the +j direction, We reqire R = (800 N)(+j) = 800j6 N g 77.6 b 1.8 a N R 800 N R = j = 1.1i + 10j - 10k + i + j + k N (b) 800j = (1.1 + )i + (10 + )j + (-10 + )k ig. 31 To satisf this eqation the i, j, k components of R mst be eqal to the corresponding i, j, k components of ( 1 + ). Hence, 0 = = -1.1 N 800 = 10 + = 60 N 0 = = 10 N The magnitde of is ths = (-1.1 N) + (60 N) + (10 N) = 700 N ns. We can se Eq. 9 to determine a, b, g. cos a = ; a = 108 ns. cos b = ; b = 1.8 ns. cos g = ; g = 77.6 ns. These reslts are shown in ig. 31b.

35 0 H P T E R O R E VETORS PRELIMINRY PROLEMS P 3. Sketch the following forces on the,, coordinate aes. Show a, b, g. a) = {0i + 60j - 10k} kn b) = {-40i - 80j + 60k} kn P 4. In each case, establish as a artesian vector, and find the magnitde of and the direction cosine of b. P. Show how to resolve each force into its,, components. Set p the calclation sed to find the magnitde of each component. 600 N kn 0 4 kn 4 kn (a) 00 N 4 3 (a) 4 3 (b) 0 N 800 N 0 N 10 N 60 (b) Prob. P 4 (c) Prob. P

36 .6 DDITION O RTESIN VETORS 1 UNDMENTL PROLEMS 13. Determine the coordinate direction angles of the force. 16. Epress the force as a artesian vector. 0 lb lb Prob Epress the force as a artesian vector. Prob Epress the force as a artesian vector. 00 N 70 N Prob Determine the resltant force acting on the hook. Prob Epress the force as a artesian vector lb 00 N lb Prob. 1 Prob. 18

37 H P T E R O R E VETORS PROLEMS * 60. The force has a magnitde of 80 lb and acts within the octant shown. Determine the magnitdes of the,, components of. 6. Determine the magnitde and coordinate direction angles of the force acting on the spport. The component of in the plane is 7 kn. 80 lb a 60 b 7 kn 40 Prob. 6 Prob The bolt is sbjected to the force, which has components acting along the,, aes as shown. If the magnitde of is 80 N, and a = 60 and g =, determine the magnitdes of its components. 63. Determine the magnitde and coordinate direction angles of the resltant force and sketch this vector on the coordinate sstem. * 64. Specif the coordinate direction angles of 1 and and epress each force as a artesian vector. a g b 1 80 lb lb Prob. 61 Probs. 63/64

38 .6 DDITION O RTESIN VETORS 3 6. The screw ee is sbjected to the two forces shown. Epress each force in artesian vector form and then determine the resltant force. ind the magnitde and coordinate direction angles of the resltant force. 66. Determine the coordinate direction angles of Determine the magnitde and coordinate direction angles of the resltant force, and sketch this vector on the coordinate sstem. 1 N N Prob N 70. Determine the magnitde and coordinate direction angles of the resltant force, and sketch this vector on the coordinate sstem. 00 N Probs. 6/ Determine the magnitde and coordinate direction angles of 3 so that the resltant of the three forces acts along the positive ais and has a magnitde of 600 lb. * 68. Determine the magnitde and coordinate direction angles of 3 so that the resltant of the three forces is ero. N N Prob Specif the magnitde and coordinate direction angles a 1, b 1, g 1 of 1 so that the resltant of the three forces acting on the bracket is R = -30k6 lb. Note that 3 lies in the plane lb lb g 1 b lb a lb Probs. 67/68 Prob. 71 1

39 4 H P T E R O R E VETORS * 7. Two forces 1 and act on the screw ee. If the resltant force R has a magnitde of 10 lb and the coordinate direction angles shown, determine the magnitde of and its coordinate direction angles. 7. The spr gear is sbjected to the two forces cased b contact with other gears. Epress each force as a artesian vector. * 76. The spr gear is sbjected to the two forces cased b contact with other gears. Determine the resltant of the two forces and epress the reslt as a artesian vector. 1 g 10 R 10 lb 1 80 lb 180 lb 60 Prob lb 73. Epress each force in artesian vector form. 74. Determine the magnitde and coordinate direction angles of the resltant force, and sketch this vector on the coordinate sstem. Probs. 7/ Determine the magnitde and coordinate direction angles of the resltant force, and sketch this vector on the coordinate sstem N 10 N 1 = 400 N N N Probs. 73/74 Prob. 77

40 .6 DDITION O RTESIN VETORS 78. The two forces 1 and acting at have a resltant force of R = -100k6 lb. Determine the magnitde and coordinate direction angles of. 79. Determine the coordinate direction angles of the force 1 and indicate them on the figre. 81. If the coordinate direction angles for 3 are a 3 = 10, b 3 = 60 and g 3 =, determine the magnitde and coordinate direction angles of the resltant force acting on the eebolt. 8. If the coordinate direction angles for 3 are a 3 = 10, b 3 =, and g 3 = 60, determine the magnitde and coordinate direction angles of the resltant force acting on the eebolt. 83. If the direction of the resltant force acting on the eebolt is defined b the nit vector R = cos j +sin k, determine the coordinate direction angles of 3 and the magnitde of R lb 0 1 = 60 lb lb Probs. 78/ lb * 80. The bracket is sbjected to the two forces shown. Epress each force in artesian vector form and then determine the resltant force R. ind the magnitde and coordinate direction angles of the resltant force N Probs. 81/8/83 * 84. The pole is sbjected to the force, which has components acting along the,, aes as shown. If the magnitde of is 3 kn, b =, and g = 7, determine the magnitdes of its three components. 8. The pole is sbjected to the force which has components = 1. kn and = 1. kn. If b = 7, determine the magnitdes of and. 10 g a b N Prob. 80 Probs. 84/8

41 6 H P T E R O R E VETORS.7 Position Vectors 4 m 1 m O m 4 m m 6 m ig. 3 In this section we will introdce the concept of a position vector. It will be shown that this vector is of importance in formlating a artesian force vector directed between two points in space.,, oordinates. Throghot the book we will se a righthanded coordinate sstem to reference the location of points in space. We will also se the convention followed in man technical books, which reqires the positive ais to be directed pward (the enith direction) so that it measres the height of an object or the altitde of a point. The, aes then lie in the horiontal plane, ig. 3. Points in space are located relative to the origin of coordinates, O, b sccessive measrements along the,, aes. or eample, the coordinates of point are obtained b starting at O and measring = +4 m along the ais, then = + m along the ais, and finall = - 6 m along the ais, so that (4 m, m, - 6 m). In a similar manner, measrements along the,, aes from O to ield the coordinates of, that is, (6 m, -1 m, 4 m). Position Vector. position vector r is defined as a fied vector which locates a point in space relative to another point. or eample, if r etends from the origin of coordinates, O, to point P(,, ), ig. 33a, then r can be epressed in artesian vector form as r = i + j + k Note how the head-to-tail vector addition of the three components ields vector r, ig. 33b. Starting at the origin O, one travels in the +i direction, then in the +j direction, and finall in the +k direction to arrive at point P(,, ). i k O r P(,, ) j i O r P(,, ) k (a) j (b) ig. 33

42 .7 POSITION VETORS 7 In the more general case, the position vector ma be directed from point to point in space, ig. 34a. This vector is also designated b the smbol r. s a matter of convention, we will sometimes refer to this vector with two sbscripts to indicate from and to the point where it is directed. Ths, r can also be designated as r. lso, note that r and r in ig. 34a are referenced with onl one sbscript since the etend from the origin of coordinates. rom ig. 34a, b the head-to-tail vector addition, sing the triangle rle, we reqire r (,, ) (,, ) r r + r = r r Solving for r and epressing r and r in artesian vector form ields r = r - r = ( i + j + k) - ( i + j + k) (a) or r = ( - )i + ( - )j + ( - )k ( 11) Ths, the i, j, k components of the position vector r ma be formed b taking the coordinates of the tail of the vector (,, ) and sbtracting them from the corresponding coordinates of the head (,, ). We can also form these components directl, ig. 34b, b starting at and moving throgh a distance of ( - ) along the positive ais (+i), then ( - ) along the positive ais (+j), and finall ( - ) along the positive ais (+k) to get to. ( )i r ( )j ( )k (b) ig. 34 r If an,, coordinate sstem is established, then the coordinates of two points and on the cable can be determined. rom this the position vector r acting along the cable can be formlated. Its magnitde represents the distance from to, and its nit vector, = r>r, gives the direction defined b a, b, g. ( Rssell. Hibbeler)

43 8 H P T E R O R E VETORS EXMPLE.10 3 m m 3 m m n elastic rbber band is attached to points and as shown in ig. 3a. Determine its length and its direction measred from toward. SOLUTION We first establish a position vector from to, ig. 3b. In accordance with Eq. 11, the coordinates of the tail (1 m, 0, -3 m) are sbtracted from the coordinates of the head (- m, m, 3 m), which ields 1 m (a) r (b) { j} m { 3 i} m {6 k} m r = [- m - 1 m]i + [ m - 0] j + [3 m - (-3 m)]k = -3i + j + 6k6 m These components of r can also be determined directl b realiing that the represent the direction and distance one mst travel along each ais in order to move from to, i.e., along the ais -3i6 m, along the ais j6 m, and finall along the ais 6k6 m. The length of the rbber band is therefore r = (-3 m) + ( m) + (6 m) = 7 m ormlating a nit vector in the direction of r, we have ns. = r r = i + 7 j k The components of this nit vector give the coordinate direction angles r 7 m g 31.0 b 73.4 a 11 (c) ig. 3 a = cos -1 a- 3 7 b = 11 b = cos -1 a 7 b = 73.4 g = cos -1 a 6 7 b = 31.0 ns. ns. ns. NOTE: These angles are measred from the positive aes of a localied coordinate sstem placed at the tail of r, as shown in ig. 3c.