Eighth Edition CHAPTER 3 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 2007 The McGraw-Hill Companies, Inc. All rights reser Eighth Edition Vector Mechanics for Engineers: Statics Contents Introduction Moment of a Force About a Given Axis External and Internal Forces Sample Problem 3.5

Principle of Transmissibility: Equivalent F orces Moment of a Couple Vector Products of Two Vectors Moment of a Force About a Point Addition of Couples Couples Can Be Represented By Vectors Varigons Theorem Resolution of a Force Into a Force at O and a Couple Rectangular Components of the Moment of a Force Sample Problem 3.6 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 3.8

Mixed Triple Product of Three Vectors Sample Problem 3.10 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 e.g. a spanner moment of a force about an axis e.g. a swing door moment due to a couple e.g. the steering wheel 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 - are distributive, P Q1 Q2 P Q1 P Q2 - 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: Suppose we have 2 vectors P and Q and we wish to find PxQ:

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 M O 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 M O rF sin Fd MO. 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 counterclockwise, 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 clockwise 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 Varignons 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 Varigons 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 x i M y j M z k i x Fx j y Fy k z

Fz yFz zF y 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 x B 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 y A yB Fy k z A zB 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 M O M Z xFy yFx Instructors note: Change Fz to Fx in the above and below M O x A xB Fy y A y B Fx k M O M Z x A xB Fy y A y B 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 lb 12 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 c) 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. 1200 lb in. F 20.8 in. 2007 The McGraw-Hill Companies, Inc. All rights reserved. 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. 1200 lb in. F 24 in. 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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 lb d 1200 lb in. d 5 in. 240 lb OB cos60 5 in. 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 3.4 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 3.4 SOLUTION: M A rC A F rC A rC rA 0.3 m i 0.08 m j rD C F F 200 N rD C 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 j k M A 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,

P Q Q P - are distributive, P Q1 Q2 P Q1 P Q2 - are not associative, 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 Q x Py Q y Pz Q z cos Px Q x Py Q y Pz Q z PQ Projection of a vector on a given axis: POL P cos projection of P along OL P Q PQ cos PQ P cos POL Q For an axis defined by a unit vector: POL P Px i Py j Pz k cos x i cos y j cos z k 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 There are six different mixed triple products formed from S, P, and Q which 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 Qz Pz Q y S y Pz Qx Px Q z S z Px Q y Py Q x Sx Px Qx Sy Py

Qy 2007 The McGraw-Hill Companies, Inc. All rights reserved. Sz Pz Qz 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, M O 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 Prove how did we get this!!

2007 The McGraw-Hill Companies, Inc. All rights reserved. Use this to help you 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 3.5 A cube of side (a) is acted on by a

force P as shown. Determine the moment of P a) about A b) about the edge AB and c) about the diagonal AG of the cube. d) Determine the perpendicular distance between AG and FC. Hint: first decide where is the origin of x- y-z axes, then find the cross product : (rxF). 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 27 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.5 Moment of P about A, M A rF A P rF A a i a j a i j

P P j k 2 P M A a i j j k 2 M A aP / 2 i j k Note: There was a mistake in the handout notes. Your task is to work out what is wrong and correct as above.

Moment of P about AB, M AB i M A i aP / 2 i j k 2007 The McGraw-Hill Companies, Inc. All rights reserved. M AB aP / 2 3 - 28 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.5 Moment of P about the diagonal AG,

M AG M A rG A ai aj ak 1 i j k rG A 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

Change rA/G to rG/A 2007 The McGraw-Hill Companies, Inc. All rights reserved. M AG aP 6 3 - 29 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.5 To find the perpendicular distance between AG and FC, we notice that: P 1 P P j k

i j k 0 1 1 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 (about O: take clockwise ve and anticlockwise +ve) , 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 i.e. have the same value 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 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 Varigons theorem M r F1 r F2 M 1 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 M O 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 3.6 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. * Notice here that no point is specified for the couple as it acts equally on the whole body 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 37 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.6 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 lb 9 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 3.6 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 lb k 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 at 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 3.8 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. 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. 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 43 Eighth Edition Vector Mechanics for Engineers: Statics

Sample Problem 3.8 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 3.8 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. M BR M AR rA B 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 3.10 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 3.10

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 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 FC 1000 N cos 45 i cos 45k 707 i 707k N FD 1200 N cos 60 i cos 30 j 600 i 1039 j N 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 47 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 3.10 Compute the equivalent force, R F

300 707 600 i 600 1039 j 200 707 k R 1607i 439 j 507k 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 i rC A F c 0.075

707 i rD A F D 0.100 600 600 200 j k 0 0.050 17.68 j 0 707 j k 0.100 0 163.9k 1039 0 R M A 30 i 17.68 j 118 .9k 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 48