STATICS. Rigid Bodies: Equivalent Systems of Forces VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

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1 Eighth E CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University Rigid Bodies: Equivalent Systems of Forces

2 Contents Introduction External and Internal Forces Principle of Transmissibility: Equivalent Forces Vector Products of Two Vectors Moment of a Force About a Point Varignon s Theorem 3-2

3 Contents Rectangular Components of the Moment of a Force Sample Problem 3.1 Scalar Product of Two Vectors Scalar Product of Two Vectors: Applications Mixed Triple Product of Three Vectors 3-3

4 Contents Moment of a Force About a Given Axis Sample Problem 3.5 Moment of a Couple Ad of Couples Couples Can Be Represented By Vectors Resolution of a Force Into a Force at O and a Couple 3-4

5 Contents Sample Problem 3.6 System of Forces: Reduction to a Force and a Couple Further Reduction of a System of Forces Sample Problem 3.8 Sample Problem

6 Introduction Treatment of a body as a single particle is not always possible. In general, the size of the body and the specific points of application of the forces must be considered. 3-6

7 Introduction Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the cons of equilibrium or motion of the body. 3-7

8 Introduction: Objectives Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system. - moment of a force about a point - moment of a force about an axis - moment due to a couple 3-8

9 Introduction Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. 3-9

10 External and Internal Forces Forces acting on rigid bodies are divided into two groups: - External forces - Internal forces 3-10

11 External and Internal Forces External forces are shown in a free-body diagram. If unopposed, each external force can impart a motion of translation or rotation, or both. 3-11

12 Principle of Transmissibility: Equivalent Forces Principle of Transmissibility - Cons of equilibrium or motion are not affected by transmitting a force along its line of action. 3-12

13 Principle of Transmissibility: Equivalent Forces NOTE: F and F are equivalent forces. 3-13

14 Principle of Transmissibility: Equivalent Forces Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. 3-14

15 Principle of Transmissibility: Equivalent Forces Principle of transmissibility may not always apply in determining internal forces and deformations. 3-15

16 Vector classifications: -Fixed or bound vectors have well defined points of application that cannot be changed -Free vectors may be freely moved in space -Sliding vectors may be applied anywhere along their line of action without affecting an analysis. 2-16

17 Moment of a Force About a Point A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. The moment of F about O is defined as M r O F 3-17

18 Moment of a Force About a Point M r O F The moment vector M O is perpendicular to the plane containing O and the force F. 3-18

19 Moment of a Force About a Point Magnitude of M O measures the tendency of the force to cause rotation of the body about an axis along M O. M O rf sinθ Fd The sense of the moment may be determined by the right-hand rule. 3-19

20 3-20

21 3-21

22 A human hand can comfortably apply a torque (torsional moment) of 1.5 Nm. For a bottle lid of diameter 4 cm how much uniformly distributed force (N/m) do we apply do we apply to open the lid? 3-22

23 Ans : f(2 * 22/7 * 2*2) 150 f3 N/cm. Total force f(2*22/7*2) 40 N or 4 kg. 3-23

24 3-24

25 How much torque in turning the steering wheel of the car? How much torque in tightening the nut of the car wheel? 3-25

26 3-26

27 3-27

28 3-28

29 Varignon s Theorem The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same r point O. ( ) F F r F r F

30 Varignon s Theorem Varigon s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F. 3-30

31 Rectangular Components of the Moment of a Force The moment of F about O, M O r F, r xi + yj + zk F F x i + F y j + F z k 3-31

32 Rectangular Components of the Moment of a Force M O M x i + M y j + M z k i j k x y z F x F y F z ( yf zf ) i + ( zf xf ) j + ( xf yf )k z y x z y x 3-32

33 ighth 3-33 Rectangular Components of the Moment of a Force The moment of F about B, F r M B A B / ( ) ( ) ( ) k F j F i F F k z z j y y i x x r r r z y x B A B A B A B A B A /

34 ighth 3-34 Rectangular Components of the Moment of a Force ( ) ( ) ( ) z y x B A B A B A B F F F z z y y x x k j i M

35 Moment of a Force About a Point Any force F that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment. 3-35

36 Sample Problem 3.1 A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a)moment about O, 3-36

37 Sample Problem 3.1 b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, 3-37

38 Sample Problem 3.1 d) location for a 240-N vertical force to produce the same moment, a)whether any of the forces from b, c, and d is equivalent to the original force. 3-38

39 Sample Problem 3.1 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper. 3-39

40 Sample Problem 3.1 M O Fd d ( 24m) cos60 12 m M O ( 100 N)( 12 cm) M O 1200 N m 3-40

41 Sample Problem 3.1 c) Horizontal force at A that produces the same moment, M d O 1200 N m ( 24 m) Fd F sin 60 ( 20.8 m) 1200 N m F 20.8 m 20.8 m F 57.7 N 3-41

42 Sample Problem 3.1 c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is M O 1200 N m Fd F ( 24 m) 1200 N m F 24 m perpendicular to OA. F 50 N 3-42

43 Sample Problem 3.1 d) To determine the point of application of a 240 lb force to produce the same moment, M O 1200 N m Fd ( 240 N) d OB d cos N m 240 N 5 cm 5 cm OB 10 m 3-43

44 Sample Problem 3.1 e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 N force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 N force. 3-44

45 Mixed Triple Product of Three Vectors Mixed triple product of three vectors, S P Q ( ) scalar result The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign, 3-45

46 ighth 3-46 Mixed Triple Product of Three Vectors ( ) ( ) ( ) ( ) z y x z y x z y x x y y x z z x x z y y z z y x Q Q Q P P P S S S Q P Q P S Q P P Q S P Q Q P S Q P S + + Evaluating the mixed triple product,

47 Moment about an axis 3-47

48 3-48

49 Moment of a Force About a Given Axis Moment M O of a force F applied at the point A about a point O, M O r F Scalar moment M OL about an axis OL is the projection of the moment vector M O onto the axis, M OL λ M λ O ( ) r F 3-49

50 Moment of a Force About a Given Axis Moments of F about the coordinate axes, M x yf z zf y M y zf x xf z M z xf y yf x 3-50

51 Moment of a Force About a Given Axis Moment of a force about an arbitrary axis, M r A BL B λ λ r A M A r The result is independent of the point B along the given axis. ( ) r F B B B 3-51

52 Sample Problem 3.4 The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. 3-52

53 Sample Problem 3.4 SOLUTION: The moment M A of the force F exerted by the wire is obtained by evaluating the M A r C A F vector product, 3-53

54 Sample Problem 3.4 SOLUTION: r r r 0.3 m i m k C A C A M A r C A F ( ) ( ) 3-54

55 Sample Problem 3.4 F rcd Fλ ( 200 N) rcd ( 0.3 m) i + ( 0.24 m) j ( 0.32 m) k ( 200 N) 0.5 m i + j k ( 120 N) ( 96 N) ( 128 N) 3-55

56 Sample Problem 3.4 M A i 0.3 j 0 k M A ( 7.68N m) i + ( 28.8N m) j + ( 28.8N m)k 3-56

57 Determine the moment about z-axis due to force at C. M A ( 7.68N m) i + ( 28.8N m) j + ( 28.8N m)k M A. k 28.8 Nm 3-57

58 Sample Problem 3.5 A cube is acted on by a force P as shown. Determine the moment of P a)about A b)about the edge AB and 3-58

59 Sample Problem 3.5 c)about the diagonal AG of the cube. d)determine the perpendicular distance between AG and FC. 3-59

60 ighth 3-60 Sample Problem 3.5 Moment of P about A, ( ) ( ) ( ) ( ) ( ) j i P j i a M j i P j i P P j i a a j ai r P r M A A F A F A ( )( ) k j i ap M A + + 2

61 Sample Problem 3.5 Moment of P M about AB, AB i i M A ( ap )( 2 i + j + k ) M AB ap

62 Sample Problem 3.5 M M Moment of P about M the diagonal AG, λ M A ra G ai λ r AG A AG A G ap ap 6 ( i + j + k ) ap ( i j k ) ( i + j + k ) ( 1 1 1) aj ak a M AG ( i j k ) ap

63 Sample Problem 3.5 Perpendicular distance between AG and FC, P λ P ( j k ) ( i j k ) ( ) P

64 3-64

65 Moment of a Couple Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. 3-65

66 Moment of a Couple Moment of the couple, M M ra F + ( r r ) A r F B r B F rf sinθ Fd ( ) F 3-66

67 Moment of a Couple The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect. 3-67

68 Equivalent couples If the wheel was connected to the shaft at a point other than the centre, the wheel would still turn as 12 Nm is a couple and a free vector. 3-68

Moment of a Couple Two couples will have equal moments if F d F d 1 1 2 2 the two couples lie in parallel

69 Moment of a Couple Two couples will have equal moments if F d F d the two couples lie in parallel planes, and the two couples have the same sense or the tendency to cause rotation in the same direction. 3-69

70 Ad of Couples Consider two intersecting planes P 1 and P 2 with each containing a couple M M 1 2 r F r F 1 2 in plane in plane P 1 P

71 Ad of Couples Resultants of the vectors also form a couple M r R r ( ) F + F

72 Ad of Couples By Varigon s theorem M r F 1 + r F 2 M 1 + M 2 Sum of two couples is also a couple that is equal to the vector sum of the two couples 3-72

73 Couples Can Be Represented by Vectors Couple vectors may be resolved into component vectors. 3-73

74 Resultant of couples (free vector) 3-74

75 The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles.

76 3-76

77 M k N.m { } M 20cos 20sin 30i 20cos 20cos30 j R + 20sin 20 k N.m 9.397i j k N.m { } M M + M i j k+ 50k N.m { } 9.397i j k N.m { } 3-77

78 M R ( ) ( ) ( ) ( ) ( ) ( ) N.m N.m59.9 N.m α cos 99.0 β cos ν cos 18.3

79 Sample Problem 3.6 Determine the components of the single couple equivalent to the couples shown. 3-79

80 Sample Problem 3.6 Compute the sum of the moments of the four forces about an arbitrary single point. D is a good choice as only the forces at C and E contribute to the moment about D. 3-80

81 Sample Problem

82 Sample Problem 3.6 M M D ( 18 m) j ( 30 N) k [ ( ) ( ) ] 9 m j 12 m k ( 20 N)i + M + ( 180 N m)k ( 540 N m) i + ( 240N m) j 3-82

83 3-83

84 3-84

85 Introduction Any force acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. 3-85

86 Resolution of a Force Into a Force at O and a Couple Force vector F can not be simply moved to O without modifying its action on the body. 3-86

87 Resolution of a Force Into a Force at O and a Couple The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system. 3-87

88 Resolution of a Force Into a Force at O and a Couple Attaching equal and opposite force vectors at O produces no net effect on the body. 3-88

89 Resolution of a Force Into a Force at O and a Couple Moving F from A to a different point O 3-89

90 Resolution of a Force Into a Force at O and a Couple Moving F from A to a different point O requires the ad of a different couple vector M O M r O ' F 3-90

91 Resolution of a Force Into a Force at O and a Couple The moments of F about O and O are related, M O ' r ' F M + s O ( r + s ) F F r F + s F 3-91

92 Resolution of a Force Into a Force at O and a Couple Moving the force-couple system from O to O requires the ad of the moment of the force at O about O. 3-92

93 Sample Problem 3.4 The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the equivalent force at A. 3-93

94 F i + j k R M A ( 120 N) ( 96 N) ( 128 N) ( 7.68N m) i + ( 28.8N m) j + ( 28.8N m)k 3-94

95 Equivalent force system at B? F i + j k R ( 120 N) ( 96 N) ( 128 N) M B M A +BA x F R M A +80 k x F R 3-95

96 Introduction Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. 3-96

97 System of Forces: Reduction to a Force and Couple 3-97

98 System of Forces: Reduction to a Force and Couple A system of forces may be replaced by a collection of force-couple systems acting a given point O 3-98

99 System of Forces: Reduction to a Force and Couple The force and couple vectors may be combined into a resultant force vector and a resultant couple vector, R F R F M O ( r ) 3-99

100 3-100

101 System of Forces: Reduction to a Force and Couple The force-couple system at O may be moved to O with the ad of the moment of R about O, M R O' M R O + s R 3-101

102 System of Forces: Reduction to a Force and Couple Two systems of forces are equivalent if they can be reduced to the same force-couple system

103 3-103

104 Special cases 3-104

105 Further Reduction of a System of Forces If the resultant force and couple at O are mutually perpendicular, they can be replaced by a single force acting along a new line of action

106 Further Reduction of a System of Forces The resultant force-couple system for a system of forces will be mutually perpendicular 1) the forces are concurrent, 2) the forces are coplanar, 3) the forces are parallel

107 Concurrent forces 3-107

108 Further Reduction of a System of Forces System of coplanar forces is reduced to a force-couple system R and R M O that is mutually perpendicular

109 Further Reduction of a System of Forces System can be reduced to a single force by moving the line of action of until its moment about O becomes R R M O 3-109

110 Further Reduction of a System of Forces In terms of rectangular coordinates, xr yr M R y x O x,y coordinates of point of application 3-110

111 Further Reduction of a System of Forces xr y M R O yr x M R O 3-111

112 Sample Problem 3.8 For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant.note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium

113 Sample Problem 3.8 SOLUTION: a)compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A

114 Sample Problem 3.8 b)find an equivalent force-couple system at B based on the force-couple system at A. c)determine the point of application for the resultant force such that its moment about A is equal to the resultant couple at A

115 Sample Problem 3.8 a)compute the resultant force and the resultant couple at A. R F 150N 600N + 100N ( ) j ( ) j ( ) j ( 250N )j R ( 600 N) j 3-115

116 Vector Mechanics for Engineers: Statics ( r F ) 1.6 i 600 j i i 250 j M R A ( ) ( ) ( ) ( j ) ( ) ( ) ( 1880 N m)k M R A 3-116

117 Sample Problem 3.8 b)find an equivalent force-couple system at B based on the forcecouple system at A

118 Sample Problem 3.8 M R B M R A + r B A R ( 1880N m) k + ( 4.8m ) i ( 600N) ( 1880N m) k + ( 2880N m)k j M R B + ( 1000 N m)k 3-118

119 Sample Problem 3.8 The couple at B is equal to the moment about B of the forcecouple system found at A

120 Sample Problem

121 Resultant of given system of forces is equal to R, and its point of application must be such that the moment of R about A is equal to. R M A 3-121

122 ( 600N ) j ( 1800N.m ) k ( 600N ) k ( 1800N.m )k xi x r R M R A R 600 ( N)j x3.13m 3-122

123 -(600 N )j The resultant of the given system of forces is equal to R, and its point of application must be such that the moment of R about A is equal to. We write A B and conclude that X 3.13 m. Thus, the single force equivalent to the given system is defined as Single Force or Resultant. R 600 N x 3.13 m

124 Sample Problem 3.10 Three cables are attached to the bracket as shown. Replace the forces with an equivalent forcecouple system at A

125 Sample Problem 3.10 SOLUTION: Determine the relative position vectors for the points of application of the cable forces with respect to A

126 Sample Problem 3.10 Resolve the forces into rectangular components. Compute the equivalent force, R F 3-126

127 Sample Problem 3.10 Compute the equivalent couple, M R A ( ) r F 3-127

128 Sample Problem 3.10 SOLUTION: Determine the relative position vectors with respect to A. r r r B C D A A A i k i 0.050k i j ( ) m ( ) m ( m) 3-128

129 Sample Problem 3.10 Resolve the forces into rectangular components. F F B λ B ( 700 N) λ re B 75i 150 j + 50k re B i j k 300i 600 j + 200k ( N) 3-129

130 Sample Problem 3.10 F C ( 1000 N)( cos 45i cos 45 j ) 707i 707k ( N) F D ( 1200 N )( cos 60 i + cos 30 j ) 600 i j ( N ) 3-130

131 Sample Problem 3.10 Compute the equivalent force, R F ( ) + ( ) + ( )k i j R 1607i + 439j 507k ( N) 3-131

132 Sample Problem 3.10 Compute the equivalent couple, R ( M ) A r F i j k r F i 45k BA B

133 M r B r C r R A A A D A F F F B c D ( r F) i i i j k i 45k j k j j k k M R A 30 i j k 3-133

134 Pauli Effect in Action It was a standing joke among Wolfgang Pauli's colleagues that the famed theoretical physicist should be kept as far away from experimental equipment as humanly possible. His mere presence in a laboratory, it was said, would cause something to go wrong: the power would fail, 3-134

135 Pauli Effect in Action vacuum tubes would suddenly leak, instruments would break or malfunction. Indeed, such was the frequency of Pauli-related incidents that the strange phenomenon came to be known as the 'Pauli Effect'

136 One day, a group of Pauli's mischievous colleagues set up a practical joke to show the strange effect in action: an elaborate contraption was designed and constructed to bring a chandelier crashing down the moment Pauli arrived at a large reception. Pauli duly appeared on schedule and the Pauli effect was demonstrated, though not as the mischievous crew had planned: 3-136

137 no sooner was the mechanism activated than a pulley jammed - and the chandelier did not come down. One day, some important experimental equipment in Professor James Frank's laboratory at the Physics Institute at the University of Gottingen unexpectedly blew up for no apparent reason. Moreover, Pauli, who was on his way to Denmark, had not even entered the building

138 Only later was it discovered that the catastrophe had taken place at the precise moment that the train carrying Pauli from Zurich to Copenhagen had stopped to collect passengers at the Gottingen railroad station 3-138

139 Equivalent force systems: The wrench 3-139

140 3-140

141 3-141

142 ses/en3/notes/statics/staticequiv/ Staticequiv.htm This is a good example on how CG shifts due to use of fuel in aircraft

143 Vector Product of Two Vectors Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product

144 Vector Product of Two Vectors Vector product of two vectors P and Q is defined as the vector V which satisfies the following cons: 1.Line of action of V is perpendicular to plane containing P and Q

145 Vector Product of Two Vectors 2. Magnitude of V is V PQ sin θ 3. Direction of V is obtained from the righthand rule

146 Vector Products: Rectangular Components i j k V P P P x y z Q Q Q x y z 3-146

147 Moment of a Force About a Point Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure

148 Moment of a Force About a Point The plane of the structure contains the point O and the force F. M O, the moment of the force about O is perpendicular to the plane

149 Moment of a Force About a Point If the force tends to rotate the structure counter clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive

150 Moment of a Force About a Point If the force tends to rotate the structure clockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative

151 Rectangular Components of the Moment of a Force M M O O [( x x ) F ( y y ) F ] M A ( x x B ) F y ( y A y B ) F z A Z B y A B z k 3-151

152 Rectangular Components of the Moment of a Force For twodimensional structures, M O ( ) xf yf y z k M O M Z xf y yf z 3-152

153 Couples Can Be Represented by Vectors A couple can be represented by a vector with magnitude and direction equal to the moment of the couple. Couple vectors obey the law of ad of vectors

STATICS. Equivalent Systems of Forces. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents & Objectives.

STATICS. Equivalent Systems of Forces. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents & Objectives. 3 Rigid CHATER VECTOR ECHANICS FOR ENGINEERS: STATICS Ferdinand. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Teas Tech Universit Bodies: Equivalent Sstems of Forces Contents & Objectives