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Tenth Edition CHAPTERECTOR MECHANICS FOR ENGINEERS: ST ATICS Ferdinand P. Beer 1 E. Russell Johnston, Jr. David F. Mazurektroduction John Chenotes: California Polytechnic State University ETtohn Vector Mechanics for Engineers: Statics Contents What is Mechanics? What Can You Do with Statics Knowledge? Systems of Units Method of Problem Solution Numerical Accuracy ETtoth Vector Mechanics for Engineers: Statics What is Mechanics? • Mechanics is the study of bodies under the action of forces. • Categories of Mechanics: - Rigid bodies - Statics – bodies at rest or at constant velocity - Dynamics – accelerating bodies - Deformable bodies - Fluids – gas and/or liquid • Mechanics is an applied science, closely related to physics, so many of the concepts will build on that prior knowledge. • Mechanics is the foundation of many engineering topics and is an indispensable prerequisite to their study. ETitohn Vector Mechanics for Engineers: Statics What Can You Do with Statics Knowledge? Calculate the force in each member of this structure (a truss) in order to design it to withstand the loads that it will experience. ETitohn Vector Mechanics for Engineers: Statics What Can You Do with Statics Knowledge? Determine the forces that this prosthetic arm will need to withstand to make exercise possible for the wearer. ETitohn Vector Mechanics for Engineers: Statics What Can You Do with Statics Knowledge? Design the joints and support of the Shuttle Remote Manipulator System (SRMS) so that it can be used to pick up and support various payloads. ETtohn Vector Mechanics for Engineers: Statics Systems of Units • International System of Units (SI): The basic units are length, time, and mass which are arbitrarily defined as the meter (m), second (s), and kilogram • Kinetic Units: length, time, mass, and force. (kg). Force is the derived unit, F ma • Three of the kinetic units, referred to m as basic units, may be defined 1N 1kg 1 2 s arbitrarily. The fourth unit, referred to as a derived unit, must have a • U.S. Customary Units: definition compatible with Newton’s The basic units are length, time, and 2nd Law, force which are arbitrarily defined as the foot (ft), second (s), and pound (lb). F ma Mass is the derived unit, m F a 1lb 1slug 1ft s ETtoth Vector Mechanics for Engineers: Statics Method of Problem Solution • Problem Statement: • Solution Check: Includes given data, specification of - Test for errors in reasoning by what is to be determined, and a figure verifying that the units of the showing all quantities involved. computed results are correct, - test for errors in computation by • Free-Body Diagrams: substituting given data and computed Create separate diagrams for each of results into previously unused the bodies involved with a clear equations based on the six principles, indication of all forces acting on - always apply experience and physical each body. intuition to assess whether results seem • Fundamental Principles: “reasonable” The six fundamental principles are applied to express the conditions of rest or motion of each body. The rules of algebra are applied to solve the equations for the unknown quantities. Tenth Edition CHAPTERECTOR MECHANICS FOR ENGINEERS: ST A TICS Ferdinand P. Beer 2 E. Russell Johnston, Jr. Lecture Notes:ekatics of Particles John Chen California Polytechnic State University ETtohn Vector Mechanics for Engineers: Statics Contents Introduction Sample Problem 2.3 Resultant of Two Forces Equilibrium of a Particle Vectors Free-Body Diagrams Addition of Vectors Sample Problem 2.4 Resultant of Several Concurrent Sample Problem 2.6 Forces Expressing a Vector in 3-D Space Sample Problem 2.1 Sample Problem 2.7 Sample Problem 2.2 Rectangular Components of a Force: Unit Vectors Addition of Forces by Summing Components ETitohn Vector Mechanics for Engineers: Statics Application The tension in the cable supporting this person can be found using the concepts in this chapter EdTtohn Vector Mechanics for Engineers: Statics Introduction • The objective for the current chapter is to investigate the effects of forces on particles: - replacing multiple forces acting on a particle with a single equivalent or resultant force, - relations between forces acting on a particle that is in a state of equilibrium. • The focus on particles does not imply a restriction to miniscule bodies. Rather, the study is restricted to analyses in which the size and shape of the bodies is not significant so that all forces may be assumed to be applied at a single point. ETitohn Vector Mechanics for Engineers: Statics Resultant of Two Forces • force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense. • Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. • Force is a vector quantity. ETtoth Vector Mechanics for Engineers: Statics Vectors • Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations. • Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature • Vector classifications: - Fixed or bound vectors have well defined points of application that cannot be changed without affecting an analysis. - Free vectors may be freely moved in space without changing their effect on an analysis. - Sliding vectors may be applied anywhere along their line of action without affecting an analysis. ETtohn Vector Mechanics for Engineers: Statics Addition of Vectors • Trapezoid rule for vector addition • Triangle rule for vector addition • Law of cosines, B C 2 2 2 R PQ 2PQcosB C R P Q • Law of sines, sin A sin B sinC B Q R A • Vector addition is commutative, P Q Q P • Vector subtraction ETitohn Vector Mechanics for Engineers: Statics Resultant of Several Concurrent Forces • Concurrent forces: set of forces which all pass through the same point. Aset of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces. • Vector force components: two or more force vectors which, together, have the same effect as a single force vector. ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.1 SOLUTION: • Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the the diagonal. The two forces act on a bolt at A. Determine their resultant. • Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant. ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.1 • Graphical solution -Aparallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured, R 98 N 35 • Graphical solution -Atriangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured, R 98 N 35 ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.1 • Trigonometric solution -Apply the triangle rule. From the Law of Cosines, R P Q 2PQcosB 2 2 40N 60N 2 40N 60N cos155 R 97.73N From the Law of Sines, sin A sinB Q R sin A sinBQ R 60N sin155 97.73N A 15.04 20 A 35.04 ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.2 SOLUTION: • Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge Abarge is pulled by two axis and length proportional to 5000 lbf. tugboats. If the resultant of the forces exerted by the • Find a trigonometric solution by tugboats is 5000 lbf directed applying the Triangle Rule for vector along the axis of the barge, addition. With the magnitude and determine the tension in each of the ropes for = 45 . direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Discuss with a neighbor how Law of Sines to find the rope tensions. you would solve this problem. ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.2 • Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides. T 1 3700lbf T 2 2600lbf • Trigonometric solution - Triangle Rule with Law of Sines T T 5000lbf 1 2 sin45 sin30 sin105 T 3660lbf T 2590lbf 1 2 ETtohn Vector Mechanics for Engineers: Statics What if…? • At what value of would the tension in rope 2 be a minimum? Hint: Use the triangle rule and think about how changing changes the magnitude of T . 2 After considering this, discuss your ideas with a neighbor. • The minimum tension in rope 2 occurs when T 1nd T a2e perpendicular. T 5000 lbfsin30 T 2 500 bf 2 2 T 5000lbf cos30 T 4330lbf 1 1 9030 60 ETtoth Vector Mechanics for Engineers: Statics Rectangular Components of a Force: Unit Vectors • It’s possible to resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. F and F are referred to as rectangular x y vector components and F F x F y • Define perpendicular unit vectors i and j which are parallel to the x and y axes. • Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. F F x F jy F xnd F aye referred to as the scalar components of F ETtohn Vector Mechanics for Engineers: Statics Addition of Forces by Summing Components • To find the resultant of 3 (or more) concurrent frce, R P Q S • Resolve each force into rectangular components, then add the components in each direction: Rxi Ryj P x P y Q ixQ j yS i xS j y P Q S i P Q S j x x x y y y • The scalar components of the resultant vector are equal to the sum of the corresponding scalar components of the given forces. R P Q S R P Q S x x x x y y y y Fx F y • To find the resultant magnitude and direction, Ry R R xR y tan1 Rx ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.3 SOLUTION: • Resolve each force into rectangular components. • Determine the components of the resultant by adding the corresponding force components in the x and y directions. Four forces act on bolt A as shown. • Calculate the magnitude and direction Determine the resultant of the force of the resultant. on the bolt. ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.3 SOLUTION: • Resolve each force into rectangular components. forcr mag xcomp ycomp F1 150 129.9 75.0 r F2 80 27.4 75.2 Fr 110 0 110.0 r3 F4 100 96.6 25.9 R x 199.1 Ry 14.3 • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction. R 199.1 14.3 2 R 199.6N tan 14.3N 4.1 199.1N ETtohn Vector Mechanics for Engineers: Statics Equilibrium of a Particle • When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. • Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line. • Particle acted upon by • Particle acted upon by three or more forces: two forces: - graphical solution yields a closed polygon - equal magnitude - algebraic solution - same line of action R F 0 - opposite sense Fx 0 Fy 0 ETtoth Vector Mechanics for Engineers: Statics Free-Body Diagrams Space Diagram: Asketch showing Free Body Diagram: Asketch showing the physical conditions of the only the forces on the selected particle. problem, usually provided with This must be created by you. the problem statement, or represented by the actual physical situation. ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.4 SOLUTION: • Construct a free body diagram for the particle at the junction of the rope and cable. • Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle. • Apply trigonometric relations to determine the unknown force In a ship-unloading operation, a magnitudes. 3500-lb automobile is supported by a cable. Arope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope? ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.4 SOLUTION: • Construct a free body diagram for the particle at A, and the associated polygon. • Apply the conditions for equilibrium and solve for the unknown force magnitudes. Law of Sines: TAB T AC 3500lb sin120 sin2 sin58 TAB 3570lb TAC 144lb ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.6 SOLUTION: • Decide what the appropriate “body” is and draw a free body diagram • The condition for equilibrium states that the sum of forces equals 0, or: It is desired to determine the drag force R F 0 at a given speed on a prototype sailboat Fx 0 Fy 0 hull. Amodel is placed in a test channel and three cables are used to • The two equations means we can solve align its bow on the channel centerline. for, at most, two unknowns. Since For a given speed, the tension is 40 lb there are 4 forces involved (tensions in in cable AB and 60 lb in cable AE. 3 cables and the drag force), it is easier to resolve all forces into components Determine the drag force exerted on the hull and the tension in cable AC. and apply the equilibrium conditions ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.6 SOLUTION: • The correct free body diagram is shown and the unknown angles are: 1.5ft tan 7ft 1.75 tan 0.375 4ft 4ft 60.25 20.56 • In vector form, the equilibrium condition requires that the resultant force (or the sum of all forces) be zero: R T T T F 0 AB AC AE D • Write each force vector above in component form. ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.6 • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. r r r T AB 0 lb in60.26i 4 lb cs60.26j 34.73 lb i 19.84 lb jr r T T sin20.56i T cos20.56j r AC AC r AC r 0.3512T AC i 0.9363T AC j r r T AE 0 lb r r FD F D r R0 r 34.730.3512T ACF D i r 9.84 0.9363T AC 60 ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.6 R 0 34.730.3512T F i AC D 19.840.9363T AC 60 j This equation is satisfied only if each component of the resultant is equal to zero Fx0 034.730.3512T AC FD Fy0 019.840.9363T AC60 T AC 42.9lb FD 19.66lb ETtohn Vector Mechanics for Engineers: Statics Expressing a Vector in 3-D Space If angles with some of the axes are given: • The vector F is • Resolve F into • Resolve F into h contained in the horizontal and vertical rectangular components plane OBAC. components. F F cos x h F y F cos y Fsin cys Fh Fsin y F y F shn Fsin syn ETtoth Vector Mechanics for Engineers: Statics Expressing a Vector in 3-D Space If the direction cosines are given: • With the angles between F and the axes, F x F cos x Fy F cos y F z F cos z F F x F y F kz F cos ixcos j yos k z F cos ixcos j ycos k z • is a unit vector along the line of action oF and cos x cs ,yand cos z are the direction cosines for F ETitohn Vector Mechanics for Engineers: Statics Expressing a Vector in 3-D Space If two points on the line of action are given: Direction of the force is defined by the location of two points, M x ,1y ,1 a1d N x , y ,2 2 2 r d vector joining M and N r r r dxi d y d k z dx x 2 x 1 d y y 2y 1 d z z 2z 1 r r F F r 1 r r r d x d jyd k z d Fd Fd Fd Fx x Fy y F z z d d d ETitohn Vector Mechanics for Engineers: Statics Sample Problem 2.7 SOLUTION: • Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. • Apply the unit vector to determine the components of the force acting on A. • Noting that the components of the unit vector are the direction cosines for the The tension in the guy wire is 2500 N. vector, calculate the corresponding Determine: angles. a) components F , F , F of the force x y z acting on the bolt at A, b) the angles x y, zdefining the direction of the force (the direction cosines) ETtohn Vector Mechanics for Engineers: Statics Sample Problem 2.7 SOLUTION: • Determine the unit vector pointing from A towards B. r r r AB 4m i m j 3 k AB 40m m 30 2 94.3 m r 40 80 30 r 94.3 r 94.3 r 94.3 r 0.424i 0.848j 0.318k • Determine the components of the force. F F r r r 500 N 424i 0.848 j 0.318k r r r 1060 N i 2120 N j 95 N k ETtoth Vector Mechanics for Engineers: Statics Sample Problem 2.7 • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. cos ixcos j yos k z 0.424i 0.848j 0.318k x115.1 y 32.0 71.5 z ETtohn Vector Mechanics for Engineers: Statics What if…? SOLUTION: F • Since the force in the guy wire must be BA the same throughout its length, the force at B (and acting towardA) must be the same magnitude but opposite in F direction to the force atA. AB r r F BAF AB r r r What are the components of the 060 N 2120 N j 795 N k force in the wire at point B? Can you find it without doing any calculations? Give this some thought and discuss this with a neighbor. Chapter 3 RIGID BODIES: EQUIVALENT FORCE SYSTEMS F The forces acting on a F’ rigid body can be separated into two groups: (1) external forces (representing the action of other rigid bodies on the rigid body under consideration) and (2) internal forces (representing the forces which hold together particles forming the rigid body. F F’ According to the principal of transmissibility, the effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. Two forces acting on the rigid body at two different points have the same effect on that body if they have the same magnitude, same direction, and same line of action. Two such forces are said to be equivalent. V = Px Q The vector product of two vectors is defined as Q q V = Px Q P The vector product of P and Q forms a vector which is perpendicular to both P and Q, of magnitude V = PQ sin q This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through q which brings vector P in line with vector Q. The three vectors P, Q, and V - taken in that order - form a right-hand triad. It follows that Q x P = - (P x Q) j It follows from the definition of the vector product of two vectors that the vector k i products of unit vectors i, j, and k are i x i = j x j = k x k = 0 i x j = k , j x k = i , k x i = j , i x k = - j , j x i = - k , k x j = - i The rectangular components of the vector product V of two vectors P and Q are determined as follows: Given P = P ix+ P j +yP k z Q = Q i x Q j + y k z The determinant containing each component of P and Q is expanded to define the vector V, as well as its scalar components P = P ix+ P j y P k z Q = Q ix+ Q j y Q k z i j k V = P x Q = P x Py P z = Vxi + V y + V kz Q x Q y Q z where V x P Qy- PzQ z y V = P Q - P Q y z x x z V z P Qx- PyQ y x M o The moment of force F about point O is defined as the vector product M = r x F F O where r is the position O r q vector drawn from point O d A to the point of application of the force F. The angle between the lines of action of r and F is q. The magnitude of the moment of F about O can be expressed as M =OrF sin q = Fd where d is the perpendicular distance from O to the line of action of F. y F y A (x , y, z ) y j The rectangular Fxi components of the r moment M of a Fzk o O x i force F are x determined by z k expanding the determinant of r x F. z i j k M = r x F = x y z = M i + M j + M k o x y z F x F y F z where M = y F - z F M = zF - x F x z y y x z M z x F - y F x y F y A (xA, A , A ) B (xB, B , B ) In the more general r Fxi case of the moment about an arbitrary Fzk point B of a force F applied at A, we have O x z i j k M = r x F = x y z B A/B A/B A/B A/B F x F F z y where r A/B i A/B j + A/B A/B and xA/Bx - A B y A/B - A B zA/Bz -Az B y F j F In the case of problems y involving only two dimensions, the force F (y - y ) j rA/B A Fxi can be assumed to lie A B in the xy plane. Its B moment about point B O (A - B ) i is perpendicular to that plane. It can be x z M =BM k B completely defined by the scalar M B (x - A )FB+ (y - y )AF B x The right-hand rule is useful for defining the direction of the moment as either into or out of the plane (positive or negative k direction). The scalar product of two Q vectors P and Q is denoted as P Q ,and is defined as q P Q = PQ cos q where q is the angle between P the two vectors The scalar product of P and Q is expressed in terms of the rectangular components of the two vectors as P Q = P Q +xP Qx+ P y y z z y L The projection of a vector qy A P on an axis OL can be obtained by forming the l q P x scalar product of P and the O x unit vector l along OL. q z z P OL = P l Using rectangular components, P = P cos q + P cos q + P cos q OL x x y y z z The mixed triple product of three vectors S, P, and Q is Sx Sy S z S (P x Q ) = P P P x y z Qx Q y z The elements of the determinant are the rectangular components of the three vectors. y L The moment of a M O force F about an axis OL is the projection C F OC on OL of the l moment M of the O A (x, y, z)orce F. This can be r written as a mixed O x triple product. z l x l y l z M OL = l M =Ol (r x F) = x y z F F F x y z lx, y ,zl = direction cosines of axis OL x, y , z = components of r Fx, y ,zF = components of F M - F d F Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. The moment of a couple is independent of the point about which it is computed; it is a vector M perpendicular to the plane of the couple and equal in magnitude to the product Fd. M y y - F y (M = Fd) d M y M O F O O Mx z x z x x z M z Two couples having the same moment M are equivalent (they have the same effect on a given rigid body). F F M O r A A O O Any force F acting at a point A of a rigid body can be replaced by a force-couple system at an arbitrary point O, consisting of the force F applied at O and a couple of Ooment M equal to the moment about point O of the force F in its original position. The force vector F and the couple vOctor M are always perpendicular to each other. F 3 F 1 R A 3 r r A1 3 1 r2 O A 2 O F 2 R M O As far as rigid bodies are concerned, two systems of forces, F1, 2 ,3F . . . , an1 F2 , 3’ , F’ . . . , are equivalent if, and only if, S F = S F’ and S M = S M ’ o o F3 F1 If the resultant force R R A 3 and the resultaRt couple vector MOare r1 A 1 perpendicular to each r2 A2 O other, the force-couple O system at O can be further reduced to a R F2 M O single resultant force. This is the case for systems consisting of either (a) concurrent forces, (b) coplanar forces, or (c) parallel forces. If the resultant force and couple are directed along the same line, the force-couple system is termed a wrench. Chapter 4 EQUILIBRIUM OF RIGID BODIES In the study of the equilibrium of rigid bodies, i.e. the situation when the external forces acting on a rigid body form a system equivalent to zero, we have S F = 0 S M = S (r x F) O Resolving each force and each moment into its rectangular components, the necessary and sufficient conditions for the equilibrium of a rigid body are expressed by six scalar equations: SF x 0 SF y 0 SF z 0 SM = 0 SM = 0 SM = 0 x y z These equations can be used to determine unknown forces applied to the rigid body or unknown reactions exerted by its supports. When solving a problem involving the equilibrium of a rigid body, it is essential to consider all of the forces acting on the body. Therefore, the first step in the solution of the problem should be to draw a free-body diagram showing the body under consideration and all of the unknown as well as known forces acting on it. In the case of the equilibrium of two-dimensional structures, each of the reactions exerted on the structure by its supports could involve one, two, or three unknowns, depending upon the type of support. In the case of a two-dimensional structure, three equilibrium equations are used, namely SF x 0 SF y 0 SM =A0 where A is an arbitrary point in the plane of the structure. SF x 0 SF y 0 SM A 0 These equations can be used to solve for three unknowns. While these three equilibrium equations cannot be augmented with additional equations, any one of them can be replaced by another equation. Therefore, we can write alternative sets of equilibrium equations, such as SF x 0 SM =A0 SM =B0 where point B is chosen in such a way that the line AB is not parallel to the y axis, or SM = 0 SM = 0 SM = 0 A B C where points A, B, and C do not lie in a straight line. Since any set of equilibrium equations can be solved for only three unknowns, the reactions at the supports of a rigid two- dimensional structure cannot be completely determined if they involve more than three unknowns; they are said to be statically indeterminate. On the other hand, if the reactions involve fewer than three unknowns, equilibrium will not be maintained under general loading conditions; the structure is said to be partially constrained. The fact that the reactions involve exactly three unknowns is no guarantee that the equilibrium equations can be solved for all three unknowns. If the supports are arranged is such a way that the reactions are either concurrent or parallel, the reactions are statically indeterminate, and the structure is said to be improperly constrained. F2 B A F1 Two particular cases of rigid body equilibrium are given special attention. A two-force body is a rigid body subjected to forces at only two points. The resultants F and F of these 1 2 two forces must have the same magnitude, the same line of action, and opposite sense. F2 F B 3 C D A F 1 A three-force body is a rigid body subjected to forces at only three points, and the resultant1 F 2 F ,an3 F of these forces must be either concurrent or parallel. This property provides an alternative approach to the solution of problems involving a three-force body. When considering the equilibrium of a three-dimensional body, each of the reactions exerted on the body by its supports can involve between one and six unknowns, depending upon the type of support. In the general case of the equilibrium of a three-dimensional body, the six scalar equilibrium equations listed at the beginning of this review should be used and solved for six unknowns. In most cases these equations are more conveniently obtained if we first write S F = 0 S M =OS (r x F) and express the forces F and position vectors r in terms of scalar components and unit vectors. The vector product can then be computed either directly or by means of determinants, and the desired scalar equations obtained by equating to zero the coefficients of the unit vectors. As many as three unknown reaction components can be eliminated from the computation of S M O through a judicious choice of point O. Also, the reactions at two points A and B can be eliminated from the solution of some problems by writing the equation S M AB = 0, which involves the computation of the moments of the forces about an axis AB joining points A and B. If the reactions involve more than six unknowns, some of the reactions are statically indeterminate; if they involve fewer than six unknowns, the rigid body is only partially constrained. Even with six or more unknowns, the rigid body will be improperly constrained if the reactions associated with the given supports either are parallel or intersect the same line. Figure 1.2 1 Figure 1.3 2 Figure 1.4 3 Figure 1.5 4 Figure 2.2 5 Figure 2.5 6 Figure 2.6 7 Figure 2.7 8 Figure 2.8 9 Figure 2.10 10 p02_01 11 p02_15 12 Figure 2.20 13 p02_43 14 Figure 2.25a 15 p02_91-92 16 p02_89-90 17 03_01 18 03_02 19 03_03 20 03_05a 21 03_05b 22 03_12a 23 03_12b 24 p03_23 25 p03_49-50 26 p03_96 27 p03_105 28 p03_114-115 29 Figure 4.1 30 p04_f01 31 p04_03 32 p04_04 33 p04_15-16 34 p04_62-63 35 Figure 4.11 36 Figure 4.12 37 Figure 5.8a 38 Figure 5.8b 39 p05_01 40 Figure 5.12 41 p05_34 42 p05_35 43 p05_66 44 Figure 6.2 45 Figure 6.7 46 Figure 6.8 47 Figure 6.13 48 p06_02 49 p06_43-44 50 Figure 6.21 51 Figure 6.22 52 Figure 6.23 53 Figure 7.9 54 Figure 7.10 55 p07_75 56 Figure 7.13 57 Figure 8.1 58 p08_f04 59 p08_18 60 p08_25-26 61 p08_28 62 p08_54-55-56 63 p08_103 64 p08_134 65 p11_09 66 p11_45 67 p11_47-48 68 p11_97 69 p11_99 70 p11_103 71 Figure 12.2 72 Figure 12.3 73 Figure 12.8 74 Figure 12.9 75 Figure 12.10 76 p12_37 77 Figure 13.3 78 Figure 13.4 79 Figure 13.5 80 p13_02 81 p13_33 82 p13_47 83 Figure 13.17 84 Figure 13.19 85 p13_139 86 Figure 13.21 87 Figure 13.22 88 p14_01 89 p14_04 90 p14_106 91 Figure 15.1 92 Figure 15.2 93 ph15_01 94 p15_248 95 Figure 19.1 96 Figure 19.2 97 p19_05 98 p19_07-08 99 p19_17 100 p19_20 101 p19_21 102 p19_22 103 p19_23 104 p19_27 105
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