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Eighth Edition Vector Mechanics for Engineers: Statics CE 102 Statics Chapter 3 Rigid Bodies: Equivalent Systems of Forces © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-1 Eighth Edition Vector Mechanics for Engineers: Statics Contents Introduction Moment of a Force About a Given Axis External and Internal Forces Sample Problem 3.2 Principle of Transmissibility: Equivalent Forces Moment of a Couple Vector Products of Two Vectors Couples Can Be Represented By Vectors Moment of a Force About a Point Resolution of a Force Into a Force at O and a Couple Varigon’s Theorem Rectangular Components of the Moment of a Force Addition of Couples Sample Problem 3.3 Sample Problem 3.1 System of Forces: Reduction to a Force and a Couple Scalar Product of Two Vectors Further Reduction of a System of Forces Scalar Product of Two Vectors: Applications Sample Problem 3.4 Sample Problem 3.5 Mixed Triple Product of Three Vectors © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-2 Eighth Edition Vector Mechanics for Engineers: Statics 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. • Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body. • 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 • 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-3 Eighth Edition Vector Mechanics for Engineers: Statics External and Internal Forces • Forces acting on rigid bodies are divided into two groups: - External forces - 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-4 Eighth Edition Vector Mechanics for Engineers: Statics Principle of Transmissibility: Equivalent Forces • Principle of Transmissibility Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are 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. • Principle of transmissibility may not always apply in determining internal forces and deformations. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-5 Eighth Edition Vector Mechanics for Engineers: Statics 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. • Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1. Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is V PQ sin 3. Direction of V is obtained from the right-hand rule. • Vector products: - are not commutative, Q P P Q P Q1 Q2 P Q1 P Q2 - are distributive, - are not associative, P Q S P Q S © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-6 Eighth Edition Vector Mechanics for Engineers: Statics Vector Products: Rectangular Components • Vector products of Cartesian unit vectors, i i 0 j i k k i j i j k j j 0 k j i i k j j k i k k 0 • Vector products in terms of rectangular coordinates V Px i Py j Pz k Qx i Q y j Qz k Py Q z Pz Q y i Pz Q x Px Q z j Px Q y Py Q x k i j k Px Py Pz Qx Q y Qz © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-7 Eighth Edition Vector Mechanics for Engineers: Statics 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 MO r F • The moment vector MO is perpendicular to the plane containing O and the force F. • Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO. M O rF sin Fd The sense of the moment may be determined by the right-hand rule. • 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-8 Eighth Edition Vector Mechanics for Engineers: Statics 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. • The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane. • If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive. • If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3-9 Eighth Edition Vector Mechanics for Engineers: Statics 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 point O. r F1 F2 r F1 r F2 • 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 10 Eighth Edition Vector Mechanics for Engineers: Statics Rectangular Components of the Moment of a Force The moment of F about O, M O r F , r xi yj zk F Fx i Fy j Fz k M O M xi M y j M z k i j k x y z Fx Fy Fz yFz zFy i zFx xFz j xFy yFx k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 11 Eighth Edition Vector Mechanics for Engineers: Statics Rectangular Components of the Moment of a Force The moment of F about B, M B rA / B F rA / B rA rB x A xB i y A y B j z A z B k F Fx i Fy j Fz k i M B x A xB Fx j k y A yB z A z B Fy Fz © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 12 Eighth Edition Vector Mechanics for Engineers: Statics Rectangular Components of the Moment of a Force For two-dimensional structures, M O xFy yFx k MO MZ xFy yFx M B x A xB Fy y A yB Fx k MB MZ x A xB Fy y A yB Fx © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 13 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.1 A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 14 Eighth Edition Vector Mechanics for Engineers: Statics 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. M O Fd d 24 in. cos 60 12 in. M O 100 lb12 in. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. M O 1200 lb in 3 - 15 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.1 b) Horizontal force at A that produces the same moment, d 24 in. sin 60 20.8 in. M O Fd 1200 lb in. F 20.8 in. F © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 1200 lb in. 20.8 in. F 57.7 lb 3 - 16 Eighth Edition Vector Mechanics for Engineers: Statics 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 perpendicular to OA. M O Fd 1200 lb in. F 24 in. F © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 1200 lb in. 24 in. F 50 lb 3 - 17 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.1 d) To determine the point of application of a 240 lb force to produce the same moment, M O Fd 1200 lb in. 240 lbd 1200 lb in. 5 in. 240 lb OB cos60 5 in. d © 2007 The McGraw-Hill Companies, Inc. All rights reserved. OB 10 in. 3 - 18 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.1 e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb 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 lb force. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 19 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.2 SOLUTION: The moment MA of the force F exerted by the wire is obtained by evaluating the vector product, M A rC A F 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 20 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.2 SOLUTION: M A rC A F rC (0,0,0.32) (0.3, 0, 0.4) A rC rA 0.3 m i 0.08 m k rC D F F 200 N rC D 0.3 m i 0.24 m j 0.32 m k 200 N 0.5 m 120 N i 96 N j 128 N k i M A xC x A Fx j k yC y A zC z A Fy Fz i j k 0.3 0 0.08 120 96 128 M A 7.68 N m i 28.8 N m j 28.8 N m k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 21 Eighth Edition Vector Mechanics for Engineers: Statics Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is defined as P Q PQ cos scalar result • Scalar products: - are commutative, - are distributive, - are not associative, PQ Q P P Q1 Q2 P Q1 P Q2 P Q S undefined • Scalar products with Cartesian unit components, P Q Px i Py j Pz k Qx i Q y j Qz k i i 1 j j 1 k k 1 i j 0 j k 0 k i 0 P Q Px Qx Py Q y Pz Qz P P Px2 Py2 Pz2 P 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 22 Eighth Edition Vector Mechanics for Engineers: Statics Scalar Product of Two Vectors: Applications • Angle between two vectors: P Q PQ cos Px Qx Py Q y Pz Qz cos Px Qx Py Q y Pz Qz PQ • Projection of a vector on a given axis: POL P cos projection of P along OL P Q PQ cos P Q P cos POL Q • For an axis defined by a unit vector: POL P Px cos x Py cos y Pz cos z © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 23 Eighth Edition Vector Mechanics for Engineers: Statics 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, S P Q P Q S Q S P S Q P P S Q Q P S • Evaluating the mixed triple product, S P Q S x Py Q z Pz Q y S y Pz Q x Px Q z S z Px Q y Py Q x Sx Sy Sz Px Py Pz Qx Qy Qz © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 24 Eighth Edition Vector Mechanics for Engineers: Statics Moment of a Force About a Given Axis • Moment MO of a force F applied at the point A about a point O, MO r F • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis, M OL M O r F • Moments of F about the coordinate axes, M x yFz zFy M y zFx xFz M z xFy yFx © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 25 Eighth Edition Vector Mechanics for Engineers: Statics Moment of a Force About a Given Axis • Moment of a force about an arbitrary axis, M BL M B rA B F rA B rA rB • The result is independent of the point B along the given axis. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 26 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.3 A cube is acted on by a force P as shown. Determine the moment of P a) b) c) d) © 2007 The McGraw-Hill Companies, Inc. All rights reserved. about A about the edge AB and about the diagonal AG of the cube. Determine the perpendicular distance between AG and FC. 3 - 27 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.3 • Moment of P about A, M A rF A P rF A a i a j a i j P P 1/ 2 j 1/ 2 k P / 2 j k M A a i j P / 2 j k MA aP / 2 i j k • Moment of P about AB, M AB i M A i aP / 2 i j k M AB aP / 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 28 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.3 • Moment of P about the diagonal AG, M AG M A rA G ai aj ak 1 i j k rA G a 3 3 aP MA i j k 2 1 aP M AG i j k i j k 3 2 aP 1 1 1 6 M AG © 2007 The McGraw-Hill Companies, Inc. All rights reserved. aP 6 3 - 29 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.3 • Perpendicular distance between AG and FC, P 1 P 0 1 1 P j k i j k 2 3 6 0 Therefore, P is perpendicular to AG. M AG aP Pd 6 d © 2007 The McGraw-Hill Companies, Inc. All rights reserved. a 6 3 - 30 Eighth Edition Vector Mechanics for Engineers: Statics 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. • Moment of the couple, M rA F rB F rA rB F r F M rF sin Fd • 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 31 Eighth Edition Vector Mechanics for Engineers: Statics Moment of a Couple Two couples will have equal moments if • F1d1 F2 d 2 • 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 32 Eighth Edition Vector Mechanics for Engineers: Statics Addition of Couples • Consider two intersecting planes P1 and P2 with each containing a couple M 1 r F1 in plane P1 M 2 r F2 in plane P2 • Resultants of the vectors also form a couple M r R r F1 F2 • By Varigon’s theorem M r F1 r F2 M1 M 2 • Sum of two couples is also a couple that is equal to the vector sum of the two couples © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 33 Eighth Edition Vector Mechanics for Engineers: Statics 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 addition of vectors. • Couple vectors are free vectors, i.e., the point of application is not significant. • Couple vectors may be resolved into component vectors. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 34 Eighth Edition Vector Mechanics for Engineers: Statics 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. • Attaching equal and opposite force vectors at O produces no net effect on the body. • The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 35 Eighth Edition Vector Mechanics for Engineers: Statics Resolution of a Force Into a Force at O and a Couple • Moving F from A to a different point O’ requires the addition of a different couple vector MO’ M O' r F • The moments of F about O and O’ are related, M O ' r 'F r s F r F s F MO s F • Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 36 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.4 SOLUTION: • Attach equal and opposite 20 lb forces in the +x direction at A, thereby producing 3 couples for which the moment components are easily computed. • Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions.. Determine the components of the single couple equivalent to the couples shown. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 37 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.4 • Attach equal and opposite 20 lb forces in the +x direction at A • The three couples may be represented by three couple vectors, M x 30 lb18 in. 540 lb in. M y 20 lb12 in. 240lb in. M z 20 lb9 in. 180 lb in. M 540 lb in. i 240lb in. j 180 lb in.k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 38 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.4 • Alternatively, compute the sum of the moments of the four forces about D. • Only the forces at C and E contribute to the moment about D. M M D 18 in. j 30 lbk 9 in. j 12 in.k 20 lb i M 540 lb in. i 240lb in. j 180 lb in.k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 39 Eighth Edition Vector Mechanics for Engineers: Statics 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 • The force and couple vectors may be combined into a resultant force vector and a resultant couple vector, R R F M O r F • The force-couple system at O may be moved to O’ with the addition of the moment of R about O’ , R R M O' M O s R • Two systems of forces are equivalent if they can be reduced to the same force-couple system. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 40 Eighth Edition Vector Mechanics for Engineers: Statics 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. • The resultant force-couple system for a system of forces will be mutually perpendicular if: 1) the forces are concurrent, 2) the forces are coplanar, or 3) the forces are parallel. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 41 Eighth Edition Vector Mechanics for Engineers: Statics Further Reduction of a System of Forces • System of coplanar forces isreduced to a force-couple system R and M OR that is mutually perpendicular. • System can be reduced to a single force by moving the line of action of R until its moment about O becomes M OR • In terms of rectangular coordinates, xR y yRx M OR © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 42 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.5 SOLUTION: a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A. 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. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. b) Find an equivalent force-couple system at B based on the forcecouple 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. 3 - 43 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.5 SOLUTION: a) Compute the resultant force and the resultant couple at A. R F 150 N j 600 N j 100 N j 250 N j R 600 N j R M A r F 1.6 i 600 j 2.8 i 100 j 4.8 i 250 j R M A 1880 N m k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 44 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.5 b) Find an equivalent force-couple system at B based on the force-couple system at A. The force is unchanged by the movement of the force-couple system from A to B. R 600 N j The couple at B is equal to the moment about B of the force-couple system found at A. R R M B M A rB A R 1880 N m k 4.8 m i 600 N j 1880 N m k 2880 N m k R M B 1000 N m k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 45 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.6 SOLUTION: • Determine the relative position vectors for the points of application of the cable forces with respect to A. • Resolve the forces into rectangular components. • Compute the equivalent force, R F Three cables are attached to the bracket as shown. Replace the forces with an equivalent forcecouple system at A. • Compute the equivalent couple, R M A r F © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 46 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.6 • Resolve the forces into rectangular components. FB 700 N rE B 75 i 150 j 50k rE B 175 0.429 i 0.857 j 0.289k FB 300 i 600 j 200k N SOLUTION: • Determine the relative position vectors with respect to A. rB A 0.075 i 0.050k m rC A 0.075 i 0.050k m rD A 0.100 i 0.100 j m © 2007 The McGraw-Hill Companies, Inc. All rights reserved. FC 1000 N cos 45 i cos 45 j 707 i 707 j N FD 1200 N cos 60 i cos 30 j 600 i 1039 j N 3 - 47 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.6 • Compute the equivalent force, R F 300 707 600 i 600 1039 j 200 707k R 1607 i 439 j 507 k N • Compute the equivalent couple, R M A r F i j k rB A F B 0.075 0 0.050 30i 45k 300 600 200 i j k rC A F c 0.075 0 0.050 17.68 j 707 0 707 i j k rD A F D 0.100 0.100 0 163.9k 600 1039 0 R M A 30 i 17.68 j 118.9k © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 48 Problem 3.7 y C 10 ft The 15-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 570 lb, determine the moment about A of the force exerted by the cable at B. 6 ft A z 15 ft B x 49 y Problem 3.7 C 10 ft Solving Problems on Your Own 6 ft A z 15 ft B x The 15-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 570 lb, determine the moment about A of the force exerted by the cable at B. 1. Determine the rectangular components of a force defined by its magnitude and direction. If the direction of the force is defined by two points located on its line of action, the force can be expressed by: F F = F = (dx i + dy j + dz k) d 50 y Problem 3.7 C 10 ft Solving Problems on Your Own 6 ft A z 15 ft B x The 15-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 570 lb, determine the moment about A of the force exerted by the cable at B. 2. Compute the moment of a force in three dimensions. If r is a position vector and F is the force the moment M is given by: M=rxF 51 y Problem 3.7 Solution C 10 ft 6 ft Determine the rectangular components of a force defined by its magnitude and direction. A 570 N First note: z 15 ft B dBC = (_15)2 + (6) 2 + (_10) 2 x dBC = 19 ft Then: TBC = 570 lb _ ( 15 i + 6 j _ 10 k) = _ (450 lb) i + (180 lb) j _ (300 lb)k 19 52 y Problem 3.7 Solution C 10 ft 6 ft Compute the moment of a force in three dimensions. A 570 N Have: z MA = rB/A x TBC B 15 ft x Where: rB/A = (15 ft) i Then: MA = 15 i x (_ 450 i + 180 j _ 300 k) MA = (4500 lb.ft) j + (2700 lb.ft) k 53 Problem 3.8 y 2.4 m 1.2 m D Knowing that the tension in cable AC is 1260 N, Determine (a) the angle between cable AC and the boom AB , (b) the projection on AB of the force exerted by cable AC at point A. 3m C 2.6 m z A B 2.4 m P 1.8 m x 54 Problem 3.8 y 2.4 m 1.2 m D Solving Problems on Your Own 3m Knowing that the tension in cable AC is 1260 N, Determine (a) the angle between cable AC and the boom AB , (b) the projection on AB of the force exerted by cable AC at point A. C 2.6 m z A B 2.4 m P 1.8 m x 1. Calculate the angle formed by two vectors. Express the vectors in terms of their components. The cosine of the desired angle is obtained by dividing the scalar product of the two vectors by the product of their magnitudes. 55 Problem 3.8 y 2.4 m 1.2 m D Solving Problems on Your Own 3m C 2.6 m z A B 2.4 m Knowing that the tension in cable AC is 1260 N, Determine (a) the angle between cable AC and the boom AB , (b) the projection on AB of the force exerted by cable AC at point A. P 1.8 m x 2. Compute the projection of a vector P on a given line OL. Express the vector P and the unit vector in the direction of the line in component form. The required projection is equal to the scalar product P.. If the angle formed by P and is known, the projection is also given by P cos . 56 Problem 3.8 Solution y 2.4 m 1.2 m D Calculate the angle formed by two vectors. 3m (a) First note: C 2.6 m AC = A B z (_ 2.4)2 + (0.8)2 + (1.2)2 = 2.8 m 2.4 m P 1.8 m x AB = (_ 2.4)2 + (_ 1.8)2 + ( 0 )2 = 3.0 m and AC = _ (2.4 m) i + (0.8 m) j + (1.2 m) k AB = _ (2.4 m) i _ (1.8 m) j 57 Problem 3.8 Solution y 2.4 m 1.2 m D 3m C 2.6 m z By definition: A B AC . AB = (AC) (AB) cos 2.4 m P 1.8 m x (_ 2.4 i + 0.8 j + 1.2 k) . (_ 2.4 i _ 1.8 j) = (2.8)(3.0) cos (_ 2.4)(_ 2.4) + (0.8)(_ 1.8) + (1.2)(0) = 8.4 cos cos = 0.51429 = 59.1o 58 Problem 3.8 Solution y 2.4 m D 1.2 m C 3m TAC (b) A 2.6 m B z Compute the projection of a vector on a given line. (TAC )AB = TAC . AB AB 2.4 m P 1.8 m x = TAC cos = (1260 N) (0.51429) (TAC )AB = 648 N 59 Problem 3.9 y 0.35 m 0.875 m G H O 0.925 m 0.75 m A z B D x 0.75 m C The frame ACD is hinged at A and D and is supported by a cable which passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. P 60 Problem 3.9 y 0.35 m Solving Problems on Your Own 0.875 m G H O 0.925 m 0.75 m A z B D x 0.75 m C P The frame ACD is hinged at A and D and is supported by a cable which passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. 1. Determine the moment MOL of a force about a given axis OL. MOL is defined as MOL = . MO = . ( r x F ) where is the unit vector along OL and r is a position vector from any point on the line OL to any point on the line of action of F. 61 Problem 3.9 Solution y 0.35 m Determine the moment MAD of a force about line AD. 0.875 m G H O 0.925 m 0.75 m TBH A AD z rB/A B MAD D x 0.75 m C MAD = AD . ( rB/A x TBH ) Where 1 AD = (4 i _ 3 k) 5 rB/A = (0.5 m ) i P dBH = ( 0.375 )2 + ( 0.75 )2 + ( _ 0.75 )2 = 1.125 m 450 N TBH = ( 0.375 i + 0.75 j _ 0.75 k ) 1.125 = ( 150 N) i + ( 300 N) j _ ( 300 N) k 62 Problem 3.9 Solution y 0.35 m 0.875 m G Finally: H MAD = AD . ( rB/A x TBH ) O 0.925 m 0.75 m TBH A AD z rB/A B MAD D x 0.75 m C P 4 0 _3 MAD = 1 0.5 0 0 5 150 300 _300 = 1 5 [(_3)(0.5)(300)] MAD = _ 90 N.m 63 Problem 3.10 C B a A 200 mm b 300 mm P The force P has a magnitude of 250 N and is applied at the end C of a 500 mm rod AC attached to a bracket at A and B. Assuming a = 30o and b = 60o, replace P with (a) an equivalent force-couple system at B , (b) an equivalent system formed by two parallel forces applied at A and B. 64 Problem 3.10 Solving Problems on Your Own C B a A 200 mm b 300 mm P The force P has a magnitude of 250 N and is applied at the end C of a 500 mm rod AC attached to a bracket at A and B. Assuming a = 30o and b = 60o, replace P with (a) an equivalent force-couple system at B , (b) an equivalent system formed by two parallel forces applied at A and B. 1. Replace a force with an equivalent force-couple system at a specified point. The force of the force-couple system is equal to the original force, while the required couple vector is equal to the moment of the original force about the given point. 65 Problem 3.10 Solution C B a A Replace a force with an equivalent force-couple system at a specified point. b 300 mm P 200 mm (a) Equivalence requires: SF: F=P or F = 250 N 60o S MB : M = _ (0.3 m)(250 N) = _75 N The equivalent force couple system at B is: F = 250 N 60o, M = 75 N . m 66 Problem 3.10 Solution C b B a A 300 mm P 200 mm (b) Require: The two force systems are equivalent. C 60o B A y 30 o 250 N x B = A C F FB F FA 67 Problem 3.10 Solution C y 60o B o 30 A 250 N A If FB F 0 = FA cos F + FB cos F FA = _ FB S Fy : F FA Equivalence then requires: S Fx : x B = C cos F = 0 or 250 = _ FA sin F _ FB sin F _ FA = _ FB Consequently then _ 250 = 0 cos F = 0 or reject F = 90o and 68 FA + FB = 250 Problem 3.10 Solution C y 60o B o 30 A 250 N x B = A C F FB F FA Also: + S MB : _ (0.3 m)( 250 N) = (0.2 m) FA or and FA = 375 N FA = _ 375 N FB = + 675 N 60o, FB = 625 N 60o 69 Problem 3.11 10 lb 30 lb 12 in M A 60o B 8 in C 45 lb A couple of magnitude M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. 70 Problem 3.11 Solving Problems on Your Own 10 lb 30 lb 12 in M A 60o B 8 in C 45 lb A couple of magnitude M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. 1. Determine the resultant of two or more forces. Determine the rectangular components of each force. Adding these components will yield the components of the resultant. 71 Problem 3.11 Solving Problems on Your Own 10 lb 30 lb A couple of magnitude 60o M = 54 lb.in. and the three forces shown are applied to A B an angle bracket. (a) Find the M 8 in resultant of this system of forces. (b) Locate the points C where the line of action of the 45 lb resultant intersects line AB and line BC. 2. Reduce a force system to a force and a couple at a given point. The force is the resultant R of the system obtained by adding the various forces. The couple is the moment resultant of the system M, obtained by adding the moments about the point of the various forces. 12 in R=SF M=S(rxF) 72 Problem 3.11 Solving Problems on Your Own 10 lb 30 lb A couple of magnitude 60o M = 54 lb.in. and the three forces shown are applied to A B an angle bracket. (a) Find the M 8 in resultant of this system of forces. (b) Locate the points C where the line of action of the 45 lb resultant intersects line AB and line BC. 3. Reduce a force and a couple at a given point to a single force. The single force is obtained by moving the force until its moment about the point (A) is equal to the couple vector MAR. A position vector r from the point, to any point on the line of action of the single force R must satisfy the equation 12 in r x R = MAR 73 10 lb 30 lb 12 in M A 60o B 8 in C (a) Problem 3.11 Solution Determine the resultant of two or more forces. 45 lb Adding the components of the forces: S F : R = ( _ 10 j ) + (_ 45 i ) + ( 30 cos 60o i + 30 sin 60o j ) = _ ( 30 lb ) i + ( 15.98 lb ) j R = 34.0 lb 28.0o 74 10 lb 30 lb 12 in M A Problem 3.11 Solution 60o B 8 in C Reduce a force system to a force and a couple at a given point. 45 lb (b) First reduce the given forces and couple to an equivalent force-couple system (R, MB) at B. + S MB : MB = ( 54 lb.in ) + ( 12 in )(10 lb) _ ( 8 in ) ( 45 lb) = _ 186 lb.in 75 Problem 3.11 Solution a A D B R c Reduce a force and a couple at a given point to a single force. E C With R at D: + S MB : _ 186 lb.in = _ a ( 15.9808 lb) or a = 11.64 in and with R at E + S MB : _ 186 lb.in = c ( 30 lb) or c = 6.20 in The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in below B. 76